Introduction to Special Relativity
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- Basic Principles
- Lorentz Transformation Equation For Position And Time
- Exercise 1
- Rest Frame And Time Dilation
- Shortening Of Length
- Change In Angle
- Exercise 2
- The Relativistic Doppler Effect
- Rod Emits Light
- Causality
- Exercise 3
- Velocity Transforms
- Aberration
- Exercise 4
- Exercise 5
- Exercise 6
- Exercise 7
- Energy And Momentum
- Exercise 8
- Exercise 9
- Momentum And Energy Transformation
- Exercise 10
- Exercise 11
- Minimum Energy To Form Particles
- Exercise 12
- Compton Scattering

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[{"Name":"Introduction to Special Relativity","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Basic Principles","Duration":"12m 40s","ChapterTopicVideoID":9259,"CourseChapterTopicPlaylistID":9398,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/9259.jpeg","UploadDate":"2017-04-02T18:04:12.0570000","DurationForVideoObject":"PT12M40S","Description":null,"MetaTitle":"Basic Principles: Video + Workbook | Proprep","MetaDescription":"Special Relativity - Introduction to Special Relativity. Watch the video made by an expert in the field. Download the workbook and maximize your learning.","Canonical":"https://www.proprep.uk/general-modules/all/physics-1-mechanics-waves-and-thermodynamics/special-relativity/introduction-to-special-relativity/vid9546","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.980","Text":"Hello. In this section,"},{"Start":"00:01.980 ","End":"00:05.160","Text":"we\u0027re going to be speaking about the basic principles of"},{"Start":"00:05.160 ","End":"00:10.785","Text":"special relativity and we\u0027re also going to be speaking about the Lorentz Transformation."},{"Start":"00:10.785 ","End":"00:15.630","Text":"Special relativity focuses on how to describe events observed from"},{"Start":"00:15.630 ","End":"00:20.790","Text":"different frames of reference. What does that mean?"},{"Start":"00:20.790 ","End":"00:24.195","Text":"Let\u0027s say that we have this frame of reference which we call"},{"Start":"00:24.195 ","End":"00:28.685","Text":"S. Now usually this is going to be the lab\u0027s frame of reference."},{"Start":"00:28.685 ","End":"00:30.950","Text":"Then we have our observer,"},{"Start":"00:30.950 ","End":"00:35.055","Text":"which is in a different frame of reference called S tag."},{"Start":"00:35.055 ","End":"00:36.875","Text":"Our frame of reference,"},{"Start":"00:36.875 ","End":"00:41.520","Text":"S tag is moving at some constant velocity v_0."},{"Start":"00:41.900 ","End":"00:45.410","Text":"Each one of these frames is a frame of reference."},{"Start":"00:45.410 ","End":"00:49.610","Text":"What we\u0027re going to be doing is we\u0027re going to be working out how to"},{"Start":"00:49.610 ","End":"00:55.480","Text":"move between these different frames of reference when we\u0027re doing calculations."},{"Start":"00:55.480 ","End":"01:02.075","Text":"Certain calculations and formulas which connect these 2 frames of reference."},{"Start":"01:02.075 ","End":"01:08.540","Text":"The frame of reference will always be inertial reference frames. What does that mean?"},{"Start":"01:08.540 ","End":"01:11.930","Text":"That means that the observers are moving at"},{"Start":"01:11.930 ","End":"01:16.850","Text":"a constant velocity relative to the size frame of reference."},{"Start":"01:16.850 ","End":"01:18.560","Text":"What does that mean?"},{"Start":"01:18.560 ","End":"01:21.530","Text":"We\u0027ll always have some frame of reference which is"},{"Start":"01:21.530 ","End":"01:27.359","Text":"stationary and another one moving at some constant velocity."},{"Start":"01:27.680 ","End":"01:31.760","Text":"I\u0027m going to also remind you that this idea of"},{"Start":"01:31.760 ","End":"01:35.450","Text":"a stationary frame of reference and another frame of reference moving at"},{"Start":"01:35.450 ","End":"01:42.735","Text":"a constant velocity is something to do with relative velocities, one to another."},{"Start":"01:42.735 ","End":"01:46.520","Text":"What does that mean? We\u0027re using this idea that our S,"},{"Start":"01:46.520 ","End":"01:51.200","Text":"for instance, is stationary and our S tag is moving at a constant velocity."},{"Start":"01:51.200 ","End":"01:54.500","Text":"However, this term is relative because"},{"Start":"01:54.500 ","End":"01:58.985","Text":"our observer standing in our S tag frame of reference,"},{"Start":"01:58.985 ","End":"02:01.325","Text":"which we\u0027re saying is moving."},{"Start":"02:01.325 ","End":"02:07.145","Text":"For him, he\u0027s stationary in this frame of reference and he can see"},{"Start":"02:07.145 ","End":"02:13.984","Text":"our S frame of reference moving away or towards him at a constant velocity."},{"Start":"02:13.984 ","End":"02:19.400","Text":"It\u0027s important to understand that this term of one of"},{"Start":"02:19.400 ","End":"02:24.150","Text":"the frames of reference is stationary isn\u0027t exactly accurate,"},{"Start":"02:24.150 ","End":"02:25.860","Text":"and it\u0027s not an absolute term,"},{"Start":"02:25.860 ","End":"02:27.765","Text":"and to understand what it means,"},{"Start":"02:27.765 ","End":"02:30.620","Text":"it means that if we\u0027re looking at this frame of reference and"},{"Start":"02:30.620 ","End":"02:33.575","Text":"we say that S is stationary,"},{"Start":"02:33.575 ","End":"02:39.395","Text":"then everything moving relative to S is going to be moving at some velocity."},{"Start":"02:39.395 ","End":"02:42.985","Text":"Now, we\u0027ve been speaking a lot about the term event."},{"Start":"02:42.985 ","End":"02:45.875","Text":"Let\u0027s define what an event is."},{"Start":"02:45.875 ","End":"02:51.725","Text":"An event is a physical occurrence defined for a specific point in space-time."},{"Start":"02:51.725 ","End":"02:55.400","Text":"Each event can be described via the vector coordinates x,"},{"Start":"02:55.400 ","End":"02:58.725","Text":"y, z, t. The 4-vector."},{"Start":"02:58.725 ","End":"03:02.180","Text":"E.g., lighting a light is an event and"},{"Start":"03:02.180 ","End":"03:06.065","Text":"switching it off is also an event. What does that mean?"},{"Start":"03:06.065 ","End":"03:10.400","Text":"Let\u0027s say I lit some bulb over here and I"},{"Start":"03:10.400 ","End":"03:15.230","Text":"know at what time and where I lit the bulb in my S frame of reference."},{"Start":"03:15.230 ","End":"03:17.915","Text":"When we\u0027re dealing with special relativity,"},{"Start":"03:17.915 ","End":"03:20.540","Text":"I\u0027m going to want to find out,"},{"Start":"03:20.540 ","End":"03:23.107","Text":"which is what we\u0027re going to speak about in this unit,"},{"Start":"03:23.107 ","End":"03:27.410","Text":"at what time and where I lit this bulb when we\u0027re"},{"Start":"03:27.410 ","End":"03:33.115","Text":"relative to my S tag frame of reference or any other frame of reference."},{"Start":"03:33.115 ","End":"03:37.250","Text":"Now, something to point out about this event,"},{"Start":"03:37.250 ","End":"03:40.910","Text":"we can notice that the lighting of light is an event,"},{"Start":"03:40.910 ","End":"03:43.265","Text":"and switching it off is an event."},{"Start":"03:43.265 ","End":"03:48.530","Text":"That means that our event isn\u0027t over an extended period of time."},{"Start":"03:48.530 ","End":"03:52.805","Text":"For instance, once we\u0027ve switched the light on, that\u0027s our event."},{"Start":"03:52.805 ","End":"03:56.675","Text":"When it\u0027s on, that\u0027s not considered part of the event."},{"Start":"03:56.675 ","End":"03:59.960","Text":"However, at the exact moment when we switch the light off,"},{"Start":"03:59.960 ","End":"04:05.405","Text":"that\u0027s an event and the time that the light is off is not considered an event."},{"Start":"04:05.405 ","End":"04:09.470","Text":"Our events don\u0027t have a period of time that they occur in."},{"Start":"04:09.470 ","End":"04:11.900","Text":"They have a short when the light is"},{"Start":"04:11.900 ","End":"04:15.245","Text":"switched off and when it\u0027s switched on, that\u0027s an event."},{"Start":"04:15.245 ","End":"04:18.455","Text":"If we consider that at t_1,"},{"Start":"04:18.455 ","End":"04:20.800","Text":"we switch on our lights,"},{"Start":"04:20.800 ","End":"04:25.890","Text":"and then our light shines for 10 seconds."},{"Start":"04:25.890 ","End":"04:30.639","Text":"Then we have at t_1 plus 10 seconds,"},{"Start":"04:30.639 ","End":"04:33.280","Text":"so let\u0027s call that t_2,"},{"Start":"04:33.280 ","End":"04:35.700","Text":"we switch our light off."},{"Start":"04:35.700 ","End":"04:42.590","Text":"That means that our first event happened at t_1 and our second event happened at t_2,"},{"Start":"04:42.590 ","End":"04:48.415","Text":"and that the time between these 2 events is 10 seconds."},{"Start":"04:48.415 ","End":"04:53.660","Text":"Now let\u0027s speak about the basic principles in special relativity."},{"Start":"04:53.660 ","End":"04:56.420","Text":"Now the first principle is that the laws of"},{"Start":"04:56.420 ","End":"04:59.810","Text":"physics apply in every inertial reference frame."},{"Start":"04:59.810 ","End":"05:05.510","Text":"The second basic principle is that light does not require a medium in order to travel."},{"Start":"05:05.510 ","End":"05:07.789","Text":"That means that unlike sound,"},{"Start":"05:07.789 ","End":"05:11.975","Text":"which needs some medium if it\u0027s air or some liquid or a solid."},{"Start":"05:11.975 ","End":"05:15.040","Text":"Light can also travel through a vacuum."},{"Start":"05:15.040 ","End":"05:19.790","Text":"The third basic principle is that the velocity of light is a constant and is the"},{"Start":"05:19.790 ","End":"05:24.845","Text":"same in every inertial reference frame. What does that mean?"},{"Start":"05:24.845 ","End":"05:30.050","Text":"Let\u0027s take a look at this frame of reference S just as an example."},{"Start":"05:30.050 ","End":"05:32.240","Text":"From the frame of reference S,"},{"Start":"05:32.240 ","End":"05:35.809","Text":"we can see some car driving at a velocity."},{"Start":"05:35.809 ","End":"05:40.580","Text":"Let\u0027s say that our inertial frame of reference S tag is moving at"},{"Start":"05:40.580 ","End":"05:46.400","Text":"some velocity v_2 relative to this frame of reference."},{"Start":"05:46.400 ","End":"05:49.055","Text":"As we\u0027ve learned up until now,"},{"Start":"05:49.055 ","End":"05:52.880","Text":"observers in our S frame of reference and on"},{"Start":"05:52.880 ","End":"05:58.355","Text":"our S tag frame of reference will see this car moving at different velocities."},{"Start":"05:58.355 ","End":"06:02.360","Text":"Now, what we\u0027ll see with the lights is that if our light is"},{"Start":"06:02.360 ","End":"06:06.860","Text":"traveling at the speed of light relative to our S tag,"},{"Start":"06:06.860 ","End":"06:10.250","Text":"our S frame of reference will see the light traveling at"},{"Start":"06:10.250 ","End":"06:14.660","Text":"the exact same velocity as is recorded in S tag,"},{"Start":"06:14.660 ","End":"06:21.125","Text":"regardless of the velocity at which S tag is traveling at relative to"},{"Start":"06:21.125 ","End":"06:28.835","Text":"S. That means that if we have the speed of light over here relative to our S tag,"},{"Start":"06:28.835 ","End":"06:33.380","Text":"our S will record the exact same speed for the speed of light."},{"Start":"06:33.380 ","End":"06:39.735","Text":"This isn\u0027t intuitive, and it goes against what we learned in previous chapters."},{"Start":"06:39.735 ","End":"06:41.975","Text":"This is only with the speed of light."},{"Start":"06:41.975 ","End":"06:45.260","Text":"The fourth basic principle is that nobody"},{"Start":"06:45.260 ","End":"06:48.980","Text":"can move faster than the speed of light in a vacuum."},{"Start":"06:48.980 ","End":"06:55.115","Text":"In other words, the speed of light is the fastest speed that anything can reach."},{"Start":"06:55.115 ","End":"06:57.690","Text":"There\u0027s nothing faster than that."},{"Start":"06:57.690 ","End":"07:01.040","Text":"As a result, time measurements are different between"},{"Start":"07:01.040 ","End":"07:05.195","Text":"different inertial reference frames. What does that mean?"},{"Start":"07:05.195 ","End":"07:09.830","Text":"If again, we measure the time in our S frame of reference,"},{"Start":"07:09.830 ","End":"07:12.425","Text":"and we measure, let\u0027s say 3 seconds."},{"Start":"07:12.425 ","End":"07:15.140","Text":"In our S tag frame of reference,"},{"Start":"07:15.140 ","End":"07:16.885","Text":"a different time,"},{"Start":"07:16.885 ","End":"07:21.010","Text":"sometime that isn\u0027t 3 seconds will be measured."},{"Start":"07:21.290 ","End":"07:27.635","Text":"What does that mean? That time becomes the fourth coordinates in our x, y, z,"},{"Start":"07:27.635 ","End":"07:32.862","Text":"t vector and experiences the Lorentz Transformation,"},{"Start":"07:32.862 ","End":"07:35.078","Text":"some kind of transformation."},{"Start":"07:35.078 ","End":"07:38.570","Text":"Which means that just like if we were in our S frame of"},{"Start":"07:38.570 ","End":"07:42.349","Text":"reference and we look at some body,"},{"Start":"07:42.349 ","End":"07:44.970","Text":"its position will be x, y,"},{"Start":"07:44.970 ","End":"07:51.500","Text":"z, and time t. Then if we move to our S tag frame of reference,"},{"Start":"07:51.500 ","End":"07:53.420","Text":"it\u0027s going to have different"},{"Start":"07:53.420 ","End":"07:57.175","Text":"x-coordinate and different y-coordinate, and different z-coordinate."},{"Start":"07:57.175 ","End":"07:59.015","Text":"Now also as we\u0027ve learned,"},{"Start":"07:59.015 ","End":"08:02.860","Text":"it will also have different t coordinate."},{"Start":"08:02.860 ","End":"08:05.600","Text":"I know that this is a little bit confusing."},{"Start":"08:05.600 ","End":"08:10.240","Text":"Let\u0027s look a little reminder of our Galilean transform."},{"Start":"08:10.240 ","End":"08:12.245","Text":"Again, we\u0027ll look at how this works,"},{"Start":"08:12.245 ","End":"08:14.120","Text":"and then we\u0027ll understand why in"},{"Start":"08:14.120 ","End":"08:18.400","Text":"special relativity we cannot use this Galilean transform."},{"Start":"08:18.400 ","End":"08:20.750","Text":"We have our frame of reference S,"},{"Start":"08:20.750 ","End":"08:24.035","Text":"which is our lab, and we have 2 bodies;"},{"Start":"08:24.035 ","End":"08:26.645","Text":"we have our car which is body number 1,"},{"Start":"08:26.645 ","End":"08:28.475","Text":"and we have a boy on a skateboard,"},{"Start":"08:28.475 ","End":"08:30.140","Text":"which is body number 2."},{"Start":"08:30.140 ","End":"08:32.870","Text":"Now, the boy and a skateboard is traveling at"},{"Start":"08:32.870 ","End":"08:38.995","Text":"some velocity v_2 relative to our lab\u0027s frame of reference."},{"Start":"08:38.995 ","End":"08:43.880","Text":"We have our car which is moving at some velocity v_1,"},{"Start":"08:43.880 ","End":"08:51.720","Text":"2, and that velocity is relative to our boy on the skateboard."},{"Start":"08:51.770 ","End":"08:55.325","Text":"As we can see from these formulas,"},{"Start":"08:55.325 ","End":"09:00.330","Text":"we can see that our position of the car relative to the boy,"},{"Start":"09:00.330 ","End":"09:03.590","Text":"the position of body 1 relative to body"},{"Start":"09:03.590 ","End":"09:08.420","Text":"2 is going to be the position of body 1 minus the position of body 2."},{"Start":"09:08.420 ","End":"09:10.460","Text":"Similarly, with the velocity,"},{"Start":"09:10.460 ","End":"09:14.780","Text":"the velocity of body 1 relative to body 2."},{"Start":"09:14.780 ","End":"09:19.235","Text":"The velocity of body number 1 relative to the other moving body,"},{"Start":"09:19.235 ","End":"09:20.840","Text":"not relative to the lab,"},{"Start":"09:20.840 ","End":"09:24.275","Text":"is going to be equal to the velocity of our body,"},{"Start":"09:24.275 ","End":"09:27.560","Text":"of our car minus the velocity of body 2,"},{"Start":"09:27.560 ","End":"09:30.955","Text":"which is the velocity of the boy in the skateboard."},{"Start":"09:30.955 ","End":"09:34.460","Text":"These are our equations for our Galilean transformation."},{"Start":"09:34.460 ","End":"09:40.085","Text":"Now let\u0027s imagine that we\u0027re working with special relativity values."},{"Start":"09:40.085 ","End":"09:46.990","Text":"That means that our bodies are moving but at some proportion of the speed of light."},{"Start":"09:46.990 ","End":"09:49.925","Text":"Not at something 10 meters per second"},{"Start":"09:49.925 ","End":"09:53.795","Text":"or velocities like that which we\u0027ve been working with up until now."},{"Start":"09:53.795 ","End":"09:56.885","Text":"But rather, 0.5 c,"},{"Start":"09:56.885 ","End":"09:59.060","Text":"which means half of the speed of light."},{"Start":"09:59.060 ","End":"10:00.500","Text":"C being the speed of light,"},{"Start":"10:00.500 ","End":"10:05.940","Text":"which is equal to 300,000 meters per second."},{"Start":"10:05.990 ","End":"10:08.603","Text":"Our velocity of the car,"},{"Start":"10:08.603 ","End":"10:12.470","Text":"relative to the boy on the skateboard,"},{"Start":"10:12.470 ","End":"10:17.900","Text":"is moving at 0.6 multiplied by the speed of light."},{"Start":"10:17.900 ","End":"10:22.268","Text":"Imagining that we have these values for our velocity,"},{"Start":"10:22.268 ","End":"10:27.099","Text":"when we use our Galilean transform to find the velocity of the car,"},{"Start":"10:27.099 ","End":"10:31.065","Text":"V_1 relative to our lab\u0027s frame of reference."},{"Start":"10:31.065 ","End":"10:33.900","Text":"We\u0027ll plug it in, so we\u0027ll have the V_1,"},{"Start":"10:33.900 ","End":"10:38.150","Text":"2 tag is going to be equal to 0.6 c,"},{"Start":"10:38.150 ","End":"10:40.870","Text":"which is going to be equal to our V_1,"},{"Start":"10:40.870 ","End":"10:44.195","Text":"which is the velocity of the car relative to our lab,"},{"Start":"10:44.195 ","End":"10:48.940","Text":"minus the velocity of the skateboarder relative to the lab."},{"Start":"10:48.940 ","End":"10:51.185","Text":"Once we plug all of that in,"},{"Start":"10:51.185 ","End":"10:56.060","Text":"we\u0027ll see that we get that the velocity of our car relative to"},{"Start":"10:56.060 ","End":"11:01.235","Text":"our lab is going to be equal to 1.1 times the speed of light,"},{"Start":"11:01.235 ","End":"11:07.460","Text":"which as we know is impossible because that defies our basic principle number 4,"},{"Start":"11:07.460 ","End":"11:11.225","Text":"that nobody can move faster than the speed of light."},{"Start":"11:11.225 ","End":"11:17.285","Text":"Here we can see that our car is moving 10% faster than the speed of light."},{"Start":"11:17.285 ","End":"11:19.675","Text":"This is impossible."},{"Start":"11:19.675 ","End":"11:21.755","Text":"Here\u0027s our problem."},{"Start":"11:21.755 ","End":"11:23.435","Text":"This shows that this cannot be,"},{"Start":"11:23.435 ","End":"11:26.045","Text":"and this shows that the problem is in"},{"Start":"11:26.045 ","End":"11:31.805","Text":"our Galilean transform for when we\u0027re working at relativistic velocities."},{"Start":"11:31.805 ","End":"11:34.640","Text":"When we\u0027re working with velocities at the speed of light,"},{"Start":"11:34.640 ","End":"11:38.635","Text":"we cannot use our Galilean transform."},{"Start":"11:38.635 ","End":"11:42.005","Text":"Now, in order to solve this conundrum,"},{"Start":"11:42.005 ","End":"11:48.710","Text":"we use when we\u0027re working at speeds of lights or things moving at much higher velocities,"},{"Start":"11:48.710 ","End":"11:51.560","Text":"velocities that come close to the speed of light, we,"},{"Start":"11:51.560 ","End":"11:53.900","Text":"instead of using our Galilean transform,"},{"Start":"11:53.900 ","End":"11:58.095","Text":"we use our Lorentz Transformations."},{"Start":"11:58.095 ","End":"12:02.450","Text":"That means that very similar to our Galilean transforms,"},{"Start":"12:02.450 ","End":"12:05.525","Text":"but they take into account relativistic speeds."},{"Start":"12:05.525 ","End":"12:09.350","Text":"Now one thing that you\u0027ll notice is that in the Galilean transform,"},{"Start":"12:09.350 ","End":"12:15.010","Text":"our equations for position and velocity are pretty similar."},{"Start":"12:15.010 ","End":"12:20.330","Text":"What we\u0027ll see different in the Lorentz transformation is that"},{"Start":"12:20.330 ","End":"12:26.275","Text":"our equations for our position and the equations for our velocity won\u0027t be similar."},{"Start":"12:26.275 ","End":"12:32.165","Text":"We have a different system for working out position and for working out velocity."},{"Start":"12:32.165 ","End":"12:34.205","Text":"That\u0027s the end of this lesson."},{"Start":"12:34.205 ","End":"12:35.390","Text":"In the next lesson,"},{"Start":"12:35.390 ","End":"12:41.280","Text":"we\u0027re going to be looking at our equations for our Lorentz Transformations."}],"ID":9546},{"Watched":false,"Name":"Lorentz Transformation Equation For Position And Time","Duration":"13m 37s","ChapterTopicVideoID":9269,"CourseChapterTopicPlaylistID":9398,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.710","Text":"Hello. In this lesson,"},{"Start":"00:01.710 ","End":"00:05.790","Text":"we\u0027re going to be speaking about the Lorentz transform for position and time."},{"Start":"00:05.790 ","End":"00:10.575","Text":"We\u0027re not yet going to be speaking about the Lorentz transform for velocity."},{"Start":"00:10.575 ","End":"00:12.315","Text":"Let\u0027s look over here."},{"Start":"00:12.315 ","End":"00:17.250","Text":"Now, here we can see that we have two frames of reference we have S and we have S tag."},{"Start":"00:17.250 ","End":"00:23.860","Text":"Let\u0027s say that our S is our frame of reference of our lab and S tag is our observer,"},{"Start":"00:23.860 ","End":"00:27.500","Text":"which is moving with a constant velocity v_0,"},{"Start":"00:27.500 ","End":"00:31.790","Text":"where v_0 denotes velocity of observer."},{"Start":"00:31.790 ","End":"00:36.830","Text":"Here we can say that it\u0027s a boy moving on a skateboard, for instance."},{"Start":"00:36.830 ","End":"00:38.630","Text":"Now our v_0,"},{"Start":"00:38.630 ","End":"00:39.995","Text":"the v_0 of the observer,"},{"Start":"00:39.995 ","End":"00:45.630","Text":"is the velocity of our boy relative to our S frame of reference."},{"Start":"00:45.630 ","End":"00:49.520","Text":"We\u0027re taking that our S over here is stationary and"},{"Start":"00:49.520 ","End":"00:54.250","Text":"our boy is moving relative to it at the velocity of v_0."},{"Start":"00:54.250 ","End":"00:58.805","Text":"Now let\u0027s imagine that in our frame of reference over here,"},{"Start":"00:58.805 ","End":"01:00.560","Text":"we have an event."},{"Start":"01:00.560 ","End":"01:02.675","Text":"A lamp was switched on."},{"Start":"01:02.675 ","End":"01:07.610","Text":"We can now say that this event where our lamp was lit happened at"},{"Start":"01:07.610 ","End":"01:13.625","Text":"some position and some time relative to our S frame of reference."},{"Start":"01:13.625 ","End":"01:16.460","Text":"If however, we want to know relative to"},{"Start":"01:16.460 ","End":"01:20.885","Text":"our S tag frame of reference when and where our lamp was lit,"},{"Start":"01:20.885 ","End":"01:23.045","Text":"when and where this event happened,"},{"Start":"01:23.045 ","End":"01:28.175","Text":"then what we can do is we can use these equations for our Lorentz transform."},{"Start":"01:28.175 ","End":"01:31.700","Text":"Our position in the S tag frame of reference in"},{"Start":"01:31.700 ","End":"01:35.235","Text":"the x coordinates will be according to this equation,"},{"Start":"01:35.235 ","End":"01:41.450","Text":"and the time in our S tag frame of reference will be according to this equation."},{"Start":"01:41.450 ","End":"01:47.260","Text":"Here, our x and our t represent our position and time,"},{"Start":"01:47.260 ","End":"01:50.420","Text":"in the labs frame of reference in S,"},{"Start":"01:50.420 ","End":"01:57.005","Text":"and our x tag and t tag represent the position and time in our S tag frame of reference,"},{"Start":"01:57.005 ","End":"01:59.570","Text":"the frame of reference of the boy."},{"Start":"01:59.570 ","End":"02:02.975","Text":"Now notice something that isn\u0027t very intuitive,"},{"Start":"02:02.975 ","End":"02:06.035","Text":"that the time at which the boy will see"},{"Start":"02:06.035 ","End":"02:09.720","Text":"the light being lit is in his frame of reference,"},{"Start":"02:09.720 ","End":"02:13.220","Text":"and S tag is different to the time"},{"Start":"02:13.220 ","End":"02:17.645","Text":"of the frame of reference of the lab at which the event occurred."},{"Start":"02:17.645 ","End":"02:19.985","Text":"That\u0027s very interesting."},{"Start":"02:19.985 ","End":"02:22.550","Text":"Also the position will be different."},{"Start":"02:22.550 ","End":"02:25.400","Text":"However, that\u0027s a bit more intuitive to us because"},{"Start":"02:25.400 ","End":"02:28.520","Text":"he\u0027s moving and he\u0027ll see it from a different angle and whatever."},{"Start":"02:28.520 ","End":"02:31.760","Text":"But the time is something that\u0027s new to us."},{"Start":"02:31.760 ","End":"02:36.845","Text":"Now let\u0027s take a look at the values that are written in these equations."},{"Start":"02:36.845 ","End":"02:39.850","Text":"What we can see here is Gamma_0."},{"Start":"02:39.850 ","End":"02:42.530","Text":"Gamma_0, I wrote over here,"},{"Start":"02:42.530 ","End":"02:46.340","Text":"it\u0027s equal to 1 divided by the square root of 1"},{"Start":"02:46.340 ","End":"02:52.335","Text":"minus the velocity of the observer divided by the speed of light squared,"},{"Start":"02:52.335 ","End":"02:57.620","Text":"and it can also be simplified to this where our Beta is simply this over here."},{"Start":"02:57.620 ","End":"03:00.935","Text":"So instead of writing this whole long expression,"},{"Start":"03:00.935 ","End":"03:03.305","Text":"we just simplify it over here to Gamma_0,"},{"Start":"03:03.305 ","End":"03:10.555","Text":"and this is a very common equation to use when we\u0027re dealing with special relativity."},{"Start":"03:10.555 ","End":"03:13.430","Text":"Now, we can see that our Gamma_0 is"},{"Start":"03:13.430 ","End":"03:17.570","Text":"dependent on the velocity of the observer on our v_0."},{"Start":"03:17.570 ","End":"03:19.445","Text":"That means that,"},{"Start":"03:19.445 ","End":"03:21.575","Text":"as soon as I know what my v_0 is,"},{"Start":"03:21.575 ","End":"03:24.530","Text":"I know my value for Gamma_0."},{"Start":"03:24.530 ","End":"03:27.290","Text":"In order to find the position that this event"},{"Start":"03:27.290 ","End":"03:30.590","Text":"happened relative to our S tag frame of reference,"},{"Start":"03:30.590 ","End":"03:32.630","Text":"we have to know what our v_0 is,"},{"Start":"03:32.630 ","End":"03:34.410","Text":"and then we can work out what our Gamma_0 is,"},{"Start":"03:34.410 ","End":"03:39.470","Text":"and then we multiply that by the position that the event happened in our S frame"},{"Start":"03:39.470 ","End":"03:44.670","Text":"of reference minus the velocity of the observer multiplied by the time,"},{"Start":"03:44.670 ","End":"03:52.550","Text":"where the time is the time measured by different observer in our S frame of reference,"},{"Start":"03:52.550 ","End":"03:54.395","Text":"the time that he measured or she,"},{"Start":"03:54.395 ","End":"03:58.195","Text":"that the light was shining in the lab,"},{"Start":"03:58.195 ","End":"04:00.934","Text":"and a similar equation for finding"},{"Start":"04:00.934 ","End":"04:06.350","Text":"the time that the lamp is lit relative to our S tag frame of reference."},{"Start":"04:06.350 ","End":"04:11.390","Text":"What is interesting about that is that the time that the lamp is lit in"},{"Start":"04:11.390 ","End":"04:13.820","Text":"our S tag frame of reference depends on"},{"Start":"04:13.820 ","End":"04:17.825","Text":"the position of the lamp in the S frame of reference."},{"Start":"04:17.825 ","End":"04:22.955","Text":"Now generally what we\u0027ll do is we\u0027ll define our axis,"},{"Start":"04:22.955 ","End":"04:28.130","Text":"such as that if this is our x-axis in our S frame of reference that"},{"Start":"04:28.130 ","End":"04:31.250","Text":"our S tag frame of reference has the x-axis"},{"Start":"04:31.250 ","End":"04:34.835","Text":"going in the same direction as the other frame of reference,"},{"Start":"04:34.835 ","End":"04:38.974","Text":"and we\u0027ll also choose that our velocity of this observer,"},{"Start":"04:38.974 ","End":"04:43.760","Text":"our v_0, is traveling only in the x direction."},{"Start":"04:43.760 ","End":"04:48.320","Text":"In that case, if we define our axes like so,"},{"Start":"04:48.320 ","End":"04:51.500","Text":"then we\u0027ll get that our y tag is equal to y."},{"Start":"04:51.500 ","End":"04:55.400","Text":"The y coordinate in our S tag is equal to the y coordinate in our S,"},{"Start":"04:55.400 ","End":"04:56.960","Text":"and the same for z."},{"Start":"04:56.960 ","End":"04:59.320","Text":"Our z tag will equal z."},{"Start":"04:59.320 ","End":"05:03.723","Text":"We spoke a little bit about our Gamma_0. It\u0027s useful."},{"Start":"05:03.723 ","End":"05:06.290","Text":"Again, I\u0027m reminding you about this Beta symbol."},{"Start":"05:06.290 ","End":"05:08.320","Text":"Our Gamma_0 is sometimes written like this."},{"Start":"05:08.320 ","End":"05:13.295","Text":"It\u0027s good to know that our Beta is equal to our v observer divided by the speed of light."},{"Start":"05:13.295 ","End":"05:17.570","Text":"Another important trait about our Gamma_0 is that it\u0027s always going to be"},{"Start":"05:17.570 ","End":"05:21.800","Text":"bigger than 1 or equal to 1,"},{"Start":"05:21.800 ","End":"05:23.105","Text":"bigger or equal to 1,"},{"Start":"05:23.105 ","End":"05:25.220","Text":"and that is because if we look,"},{"Start":"05:25.220 ","End":"05:29.460","Text":"if our S and S tag are the same system,"},{"Start":"05:29.460 ","End":"05:33.065","Text":"then our v_0 will be equal to 0,"},{"Start":"05:33.065 ","End":"05:37.955","Text":"and then we\u0027ll have 1 divided by the square root of 1, which is equal to 1."},{"Start":"05:37.955 ","End":"05:42.320","Text":"Alternatively, our v_0 will be some number bigger than 0,"},{"Start":"05:42.320 ","End":"05:45.305","Text":"in which case when it\u0027s being divided by c,"},{"Start":"05:45.305 ","End":"05:47.465","Text":"which is 3 times 10^8,"},{"Start":"05:47.465 ","End":"05:49.010","Text":"it\u0027s a very large number,"},{"Start":"05:49.010 ","End":"05:56.040","Text":"that we\u0027ll have some positive integer here and then we\u0027ll have 1 minus that,"},{"Start":"05:56.040 ","End":"05:58.595","Text":"and then 1 divided by the square root,"},{"Start":"05:58.595 ","End":"06:05.245","Text":"which is a number smaller than 1 is going to be a larger number than 1."},{"Start":"06:05.245 ","End":"06:10.475","Text":"Now what we\u0027re going to do is we\u0027re going to look at the case where our v_0,"},{"Start":"06:10.475 ","End":"06:12.500","Text":"the velocity of our observer,"},{"Start":"06:12.500 ","End":"06:16.355","Text":"is significantly smaller than the speed of light."},{"Start":"06:16.355 ","End":"06:20.765","Text":"Then when we substitute this value into our equations,"},{"Start":"06:20.765 ","End":"06:23.675","Text":"we will see what will happen to our equations,"},{"Start":"06:23.675 ","End":"06:26.810","Text":"and what we can expect is that our equations for"},{"Start":"06:26.810 ","End":"06:32.765","Text":"the Lorentz transform will turn in to our equations for our Galilean transform."},{"Start":"06:32.765 ","End":"06:35.570","Text":"This is to be expected because we learned how Galilean"},{"Start":"06:35.570 ","End":"06:38.495","Text":"transforms for a reason and we know that they\u0027re correct."},{"Start":"06:38.495 ","End":"06:40.580","Text":"However, they\u0027re correct for velocities,"},{"Start":"06:40.580 ","End":"06:43.025","Text":"which are smaller than the speed of light."},{"Start":"06:43.025 ","End":"06:46.295","Text":"When we\u0027re dealing with velocities close to the speed of light,"},{"Start":"06:46.295 ","End":"06:49.785","Text":"then we get something slightly different."},{"Start":"06:49.785 ","End":"06:51.735","Text":"Let\u0027s see what happens."},{"Start":"06:51.735 ","End":"06:55.985","Text":"Let\u0027s substitute this into our equation for t tag."},{"Start":"06:55.985 ","End":"07:00.050","Text":"If our v_0 is significantly smaller than our c,"},{"Start":"07:00.050 ","End":"07:03.210","Text":"let\u0027s take a look at what will happen to our Gamma_0."},{"Start":"07:03.500 ","End":"07:11.735","Text":"A small number divided by an infinitely large number will be approximately equal to 0."},{"Start":"07:11.735 ","End":"07:17.705","Text":"In that case, we\u0027ll have in our denominator the square root of 1 minus approximately 0,"},{"Start":"07:17.705 ","End":"07:20.730","Text":"which is, the square root of that will be approximately 1,"},{"Start":"07:20.730 ","End":"07:23.590","Text":"and then 1 divided by 1 is 1."},{"Start":"07:23.600 ","End":"07:29.780","Text":"We\u0027ll get that our Gamma_0 is approximately equal to 1."},{"Start":"07:29.780 ","End":"07:31.865","Text":"Then in that case,"},{"Start":"07:31.865 ","End":"07:33.470","Text":"when we look over here,"},{"Start":"07:33.470 ","End":"07:37.730","Text":"we\u0027ll have 1 multiplied by t minus, and then again,"},{"Start":"07:37.730 ","End":"07:43.415","Text":"a small number divided by a very large number is again approximately equal to 0."},{"Start":"07:43.415 ","End":"07:50.020","Text":"We\u0027ll have that our t tag is going to be equal to 1 multiplied by t minus 0."},{"Start":"07:50.020 ","End":"07:52.835","Text":"We\u0027ll see that our t tag is equal to t,"},{"Start":"07:52.835 ","End":"07:56.150","Text":"which is what we expect it to get in our Galilean transform,"},{"Start":"07:56.150 ","End":"07:59.645","Text":"where the time observed in a frame of reference"},{"Start":"07:59.645 ","End":"08:03.865","Text":"S will be the same time observed in our frame of reference S tag."},{"Start":"08:03.865 ","End":"08:06.956","Text":"Similarly for our x tag,"},{"Start":"08:06.956 ","End":"08:09.830","Text":"we\u0027ll again have that our Gamma_0 is equal to 1."},{"Start":"08:09.830 ","End":"08:18.665","Text":"We\u0027ll get that our x tag is equal to x minus v_0t. That makes sense."},{"Start":"08:18.665 ","End":"08:24.470","Text":"That means that the position that we see the event happening in when we\u0027re in our S tag,"},{"Start":"08:24.470 ","End":"08:28.730","Text":"it\u0027s going to be equal to the position that it is happening in"},{"Start":"08:28.730 ","End":"08:35.605","Text":"our S minus the distance traveled by our observer in S tag."},{"Start":"08:35.605 ","End":"08:39.720","Text":"What we can do is we can call our v_0t."},{"Start":"08:39.720 ","End":"08:43.400","Text":"We can say that that\u0027s the position of the observer, x_0,"},{"Start":"08:43.400 ","End":"08:48.560","Text":"and then we get the exact same equation that we got in our Galilean transform."},{"Start":"08:48.560 ","End":"08:51.800","Text":"Now we\u0027ve seen that when we\u0027re working"},{"Start":"08:51.800 ","End":"08:55.670","Text":"with velocity which are significantly smaller than the speed of light,"},{"Start":"08:55.670 ","End":"08:59.360","Text":"our Lorentz transform is still correct and it simply"},{"Start":"08:59.360 ","End":"09:03.155","Text":"cancels out into our Galilean transform,"},{"Start":"09:03.155 ","End":"09:04.739","Text":"which we know is correct."},{"Start":"09:04.739 ","End":"09:07.550","Text":"Also, all of the or most of the velocities"},{"Start":"09:07.550 ","End":"09:10.484","Text":"that we\u0027ll be measuring on a day-to-day basis,"},{"Start":"09:10.484 ","End":"09:14.560","Text":"will be velocities which are significantly smaller than our c,"},{"Start":"09:14.560 ","End":"09:16.760","Text":"so we can see that it all works out."},{"Start":"09:16.760 ","End":"09:20.405","Text":"Now let\u0027s take a look at our conditions for use for"},{"Start":"09:20.405 ","End":"09:25.735","Text":"the Lorentz transform in order for us to use it correctly and get the correct answers."},{"Start":"09:25.735 ","End":"09:30.560","Text":"Number 1 is that the axis of reference frames are parallel."},{"Start":"09:30.560 ","End":"09:32.075","Text":"What does this mean?"},{"Start":"09:32.075 ","End":"09:33.910","Text":"We\u0027ve already spoken about this."},{"Start":"09:33.910 ","End":"09:39.500","Text":"That means that our x-axis must be parallel in"},{"Start":"09:39.500 ","End":"09:45.875","Text":"our S frame of reference to the x-axis in our S tag frame of reference."},{"Start":"09:45.875 ","End":"09:50.210","Text":"They have to be pointing in the same direction as we already said."},{"Start":"09:50.210 ","End":"09:53.360","Text":"This also includes our explanation when we\u0027re"},{"Start":"09:53.360 ","End":"09:56.720","Text":"defining in which direction our x-axis is in."},{"Start":"09:56.720 ","End":"10:02.595","Text":"That always has to be in the direction of our velocity,"},{"Start":"10:02.595 ","End":"10:04.310","Text":"the direction of travel."},{"Start":"10:04.310 ","End":"10:10.220","Text":"Now the second condition for use is that at time t is equal to t tag,"},{"Start":"10:10.220 ","End":"10:12.460","Text":"which is equal to 0."},{"Start":"10:12.460 ","End":"10:16.190","Text":"That means when our absolute time is equal to 0,"},{"Start":"10:16.190 ","End":"10:18.185","Text":"right at the beginning of our experiment,"},{"Start":"10:18.185 ","End":"10:21.620","Text":"the origins of both our S frame of reference"},{"Start":"10:21.620 ","End":"10:25.315","Text":"and our S tag frame of reference must coalesce."},{"Start":"10:25.315 ","End":"10:29.690","Text":"What does that mean? That means the origin of our S frame of reference and"},{"Start":"10:29.690 ","End":"10:34.310","Text":"our S tag frame of reference began at the exact same point."},{"Start":"10:34.310 ","End":"10:36.070","Text":"They started at the exact same points,"},{"Start":"10:36.070 ","End":"10:39.470","Text":"and then 1 of the frames of reference in this specific example,"},{"Start":"10:39.470 ","End":"10:46.115","Text":"our S tag, started moving at a certain velocity and moved away from that initial point."},{"Start":"10:46.115 ","End":"10:48.395","Text":"But at t is equal to 0,"},{"Start":"10:48.395 ","End":"10:54.125","Text":"that specific point has to be the same for both origins."},{"Start":"10:54.125 ","End":"10:58.280","Text":"Now the next set of equations which are important to note,"},{"Start":"10:58.280 ","End":"11:01.760","Text":"is our reverse transformation. What does that mean?"},{"Start":"11:01.760 ","End":"11:06.500","Text":"Here we saw that our event happened in our S frame of reference and we want to"},{"Start":"11:06.500 ","End":"11:11.440","Text":"know what that event looks like in our S tag frame of reference,"},{"Start":"11:11.440 ","End":"11:13.909","Text":"and then we use our Lorentz transform,"},{"Start":"11:13.909 ","End":"11:15.680","Text":"which is these equations."},{"Start":"11:15.680 ","End":"11:21.020","Text":"Now, what happens if our light bulb is let in our S tag frame of"},{"Start":"11:21.020 ","End":"11:29.070","Text":"reference and we want to know what this event looks like in our S frame of reference."},{"Start":"11:30.320 ","End":"11:34.565","Text":"What we use is our reverse transformation."},{"Start":"11:34.565 ","End":"11:36.980","Text":"Then we have, it\u0027s a very similar equation"},{"Start":"11:36.980 ","End":"11:40.355","Text":"to our transformation that we just learned about."},{"Start":"11:40.355 ","End":"11:44.375","Text":"However, it\u0027s using our coordinates from our S tag frame of reference,"},{"Start":"11:44.375 ","End":"11:48.880","Text":"and then calculating them for our S frame of reference."},{"Start":"11:48.880 ","End":"11:52.350","Text":"How did we get to these equations?"},{"Start":"11:52.350 ","End":"11:56.420","Text":"What we did is from our Lorentz transform,"},{"Start":"11:56.420 ","End":"11:59.360","Text":"we isolated out our x and our t and then"},{"Start":"11:59.360 ","End":"12:02.390","Text":"did some algebra in order to get to these equations."},{"Start":"12:02.390 ","End":"12:06.565","Text":"Now another way to get this reverse transformation is that"},{"Start":"12:06.565 ","End":"12:11.390","Text":"if we\u0027re going from our S being our frame of reference,"},{"Start":"12:11.390 ","End":"12:13.640","Text":"which is stationary, and our S tag is moving"},{"Start":"12:13.640 ","End":"12:17.645","Text":"away in this direction at a velocity of v_0."},{"Start":"12:17.645 ","End":"12:21.830","Text":"We know that if we\u0027re starting from our S tag frame of reference and we say that we\u0027re"},{"Start":"12:21.830 ","End":"12:26.045","Text":"going to take this frame of reference as the stationary frame of reference,"},{"Start":"12:26.045 ","End":"12:30.520","Text":"then that means that our S frame of reference is going to be moving"},{"Start":"12:30.520 ","End":"12:35.675","Text":"away from our S tag frame of reference at the exact same velocity of v_0,"},{"Start":"12:35.675 ","End":"12:38.275","Text":"just in the opposite direction."},{"Start":"12:38.275 ","End":"12:44.870","Text":"That means that all of our equations have to be kept because they\u0027re continued."},{"Start":"12:44.870 ","End":"12:47.795","Text":"The physics doesn\u0027t change just because the direction"},{"Start":"12:47.795 ","End":"12:51.399","Text":"of the velocity vector is in the opposite direction."},{"Start":"12:51.399 ","End":"12:54.650","Text":"In that case, I just switch over."},{"Start":"12:54.650 ","End":"12:57.035","Text":"Instead of x tag is equal to,"},{"Start":"12:57.035 ","End":"12:58.670","Text":"I\u0027ll have x is equal to."},{"Start":"12:58.670 ","End":"13:01.490","Text":"Then because over here I have"},{"Start":"13:01.490 ","End":"13:05.990","Text":"my v_0 traveling in this direction and here I have negative v_0."},{"Start":"13:05.990 ","End":"13:08.735","Text":"In my equation, wherever I see a v_0,"},{"Start":"13:08.735 ","End":"13:11.245","Text":"I\u0027ll substitute a negative v_0."},{"Start":"13:11.245 ","End":"13:14.250","Text":"Here we can see the sign switch."},{"Start":"13:14.250 ","End":"13:17.285","Text":"Here we have negative v_0 and here we have negative negative v_0,"},{"Start":"13:17.285 ","End":"13:18.790","Text":"which is positive v_0,"},{"Start":"13:18.790 ","End":"13:24.305","Text":"and the similar idea by switching t and t tag in the equations."},{"Start":"13:24.305 ","End":"13:26.695","Text":"That\u0027s all we have to do."},{"Start":"13:26.695 ","End":"13:30.995","Text":"Then we can see that when we\u0027re going from S tag to S,"},{"Start":"13:30.995 ","End":"13:35.135","Text":"we just have to reverse our Lorentz transform."},{"Start":"13:35.135 ","End":"13:38.940","Text":"That\u0027s the end of this lesson."}],"ID":9547},{"Watched":false,"Name":"Exercise 1","Duration":"20m 40s","ChapterTopicVideoID":9267,"CourseChapterTopicPlaylistID":9398,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.025","Text":"Hello. In this question,"},{"Start":"00:02.025 ","End":"00:04.410","Text":"we have a motorbike which we\u0027re being told"},{"Start":"00:04.410 ","End":"00:08.010","Text":"travels at a constant velocity in a straight line."},{"Start":"00:08.010 ","End":"00:11.145","Text":"An observer on the ground records that the motorbike"},{"Start":"00:11.145 ","End":"00:15.270","Text":"traveled a distance of 540 kilometers."},{"Start":"00:15.270 ","End":"00:20.445","Text":"Another observer is in an airplane which is traveling really fast"},{"Start":"00:20.445 ","End":"00:26.100","Text":"at v is equal to half the speed of light in the same direction as the motorbike."},{"Start":"00:26.100 ","End":"00:29.850","Text":"This observer, the observer in the plane records that"},{"Start":"00:29.850 ","End":"00:34.139","Text":"the motorbikes travels a total of 0.01 seconds."},{"Start":"00:34.139 ","End":"00:35.805","Text":"Our first question is,"},{"Start":"00:35.805 ","End":"00:39.710","Text":"what is the velocity of the motorbike in the Earth\u0027s reference frame,"},{"Start":"00:39.710 ","End":"00:42.500","Text":"and our second one is what is the distance traveled by"},{"Start":"00:42.500 ","End":"00:47.215","Text":"the motorbike according to the observer in the airplane?"},{"Start":"00:47.215 ","End":"00:49.580","Text":"Let\u0027s take a look."},{"Start":"00:49.580 ","End":"00:52.595","Text":"First of all, we have our frame of reference,"},{"Start":"00:52.595 ","End":"00:54.575","Text":"which is our observe on the ground,"},{"Start":"00:54.575 ","End":"00:58.610","Text":"and let\u0027s call this frame of reference our s. What we\u0027re"},{"Start":"00:58.610 ","End":"01:02.840","Text":"trying to find as the velocity of the motorbike on the ground."},{"Start":"01:02.840 ","End":"01:05.430","Text":"Let\u0027s call this v_1."},{"Start":"01:05.710 ","End":"01:12.695","Text":"Our next frame of reference is our frame of reference for the observer in the airplane."},{"Start":"01:12.695 ","End":"01:14.915","Text":"Let\u0027s call this S tag."},{"Start":"01:14.915 ","End":"01:18.575","Text":"And we\u0027re trying to find the distance traveled."},{"Start":"01:18.575 ","End":"01:26.315","Text":"So we\u0027re trying to find displacement in s\u0027 frame of reference."},{"Start":"01:26.315 ","End":"01:30.535","Text":"Let\u0027s see how we answer this question."},{"Start":"01:30.535 ","End":"01:34.820","Text":"The first thing that we have to do is we have to notice that we have"},{"Start":"01:34.820 ","End":"01:38.840","Text":"a question here which is referring to half the speed of light,"},{"Start":"01:38.840 ","End":"01:41.510","Text":"which is already relatively close to the speed of light,"},{"Start":"01:41.510 ","End":"01:45.910","Text":"which means that we\u0027re dealing with relativistic motion."},{"Start":"01:45.910 ","End":"01:49.040","Text":"In that case we have to write down our equations which are"},{"Start":"01:49.040 ","End":"01:52.115","Text":"different to our normal motion."},{"Start":"01:52.115 ","End":"01:54.410","Text":"Let\u0027s write down all of our equations."},{"Start":"01:54.410 ","End":"01:57.410","Text":"We know that our x-coordinate and"},{"Start":"01:57.410 ","End":"02:02.565","Text":"our s\u0027 frame of reference is going to be equal to Gamma 0,"},{"Start":"02:02.565 ","End":"02:04.830","Text":"which is our Lorentz transformation,"},{"Start":"02:04.830 ","End":"02:08.705","Text":"multiplied by our x coordinate and our s frame of reference"},{"Start":"02:08.705 ","End":"02:15.780","Text":"minus our velocity being traveled multiplied by t."},{"Start":"02:16.120 ","End":"02:24.950","Text":"Then we have that our time is going to be equal to Gamma 0 multiplied by t minus"},{"Start":"02:24.950 ","End":"02:30.915","Text":"our v multiplied by"},{"Start":"02:30.915 ","End":"02:36.935","Text":"our x coordinate and our s frame of reference divided by the speed of light squared."},{"Start":"02:36.935 ","End":"02:40.894","Text":"Then we have that y\u0027 is equal to y,"},{"Start":"02:40.894 ","End":"02:45.230","Text":"and our z\u0027 is equal to z."},{"Start":"02:45.230 ","End":"02:53.045","Text":"Then we have that our Beta is equal to our velocity divided by our speed of light,"},{"Start":"02:53.045 ","End":"02:59.160","Text":"and that our Gamma 0 is simply equal to our"},{"Start":"02:59.160 ","End":"03:07.590","Text":"1 divided by the square root of 1 minus Beta squared."},{"Start":"03:07.720 ","End":"03:11.250","Text":"Sorry, that\u0027s a bit tilted."},{"Start":"03:12.740 ","End":"03:15.020","Text":"These are our equations."},{"Start":"03:15.020 ","End":"03:18.965","Text":"Now let\u0027s go on to seeing how we are going to answer this question."},{"Start":"03:18.965 ","End":"03:22.010","Text":"Now, when we\u0027re working with relativistic motion,"},{"Start":"03:22.010 ","End":"03:25.875","Text":"it\u0027s very easy to work with our intuition,"},{"Start":"03:25.875 ","End":"03:29.165","Text":"and our intuition in this case is something"},{"Start":"03:29.165 ","End":"03:32.990","Text":"that can confuse us and can make us write the wrong answer."},{"Start":"03:32.990 ","End":"03:37.550","Text":"The way that we\u0027re going to answer this question is by using a table and"},{"Start":"03:37.550 ","End":"03:41.820","Text":"we\u0027re going to work in a logical order,"},{"Start":"03:41.820 ","End":"03:45.905","Text":"and that way we\u0027re not going to make any stupid mistakes."},{"Start":"03:45.905 ","End":"03:48.290","Text":"Let\u0027s draw out this table."},{"Start":"03:48.290 ","End":"03:53.240","Text":"Here we\u0027re going to have our event over here."},{"Start":"03:53.240 ","End":"03:58.730","Text":"Then here we\u0027re going to have our s frame of reference and what\u0027s happening there."},{"Start":"03:58.730 ","End":"04:04.025","Text":"Then here we\u0027re going to have our s\u0027 frame of reference."},{"Start":"04:04.025 ","End":"04:11.410","Text":"Our first event, number 1 is our motivate begins motion,"},{"Start":"04:15.110 ","End":"04:23.670","Text":"and then our second event is when our motorbike reaches its destination."},{"Start":"04:24.110 ","End":"04:27.290","Text":"Now in our s\u0027 frame of reference,"},{"Start":"04:27.290 ","End":"04:31.370","Text":"we just have to mention that the velocity that our frame of reference is"},{"Start":"04:31.370 ","End":"04:36.295","Text":"traveling in is equal to half of the speed of light."},{"Start":"04:36.295 ","End":"04:38.675","Text":"Now for each frame of reference,"},{"Start":"04:38.675 ","End":"04:44.539","Text":"we\u0027re going to describe what event 1 and event 2 looks like in that frame of reference."},{"Start":"04:44.539 ","End":"04:49.970","Text":"Then when we\u0027re speaking simply within our s frame of reference on Earth,"},{"Start":"04:49.970 ","End":"04:53.900","Text":"we can simply describe its motion within our normal terms."},{"Start":"04:53.900 ","End":"04:56.630","Text":"All we have to do is when we move between R,"},{"Start":"04:56.630 ","End":"05:02.440","Text":"s and s\u0027, we have to use these equations for a transform."},{"Start":"05:02.440 ","End":"05:05.890","Text":"Let\u0027s see how we work."},{"Start":"05:05.990 ","End":"05:10.350","Text":"The first thing that we want to do and the first rule of how we can"},{"Start":"05:10.350 ","End":"05:17.044","Text":"use our different frame of references is to say that at some point within the motion,"},{"Start":"05:17.044 ","End":"05:20.945","Text":"ideally right at the beginning our two origins for"},{"Start":"05:20.945 ","End":"05:26.430","Text":"both frame of references will be at the exact same points."},{"Start":"05:26.690 ","End":"05:30.320","Text":"That means that in our s frame of reference,"},{"Start":"05:30.320 ","End":"05:33.440","Text":"we can say that our x-coordinate it will"},{"Start":"05:33.440 ","End":"05:41.490","Text":"start at 0 and our time there is also equal to 0."},{"Start":"05:41.490 ","End":"05:44.480","Text":"It\u0027s right at the beginning of the motion of the motorbike."},{"Start":"05:44.480 ","End":"05:52.220","Text":"In that case, once we substitute in our x=0 or t=0 into this equation over here,"},{"Start":"05:52.220 ","End":"05:57.420","Text":"for our transform, we\u0027ll get that our x\u0027 is therefore equal to 0,"},{"Start":"05:57.740 ","End":"06:01.960","Text":"and so is our t\u0027=0."},{"Start":"06:04.010 ","End":"06:09.665","Text":"What we first have to do is that both of our origins will be right at the beginning,"},{"Start":"06:09.665 ","End":"06:12.870","Text":"lined up at the exact same point."},{"Start":"06:13.820 ","End":"06:18.410","Text":"Now, let\u0027s speak about our next point or our next event,"},{"Start":"06:18.410 ","End":"06:22.360","Text":"which is when our motivate reaches its destination."},{"Start":"06:22.360 ","End":"06:27.425","Text":"That means that its position x_2 is equal to,"},{"Start":"06:27.425 ","End":"06:31.475","Text":"we\u0027re told in the question, 540 kilometers away."},{"Start":"06:31.475 ","End":"06:39.035","Text":"That\u0027s 540 times 10^3 meters."},{"Start":"06:39.035 ","End":"06:43.970","Text":"Then we want to find the time that this happens, t_2."},{"Start":"06:43.970 ","End":"06:46.115","Text":"Now, this is something that we don\u0027t know."},{"Start":"06:46.115 ","End":"06:51.175","Text":"We have no idea how long it took to travel from 0-540 kilometers,"},{"Start":"06:51.175 ","End":"06:53.500","Text":"so we\u0027ll put a question mark here."},{"Start":"06:53.500 ","End":"06:56.705","Text":"Then we know that its velocity over here,"},{"Start":"06:56.705 ","End":"07:00.815","Text":"which is what we\u0027re trying to find from over here,"},{"Start":"07:00.815 ","End":"07:03.890","Text":"is going to be our distance,"},{"Start":"07:03.890 ","End":"07:05.030","Text":"which is our x_2,"},{"Start":"07:05.030 ","End":"07:09.470","Text":"which is what we know, divided by our time,"},{"Start":"07:09.470 ","End":"07:11.210","Text":"which is what we don\u0027t know."},{"Start":"07:11.210 ","End":"07:14.600","Text":"This is in fact what we\u0027re being asked in question 1."},{"Start":"07:14.600 ","End":"07:16.430","Text":"What is the velocity of the motorbike?"},{"Start":"07:16.430 ","End":"07:18.455","Text":"In order to find the velocity,"},{"Start":"07:18.455 ","End":"07:24.240","Text":"what we have to do is we have to find out what our t_2 is equal to."},{"Start":"07:24.280 ","End":"07:30.320","Text":"Now let\u0027s go on to describe event number 2 when a motorbike reaches its destination,"},{"Start":"07:30.320 ","End":"07:35.600","Text":"but for our s\u0027 frame of reference. Let\u0027s take a look."},{"Start":"07:35.600 ","End":"07:39.735","Text":"Our x\u0027_2 is unknown,"},{"Start":"07:39.735 ","End":"07:43.250","Text":"that\u0027s in fact what we\u0027re being asked in question number 2,"},{"Start":"07:43.250 ","End":"07:48.680","Text":"what\u0027s the distance traveled by the motorbike according to our s\u0027 frame of reference."},{"Start":"07:48.680 ","End":"07:50.350","Text":"So that we don\u0027t know."},{"Start":"07:50.350 ","End":"07:53.400","Text":"Then we have our t\u0027_2."},{"Start":"07:53.400 ","End":"07:57.500","Text":"Here, this detail we\u0027re told in our question."},{"Start":"07:57.500 ","End":"08:01.160","Text":"We\u0027re told that it\u0027s 0.01 seconds,"},{"Start":"08:01.160 ","End":"08:04.535","Text":"so 0.01 seconds,"},{"Start":"08:04.535 ","End":"08:11.265","Text":"and notice that this t\u0027_2 is definitely not equal to our t_2 over here,"},{"Start":"08:11.265 ","End":"08:17.520","Text":"and that our x\u0027_2 here is definitely not equal to our x_2 over here."},{"Start":"08:17.840 ","End":"08:23.570","Text":"Then the next thing that we want to know is our velocity,"},{"Start":"08:23.570 ","End":"08:25.869","Text":"and that were given in our question."},{"Start":"08:25.869 ","End":"08:29.190","Text":"We can write that, so it\u0027s equal to half of"},{"Start":"08:29.190 ","End":"08:38.270","Text":"c. So now let\u0027s see how we can find out what our t_2 is equal to."},{"Start":"08:38.390 ","End":"08:42.785","Text":"Now what we\u0027re going to do is we\u0027re going to try and find out,"},{"Start":"08:42.785 ","End":"08:45.835","Text":"first of all to answer question number 1."},{"Start":"08:45.835 ","End":"08:50.920","Text":"That means we\u0027re going to try and find out what our t_2 equal to,"},{"Start":"08:50.920 ","End":"08:53.780","Text":"and then we can find out our velocity."},{"Start":"08:53.780 ","End":"08:58.340","Text":"As we know, we have our value for t\u0027_2,"},{"Start":"08:58.340 ","End":"09:01.230","Text":"and we have our Lorentz transform,"},{"Start":"09:01.230 ","End":"09:04.575","Text":"transforming between our t\u0027 to"},{"Start":"09:04.575 ","End":"09:11.965","Text":"our t. Now what we\u0027re going to do is we\u0027re going to try and solve this equation."},{"Start":"09:11.965 ","End":"09:14.595","Text":"What we\u0027re going to have is"},{"Start":"09:14.595 ","End":"09:22.830","Text":"our t\u0027 is going to be equal to 0.01 and that\u0027s going to be equal to Gamma 0."},{"Start":"09:22.830 ","End":"09:30.545","Text":"Now our Gamma 0 is equal to 1 divided by the square root of 1 minus Beta squared,"},{"Start":"09:30.545 ","End":"09:34.090","Text":"where Beta is our v/c,"},{"Start":"09:34.090 ","End":"09:38.615","Text":"so this is going to be 1 minus Beta squared,"},{"Start":"09:38.615 ","End":"09:40.985","Text":"so that\u0027s going to be our v^2."},{"Start":"09:40.985 ","End":"09:43.085","Text":"Now what\u0027s our v it\u0027s half c,"},{"Start":"09:43.085 ","End":"09:50.620","Text":"so it\u0027s going to be 1/4 c^2 divided by c^2,"},{"Start":"09:50.620 ","End":"09:53.825","Text":"and the square root of all of that,"},{"Start":"09:53.825 ","End":"09:58.490","Text":"then that is going to be multiplied by our time,"},{"Start":"09:58.490 ","End":"10:00.050","Text":"which is what we\u0027re trying to find,"},{"Start":"10:00.050 ","End":"10:02.930","Text":"so that\u0027s going to be multiplied by t_2."},{"Start":"10:02.930 ","End":"10:04.520","Text":"Because here we have our t\u0027_2,"},{"Start":"10:04.520 ","End":"10:06.985","Text":"which means that here goes our t,"},{"Start":"10:06.985 ","End":"10:14.630","Text":"and then minus the velocity in our s tag frame of reference,"},{"Start":"10:14.630 ","End":"10:17.465","Text":"which is equal to a half c,"},{"Start":"10:17.465 ","End":"10:23.785","Text":"so half of c multiplied by our position,"},{"Start":"10:23.785 ","End":"10:26.000","Text":"so that\u0027s our x_2 position,"},{"Start":"10:26.000 ","End":"10:32.405","Text":"which is multiplied by 540 times 10^3,"},{"Start":"10:32.405 ","End":"10:40.170","Text":"and then divided by our speed of light squared, so c^2."},{"Start":"10:40.480 ","End":"10:44.890","Text":"Now let\u0027s try and solve this."},{"Start":"10:44.890 ","End":"10:50.775","Text":"The first thing is we can do is cross out this c^2 and this c^2."},{"Start":"10:50.775 ","End":"10:56.070","Text":"Then we can also cross out this c and this square root over here,"},{"Start":"10:56.070 ","End":"11:01.250","Text":"and then we\u0027ll be left with 0.01 is equal"},{"Start":"11:01.250 ","End":"11:07.145","Text":"to 1 divided by the square root of 1 minus 1/4."},{"Start":"11:07.145 ","End":"11:14.035","Text":"That\u0027s going to equal to 1 over the square root 3/4."},{"Start":"11:14.035 ","End":"11:16.295","Text":"Which when we simplify that,"},{"Start":"11:16.295 ","End":"11:22.215","Text":"is going to be equal to 2 divided by the square root of 3,"},{"Start":"11:22.215 ","End":"11:27.590","Text":"and then that\u0027s going to be multiplied by t_2 minus and then we have"},{"Start":"11:27.590 ","End":"11:33.720","Text":"1/2 times 540 times 10^3."},{"Start":"11:33.720 ","End":"11:36.370","Text":"That\u0027s going to be equal to"},{"Start":"11:38.330 ","End":"11:45.765","Text":"270 times 10^3 divided by our c,"},{"Start":"11:45.765 ","End":"11:47.010","Text":"which as we know,"},{"Start":"11:47.010 ","End":"11:55.078","Text":"is 3 times 10^8 meters per second."},{"Start":"11:55.078 ","End":"11:58.840","Text":"Now let\u0027s simplify that even more."},{"Start":"11:58.840 ","End":"12:02.800","Text":"We have 2 divided by root 3,"},{"Start":"12:02.800 ","End":"12:05.060","Text":"and then we\u0027ll have that multiple,"},{"Start":"12:05.060 ","End":"12:13.755","Text":"t_2 minus 270 divided by 3 is going to be 90 times 10."},{"Start":"12:13.755 ","End":"12:21.515","Text":"Then 10^ 3 divided by 10^ 8 will be 10^ minus 5."},{"Start":"12:21.515 ","End":"12:32.260","Text":"Then we can write 0.01 is equal 2 over root 3 multiplied by t_2 minus,"},{"Start":"12:32.260 ","End":"12:35.909","Text":"and let\u0027s simplify this even more,"},{"Start":"12:35.909 ","End":"12:42.410","Text":"0.9 times 10^ negative 3."},{"Start":"12:43.530 ","End":"12:48.130","Text":"Now we\u0027re remembering that we have this 1 unknown, which is t_2."},{"Start":"12:48.130 ","End":"12:50.470","Text":"We want to isolate this out."},{"Start":"12:50.470 ","End":"12:54.325","Text":"Once we rearrange this equation through simple algebra,"},{"Start":"12:54.325 ","End":"13:01.695","Text":"we\u0027ll get that our t_2 is simply going to be equal to root 3 divided by 2"},{"Start":"13:01.695 ","End":"13:11.640","Text":"multiplied by 0.01 plus 0.9 times 10^ minus 3."},{"Start":"13:11.640 ","End":"13:14.955","Text":"Then once you plug this into a calculator,"},{"Start":"13:14.955 ","End":"13:25.105","Text":"you\u0027ll get that our t_2 is equal to 9.56 times 10^ minus 3 seconds."},{"Start":"13:25.105 ","End":"13:29.930","Text":"Now we found our t_2."},{"Start":"13:30.750 ","End":"13:33.430","Text":"We can remember what our question is,"},{"Start":"13:33.430 ","End":"13:37.120","Text":"which is to find the velocity of the motorbike."},{"Start":"13:37.120 ","End":"13:39.760","Text":"We found what our t_2 is equal to,"},{"Start":"13:39.760 ","End":"13:41.500","Text":"let\u0027s write this again over here,"},{"Start":"13:41.500 ","End":"13:49.540","Text":"9.56 times 10^ negative 3 seconds."},{"Start":"13:49.540 ","End":"13:56.230","Text":"Now we simply have to plug this in to our equation here for v_1."},{"Start":"13:56.230 ","End":"13:59.560","Text":"Then we can get that our v_1,"},{"Start":"13:59.560 ","End":"14:01.210","Text":"which is what we\u0027re trying to find,"},{"Start":"14:01.210 ","End":"14:06.100","Text":"is equal to x_2 divided by t_2,"},{"Start":"14:06.100 ","End":"14:16.335","Text":"which is equal to 540 times 10^ 3 meters divided by our t_2,"},{"Start":"14:16.335 ","End":"14:22.920","Text":"which is 9.56 times 10^ negative 3."},{"Start":"14:22.920 ","End":"14:28.410","Text":"Then we will get that this is equal to 5.65,"},{"Start":"14:28.410 ","End":"14:30.365","Text":"just put this into a calculator,"},{"Start":"14:30.365 ","End":"14:36.470","Text":"times 10^ 7 meters per second."},{"Start":"14:37.370 ","End":"14:41.835","Text":"Now what we\u0027ve done is we\u0027ve solved question number 1."},{"Start":"14:41.835 ","End":"14:44.940","Text":"We got our answer for t_2 and then we could"},{"Start":"14:44.940 ","End":"14:47.670","Text":"substitute it in to final answer for our velocity,"},{"Start":"14:47.670 ","End":"14:49.050","Text":"which was the question."},{"Start":"14:49.050 ","End":"14:51.820","Text":"I\u0027ve put the answers in our table."},{"Start":"14:51.820 ","End":"14:57.625","Text":"Let\u0027s go back 1 second and see what we did in order to answer question number 1."},{"Start":"14:57.625 ","End":"15:00.070","Text":"The first thing that we did is we"},{"Start":"15:00.070 ","End":"15:03.265","Text":"understood that there were 2 different frames of reference over here,"},{"Start":"15:03.265 ","End":"15:06.430","Text":"and that we\u0027re dealing with relativistic velocities."},{"Start":"15:06.430 ","End":"15:09.220","Text":"Then we said, that\u0027s relativity."},{"Start":"15:09.220 ","End":"15:13.225","Text":"We\u0027re going to write down our equations for our Lorentz transforms."},{"Start":"15:13.225 ","End":"15:17.815","Text":"Then we remembered that in order to use our Lorentz transforms,"},{"Start":"15:17.815 ","End":"15:19.825","Text":"that means that at some point in time,"},{"Start":"15:19.825 ","End":"15:23.410","Text":"ideally right at the beginning of the motion,"},{"Start":"15:23.410 ","End":"15:27.070","Text":"our s frame of reference and our s tag frame of reference have to be"},{"Start":"15:27.070 ","End":"15:31.435","Text":"at the exact same point in time and in space."},{"Start":"15:31.435 ","End":"15:35.950","Text":"Then what we said is that we\u0027ll write out a table which"},{"Start":"15:35.950 ","End":"15:40.570","Text":"describes our events and how they take place in each frame of reference,"},{"Start":"15:40.570 ","End":"15:42.595","Text":"in our S frame of reference on earth,"},{"Start":"15:42.595 ","End":"15:44.575","Text":"and our S tag frame of reference,"},{"Start":"15:44.575 ","End":"15:49.225","Text":"which is our relativistic frame of reference of the airplane over here."},{"Start":"15:49.225 ","End":"15:53.710","Text":"Then what we did is we said that both of our frames of reference"},{"Start":"15:53.710 ","End":"15:58.060","Text":"will be at the exact same time and place right at the beginning."},{"Start":"15:58.060 ","End":"15:59.980","Text":"Then we could carry on."},{"Start":"15:59.980 ","End":"16:04.284","Text":"We described each event in its frame of reference,"},{"Start":"16:04.284 ","End":"16:06.595","Text":"and works out what our unknowns were,"},{"Start":"16:06.595 ","End":"16:12.160","Text":"and then use our Lorentz transform to transform between our time that we"},{"Start":"16:12.160 ","End":"16:14.710","Text":"knew in our S tag frame of reference in"},{"Start":"16:14.710 ","End":"16:17.679","Text":"order to get our time in our S frame of reference,"},{"Start":"16:17.679 ","End":"16:19.885","Text":"and then we could use our symbol,"},{"Start":"16:19.885 ","End":"16:24.280","Text":"speed is equal to distance over time in order to work"},{"Start":"16:24.280 ","End":"16:29.455","Text":"out the velocity of the motorbike in our S frame of reference,"},{"Start":"16:29.455 ","End":"16:32.450","Text":"which is what we were being asked to do."},{"Start":"16:32.480 ","End":"16:36.540","Text":"Now let\u0027s move on to what is the distance traveled"},{"Start":"16:36.540 ","End":"16:39.960","Text":"by the motorbike according to the observer in the airplane."},{"Start":"16:39.960 ","End":"16:45.760","Text":"That means according to our observer in our S tag frame of reference."},{"Start":"16:46.590 ","End":"16:51.520","Text":"What I wanted to do, is I want to find my x_2 tag."},{"Start":"16:51.520 ","End":"16:54.445","Text":"Now looking at my Lorentz transform,"},{"Start":"16:54.445 ","End":"16:56.515","Text":"I can see that my equation at"},{"Start":"16:56.515 ","End":"17:03.639","Text":"my position in my S tag frame of reference is equal to my Gamma 0"},{"Start":"17:03.639 ","End":"17:08.440","Text":"multiplied by my position in my S frame of reference minus"},{"Start":"17:08.440 ","End":"17:16.040","Text":"my relativistic velocity times my time in my S frame of reference."},{"Start":"17:16.080 ","End":"17:19.855","Text":"We\u0027ll notice that I have my position,"},{"Start":"17:19.855 ","End":"17:23.035","Text":"which is just x_2, I have my v,"},{"Start":"17:23.035 ","End":"17:28.750","Text":"which is 1/2c, and I have my t because I just worked that out in the previous question,"},{"Start":"17:28.750 ","End":"17:30.250","Text":"and that\u0027s this over here."},{"Start":"17:30.250 ","End":"17:34.765","Text":"All I have to do is plug in everything into that equation."},{"Start":"17:34.765 ","End":"17:40.045","Text":"I have my x_2 tag is equal to my Gamma 0,"},{"Start":"17:40.045 ","End":"17:42.565","Text":"which as discussed before, it\u0027s this,"},{"Start":"17:42.565 ","End":"17:48.295","Text":"which we saw was equal to 2 divided by root 3."},{"Start":"17:48.295 ","End":"17:54.160","Text":"Then that is going to be multiplied by my position in my S frame of reference,"},{"Start":"17:54.160 ","End":"17:59.710","Text":"which is 540 times 10^ 3 meters."},{"Start":"17:59.710 ","End":"18:04.795","Text":"Then it\u0027s going to be minus my velocity."},{"Start":"18:04.795 ","End":"18:11.245","Text":"That\u0027s equal to 1/2C multiplied by my time."},{"Start":"18:11.245 ","End":"18:13.570","Text":"Because I\u0027m doing x_2,"},{"Start":"18:13.570 ","End":"18:15.475","Text":"so I have to find my t_2,"},{"Start":"18:15.475 ","End":"18:17.395","Text":"which as we just found out,"},{"Start":"18:17.395 ","End":"18:24.410","Text":"was 9.56 times 10^ minus 3."},{"Start":"18:24.840 ","End":"18:32.125","Text":"Then that is going to be equal to 2 divided by root 3,"},{"Start":"18:32.125 ","End":"18:34.315","Text":"and then we\u0027ll have multiplied,"},{"Start":"18:34.315 ","End":"18:40.460","Text":"and then it will be 5.4 times 10^ 5."},{"Start":"18:40.770 ","End":"18:43.720","Text":"See how I got from here to here."},{"Start":"18:43.720 ","End":"18:48.625","Text":"Then that\u0027s going to be minus 1/2 times 9.56."},{"Start":"18:48.625 ","End":"18:54.325","Text":"That\u0027s going to be 14.34."},{"Start":"18:54.325 ","End":"19:00.625","Text":"Then my speed of light is 3 times 10^ 8 multiplied by 10^ minus 3."},{"Start":"19:00.625 ","End":"19:06.295","Text":"Multiplied by all of that is going to be times 10^ 5."},{"Start":"19:06.295 ","End":"19:11.470","Text":"You could have also just plugged everything over here into a calculator,"},{"Start":"19:11.470 ","End":"19:16.135","Text":"where remembering that C is 3 times 10^ 8 meters per second."},{"Start":"19:16.135 ","End":"19:18.265","Text":"Then once you solve all of this,"},{"Start":"19:18.265 ","End":"19:20.155","Text":"also in a calculator you can,"},{"Start":"19:20.155 ","End":"19:29.990","Text":"it\u0027s going to be equal to negative 10.32 times 10^ 5 meters."},{"Start":"19:31.710 ","End":"19:39.220","Text":"What we can notice is that we got a negative position that the motorbike will be in."},{"Start":"19:39.220 ","End":"19:44.935","Text":"In other words, if our airplane is over here,"},{"Start":"19:44.935 ","End":"19:47.215","Text":"here\u0027s our airplane,"},{"Start":"19:47.215 ","End":"19:53.090","Text":"then that means that our motorbike is going to be somewhere over here."},{"Start":"19:53.190 ","End":"19:59.124","Text":"What does this mean? We can see that our airplane"},{"Start":"19:59.124 ","End":"20:04.525","Text":"is flying at a much higher velocity than what a motorbike is traveling at."},{"Start":"20:04.525 ","End":"20:08.095","Text":"Which means that after a certain amount of time,"},{"Start":"20:08.095 ","End":"20:11.365","Text":"our airplane which is traveling at a faster velocity,"},{"Start":"20:11.365 ","End":"20:16.375","Text":"is going to overtake our motorbike and travel further ahead,"},{"Start":"20:16.375 ","End":"20:18.235","Text":"and our motorbike will be left behind,"},{"Start":"20:18.235 ","End":"20:20.800","Text":"which means that relative to the airplane,"},{"Start":"20:20.800 ","End":"20:29.899","Text":"it\u0027s behind the origin"},{"Start":"20:29.899 ","End":"20:31.330","Text":"S tag frame of reference."},{"Start":"20:31.330 ","End":"20:37.465","Text":"Which means that it\u0027s going to have some negative coordinate for its position."},{"Start":"20:37.465 ","End":"20:41.360","Text":"That is the end of the lesson."}],"ID":9548},{"Watched":false,"Name":"Rest Frame And Time Dilation","Duration":"9m 24s","ChapterTopicVideoID":9252,"CourseChapterTopicPlaylistID":9398,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.145","Text":"Hello. In this lesson,"},{"Start":"00:02.145 ","End":"00:05.940","Text":"we\u0027re going to be defining what rest frame means."},{"Start":"00:05.940 ","End":"00:12.060","Text":"We\u0027re also going to be having a little introduction into the idea of time dilation."},{"Start":"00:12.060 ","End":"00:19.980","Text":"A rest frame is the frame of reference where the event being observed is always at rest."},{"Start":"00:19.980 ","End":"00:24.005","Text":"That means that if we have an object some point mass,"},{"Start":"00:24.005 ","End":"00:28.160","Text":"then our frame of reference is moving with that point-mass."},{"Start":"00:28.160 ","End":"00:33.890","Text":"The point mass is always in the same place in that frame of reference and moves with it."},{"Start":"00:33.890 ","End":"00:35.980","Text":"It always looks stationary."},{"Start":"00:35.980 ","End":"00:38.540","Text":"A rest frame of certain events,"},{"Start":"00:38.540 ","End":"00:41.240","Text":"of multiple events in plural is"},{"Start":"00:41.240 ","End":"00:44.720","Text":"a reference frame where all of the different events always"},{"Start":"00:44.720 ","End":"00:51.500","Text":"take place in the same place or at the exact same point in space."},{"Start":"00:51.500 ","End":"00:53.840","Text":"That basically means the same thing."},{"Start":"00:53.840 ","End":"00:58.925","Text":"It just means that there are now a few events occurring and they all are occurring"},{"Start":"00:58.925 ","End":"01:05.820","Text":"in stationary points relative to that certain frame of reference, the rest frame."},{"Start":"01:05.820 ","End":"01:08.510","Text":"Someone in that rest frame,"},{"Start":"01:08.510 ","End":"01:11.210","Text":"that frame of reference will see all of the events"},{"Start":"01:11.210 ","End":"01:15.605","Text":"happening at one specific point in space and they won\u0027t be moving."},{"Start":"01:15.605 ","End":"01:18.790","Text":"Now, let\u0027s defined rest time."},{"Start":"01:18.790 ","End":"01:21.950","Text":"Rest time is the difference in times recorded"},{"Start":"01:21.950 ","End":"01:25.820","Text":"between each observed event in the rest frame."},{"Start":"01:25.820 ","End":"01:30.110","Text":"Now, the rest time is denoted by the Greek letter Tau."},{"Start":"01:30.110 ","End":"01:32.915","Text":"Let\u0027s give an example of what this means."},{"Start":"01:32.915 ","End":"01:36.605","Text":"Let\u0027s say that we have our rest frame,"},{"Start":"01:36.605 ","End":"01:42.465","Text":"which is over here and we have Ed standing over here at the origin."},{"Start":"01:42.465 ","End":"01:46.575","Text":"He switches on a light in the room."},{"Start":"01:46.575 ","End":"01:48.230","Text":"Then after 2 seconds,"},{"Start":"01:48.230 ","End":"01:50.495","Text":"he switches the light back off."},{"Start":"01:50.495 ","End":"01:53.645","Text":"We\u0027re being told that Ed doesn\u0027t move relative to the light."},{"Start":"01:53.645 ","End":"01:56.405","Text":"He\u0027s always standing stationary over here."},{"Start":"01:56.405 ","End":"02:02.330","Text":"Then we call that time between switching on the light and switching off the lights,"},{"Start":"02:02.330 ","End":"02:09.320","Text":"those 2 seconds, that is called the rest time for switching on the light for that event."},{"Start":"02:09.320 ","End":"02:11.510","Text":"However, if we have Sally,"},{"Start":"02:11.510 ","End":"02:14.080","Text":"who\u0027s traveling in a spaceship next to Ed,"},{"Start":"02:14.080 ","End":"02:19.460","Text":"so in some different frame of reference. Here\u0027s Sally."},{"Start":"02:20.750 ","End":"02:24.125","Text":"She will measure a different amount of time"},{"Start":"02:24.125 ","End":"02:27.065","Text":"between the lights being turned on and turned off,"},{"Start":"02:27.065 ","End":"02:33.010","Text":"because she is moving at some velocity relative to this rest frame."},{"Start":"02:33.010 ","End":"02:35.794","Text":"Those 2 seconds that Ed observed,"},{"Start":"02:35.794 ","End":"02:39.065","Text":"Sally will observe a different time."},{"Start":"02:39.065 ","End":"02:41.945","Text":"Now we can speak about the rest length,"},{"Start":"02:41.945 ","End":"02:46.640","Text":"which is the measured length of the body when it is in its rest frame."},{"Start":"02:46.640 ","End":"02:52.190","Text":"For example, the distance between Ed and the light is 2 meters,"},{"Start":"02:52.190 ","End":"02:55.445","Text":"so this distance over here is 2 meters."},{"Start":"02:55.445 ","End":"02:58.520","Text":"Sally, who is in the spaceship over here,"},{"Start":"02:58.520 ","End":"03:06.420","Text":"we\u0027ll measure a different difference between Ed and the light or a different distance."},{"Start":"03:06.420 ","End":"03:10.105","Text":"Soon, we\u0027re going to speak about how we can work out"},{"Start":"03:10.105 ","End":"03:12.820","Text":"the different times that will be"},{"Start":"03:12.820 ","End":"03:17.090","Text":"measured and the different distances which will be measured."},{"Start":"03:17.190 ","End":"03:22.235","Text":"Now, let\u0027s take a look at the effect of time dilation."},{"Start":"03:22.235 ","End":"03:26.985","Text":"This time what we have is some lamp which is lit,"},{"Start":"03:26.985 ","End":"03:31.210","Text":"but it is lit in the S\u0027 frame of reference."},{"Start":"03:31.210 ","End":"03:36.515","Text":"This is the frame of reference which is moving at a velocity of v_0."},{"Start":"03:36.515 ","End":"03:41.140","Text":"We\u0027re told that the lamp shines for Tau seconds in"},{"Start":"03:41.140 ","End":"03:45.505","Text":"the system traveling with velocity v_0 in the x-direction."},{"Start":"03:45.505 ","End":"03:47.600","Text":"What we we want to know is,"},{"Start":"03:47.600 ","End":"03:52.110","Text":"how long were the observer in the lab\u0027s frame of reference,"},{"Start":"03:52.110 ","End":"03:56.315","Text":"in the S\u0027 frame of reference will see the lamp shining?"},{"Start":"03:56.315 ","End":"04:02.480","Text":"What we\u0027re trying to find is our Delta t. Here we can see"},{"Start":"04:02.480 ","End":"04:04.520","Text":"that we have all of our equations for"},{"Start":"04:04.520 ","End":"04:09.350","Text":"the Lorentz transformation and for the opposite transformation."},{"Start":"04:09.350 ","End":"04:12.955","Text":"To undo the Lorentz transformation."},{"Start":"04:12.955 ","End":"04:16.400","Text":"Now let\u0027s write down our equation for how"},{"Start":"04:16.400 ","End":"04:20.700","Text":"long the lamp is lit in its own frame of reference."},{"Start":"04:22.250 ","End":"04:26.570","Text":"Tau it\u0027s rest time. In the lamps frame of reference,"},{"Start":"04:26.570 ","End":"04:31.490","Text":"we have that our Tau is equal to t\u0027_off - t\u0027_on."},{"Start":"04:31.490 ","End":"04:34.945","Text":"It has a tag because we\u0027re in the S\u0027 frame of reference,"},{"Start":"04:34.945 ","End":"04:39.880","Text":"where t\u0027_on is the moment in time in the rest frame,"},{"Start":"04:39.880 ","End":"04:42.720","Text":"S\u0027 that the lamp is switched on and"},{"Start":"04:42.720 ","End":"04:46.860","Text":"t\u0027_off is the time in the rest frame that the lamp was switched off."},{"Start":"04:46.860 ","End":"04:51.710","Text":"Then the same its position when the lamp is switched off is equal to"},{"Start":"04:51.710 ","End":"04:59.150","Text":"the position when the lamp is switched on relative to the S\u0027 frame of reference."},{"Start":"04:59.750 ","End":"05:05.690","Text":"Now we have our equations in the lamps frame of reference in our S\u0027."},{"Start":"05:05.690 ","End":"05:08.465","Text":"Now what we want to do is we want to know what"},{"Start":"05:08.465 ","End":"05:13.100","Text":"those actions look like in our S frame of reference,"},{"Start":"05:13.100 ","End":"05:15.970","Text":"which is our lab\u0027s frame of reference."},{"Start":"05:15.970 ","End":"05:19.190","Text":"We have that the time at which the lamp is"},{"Start":"05:19.190 ","End":"05:22.510","Text":"switched on in our lab frame of reference is equal to."},{"Start":"05:22.510 ","End":"05:27.040","Text":"Then we\u0027re using simply this equation over here,"},{"Start":"05:27.040 ","End":"05:29.915","Text":"and we\u0027re just plugging in our numbers."},{"Start":"05:29.915 ","End":"05:32.075","Text":"We\u0027re using the opposite transformation"},{"Start":"05:32.075 ","End":"05:35.045","Text":"because our Lorentz transformation is taking us from"},{"Start":"05:35.045 ","End":"05:40.910","Text":"a lab frame of reference into some other frame of reference into the S\u0027."},{"Start":"05:40.910 ","End":"05:43.400","Text":"But right now, we\u0027re trying to convert equations from"},{"Start":"05:43.400 ","End":"05:47.075","Text":"our S\u0027 frame of reference into our S frame of reference."},{"Start":"05:47.075 ","End":"05:50.165","Text":"We\u0027re using the opposite transformation and simply"},{"Start":"05:50.165 ","End":"05:54.945","Text":"substituting in numbers for this equation into here."},{"Start":"05:54.945 ","End":"05:58.820","Text":"Now we know in our S frame of reference,"},{"Start":"05:58.820 ","End":"06:02.225","Text":"the time that relative to our S frame of reference,"},{"Start":"06:02.225 ","End":"06:07.615","Text":"the time that our lamp was switched on and the time that our lamp was switched off."},{"Start":"06:07.615 ","End":"06:12.650","Text":"Now, what we want to do is we want to find the amount of time that the lamp was switched"},{"Start":"06:12.650 ","End":"06:17.610","Text":"on relative to our S frame of reference."},{"Start":"06:17.610 ","End":"06:20.250","Text":"What the observer in our S frame of reference will see."},{"Start":"06:20.250 ","End":"06:25.600","Text":"What do we want to do is we want to find our Delta t. This is what we wanted to find."},{"Start":"06:25.600 ","End":"06:30.180","Text":"That\u0027s equal to our t_off - our t_on."},{"Start":"06:30.180 ","End":"06:33.845","Text":"Very similar to the equation that we saw over here."},{"Start":"06:33.845 ","End":"06:38.270","Text":"Then once we work all of that out and substitute in all of our algebra,"},{"Start":"06:38.270 ","End":"06:40.910","Text":"will get this expression over here,"},{"Start":"06:40.910 ","End":"06:45.680","Text":"which is Gamma_0 multiply by t_off - t_on."},{"Start":"06:45.680 ","End":"06:48.200","Text":"When we substitute all of that in,"},{"Start":"06:48.200 ","End":"06:52.145","Text":"we will get that our Delta t and our S frame of reference is"},{"Start":"06:52.145 ","End":"06:57.085","Text":"equal to Gamma_0 multiply by Tau."},{"Start":"06:57.085 ","End":"06:59.180","Text":"Something that we\u0027ll notice,"},{"Start":"06:59.180 ","End":"07:00.515","Text":"let\u0027s write it over here."},{"Start":"07:00.515 ","End":"07:02.315","Text":"We\u0027ll get that our Delta T,"},{"Start":"07:02.315 ","End":"07:06.140","Text":"which is equal to Gamma_0 multiply Tau."},{"Start":"07:06.140 ","End":"07:09.330","Text":"Now what do we remember about our Gamma_0?"},{"Start":"07:09.740 ","End":"07:15.875","Text":"What we remember is that our Gamma_0 is always going to be bigger than 1."},{"Start":"07:15.875 ","End":"07:17.990","Text":"Which means that whatever happens,"},{"Start":"07:17.990 ","End":"07:22.210","Text":"this expression over here is going to be bigger than our Tau,"},{"Start":"07:22.210 ","End":"07:24.405","Text":"which was our rest time."},{"Start":"07:24.405 ","End":"07:28.865","Text":"The amount of time that our lamp was lit in the S\u0027 frame of reference."},{"Start":"07:28.865 ","End":"07:32.270","Text":"The time measured in the lab frame of reference"},{"Start":"07:32.270 ","End":"07:36.215","Text":"is always going to be bigger than the rest time."},{"Start":"07:36.215 ","End":"07:38.765","Text":"This is called time dilation."},{"Start":"07:38.765 ","End":"07:43.175","Text":"Time dilation means that if you have Tau, your rest time,"},{"Start":"07:43.175 ","End":"07:47.690","Text":"any other observer at another point in time is going"},{"Start":"07:47.690 ","End":"07:52.595","Text":"to measure a longer time than the rest time."},{"Start":"07:52.595 ","End":"07:55.130","Text":"That is called time dilation."},{"Start":"07:55.130 ","End":"07:58.160","Text":"Now a quick little comment which will be expanded"},{"Start":"07:58.160 ","End":"08:02.620","Text":"on in future questions is that this time,"},{"Start":"08:02.620 ","End":"08:05.315","Text":"our rest time or the time that our lamp was"},{"Start":"08:05.315 ","End":"08:11.690","Text":"lit refers to the lamp being lit at some position and our S\u0027 carrying on"},{"Start":"08:11.690 ","End":"08:16.310","Text":"moving and then the lamp being switched off at some other position relative"},{"Start":"08:16.310 ","End":"08:22.530","Text":"to our S frame of reference at some different space over here."},{"Start":"08:22.530 ","End":"08:29.275","Text":"It\u0027s not the time that the observer in our S frame of reference sees."},{"Start":"08:29.275 ","End":"08:35.030","Text":"This time over here is the time in our S frame of reference at"},{"Start":"08:35.030 ","End":"08:37.970","Text":"which these events of the lamp being"},{"Start":"08:37.970 ","End":"08:41.435","Text":"switched on and off occur in the S frame of reference."},{"Start":"08:41.435 ","End":"08:48.035","Text":"But it\u0027s not what is observed for someone standing in the S frame of reference watching."},{"Start":"08:48.035 ","End":"08:52.950","Text":"In order to know what a person standing over here we\u0027ll observe,"},{"Start":"08:52.950 ","End":"08:55.925","Text":"we have to also take into account the speed of light"},{"Start":"08:55.925 ","End":"08:59.960","Text":"and how long it will take to see the lamp being"},{"Start":"08:59.960 ","End":"09:07.835","Text":"lit when it\u0027s a distance x away from our observer and when our lamp is switched off,"},{"Start":"09:07.835 ","End":"09:09.590","Text":"let\u0027s say somewhere around here."},{"Start":"09:09.590 ","End":"09:15.225","Text":"When it\u0027s the distance of x plus x_1 away from the observer."},{"Start":"09:15.225 ","End":"09:17.930","Text":"We have to take into account that the light has to"},{"Start":"09:17.930 ","End":"09:21.785","Text":"travel to the observer in both of these cases."},{"Start":"09:21.785 ","End":"09:25.080","Text":"That\u0027s the end of this lesson."}],"ID":9549},{"Watched":false,"Name":"Shortening Of Length","Duration":"7m 16s","ChapterTopicVideoID":9254,"CourseChapterTopicPlaylistID":9398,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.650","Text":"Hello. In this video,"},{"Start":"00:01.650 ","End":"00:07.755","Text":"we\u0027re going to be speaking about the shortening or lengthening of some length measured."},{"Start":"00:07.755 ","End":"00:11.670","Text":"What we have over here is our S frame of reference."},{"Start":"00:11.670 ","End":"00:15.960","Text":"This is our labs frame of reference and it is stationary."},{"Start":"00:15.960 ","End":"00:21.285","Text":"Then we have some kind of observer standing and our S tag frame of reference."},{"Start":"00:21.285 ","End":"00:25.530","Text":"The whole S tag frame of reference is moving at a velocity of v 0."},{"Start":"00:25.530 ","End":"00:31.170","Text":"The observer standing in the S tag frame of reference is looking at this rod"},{"Start":"00:31.170 ","End":"00:37.725","Text":"of length L0 and relative to the observer in the S tag frame of reference,"},{"Start":"00:37.725 ","End":"00:41.775","Text":"the rod is stationary and is of length L0."},{"Start":"00:41.775 ","End":"00:46.820","Text":"Now this length L0 is called the rest length."},{"Start":"00:46.820 ","End":"00:52.550","Text":"Because its length is at rest and unchanging relative to the observer in"},{"Start":"00:52.550 ","End":"00:59.405","Text":"the S tag frame of reference in this rods rest frame of reference."},{"Start":"00:59.405 ","End":"01:04.055","Text":"Now what we\u0027re going to try and do in this question is to work"},{"Start":"01:04.055 ","End":"01:08.390","Text":"out what length and observer in a Labs frame of reference,"},{"Start":"01:08.390 ","End":"01:10.520","Text":"S frame of reference."},{"Start":"01:10.520 ","End":"01:14.315","Text":"We\u0027ll measure this rod to be."},{"Start":"01:14.315 ","End":"01:17.810","Text":"Now what we will see at the end is that"},{"Start":"01:17.810 ","End":"01:21.950","Text":"the observer in our S frame of reference will see that"},{"Start":"01:21.950 ","End":"01:26.090","Text":"this rod will appear shorter in S frame of"},{"Start":"01:26.090 ","End":"01:31.519","Text":"reference than it does to the observer in the S tag frame of reference."},{"Start":"01:31.519 ","End":"01:38.195","Text":"That means that an observer in the rods rest frame will record"},{"Start":"01:38.195 ","End":"01:42.200","Text":"some length and any observer in any other frame"},{"Start":"01:42.200 ","End":"01:46.475","Text":"of reference aside from the S tag frame of reference,"},{"Start":"01:46.475 ","End":"01:48.650","Text":"aside from the rest frame of the rod,"},{"Start":"01:48.650 ","End":"01:52.480","Text":"will observe always a shorter length."},{"Start":"01:52.480 ","End":"01:55.100","Text":"Let\u0027s see how we do this."},{"Start":"01:55.100 ","End":"01:57.350","Text":"Now we can see that the length of the rod,"},{"Start":"01:57.350 ","End":"02:02.435","Text":"relative to the observer and our S tag frame of reference."},{"Start":"02:02.435 ","End":"02:08.030","Text":"So L0, is going to be equal to this distance over here."},{"Start":"02:08.030 ","End":"02:15.520","Text":"That\u0027s our x2 tag position minus our x1 tag position."},{"Start":"02:15.520 ","End":"02:19.815","Text":"This length over here is our rest length."},{"Start":"02:19.815 ","End":"02:28.265","Text":"Now for the observer standing in our S frame of reference,"},{"Start":"02:28.265 ","End":"02:31.125","Text":"that person will measure,"},{"Start":"02:31.125 ","End":"02:33.690","Text":"this is in our S tag."},{"Start":"02:33.690 ","End":"02:35.480","Text":"Now in our S frame of reference,"},{"Start":"02:35.480 ","End":"02:40.635","Text":"the observer will measure our position x_1 tag at time"},{"Start":"02:40.635 ","End":"02:47.630","Text":"t_1 and she will measure the end of the rod,"},{"Start":"02:47.630 ","End":"02:52.295","Text":"the position of x_2 at time t_2."},{"Start":"02:52.295 ","End":"02:56.245","Text":"Our observer in S frame of reference,"},{"Start":"02:56.245 ","End":"02:59.145","Text":"sees a t_1,"},{"Start":"02:59.145 ","End":"03:09.060","Text":"this the left edge of the rod and at time t_2 sees or measures the right edge of the rod."},{"Start":"03:09.060 ","End":"03:12.455","Text":"Now, in order to define"},{"Start":"03:12.455 ","End":"03:16.820","Text":"the measurement of any length when we\u0027re dealing with different frames of"},{"Start":"03:16.820 ","End":"03:25.230","Text":"references is defined as taking both of these measurements simultaneously."},{"Start":"03:25.330 ","End":"03:30.485","Text":"What does that mean that, t_1=t_2."},{"Start":"03:30.485 ","End":"03:35.315","Text":"In order to measure some length,"},{"Start":"03:35.315 ","End":"03:44.090","Text":"the measurements of both edges of the length must be taken at the exact same time,"},{"Start":"03:44.090 ","End":"03:46.670","Text":"must be done simultaneously."},{"Start":"03:46.670 ","End":"03:55.755","Text":"This point in red is very important and must be remembered."},{"Start":"03:55.755 ","End":"03:58.055","Text":"Now that we know this,"},{"Start":"03:58.055 ","End":"04:00.635","Text":"we can begin our calculations."},{"Start":"04:00.635 ","End":"04:08.925","Text":"We know that our length l0=x_2-x_1 tag."},{"Start":"04:08.925 ","End":"04:12.080","Text":"Then using our Lawrence transformation,"},{"Start":"04:12.080 ","End":"04:19.055","Text":"we can say that this is equal to our gamma o multiplied by our X position,"},{"Start":"04:19.055 ","End":"04:21.845","Text":"which will be x2 without the tag,"},{"Start":"04:21.845 ","End":"04:25.715","Text":"minus our velocity v_0,"},{"Start":"04:25.715 ","End":"04:29.870","Text":"multiplied by the time that this measurement was taken in,"},{"Start":"04:29.870 ","End":"04:35.480","Text":"which is at time t_2 and then minus our x_1 tag, which again,"},{"Start":"04:35.480 ","End":"04:38.240","Text":"using the Lawrence transmitted transformation,"},{"Start":"04:38.240 ","End":"04:47.750","Text":"it\u0027s going to be gamma 0 x x_1 - v_0x t_1."},{"Start":"04:47.750 ","End":"04:50.135","Text":"Now this is going to be equal to,"},{"Start":"04:50.135 ","End":"04:54.640","Text":"because we know that our t_1= t_2."},{"Start":"04:54.640 ","End":"04:59.675","Text":"The position measurements were taken at the same time simultaneously."},{"Start":"04:59.675 ","End":"05:04.985","Text":"Then when we substitute in either t_2 as t_1,"},{"Start":"05:04.985 ","End":"05:07.220","Text":"or our t_1 over here as t_2."},{"Start":"05:07.220 ","End":"05:13.225","Text":"We will get that this is equal to gamma 0x X x_2 - x_1."},{"Start":"05:13.225 ","End":"05:19.715","Text":"This is exactly the length"},{"Start":"05:19.715 ","End":"05:26.214","Text":"of the rod relative to the observer in the labs frame of reference."},{"Start":"05:26.214 ","End":"05:30.125","Text":"What we\u0027re going to do is we\u0027re going to call this x_2 - x_1."},{"Start":"05:30.125 ","End":"05:37.370","Text":"We\u0027ll just call it L. Then we\u0027ll get that our length in our S tag frame of"},{"Start":"05:37.370 ","End":"05:44.930","Text":"reference for the rod l0=Gamma o X L,"},{"Start":"05:44.930 ","End":"05:49.745","Text":"which is the length relative to our labs frame of reference."},{"Start":"05:49.745 ","End":"05:52.505","Text":"Or we can rewrite this and say,"},{"Start":"05:52.505 ","End":"05:57.530","Text":"that the length of the rod relative to our labs frame of reference is simply equal"},{"Start":"05:57.530 ","End":"06:03.905","Text":"to the rest length in the S tag frame of reference divided by Gamma o."},{"Start":"06:03.905 ","End":"06:08.660","Text":"Now from this, what we can see over"},{"Start":"06:08.660 ","End":"06:13.985","Text":"here is that our length when viewed from our lab frame of reference,"},{"Start":"06:13.985 ","End":"06:21.470","Text":"is always going to be less than the original length in our rest frame."},{"Start":"06:21.470 ","End":"06:28.220","Text":"As we know, Gamma 0 over here is always going to be bigger than 1."},{"Start":"06:28.220 ","End":"06:31.430","Text":"Any number divided by a number bigger than 1"},{"Start":"06:31.430 ","End":"06:34.940","Text":"is going to be smaller than the original number."},{"Start":"06:34.940 ","End":"06:40.760","Text":"If we have any length in some frame of"},{"Start":"06:40.760 ","End":"06:47.645","Text":"reference where this length is stationary relative to this frame of reference."},{"Start":"06:47.645 ","End":"06:51.380","Text":"This length is going to be called the rest length."},{"Start":"06:51.380 ","End":"06:56.195","Text":"Now, when observed from a lab frame of reference,"},{"Start":"06:56.195 ","End":"06:59.105","Text":"this length is going to appear"},{"Start":"06:59.105 ","End":"07:05.870","Text":"smaller so that we can see it as different to what we saw with the time,"},{"Start":"07:05.870 ","End":"07:09.245","Text":"where our time appears to be longer."},{"Start":"07:09.245 ","End":"07:13.110","Text":"Here our length appears to be shorter."},{"Start":"07:13.400 ","End":"07:16.810","Text":"That\u0027s the end of this lesson."}],"ID":9550},{"Watched":false,"Name":"Change In Angle","Duration":"4m 50s","ChapterTopicVideoID":9262,"CourseChapterTopicPlaylistID":9398,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.965","Text":"Hello. In this video,"},{"Start":"00:01.965 ","End":"00:05.475","Text":"we\u0027re going to be speaking about the change of the angle"},{"Start":"00:05.475 ","End":"00:09.705","Text":"between the x-axis and the position vector."},{"Start":"00:09.705 ","End":"00:14.100","Text":"What we can see over here is that we have our S frame of reference,"},{"Start":"00:14.100 ","End":"00:18.630","Text":"which is our lab\u0027s frame of reference and we have our S\u0027 frame of reference."},{"Start":"00:18.630 ","End":"00:21.270","Text":"Now our S\u0027 frame of reference has some kind of"},{"Start":"00:21.270 ","End":"00:24.850","Text":"observer standing somewhere in the frame of reference,"},{"Start":"00:24.850 ","End":"00:27.240","Text":"and the whole frame of reference, including the observer,"},{"Start":"00:27.240 ","End":"00:32.924","Text":"is moving at a velocity of v_0 relative to our lab\u0027s frame of reference."},{"Start":"00:32.924 ","End":"00:36.555","Text":"Now, what we will see is that our position vector,"},{"Start":"00:36.555 ","End":"00:41.000","Text":"which we will measure or the angle of the position vector,"},{"Start":"00:41.000 ","End":"00:44.390","Text":"which we will measure when we are in our S\u0027 frame of reference will be"},{"Start":"00:44.390 ","End":"00:49.295","Text":"different to the angle measured in our lab\u0027s frame of reference."},{"Start":"00:49.295 ","End":"00:51.365","Text":"Let\u0027s mark this out."},{"Start":"00:51.365 ","End":"00:55.925","Text":"We can see that from this tip all the way over here,"},{"Start":"00:55.925 ","End":"01:02.070","Text":"we can see that this length over here is our Delta y\u0027."},{"Start":"01:02.070 ","End":"01:07.370","Text":"Then from this point over here until somewhere over here,"},{"Start":"01:07.370 ","End":"01:10.460","Text":"we have our Delta x\u0027,"},{"Start":"01:10.460 ","End":"01:14.930","Text":"and then we have the same over here until the tip over here."},{"Start":"01:14.930 ","End":"01:17.900","Text":"This is going to be our Delta x."},{"Start":"01:17.900 ","End":"01:22.345","Text":"From our tip over here until the bottom,"},{"Start":"01:22.345 ","End":"01:24.570","Text":"we have our Delta y."},{"Start":"01:24.570 ","End":"01:28.080","Text":"Over here we have our Lorentz transformation equation,"},{"Start":"01:28.080 ","End":"01:32.285","Text":"and in order to undo our Lorentz transformation in the opposite direction,"},{"Start":"01:32.285 ","End":"01:36.115","Text":"we have our opposite transformation and now let\u0027s see what we have."},{"Start":"01:36.115 ","End":"01:38.990","Text":"What we can tell is that our Delta y\u0027,"},{"Start":"01:38.990 ","End":"01:41.585","Text":"Let\u0027s see what this is equal to."},{"Start":"01:41.585 ","End":"01:44.045","Text":"This is equal to our top corner,"},{"Start":"01:44.045 ","End":"01:48.125","Text":"which let\u0027s call that y_2\u0027 minus our bottom corner,"},{"Start":"01:48.125 ","End":"01:50.165","Text":"which is our y_1\u0027."},{"Start":"01:50.165 ","End":"01:53.465","Text":"Then, once we stick that into our transform,"},{"Start":"01:53.465 ","End":"01:56.810","Text":"we can see that our y\u0027 is equal to our y."},{"Start":"01:56.810 ","End":"02:01.905","Text":"Then we\u0027ll get that, that is simply equal to y_2 minus y_1,"},{"Start":"02:01.905 ","End":"02:05.385","Text":"which is the exact same thing as our Delta y."},{"Start":"02:05.385 ","End":"02:07.805","Text":"That\u0027s just going to be equal to our Delta y,"},{"Start":"02:07.805 ","End":"02:11.315","Text":"the same as in our lab\u0027s frame of reference."},{"Start":"02:11.315 ","End":"02:15.860","Text":"That means that our y value in our S\u0027 is equal"},{"Start":"02:15.860 ","End":"02:21.395","Text":"to a y value in our S. That\u0027s far our y. Now what about our x?"},{"Start":"02:21.395 ","End":"02:26.270","Text":"We saw in the previous video that we have a shortening of length when we\u0027re"},{"Start":"02:26.270 ","End":"02:28.790","Text":"working with our S and S\u0027 frame of"},{"Start":"02:28.790 ","End":"02:32.335","Text":"references when we\u0027re dealing with relativistic speeds."},{"Start":"02:32.335 ","End":"02:36.350","Text":"Now we can see that we\u0027re going to have a shortening of length in our x-axis,"},{"Start":"02:36.350 ","End":"02:37.745","Text":"and from our previous video,"},{"Start":"02:37.745 ","End":"02:46.480","Text":"we saw that our Delta X is equal to our Delta X\u0027 divided by our Gamma 0."},{"Start":"02:46.480 ","End":"02:50.270","Text":"To also remind you when we\u0027re talking about our x-axis,"},{"Start":"02:50.270 ","End":"02:55.100","Text":"that it\u0027s always going to be in the direction of our relative velocity."},{"Start":"02:55.100 ","End":"02:59.360","Text":"If our relative velocity was going upwards in this direction,"},{"Start":"02:59.360 ","End":"03:03.200","Text":"then our x-axis would be this axis over here."},{"Start":"03:03.200 ","End":"03:06.935","Text":"The direction that the relative velocity is pointing in,"},{"Start":"03:06.935 ","End":"03:13.075","Text":"that is the x-direction and that is the direction that we have our shortening of length."},{"Start":"03:13.075 ","End":"03:16.020","Text":"That\u0027s what our x value is going to be."},{"Start":"03:16.020 ","End":"03:21.920","Text":"Now we\u0027re trying to see what\u0027s going to happen to the angle. Let\u0027s take a look."},{"Start":"03:21.920 ","End":"03:27.165","Text":"If we say that we have tan of Theta, and as we know,"},{"Start":"03:27.165 ","End":"03:32.360","Text":"that is equal to the change in y divided by the change in x."},{"Start":"03:32.360 ","End":"03:34.640","Text":"As we know from our previous equations,"},{"Start":"03:34.640 ","End":"03:39.170","Text":"the change in y is also equal to the change in y\u0027 and the change in"},{"Start":"03:39.170 ","End":"03:45.425","Text":"x is equal to the change in x\u0027 divided by Gamma 0,"},{"Start":"03:45.425 ","End":"03:53.760","Text":"which is simply going to be equal to Gamma 0 multiplied by tan of Theta\u0027."},{"Start":"03:53.760 ","End":"03:58.984","Text":"Now we can see that if we know our angle in our frame of reference,"},{"Start":"03:58.984 ","End":"04:00.755","Text":"our lab\u0027s frame of reference,"},{"Start":"04:00.755 ","End":"04:04.295","Text":"then we can use this equation to find"},{"Start":"04:04.295 ","End":"04:09.110","Text":"our angle over here between the x-axis and our position vector,"},{"Start":"04:09.110 ","End":"04:11.645","Text":"in our S\u0027 frame of reference."},{"Start":"04:11.645 ","End":"04:17.460","Text":"What we can see is that our angle Theta is also changing."},{"Start":"04:17.460 ","End":"04:21.575","Text":"This is the equation to know the angle that is changed"},{"Start":"04:21.575 ","End":"04:25.460","Text":"when we have some system moving at a relative velocity,"},{"Start":"04:25.460 ","End":"04:29.435","Text":"close to the speed of light away from our lab\u0027s frame of reference."},{"Start":"04:29.435 ","End":"04:35.270","Text":"But it\u0027s important to note that if we\u0027re going to change the velocity vector,"},{"Start":"04:35.270 ","End":"04:36.755","Text":"the direction that that is in,"},{"Start":"04:36.755 ","End":"04:39.680","Text":"then we\u0027ll get a different equation to this."},{"Start":"04:39.680 ","End":"04:44.090","Text":"This equation only stands when we have one of our lengths,"},{"Start":"04:44.090 ","End":"04:49.040","Text":"our length along the x-axis, which is changing."},{"Start":"04:49.040 ","End":"04:51.540","Text":"Okay, that\u0027s the end of this lesson."}],"ID":9551},{"Watched":false,"Name":"Exercise 2","Duration":"4m 53s","ChapterTopicVideoID":9260,"CourseChapterTopicPlaylistID":9398,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.410","Text":"Hello. In this video,"},{"Start":"00:01.410 ","End":"00:04.245","Text":"we\u0027re going to be speaking about the following paradox."},{"Start":"00:04.245 ","End":"00:08.055","Text":"We have 2 rockets with identical rest lengths that fly past each other."},{"Start":"00:08.055 ","End":"00:11.820","Text":"The S rocket has a cannon at its base positioned perpendicularly to"},{"Start":"00:11.820 ","End":"00:16.590","Text":"its direction of travel and into the direction of rocket S tag."},{"Start":"00:16.590 ","End":"00:18.120","Text":"This is Diagram 1."},{"Start":"00:18.120 ","End":"00:23.250","Text":"The S rocket then fires its cannon at the exact moment that the point a over"},{"Start":"00:23.250 ","End":"00:28.476","Text":"here on its head reaches point a tag at the base of the S tag rocket."},{"Start":"00:28.476 ","End":"00:33.090","Text":"Due to rocket S tag having a shorter length than the rest length of S,"},{"Start":"00:33.090 ","End":"00:38.355","Text":"it is assumed that the shot fired by the cannon on S will miss S tag."},{"Start":"00:38.355 ","End":"00:41.140","Text":"As we can see here in Diagram 2,"},{"Start":"00:41.140 ","End":"00:44.285","Text":"because both of the rockets are traveling at relativistic speeds."},{"Start":"00:44.285 ","End":"00:46.835","Text":"However, in the S tag reference frame,"},{"Start":"00:46.835 ","End":"00:51.290","Text":"rocket S has a shorter length than the rest length of rocket S tag,"},{"Start":"00:51.290 ","End":"00:56.885","Text":"and so the cannon shot by S will hit S tag as we can see in this diagram."},{"Start":"00:56.885 ","End":"01:01.510","Text":"What we\u0027re going to do in this video is we\u0027re going to resolve this paradox."},{"Start":"01:01.510 ","End":"01:03.185","Text":"Let\u0027s begin to resolve this."},{"Start":"01:03.185 ","End":"01:06.020","Text":"We know that usually when dealing with these types of questions,"},{"Start":"01:06.020 ","End":"01:08.705","Text":"it\u0027s easiest to work with a table."},{"Start":"01:08.705 ","End":"01:10.520","Text":"What we\u0027re going to do is we\u0027re going to write all of"},{"Start":"01:10.520 ","End":"01:13.070","Text":"the details when we\u0027re looking from the S frame of"},{"Start":"01:13.070 ","End":"01:17.494","Text":"reference and all of the details when we\u0027re looking from the S tag frame of reference."},{"Start":"01:17.494 ","End":"01:19.970","Text":"Now, we know that at the exact point,"},{"Start":"01:19.970 ","End":"01:23.645","Text":"when our a is at the exact point as our a tag,"},{"Start":"01:23.645 ","End":"01:25.910","Text":"that is when a cannon is shot."},{"Start":"01:25.910 ","End":"01:33.605","Text":"If we say that this is our axis for our S frame of reference,"},{"Start":"01:33.605 ","End":"01:40.190","Text":"and that this over here is our axis for our S tag frame of reference,"},{"Start":"01:40.190 ","End":"01:43.115","Text":"then we know that at t=0,"},{"Start":"01:43.115 ","End":"01:44.630","Text":"both of our axes,"},{"Start":"01:44.630 ","End":"01:47.260","Text":"the origins must be 1 on top of the other."},{"Start":"01:47.260 ","End":"01:50.975","Text":"Now what we can do is we can define our event."},{"Start":"01:50.975 ","End":"01:53.990","Text":"We know that our event is the cannon being shot,"},{"Start":"01:53.990 ","End":"01:58.320","Text":"and it\u0027s being shot at the position X_1 is"},{"Start":"01:58.320 ","End":"02:02.910","Text":"equal to l. We can say that this height over here,"},{"Start":"02:02.910 ","End":"02:12.035","Text":"up until where the canon is l. We know that the event is happening at time t_1=0."},{"Start":"02:12.035 ","End":"02:15.035","Text":"Now, when we use our Lorentz transformation,"},{"Start":"02:15.035 ","End":"02:21.110","Text":"we can find out that our X_1 in our S tag frame of reference,"},{"Start":"02:21.110 ","End":"02:22.775","Text":"which is this rocket over here,"},{"Start":"02:22.775 ","End":"02:31.875","Text":"is simply going to be equal to gamma 0 multiplied by l and our t_1 tag through"},{"Start":"02:31.875 ","End":"02:35.645","Text":"our Lorentz transformation is simply going to be equal to"},{"Start":"02:35.645 ","End":"02:39.995","Text":"gamma 0 multiplied by negative v_0"},{"Start":"02:39.995 ","End":"02:44.600","Text":"divided by c^2 times l. Now what"},{"Start":"02:44.600 ","End":"02:49.045","Text":"we can see over here is that our t_1 tag has a negative value,"},{"Start":"02:49.045 ","End":"02:52.309","Text":"and we can see that throughout the motion,"},{"Start":"02:52.309 ","End":"02:56.810","Text":"our t_1 tag is always going to be a negative number because we see"},{"Start":"02:56.810 ","End":"02:58.370","Text":"that our gamma naught and all of"},{"Start":"02:58.370 ","End":"03:01.610","Text":"the other values that we have over here are bigger than 0."},{"Start":"03:01.610 ","End":"03:03.065","Text":"What does that mean?"},{"Start":"03:03.065 ","End":"03:09.050","Text":"That means that the event seen from our S tag frame of reference happens in negative time"},{"Start":"03:09.050 ","End":"03:15.335","Text":"before our origins of the S tag and S frame of references are at the same point,"},{"Start":"03:15.335 ","End":"03:18.725","Text":"which means that when the cannon fire that shot"},{"Start":"03:18.725 ","End":"03:24.460","Text":"rocket S tag wasn\u0027t in a position where the cannon shot by S can hit it."},{"Start":"03:24.460 ","End":"03:26.690","Text":"That\u0027s with regards to the time."},{"Start":"03:26.690 ","End":"03:29.975","Text":"But we can also see that with regards to the position,"},{"Start":"03:29.975 ","End":"03:34.250","Text":"the cannon shot by rocket S also won\u0027t hit rocket S tag."},{"Start":"03:34.250 ","End":"03:36.710","Text":"As we know, as we\u0027ve seen in previous videos,"},{"Start":"03:36.710 ","End":"03:41.245","Text":"our value for gamma 0 is always going to be bigger than 1."},{"Start":"03:41.245 ","End":"03:46.520","Text":"That means that if we have our rocket S tag over here,"},{"Start":"03:46.520 ","End":"03:51.275","Text":"excuse my drawing, then the length or the distance, as we know,"},{"Start":"03:51.275 ","End":"03:54.690","Text":"this distance is l,"},{"Start":"03:55.220 ","End":"04:00.020","Text":"and we know that the cannon is fired at"},{"Start":"04:00.020 ","End":"04:04.640","Text":"gamma naught multiplied by l. Because our gamma naught is bigger than 1,"},{"Start":"04:04.640 ","End":"04:10.255","Text":"we can see that the rocket is going to fire the cannon over here at this height,"},{"Start":"04:10.255 ","End":"04:13.820","Text":"and this is gamma naught multiplied by l,"},{"Start":"04:13.820 ","End":"04:14.945","Text":"which means that again,"},{"Start":"04:14.945 ","End":"04:16.850","Text":"we can see that our cannon,"},{"Start":"04:16.850 ","End":"04:20.020","Text":"is going to miss our rocket S tag."},{"Start":"04:20.020 ","End":"04:23.375","Text":"Now if we\u0027re going to look at our S tag,"},{"Start":"04:23.375 ","End":"04:25.505","Text":"as we said to look in Diagram 3,"},{"Start":"04:25.505 ","End":"04:30.965","Text":"because we know that the cannon from our S frame of reference, our S rocket,"},{"Start":"04:30.965 ","End":"04:36.306","Text":"is shot at a negative time when we\u0027re looking at our S tag frame of reference,"},{"Start":"04:36.306 ","End":"04:39.690","Text":"we know that again, the canon will miss this rocket."},{"Start":"04:39.690 ","End":"04:42.980","Text":"That means that from whichever frame of reference we\u0027re looking"},{"Start":"04:42.980 ","End":"04:46.670","Text":"be it from our S tag rocket or from our S rocket,"},{"Start":"04:46.670 ","End":"04:50.945","Text":"the cannon is shot and will not hit a rocket S tag."},{"Start":"04:50.945 ","End":"04:53.550","Text":"That\u0027s the end of this lesson."}],"ID":9552},{"Watched":false,"Name":"The Relativistic Doppler Effect","Duration":"9m 57s","ChapterTopicVideoID":9255,"CourseChapterTopicPlaylistID":9398,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.680","Text":"Hello. In this lesson we\u0027re going to be learning about the Relativistic Doppler Effect."},{"Start":"00:04.680 ","End":"00:08.190","Text":"Which means that we have a light source in a frame,"},{"Start":"00:08.190 ","End":"00:10.095","Text":"our S tag frame of reference,"},{"Start":"00:10.095 ","End":"00:12.795","Text":"which is moving with a velocity, V_ 0."},{"Start":"00:12.795 ","End":"00:15.630","Text":"Our light source emits a pulse of light,"},{"Start":"00:15.630 ","End":"00:20.115","Text":"which has some wavelength or frequency or period, however you want to put it."},{"Start":"00:20.115 ","End":"00:21.615","Text":"Now what do we want to do,"},{"Start":"00:21.615 ","End":"00:24.285","Text":"is we want to see how this wavelength,"},{"Start":"00:24.285 ","End":"00:25.860","Text":"or frequency, or period,"},{"Start":"00:25.860 ","End":"00:29.715","Text":"changes as our S tag frame of reference moves away."},{"Start":"00:29.715 ","End":"00:35.750","Text":"Now we\u0027re being told in the question that the duration of the light emission and the in"},{"Start":"00:35.750 ","End":"00:42.000","Text":"the S tag frame is Tau and we\u0027re being asked to find the relation between Tau and T,"},{"Start":"00:42.000 ","End":"00:45.109","Text":"where T is the duration of the light emission"},{"Start":"00:45.109 ","End":"00:48.635","Text":"when it\u0027s being observed from our S frame of reference."},{"Start":"00:48.635 ","End":"00:49.910","Text":"Now at later on,"},{"Start":"00:49.910 ","End":"00:54.800","Text":"a way using Dynamics is going to be used in order to solve this exact same question."},{"Start":"00:54.800 ","End":"00:57.680","Text":"But right now we are going to be working with our table for"},{"Start":"00:57.680 ","End":"01:00.380","Text":"our S and S tag frame of reference because"},{"Start":"01:00.380 ","End":"01:03.080","Text":"it\u0027s a way that really explains what\u0027s going on and it"},{"Start":"01:03.080 ","End":"01:05.990","Text":"will help you solve more complicated questions later on."},{"Start":"01:05.990 ","End":"01:07.895","Text":"Let\u0027s draw the table."},{"Start":"01:07.895 ","End":"01:10.235","Text":"Let\u0027s begin with the first event,"},{"Start":"01:10.235 ","End":"01:12.625","Text":"which was when our pulse is emitted."},{"Start":"01:12.625 ","End":"01:16.084","Text":"If we\u0027re talking relative to our S frame of reference,"},{"Start":"01:16.084 ","End":"01:18.560","Text":"we can say that the pulse of light is emitted at"},{"Start":"01:18.560 ","End":"01:23.495","Text":"some space relative to the origin of our S frame of reference."},{"Start":"01:23.495 ","End":"01:28.195","Text":"We can say that it\u0027s at position L. Then"},{"Start":"01:28.195 ","End":"01:33.995","Text":"we can say that it happens at time is equal to 0 right at the beginning."},{"Start":"01:33.995 ","End":"01:40.090","Text":"Now in order to see what is happening in our S tag frame of reference,"},{"Start":"01:40.090 ","End":"01:45.760","Text":"we\u0027re simply going to use our Lawrence transformation for x and t tag."},{"Start":"01:45.760 ","End":"01:53.865","Text":"Then we\u0027ll get that our x_1 tag is equal to Gamma 0 multiplied by L,"},{"Start":"01:53.865 ","End":"02:00.300","Text":"and that our t_1 tag is going to be equal to Gamma 0 multiplied"},{"Start":"02:00.300 ","End":"02:07.300","Text":"by negative v_0 L divided by c^2."},{"Start":"02:07.300 ","End":"02:09.875","Text":"Now let\u0027s speak about the second event,"},{"Start":"02:09.875 ","End":"02:13.505","Text":"which is when we finish emitting the pulse of light."},{"Start":"02:13.505 ","End":"02:17.660","Text":"Now, because this happens in our S tag frame of reference and"},{"Start":"02:17.660 ","End":"02:21.965","Text":"we\u0027re being told that the duration of the light emission in its rest frame,"},{"Start":"02:21.965 ","End":"02:25.415","Text":"which means the S tag frame is Tau."},{"Start":"02:25.415 ","End":"02:30.455","Text":"That means that we\u0027ll have that T_2 tag is going to"},{"Start":"02:30.455 ","End":"02:35.420","Text":"be equal to the time at which our pulse was emitted originally,"},{"Start":"02:35.420 ","End":"02:37.400","Text":"which is T_1 tag,"},{"Start":"02:37.400 ","End":"02:41.435","Text":"plus the duration that it was emitted, which is Tau."},{"Start":"02:41.435 ","End":"02:44.135","Text":"Then once we plug in all of these values,"},{"Start":"02:44.135 ","End":"02:49.250","Text":"will get that this is equal to negative Gamma naught multiplied by"},{"Start":"02:49.250 ","End":"02:54.745","Text":"V_0 L divided by c^2 plus Tau."},{"Start":"02:54.745 ","End":"02:56.360","Text":"Now for the position,"},{"Start":"02:56.360 ","End":"02:58.925","Text":"because the light is being lit in its rest frame,"},{"Start":"02:58.925 ","End":"03:01.910","Text":"the S tag frame, it stationary there."},{"Start":"03:01.910 ","End":"03:05.630","Text":"Its position, that\u0027s x_2 tag is going"},{"Start":"03:05.630 ","End":"03:09.410","Text":"to be the exact same position as when the pulse was emitted,"},{"Start":"03:09.410 ","End":"03:12.370","Text":"so that\u0027s going to be equal to X_1 tag,"},{"Start":"03:12.370 ","End":"03:16.775","Text":"which is equal to Gamma 0 multiplied by L. Now,"},{"Start":"03:16.775 ","End":"03:21.020","Text":"in order to know what we will get in S frame of reference in our lab,"},{"Start":"03:21.020 ","End":"03:24.080","Text":"we simply do the opposite transformation,"},{"Start":"03:24.080 ","End":"03:30.515","Text":"and we will get that our x_2 is going to be equal to this over here."},{"Start":"03:30.515 ","End":"03:39.900","Text":"It\u0027s equal to Gamma 0 multiplied by Gamma 0_ L plus V_ 0 multiplied by negative Gamma 0,"},{"Start":"03:39.900 ","End":"03:44.085","Text":"v_0 L divided by c^2 plus Tau."},{"Start":"03:44.085 ","End":"03:47.175","Text":"Then for our time T_2,"},{"Start":"03:47.175 ","End":"03:53.085","Text":"this is going to be equal to Gamma 0 multiplied by negative Gamma 0,"},{"Start":"03:53.085 ","End":"04:01.840","Text":"v_0 L divided by c^2 plus Tau plus V_0 L divided by c squared multiplied by Gamma 0."},{"Start":"04:01.840 ","End":"04:07.415","Text":"Now comes in something slightly more complicated that we mentioned earlier that"},{"Start":"04:07.415 ","End":"04:10.760","Text":"the time difference between these 2"},{"Start":"04:10.760 ","End":"04:14.435","Text":"is not the same time difference that the observer in the lab,"},{"Start":"04:14.435 ","End":"04:16.130","Text":"we\u0027ll observe the light in."},{"Start":"04:16.130 ","End":"04:19.040","Text":"This is because the beginning of emission starts at"},{"Start":"04:19.040 ","End":"04:23.715","Text":"some point L and because our light source,"},{"Start":"04:23.715 ","End":"04:26.165","Text":"our S tag frame of reference is moving."},{"Start":"04:26.165 ","End":"04:28.310","Text":"Once the emission has ended,"},{"Start":"04:28.310 ","End":"04:30.545","Text":"it\u0027s going to be at a different point,"},{"Start":"04:30.545 ","End":"04:32.205","Text":"which is further away."},{"Start":"04:32.205 ","End":"04:34.075","Text":"We need to take into account,"},{"Start":"04:34.075 ","End":"04:37.250","Text":"that it takes the light time to get from"},{"Start":"04:37.250 ","End":"04:43.075","Text":"its new position back to the observer in the S frame of reference."},{"Start":"04:43.075 ","End":"04:48.470","Text":"Now what we\u0027re going to do is we\u0027re going to work out how long it"},{"Start":"04:48.470 ","End":"04:53.680","Text":"takes relative to the lab\u0027s frame of reference to record this pulse?"},{"Start":"04:53.680 ","End":"04:58.190","Text":"How long this pulse emission is relative to the lab?"},{"Start":"04:58.190 ","End":"04:59.780","Text":"Let\u0027s see how we do this."},{"Start":"04:59.780 ","End":"05:02.720","Text":"Here we\u0027re going to take into account that light"},{"Start":"05:02.720 ","End":"05:06.260","Text":"takes time to travel from 1 point to another."},{"Start":"05:06.260 ","End":"05:09.005","Text":"Our time, let\u0027s call it, t_3,"},{"Start":"05:09.005 ","End":"05:15.050","Text":"is going to be equal to the time in the lab that this happens, that\u0027s t_1,"},{"Start":"05:15.050 ","End":"05:19.235","Text":"which is equal to 0, plus the time that it takes"},{"Start":"05:19.235 ","End":"05:24.095","Text":"for the light to travel from whichever point it is relative to the lab,"},{"Start":"05:24.095 ","End":"05:25.580","Text":"back until the lab."},{"Start":"05:25.580 ","End":"05:32.075","Text":"That\u0027s going to be its distance away divided by the speed of light,"},{"Start":"05:32.075 ","End":"05:37.990","Text":"which is c. Then we\u0027re going to have that this is equal to 0 plus l,"},{"Start":"05:37.990 ","End":"05:43.730","Text":"divided by c. Now and we\u0027re going to do is the exact same thing."},{"Start":"05:43.730 ","End":"05:47.510","Text":"But when the lab is finished receiving the pulse,"},{"Start":"05:47.510 ","End":"05:54.710","Text":"so let\u0027s call that time t_4 when light at the end of the pulse is received by the lab,"},{"Start":"05:54.710 ","End":"05:57.860","Text":"and that\u0027s simply going to be equal to the time that it"},{"Start":"05:57.860 ","End":"06:01.600","Text":"occurs relative to the lab, so that\u0027s t_2."},{"Start":"06:01.600 ","End":"06:04.715","Text":"Plus we\u0027re going to add the distance it is"},{"Start":"06:04.715 ","End":"06:08.615","Text":"away from the observer and the lab from the origin,"},{"Start":"06:08.615 ","End":"06:13.580","Text":"divided by the speed that it\u0027s traveling at."},{"Start":"06:13.580 ","End":"06:20.135","Text":"That\u0027s going to be equal to x_2 divided by c. Then when we substitute everything in,"},{"Start":"06:20.135 ","End":"06:25.990","Text":"we get that our t_4 is equal to L divided by c plus"},{"Start":"06:25.990 ","End":"06:33.930","Text":"Gamma naught 1 plus Beta multiplied by Tau."},{"Start":"06:33.930 ","End":"06:37.610","Text":"I\u0027ll just write on the side to remind you that Beta is"},{"Start":"06:37.610 ","End":"06:41.930","Text":"equal to V_ 0 divided by c. In our question,"},{"Start":"06:41.930 ","End":"06:45.500","Text":"we were asked to find the duration of light emission in the lab,"},{"Start":"06:45.500 ","End":"06:52.035","Text":"relative to the lab and we know that it\u0027s going to be a relationship between T and Tau."},{"Start":"06:52.035 ","End":"06:56.840","Text":"What we\u0027re going to say is that our duration that the light is on"},{"Start":"06:56.840 ","End":"07:02.771","Text":"relative to the lab is going to be equal to when the lab reception of the pulse ends,"},{"Start":"07:02.771 ","End":"07:07.970","Text":"so that\u0027s t_4 minus the beginning of the reception in the labs."},{"Start":"07:07.970 ","End":"07:11.720","Text":"Lab reception of pulse begins, minus t_3."},{"Start":"07:11.720 ","End":"07:14.045","Text":"Now, when we substitute all of that in,"},{"Start":"07:14.045 ","End":"07:16.280","Text":"we\u0027re going to get this equation,"},{"Start":"07:16.280 ","End":"07:22.430","Text":"1 plus Beta divided by 1 minus Beta,"},{"Start":"07:22.430 ","End":"07:27.005","Text":"the square root of all of that multiplied by Tau."},{"Start":"07:27.005 ","End":"07:29.480","Text":"Just substitute in these 2."},{"Start":"07:29.480 ","End":"07:32.435","Text":"Now, if we want to find out our wavelength."},{"Start":"07:32.435 ","End":"07:35.810","Text":"Our wavelength is equal to Lambda,"},{"Start":"07:35.810 ","End":"07:37.160","Text":"which is equal to c,"},{"Start":"07:37.160 ","End":"07:40.820","Text":"the speed of light multiplied by our T. That is"},{"Start":"07:40.820 ","End":"07:45.485","Text":"simply going to be equal to the square root of"},{"Start":"07:45.485 ","End":"07:54.885","Text":"1 plus Beta divided by 1 minus Beta multiplied by Lambda tag,"},{"Start":"07:54.885 ","End":"07:58.415","Text":"where lambda tag, over here,"},{"Start":"07:58.415 ","End":"08:00.800","Text":"is simply equal to the speed of light,"},{"Start":"08:00.800 ","End":"08:03.575","Text":"c, multiplied by our Tau."},{"Start":"08:03.575 ","End":"08:06.100","Text":"For our frequency,"},{"Start":"08:06.100 ","End":"08:10.610","Text":"we know that our frequency is equal to 1 divided"},{"Start":"08:10.610 ","End":"08:15.050","Text":"by T. What we\u0027ll get is that it\u0027s equal to"},{"Start":"08:15.050 ","End":"08:25.245","Text":"the square root of 1 minus Beta divided by 1 plus beta multiplied by f tag."},{"Start":"08:25.245 ","End":"08:29.715","Text":"Of course, where f tag is 1 divided by Tau."},{"Start":"08:29.715 ","End":"08:32.540","Text":"That\u0027s it, these are the equations that you need to"},{"Start":"08:32.540 ","End":"08:35.440","Text":"know for the Relativistic Doppler Effect."},{"Start":"08:35.440 ","End":"08:36.925","Text":"All we did was,"},{"Start":"08:36.925 ","End":"08:43.865","Text":"we found the details of when the first event occurs relative to our S frame of reference."},{"Start":"08:43.865 ","End":"08:46.310","Text":"Then we use the Lawrence transformation to get"},{"Start":"08:46.310 ","End":"08:49.445","Text":"that with respect to S tag frame of reference."},{"Start":"08:49.445 ","End":"08:55.802","Text":"Then we used our duration of the light emission and the rest frame being Tau"},{"Start":"08:55.802 ","End":"08:58.895","Text":"and we found out at what time"},{"Start":"08:58.895 ","End":"09:02.525","Text":"the light emission will end in our S tag frame of reference."},{"Start":"09:02.525 ","End":"09:05.720","Text":"Then we did the opposite to the Lawrence transformation"},{"Start":"09:05.720 ","End":"09:09.619","Text":"to get all of those details back into our S frame of reference."},{"Start":"09:09.619 ","End":"09:13.595","Text":"Then we simply used these equations,"},{"Start":"09:13.595 ","End":"09:16.565","Text":"what we have found previously in order to find"},{"Start":"09:16.565 ","End":"09:20.300","Text":"out when our lab will start receiving the pulse,"},{"Start":"09:20.300 ","End":"09:23.510","Text":"which means that we took into account when it began,"},{"Start":"09:23.510 ","End":"09:25.700","Text":"and the distance that the pulse was when it was"},{"Start":"09:25.700 ","End":"09:28.955","Text":"emitted away from the origin in the S frame of reference."},{"Start":"09:28.955 ","End":"09:32.150","Text":"Then we did the exact same thing for when"},{"Start":"09:32.150 ","End":"09:35.660","Text":"the reception of the pulse will end in the lab\u0027s frame of reference."},{"Start":"09:35.660 ","End":"09:40.295","Text":"Again, we used our time that the emission ends plus"},{"Start":"09:40.295 ","End":"09:46.460","Text":"the distance traveled by the light wave to get back to the origin."},{"Start":"09:46.460 ","End":"09:49.550","Text":"Then we simply plugged everything in,"},{"Start":"09:49.550 ","End":"09:55.040","Text":"and found out a big T and all of our equations for the Doppler Effect."},{"Start":"09:55.040 ","End":"09:58.050","Text":"That\u0027s the end of this lesson."}],"ID":9553},{"Watched":false,"Name":"Rod Emits Light","Duration":"8m ","ChapterTopicVideoID":9253,"CourseChapterTopicPlaylistID":9398,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.700","Text":"Hello, here we have a rod,"},{"Start":"00:02.700 ","End":"00:05.880","Text":"this over here in pink of rest length L_0,"},{"Start":"00:05.880 ","End":"00:11.625","Text":"which is moving with a velocity v relative to Earth, at time t=0."},{"Start":"00:11.625 ","End":"00:18.045","Text":"The left edge of the rod is at x=x tag, which is equal to 0."},{"Start":"00:18.045 ","End":"00:23.100","Text":"Which means that the origin of 2 of the systems are at the exact same point."},{"Start":"00:23.100 ","End":"00:27.105","Text":"At this moment, the rod emits light from its right edge,"},{"Start":"00:27.105 ","End":"00:28.785","Text":"from Point B over here."},{"Start":"00:28.785 ","End":"00:32.445","Text":"Then sometime later at t=Tau,"},{"Start":"00:32.445 ","End":"00:36.660","Text":"the rod emits light from its left edge from over here at Point A."},{"Start":"00:36.660 ","End":"00:39.450","Text":"Now we\u0027re being asked to calculate the time difference between"},{"Start":"00:39.450 ","End":"00:43.020","Text":"the 2 events as seen by an observer on Earth,"},{"Start":"00:43.020 ","End":"00:47.055","Text":"at the origin of Earth\u0027s frame of reference."},{"Start":"00:47.055 ","End":"00:51.755","Text":"Note; consider the time taken for the light to reach the observer."},{"Start":"00:51.755 ","End":"00:57.065","Text":"That means that an observer standing at the origin of the Earth\u0027s frame of reference."},{"Start":"00:57.065 ","End":"01:02.120","Text":"We\u0027re trying to see the time difference between receiving the light wave"},{"Start":"01:02.120 ","End":"01:07.860","Text":"coming in from Point B and the light wave coming in from Point A."},{"Start":"01:07.860 ","End":"01:12.005","Text":"As per usual, what we\u0027re going to do is we\u0027re going to use our table"},{"Start":"01:12.005 ","End":"01:16.060","Text":"with our events happening on our S frame of reference so that\u0027s our Earth."},{"Start":"01:16.060 ","End":"01:17.540","Text":"Our S tag frame of reference,"},{"Start":"01:17.540 ","End":"01:19.355","Text":"which is the rest frame of the rod."},{"Start":"01:19.355 ","End":"01:22.160","Text":"We\u0027re going to write out all of our equations and here we"},{"Start":"01:22.160 ","End":"01:25.220","Text":"have our equations for our Lorentz transform."},{"Start":"01:25.220 ","End":"01:29.920","Text":"Our first event is when our light is emitted from B,"},{"Start":"01:29.920 ","End":"01:31.695","Text":"from the right side of the rod."},{"Start":"01:31.695 ","End":"01:35.690","Text":"This happens in our rod\u0027s frame of reference at S tag,"},{"Start":"01:35.690 ","End":"01:37.820","Text":"the position that this happens,"},{"Start":"01:37.820 ","End":"01:43.590","Text":"x_1 tag= L_0 as given in the question."},{"Start":"01:43.590 ","End":"01:45.690","Text":"The time that this happens,"},{"Start":"01:45.690 ","End":"01:49.650","Text":"t_1 tag= 0,"},{"Start":"01:49.650 ","End":"01:52.845","Text":"also given in the question."},{"Start":"01:52.845 ","End":"01:57.110","Text":"Now what we\u0027re going to do in order to get these equations"},{"Start":"01:57.110 ","End":"02:01.835","Text":"into an equation that matches our Earth\u0027s frame of reference,"},{"Start":"02:01.835 ","End":"02:05.345","Text":"our S. We\u0027re going to use these equations here."},{"Start":"02:05.345 ","End":"02:08.165","Text":"Now these equations are for the Lorentz transform."},{"Start":"02:08.165 ","End":"02:09.950","Text":"In order to get the opposite transform,"},{"Start":"02:09.950 ","End":"02:15.165","Text":"we just have to rearrange to get our x over here instead of our x tag."},{"Start":"02:15.165 ","End":"02:21.540","Text":"Once we do that, we\u0027ll get that our x_1 is going to be equal to Gamma_naught multiplied"},{"Start":"02:21.540 ","End":"02:30.995","Text":"by L_0 and our t_1 is going to be equal to Gamma_naught multiplied by v_0,"},{"Start":"02:30.995 ","End":"02:34.800","Text":"where v_0 is the velocity of the rod,"},{"Start":"02:34.800 ","End":"02:40.240","Text":"v_0 divided by c^2 multiplied by L_0."},{"Start":"02:40.430 ","End":"02:44.050","Text":"Now we\u0027ll move on to our second event,"},{"Start":"02:44.050 ","End":"02:47.905","Text":"which is when the light is emitted from A, from over here."},{"Start":"02:47.905 ","End":"02:52.900","Text":"Again, we\u0027re speaking with reference to our rest frame of the rod which is S tag."},{"Start":"02:52.900 ","End":"03:00.280","Text":"We know that the position that A is emitted is going to be called x_2 tag."},{"Start":"03:00.280 ","End":"03:04.390","Text":"Because we know that A is at the origin of the rest frame,"},{"Start":"03:04.390 ","End":"03:08.610","Text":"so it\u0027s at Point 0 still relative to the rest frame."},{"Start":"03:08.610 ","End":"03:12.010","Text":"This happens at t_2 tag=Tau,"},{"Start":"03:12.450 ","End":"03:17.345","Text":"which is also given to us in the question, at t=Tau."},{"Start":"03:17.345 ","End":"03:22.915","Text":"Now again, we\u0027re going to use the transform which is opposite to the Lorentz transform"},{"Start":"03:22.915 ","End":"03:28.825","Text":"in order to find out what x_2 tag and t_2 tag means in the Earth\u0027s frame of reference."},{"Start":"03:28.825 ","End":"03:30.665","Text":"Once we plug everything in,"},{"Start":"03:30.665 ","End":"03:37.290","Text":"we\u0027ll get that our x_2 is equal to Gamma_naught multiplied by v_0 multiplied by"},{"Start":"03:37.290 ","End":"03:44.865","Text":"Tau and that our t_2 is going to be equal to Gamma_naught multiplied by Tau."},{"Start":"03:44.865 ","End":"03:50.180","Text":"Now we have the time that these 2 events and the position that"},{"Start":"03:50.180 ","End":"03:54.995","Text":"these 2 events took place in relative to our Earth\u0027s frame of reference."},{"Start":"03:54.995 ","End":"03:57.560","Text":"But now what we want to take into account is that"},{"Start":"03:57.560 ","End":"04:00.380","Text":"the light takes time to travel from wherever"},{"Start":"04:00.380 ","End":"04:05.990","Text":"it is in the Earth\u0027s frame of reference to the origin where our observer is."},{"Start":"04:05.990 ","End":"04:09.335","Text":"If we say that our observer is over here, at the beginning,"},{"Start":"04:09.335 ","End":"04:15.680","Text":"we know that our S and S tag frame of reference is at the exact same point."},{"Start":"04:15.680 ","End":"04:19.505","Text":"Their origins are one on top of each other when we begin."},{"Start":"04:19.505 ","End":"04:21.695","Text":"Then when B is off,"},{"Start":"04:21.695 ","End":"04:25.295","Text":"because we know that our rod is moving with a constant velocity v,"},{"Start":"04:25.295 ","End":"04:29.040","Text":"we know that B emits light and it\u0027s also moving"},{"Start":"04:29.040 ","End":"04:33.510","Text":"simultaneously so its emission will happen at around about here,"},{"Start":"04:33.510 ","End":"04:41.415","Text":"at t=0, it\u0027s still a distance of L_0 away from our S tag frame of reference."},{"Start":"04:41.415 ","End":"04:44.055","Text":"This distance is L_0,"},{"Start":"04:44.055 ","End":"04:49.085","Text":"but this distance away from our now S-only S frame of reference,"},{"Start":"04:49.085 ","End":"04:52.865","Text":"which is stationary, is going to be greater."},{"Start":"04:52.865 ","End":"04:57.050","Text":"Then when our A side begins to emit light,"},{"Start":"04:57.050 ","End":"04:59.780","Text":"this will also happen at a different area,"},{"Start":"04:59.780 ","End":"05:02.725","Text":"let\u0027s say somewhere around here."},{"Start":"05:02.725 ","End":"05:09.500","Text":"Then it\u0027s still at the origin of our S tag where A emits light, however,"},{"Start":"05:09.500 ","End":"05:12.080","Text":"it\u0027s now also further away from our S frame of"},{"Start":"05:12.080 ","End":"05:14.870","Text":"reference and the amount of time it takes to get"},{"Start":"05:14.870 ","End":"05:20.520","Text":"from this position to our observer standing at the origin is also going to take time."},{"Start":"05:20.520 ","End":"05:25.620","Text":"What we want to do is we want to find the time that this takes."},{"Start":"05:25.620 ","End":"05:31.190","Text":"The third thing that we\u0027re going to do is we\u0027re going to see when the light from"},{"Start":"05:31.190 ","End":"05:37.310","Text":"our Point B in a rod reaches the origin in our S frame of reference."},{"Start":"05:37.310 ","End":"05:42.915","Text":"We already know that this happens at time t=1."},{"Start":"05:42.915 ","End":"05:46.710","Text":"Let\u0027s call this total time taken t_3."},{"Start":"05:46.710 ","End":"05:51.740","Text":"It\u0027s going to be equal to t_1 plus"},{"Start":"05:51.740 ","End":"05:57.950","Text":"the time taken for the speed of light to travel that distance."},{"Start":"05:57.950 ","End":"06:02.020","Text":"That\u0027s going to be the distance or x_1."},{"Start":"06:02.020 ","End":"06:10.630","Text":"The distance that Point B is away from the origin and S divided by the speed of light."},{"Start":"06:10.630 ","End":"06:14.419","Text":"Then we will see after doing some algebra,"},{"Start":"06:14.419 ","End":"06:19.130","Text":"that this is equal to Gamma_naught divided by c"},{"Start":"06:19.130 ","End":"06:25.445","Text":"multiplied by 1 plus Beta multiplied by L_0."},{"Start":"06:25.445 ","End":"06:31.130","Text":"Now, we\u0027re going to do the same to find out the time it takes for the light"},{"Start":"06:31.130 ","End":"06:37.265","Text":"emitted by Point A to reach the origin of the S frame of reference."},{"Start":"06:37.265 ","End":"06:42.575","Text":"This time that this is going to happen is going to be equal to t_4."},{"Start":"06:42.575 ","End":"06:47.599","Text":"That\u0027s going to be equal to the time at which it\u0027s emitted,"},{"Start":"06:47.599 ","End":"06:49.160","Text":"which is t_2,"},{"Start":"06:49.160 ","End":"06:51.454","Text":"relative to the S frame of reference,"},{"Start":"06:51.454 ","End":"06:57.830","Text":"plus the time it takes for it to reach the origin from whichever Point it"},{"Start":"06:57.830 ","End":"07:04.430","Text":"is in the system relative to the origin in the S frame of reference."},{"Start":"07:04.430 ","End":"07:09.950","Text":"That again is going to be its position relative to the S frame of reference,"},{"Start":"07:09.950 ","End":"07:13.985","Text":"which is x_2 divided by the speed of light."},{"Start":"07:13.985 ","End":"07:17.345","Text":"Then, once we plug in all the numbers and do some algebra,"},{"Start":"07:17.345 ","End":"07:24.910","Text":"we\u0027ll get that this is equal to Gamma_naught 1 plus Beta multiplied by Tau."},{"Start":"07:24.910 ","End":"07:26.164","Text":"Now, in the question,"},{"Start":"07:26.164 ","End":"07:28.670","Text":"we\u0027re being told to calculate the time difference"},{"Start":"07:28.670 ","End":"07:32.255","Text":"between the 2 events as seen by the observer on Earth."},{"Start":"07:32.255 ","End":"07:37.025","Text":"All we have to do is we want to find our Delta t. Our Delta t,"},{"Start":"07:37.025 ","End":"07:43.115","Text":"which is simply going to be equal to our t_4 minus our t_3,"},{"Start":"07:43.115 ","End":"07:45.485","Text":"which is simply going to be equal to,"},{"Start":"07:45.485 ","End":"07:47.300","Text":"once we do all the algebra,"},{"Start":"07:47.300 ","End":"07:52.590","Text":"Gamma_naught 1 plus Beta multiplied by"},{"Start":"07:52.590 ","End":"07:58.715","Text":"Tau minus L_0 divided by C. That\u0027s the answer."},{"Start":"07:58.715 ","End":"08:01.110","Text":"That\u0027s the end of this question."}],"ID":9554},{"Watched":false,"Name":"Causality","Duration":"8m 32s","ChapterTopicVideoID":9261,"CourseChapterTopicPlaylistID":9398,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.680","Text":"Hello. In this question,"},{"Start":"00:01.680 ","End":"00:05.520","Text":"we have 2 events which take place in the S\u0027 system,"},{"Start":"00:05.520 ","End":"00:08.100","Text":"which moves with the velocity v. Now,"},{"Start":"00:08.100 ","End":"00:11.640","Text":"we\u0027re being told that the first event was the lifting of"},{"Start":"00:11.640 ","End":"00:17.190","Text":"a switch at time t_1\u0027 and in position x_1\u0027."},{"Start":"00:17.190 ","End":"00:20.790","Text":"Both of these relative to the S\u0027 frame of reference,"},{"Start":"00:20.790 ","End":"00:24.120","Text":"and that the second event was an emission of light."},{"Start":"00:24.120 ","End":"00:28.890","Text":"A light bulb was switched on at time t_2\u0027 and position x_2\u0027,"},{"Start":"00:28.890 ","End":"00:32.465","Text":"again, relative to our S\u0027 frame of reference."},{"Start":"00:32.465 ","End":"00:38.580","Text":"Now let\u0027s assume that event number 1 took place before event number 2 in the S\u0027 system."},{"Start":"00:38.580 ","End":"00:41.850","Text":"That means that t_1\u0027 is smaller than t_2\u0027."},{"Start":"00:41.850 ","End":"00:47.165","Text":"We\u0027re being asked to find the times at which the events took place in the S system."},{"Start":"00:47.165 ","End":"00:53.555","Text":"Now, one of the goals of this lesson is to speak about the term cause and to see that"},{"Start":"00:53.555 ","End":"00:59.959","Text":"if one event takes place before another event in one system relative to another system,"},{"Start":"00:59.959 ","End":"01:04.765","Text":"the second event could have taken place before the first event."},{"Start":"01:04.765 ","End":"01:09.740","Text":"Let\u0027s take a look. The first thing that we\u0027re going to do is we\u0027re going to"},{"Start":"01:09.740 ","End":"01:15.320","Text":"translate our t_1\u0027 and t_2\u0027 into terms within our S frame of reference."},{"Start":"01:15.320 ","End":"01:20.300","Text":"What we\u0027re going to do is we\u0027re going to use our opposite Lorentz transformation."},{"Start":"01:20.300 ","End":"01:26.595","Text":"That means that our t_1 is going to be equal to Gamma 0 multiplied by"},{"Start":"01:26.595 ","End":"01:35.100","Text":"t_1\u0027 plus v_0 divided by c multiplied by x_1\u0027."},{"Start":"01:35.100 ","End":"01:43.140","Text":"Then our t_2 is going to be equal to Gamma 0 multiplied by"},{"Start":"01:43.140 ","End":"01:51.670","Text":"t_2\u0027 plus v_0 divided by c multiplied by x_2\u0027."},{"Start":"01:51.670 ","End":"01:54.665","Text":"Here we\u0027re assuming that we know our values for"},{"Start":"01:54.665 ","End":"01:59.615","Text":"x_1\u0027 and t_1\u0027 and the same for our event number 2."},{"Start":"01:59.615 ","End":"02:03.890","Text":"Now what we want to do is we want to check if"},{"Start":"02:03.890 ","End":"02:09.195","Text":"potentially our t_2 could be smaller than our t_1,"},{"Start":"02:09.195 ","End":"02:12.270","Text":"if our event number 2 that happened after"},{"Start":"02:12.270 ","End":"02:16.125","Text":"event number 1 relative to our S\u0027 frame of reference."},{"Start":"02:16.125 ","End":"02:19.885","Text":"If this events could have happened then before,"},{"Start":"02:19.885 ","End":"02:23.000","Text":"if we\u0027re looking in our S frame of reference."},{"Start":"02:23.000 ","End":"02:24.755","Text":"Let\u0027s check this."},{"Start":"02:24.755 ","End":"02:26.915","Text":"Let\u0027s substitute in all of our values."},{"Start":"02:26.915 ","End":"02:37.544","Text":"We will have that our Gamma 0 t_2\u0027 plus our v_0 divided by c"},{"Start":"02:37.544 ","End":"02:43.230","Text":"multiplied by x_2\u0027 would then be smaller than our Gamma"},{"Start":"02:43.230 ","End":"02:50.550","Text":"0 t_1\u0027 plus v_0 divided by c multiplied by x_1\u0027."},{"Start":"02:50.550 ","End":"02:52.365","Text":"When we do our algebra,"},{"Start":"02:52.365 ","End":"02:57.145","Text":"we\u0027ll get that if t_2 is smaller than t_1,"},{"Start":"02:57.145 ","End":"03:02.160","Text":"that means that the difference between t_2\u0027 and"},{"Start":"03:02.160 ","End":"03:07.250","Text":"t_1\u0027 has to be smaller than this relationship over here,"},{"Start":"03:07.250 ","End":"03:15.980","Text":"which is v_0 divided by c^2 multiplied by x_1\u0027 minus x_2\u0027."},{"Start":"03:15.980 ","End":"03:23.360","Text":"Now, another way of looking at this equation is if our x_1\u0027 is bigger"},{"Start":"03:23.360 ","End":"03:31.595","Text":"than our x_2\u0027 plus c^2 divided by v_0 multiplied by Delta t\u0027."},{"Start":"03:31.595 ","End":"03:37.790","Text":"What does that mean? Simply that our position at which our first event in"},{"Start":"03:37.790 ","End":"03:45.104","Text":"our S\u0027 frame of reference happens is further away than where the second event occurs."},{"Start":"03:45.104 ","End":"03:48.934","Text":"That means that if we\u0027re looking at our S frame of reference,"},{"Start":"03:48.934 ","End":"03:50.160","Text":"that our light,"},{"Start":"03:50.160 ","End":"03:52.290","Text":"which was our event number 2,"},{"Start":"03:52.290 ","End":"03:56.820","Text":"is closer to the origin than our event number 1,"},{"Start":"03:56.820 ","End":"04:00.475","Text":"which was the lifting of some switch."},{"Start":"04:00.475 ","End":"04:04.070","Text":"What we can see is that one event that"},{"Start":"04:04.070 ","End":"04:07.655","Text":"takes place before another event in one frame of reference,"},{"Start":"04:07.655 ","End":"04:13.030","Text":"they can still take place after the first event in a different frame of reference."},{"Start":"04:13.030 ","End":"04:16.550","Text":"As in, the order that these 2 events can occur can"},{"Start":"04:16.550 ","End":"04:20.315","Text":"switch when we move between different frames of reference."},{"Start":"04:20.315 ","End":"04:21.950","Text":"Now in the next section,"},{"Start":"04:21.950 ","End":"04:24.595","Text":"we\u0027re going to be speaking about causality."},{"Start":"04:24.595 ","End":"04:26.525","Text":"As we can see over here,"},{"Start":"04:26.525 ","End":"04:31.400","Text":"the condition that we needed to meet in order for us to see the order of"},{"Start":"04:31.400 ","End":"04:33.410","Text":"our events to take place in"},{"Start":"04:33.410 ","End":"04:37.535","Text":"a different order when we\u0027re moving to a different frame of reference,"},{"Start":"04:37.535 ","End":"04:41.190","Text":"was that the difference between the 2 events taking place;"},{"Start":"04:41.190 ","End":"04:43.270","Text":"t_2\u0027 minus t_1\u0027,"},{"Start":"04:43.270 ","End":"04:51.115","Text":"had to be smaller than the difference between their positions, x_1\u0027 minus x_2\u0027."},{"Start":"04:51.115 ","End":"04:56.565","Text":"Now let\u0027s look at our term causality and see what its definition is."},{"Start":"04:56.565 ","End":"05:00.320","Text":"Causality, if one event is the cause of another event,"},{"Start":"05:00.320 ","End":"05:04.850","Text":"the order in which the events take place cannot be changed."},{"Start":"05:04.850 ","End":"05:07.490","Text":"That means that, for instance,"},{"Start":"05:07.490 ","End":"05:09.905","Text":"with the flipping of the switch,"},{"Start":"05:09.905 ","End":"05:14.705","Text":"if that was the cause for the light bulb to start evincing light,"},{"Start":"05:14.705 ","End":"05:20.585","Text":"to turn on, then that means that no matter which frame of reference we\u0027re looking at,"},{"Start":"05:20.585 ","End":"05:22.860","Text":"the order can never be changed,"},{"Start":"05:22.860 ","End":"05:28.595","Text":"as in the observer will always observe the light switch being switched on first,"},{"Start":"05:28.595 ","End":"05:30.155","Text":"and only after that,"},{"Start":"05:30.155 ","End":"05:32.885","Text":"they\u0027ll see the bulb switching on."},{"Start":"05:32.885 ","End":"05:36.595","Text":"Now, this notion, causality,"},{"Start":"05:36.595 ","End":"05:41.330","Text":"comes from the notion that nothing can travel faster than the speed of light."},{"Start":"05:41.330 ","End":"05:46.055","Text":"That also means the power transmitted in electricity, electric currents,"},{"Start":"05:46.055 ","End":"05:49.310","Text":"or anything else that travels quickly, electrons,"},{"Start":"05:49.310 ","End":"05:52.910","Text":"whatever it might be, nothing can travel faster than the speed of light."},{"Start":"05:52.910 ","End":"05:55.415","Text":"Now, if we look back at our example,"},{"Start":"05:55.415 ","End":"06:00.815","Text":"we can see that if flipping the light switch cause the light bulb to turn on,"},{"Start":"06:00.815 ","End":"06:05.495","Text":"then our t_2\u0027 which is the time that our light bulb was switched on,"},{"Start":"06:05.495 ","End":"06:12.230","Text":"has to be greater or equal to our t_1\u0027 which was the time that the switch was flipped,"},{"Start":"06:12.230 ","End":"06:14.695","Text":"plus this thing over here."},{"Start":"06:14.695 ","End":"06:18.020","Text":"This thing over here in the red circle is the time"},{"Start":"06:18.020 ","End":"06:21.590","Text":"taken for the transmitted power to reach the bulb."},{"Start":"06:21.590 ","End":"06:27.500","Text":"That means from the light switch until the light bulb. What does that mean?"},{"Start":"06:27.500 ","End":"06:33.275","Text":"That means that if our flipping of the switch happened at this position in this time,"},{"Start":"06:33.275 ","End":"06:36.020","Text":"and that caused our light to switch on,"},{"Start":"06:36.020 ","End":"06:42.020","Text":"the amount of time that it takes for the electricity or the power from the switch to"},{"Start":"06:42.020 ","End":"06:44.900","Text":"reach our light bulb is always going to be"},{"Start":"06:44.900 ","End":"06:48.910","Text":"slower than the speed of light or equal to the speed of light,"},{"Start":"06:48.910 ","End":"06:50.765","Text":"but it\u0027s never going to be faster."},{"Start":"06:50.765 ","End":"06:53.990","Text":"Which means that in order for"},{"Start":"06:53.990 ","End":"07:00.230","Text":"that event to then be seen over here or in our S frame of reference,"},{"Start":"07:00.230 ","End":"07:02.644","Text":"wherever that might be, a different frame of reference,"},{"Start":"07:02.644 ","End":"07:06.600","Text":"because the speed of light is so much faster than this,"},{"Start":"07:06.600 ","End":"07:11.709","Text":"so our first event is always going to be seen first."},{"Start":"07:11.709 ","End":"07:13.580","Text":"That means, again,"},{"Start":"07:13.580 ","End":"07:19.250","Text":"going back to our equation that the time that we see our second event happening is"},{"Start":"07:19.250 ","End":"07:25.040","Text":"always going to be at a time greater or equal to the time that our first event happened,"},{"Start":"07:25.040 ","End":"07:27.320","Text":"plus this over here,"},{"Start":"07:27.320 ","End":"07:30.890","Text":"which is the time taken for the information to get"},{"Start":"07:30.890 ","End":"07:34.700","Text":"passed on from our first event until our second event."},{"Start":"07:34.700 ","End":"07:37.295","Text":"That means when we play around with this,"},{"Start":"07:37.295 ","End":"07:44.225","Text":"that our Delta t is always going to have to be greater or equal to this over here,"},{"Start":"07:44.225 ","End":"07:49.580","Text":"which is the distance between the 2 events divided by the speed of light."},{"Start":"07:49.580 ","End":"07:54.875","Text":"Which means that never ever will we be able to"},{"Start":"07:54.875 ","End":"08:00.845","Text":"observe 2 events such that we see the second event first."},{"Start":"08:00.845 ","End":"08:04.480","Text":"Therefore, what we found earlier in this video,"},{"Start":"08:04.480 ","End":"08:07.880","Text":"this condition over here can never ever occur."},{"Start":"08:07.880 ","End":"08:09.995","Text":"This is the law that we have to use,"},{"Start":"08:09.995 ","End":"08:11.450","Text":"and it\u0027s called causality."},{"Start":"08:11.450 ","End":"08:15.125","Text":"That means that when you\u0027re looking from any frame of reference,"},{"Start":"08:15.125 ","End":"08:18.155","Text":"if one event occurred before the other event,"},{"Start":"08:18.155 ","End":"08:22.870","Text":"in every single other frame of reference in which you observe these 2 events,"},{"Start":"08:22.870 ","End":"08:26.975","Text":"the cause event will always be seen first,"},{"Start":"08:26.975 ","End":"08:28.805","Text":"followed by the event,"},{"Start":"08:28.805 ","End":"08:30.140","Text":"which is the effect."},{"Start":"08:30.140 ","End":"08:33.150","Text":"Okay, that\u0027s the end of this lesson."}],"ID":9555},{"Watched":false,"Name":"Exercise 3","Duration":"19m 8s","ChapterTopicVideoID":9268,"CourseChapterTopicPlaylistID":9398,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.100","Text":"A rocket leaves the star Alpha and travels back to Earth."},{"Start":"00:05.100 ","End":"00:09.090","Text":"On the way, the rocket passes next to Alpha\u0027s moon and"},{"Start":"00:09.090 ","End":"00:13.965","Text":"observes a strong electromagnetic pulse heading towards the star."},{"Start":"00:13.965 ","End":"00:17.355","Text":"It is known that aggressive aliens live on the moon."},{"Start":"00:17.355 ","End":"00:22.305","Text":"Now 1.3 seconds later in the rocket\u0027s frame of reference,"},{"Start":"00:22.305 ","End":"00:26.070","Text":"the rocket sees a huge explosion go off on Alpha."},{"Start":"00:26.070 ","End":"00:29.445","Text":"The distance between Alpha and its moon,"},{"Start":"00:29.445 ","End":"00:32.040","Text":"so the distance between here to here is"},{"Start":"00:32.040 ","End":"00:37.060","Text":"500 million meters as recorded in the rocket system."},{"Start":"00:37.060 ","End":"00:42.710","Text":"The velocity of the rocket relative to Alpha and its moon is 0.9c,"},{"Start":"00:42.710 ","End":"00:45.365","Text":"where c is the speed of light."},{"Start":"00:45.365 ","End":"00:49.009","Text":"Now before we go on to answering the questions,"},{"Start":"00:49.009 ","End":"00:52.715","Text":"let\u0027s just define some basic things."},{"Start":"00:52.715 ","End":"00:55.475","Text":"Let\u0027s say Beta."},{"Start":"00:55.475 ","End":"00:59.990","Text":"Beta is going to be the relative velocity of our system,"},{"Start":"00:59.990 ","End":"01:02.640","Text":"of our S tag system."},{"Start":"01:02.640 ","End":"01:08.795","Text":"That is going to be equal to the velocity of the system divided by the speed of light,"},{"Start":"01:08.795 ","End":"01:14.405","Text":"which is going to be equal to 0.9c divided by c,"},{"Start":"01:14.405 ","End":"01:18.455","Text":"which is simply going to be equal to 0.9."},{"Start":"01:18.455 ","End":"01:23.960","Text":"Next, we\u0027re going to see what Gamma 0 is equal to."},{"Start":"01:23.960 ","End":"01:27.695","Text":"So that\u0027s going to be equal to 1 minus"},{"Start":"01:27.695 ","End":"01:34.550","Text":"0.9^2 to the power of negative 1/2,"},{"Start":"01:34.550 ","End":"01:37.025","Text":"and that when plugged into a calculator,"},{"Start":"01:37.025 ","End":"01:41.795","Text":"is going to yield 2.294."},{"Start":"01:41.795 ","End":"01:43.835","Text":"Then we\u0027re going to have c,"},{"Start":"01:43.835 ","End":"01:45.275","Text":"which is the speed of light,"},{"Start":"01:45.275 ","End":"01:53.229","Text":"and that is equal to 3 times 10^8 meters per second."},{"Start":"01:53.680 ","End":"01:57.545","Text":"Now let\u0027s begin answering our questions."},{"Start":"01:57.545 ","End":"01:59.090","Text":"Question number 1 is,"},{"Start":"01:59.090 ","End":"02:02.255","Text":"what is the time span between observing"},{"Start":"02:02.255 ","End":"02:08.455","Text":"the electromagnetic wave and the actual explosion relative to Alpha and its moon?"},{"Start":"02:08.455 ","End":"02:13.070","Text":"Relative to our S frame of reference over here,"},{"Start":"02:13.070 ","End":"02:17.525","Text":"and we can see that the moon is right at the origin."},{"Start":"02:17.525 ","End":"02:21.140","Text":"How we\u0027re going to do this is by drawing our table."},{"Start":"02:21.140 ","End":"02:24.455","Text":"We have our event."},{"Start":"02:24.455 ","End":"02:28.510","Text":"Event number 1 is the pulse."},{"Start":"02:28.790 ","End":"02:32.150","Text":"Seeing the pulse and then relative to"},{"Start":"02:32.150 ","End":"02:37.790","Text":"our S frame of reference and relative to our S tag frame of reference."},{"Start":"02:37.790 ","End":"02:40.460","Text":"Now let\u0027s begin filling this in."},{"Start":"02:40.460 ","End":"02:43.010","Text":"In our S frame of reference,"},{"Start":"02:43.010 ","End":"02:48.410","Text":"the position is at x_1 is equal to 0 because it\u0027s at the origin,"},{"Start":"02:48.410 ","End":"02:50.135","Text":"and our t_1,"},{"Start":"02:50.135 ","End":"02:53.105","Text":"the time that this happens is also equal to 0."},{"Start":"02:53.105 ","End":"02:54.770","Text":"This is when the origin of"},{"Start":"02:54.770 ","End":"03:00.140","Text":"our S and S tag frame of references are right at the same point at the moon."},{"Start":"03:00.140 ","End":"03:02.945","Text":"Now when we do the Lorentz transform,"},{"Start":"03:02.945 ","End":"03:10.730","Text":"we\u0027ll get that our x tag_1 is also equal to 0 and our t tag_1 is also equal to 0."},{"Start":"03:10.730 ","End":"03:13.999","Text":"Now let\u0027s speak about the second event,"},{"Start":"03:13.999 ","End":"03:17.135","Text":"which is discovering the pulse."},{"Start":"03:17.135 ","End":"03:21.530","Text":"Now this discovery only happens in our S tag frame of reference,"},{"Start":"03:21.530 ","End":"03:24.740","Text":"so we\u0027re not going to fill out the column for"},{"Start":"03:24.740 ","End":"03:28.160","Text":"our S frame of reference and soon we\u0027ll speak about that."},{"Start":"03:28.160 ","End":"03:32.705","Text":"The discovery of the pulse in our S tag frame of reference is"},{"Start":"03:32.705 ","End":"03:39.150","Text":"at t_2 tag is equal to 1.3 from the question,"},{"Start":"03:39.150 ","End":"03:44.225","Text":"and our position, our x_2 tag is equal to"},{"Start":"03:44.225 ","End":"03:50.180","Text":"0 because it\u0027s the person on the rocket which is discovering the pulse."},{"Start":"03:50.180 ","End":"03:56.180","Text":"The person on the rocket is located right at the origin of our S tag frame of reference,"},{"Start":"03:56.180 ","End":"03:58.420","Text":"so that means it\u0027s at 0."},{"Start":"03:58.420 ","End":"04:01.849","Text":"Now we\u0027re going to speak about our third event,"},{"Start":"04:01.849 ","End":"04:04.325","Text":"which is the explosion."},{"Start":"04:04.325 ","End":"04:07.010","Text":"Before we fill in this row,"},{"Start":"04:07.010 ","End":"04:13.375","Text":"why do I differentiate between the discovery of the pulse and the explosion?"},{"Start":"04:13.375 ","End":"04:16.745","Text":"Right now we\u0027re dealing with relativity,"},{"Start":"04:16.745 ","End":"04:20.630","Text":"which means that from the moment that there is the explosion,"},{"Start":"04:20.630 ","End":"04:24.335","Text":"it takes time for the light to travel"},{"Start":"04:24.335 ","End":"04:29.065","Text":"back to the rocket in order for the rocket to see the explosion."},{"Start":"04:29.065 ","End":"04:34.520","Text":"That means that there\u0027s a different event for when the observer on"},{"Start":"04:34.520 ","End":"04:39.410","Text":"the rocket sees or discovers the pulse and another event"},{"Start":"04:39.410 ","End":"04:44.000","Text":"for when the explosion actually took place relative"},{"Start":"04:44.000 ","End":"04:49.490","Text":"to our S frame of reference and also relative to our S tag frame of reference,"},{"Start":"04:49.490 ","End":"04:52.760","Text":"and that is event number 3."},{"Start":"04:53.360 ","End":"04:57.680","Text":"Relative to our S tag frame of reference,"},{"Start":"04:57.680 ","End":"05:00.470","Text":"we can say that the time at which"},{"Start":"05:00.470 ","End":"05:04.850","Text":"the explosion was observed in our S tag frame of reference,"},{"Start":"05:04.850 ","End":"05:08.240","Text":"so let\u0027s call that t_3 tag,"},{"Start":"05:08.240 ","End":"05:15.175","Text":"is going to be equal to the time that the explosion was discovered."},{"Start":"05:15.175 ","End":"05:22.100","Text":"That\u0027s t_2 tag minus the time it took for the light"},{"Start":"05:22.100 ","End":"05:29.370","Text":"from the explosion to travel from the star to our rocket in our S tag frame of reference."},{"Start":"05:29.370 ","End":"05:37.355","Text":"That\u0027s going to be some distance x_Alpha tag divided by the speed of light."},{"Start":"05:37.355 ","End":"05:43.505","Text":"Let\u0027s give ourselves some intuition at what is going on over here in this equation."},{"Start":"05:43.505 ","End":"05:48.004","Text":"Over here, we\u0027re going to have our S tag frame of reference,"},{"Start":"05:48.004 ","End":"05:51.055","Text":"which is stationary to itself."},{"Start":"05:51.055 ","End":"05:54.980","Text":"That means that our S frame of reference is moving"},{"Start":"05:54.980 ","End":"05:59.840","Text":"away from our S tag frame of reference relative to our S tag."},{"Start":"05:59.840 ","End":"06:02.420","Text":"At our time t_2,"},{"Start":"06:02.420 ","End":"06:05.845","Text":"we\u0027ve discovered that there was an explosion."},{"Start":"06:05.845 ","End":"06:11.030","Text":"That means that our S frame of reference is somewhere over here."},{"Start":"06:11.030 ","End":"06:15.440","Text":"We\u0027ve had the explosion and the information of the explosion"},{"Start":"06:15.440 ","End":"06:20.780","Text":"has managed to reach our astronauts on our rocket."},{"Start":"06:20.780 ","End":"06:28.670","Text":"Now because we know that in order for us to receive this information from wherever it is,"},{"Start":"06:28.670 ","End":"06:32.330","Text":"we need to take into account that the information takes"},{"Start":"06:32.330 ","End":"06:35.900","Text":"time to reach our astronauts on our rocket."},{"Start":"06:35.900 ","End":"06:41.540","Text":"In order to find out when the explosion actually took place over here was at t_2 tag."},{"Start":"06:41.540 ","End":"06:47.660","Text":"We know that the explosion took place at time t_2"},{"Start":"06:47.660 ","End":"06:54.890","Text":"minus the amount of time it took for the information to reach our astronauts."},{"Start":"06:54.890 ","End":"06:59.510","Text":"Then we can say that if our S frame of reference was over here,"},{"Start":"06:59.510 ","End":"07:03.755","Text":"at t_3 tag when the explosion actually took place,"},{"Start":"07:03.755 ","End":"07:07.370","Text":"this distance traveled over here by"},{"Start":"07:07.370 ","End":"07:11.090","Text":"our S frame of reference relative to our S tag frame of reference,"},{"Start":"07:11.090 ","End":"07:15.365","Text":"we can call this distance x_Alpha tag,"},{"Start":"07:15.365 ","End":"07:21.395","Text":"which means that the time taken to travel this distance is simply going to be"},{"Start":"07:21.395 ","End":"07:28.595","Text":"this distance divided by the speed at which the information is traveling,"},{"Start":"07:28.595 ","End":"07:30.785","Text":"which is c, the speed of light."},{"Start":"07:30.785 ","End":"07:35.555","Text":"Then we can find that the explosion actually took place at t_3 tag,"},{"Start":"07:35.555 ","End":"07:42.380","Text":"which is equal to the time at which we discovered the explosion minus the time"},{"Start":"07:42.380 ","End":"07:45.530","Text":"taken for the information to travel from"},{"Start":"07:45.530 ","End":"07:50.215","Text":"this location until our S tag frame of reference."},{"Start":"07:50.215 ","End":"07:55.520","Text":"Then we can say that the position that the explosion was at"},{"Start":"07:55.520 ","End":"08:03.150","Text":"relative to our S tag frame of reference is simply going to be our initial position."},{"Start":"08:03.150 ","End":"08:06.545","Text":"That\u0027s going to be x_Alpha tag,"},{"Start":"08:06.545 ","End":"08:10.985","Text":"our initial position when t=0,"},{"Start":"08:10.985 ","End":"08:17.725","Text":"let\u0027s actually write that when t=0 and then minus our velocity."},{"Start":"08:17.725 ","End":"08:25.849","Text":"That is going to be our 0.9c multiplied by the time that this happens,"},{"Start":"08:25.849 ","End":"08:28.060","Text":"which is at t_3 tag."},{"Start":"08:28.060 ","End":"08:33.965","Text":"Then in order to find when the explosion occurred in our S frame of reference,"},{"Start":"08:33.965 ","End":"08:36.530","Text":"we simply use our Lorentz transform."},{"Start":"08:36.530 ","End":"08:41.200","Text":"Then we can say that our t_3 is simply equal to our Gamma naught,"},{"Start":"08:41.200 ","End":"08:42.920","Text":"which we defined over here,"},{"Start":"08:42.920 ","End":"08:50.105","Text":"multiplied by t_3 tag plus v_0,"},{"Start":"08:50.105 ","End":"09:00.675","Text":"which as we know is simply 0.9c divided by c^2 multiplied by our x tag_3."},{"Start":"09:00.675 ","End":"09:08.865","Text":"Now let\u0027s scroll down and let\u0027s plug in our values. Let\u0027s see."},{"Start":"09:08.865 ","End":"09:15.995","Text":"Our x tag_3 is simply going to be equal to our beginning distance,"},{"Start":"09:15.995 ","End":"09:17.300","Text":"right at the beginning,"},{"Start":"09:17.300 ","End":"09:21.365","Text":"the distance between the rocket and the star,"},{"Start":"09:21.365 ","End":"09:23.015","Text":"which we\u0027re given in the question,"},{"Start":"09:23.015 ","End":"09:26.945","Text":"is 500 million meters away."},{"Start":"09:26.945 ","End":"09:33.225","Text":"That is going to be 5 times 10^8 meters."},{"Start":"09:33.225 ","End":"09:37.790","Text":"Then we\u0027re going to have plus, because here,"},{"Start":"09:37.790 ","End":"09:44.795","Text":"if our S frame of reference is moving relative to our S tag at negative 0.9c,"},{"Start":"09:44.795 ","End":"09:50.495","Text":"that means that our S\u0027 is moving relative to our S at 0.9c,"},{"Start":"09:50.495 ","End":"09:53.175","Text":"that\u0027s also what we\u0027re given in the question."},{"Start":"09:53.175 ","End":"10:00.410","Text":"So plus 0.9c multiplied by t_3 tag."},{"Start":"10:00.410 ","End":"10:02.825","Text":"Now we know what our x tag_3 is equal to,"},{"Start":"10:02.825 ","End":"10:06.740","Text":"but we still need to find out what our t_3 tag is equal to."},{"Start":"10:06.740 ","End":"10:12.050","Text":"Our t_3 tag is simply going to be equal to our t_2 tag,"},{"Start":"10:12.050 ","End":"10:14.180","Text":"which as we know already from here,"},{"Start":"10:14.180 ","End":"10:19.580","Text":"our t_2 tag is equal to 1.3 minus our x_3"},{"Start":"10:19.580 ","End":"10:26.056","Text":"is also being used as our x_Alpha, so minus."},{"Start":"10:26.056 ","End":"10:29.920","Text":"Then we\u0027re going to have 5 times 10^8,"},{"Start":"10:29.920 ","End":"10:32.425","Text":"divided by the speed of light,"},{"Start":"10:32.425 ","End":"10:36.160","Text":"which 5 times 10^8 divided by the speed of light,"},{"Start":"10:36.160 ","End":"10:38.800","Text":"which is 3 times 10^8."},{"Start":"10:38.800 ","End":"10:43.315","Text":"We\u0027re going to have simply 5 divided by 3."},{"Start":"10:43.315 ","End":"10:49.330","Text":"Then minus 0.9c multiplied by"},{"Start":"10:49.330 ","End":"10:57.670","Text":"t_3 tag divided by c is going to be simply 0.9 t_3 tag."},{"Start":"10:57.670 ","End":"11:04.870","Text":"Note here that I\u0027ve said that my x_3 tag is"},{"Start":"11:04.870 ","End":"11:13.660","Text":"simply equal to my x_Alpha tag at time t_3 tag."},{"Start":"11:13.660 ","End":"11:20.199","Text":"Now you can see that our unknown is our t_3 tag on both sides of the equation."},{"Start":"11:20.199 ","End":"11:25.600","Text":"We can say that our 1.9 multiplied by"},{"Start":"11:25.600 ","End":"11:32.290","Text":"t_3 tag is equal to negative 1.1 divided by 3."},{"Start":"11:32.290 ","End":"11:36.250","Text":"Then when we plug everything in to the calculator,"},{"Start":"11:36.250 ","End":"11:42.110","Text":"we\u0027ll get that our t_3 tag is equal to negative 0.113."},{"Start":"11:42.360 ","End":"11:45.760","Text":"We found our t_3 tag,"},{"Start":"11:45.760 ","End":"11:50.040","Text":"and now we can finally find out what our x_3 tag is."},{"Start":"11:50.040 ","End":"11:55.585","Text":"Our x_3 tag, when we plug in what our t_3 tag is going to be equal to,"},{"Start":"11:55.585 ","End":"12:05.755","Text":"we get that it\u0027s equal to negative 4.479 times 10^8 meters."},{"Start":"12:05.755 ","End":"12:12.205","Text":"Now, all we have to do is do the Lorentz transform to find our t_3,"},{"Start":"12:12.205 ","End":"12:15.040","Text":"which was what we want to find in the end."},{"Start":"12:15.040 ","End":"12:19.435","Text":"So what is the timespan between observing the electromagnetic wave"},{"Start":"12:19.435 ","End":"12:23.950","Text":"or pulse and the actual explosion relative to Alpha and its moons?"},{"Start":"12:23.950 ","End":"12:26.515","Text":"So relative to our S frame of reference,"},{"Start":"12:26.515 ","End":"12:30.535","Text":"that is simply going to be plugging in our Gamma 0,"},{"Start":"12:30.535 ","End":"12:35.830","Text":"which is 2.294 multiplied by our t_3 tag,"},{"Start":"12:35.830 ","End":"12:39.085","Text":"which we got was this over here,"},{"Start":"12:39.085 ","End":"12:44.705","Text":"negative 0.113 plus 0.9c."},{"Start":"12:44.705 ","End":"12:48.645","Text":"This crosses out multiplied by our x_3 tag,"},{"Start":"12:48.645 ","End":"12:50.115","Text":"which we have over here."},{"Start":"12:50.115 ","End":"12:59.770","Text":"Then we get that our t_3 is equal to negative 3.5 to 5 seconds."},{"Start":"12:59.770 ","End":"13:05.905","Text":"Now, this is the time that the explosion took place relative to our S frame of reference."},{"Start":"13:05.905 ","End":"13:09.010","Text":"We\u0027re being asked what is the timespan between"},{"Start":"13:09.010 ","End":"13:12.235","Text":"observing the electromagnetic wave in the actual explosion,"},{"Start":"13:12.235 ","End":"13:14.440","Text":"we begin at t=0."},{"Start":"13:14.440 ","End":"13:18.610","Text":"This answer is remaining the same."},{"Start":"13:18.610 ","End":"13:21.310","Text":"This is also equal to our timespan."},{"Start":"13:21.310 ","End":"13:23.245","Text":"Now for the second question,"},{"Start":"13:23.245 ","End":"13:28.210","Text":"what is the significance of our answer to question number 1?"},{"Start":"13:28.210 ","End":"13:33.100","Text":"The significance is that here we have a negative sign."},{"Start":"13:33.100 ","End":"13:34.915","Text":"Now we\u0027re answering question number 2,"},{"Start":"13:34.915 ","End":"13:37.765","Text":"we have a negative answer for our timespan."},{"Start":"13:37.765 ","End":"13:39.085","Text":"What does that mean?"},{"Start":"13:39.085 ","End":"13:42.745","Text":"That means that our explosion was before"},{"Start":"13:42.745 ","End":"13:47.800","Text":"the pulse reached Alpha and before the pulse was even shot."},{"Start":"13:47.800 ","End":"13:53.035","Text":"The explosion happened well before these two things."},{"Start":"13:53.035 ","End":"13:54.910","Text":"Now on to question 3,"},{"Start":"13:54.910 ","End":"13:58.720","Text":"did the pulse cause the explosion or was it the other way round?"},{"Start":"13:58.720 ","End":"14:03.100","Text":"We already know that the pulse didn\u0027t cause the explosion because we know"},{"Start":"14:03.100 ","End":"14:07.525","Text":"from question number 2 that the explosion was prior to the pulse being shot."},{"Start":"14:07.525 ","End":"14:15.130","Text":"Now what we want to do is check if the pulse was shot due to the explosion."},{"Start":"14:15.130 ","End":"14:20.365","Text":"Maybe the aliens saw the explosion taking place and therefore shot the pulse."},{"Start":"14:20.365 ","End":"14:22.420","Text":"Let\u0027s take a look at that."},{"Start":"14:22.420 ","End":"14:29.020","Text":"We know from our lesson on causality that there is no way that this can be."},{"Start":"14:29.020 ","End":"14:33.400","Text":"The pulse wasn\u0027t sent due the explosion,"},{"Start":"14:33.400 ","End":"14:35.500","Text":"it wasn\u0027t caused by the explosion."},{"Start":"14:35.500 ","End":"14:39.610","Text":"That\u0027s because we can see in our S tag frame of reference that"},{"Start":"14:39.610 ","End":"14:46.195","Text":"the pulse occurs before the discovery of the explosion and before the explosion."},{"Start":"14:46.195 ","End":"14:48.745","Text":"When we learned our lesson on causality,"},{"Start":"14:48.745 ","End":"14:52.380","Text":"we saw that if one event is caused by another,"},{"Start":"14:52.380 ","End":"14:55.650","Text":"that means that the order in which the events occurred"},{"Start":"14:55.650 ","End":"15:00.230","Text":"can\u0027t be switched no matter which frame of reference we\u0027re looking in."},{"Start":"15:00.230 ","End":"15:03.805","Text":"That means that it cannot be that in one frame of reference,"},{"Start":"15:03.805 ","End":"15:06.835","Text":"the pulse was an effect of the explosion,"},{"Start":"15:06.835 ","End":"15:09.235","Text":"i.e in our S frame of reference."},{"Start":"15:09.235 ","End":"15:12.080","Text":"However, in our S tag frame of reference,"},{"Start":"15:12.080 ","End":"15:19.465","Text":"we suddenly see an opposite order as in first the pulse, then the explosion."},{"Start":"15:19.465 ","End":"15:23.424","Text":"That means that this cannot be the case."},{"Start":"15:23.424 ","End":"15:28.520","Text":"But let\u0027s just prove this by looking at our equations."},{"Start":"15:28.520 ","End":"15:34.830","Text":"Now what we\u0027re going to do is we\u0027re going to assume that there was this explosion and"},{"Start":"15:34.830 ","End":"15:41.635","Text":"then that the information of the explosion managed to reach our moon."},{"Start":"15:41.635 ","End":"15:46.465","Text":"We want to see if this time is greater or less than"},{"Start":"15:46.465 ","End":"15:52.345","Text":"the time at which the pulse was sent towards Alpha from the moon."},{"Start":"15:52.345 ","End":"15:59.230","Text":"The distance traveled by the light from the moment of the explosion until the release of"},{"Start":"15:59.230 ","End":"16:07.255","Text":"pulse is going to be the speed of light multiplied by the time it took,"},{"Start":"16:07.255 ","End":"16:11.560","Text":"the time span, which is simply going to be equal to"},{"Start":"16:11.560 ","End":"16:20.050","Text":"3 times 10^8 multiplied by 3.5 to 5,"},{"Start":"16:20.050 ","End":"16:23.725","Text":"to the absolute value of this over here."},{"Start":"16:23.725 ","End":"16:27.040","Text":"Then once we plug that into our calculator,"},{"Start":"16:27.040 ","End":"16:35.995","Text":"we\u0027re going to get 10.575 multiplied by 10^8 meters."},{"Start":"16:35.995 ","End":"16:43.210","Text":"Now, what do we want to do is we want to find the distance between Alpha and our moon."},{"Start":"16:43.210 ","End":"16:45.715","Text":"Let\u0027s call that distance simply x."},{"Start":"16:45.715 ","End":"16:48.235","Text":"This is the rest length,"},{"Start":"16:48.235 ","End":"16:52.450","Text":"because this length between our Alpha and its moon is"},{"Start":"16:52.450 ","End":"16:56.830","Text":"always the same relative to the S frame of reference,"},{"Start":"16:56.830 ","End":"16:58.285","Text":"which is its rest frame."},{"Start":"16:58.285 ","End":"17:01.330","Text":"That means that it\u0027s going to be equal too."},{"Start":"17:01.330 ","End":"17:02.905","Text":"Now in the question,"},{"Start":"17:02.905 ","End":"17:11.245","Text":"we\u0027re given that the distance between Alpha and its moon is 500 million meters."},{"Start":"17:11.245 ","End":"17:14.125","Text":"But we\u0027re being told that it\u0027s recorded in"},{"Start":"17:14.125 ","End":"17:19.000","Text":"the rockets system as in in the S tag frame of reference."},{"Start":"17:19.000 ","End":"17:21.310","Text":"In order to find the rest length,"},{"Start":"17:21.310 ","End":"17:25.735","Text":"in order to find the distance between these two in our S frame of reference,"},{"Start":"17:25.735 ","End":"17:29.530","Text":"we will say 500 million meters,"},{"Start":"17:29.530 ","End":"17:33.610","Text":"which is 5 times 10^8 multiplied by."},{"Start":"17:33.610 ","End":"17:36.790","Text":"In order to get this into our S frame of reference,"},{"Start":"17:36.790 ","End":"17:39.850","Text":"we\u0027re going to multiply this by Gamma naught,"},{"Start":"17:39.850 ","End":"17:42.625","Text":"which is this number over here."},{"Start":"17:42.625 ","End":"17:45.850","Text":"Then once we plug that into our calculator,"},{"Start":"17:45.850 ","End":"17:53.875","Text":"we\u0027ll get that the answer is equal to 11.47 times 10^8 meters."},{"Start":"17:53.875 ","End":"18:01.180","Text":"Now what we want to see is what is the size of this relative to this over here."},{"Start":"18:01.180 ","End":"18:10.884","Text":"We can see that it\u0027s bigger than 10.575 times 10^8 meters."},{"Start":"18:10.884 ","End":"18:12.520","Text":"What does this mean?"},{"Start":"18:12.520 ","End":"18:15.820","Text":"We can see that the distance traveled by"},{"Start":"18:15.820 ","End":"18:19.900","Text":"the light from the moment of explosion until the release of the pulse,"},{"Start":"18:19.900 ","End":"18:27.490","Text":"was this distance, which is less than the distance between Alpha and its moon."},{"Start":"18:27.490 ","End":"18:33.670","Text":"Which means that the information that an explosion took place didn\u0027t"},{"Start":"18:33.670 ","End":"18:40.450","Text":"have enough time in order to reach the aliens on the moon and they shot the pulse out."},{"Start":"18:40.450 ","End":"18:44.170","Text":"That means before the information reached them,"},{"Start":"18:44.170 ","End":"18:51.835","Text":"which means that the pulse wasn\u0027t shot as a response to the explosion."},{"Start":"18:51.835 ","End":"18:54.520","Text":"So the two events were independent,"},{"Start":"18:54.520 ","End":"18:58.405","Text":"which means that as an answer to our question 3,"},{"Start":"18:58.405 ","End":"19:01.780","Text":"did the pulse cause the explosion or was it the other way round?"},{"Start":"19:01.780 ","End":"19:05.920","Text":"It was neither. Both of the events were independent."},{"Start":"19:05.920 ","End":"19:09.440","Text":"Okay. That\u0027s the end of this lesson."}],"ID":9556},{"Watched":false,"Name":"Velocity Transforms","Duration":"6m 9s","ChapterTopicVideoID":9256,"CourseChapterTopicPlaylistID":9398,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.815","Text":"Hello. In this lesson,"},{"Start":"00:01.815 ","End":"00:06.810","Text":"we\u0027re going to be speaking about transforming the velocity equations."},{"Start":"00:06.810 ","End":"00:09.150","Text":"The first thing that we\u0027re going to look at is"},{"Start":"00:09.150 ","End":"00:15.180","Text":"our velocity definitions when we\u0027re working in some S frame of reference,"},{"Start":"00:15.180 ","End":"00:18.885","Text":"which as we know our v_x is going to be d_x by dt,"},{"Start":"00:18.885 ","End":"00:22.380","Text":"v_y is going to be dy by dt, and similarly,"},{"Start":"00:22.380 ","End":"00:26.805","Text":"if we\u0027re in some other frame of reference or a stack frame of reference."},{"Start":"00:26.805 ","End":"00:31.125","Text":"Our v_x tag will be equal to d_x tagged by dt tag,"},{"Start":"00:31.125 ","End":"00:33.765","Text":"and the same for y\u0027s and z\u0027s."},{"Start":"00:33.765 ","End":"00:37.110","Text":"Now what we\u0027re going to do is we want to work out"},{"Start":"00:37.110 ","End":"00:41.145","Text":"the transform for our S tag frame of references."},{"Start":"00:41.145 ","End":"00:45.450","Text":"We\u0027re going to try and find out what our dx tag is equal"},{"Start":"00:45.450 ","End":"00:49.755","Text":"to and what our dt tag is equal to. Let\u0027s begin."},{"Start":"00:49.755 ","End":"00:55.350","Text":"We\u0027re going to take our Lorentz transform equation for r_x tag,"},{"Start":"00:55.350 ","End":"01:00.495","Text":"which we know is equal to Gamma_o multiplied by x"},{"Start":"01:00.495 ","End":"01:06.255","Text":"minus v_o multiplied by t. Then,"},{"Start":"01:06.255 ","End":"01:07.755","Text":"if we want to find out,"},{"Start":"01:07.755 ","End":"01:11.560","Text":"therefore what our d_x tag is equal to."},{"Start":"01:11.560 ","End":"01:15.435","Text":"We can see that there are 2 variables in here."},{"Start":"01:15.435 ","End":"01:20.490","Text":"We have our x and we have t. We will get that this has Gamma_o and then we\u0027ll"},{"Start":"01:20.490 ","End":"01:28.305","Text":"have that this is equal to d_x minus v_o and then dt."},{"Start":"01:28.305 ","End":"01:32.640","Text":"Now the same for our equation for r_t tag,"},{"Start":"01:32.640 ","End":"01:37.350","Text":"which we know is equal to Gamma_o multiplied by t"},{"Start":"01:37.350 ","End":"01:43.110","Text":"minus v_ox divided by c^2."},{"Start":"01:43.110 ","End":"01:47.040","Text":"Then we work out our d _t tag again,"},{"Start":"01:47.040 ","End":"01:48.390","Text":"we have our 2 variables."},{"Start":"01:48.390 ","End":"01:50.205","Text":"We have our t and our x."},{"Start":"01:50.205 ","End":"01:57.465","Text":"This is going to be equal to Gamma_o dt minus v_o"},{"Start":"01:57.465 ","End":"02:07.305","Text":"multiplied by d_x divided by our c^2 and from our other Lorentz transformations,"},{"Start":"02:07.305 ","End":"02:13.890","Text":"we know that our y tag is equal to y and then our z tag is equal to z,"},{"Start":"02:13.890 ","End":"02:23.400","Text":"which means that our dy tag is equal to dy and our dz tag is equal to d set."},{"Start":"02:23.400 ","End":"02:27.570","Text":"Now what we\u0027re going to do is we\u0027re going to try and find out what"},{"Start":"02:27.570 ","End":"02:32.190","Text":"our v_x tag is equal to and as we know from here,"},{"Start":"02:32.190 ","End":"02:37.170","Text":"the definition, it\u0027s equal to dx tag divided by dt tag."},{"Start":"02:37.170 ","End":"02:39.665","Text":"What we\u0027re going to do is we\u0027re going to sub in"},{"Start":"02:39.665 ","End":"02:43.100","Text":"our equation here for dx tag and here for dt tag,"},{"Start":"02:43.100 ","End":"02:48.335","Text":"that\u0027s going to be equal to Gamma_o multiplied by dx"},{"Start":"02:48.335 ","End":"02:55.230","Text":"minus v_o dt divided by our dt tag,"},{"Start":"02:55.230 ","End":"02:59.550","Text":"which is Gamma_o multiplied by dt"},{"Start":"02:59.550 ","End":"03:05.700","Text":"minus v_o divided by c^2 multiplied by dx."},{"Start":"03:05.700 ","End":"03:09.150","Text":"Now, once we cancel out,"},{"Start":"03:09.150 ","End":"03:10.950","Text":"our like terms and rearrange,"},{"Start":"03:10.950 ","End":"03:14.460","Text":"everything will get that v_x"},{"Start":"03:14.460 ","End":"03:21.825","Text":"minus v_o divided by 1 minus v_o,"},{"Start":"03:21.825 ","End":"03:25.485","Text":"v_x divided by c^2,"},{"Start":"03:25.485 ","End":"03:28.815","Text":"that this is equal to v_x tag."},{"Start":"03:28.815 ","End":"03:32.820","Text":"Now let\u0027s do the same for our v_y tag."},{"Start":"03:32.820 ","End":"03:39.480","Text":"We know that our v_y tag is going to be equal to our d_y tag divided by our dt tag,"},{"Start":"03:39.480 ","End":"03:48.225","Text":"so our d_y tag is simply going to be equal to dy divided by dt tag,"},{"Start":"03:48.225 ","End":"03:53.235","Text":"which is again Gamma naught multiplied by dt"},{"Start":"03:53.235 ","End":"03:59.010","Text":"minus v_o divided by c^2 multiplied by d_x."},{"Start":"03:59.010 ","End":"04:01.200","Text":"Then once again,"},{"Start":"04:01.200 ","End":"04:03.300","Text":"when we rearrange everything,"},{"Start":"04:03.300 ","End":"04:08.895","Text":"will get that our v_y tag is equal to v_y divided by"},{"Start":"04:08.895 ","End":"04:17.100","Text":"Gamma_o 1 minus v_o v_x divided by c^2,"},{"Start":"04:17.100 ","End":"04:22.560","Text":"and similarly, for our v_z tag,"},{"Start":"04:22.560 ","End":"04:24.210","Text":"I\u0027m just going to write the final answer,"},{"Start":"04:24.210 ","End":"04:27.990","Text":"but we do it the exact same way so I get that it\u0027s equal to v_z"},{"Start":"04:27.990 ","End":"04:34.485","Text":"divided by Gamma_o 1 minus v_o,"},{"Start":"04:34.485 ","End":"04:43.100","Text":"v_x divided by c^2 and just to clarify how he got our v_x over here."},{"Start":"04:43.100 ","End":"04:46.565","Text":"I took out from all of the denominators,"},{"Start":"04:46.565 ","End":"04:52.130","Text":"the common factor of dt and then we get d_x divided by dt,"},{"Start":"04:52.130 ","End":"04:57.910","Text":"which as we know as v_x and here we\u0027ll get again d_y divided by dt, which is v_y,"},{"Start":"04:57.910 ","End":"05:01.145","Text":"and then similarly over here,"},{"Start":"05:01.145 ","End":"05:03.095","Text":"there was dt,"},{"Start":"05:03.095 ","End":"05:06.440","Text":"but we didn\u0027t write this in and then we get a v_z."},{"Start":"05:06.440 ","End":"05:10.085","Text":"Also when we take out our common dt over here,"},{"Start":"05:10.085 ","End":"05:13.205","Text":"we get our dx divided by dt over here,"},{"Start":"05:13.205 ","End":"05:15.740","Text":"which gives us our v_x,"},{"Start":"05:15.740 ","End":"05:18.350","Text":"and here again divided by our dt,"},{"Start":"05:18.350 ","End":"05:21.190","Text":"which gives us our v_x and similarly there."},{"Start":"05:21.190 ","End":"05:26.540","Text":"Here are our equations for the Lorentz transforms for velocity."},{"Start":"05:26.540 ","End":"05:28.580","Text":"We have our equation for v_x tag,"},{"Start":"05:28.580 ","End":"05:30.185","Text":"v_y tag, and v_z tag,"},{"Start":"05:30.185 ","End":"05:34.910","Text":"which will be the velocities in some S tag frame of reference,"},{"Start":"05:34.910 ","End":"05:41.600","Text":"which is moving at some velocity v_o relative to the S frame of reference."},{"Start":"05:41.600 ","End":"05:45.840","Text":"Now, this v_x over here is the velocity"},{"Start":"05:45.840 ","End":"05:50.180","Text":"of some third body which is moving at velocity v_x,"},{"Start":"05:50.180 ","End":"05:54.680","Text":"which both the observer in the S frame of reference and the observer in"},{"Start":"05:54.680 ","End":"06:00.465","Text":"the S tag frame of reference can see this object moving."},{"Start":"06:00.465 ","End":"06:02.660","Text":"That\u0027s the end of this lesson."},{"Start":"06:02.660 ","End":"06:05.405","Text":"Don\u0027t forget to write down these equations"},{"Start":"06:05.405 ","End":"06:09.360","Text":"in your equation sheet as they\u0027re very important."}],"ID":9557},{"Watched":false,"Name":"Aberration","Duration":"4m 44s","ChapterTopicVideoID":9257,"CourseChapterTopicPlaylistID":9398,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.280","Text":"Hello. In this lesson we\u0027re going to be"},{"Start":"00:02.280 ","End":"00:06.180","Text":"discussing the change in the angle of the velocity vector."},{"Start":"00:06.180 ","End":"00:14.340","Text":"That means that if we have some velocity vector in our S frame of reference, some angle."},{"Start":"00:14.340 ","End":"00:20.805","Text":"We\u0027re going to see how that angle changes when we move to a different frame of reference."},{"Start":"00:20.805 ","End":"00:25.935","Text":"Let\u0027s imagine that we have a particle which moves in our S frame of reference"},{"Start":"00:25.935 ","End":"00:31.455","Text":"with a velocity v at an angle Theta relative to the x-axis."},{"Start":"00:31.455 ","End":"00:38.025","Text":"Then here we have that our v vector is equal to v cosine of Theta v sine of Theta, 0."},{"Start":"00:38.025 ","End":"00:40.665","Text":"We\u0027re being asked to find the direction of its speed"},{"Start":"00:40.665 ","End":"00:44.525","Text":"relative to our S tag frame of reference."},{"Start":"00:44.525 ","End":"00:47.765","Text":"Now, notice that what we\u0027re doing here is we\u0027re finding"},{"Start":"00:47.765 ","End":"00:50.975","Text":"the angle for our velocity vector."},{"Start":"00:50.975 ","End":"00:54.740","Text":"That means that we\u0027re going to get a different expression to what we"},{"Start":"00:54.740 ","End":"01:00.065","Text":"found when we found how the angle changes when we move between reference frames,"},{"Start":"01:00.065 ","End":"01:04.270","Text":"when using the angle between the position vector."},{"Start":"01:04.270 ","End":"01:07.085","Text":"Here, when using the velocity vector,"},{"Start":"01:07.085 ","End":"01:11.765","Text":"will get a different expression because we\u0027re using the Lorentz transform for"},{"Start":"01:11.765 ","End":"01:17.035","Text":"our velocity expressions rather than for our position expressions."},{"Start":"01:17.035 ","End":"01:18.575","Text":"Let\u0027s begin."},{"Start":"01:18.575 ","End":"01:24.890","Text":"We know that we\u0027ll have in our S tag frame of reference that tan of"},{"Start":"01:24.890 ","End":"01:32.285","Text":"Theta tag is going to be equal to our V_y tag divided by our V_x tag."},{"Start":"01:32.285 ","End":"01:34.235","Text":"Which is going to be equal to,"},{"Start":"01:34.235 ","End":"01:35.705","Text":"we know our expressions."},{"Start":"01:35.705 ","End":"01:37.040","Text":"For our V_y tag,"},{"Start":"01:37.040 ","End":"01:45.765","Text":"the expression is V_y divided by Gamma naught multiplied by 1 minus V_0,"},{"Start":"01:45.765 ","End":"01:49.365","Text":"V_ x divided by C^2."},{"Start":"01:49.365 ","End":"01:52.200","Text":"Then divide it by our V_x tag,"},{"Start":"01:52.200 ","End":"01:56.340","Text":"which is equal to V_x minus V_0"},{"Start":"01:56.340 ","End":"02:02.445","Text":"divided by 1 minus V_0,"},{"Start":"02:02.445 ","End":"02:05.710","Text":"V_x divided by C^2."},{"Start":"02:05.710 ","End":"02:11.465","Text":"Then we\u0027ll get that our answer is equal to once we simplify"},{"Start":"02:11.465 ","End":"02:17.370","Text":"everything and substitute in where we see V_y we\u0027ll substitute v sine of Theta,"},{"Start":"02:17.370 ","End":"02:18.970","Text":"and where we see V_x,"},{"Start":"02:18.970 ","End":"02:22.280","Text":"we\u0027ll substitute in v cosine of Theta."},{"Start":"02:22.280 ","End":"02:32.045","Text":"We\u0027ll therefore get that our tan of Theta tag is going to be equal to v sine"},{"Start":"02:32.045 ","End":"02:37.890","Text":"of Theta divided by v cosine of"},{"Start":"02:37.890 ","End":"02:45.190","Text":"Theta minus V_0 multiplied by Gamma naught."},{"Start":"02:45.200 ","End":"02:50.765","Text":"What we have over here is an equation describing how"},{"Start":"02:50.765 ","End":"02:55.510","Text":"the angle will change when we move from our labs reference frame,"},{"Start":"02:55.510 ","End":"02:57.105","Text":"our S reference frame,"},{"Start":"02:57.105 ","End":"03:01.385","Text":"to some moving reference frame S tag."},{"Start":"03:01.385 ","End":"03:08.710","Text":"Which is moving with a velocity V_0 relative to our labs frame of reference."},{"Start":"03:08.710 ","End":"03:14.045","Text":"This is specifically for the angle changing when we\u0027re dealing with a velocity vector."},{"Start":"03:14.045 ","End":"03:15.455","Text":"Just as a little note,"},{"Start":"03:15.455 ","End":"03:18.410","Text":"when we\u0027re moving between reference frames,"},{"Start":"03:18.410 ","End":"03:21.160","Text":"from our labs reference frame until the S tag frame,"},{"Start":"03:21.160 ","End":"03:26.810","Text":"and when we\u0027re using our position vector we\u0027ll get that our tan of Theta tag for"},{"Start":"03:26.810 ","End":"03:33.225","Text":"our position vector will equal to Gamma naught multiplied by tan of Theta."},{"Start":"03:33.225 ","End":"03:37.400","Text":"We can see that we\u0027re getting a very different expression over here."},{"Start":"03:37.400 ","End":"03:40.325","Text":"Let\u0027s deal with a specific case."},{"Start":"03:40.325 ","End":"03:42.620","Text":"When we\u0027re dealing with photons,"},{"Start":"03:42.620 ","End":"03:46.565","Text":"we know that photons travel at the speed of light."},{"Start":"03:46.565 ","End":"03:50.990","Text":"That means that the V is equal to C. What would"},{"Start":"03:50.990 ","End":"03:55.920","Text":"happen when we substitute that into this equation, we\u0027ll get, therefore,"},{"Start":"03:55.920 ","End":"04:04.190","Text":"that our tan of Theta tag will simply be equal to sine of Theta divided by"},{"Start":"04:04.190 ","End":"04:09.225","Text":"Gamma naught multiplied by cosine Theta"},{"Start":"04:09.225 ","End":"04:14.615","Text":"minus V_0 divided by C. Because everywhere we see a V,"},{"Start":"04:14.615 ","End":"04:20.655","Text":"we substitute in our C. Then once we take out our common factors we\u0027ll get this."},{"Start":"04:20.655 ","End":"04:23.300","Text":"This specific situation when we\u0027re"},{"Start":"04:23.300 ","End":"04:26.150","Text":"dealing with a photon which is traveling at the speed of light."},{"Start":"04:26.150 ","End":"04:29.095","Text":"This case over here is called aberration."},{"Start":"04:29.095 ","End":"04:32.210","Text":"Aberration refers only to the case when"},{"Start":"04:32.210 ","End":"04:35.210","Text":"we have a photon and how it is viewed in 1 frame of"},{"Start":"04:35.210 ","End":"04:38.510","Text":"reference and in another frame of reference with"},{"Start":"04:38.510 ","End":"04:41.900","Text":"regards to the direction in which it is traveling."},{"Start":"04:41.900 ","End":"04:44.340","Text":"That\u0027s the end of this lesson."}],"ID":9558},{"Watched":false,"Name":"Exercise 4","Duration":"5m 37s","ChapterTopicVideoID":9277,"CourseChapterTopicPlaylistID":9398,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.950","Text":"Hello. In this question,"},{"Start":"00:01.950 ","End":"00:05.760","Text":"we have to cars which travel perpendicularly to one another."},{"Start":"00:05.760 ","End":"00:09.540","Text":"The velocity of the first car is 0.6c,"},{"Start":"00:09.540 ","End":"00:13.260","Text":"traveling in the positive x-direction relative to"},{"Start":"00:13.260 ","End":"00:18.465","Text":"our S tag frame of reference and the velocity of the second car is 0.9 c,"},{"Start":"00:18.465 ","End":"00:24.224","Text":"traveling in the negative y-direction relative to the S frame of reference."},{"Start":"00:24.224 ","End":"00:28.730","Text":"In our question, we\u0027re being told to find the relative velocity between the 2."},{"Start":"00:28.730 ","End":"00:36.755","Text":"The way we\u0027re going to do this is we\u0027re going to define the system of the first car to be"},{"Start":"00:36.755 ","End":"00:40.565","Text":"our reference frame which is moving and then we\u0027ll find"},{"Start":"00:40.565 ","End":"00:47.390","Text":"the relative velocity of Car number 2 relative to this moving reference frame."},{"Start":"00:47.390 ","End":"00:52.805","Text":"Here we can see the equations for our v_x tag and v_y tag."},{"Start":"00:52.805 ","End":"00:56.290","Text":"The Lorentz transform is for the velocity."},{"Start":"00:56.290 ","End":"01:00.830","Text":"If we define our S-tag frame of reference to be"},{"Start":"01:00.830 ","End":"01:06.050","Text":"the frame of reference of this Car number 1 over here, which is moving,"},{"Start":"01:06.050 ","End":"01:08.525","Text":"then that means that our v_0,"},{"Start":"01:08.525 ","End":"01:14.135","Text":"the velocity of our S tag frame of reference relative to our S frame of reference,"},{"Start":"01:14.135 ","End":"01:17.525","Text":"is going to be equal to the velocity of the car,"},{"Start":"01:17.525 ","End":"01:23.490","Text":"which is v_1, which is equal to 0.6c."},{"Start":"01:23.490 ","End":"01:28.080","Text":"A little reminder of what our Gamma_0 is equal to."},{"Start":"01:28.080 ","End":"01:32.885","Text":"We have Gamma_0 is equal to 1 divided by"},{"Start":"01:32.885 ","End":"01:40.630","Text":"the square root of 1 minus v_0 divided by c^2."},{"Start":"01:40.630 ","End":"01:46.760","Text":"That means therefore that our Gamma_0 over here is going to be"},{"Start":"01:46.760 ","End":"01:52.820","Text":"equal to 1 divided by the square root of 1 minus, now what\u0027s our v_0?"},{"Start":"01:52.820 ","End":"02:02.135","Text":"It\u0027s 0.6c divided by c. It\u0027s just going to be 0.6^2 and the square root of that."},{"Start":"02:02.135 ","End":"02:06.605","Text":"Then once we plug all of this into our calculator,"},{"Start":"02:06.605 ","End":"02:11.495","Text":"we\u0027ll get that that is equal to 1.25."},{"Start":"02:11.495 ","End":"02:15.905","Text":"Now let\u0027s write down the velocities of"},{"Start":"02:15.905 ","End":"02:20.825","Text":"our Car number 2 relative to our S frame of reference."},{"Start":"02:20.825 ","End":"02:27.515","Text":"We\u0027ll have that the velocity of car 2 in the x-direction is of course equal to 0."},{"Start":"02:27.515 ","End":"02:29.630","Text":"It has no x component."},{"Start":"02:29.630 ","End":"02:33.680","Text":"The velocity of Car number 2 in the y-direction,"},{"Start":"02:33.680 ","End":"02:36.050","Text":"because it\u0027s traveling in the negative y-direction,"},{"Start":"02:36.050 ","End":"02:39.185","Text":"is going to be negative 0.9c."},{"Start":"02:39.185 ","End":"02:43.819","Text":"Now let\u0027s work out the components of v_2x"},{"Start":"02:43.819 ","End":"02:49.270","Text":"and our v_2y relative to our S tag frame of reference."},{"Start":"02:49.270 ","End":"02:55.065","Text":"We\u0027ll have that our v_2x tag is going to be equal to,"},{"Start":"02:55.065 ","End":"02:57.060","Text":"so let\u0027s look at this equation."},{"Start":"02:57.060 ","End":"03:06.270","Text":"Our v_2x is equal to 0 minus our v_0. What is our v_0?"},{"Start":"03:06.270 ","End":"03:13.140","Text":"It\u0027s 0.6c divided by 1 minus v_0,"},{"Start":"03:13.140 ","End":"03:15.045","Text":"which is 0.6c,"},{"Start":"03:15.045 ","End":"03:18.810","Text":"multiplied by our v_2x, which is 0."},{"Start":"03:18.810 ","End":"03:20.825","Text":"That means that all of this,"},{"Start":"03:20.825 ","End":"03:24.230","Text":"this whole little section becomes 0 and then we\u0027ll get that"},{"Start":"03:24.230 ","End":"03:30.140","Text":"our v_2x tag is equal to negative 0.6c."},{"Start":"03:30.140 ","End":"03:33.755","Text":"Now let\u0027s do the same for our v_2y tag."},{"Start":"03:33.755 ","End":"03:39.125","Text":"We\u0027re finding this component relative to a moving frame of reference."},{"Start":"03:39.125 ","End":"03:47.960","Text":"We have our v_y, which as we know is negative 0.9c divided by our Gamma_0,"},{"Start":"03:47.960 ","End":"03:54.130","Text":"where we discovered that our Gamma_0 was 1.25 multiplied"},{"Start":"03:54.130 ","End":"04:00.570","Text":"by 1 minus our v_0 multiplied by our v_x,"},{"Start":"04:00.570 ","End":"04:02.790","Text":"which is again equal to 0,"},{"Start":"04:02.790 ","End":"04:05.265","Text":"so 1 minus 0."},{"Start":"04:05.265 ","End":"04:14.750","Text":"Then we\u0027ll get this expression is equal to negative 0.72c."},{"Start":"04:14.750 ","End":"04:17.820","Text":"Now we have for each component,"},{"Start":"04:17.820 ","End":"04:20.050","Text":"our x and our y component."},{"Start":"04:20.050 ","End":"04:23.740","Text":"We know the relative velocity between Car number 1 and"},{"Start":"04:23.740 ","End":"04:29.100","Text":"Car number 2 relative to Car number 1\u0027s frame of reference."},{"Start":"04:29.100 ","End":"04:34.550","Text":"Now, if we want to find the size of this relativistic velocity,"},{"Start":"04:34.550 ","End":"04:37.430","Text":"so what we\u0027re going to do is this."},{"Start":"04:37.430 ","End":"04:41.510","Text":"We say that our v_2 tag, the size of it,"},{"Start":"04:41.510 ","End":"04:46.700","Text":"simply equal to the square root of (v_2x"},{"Start":"04:46.700 ","End":"04:54.900","Text":"tag)^2 plus (v_2y tag)^2."},{"Start":"04:54.900 ","End":"04:59.330","Text":"That is going to be equal to the square root"},{"Start":"04:59.330 ","End":"05:08.430","Text":"of (0.6c)^2 plus (0.72c)^2."},{"Start":"05:08.430 ","End":"05:11.675","Text":"Then that is simply going to be equal to,"},{"Start":"05:11.675 ","End":"05:15.245","Text":"we can take out the cs from the square root sign,"},{"Start":"05:15.245 ","End":"05:25.490","Text":"we\u0027ll get that the answer is equal to 0.9372 multiplied by"},{"Start":"05:25.490 ","End":"05:30.030","Text":"c. Notice that this is smaller than c."},{"Start":"05:30.030 ","End":"05:35.765","Text":"The answer that we\u0027re going to get is always going to be smaller than the speed of light."},{"Start":"05:35.765 ","End":"05:38.310","Text":"That\u0027s the end of this lesson."}],"ID":9559},{"Watched":false,"Name":"Exercise 5","Duration":"7m 26s","ChapterTopicVideoID":9266,"CourseChapterTopicPlaylistID":9398,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.730","Text":"The physics exam began at 9:00 AM,"},{"Start":"00:03.730 ","End":"00:09.080","Text":"at which time the examiner went on a walk at a velocity of 0.8c."},{"Start":"00:09.080 ","End":"00:13.915","Text":"After an hour relative to her watch, the examiner\u0027s watch,"},{"Start":"00:13.915 ","End":"00:16.375","Text":"she sent a radio signal to the students"},{"Start":"00:16.375 ","End":"00:20.064","Text":"saying that time is up and to put down their pencils."},{"Start":"00:20.064 ","End":"00:24.950","Text":"The question is, how long was the exam according to the students?"},{"Start":"00:24.950 ","End":"00:30.310","Text":"Here, we can see that the exam is taking place in the lab\u0027s frame of reference,"},{"Start":"00:30.310 ","End":"00:32.080","Text":"our S frame of reference."},{"Start":"00:32.080 ","End":"00:35.635","Text":"Here\u0027s our examiner in our S\u0027 frame of reference,"},{"Start":"00:35.635 ","End":"00:40.055","Text":"which was walking away at a velocity of 0.8c."},{"Start":"00:40.055 ","End":"00:45.440","Text":"Then after an hour relative to our S\u0027 frame of reference,"},{"Start":"00:45.440 ","End":"00:49.525","Text":"she sends out a radio signal to tell them that the exam is over."},{"Start":"00:49.525 ","End":"00:51.920","Text":"Now, what we want to do is we want to see how long"},{"Start":"00:51.920 ","End":"00:54.770","Text":"the exam was relative to the students."},{"Start":"00:54.770 ","End":"00:59.425","Text":"The first thing that we\u0027re going to do is we\u0027re going to draw our table."},{"Start":"00:59.425 ","End":"01:04.655","Text":"As we remember, we have our events that took place,"},{"Start":"01:04.655 ","End":"01:10.205","Text":"then we have relative to our S frame of reference,"},{"Start":"01:10.205 ","End":"01:15.995","Text":"and we have the events relative to our S\u0027 frame of reference."},{"Start":"01:15.995 ","End":"01:22.205","Text":"Let\u0027s begin. The first event that we have is that the examiner,"},{"Start":"01:22.205 ","End":"01:25.080","Text":"so E goes on a walk."},{"Start":"01:25.250 ","End":"01:27.805","Text":"This is our first event,"},{"Start":"01:27.805 ","End":"01:33.130","Text":"and that means that both of our reference frames are beginning from the exact same point."},{"Start":"01:33.130 ","End":"01:36.480","Text":"The exam and the examiner are in the same place."},{"Start":"01:36.480 ","End":"01:40.750","Text":"She tells the students to begin the exam and begins walking."},{"Start":"01:40.750 ","End":"01:42.595","Text":"Right at the beginning,"},{"Start":"01:42.595 ","End":"01:48.705","Text":"we have that this happens in x_1=0 and t_1=0,"},{"Start":"01:48.705 ","End":"01:56.330","Text":"and similarly, x_1\u0027 is equal to 0 and t_1\u0027 is also equal to 0."},{"Start":"01:56.330 ","End":"02:02.780","Text":"The second event occurs when E sends the radio signal."},{"Start":"02:02.850 ","End":"02:07.895","Text":"Now, relative to the examiner in the S\u0027 frame of reference,"},{"Start":"02:07.895 ","End":"02:10.790","Text":"the position that she\u0027s in when she sends"},{"Start":"02:10.790 ","End":"02:16.595","Text":"the radio signal is still at the origin of our S\u0027 frame of reference,"},{"Start":"02:16.595 ","End":"02:19.445","Text":"because the S\u0027 frame of reference is her,"},{"Start":"02:19.445 ","End":"02:21.250","Text":"she\u0027s always at the origin."},{"Start":"02:21.250 ","End":"02:30.794","Text":"That happens, so x_2\u0027 is equal to 0 and t_2\u0027 when she sends this radio signal,"},{"Start":"02:30.794 ","End":"02:35.260","Text":"we\u0027re told that it\u0027s after 1 hour, so 1 hour."},{"Start":"02:36.320 ","End":"02:40.265","Text":"Now, what we\u0027re going to do is we\u0027re going to find out"},{"Start":"02:40.265 ","End":"02:43.481","Text":"what this means in our S frame of reference,"},{"Start":"02:43.481 ","End":"02:45.590","Text":"our lab\u0027s frame of reference."},{"Start":"02:45.590 ","End":"02:50.270","Text":"What we\u0027re going to do is we\u0027re going to use the Lorentz transform."},{"Start":"02:50.270 ","End":"02:54.720","Text":"We\u0027ll have, sorry, without a tag,"},{"Start":"02:54.720 ","End":"03:01.275","Text":"our x_2 is equal to Gamma 0 multiplied by our x_2\u0027,"},{"Start":"03:01.275 ","End":"03:04.770","Text":"which is 0, plus our v_0,"},{"Start":"03:04.770 ","End":"03:12.150","Text":"which is 0.8c multiplied by our t_2\u0027."},{"Start":"03:12.150 ","End":"03:14.490","Text":"Our t_2\u0027 is 1 hour,"},{"Start":"03:14.490 ","End":"03:20.500","Text":"which is equal to 3,600 seconds."},{"Start":"03:20.500 ","End":"03:23.660","Text":"Then we\u0027ll do the same transform,"},{"Start":"03:23.660 ","End":"03:25.025","Text":"but for our t_2,"},{"Start":"03:25.025 ","End":"03:31.545","Text":"so that is equal to Gamma 0 multiplied by our t_2\u0027,"},{"Start":"03:31.545 ","End":"03:32.835","Text":"which is 1 hour,"},{"Start":"03:32.835 ","End":"03:36.780","Text":"which is 3,600 seconds,"},{"Start":"03:36.780 ","End":"03:40.710","Text":"plus our x_2\u0027, which is 0."},{"Start":"03:40.710 ","End":"03:42.330","Text":"Those are our 2 events,"},{"Start":"03:42.330 ","End":"03:46.820","Text":"and our third event is when the radio signal that was"},{"Start":"03:46.820 ","End":"03:51.405","Text":"sent by the examiner reaches our S frame of reference,"},{"Start":"03:51.405 ","End":"03:53.385","Text":"so it reaches our exam."},{"Start":"03:53.385 ","End":"03:56.990","Text":"As we know, when the signal is sent out originally,"},{"Start":"03:56.990 ","End":"04:01.970","Text":"it takes a few moments for it to travel to its destination."},{"Start":"04:01.970 ","End":"04:05.990","Text":"What we\u0027re doing here is we\u0027re finding when that happens."},{"Start":"04:05.990 ","End":"04:11.390","Text":"The position that it reaches relative to the S frame of"},{"Start":"04:11.390 ","End":"04:13.623","Text":"reference is going to be equal to 0"},{"Start":"04:13.623 ","End":"04:17.695","Text":"because it reaches our S frame of reference where the exam is,"},{"Start":"04:17.695 ","End":"04:19.575","Text":"so relative to the S frame of reference,"},{"Start":"04:19.575 ","End":"04:20.900","Text":"it\u0027s at the origin."},{"Start":"04:20.900 ","End":"04:25.790","Text":"But what interests us right now is the time, our t_3."},{"Start":"04:25.790 ","End":"04:29.300","Text":"That\u0027s going to be equal to the time at which it"},{"Start":"04:29.300 ","End":"04:32.690","Text":"is sent relative to our S frame of reference,"},{"Start":"04:32.690 ","End":"04:34.535","Text":"so that\u0027s t_2,"},{"Start":"04:34.535 ","End":"04:39.740","Text":"plus the time it takes for it to"},{"Start":"04:39.740 ","End":"04:46.925","Text":"reach our S frame of reference from the position that it was sent from."},{"Start":"04:46.925 ","End":"04:50.345","Text":"It was sent from, relative to the S frame of reference,"},{"Start":"04:50.345 ","End":"04:52.145","Text":"from position x_2,"},{"Start":"04:52.145 ","End":"04:54.380","Text":"and it\u0027s traveling at the speed of light,"},{"Start":"04:54.380 ","End":"04:58.520","Text":"so divided by c. This will be the time that it takes to travel from"},{"Start":"04:58.520 ","End":"05:03.725","Text":"x_2 to our origin, to point x_3."},{"Start":"05:03.725 ","End":"05:05.690","Text":"Once we solve this,"},{"Start":"05:05.690 ","End":"05:13.285","Text":"we\u0027ll get that this is equal to Gamma 0 multiplied by 3,600,"},{"Start":"05:13.285 ","End":"05:19.930","Text":"multiplied by 1 plus 0.8."},{"Start":"05:21.320 ","End":"05:23.805","Text":"Now, we\u0027ve filled out our table,"},{"Start":"05:23.805 ","End":"05:25.065","Text":"so I\u0027m moving up here."},{"Start":"05:25.065 ","End":"05:30.235","Text":"What we want to do is we want to see how long the exam was according to the students."},{"Start":"05:30.235 ","End":"05:33.745","Text":"That means that we want to find our Delta t,"},{"Start":"05:33.745 ","End":"05:37.315","Text":"which is the difference in time between when the radio signal"},{"Start":"05:37.315 ","End":"05:40.960","Text":"reaches the exam room to stop writing the exam,"},{"Start":"05:40.960 ","End":"05:43.090","Text":"and from the beginning of the exam,"},{"Start":"05:43.090 ","End":"05:44.650","Text":"which was at 9:00 AM."},{"Start":"05:44.650 ","End":"05:52.120","Text":"We\u0027re going to do Delta t is equal to t_3 minus t_1, because at t_1,"},{"Start":"05:52.120 ","End":"05:53.350","Text":"the examiner goes on a walk,"},{"Start":"05:53.350 ","End":"05:56.185","Text":"which is also when the exam begins and"},{"Start":"05:56.185 ","End":"06:00.620","Text":"the radio signal reaches the exam room signaling the end of the exam."},{"Start":"06:00.620 ","End":"06:03.635","Text":"Once we work this out,"},{"Start":"06:03.635 ","End":"06:08.060","Text":"we\u0027ll get that this is equal to Gamma 0 multiplied by"},{"Start":"06:08.060 ","End":"06:15.465","Text":"1.8 multiplied by 3,600 seconds."},{"Start":"06:15.465 ","End":"06:19.655","Text":"Now, what we want to do, is we want to know what our Gamma 0 is equal to."},{"Start":"06:19.655 ","End":"06:25.100","Text":"As we know, that\u0027s going to be equal to 1 divided by the square root of"},{"Start":"06:25.100 ","End":"06:31.470","Text":"1 minus 0.8c divided by c^2,"},{"Start":"06:32.300 ","End":"06:35.480","Text":"and that is going to be equal to,"},{"Start":"06:35.480 ","End":"06:38.720","Text":"once we plug everything into our calculators,"},{"Start":"06:38.720 ","End":"06:42.510","Text":"it\u0027s going to be equal to 5.27."},{"Start":"06:43.180 ","End":"06:48.245","Text":"Then, once we plug in our Gamma 0 into here,"},{"Start":"06:48.245 ","End":"06:53.150","Text":"and we multiply it by 1.8 and then again by 3,600,"},{"Start":"06:53.150 ","End":"06:59.050","Text":"we\u0027ll get therefore that our Delta t is equal to"},{"Start":"06:59.050 ","End":"07:07.945","Text":"3.415 times 10^4 seconds."},{"Start":"07:07.945 ","End":"07:11.090","Text":"This is our answer to the question."},{"Start":"07:11.090 ","End":"07:12.620","Text":"Relative to the students,"},{"Start":"07:12.620 ","End":"07:18.545","Text":"the exam was 3.415 times 10^4 seconds long,"},{"Start":"07:18.545 ","End":"07:24.045","Text":"and we can see that is significantly longer than 1 hour."},{"Start":"07:24.045 ","End":"07:27.070","Text":"Okay, that\u0027s the end of this lesson."}],"ID":9560},{"Watched":false,"Name":"Exercise 6","Duration":"14m 56s","ChapterTopicVideoID":9258,"CourseChapterTopicPlaylistID":9398,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.905","Text":"A rocket with a rest length of 200 meters travels"},{"Start":"00:04.905 ","End":"00:09.855","Text":"at 0.9c relative to an inertial reference frame S. A"},{"Start":"00:09.855 ","End":"00:15.390","Text":"small ball rolls along the rocket\u0027s length at a velocity of"},{"Start":"00:15.390 ","End":"00:23.055","Text":"u’=0.4c in the x direction as recorded by an observer inside the rocket."},{"Start":"00:23.055 ","End":"00:25.110","Text":"Now, question number 1 is,"},{"Start":"00:25.110 ","End":"00:32.950","Text":"what is the velocity of the ball relative to an observer in the S reference frame?"},{"Start":"00:32.950 ","End":"00:35.505","Text":"Let\u0027s begin by answering this question."},{"Start":"00:35.505 ","End":"00:40.850","Text":"Here, we have all of our relevant Lorentz transform equations."},{"Start":"00:40.850 ","End":"00:44.120","Text":"Let\u0027s begin with question number 1."},{"Start":"00:44.120 ","End":"00:50.715","Text":"We want the velocity of the ball relative to our S reference frame."},{"Start":"00:50.715 ","End":"00:54.240","Text":"Here we can see in this equation,"},{"Start":"00:54.240 ","End":"00:57.665","Text":"if we had the velocity in our S reference frame,"},{"Start":"00:57.665 ","End":"01:02.580","Text":"we could find the velocity in our S\u0027 or rocket\u0027s reference frame."},{"Start":"01:02.580 ","End":"01:06.140","Text":"We want the opposite transformation to this."},{"Start":"01:06.140 ","End":"01:11.015","Text":"In that case, we can say that V_x is equal to."},{"Start":"01:11.015 ","End":"01:13.175","Text":"Then here instead of v_x,"},{"Start":"01:13.175 ","End":"01:15.875","Text":"we substitute in v_x\u0027,"},{"Start":"01:15.875 ","End":"01:18.050","Text":"and then instead of v_o,"},{"Start":"01:18.050 ","End":"01:20.778","Text":"we substitute in negative v_o,"},{"Start":"01:20.778 ","End":"01:29.700","Text":"so negative negative v_o becomes positive v_o divided by 1 minus."},{"Start":"01:29.700 ","End":"01:31.680","Text":"Then again here negative v_o,"},{"Start":"01:31.680 ","End":"01:34.440","Text":"so negative negative v_o,"},{"Start":"01:34.440 ","End":"01:43.105","Text":"so it\u0027s plus v_o multiplied by v_x tag divided by c^2."},{"Start":"01:43.105 ","End":"01:48.785","Text":"Now, what\u0027s left is to plug in our numbers."},{"Start":"01:48.785 ","End":"01:50.525","Text":"We know what our v_x\u0027 is."},{"Start":"01:50.525 ","End":"01:51.935","Text":"It\u0027s our u\u0027 over here."},{"Start":"01:51.935 ","End":"01:54.350","Text":"Now, notice I corrected this,"},{"Start":"01:54.350 ","End":"01:58.810","Text":"it\u0027s 0.04c and before it was 0.4c."},{"Start":"01:58.810 ","End":"02:01.065","Text":"Just note that difference."},{"Start":"02:01.065 ","End":"02:07.410","Text":"Our v_x\u0027 is 0.04c plus our v_o"},{"Start":"02:07.410 ","End":"02:15.120","Text":"is 0.9c divided by 1 plus our v_o,"},{"Start":"02:15.120 ","End":"02:21.555","Text":"which is our 0.9c multiplied by our v_x\u0027,"},{"Start":"02:21.555 ","End":"02:25.770","Text":"which is 0.04c,"},{"Start":"02:25.770 ","End":"02:29.565","Text":"and then divided by c^2."},{"Start":"02:29.565 ","End":"02:33.165","Text":"We can cross out the c^2 with this c and this c,"},{"Start":"02:33.165 ","End":"02:38.420","Text":"and then we can simply work this out and we\u0027ll get"},{"Start":"02:38.420 ","End":"02:46.020","Text":"0.94c on top divided by 1.036,"},{"Start":"02:46.020 ","End":"02:52.560","Text":"which after plugging this into a calculator, we\u0027ll get 0.907c."},{"Start":"02:57.200 ","End":"03:00.660","Text":"This is our answer to question 1,"},{"Start":"03:00.660 ","End":"03:05.990","Text":"and we can see that it\u0027s slightly faster than the velocity of the rocket,"},{"Start":"03:05.990 ","End":"03:09.890","Text":"which makes sense because the ball is moving with the velocity of the rocket"},{"Start":"03:09.890 ","End":"03:15.845","Text":"plus its own velocity when it\u0027s rolling inside the rocket."},{"Start":"03:15.845 ","End":"03:20.095","Text":"Let\u0027s move on to question number 2."},{"Start":"03:20.095 ","End":"03:22.685","Text":"Question number 2 is,"},{"Start":"03:22.685 ","End":"03:25.880","Text":"how long will it take the ball to travel from one end of"},{"Start":"03:25.880 ","End":"03:30.395","Text":"the rocket to the other relative to S?"},{"Start":"03:30.395 ","End":"03:36.280","Text":"Now, we\u0027re going to work this out by using our table."},{"Start":"03:36.590 ","End":"03:41.990","Text":"The first event that we have is when our ball was"},{"Start":"03:41.990 ","End":"03:48.205","Text":"released and started rolling from one end of the rocket."},{"Start":"03:48.205 ","End":"03:51.625","Text":"Now, this event will say,"},{"Start":"03:51.625 ","End":"03:56.380","Text":"relative to our S frame of reference happens at"},{"Start":"03:56.380 ","End":"04:03.735","Text":"the position x_0=0 and at time t_0=0."},{"Start":"04:03.735 ","End":"04:05.730","Text":"This is right at the beginning,"},{"Start":"04:05.730 ","End":"04:09.250","Text":"and we\u0027re saying that this is the first thing that happens."},{"Start":"04:09.250 ","End":"04:12.130","Text":"Then in our S\u0027 frame of reference,"},{"Start":"04:12.130 ","End":"04:17.350","Text":"the origin of the S\u0027 frame of reference is right now at the starting points,"},{"Start":"04:17.350 ","End":"04:20.885","Text":"the same as the origin in our S frame of reference."},{"Start":"04:20.885 ","End":"04:29.850","Text":"We can say that our x_0=0 and that our t_0\u0027=0."},{"Start":"04:29.850 ","End":"04:31.125","Text":"Right at the beginning,"},{"Start":"04:31.125 ","End":"04:33.460","Text":"the ball begins rolling."},{"Start":"04:35.020 ","End":"04:41.210","Text":"The second event is when the ball reaches the other end of the rocket."},{"Start":"04:41.210 ","End":"04:42.890","Text":"Our second question is,"},{"Start":"04:42.890 ","End":"04:45.500","Text":"how long will it take the ball to travel from one end of"},{"Start":"04:45.500 ","End":"04:49.430","Text":"the rocket to the other relative to S?"},{"Start":"04:49.430 ","End":"04:53.000","Text":"Soon we\u0027ll work out the relative to S. But let\u0027s see what\u0027s"},{"Start":"04:53.000 ","End":"04:57.655","Text":"happening in our S\u0027 frame of reference."},{"Start":"04:57.655 ","End":"05:00.455","Text":"In our S\u0027 frame of reference,"},{"Start":"05:00.455 ","End":"05:05.105","Text":"we don\u0027t have to use all of our Lorentz\u0027s transforms and theory of relativity."},{"Start":"05:05.105 ","End":"05:09.725","Text":"We can just use our classic mechanics for this."},{"Start":"05:09.725 ","End":"05:12.665","Text":"We can say that a t_1\u0027,"},{"Start":"05:12.665 ","End":"05:18.365","Text":"the ball has reached the other end of the rocket. What does that mean?"},{"Start":"05:18.365 ","End":"05:23.845","Text":"From velocity is equal to distance divided by time,"},{"Start":"05:23.845 ","End":"05:27.725","Text":"so we can say therefore that our time is equal to"},{"Start":"05:27.725 ","End":"05:33.225","Text":"the distance traveled Delta x divided by the velocity,"},{"Start":"05:33.225 ","End":"05:37.640","Text":"and we know that the ball is rolling at a velocity relative"},{"Start":"05:37.640 ","End":"05:42.690","Text":"to our S\u0027 frame of reference at 0.04c."},{"Start":"05:43.250 ","End":"05:47.119","Text":"We know that the distance that was traveled,"},{"Start":"05:47.119 ","End":"05:50.470","Text":"Delta x is from one end to the other,"},{"Start":"05:50.470 ","End":"05:54.195","Text":"is the rest length, which is 200 meters."},{"Start":"05:54.195 ","End":"06:00.610","Text":"We have 200 meters divided by 0.04c,"},{"Start":"06:01.610 ","End":"06:07.485","Text":"and c is 3 times 10^8,"},{"Start":"06:07.485 ","End":"06:13.770","Text":"and then here we\u0027ll have simply meters per second."},{"Start":"06:14.000 ","End":"06:17.929","Text":"Then once we plug in all of these numbers,"},{"Start":"06:17.929 ","End":"06:27.770","Text":"we will get the answer of 5 divided by 3 times 10^-8 seconds."},{"Start":"06:27.770 ","End":"06:29.630","Text":"That\u0027s the time that this happens."},{"Start":"06:29.630 ","End":"06:33.950","Text":"Then in order to have a slightly nicer number to look at,"},{"Start":"06:33.950 ","End":"06:37.690","Text":"we can change this into microseconds."},{"Start":"06:37.690 ","End":"06:42.095","Text":"We just multiply this by 10^6,"},{"Start":"06:42.095 ","End":"06:46.520","Text":"and then we write in milliseconds."},{"Start":"06:46.520 ","End":"06:55.620","Text":"This will give us 16.67 microseconds."},{"Start":"06:56.780 ","End":"07:02.250","Text":"That\u0027s the time, and now let\u0027s see our position x_1\u0027."},{"Start":"07:02.250 ","End":"07:05.960","Text":"The position that the ball is in at this time,"},{"Start":"07:05.960 ","End":"07:13.010","Text":"t_1\u0027 is simply at the other end of the rocket relative to the rocket\u0027s reference frame,"},{"Start":"07:13.010 ","End":"07:19.230","Text":"which simply means that it\u0027s at 200 meters."},{"Start":"07:22.530 ","End":"07:28.600","Text":"Now, we have all of our information in our S\u0027 frame of reference."},{"Start":"07:28.600 ","End":"07:32.050","Text":"But what we want is to have the information in our S frame of"},{"Start":"07:32.050 ","End":"07:36.340","Text":"reference and then to answer our question number 2."},{"Start":"07:36.340 ","End":"07:43.135","Text":"The time taken for the ball to reach the end relative to our S frame of reference."},{"Start":"07:43.135 ","End":"07:47.320","Text":"Now, we\u0027re going to use our opposite Lorentz transform."},{"Start":"07:47.320 ","End":"07:53.155","Text":"That means that our t_1 is simply going to be equal to"},{"Start":"07:53.155 ","End":"08:00.400","Text":"Gamma 0 multiplied by t_1\u0027 negative, negative v_0."},{"Start":"08:00.400 ","End":"08:10.510","Text":"That\u0027s going to be plus our v_0 multiplied by x_1\u0027 divided by c^2."},{"Start":"08:10.510 ","End":"08:13.765","Text":"Let\u0027s take a look on the side what that is."},{"Start":"08:13.765 ","End":"08:24.490","Text":"Our Gamma 0 is equal to 1 divided by the square root of 1 minus our v_0 divided by c^2."},{"Start":"08:24.490 ","End":"08:27.490","Text":"Our v_0 is 0.9,"},{"Start":"08:27.490 ","End":"08:30.890","Text":"c divided by c^2,"},{"Start":"08:31.260 ","End":"08:35.350","Text":"and the square root of all of that."},{"Start":"08:35.350 ","End":"08:39.745","Text":"Once we plug all of this into our calculator,"},{"Start":"08:39.745 ","End":"08:44.620","Text":"we\u0027ll get that our Gamma 0 is simply equal to 2.294."},{"Start":"08:44.620 ","End":"08:50.965","Text":"Now, let\u0027s work out what our v_0"},{"Start":"08:50.965 ","End":"08:58.780","Text":"x_1\u0027 divided by c^2 is equal to."},{"Start":"08:58.780 ","End":"09:03.340","Text":"Our v_0 is 0.9c,"},{"Start":"09:03.340 ","End":"09:13.255","Text":"and our x_1\u0027 is times 200 meters divided by our c^2."},{"Start":"09:13.255 ","End":"09:17.005","Text":"We can cross off this c and one of the c\u0027s over here."},{"Start":"09:17.005 ","End":"09:24.610","Text":"Then once we plug in that our c=3 times 10^8,"},{"Start":"09:24.610 ","End":"09:33.850","Text":"then we will get 60 times 10^-8 seconds."},{"Start":"09:33.850 ","End":"09:38.255","Text":"But because here we\u0027re using our microseconds,"},{"Start":"09:38.255 ","End":"09:40.570","Text":"so let\u0027s change this into microseconds,"},{"Start":"09:40.570 ","End":"09:46.720","Text":"which means multiplying this expression over here by 10^6."},{"Start":"09:46.720 ","End":"09:53.510","Text":"We\u0027ll get that this is equal to 0.6 microseconds."},{"Start":"09:53.510 ","End":"09:57.750","Text":"Now, what we can do is we can plug this in for t_1."},{"Start":"09:57.750 ","End":"10:00.750","Text":"We\u0027ll get that t_1 is equal to Gamma 0,"},{"Start":"10:00.750 ","End":"10:05.085","Text":"which is 2.294 multiplied by t_1\u0027,"},{"Start":"10:05.085 ","End":"10:09.445","Text":"which is 16.67 microseconds,"},{"Start":"10:09.445 ","End":"10:11.800","Text":"plus this expression over here,"},{"Start":"10:11.800 ","End":"10:14.665","Text":"which we saw was 0.6 microseconds."},{"Start":"10:14.665 ","End":"10:17.980","Text":"Once we plug all of that into our calculator,"},{"Start":"10:17.980 ","End":"10:26.480","Text":"we\u0027ll get the t_1 is equal to 39.62 microseconds."},{"Start":"10:26.790 ","End":"10:29.200","Text":"This is, of course,"},{"Start":"10:29.200 ","End":"10:33.680","Text":"our answer to question number 2."},{"Start":"10:33.810 ","End":"10:38.890","Text":"Now, what we\u0027re going to do is to answer question number 3."},{"Start":"10:38.890 ","End":"10:45.760","Text":"What distance has the ball traveled but relative to our S frame of reference?"},{"Start":"10:45.760 ","End":"10:49.015","Text":"Again, we\u0027re going to need to use this table."},{"Start":"10:49.015 ","End":"10:53.695","Text":"Let\u0027s just write that this table is being used to answer 2 and 3,"},{"Start":"10:53.695 ","End":"10:57.010","Text":"and that this answer over here was for question 2."},{"Start":"10:57.010 ","End":"11:01.000","Text":"Now, we\u0027re going to answer question number 3."},{"Start":"11:01.000 ","End":"11:04.735","Text":"In actual fact, which is question number 3 asking?"},{"Start":"11:04.735 ","End":"11:08.680","Text":"It\u0027s what is x_1 equal to?"},{"Start":"11:08.680 ","End":"11:13.555","Text":"This is our question number 3."},{"Start":"11:13.555 ","End":"11:16.375","Text":"Let\u0027s work this out on the side."},{"Start":"11:16.375 ","End":"11:20.005","Text":"Our x_1, we simply have to use"},{"Start":"11:20.005 ","End":"11:24.295","Text":"our opposite Lorentz transform to translate this over here."},{"Start":"11:24.295 ","End":"11:32.485","Text":"Our x_1 is going to be equal to our Gamma naught multiplied by instead of x_1,"},{"Start":"11:32.485 ","End":"11:34.150","Text":"it\u0027s going to be x_1\u0027,"},{"Start":"11:34.150 ","End":"11:36.520","Text":"negative, negative v_0,"},{"Start":"11:36.520 ","End":"11:42.620","Text":"so plus v_0 multiplied by t_1\u0027."},{"Start":"11:44.300 ","End":"11:48.465","Text":"We\u0027re doing the opposite Lorentz transform."},{"Start":"11:48.465 ","End":"11:52.495","Text":"Let\u0027s then see what this is equal to."},{"Start":"11:52.495 ","End":"12:00.990","Text":"We have our Gamma 0 is equal to 2.294."},{"Start":"12:00.990 ","End":"12:03.802","Text":"I robbed it out from earlier,"},{"Start":"12:03.802 ","End":"12:05.920","Text":"but this is what it was equal to."},{"Start":"12:05.920 ","End":"12:09.310","Text":"Then multiplied by our x_1\u0027,"},{"Start":"12:09.310 ","End":"12:12.880","Text":"which is 200 meters,"},{"Start":"12:12.880 ","End":"12:14.395","Text":"plus our v_0,"},{"Start":"12:14.395 ","End":"12:16.660","Text":"which is 0.9c,"},{"Start":"12:16.660 ","End":"12:19.430","Text":"multiplied by our t_1\u0027,"},{"Start":"12:19.430 ","End":"12:25.560","Text":"which is 16.67 microseconds."},{"Start":"12:25.560 ","End":"12:31.615","Text":"Remember this. Because our 0.9c,"},{"Start":"12:31.615 ","End":"12:33.700","Text":"because c is in meters per second,"},{"Start":"12:33.700 ","End":"12:37.255","Text":"we have to convert this back to seconds."},{"Start":"12:37.255 ","End":"12:43.030","Text":"All we have to do is multiply it by 10^-6 and then it\u0027s in seconds,"},{"Start":"12:43.030 ","End":"12:49.105","Text":"and then our units work out between our c in this and we can carry on our calculations."},{"Start":"12:49.105 ","End":"12:53.035","Text":"Once we plug in all of this into our calculator,"},{"Start":"12:53.035 ","End":"13:00.820","Text":"we\u0027ll get that this is equal to 10,780 meters."},{"Start":"13:00.820 ","End":"13:10.405","Text":"We can also say instead of in meters that this is equal to 10.78 kilometers."},{"Start":"13:10.405 ","End":"13:20.030","Text":"This is what our x_1 is equal to 10.78 kilometers."},{"Start":"13:21.360 ","End":"13:24.220","Text":"That\u0027s the end of the question."},{"Start":"13:24.220 ","End":"13:27.190","Text":"It was pretty straightforward, especially again,"},{"Start":"13:27.190 ","End":"13:30.700","Text":"when we see how using this table makes everything easier."},{"Start":"13:30.700 ","End":"13:34.090","Text":"The most important things to remember is that"},{"Start":"13:34.090 ","End":"13:37.615","Text":"these are the equations for the Lorentz transform."},{"Start":"13:37.615 ","End":"13:40.600","Text":"If you want to go from your S\u0027 frame of"},{"Start":"13:40.600 ","End":"13:44.545","Text":"reference and transform it into your S frame of reference,"},{"Start":"13:44.545 ","End":"13:46.885","Text":"then you have to do the opposite transform."},{"Start":"13:46.885 ","End":"13:51.520","Text":"Which means everywhere in the Lorentz transform where there\u0027s a tag,"},{"Start":"13:51.520 ","End":"13:56.125","Text":"you write it without a tag and everywhere where there\u0027s no tag,"},{"Start":"13:56.125 ","End":"13:58.210","Text":"you add in a tag."},{"Start":"13:58.210 ","End":"14:01.315","Text":"Then instead of having negative v_0,"},{"Start":"14:01.315 ","End":"14:02.950","Text":"you write a negative,"},{"Start":"14:02.950 ","End":"14:08.485","Text":"negative v_0 and then you get a positive like what we did."},{"Start":"14:08.485 ","End":"14:11.500","Text":"The second most important thing is if you\u0027ve"},{"Start":"14:11.500 ","End":"14:18.025","Text":"converted to other units rather than seconds, microseconds,"},{"Start":"14:18.025 ","End":"14:20.650","Text":"or rather than meters to kilometers,"},{"Start":"14:20.650 ","End":"14:29.425","Text":"you must remember to keep all of the units when working out the equations to be the same."},{"Start":"14:29.425 ","End":"14:35.185","Text":"For instance, the most important thing is to note that here you have 0.9c,"},{"Start":"14:35.185 ","End":"14:37.600","Text":"and c is in meters per second."},{"Start":"14:37.600 ","End":"14:42.940","Text":"Which means that if you wrote your t_1\u0027 in microseconds,"},{"Start":"14:42.940 ","End":"14:45.530","Text":"you have to convert it back to seconds."},{"Start":"14:45.530 ","End":"14:53.385","Text":"Or alternatively, write your c in terms of meters per microsecond, whichever you prefer."},{"Start":"14:53.385 ","End":"14:56.540","Text":"That\u0027s the end of this lesson."}],"ID":9561},{"Watched":false,"Name":"Exercise 7","Duration":"19m 45s","ChapterTopicVideoID":9274,"CourseChapterTopicPlaylistID":9398,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.115","Text":"2 particles are created at a height h above the ground."},{"Start":"00:05.115 ","End":"00:11.760","Text":"Particle a is emitted at an angle of 225 degrees to the x-axis."},{"Start":"00:11.760 ","End":"00:16.410","Text":"That means this is 225 degrees,"},{"Start":"00:16.410 ","End":"00:19.740","Text":"so over here it\u0027s 45 degrees in this direction,"},{"Start":"00:19.740 ","End":"00:24.675","Text":"and particle a decays at time t. Over here,"},{"Start":"00:24.675 ","End":"00:29.280","Text":"we have a decay time t and the height is 3/4(h)."},{"Start":"00:29.280 ","End":"00:36.495","Text":"Then we\u0027re being told that particle b is emitted at negative 45 degrees to the x-axis,"},{"Start":"00:36.495 ","End":"00:41.430","Text":"so that\u0027s going to be that over here and that particle b decays at 2t."},{"Start":"00:41.430 ","End":"00:46.855","Text":"This over here is 2t and its height is 1/4h."},{"Start":"00:46.855 ","End":"00:50.600","Text":"Now we\u0027re being told in the question to ignore gravity."},{"Start":"00:50.600 ","End":"00:56.645","Text":"The question Number 1 is express the velocities of the 2 particles as a function of"},{"Start":"00:56.645 ","End":"01:04.385","Text":"h and t. Let\u0027s begin by answering this question for our particle a."},{"Start":"01:04.385 ","End":"01:07.565","Text":"Now, because we are dealing with both of these particles in"},{"Start":"01:07.565 ","End":"01:11.000","Text":"the same frame of reference, the rest frame."},{"Start":"01:11.000 ","End":"01:14.870","Text":"We can use our usual equation for velocity."},{"Start":"01:14.870 ","End":"01:22.655","Text":"The velocity of particle a is going to be equal to the distance divided by the time."},{"Start":"01:22.655 ","End":"01:26.765","Text":"Now we already know that our time Delta t is equal to capital T,"},{"Start":"01:26.765 ","End":"01:29.690","Text":"that\u0027s when the decay happens and now we simply have"},{"Start":"01:29.690 ","End":"01:32.975","Text":"to find the distance that it is a way."},{"Start":"01:32.975 ","End":"01:37.850","Text":"What is l? We can say that this length over here"},{"Start":"01:37.850 ","End":"01:42.845","Text":"is equal to l. Then we can say that l is simply equal to,"},{"Start":"01:42.845 ","End":"01:44.990","Text":"we can find it out using Pythagoras."},{"Start":"01:44.990 ","End":"01:52.925","Text":"This length over here is simply going to be h minus 3/4h."},{"Start":"01:52.925 ","End":"01:55.580","Text":"This is going to be 1/4h,"},{"Start":"01:55.580 ","End":"01:58.490","Text":"we can add that in here."},{"Start":"01:58.490 ","End":"02:04.505","Text":"Then we know that 1/4h divided by"},{"Start":"02:04.505 ","End":"02:10.720","Text":"l is going to be equal to sine of 45 degrees."},{"Start":"02:10.720 ","End":"02:15.855","Text":"That means that l is going to be 1/4h divided by"},{"Start":"02:15.855 ","End":"02:22.255","Text":"sine of 45 degrees so add that in here."},{"Start":"02:22.255 ","End":"02:24.530","Text":"Now, as we know,"},{"Start":"02:24.530 ","End":"02:32.240","Text":"sine of 45 degrees is equal to 1 over root 2 and so we\u0027ll get"},{"Start":"02:32.240 ","End":"02:40.575","Text":"over here that l is therefore equal to h divided by 2 root 2."},{"Start":"02:40.575 ","End":"02:45.050","Text":"Therefore, we can say that the velocity for particle a,"},{"Start":"02:45.050 ","End":"02:46.355","Text":"or particle 1,"},{"Start":"02:46.355 ","End":"02:47.705","Text":"it doesn\u0027t really matter,"},{"Start":"02:47.705 ","End":"02:54.430","Text":"it\u0027s going to be equal to h divided by 2 root 2t."},{"Start":"02:55.390 ","End":"02:59.315","Text":"Now we can do the exact same thing for particle b."},{"Start":"02:59.315 ","End":"03:05.000","Text":"We know that our v_b is going to be l divided by Delta t. However,"},{"Start":"03:05.000 ","End":"03:10.055","Text":"this time we know that our Delta t is equal to 2t."},{"Start":"03:10.055 ","End":"03:14.750","Text":"Then we can say again similarly that are l is equal 2."},{"Start":"03:14.750 ","End":"03:18.110","Text":"We\u0027re trying to find this length over here."},{"Start":"03:18.110 ","End":"03:23.180","Text":"It\u0027s going to be h minus 1/4h,"},{"Start":"03:23.180 ","End":"03:29.400","Text":"we\u0027re going to have 3/4(h) divided by,"},{"Start":"03:29.400 ","End":"03:36.785","Text":"again, 1 divided by sine of 45 degrees."},{"Start":"03:36.785 ","End":"03:45.180","Text":"Now we\u0027re going to get that our l is equal to 3h divided by 2 root 2."},{"Start":"03:45.180 ","End":"03:51.380","Text":"Therefore will get that our v_b or a v2 is simply going to be"},{"Start":"03:51.380 ","End":"03:58.675","Text":"equal to 3h divided by 4 root 2t."},{"Start":"03:58.675 ","End":"04:04.000","Text":"When we substitute in our 2t from over here."},{"Start":"04:04.490 ","End":"04:08.190","Text":"These are the answers to question Number 1,"},{"Start":"04:08.190 ","End":"04:11.480","Text":"Now let\u0027s go on to question Number 2."},{"Start":"04:12.080 ","End":"04:20.300","Text":"Question number 2 is to find the lifespan of each particle in the rest frame."},{"Start":"04:21.020 ","End":"04:24.190","Text":"Let\u0027s write it over here, Question 2."},{"Start":"04:24.190 ","End":"04:28.300","Text":"Now I\u0027m going to label the lifespan of each particle is Tau."},{"Start":"04:28.300 ","End":"04:32.675","Text":"If particle a, we\u0027re going to have Tau a."},{"Start":"04:32.675 ","End":"04:37.360","Text":"Now, we\u0027re going to use our equation for time dilation,"},{"Start":"04:37.360 ","End":"04:46.610","Text":"which is Tau a multiplied by Gamma a is equal to Delta t_a."},{"Start":"04:47.310 ","End":"04:49.720","Text":"Now just a reminder,"},{"Start":"04:49.720 ","End":"04:55.120","Text":"we can use this Tau only when we\u0027re using it in the rest frame."},{"Start":"04:55.120 ","End":"04:56.560","Text":"What does that mean?"},{"Start":"04:56.560 ","End":"05:02.240","Text":"It means that our particle is at rest in that specific frame."},{"Start":"05:02.690 ","End":"05:06.970","Text":"Again, we can only use this equation when"},{"Start":"05:06.970 ","End":"05:13.625","Text":"our particle or our event is always happening in the same position and is not moving."},{"Start":"05:13.625 ","End":"05:19.345","Text":"It\u0027s stationary relative to this frame and this frame is called the rest frame."},{"Start":"05:19.345 ","End":"05:22.030","Text":"If we aren\u0027t in our rest frame and the particles in"},{"Start":"05:22.030 ","End":"05:25.105","Text":"fact moving in this specific frame of reference,"},{"Start":"05:25.105 ","End":"05:30.830","Text":"then we\u0027re going to have to add on to this equation over here."},{"Start":"05:31.580 ","End":"05:37.560","Text":"Let\u0027s see. We want to find what our Tau a is equal to,"},{"Start":"05:37.560 ","End":"05:44.045","Text":"that\u0027s going to be equal to our Delta t_a divided by our Gamma a."},{"Start":"05:44.045 ","End":"05:46.790","Text":"Now our Delta t_a as we know,"},{"Start":"05:46.790 ","End":"05:50.250","Text":"it\u0027s our t over here,"},{"Start":"05:50.250 ","End":"05:54.780","Text":"so we can write t up top over here and our Gamma, as we know,"},{"Start":"05:54.780 ","End":"06:02.740","Text":"is equal to 1 minus v_a divided by c^2."},{"Start":"06:04.400 ","End":"06:08.790","Text":"That is that and we know what our v_a is,"},{"Start":"06:08.790 ","End":"06:12.140","Text":"our v_a is what we worked out in this question over here."},{"Start":"06:12.140 ","End":"06:16.955","Text":"Now all we have to do is substitute everything in,"},{"Start":"06:16.955 ","End":"06:23.090","Text":"and then we\u0027ll get that our Tau a is equal to t multiplied by"},{"Start":"06:23.090 ","End":"06:30.330","Text":"the square root of 1 minus h^2 divided by 8t^2c^2."},{"Start":"06:34.490 ","End":"06:41.725","Text":"We will have the exact same thing or exact same process for our Tau b."},{"Start":"06:41.725 ","End":"06:47.960","Text":"The only difference is that our Delta t for b is equal to 2t."},{"Start":"06:48.620 ","End":"06:58.080","Text":"That is going to be equal to 2t divided by Gamma b,"},{"Start":"06:58.080 ","End":"07:00.990","Text":"for particle b, and then again,"},{"Start":"07:00.990 ","End":"07:03.120","Text":"we substitute in 1 minus,"},{"Start":"07:03.120 ","End":"07:06.295","Text":"but this time v_b divided by c^2."},{"Start":"07:06.295 ","End":"07:08.755","Text":"Then once we do that,"},{"Start":"07:08.755 ","End":"07:17.715","Text":"we\u0027ll get that our Tau for particle b is equal to 2t multiplied by the square root of 1"},{"Start":"07:17.715 ","End":"07:27.660","Text":"minus 9h^2 divided by 64t^2c^2."},{"Start":"07:27.660 ","End":"07:31.010","Text":"Those are the answers to our Question 2,"},{"Start":"07:31.010 ","End":"07:33.475","Text":"let\u0027s move on to Question 3."},{"Start":"07:33.475 ","End":"07:38.405","Text":"Question Number 3 is to find a reference frame s tag,"},{"Start":"07:38.405 ","End":"07:46.480","Text":"which moves in the positive x-direction such that both particles decay the same time."},{"Start":"07:46.480 ","End":"07:56.600","Text":"That means this t will equal to this t. We\u0027re going to use our trusty table that we love."},{"Start":"07:56.600 ","End":"07:59.730","Text":"Here we have our event,"},{"Start":"07:59.730 ","End":"08:02.600","Text":"here is what it looks like in our s frame of reference,"},{"Start":"08:02.600 ","End":"08:06.415","Text":"and here\u0027s what it looks like in our s tag frame of reference."},{"Start":"08:06.415 ","End":"08:14.359","Text":"Our first event is when the particles are created, pc particles created."},{"Start":"08:14.359 ","End":"08:19.790","Text":"We\u0027re not s frame of reference we know that this is right at the beginning,"},{"Start":"08:19.790 ","End":"08:24.590","Text":"so both of the particles are created at the same exact position."},{"Start":"08:24.590 ","End":"08:28.695","Text":"That means that our x_a,"},{"Start":"08:28.695 ","End":"08:31.455","Text":"is going to be equal to our y_a,"},{"Start":"08:31.455 ","End":"08:34.215","Text":"which is going to be equal to our x_b,"},{"Start":"08:34.215 ","End":"08:37.635","Text":"which is going to be equal to our y_b."},{"Start":"08:37.635 ","End":"08:40.534","Text":"That is all going to be at the origin,"},{"Start":"08:40.534 ","End":"08:42.820","Text":"so it\u0027s equal to 0."},{"Start":"08:42.820 ","End":"08:45.305","Text":"Now, what about the s tag?"},{"Start":"08:45.305 ","End":"08:50.210","Text":"Now, if we substitute in 0 into our Lorentz transform,"},{"Start":"08:50.210 ","End":"08:52.190","Text":"to get from here to here,"},{"Start":"08:52.190 ","End":"08:54.410","Text":"we use our Lorentz transform."},{"Start":"08:54.410 ","End":"08:58.265","Text":"We\u0027re just going to get the exact same thing."},{"Start":"08:58.265 ","End":"09:03.210","Text":"I\u0027m just going to be writing here same."},{"Start":"09:04.240 ","End":"09:10.040","Text":"Now let\u0027s move on to our second event."},{"Start":"09:10.040 ","End":"09:12.455","Text":"Before I do that, of course,"},{"Start":"09:12.455 ","End":"09:20.740","Text":"we have to say that our times are also equal to 0."},{"Start":"09:20.780 ","End":"09:28.170","Text":"Again, our second event is when particle a decays."},{"Start":"09:29.330 ","End":"09:31.725","Text":"When does that happen?"},{"Start":"09:31.725 ","End":"09:34.380","Text":"Our position x_a,"},{"Start":"09:34.380 ","End":"09:35.910","Text":"and let\u0027s call this 2,"},{"Start":"09:35.910 ","End":"09:39.210","Text":"so here we can say that all of this is 1."},{"Start":"09:39.560 ","End":"09:45.035","Text":"So x_a2 is, as we know, its height,"},{"Start":"09:45.035 ","End":"09:51.150","Text":"is that h divided by 1/4h."},{"Start":"09:51.150 ","End":"09:55.800","Text":"Let\u0027s say that it\u0027s h divided by 4,"},{"Start":"09:55.800 ","End":"10:00.005","Text":"and we know that its y-coordinate is also,"},{"Start":"10:00.005 ","End":"10:03.080","Text":"and this is a minus because it\u0027s in this direction."},{"Start":"10:03.080 ","End":"10:06.980","Text":"Its y-coordinate is also a negative coordinate,"},{"Start":"10:06.980 ","End":"10:11.010","Text":"which is also here."},{"Start":"10:11.110 ","End":"10:18.665","Text":"Our y_a2 is also equal to negative h divided by 4."},{"Start":"10:18.665 ","End":"10:23.805","Text":"As we know that our t for our a"},{"Start":"10:23.805 ","End":"10:30.030","Text":"is equal to t. That\u0027s when our particle a decays."},{"Start":"10:30.030 ","End":"10:38.925","Text":"Now, what we can do is our Lorentz transform to get it into our s tag frame of reference."},{"Start":"10:38.925 ","End":"10:45.290","Text":"As we know, we don\u0027t really need the positions right now in our s tag frame of reference."},{"Start":"10:45.290 ","End":"10:48.800","Text":"Because all we\u0027re being asked is to find this s tag frame of"},{"Start":"10:48.800 ","End":"10:52.550","Text":"reference such that both the particles decay at the same time."},{"Start":"10:52.550 ","End":"10:57.115","Text":"What\u0027s interesting to us right now is the time."},{"Start":"10:57.115 ","End":"11:01.605","Text":"Let\u0027s take a look so we can see that our t tag"},{"Start":"11:01.605 ","End":"11:07.640","Text":"a_1 is simply going to be equal to our Gamma naught multiplied"},{"Start":"11:07.640 ","End":"11:13.550","Text":"by our t minus v_0 divided by"},{"Start":"11:13.550 ","End":"11:21.840","Text":"c^2 multiplied by negative h/4."},{"Start":"11:21.840 ","End":"11:26.665","Text":"Next, we\u0027re going to have our third event,"},{"Start":"11:26.665 ","End":"11:34.585","Text":"which is when particle B decays That happens of course afterwards because it\u0027s at 2T."},{"Start":"11:34.585 ","End":"11:42.325","Text":"So we can say that our x_B2 is simply going to be equal to this distance over here."},{"Start":"11:42.325 ","End":"11:45.220","Text":"So first of all, it\u0027s in the positive X direction."},{"Start":"11:45.220 ","End":"11:50.740","Text":"We\u0027ll see that\u0027s equal to 3/4 h and similarly"},{"Start":"11:50.740 ","End":"11:56.560","Text":"our y_B2 is also going to be this height over here."},{"Start":"11:56.560 ","End":"11:58.480","Text":"This is negative."},{"Start":"11:58.480 ","End":"12:03.925","Text":"It\u0027s in the negative y-direction, again, 3/4h."},{"Start":"12:03.925 ","End":"12:09.670","Text":"Now let\u0027s see what our t_B is going to be equal to."},{"Start":"12:09.670 ","End":"12:15.775","Text":"This is, as we know over here, equal to 2T."},{"Start":"12:15.775 ","End":"12:20.500","Text":"Now, again, we\u0027re going to transform this via the Lorentz"},{"Start":"12:20.500 ","End":"12:24.805","Text":"transform into what this means relative to R,"},{"Start":"12:24.805 ","End":"12:27.040","Text":"S tag frame of reference."},{"Start":"12:27.040 ","End":"12:31.600","Text":"So we\u0027ll get that our t_B1 tag"},{"Start":"12:31.600 ","End":"12:37.675","Text":"is going to be equal to our Gamma naught multiplied by this time 2T"},{"Start":"12:37.675 ","End":"12:43.675","Text":"minus V_0 divided by C^2 multiplied by"},{"Start":"12:43.675 ","End":"12:51.775","Text":"3h divided by 4, our x value."},{"Start":"12:51.775 ","End":"12:55.690","Text":"Now what we wanna do is we want to find that"},{"Start":"12:55.690 ","End":"13:01.467","Text":"both particles decay at the same time in this S tag frame of reference."},{"Start":"13:01.467 ","End":"13:02.830","Text":"What does that mean?"},{"Start":"13:02.830 ","End":"13:12.170","Text":"That means therefore that our t\u0027_A1 must be equal to our t\u0027_B1."},{"Start":"13:13.170 ","End":"13:20.455","Text":"That is the time that they both decay and we want them to both be equal."},{"Start":"13:20.455 ","End":"13:25.030","Text":"Then I can cross out my Gamma naught from both sides and I\u0027ll get"},{"Start":"13:25.030 ","End":"13:31.420","Text":"that T plus V_0 divided by C^2,"},{"Start":"13:31.420 ","End":"13:33.595","Text":"where a visa is what we\u0027re trying to find."},{"Start":"13:33.595 ","End":"13:38.335","Text":"Multiplied by h divided by 4 is going to be equal to"},{"Start":"13:38.335 ","End":"13:48.145","Text":"our 2T minus V_0 multiplied by 3h divided by 4C^2."},{"Start":"13:48.145 ","End":"13:55.165","Text":"Once we play around with the 2 sides and move and shift things around,"},{"Start":"13:55.165 ","End":"14:02.274","Text":"we\u0027ll get that our T is equal to V_0h divided by C^2."},{"Start":"14:02.274 ","End":"14:05.050","Text":"But because it\u0027s our V_0 that we\u0027re looking for,"},{"Start":"14:05.050 ","End":"14:10.060","Text":"because that is what\u0027s describing the motion of our S tag frame of reference."},{"Start":"14:10.060 ","End":"14:13.135","Text":"We know that it\u0027s in the positive x-direction."},{"Start":"14:13.135 ","End":"14:19.569","Text":"We just now need to find its velocity relative to our observer."},{"Start":"14:19.569 ","End":"14:23.530","Text":"We can see that our V_0 is equal to C^2T"},{"Start":"14:23.530 ","End":"14:29.090","Text":"divided by h. This is the answer to question number 3."},{"Start":"14:29.090 ","End":"14:34.255","Text":"This is the onset and now let\u0027s take a look at question number 4."},{"Start":"14:34.255 ","End":"14:42.460","Text":"What is the distance between the 2 decays in our S\u0027 frame of reference?"},{"Start":"14:42.460 ","End":"14:47.650","Text":"Let\u0027s take a look at how we\u0027re going to answer this question."},{"Start":"14:47.650 ","End":"14:53.140","Text":"The distance between this decay and this decay,"},{"Start":"14:53.140 ","End":"14:55.540","Text":"let\u0027s call this D\u0027."},{"Start":"14:55.540 ","End":"15:00.925","Text":"Now, a common misconception is to use this equation,"},{"Start":"15:00.925 ","End":"15:03.565","Text":"which is an equation for our rest length,"},{"Start":"15:03.565 ","End":"15:11.995","Text":"which is that Delta x is equal to Delta x naught divided by Gamma 0."},{"Start":"15:11.995 ","End":"15:18.700","Text":"Now the reason we can\u0027t use this equation over here is because this equation can only be"},{"Start":"15:18.700 ","End":"15:26.360","Text":"used when this distance d is constant in the rest frame."},{"Start":"15:26.490 ","End":"15:31.000","Text":"Now we know that this distance d is not constant in the rest frame."},{"Start":"15:31.000 ","End":"15:34.855","Text":"Because we know that both of our particles are constantly moving,"},{"Start":"15:34.855 ","End":"15:39.740","Text":"meaning they\u0027re not stationary in this S frame."},{"Start":"15:40.280 ","End":"15:45.344","Text":"That means we cannot use this equation and instead,"},{"Start":"15:45.344 ","End":"15:49.835","Text":"we\u0027re going to have to use our normal Lawrence transform."},{"Start":"15:49.835 ","End":"15:54.220","Text":"Let\u0027s do this now."},{"Start":"15:54.220 ","End":"15:58.060","Text":"We can see that for our particle a,"},{"Start":"15:58.060 ","End":"16:00.145","Text":"when it begins to decay,"},{"Start":"16:00.145 ","End":"16:06.100","Text":"this event is at x\u0027A,"},{"Start":"16:06.100 ","End":"16:09.190","Text":"What is this equal to?"},{"Start":"16:09.190 ","End":"16:16.510","Text":"This is equal to Gamma naught multiplied by negative h divided by 4."},{"Start":"16:16.510 ","End":"16:20.334","Text":"Its x position, negative x position,"},{"Start":"16:20.334 ","End":"16:25.360","Text":"negative V_0, which we figured out already,"},{"Start":"16:25.360 ","End":"16:28.345","Text":"multiplied by the time in which this happens,"},{"Start":"16:28.345 ","End":"16:30.550","Text":"which is capital T,"},{"Start":"16:30.550 ","End":"16:37.450","Text":"then our y\u0027A is simply going to be equal to,"},{"Start":"16:37.450 ","End":"16:39.925","Text":"as we know from our Lawrence transform,"},{"Start":"16:39.925 ","End":"16:42.445","Text":"simply our y_A,"},{"Start":"16:42.445 ","End":"16:45.250","Text":"same as in our S frame of reference."},{"Start":"16:45.250 ","End":"16:50.120","Text":"Our y and z transforms are exactly the same."},{"Start":"16:50.790 ","End":"16:57.580","Text":"Now, we\u0027re going to do the exact same thing for event number 3."},{"Start":"16:57.580 ","End":"17:01.810","Text":"We can say that our x_B, sorry,"},{"Start":"17:01.810 ","End":"17:07.570","Text":"tag is going to be equal to, let\u0027s see."},{"Start":"17:07.570 ","End":"17:13.180","Text":"It will be Gamma naught multiplied by 3h divided by"},{"Start":"17:13.180 ","End":"17:20.425","Text":"4 minus the velocity of our S tag multiplied by the time at which this happens,"},{"Start":"17:20.425 ","End":"17:23.425","Text":"the time of the decay, which is 2T."},{"Start":"17:23.425 ","End":"17:28.135","Text":"Then our y\u0027_B well, again,"},{"Start":"17:28.135 ","End":"17:36.340","Text":"just like when our aid decayed will be equal to our y_B without"},{"Start":"17:36.340 ","End":"17:41.170","Text":"a tag that is going to be simply equal to"},{"Start":"17:41.170 ","End":"17:47.905","Text":"negative 3h divided by 4."},{"Start":"17:47.905 ","End":"17:49.855","Text":"That\u0027s its y position."},{"Start":"17:49.855 ","End":"17:52.135","Text":"Sorry, I didn\u0027t write this y position."},{"Start":"17:52.135 ","End":"17:55.495","Text":"This is negative h divided by 4."},{"Start":"17:55.495 ","End":"18:01.855","Text":"Now we have x_A\u0027 y_A\u0027, x_B\u0027 and y_B\u0027."},{"Start":"18:01.855 ","End":"18:06.265","Text":"We have the positions of our particles in our S\u0027 frame of reference."},{"Start":"18:06.265 ","End":"18:11.935","Text":"Now what we want to do is we want to find the distance d\u0027 between them."},{"Start":"18:11.935 ","End":"18:18.730","Text":"We can use Pythagoras and we can say that d\u0027^2 is going to be equal"},{"Start":"18:18.730 ","End":"18:25.435","Text":"to x _B\u0027 minus x_A\u0027^2"},{"Start":"18:25.435 ","End":"18:34.015","Text":"plus y_B\u0027^2 minus y_A\u0027^2."},{"Start":"18:34.015 ","End":"18:36.520","Text":"Then we\u0027ll substitute everything in."},{"Start":"18:36.520 ","End":"18:44.905","Text":"So that will be Gamma naught multiplied by h minus V_0t^2 plus"},{"Start":"18:44.905 ","End":"18:50.650","Text":"negative 1/2h^2 which is going to be"},{"Start":"18:50.650 ","End":"18:57.010","Text":"equal to after substituting in here what our V_0 is equal to."},{"Start":"18:57.010 ","End":"19:03.400","Text":"So we\u0027ll get 5h^4 divided by"},{"Start":"19:03.400 ","End":"19:05.215","Text":"4 minus"},{"Start":"19:05.215 ","End":"19:15.055","Text":"3c^2T^2h^2 plus C^4T^4."},{"Start":"19:15.055 ","End":"19:22.720","Text":"Then all of this is going to be divided by h^2 minus c^2T^2."},{"Start":"19:22.720 ","End":"19:26.875","Text":"Remember, this is the equation to find the distance."},{"Start":"19:26.875 ","End":"19:29.740","Text":"It\u0027s our distance squared is equal to,"},{"Start":"19:29.740 ","End":"19:36.050","Text":"our x_B minus x_A squared plus our Y_B minus y_A squared."},{"Start":"19:36.510 ","End":"19:45.889","Text":"This is the final answer to our question number 4 and that is the end of our lesson."}],"ID":9562},{"Watched":false,"Name":"Energy And Momentum","Duration":"8m 12s","ChapterTopicVideoID":9265,"CourseChapterTopicPlaylistID":9398,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.830","Text":"Hello, in this topic we\u0027re going to be learning about special relativity and,"},{"Start":"00:04.830 ","End":"00:08.250","Text":"specifically, the dynamics of special relativity."},{"Start":"00:08.250 ","End":"00:10.875","Text":"In the dynamics of special relativity,"},{"Start":"00:10.875 ","End":"00:14.280","Text":"we\u0027re going to be dealing a lot with momentum and energy."},{"Start":"00:14.280 ","End":"00:18.150","Text":"We\u0027re going to have different expressions for our momentum"},{"Start":"00:18.150 ","End":"00:23.110","Text":"and for our energy than what we\u0027re used to from our classical mechanics."},{"Start":"00:23.300 ","End":"00:26.595","Text":"What Einstein did was that he showed"},{"Start":"00:26.595 ","End":"00:29.520","Text":"that in order for us to still look at our momentum and"},{"Start":"00:29.520 ","End":"00:32.610","Text":"our energy as sizes which are"},{"Start":"00:32.610 ","End":"00:36.960","Text":"fixed or constant or unchanging throughout our calculation,"},{"Start":"00:36.960 ","End":"00:41.450","Text":"we have to include another variable or"},{"Start":"00:41.450 ","End":"00:47.520","Text":"another component in our equations in order to work out our equations."},{"Start":"00:47.520 ","End":"00:49.430","Text":"Like in classical mechanics,"},{"Start":"00:49.430 ","End":"00:51.440","Text":"we\u0027ve seen that we can use the idea of conservation"},{"Start":"00:51.440 ","End":"00:54.055","Text":"of energy or conservation of momentum."},{"Start":"00:54.055 ","End":"00:56.750","Text":"With special relativity, we can still use"},{"Start":"00:56.750 ","End":"00:59.495","Text":"those ideas of conservation of momentum and energy."},{"Start":"00:59.495 ","End":"01:01.460","Text":"However, in order to do that,"},{"Start":"01:01.460 ","End":"01:08.600","Text":"we have to put this Gamma symbol over here and multiply our expression."},{"Start":"01:08.600 ","End":"01:14.090","Text":"We use for our momentum equation being our mass multiplied by our velocity."},{"Start":"01:14.090 ","End":"01:16.020","Text":"Now in special relativity,"},{"Start":"01:16.020 ","End":"01:19.280","Text":"we\u0027re also going to be multiplying it by Gamma."},{"Start":"01:19.280 ","End":"01:24.440","Text":"Now the Gamma that we\u0027re substituting in is this Gamma over here."},{"Start":"01:24.440 ","End":"01:30.470","Text":"It\u0027s 1 divided by the square root of 1 minus v divided by c squared,"},{"Start":"01:30.470 ","End":"01:32.855","Text":"where c is our speed of light."},{"Start":"01:32.855 ","End":"01:41.315","Text":"Now here, our Gamma doesn\u0027t represent our move from different frames of reference."},{"Start":"01:41.315 ","End":"01:44.710","Text":"Here it means something else."},{"Start":"01:44.710 ","End":"01:48.440","Text":"Now what we\u0027re doing is we\u0027re saying that the Gamma has to"},{"Start":"01:48.440 ","End":"01:52.610","Text":"do with the velocity of our particle,"},{"Start":"01:52.610 ","End":"01:56.770","Text":"and our Gamma is also called our Lorentz factor."},{"Start":"01:56.770 ","End":"02:01.340","Text":"So when I multiply by my momentum or my energy by this Gamma,"},{"Start":"02:01.340 ","End":"02:04.850","Text":"I\u0027m still in the same frame of reference that I was before,"},{"Start":"02:04.850 ","End":"02:07.390","Text":"such as the frame of reference of my lab."},{"Start":"02:07.390 ","End":"02:09.860","Text":"But I have to include this Gamma,"},{"Start":"02:09.860 ","End":"02:13.670","Text":"which has to do with the velocity of the particle in order to use"},{"Start":"02:13.670 ","End":"02:18.635","Text":"the idea of conservation of momentum and energy when dealing with special relativity."},{"Start":"02:18.635 ","End":"02:23.870","Text":"Now when I do want to represent changing frame of reference,"},{"Start":"02:23.870 ","End":"02:27.730","Text":"I\u0027ll use my symbol of Gamma_0,"},{"Start":"02:27.730 ","End":"02:30.650","Text":"and this will represent that I\u0027m changing my frame of reference"},{"Start":"02:30.650 ","End":"02:34.610","Text":"from the lab frame of reference to something else."},{"Start":"02:34.610 ","End":"02:39.065","Text":"That is not to be confused with this Gamma over here,"},{"Start":"02:39.065 ","End":"02:42.330","Text":"which has to do with the velocity of the particle."},{"Start":"02:42.460 ","End":"02:46.745","Text":"Our momentum is going to be our normal momentum,"},{"Start":"02:46.745 ","End":"02:50.555","Text":"mv, and then multiply it by our Gamma for special relativity,"},{"Start":"02:50.555 ","End":"02:53.990","Text":"and our energy is going to be m c squared,"},{"Start":"02:53.990 ","End":"03:00.115","Text":"where c is the velocity of light and again multiplied by our Gamma."},{"Start":"03:00.115 ","End":"03:04.820","Text":"These are the 2 basic equations that are very important to know."},{"Start":"03:04.820 ","End":"03:07.970","Text":"Write them out in your equation sheets."},{"Start":"03:07.970 ","End":"03:15.205","Text":"Now 2 other equations which come from playing around with these 2 equations are these 2."},{"Start":"03:15.205 ","End":"03:17.090","Text":"This equation over here,"},{"Start":"03:17.090 ","End":"03:22.790","Text":"E^2 is equal to the absolute value of our momentum squared multiplied"},{"Start":"03:22.790 ","End":"03:29.550","Text":"by c^2 plus our m^2 multiplied by c^4."},{"Start":"03:29.960 ","End":"03:32.630","Text":"So that\u0027s another way of writing our E,"},{"Start":"03:32.630 ","End":"03:34.405","Text":"and then we can just square root it,"},{"Start":"03:34.405 ","End":"03:36.750","Text":"and it equals to this over here."},{"Start":"03:36.750 ","End":"03:39.525","Text":"Our absolute value for our momentum,"},{"Start":"03:39.525 ","End":"03:41.225","Text":"so the size of our momentum,"},{"Start":"03:41.225 ","End":"03:48.715","Text":"is going to be equal to the square root of our Gamma^2 minus 1 multiplied by mc."},{"Start":"03:48.715 ","End":"03:52.310","Text":"So these equations are essentially the same as this."},{"Start":"03:52.310 ","End":"03:54.620","Text":"However, they\u0027re also useful,"},{"Start":"03:54.620 ","End":"03:58.445","Text":"and they\u0027re written out in a useful format that it might make it"},{"Start":"03:58.445 ","End":"04:03.310","Text":"easier to solve certain questions depending on what we\u0027re given."},{"Start":"04:03.310 ","End":"04:06.215","Text":"So these are the basic equations,"},{"Start":"04:06.215 ","End":"04:08.509","Text":"these being the most important."},{"Start":"04:08.509 ","End":"04:11.990","Text":"Now what we\u0027re going to do is we\u0027re going to define"},{"Start":"04:11.990 ","End":"04:16.720","Text":"a new term which is called rest mass energy."},{"Start":"04:16.720 ","End":"04:20.715","Text":"This is our equation for rest mass energy,"},{"Start":"04:20.715 ","End":"04:23.090","Text":"and what we\u0027re speaking about over here,"},{"Start":"04:23.090 ","End":"04:31.070","Text":"we\u0027re referring to when our mass or a particle has 0 velocity,"},{"Start":"04:31.070 ","End":"04:33.530","Text":"that means that either it\u0027s not moving,"},{"Start":"04:33.530 ","End":"04:36.890","Text":"or we\u0027re calculating it in a frame of reference,"},{"Start":"04:36.890 ","End":"04:41.190","Text":"where in the frame of reference it is stationary."},{"Start":"04:41.190 ","End":"04:44.840","Text":"So that means that its velocity is equal to 0."},{"Start":"04:44.840 ","End":"04:50.555","Text":"Now, if we substitute in v is equal to 0 into our equation for Gamma,"},{"Start":"04:50.555 ","End":"04:56.195","Text":"so we\u0027ll see that our Gamma will therefore be equal to 1."},{"Start":"04:56.195 ","End":"05:00.950","Text":"Now if we substitute this into our equation for energy,"},{"Start":"05:00.950 ","End":"05:06.120","Text":"so our equation for energy is Gamma multiplied by mass multiplied by c^2."},{"Start":"05:06.120 ","End":"05:07.720","Text":"So our Gamma is now 1,"},{"Start":"05:07.720 ","End":"05:10.175","Text":"and then we\u0027ll be left with this equation."},{"Start":"05:10.175 ","End":"05:12.830","Text":"E_0 is our rest mass energy."},{"Start":"05:12.830 ","End":"05:15.455","Text":"So when a particle has no velocity,"},{"Start":"05:15.455 ","End":"05:20.425","Text":"that is simply going to be equal to mc^2."},{"Start":"05:20.425 ","End":"05:26.420","Text":"This is that famous equation that you\u0027ll find on t-shirts and absolutely everywhere,"},{"Start":"05:26.420 ","End":"05:29.365","Text":"E is equal to mc^2."},{"Start":"05:29.365 ","End":"05:33.574","Text":"Now the important thing to notice here is that our energy,"},{"Start":"05:33.574 ","End":"05:36.305","Text":"even when our particle is at rest,"},{"Start":"05:36.305 ","End":"05:37.855","Text":"v is equal to 0,"},{"Start":"05:37.855 ","End":"05:43.265","Text":"our energy does not equal 0 because then we can see that our energy is equal to our mass,"},{"Start":"05:43.265 ","End":"05:46.730","Text":"which is not 0, multiplied by the speed of light squared,"},{"Start":"05:46.730 ","End":"05:49.310","Text":"which is definitely not equal to 0."},{"Start":"05:49.310 ","End":"05:53.576","Text":"So that means that mass is energy."},{"Start":"05:53.576 ","End":"05:55.745","Text":"This is what this equation means."},{"Start":"05:55.745 ","End":"05:57.935","Text":"From this equation, we can see that"},{"Start":"05:57.935 ","End":"06:03.485","Text":"even an object which is stationary is going to have energy,"},{"Start":"06:03.485 ","End":"06:07.970","Text":"and then that means that our mass is another type of energy."},{"Start":"06:07.970 ","End":"06:10.535","Text":"That mass and energy are interchangeable."},{"Start":"06:10.535 ","End":"06:12.275","Text":"If we have some mass,"},{"Start":"06:12.275 ","End":"06:14.900","Text":"we can convert it into energy."},{"Start":"06:14.900 ","End":"06:16.295","Text":"If we have some energy,"},{"Start":"06:16.295 ","End":"06:19.250","Text":"we can convert it into mass."},{"Start":"06:19.250 ","End":"06:24.890","Text":"So therefore, that also means that mass is not a constant."},{"Start":"06:24.890 ","End":"06:30.530","Text":"It\u0027s a variable which is changing as a function of time or whatever it might be."},{"Start":"06:30.530 ","End":"06:33.410","Text":"When we\u0027re speaking about special relativity,"},{"Start":"06:33.410 ","End":"06:35.885","Text":"it\u0027s a variable which is constantly changing,"},{"Start":"06:35.885 ","End":"06:38.435","Text":"which means it\u0027s not a fixed value."},{"Start":"06:38.435 ","End":"06:43.175","Text":"Now we\u0027re going to move on to the idea of our kinetic energy."},{"Start":"06:43.175 ","End":"06:46.115","Text":"Our kinetic energy, or E_k,"},{"Start":"06:46.115 ","End":"06:48.920","Text":"is worked out like so."},{"Start":"06:48.920 ","End":"06:53.000","Text":"It equals the total energy, E total,"},{"Start":"06:53.000 ","End":"06:56.420","Text":"minus our rest mass energy,"},{"Start":"06:56.420 ","End":"06:59.270","Text":"which is this over here."},{"Start":"06:59.270 ","End":"07:03.600","Text":"Then our E total is what we\u0027re given over here,"},{"Start":"07:03.600 ","End":"07:10.700","Text":"so that\u0027s Gamma multiplied by m multiplied by c^2 minus our rest mass energy,"},{"Start":"07:10.700 ","End":"07:13.145","Text":"which is mc^2 squared."},{"Start":"07:13.145 ","End":"07:21.115","Text":"Then that is going to be equal to our mc^2Gamma minus 1,"},{"Start":"07:21.115 ","End":"07:23.765","Text":"which is exactly what we have written over here."},{"Start":"07:23.765 ","End":"07:28.615","Text":"This is our kinetic energy when we\u0027re speaking about special relativity."},{"Start":"07:28.615 ","End":"07:35.450","Text":"All of the equations which are squared in red should be written in your notes and in"},{"Start":"07:35.450 ","End":"07:37.580","Text":"your equation sheets including also maybe"},{"Start":"07:37.580 ","End":"07:41.810","Text":"this Gamma so that you can remember what this is over here."},{"Start":"07:41.810 ","End":"07:44.090","Text":"We can see that our equations when dealing with"},{"Start":"07:44.090 ","End":"07:47.795","Text":"special relativity for momentum and energy change,"},{"Start":"07:47.795 ","End":"07:51.819","Text":"they look different to what we have in classical mechanics."},{"Start":"07:51.819 ","End":"07:57.635","Text":"The important thing also to remember is that even when our particle is at rest,"},{"Start":"07:57.635 ","End":"07:59.660","Text":"when we\u0027re dealing with special relativity,"},{"Start":"07:59.660 ","End":"08:02.410","Text":"it still has energy."},{"Start":"08:02.410 ","End":"08:09.875","Text":"That\u0027s given by the famous equation E equals mc^2 that Einstein gave us."},{"Start":"08:09.875 ","End":"08:13.170","Text":"That is the end of this lesson."}],"ID":9563},{"Watched":false,"Name":"Exercise 8","Duration":"12m 1s","ChapterTopicVideoID":9272,"CourseChapterTopicPlaylistID":9398,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.830","Text":"Hello. In this question,"},{"Start":"00:01.830 ","End":"00:05.370","Text":"we\u0027re being told to find the energy needed for a neutron to"},{"Start":"00:05.370 ","End":"00:09.210","Text":"reach Earth from a distance of 5 light years away,"},{"Start":"00:09.210 ","End":"00:14.070","Text":"given that the lifespan of a neutron is 881 seconds,"},{"Start":"00:14.070 ","End":"00:23.585","Text":"and then mass of the neutron is equal to 940 Mega electron volts divided by c^2."},{"Start":"00:23.585 ","End":"00:32.250","Text":"Mega is 10^6 electron volts,"},{"Start":"00:32.250 ","End":"00:34.575","Text":"so that\u0027s a unit of energy."},{"Start":"00:34.575 ","End":"00:38.370","Text":"We have the basic information and we\u0027re being told that the distance"},{"Start":"00:38.370 ","End":"00:42.450","Text":"that the neutrons from Earth is 5 light years away."},{"Start":"00:42.450 ","End":"00:44.700","Text":"Let\u0027s see what that means."},{"Start":"00:44.700 ","End":"00:50.570","Text":"If we call our variable x and that is equal to 5 light years,"},{"Start":"00:50.570 ","End":"00:52.545","Text":"let\u0027s see what that means."},{"Start":"00:52.545 ","End":"00:56.225","Text":"Five light years means it\u0027s the distance"},{"Start":"00:56.225 ","End":"01:00.065","Text":"that light will travel in five years. What does that mean?"},{"Start":"01:00.065 ","End":"01:05.530","Text":"We have five for the 5 years times c,"},{"Start":"01:05.530 ","End":"01:07.970","Text":"where c is our speed of light,"},{"Start":"01:07.970 ","End":"01:11.585","Text":"multiplied by the time that we have in 1 year."},{"Start":"01:11.585 ","End":"01:14.690","Text":"We have 60 seconds in a minute,"},{"Start":"01:14.690 ","End":"01:16.925","Text":"60 minutes in an hour,"},{"Start":"01:16.925 ","End":"01:20.030","Text":"24 hours in a day,"},{"Start":"01:20.030 ","End":"01:26.595","Text":"and approximately 365 days per year."},{"Start":"01:26.595 ","End":"01:28.880","Text":"Then once you calculate this,"},{"Start":"01:28.880 ","End":"01:30.650","Text":"you can also do it on your calculator,"},{"Start":"01:30.650 ","End":"01:39.250","Text":"you\u0027ll get 1.577 times 10^8c."},{"Start":"01:39.250 ","End":"01:43.435","Text":"What do we actually have to do in order to solve our distance,"},{"Start":"01:43.435 ","End":"01:46.889","Text":"our x, in 5 light years?"},{"Start":"01:46.889 ","End":"01:50.345","Text":"We have to work out the time that we have in five light years,"},{"Start":"01:50.345 ","End":"01:55.954","Text":"so that\u0027s going to be 5 times the time that we have per year in seconds."},{"Start":"01:55.954 ","End":"01:59.695","Text":"Then multiplied by the speed of light."},{"Start":"01:59.695 ","End":"02:03.095","Text":"Then we\u0027ll get this answer where over here,"},{"Start":"02:03.095 ","End":"02:12.260","Text":"this is the speed of light c. What we have in this section is our seconds per year,"},{"Start":"02:12.260 ","End":"02:16.120","Text":"then we have to multiply this by the amount of time that we\u0027re given,"},{"Start":"02:16.120 ","End":"02:17.515","Text":"which is five years."},{"Start":"02:17.515 ","End":"02:20.080","Text":"We have seconds per year times 5 years,"},{"Start":"02:20.080 ","End":"02:24.020","Text":"and then multiplied by the speed of light."},{"Start":"02:24.230 ","End":"02:30.220","Text":"This, our answer is the distance that our neutron has to travel."},{"Start":"02:30.380 ","End":"02:35.575","Text":"Now we have to see how much time it has to make this journey."},{"Start":"02:35.575 ","End":"02:40.555","Text":"How much time it has when we\u0027re speaking about time on earth"},{"Start":"02:40.555 ","End":"02:45.680","Text":"is going to be equal to Gamma multiplied by Tau."},{"Start":"02:45.680 ","End":"02:49.515","Text":"Now, Tau represents our lifespan,"},{"Start":"02:49.515 ","End":"02:52.290","Text":"that\u0027s over here. Let\u0027s write it."},{"Start":"02:52.290 ","End":"02:55.685","Text":"Lifespan, which is over here,"},{"Start":"02:55.685 ","End":"03:00.055","Text":"equal to 881 seconds."},{"Start":"03:00.055 ","End":"03:03.350","Text":"The reason I can use this equation is because"},{"Start":"03:03.350 ","End":"03:08.990","Text":"our neutron is located at rest in its frame of reference."},{"Start":"03:08.990 ","End":"03:12.050","Text":"If however, I was checking"},{"Start":"03:12.050 ","End":"03:16.715","Text":"the time difference between two different systems or two different frames of"},{"Start":"03:16.715 ","End":"03:24.865","Text":"reference and also the position would be changing in one of the systems,"},{"Start":"03:24.865 ","End":"03:28.220","Text":"then I wouldn\u0027t be able to use this equation."},{"Start":"03:28.220 ","End":"03:33.815","Text":"But because the neutron is always located at the origin of its frame of reference,"},{"Start":"03:33.815 ","End":"03:37.975","Text":"then we can use this equation for time dilation."},{"Start":"03:37.975 ","End":"03:43.310","Text":"That means that the time taken on earth is going to be equal to our Gamma"},{"Start":"03:43.310 ","End":"03:49.845","Text":"multiplied by our proper lifetime or our lifespan of our neutron."},{"Start":"03:49.845 ","End":"03:56.005","Text":"Then we can take a look and we can see that our Gamma is going to be bigger than 1,"},{"Start":"03:56.005 ","End":"04:01.850","Text":"which means that our time that we\u0027re going to measure on earth is going to be longer"},{"Start":"04:01.850 ","End":"04:08.230","Text":"than this time over here I have 881 seconds longer than the proper lifetime."},{"Start":"04:08.230 ","End":"04:12.910","Text":"We can see that we\u0027re going to have a case of time dilation."},{"Start":"04:12.910 ","End":"04:19.055","Text":"Now what we can do is we can say that our Delta t is going to be equal to also"},{"Start":"04:19.055 ","End":"04:27.490","Text":"our distance divided by our normal velocity time equation."},{"Start":"04:28.760 ","End":"04:32.385","Text":"Now I\u0027m going to look at my question."},{"Start":"04:32.385 ","End":"04:35.105","Text":"My question is to find the energy needed."},{"Start":"04:35.105 ","End":"04:36.755","Text":"I\u0027m looking for energy."},{"Start":"04:36.755 ","End":"04:38.720","Text":"Now how am I going to come up with this?"},{"Start":"04:38.720 ","End":"04:44.165","Text":"I\u0027m going to use my idea of energy and kinetic energy."},{"Start":"04:44.165 ","End":"04:47.195","Text":"I\u0027m going to use my velocity."},{"Start":"04:47.195 ","End":"04:49.580","Text":"I need my velocity and of course we\u0027re going to"},{"Start":"04:49.580 ","End":"04:53.525","Text":"remember that when I\u0027m dealing with relative energy,"},{"Start":"04:53.525 ","End":"04:56.480","Text":"so my velocity is going to have to have"},{"Start":"04:56.480 ","End":"05:00.725","Text":"my relativistic correction, my Lawrence transform."},{"Start":"05:00.725 ","End":"05:04.960","Text":"I\u0027m going to have to isolate out my term v Gamma."},{"Start":"05:04.960 ","End":"05:09.890","Text":"The way I\u0027m going to do that is I\u0027m going to do this,"},{"Start":"05:09.890 ","End":"05:17.490","Text":"multiply both sides by v and divide both sides by my Tau."},{"Start":"05:18.440 ","End":"05:23.630","Text":"Then I\u0027m going to get that my v multiplied by"},{"Start":"05:23.630 ","End":"05:30.120","Text":"my Gamma is going to be equal to my x divided by my Tau."},{"Start":"05:30.120 ","End":"05:32.315","Text":"That\u0027s going to be equal to"},{"Start":"05:32.315 ","End":"05:42.105","Text":"1.577 times 10^8 times c divided by my Tau,"},{"Start":"05:42.105 ","End":"05:46.290","Text":"which is 881 seconds."},{"Start":"05:46.290 ","End":"05:50.165","Text":"Then when you plug everything in to the calculator,"},{"Start":"05:50.165 ","End":"05:56.625","Text":"you\u0027ll get 1.798 times 10^5"},{"Start":"05:56.625 ","End":"06:05.210","Text":"times c. Now we have a value for v Gamma and we want to know our energy needed."},{"Start":"06:05.210 ","End":"06:09.530","Text":"Our energy needed for the neutron is as we know,"},{"Start":"06:09.530 ","End":"06:12.845","Text":"the equation for the resting mass of the neutron,"},{"Start":"06:12.845 ","End":"06:19.675","Text":"which were given multiplied by c^2 and then multiplied by Gamma."},{"Start":"06:19.675 ","End":"06:24.755","Text":"Let\u0027s take a look. Our resting mass of the neutron is going to be"},{"Start":"06:24.755 ","End":"06:31.020","Text":"940 times 10^6,"},{"Start":"06:31.020 ","End":"06:35.790","Text":"because it\u0027s Mega, and then electron volts is the units and then divided"},{"Start":"06:35.790 ","End":"06:41.790","Text":"by c^2 and then multiplied by c^2 and then multiplied by Gamma."},{"Start":"06:41.790 ","End":"06:43.630","Text":"We can cross off our c^2,"},{"Start":"06:43.630 ","End":"06:45.275","Text":"they cancel each other out."},{"Start":"06:45.275 ","End":"06:50.375","Text":"Now the problem here is that we don\u0027t know what our Gamma is equal to."},{"Start":"06:50.375 ","End":"06:52.730","Text":"We know what our v Gamma is equal to,"},{"Start":"06:52.730 ","End":"06:56.060","Text":"but we don\u0027t know exactly what our Gamma is"},{"Start":"06:56.060 ","End":"07:00.920","Text":"and if we\u0027re going to look at our equation for what Gamma is,"},{"Start":"07:00.920 ","End":"07:09.215","Text":"so we\u0027ll see that our Gamma is equal to"},{"Start":"07:09.215 ","End":"07:18.715","Text":"1 over the square root of 1 minus v^2 divided by c^2."},{"Start":"07:18.715 ","End":"07:22.395","Text":"We know what v multiplied by Gamma is equal to,"},{"Start":"07:22.395 ","End":"07:24.320","Text":"but we don\u0027t know what our Gamma is equal to,"},{"Start":"07:24.320 ","End":"07:27.050","Text":"and we don\u0027t know what our v is equal to."},{"Start":"07:27.050 ","End":"07:29.620","Text":"We have 1 equation and 2 unknowns."},{"Start":"07:29.620 ","End":"07:32.420","Text":"We still can\u0027t find out what our energy is because"},{"Start":"07:32.420 ","End":"07:35.720","Text":"we\u0027re missing this information over here."},{"Start":"07:35.720 ","End":"07:41.995","Text":"Now what do we have to do is we have to try it and find a way around this over here."},{"Start":"07:41.995 ","End":"07:43.805","Text":"Let\u0027s try and think."},{"Start":"07:43.805 ","End":"07:53.665","Text":"We can remember that our equation for relativistic momentum is equal to mv Gamma."},{"Start":"07:53.665 ","End":"07:57.380","Text":"Now, that\u0027s our basic equation and we also saw from"},{"Start":"07:57.380 ","End":"07:59.690","Text":"our previous lesson that if we play around"},{"Start":"07:59.690 ","End":"08:02.330","Text":"with this equation and also our equation for energy,"},{"Start":"08:02.330 ","End":"08:09.530","Text":"another equation for relativistic momentum is equal to the square root of Gamma^2 minus"},{"Start":"08:09.530 ","End":"08:11.040","Text":"1"},{"Start":"08:17.660 ","End":"08:20.190","Text":"multiplied by mc."},{"Start":"08:20.190 ","End":"08:23.630","Text":"Now what we can do is we can divide both sides by our m and"},{"Start":"08:23.630 ","End":"08:27.305","Text":"then we\u0027re left with our v Gamma,"},{"Start":"08:27.305 ","End":"08:34.820","Text":"which is equal to the square root of Gamma^2 minus 1 multiplied by"},{"Start":"08:34.820 ","End":"08:43.925","Text":"c. Now we know that our v Gamma is equal to this equation over here."},{"Start":"08:43.925 ","End":"08:45.515","Text":"Let\u0027s write that out,"},{"Start":"08:45.515 ","End":"08:54.410","Text":"v Gamma so it\u0027s equal to 1.798 times 10^5 multiplied by c,"},{"Start":"08:54.410 ","End":"09:01.760","Text":"which is equal to the square root of Gamma^2 minus 1 multiplied by c. Again,"},{"Start":"09:01.760 ","End":"09:07.660","Text":"we can divide both sides by c. Now we have some number in this expression over here."},{"Start":"09:07.660 ","End":"09:13.340","Text":"Now when we isolate out our Gamma from this expression with simple algebra,"},{"Start":"09:13.340 ","End":"09:21.900","Text":"we\u0027ll see that our Gamma is approximately equal to 1.798 times 10^5."},{"Start":"09:23.210 ","End":"09:26.040","Text":"This is our value for Gamma."},{"Start":"09:26.040 ","End":"09:30.660","Text":"Now when we go back to our equation over here,"},{"Start":"09:31.370 ","End":"09:36.470","Text":"we can say that our energy is equal to our"},{"Start":"09:36.470 ","End":"09:42.695","Text":"940 times 10^6 multiplied by our Gamma,"},{"Start":"09:42.695 ","End":"09:50.670","Text":"which is going to be multiplied by 1.798 times 10^5."},{"Start":"09:51.400 ","End":"09:54.635","Text":"Now instead of writing 940,"},{"Start":"09:54.635 ","End":"10:02.895","Text":"we can actually times 10^6 we can just leave it as 940 Mega electron volts."},{"Start":"10:02.895 ","End":"10:04.350","Text":"Instead of times 10^6,"},{"Start":"10:04.350 ","End":"10:07.990","Text":"we\u0027ll just write this capital M. Now,"},{"Start":"10:07.990 ","End":"10:11.375","Text":"when we do this calculation in our calculator,"},{"Start":"10:11.375 ","End":"10:17.390","Text":"we\u0027ll get that our energy is therefore going to be equal"},{"Start":"10:17.390 ","End":"10:25.830","Text":"to 1.69 times 10^8 MeV."},{"Start":"10:28.310 ","End":"10:30.995","Text":"This is our final answer."},{"Start":"10:30.995 ","End":"10:33.680","Text":"As we can see, we were being asked to find the energy"},{"Start":"10:33.680 ","End":"10:36.470","Text":"needed for a neutron to reach Earth from"},{"Start":"10:36.470 ","End":"10:42.350","Text":"a certain distance given a certain lifespan and with a certain resting mass."},{"Start":"10:42.350 ","End":"10:50.075","Text":"How we did this was we worked out what the lifespan in light years meant in seconds."},{"Start":"10:50.075 ","End":"10:54.950","Text":"What we did is we found the amount of seconds that we have per year and then we"},{"Start":"10:54.950 ","End":"10:57.230","Text":"multiplied it by 5 because we\u0027re being told we have"},{"Start":"10:57.230 ","End":"10:59.810","Text":"five light years and then multiply it by c,"},{"Start":"10:59.810 ","End":"11:01.400","Text":"the speed of light."},{"Start":"11:01.400 ","End":"11:03.710","Text":"Then we use our equation for"},{"Start":"11:03.710 ","End":"11:08.900","Text":"a time dilation because we\u0027re being told that the neutron has a resting mass of this."},{"Start":"11:08.900 ","End":"11:12.425","Text":"We know that our neutron is at rest and it\u0027s located"},{"Start":"11:12.425 ","End":"11:16.163","Text":"at rest relative to its frame of reference."},{"Start":"11:16.163 ","End":"11:19.160","Text":"So that meant that we could use this equation over here."},{"Start":"11:19.160 ","End":"11:24.155","Text":"Then by rearranging, we got our equation for v Gamma."},{"Start":"11:24.155 ","End":"11:30.755","Text":"Then we realized that our energy equation required us to know our value for Gamma."},{"Start":"11:30.755 ","End":"11:34.550","Text":"What we did is that we used our two equations that we have for"},{"Start":"11:34.550 ","End":"11:39.860","Text":"relativistic momentum in order to isolate out our Gamma and then the end,"},{"Start":"11:39.860 ","End":"11:44.540","Text":"we found a value for Gamma to be approximately this."},{"Start":"11:44.540 ","End":"11:48.980","Text":"Then what we did is we substituted in a value for a Gamma into"},{"Start":"11:48.980 ","End":"11:54.470","Text":"our equation and multiplied it by our mass of the neutron,"},{"Start":"11:54.470 ","End":"11:58.575","Text":"multiplied by c^2 and we got our energy in the end."},{"Start":"11:58.575 ","End":"12:01.570","Text":"That\u0027s the end of this lesson."}],"ID":9564},{"Watched":false,"Name":"Exercise 9","Duration":"12m 58s","ChapterTopicVideoID":9263,"CourseChapterTopicPlaylistID":9398,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.280","Text":"Hello. In this question,"},{"Start":"00:02.280 ","End":"00:08.715","Text":"we\u0027re being told that a particle of mass m collides with another particle of mass 3m."},{"Start":"00:08.715 ","End":"00:14.670","Text":"The first particle had an initial energy of 5mc^2 before the collision."},{"Start":"00:14.670 ","End":"00:17.730","Text":"Then we\u0027re being told that the total momentum of"},{"Start":"00:17.730 ","End":"00:22.120","Text":"the two particles relative to the lab is 0."},{"Start":"00:22.250 ","End":"00:26.820","Text":"We can write P_T is equal to 0."},{"Start":"00:26.820 ","End":"00:28.425","Text":"As a result of the collision,"},{"Start":"00:28.425 ","End":"00:35.310","Text":"the two particles are destroyed and become a new third particle which is located at rest."},{"Start":"00:35.310 ","End":"00:36.795","Text":"Now the first question is,"},{"Start":"00:36.795 ","End":"00:40.005","Text":"what is the kinetic energy of particle number 1?"},{"Start":"00:40.005 ","End":"00:43.340","Text":"The second question is to calculate the Lorentz factor,"},{"Start":"00:43.340 ","End":"00:47.105","Text":"which is our Gamma of the particles before the collision,"},{"Start":"00:47.105 ","End":"00:49.865","Text":"and the kinetic energy of the second particle,"},{"Start":"00:49.865 ","End":"00:51.485","Text":"which is this one over here."},{"Start":"00:51.485 ","End":"00:52.985","Text":"Our third question is,"},{"Start":"00:52.985 ","End":"00:59.040","Text":"what is the mass of the third particle formed after the collision?"},{"Start":"00:59.040 ","End":"01:01.320","Text":"Let\u0027s see how we do this."},{"Start":"01:01.320 ","End":"01:05.505","Text":"Let\u0027s begin by answering question number 1."},{"Start":"01:05.505 ","End":"01:09.485","Text":"We know that our equation for"},{"Start":"01:09.485 ","End":"01:16.085","Text":"the total kinetic energy of a particle is going to be equal to the resting energy,"},{"Start":"01:16.085 ","End":"01:22.790","Text":"our E_0, our rest mass energy plus our kinetic energy."},{"Start":"01:22.790 ","End":"01:25.400","Text":"In here of particle number 1,"},{"Start":"01:25.400 ","End":"01:27.395","Text":"because that\u0027s what we\u0027re working out."},{"Start":"01:27.395 ","End":"01:32.075","Text":"We know that our total energy is what we are given."},{"Start":"01:32.075 ","End":"01:37.520","Text":"The energy of the particle before the collision is 5mc^2."},{"Start":"01:37.520 ","End":"01:44.345","Text":"We know that 5mc^2 is going to be equal to our rest mass energy plus our kinetic energy."},{"Start":"01:44.345 ","End":"01:52.915","Text":"Now our rest mass energy is always equal to mc^2 and then plus our E_k,"},{"Start":"01:52.915 ","End":"01:55.425","Text":"which is what we\u0027re trying to find."},{"Start":"01:55.425 ","End":"01:58.465","Text":"Once we isolate out our E_k,"},{"Start":"01:58.465 ","End":"02:00.170","Text":"which is our kinetic energy,"},{"Start":"02:00.170 ","End":"02:04.510","Text":"we\u0027ll see that it\u0027s equal to 4mc^2."},{"Start":"02:04.510 ","End":"02:06.765","Text":"That\u0027s the answer to question number 1."},{"Start":"02:06.765 ","End":"02:08.910","Text":"Now let\u0027s look at question number 2."},{"Start":"02:08.910 ","End":"02:10.855","Text":"In question number 2,"},{"Start":"02:10.855 ","End":"02:15.500","Text":"we\u0027re being asked to calculate the Lorentz factor of the particles before the collision,"},{"Start":"02:15.500 ","End":"02:18.530","Text":"so that\u0027s our particle 1 and 2 and"},{"Start":"02:18.530 ","End":"02:22.540","Text":"the kinetic energy of our second particle, this one over here."},{"Start":"02:22.540 ","End":"02:25.835","Text":"Let\u0027s write this out. We know that"},{"Start":"02:25.835 ","End":"02:32.220","Text":"the total energy of particle number 1 is equal to 5mc^2."},{"Start":"02:32.530 ","End":"02:37.460","Text":"Now, we know that an equation aside from this equation over here,"},{"Start":"02:37.460 ","End":"02:43.125","Text":"in order to express our total energy is equal to simply"},{"Start":"02:43.125 ","End":"02:49.605","Text":"our mc^2 multiplied by our Lorentz factor Gamma."},{"Start":"02:49.605 ","End":"02:52.460","Text":"This one is specifically for particle number 1."},{"Start":"02:52.460 ","End":"02:56.705","Text":"If we cancel our m\u0027s on both sides and our c^2 from both sides,"},{"Start":"02:56.705 ","End":"03:04.510","Text":"we\u0027ll see that our Gamma or Lorentz factor for a particle number 1 is simply equal to 5."},{"Start":"03:04.510 ","End":"03:11.660","Text":"Now we want to work out our Gamma for our second particle and also our kinetic energy."},{"Start":"03:11.660 ","End":"03:16.220","Text":"What we\u0027re going to do is we\u0027re going to use this fact over here that"},{"Start":"03:16.220 ","End":"03:22.465","Text":"the total momentum of the two particles relative to the lab is equal to 0."},{"Start":"03:22.465 ","End":"03:27.125","Text":"That means that 0 is equal to our momentum of particle 1"},{"Start":"03:27.125 ","End":"03:32.120","Text":"plus our momentum for particle 2."},{"Start":"03:32.120 ","End":"03:34.115","Text":"Then once we rearrange it,"},{"Start":"03:34.115 ","End":"03:39.890","Text":"we\u0027ll see that the size of both of the momentums are equal."},{"Start":"03:39.890 ","End":"03:44.870","Text":"The size of momentum 1 is equal to the size of momentum 2."},{"Start":"03:44.870 ","End":"03:48.950","Text":"Yes, one will have a negative sign and the other a positive sign because"},{"Start":"03:48.950 ","End":"03:53.610","Text":"they\u0027re pointing in different directions but the size of them is equal."},{"Start":"03:53.610 ","End":"03:57.500","Text":"Now we can use this equation over here,"},{"Start":"03:57.500 ","End":"04:01.850","Text":"which I\u0027m going to write in red because it\u0027s very useful and you should remember it."},{"Start":"04:01.850 ","End":"04:08.100","Text":"It means that the size of our momentum is equal to the square root of Gamma"},{"Start":"04:08.100 ","End":"04:14.585","Text":"squared minus 1 multiplied by the mass of the particle,"},{"Start":"04:14.585 ","End":"04:17.235","Text":"multiplied by c,"},{"Start":"04:17.235 ","End":"04:18.915","Text":"our speed of light."},{"Start":"04:18.915 ","End":"04:22.400","Text":"That means that if we can work out our momentum,"},{"Start":"04:22.400 ","End":"04:26.515","Text":"then we can isolate out our Gamma from over here."},{"Start":"04:26.515 ","End":"04:29.405","Text":"Let\u0027s see how we do this."},{"Start":"04:29.405 ","End":"04:35.824","Text":"So what I actually want to do is I want to find out my momentum for the second particle."},{"Start":"04:35.824 ","End":"04:41.205","Text":"Then I\u0027ll be able to find out my Gamma and I\u0027ll be able to find its kinetic energy."},{"Start":"04:41.205 ","End":"04:42.815","Text":"In order for me to do that,"},{"Start":"04:42.815 ","End":"04:49.305","Text":"I\u0027m first going to have to work out the momentum of my first particle. Let\u0027s see."},{"Start":"04:49.305 ","End":"04:53.900","Text":"My momentum of my first particle is going to be equal to the square root of"},{"Start":"04:53.900 ","End":"04:58.819","Text":"Gamma 1^2 minus 1 multiplied by its mass,"},{"Start":"04:58.819 ","End":"05:01.325","Text":"which is m, multiplied by c,"},{"Start":"05:01.325 ","End":"05:02.915","Text":"the speed of light."},{"Start":"05:02.915 ","End":"05:05.915","Text":"As we saw, my Gamma 1,"},{"Start":"05:05.915 ","End":"05:07.085","Text":"which is this over here,"},{"Start":"05:07.085 ","End":"05:08.735","Text":"is equal to 5."},{"Start":"05:08.735 ","End":"05:17.550","Text":"It\u0027s going to be equal to the square root of 25 minus 1 multiplied by mc."},{"Start":"05:18.130 ","End":"05:25.100","Text":"Then we know that the size of this is also equal to the size of"},{"Start":"05:25.100 ","End":"05:31.690","Text":"my momentum for particle number 2 from this equation over here."},{"Start":"05:31.690 ","End":"05:33.915","Text":"Let\u0027s just underline these."},{"Start":"05:33.915 ","End":"05:35.420","Text":"From these 2 equations,"},{"Start":"05:35.420 ","End":"05:37.385","Text":"we\u0027ve come up with this over here."},{"Start":"05:37.385 ","End":"05:41.270","Text":"Now what we can do is we can therefore see that"},{"Start":"05:41.270 ","End":"05:45.110","Text":"the size of our momentum for particle 2 is equal"},{"Start":"05:45.110 ","End":"05:53.675","Text":"to the square root of 24 multiplied by m multiplied by c. That\u0027s from this equation."},{"Start":"05:53.675 ","End":"05:59.785","Text":"But we also know that that is going to be equal to this same equation,"},{"Start":"05:59.785 ","End":"06:02.355","Text":"but for particle number 2."},{"Start":"06:02.355 ","End":"06:06.290","Text":"It\u0027s also equal to the square root of Gamma,"},{"Start":"06:06.290 ","End":"06:13.460","Text":"this time 2^2 minus 1 multiplied by our mass of particle number 2,"},{"Start":"06:13.460 ","End":"06:18.220","Text":"which is 3m multiplied by C."},{"Start":"06:18.220 ","End":"06:24.545","Text":"Now all we have to do is we have to rearrange this in order to isolate out our Gamma 2."},{"Start":"06:24.545 ","End":"06:30.410","Text":"We can see that our m\u0027s cross out from both sides and so does our c. Then we\u0027re"},{"Start":"06:30.410 ","End":"06:36.350","Text":"going to be left with the square root of 24 divided by 3."},{"Start":"06:36.350 ","End":"06:41.545","Text":"It\u0027s being equal to the square root of Gamma 2 squared minus 1."},{"Start":"06:41.545 ","End":"06:44.990","Text":"Now what we\u0027re going to do is we\u0027re going to square both sides."},{"Start":"06:44.990 ","End":"06:51.015","Text":"Then we\u0027ll get 24 divided by 9 is equal to Gamma 2^2 minus 1."},{"Start":"06:51.015 ","End":"06:53.810","Text":"When we add 1 to both sides and then"},{"Start":"06:53.810 ","End":"06:57.260","Text":"square root both sides in order to isolate out our Gamma 2,"},{"Start":"06:57.260 ","End":"07:06.645","Text":"we\u0027ll get that our Gamma 2 is simply equal to the square root of 11 divided by 3."},{"Start":"07:06.645 ","End":"07:10.415","Text":"Now we\u0027ve found what our Gamma 1 and Gamma 2 are equal to."},{"Start":"07:10.415 ","End":"07:15.835","Text":"The next part is that we want to find the kinetic energy of the second particle."},{"Start":"07:15.835 ","End":"07:19.670","Text":"We can use our equation for energy, which again,"},{"Start":"07:19.670 ","End":"07:25.485","Text":"I\u0027m writing in red because you should include this in your equation sheets."},{"Start":"07:25.485 ","End":"07:32.745","Text":"That is equal to mc^2 multiplied by Gamma minus 1."},{"Start":"07:32.745 ","End":"07:38.465","Text":"Now, in order to work out our kinetic energy for our particle number 2,"},{"Start":"07:38.465 ","End":"07:46.785","Text":"its mass is 3m multiplied by c^2 multiplied by its Gamma, which is Gamma 2."},{"Start":"07:46.785 ","End":"07:56.340","Text":"That\u0027s the square root of 11 divided by 3 minus 1. There we have it."},{"Start":"07:56.340 ","End":"07:58.695","Text":"That\u0027s our second answer."},{"Start":"07:58.695 ","End":"08:02.340","Text":"We found our Lorentz transform for our particle number 1,"},{"Start":"08:02.340 ","End":"08:03.625","Text":"particle number 2,"},{"Start":"08:03.625 ","End":"08:07.000","Text":"and our kinetic energy for our particle number 2."},{"Start":"08:07.000 ","End":"08:09.385","Text":"Now let\u0027s look at question number 3."},{"Start":"08:09.385 ","End":"08:13.505","Text":"What is the mass of the third particle formed after the collision?"},{"Start":"08:13.505 ","End":"08:15.185","Text":"In order to solve this,"},{"Start":"08:15.185 ","End":"08:22.410","Text":"we\u0027re going to take that there is conservation of momentum and of energy."},{"Start":"08:22.410 ","End":"08:26.710","Text":"This is something that we\u0027re going to be using in all of the questions from now on."},{"Start":"08:26.710 ","End":"08:30.695","Text":"We\u0027re always going to have conservation of energy and of momentum."},{"Start":"08:30.695 ","End":"08:33.275","Text":"Right now whilst we\u0027re dealing with relativity,"},{"Start":"08:33.275 ","End":"08:36.960","Text":"we\u0027re not going to be dealing with forces."},{"Start":"08:36.960 ","End":"08:42.110","Text":"Let\u0027s go back to our question where we\u0027re being told that the total momentum of"},{"Start":"08:42.110 ","End":"08:46.910","Text":"the two particles before the collision relative to the lab is equal to 0."},{"Start":"08:46.910 ","End":"08:49.370","Text":"Because of conservation of momentum,"},{"Start":"08:49.370 ","End":"08:53.165","Text":"that means if the momentum before was equal to 0,"},{"Start":"08:53.165 ","End":"09:01.385","Text":"that means that our P total final after the collision is also going to be equal to 0."},{"Start":"09:01.385 ","End":"09:05.600","Text":"We\u0027re also being told that as a result of the collision,"},{"Start":"09:05.600 ","End":"09:11.025","Text":"the two particles become a new third particle at rest."},{"Start":"09:11.025 ","End":"09:13.395","Text":"Let\u0027s remember all of that."},{"Start":"09:13.395 ","End":"09:17.030","Text":"We\u0027re dealing with conservation of momentum and energy."},{"Start":"09:17.030 ","End":"09:21.215","Text":"That means that our total momentum afterwards,"},{"Start":"09:21.215 ","End":"09:24.185","Text":"after the collision is going to be equal to 0."},{"Start":"09:24.185 ","End":"09:28.115","Text":"That means that the momentum of particle 3,"},{"Start":"09:28.115 ","End":"09:31.355","Text":"because that\u0027s the only particle in the system after the collision,"},{"Start":"09:31.355 ","End":"09:34.950","Text":"is going to be equal to 0."},{"Start":"09:35.150 ","End":"09:38.610","Text":"Let\u0027s see what that means for our Gamma."},{"Start":"09:38.610 ","End":"09:45.660","Text":"Now, our Gamma we know is going to be equal to."},{"Start":"09:45.660 ","End":"09:49.935","Text":"Let\u0027s see what this means for our Gamma."},{"Start":"09:49.935 ","End":"09:51.890","Text":"Our Gamma, as we know,"},{"Start":"09:51.890 ","End":"10:00.180","Text":"is equal to 1 divided by the square root of 1 minus v divided by c^2."},{"Start":"10:02.530 ","End":"10:09.140","Text":"Because we were also told that our third particle is at rest after the collision,"},{"Start":"10:09.140 ","End":"10:12.095","Text":"that means that the V,"},{"Start":"10:12.095 ","End":"10:16.090","Text":"the velocity of our third particle is going to be equal to 0."},{"Start":"10:16.090 ","End":"10:19.550","Text":"If the velocity of our third particle is equal to 0,"},{"Start":"10:19.550 ","End":"10:23.195","Text":"that means that our Gamma for our third particle,"},{"Start":"10:23.195 ","End":"10:25.770","Text":"once we substitute it in,"},{"Start":"10:25.930 ","End":"10:30.980","Text":"is going to be simply equal to 1."},{"Start":"10:30.980 ","End":"10:36.050","Text":"Now, the next thing that we want to look at is what is the energy?"},{"Start":"10:36.050 ","End":"10:38.495","Text":"Our equation for energy,"},{"Start":"10:38.495 ","End":"10:41.000","Text":"the total energy of particle Number 3."},{"Start":"10:41.000 ","End":"10:45.245","Text":"If we go back to our equation over here,"},{"Start":"10:45.245 ","End":"10:48.500","Text":"we can see that our total energy is equal to"},{"Start":"10:48.500 ","End":"10:52.385","Text":"our rest mass energy plus our kinetic energy."},{"Start":"10:52.385 ","End":"10:59.540","Text":"It\u0027s equal to our rest mass energy plus our kinetic energy."},{"Start":"10:59.540 ","End":"11:02.660","Text":"Now we know that because our third particle is at rest,"},{"Start":"11:02.660 ","End":"11:04.338","Text":"it was given to us in the question,"},{"Start":"11:04.338 ","End":"11:06.200","Text":"so it\u0027s going to have no kinetic energy."},{"Start":"11:06.200 ","End":"11:07.580","Text":"This is going to be equal to 0."},{"Start":"11:07.580 ","End":"11:12.770","Text":"That means that our total energy is just equal to our rest mass energy,"},{"Start":"11:12.770 ","End":"11:19.010","Text":"which we know is simply equal to the mass of our third particle multiplied"},{"Start":"11:19.010 ","End":"11:26.210","Text":"by c^2 where I\u0027m reminding you our m3 is what we\u0027re trying to find."},{"Start":"11:26.210 ","End":"11:27.920","Text":"Now, in order to find this,"},{"Start":"11:27.920 ","End":"11:31.615","Text":"we\u0027re going to use the idea of conservation of energy."},{"Start":"11:31.615 ","End":"11:34.715","Text":"That means that our energy after,"},{"Start":"11:34.715 ","End":"11:37.850","Text":"so our energy of a third particle is going to be equal"},{"Start":"11:37.850 ","End":"11:41.060","Text":"to the sum of the energies before the collision,"},{"Start":"11:41.060 ","End":"11:45.280","Text":"which is the sum of the energy of particle 1 and of particle 2."},{"Start":"11:45.280 ","End":"11:50.495","Text":"Let\u0027s write that out. That means that our E_T_3 is going to be equal to"},{"Start":"11:50.495 ","End":"11:56.390","Text":"our E_T of particle 1 plus our E_T of particle 2."},{"Start":"11:56.390 ","End":"12:04.455","Text":"We already know that our E_T of particle 1 is equal to 5mc^2."},{"Start":"12:04.455 ","End":"12:06.360","Text":"That was given to us in the question."},{"Start":"12:06.360 ","End":"12:09.700","Text":"Then our E_T of particle number 2,"},{"Start":"12:09.700 ","End":"12:12.920","Text":"we already know it\u0027s going to be the mass,"},{"Start":"12:12.920 ","End":"12:20.565","Text":"so that\u0027s going to be 3m multiplied by c^2 and then multiplied by Gamma."},{"Start":"12:20.565 ","End":"12:22.940","Text":"Our Gamma for particle number 2,"},{"Start":"12:22.940 ","End":"12:24.950","Text":"we already found in the previous question."},{"Start":"12:24.950 ","End":"12:29.410","Text":"It\u0027s equal to the square root of 11 divided by 3."},{"Start":"12:29.410 ","End":"12:33.690","Text":"E_T_3 is our m3 multiplied by c^2,"},{"Start":"12:33.690 ","End":"12:35.190","Text":"where m3 is our unknown."},{"Start":"12:35.190 ","End":"12:38.900","Text":"Now what we can do is we can divide both sides by c^2."},{"Start":"12:38.900 ","End":"12:42.725","Text":"Then all we have to do is isolate out our m3."},{"Start":"12:42.725 ","End":"12:51.735","Text":"What we\u0027ll get is that our mass of particle number 3 is equal to 6.91 multiplied by m,"},{"Start":"12:51.735 ","End":"12:56.165","Text":"where m I\u0027m reminding is the mass of the first particle."},{"Start":"12:56.165 ","End":"12:59.460","Text":"That\u0027s the end of this exercise."}],"ID":9565},{"Watched":false,"Name":"Momentum And Energy Transformation","Duration":"17m 15s","ChapterTopicVideoID":9271,"CourseChapterTopicPlaylistID":9398,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.890","Text":"In this lesson, we\u0027re going to speak about the transformation of energy and momentum."},{"Start":"00:04.890 ","End":"00:10.830","Text":"Up until now, we\u0027ve been speaking of energy and momentum in 1 reference frame."},{"Start":"00:10.830 ","End":"00:13.830","Text":"We haven\u0027t spoken about energy and"},{"Start":"00:13.830 ","End":"00:17.400","Text":"momentum when we\u0027re dealing with different reference frames."},{"Start":"00:17.400 ","End":"00:20.520","Text":"What we\u0027re going to do now is we\u0027re going to look in"},{"Start":"00:20.520 ","End":"00:24.885","Text":"the same way as how we looked when we were dealing with the Lawrence transforms,"},{"Start":"00:24.885 ","End":"00:31.490","Text":"where we have a frame of reference s and another frame of reference s tag."},{"Start":"00:31.490 ","End":"00:35.430","Text":"We\u0027re going to say that there\u0027s some body with"},{"Start":"00:35.430 ","End":"00:40.930","Text":"some momentum p and then energy in our s reference frame."},{"Start":"00:40.930 ","End":"00:43.610","Text":"We\u0027re going to see what an observer in"},{"Start":"00:43.610 ","End":"00:49.950","Text":"the s tag reference frame will see as the particle\u0027s momentum and energy."},{"Start":"00:49.950 ","End":"00:58.045","Text":"What the observer in s tag we\u0027ll see is a momentum of p tag and an energy of E tag."},{"Start":"00:58.045 ","End":"01:02.330","Text":"These are our equations for energy and for momentum."},{"Start":"01:02.330 ","End":"01:06.110","Text":"Now, notice that they\u0027re exact same equations"},{"Start":"01:06.110 ","End":"01:10.685","Text":"as we have for our transformation between our position and our time."},{"Start":"01:10.685 ","End":"01:14.960","Text":"The only difference is that instead of our x for our position,"},{"Start":"01:14.960 ","End":"01:19.530","Text":"we have E and instead of our t we have px."},{"Start":"01:19.530 ","End":"01:26.005","Text":"As we can see, our py and pz have the same relationship as our y and our z."},{"Start":"01:26.005 ","End":"01:31.400","Text":"This relationship will maybe make it easier for you to remember these equations."},{"Start":"01:31.400 ","End":"01:34.625","Text":"However, they should be written in your equation sheets."},{"Start":"01:34.625 ","End":"01:39.710","Text":"Now, an important note is that I\u0027m speaking about my Gamma_0 over here"},{"Start":"01:39.710 ","End":"01:44.760","Text":"and that has a relationship with my v_0 over here,"},{"Start":"01:44.760 ","End":"01:53.110","Text":"where my v_0 is to do with the velocity of my observer if we remember these two graphs."},{"Start":"01:53.110 ","End":"02:00.290","Text":"My v_0 or my Gamma_0 are to do with the velocity of my observer which is moving."},{"Start":"02:00.290 ","End":"02:03.635","Text":"Now, it\u0027s very important to differentiate between"},{"Start":"02:03.635 ","End":"02:10.100","Text":"our Gamma of the body itself which is moving and our Gamma_0 or v_0,"},{"Start":"02:10.100 ","End":"02:12.505","Text":"which is of the observer."},{"Start":"02:12.505 ","End":"02:18.815","Text":"Let\u0027s imagine that we\u0027re in our lab and we\u0027re looking at some particle which is moving."},{"Start":"02:18.815 ","End":"02:27.680","Text":"Then we know that its momentum p is going to be equal to mv Gamma,"},{"Start":"02:27.680 ","End":"02:31.745","Text":"where this v over here is the velocity of the particle."},{"Start":"02:31.745 ","End":"02:34.880","Text":"This is our equation for momentum of the particle"},{"Start":"02:34.880 ","End":"02:38.120","Text":"and we know that another way of writing this equation"},{"Start":"02:38.120 ","End":"02:44.590","Text":"is mc multiplied by the square root of Gamma^2 minus 1."},{"Start":"02:44.590 ","End":"02:46.895","Text":"Now, if I have my observer,"},{"Start":"02:46.895 ","End":"02:48.515","Text":"which is looking at this,"},{"Start":"02:48.515 ","End":"02:51.530","Text":"and my observer is moving with a velocity of v_0,"},{"Start":"02:51.530 ","End":"03:00.365","Text":"then I\u0027m going to have to utilize my Gamma_0 over here and not my Gamma of the particle."},{"Start":"03:00.365 ","End":"03:02.840","Text":"Unless my observer is in"},{"Start":"03:02.840 ","End":"03:06.530","Text":"the same reference frame as the particle but in most of the time,"},{"Start":"03:06.530 ","End":"03:10.254","Text":"we\u0027re going to be dealing with different frames of reference."},{"Start":"03:10.254 ","End":"03:15.440","Text":"We\u0027re going to want to move or translate or transform"},{"Start":"03:15.440 ","End":"03:21.635","Text":"our momentum from this reference frame to the reference frame of the observer."},{"Start":"03:21.635 ","End":"03:22.850","Text":"Then when we do this,"},{"Start":"03:22.850 ","End":"03:26.930","Text":"we\u0027re going to have the Gamma that moves from 1 reference frame to the other,"},{"Start":"03:26.930 ","End":"03:29.780","Text":"and the Gamma of the particle."},{"Start":"03:29.780 ","End":"03:31.370","Text":"As a rule of thumb,"},{"Start":"03:31.370 ","End":"03:34.850","Text":"we\u0027re going to symbolize the Gamma when we\u0027re moving from"},{"Start":"03:34.850 ","End":"03:38.948","Text":"1 reference frame to another as Gamma_0."},{"Start":"03:38.948 ","End":"03:42.215","Text":"Our Gamma, which is relating to our particle,"},{"Start":"03:42.215 ","End":"03:45.460","Text":"is just going to be Gamma with no subscript."},{"Start":"03:45.460 ","End":"03:48.720","Text":"That\u0027s enough about the transform itself."},{"Start":"03:48.720 ","End":"03:52.280","Text":"Now, we\u0027re going to speak about another super important thing,"},{"Start":"03:52.280 ","End":"03:54.940","Text":"which is called the 4-vector."},{"Start":"03:54.940 ","End":"03:58.615","Text":"The 4-vector is written like so."},{"Start":"03:58.615 ","End":"04:03.600","Text":"We have our px, our py, and our pz,"},{"Start":"04:03.600 ","End":"04:13.815","Text":"and then E divided by c. What we do is if we have our frame of reference in the lab,"},{"Start":"04:13.815 ","End":"04:17.760","Text":"our s, and we have some particle moving."},{"Start":"04:17.760 ","End":"04:21.544","Text":"What do we do for this particle is we work out its momentum"},{"Start":"04:21.544 ","End":"04:25.505","Text":"and separate the momentum out into the different components,"},{"Start":"04:25.505 ","End":"04:27.590","Text":"the x, y, and z components."},{"Start":"04:27.590 ","End":"04:32.780","Text":"Then our final term in the vector is our E divided by c. Now,"},{"Start":"04:32.780 ","End":"04:36.710","Text":"we\u0027re soon going to understand what this means, again,"},{"Start":"04:36.710 ","End":"04:40.490","Text":"why this is important but if we\u0027re going to look at the units,"},{"Start":"04:40.490 ","End":"04:44.900","Text":"so our units are still going to be the units of momentum."},{"Start":"04:44.900 ","End":"04:48.230","Text":"This is a 4-vector momentum."},{"Start":"04:48.230 ","End":"04:54.515","Text":"Now, of course, this 4-vector is in reference only to this frame of reference."},{"Start":"04:54.515 ","End":"04:59.432","Text":"If I move to my other frame of reference, my s tag."},{"Start":"04:59.432 ","End":"05:03.230","Text":"So what I\u0027m going to have is I\u0027ll have my px tag,"},{"Start":"05:03.230 ","End":"05:04.940","Text":"py tag, pz tag,"},{"Start":"05:04.940 ","End":"05:13.010","Text":"and E tag divided by c. My 4-vector without the tag is in"},{"Start":"05:13.010 ","End":"05:16.940","Text":"reference to my particle in its frame of"},{"Start":"05:16.940 ","End":"05:22.630","Text":"reference and my 4-vector with the tags is my transform,"},{"Start":"05:22.630 ","End":"05:27.335","Text":"so it\u0027s my 4-vector relative to this new reference frame."},{"Start":"05:27.335 ","End":"05:29.090","Text":"Now, an interesting traits of"},{"Start":"05:29.090 ","End":"05:33.620","Text":"this 4-vector is that if I work out the size of the 4-vector,"},{"Start":"05:33.620 ","End":"05:35.464","Text":"now it\u0027s not exactly the size,"},{"Start":"05:35.464 ","End":"05:38.495","Text":"but let\u0027s say the size of the 4-vector,"},{"Start":"05:38.495 ","End":"05:42.955","Text":"then what we\u0027ll see is that I will come up with a constant value."},{"Start":"05:42.955 ","End":"05:46.385","Text":"This constant is going to be the same for"},{"Start":"05:46.385 ","End":"05:51.415","Text":"my s frame of reference and also for my s tag frame of reference."},{"Start":"05:51.415 ","End":"05:53.850","Text":"Let\u0027s take a look at what that means."},{"Start":"05:53.850 ","End":"05:56.165","Text":"I\u0027m going to take the size of my 4-vector,"},{"Start":"05:56.165 ","End":"06:00.140","Text":"which I said isn\u0027t exactly the size and this is why."},{"Start":"06:00.140 ","End":"06:07.335","Text":"I\u0027ll start with my px^2 plus my py^2 plus my pz^2."},{"Start":"06:07.335 ","End":"06:09.425","Text":"Then if I was taking the size,"},{"Start":"06:09.425 ","End":"06:13.175","Text":"I would have to do plus my E divided by c^2."},{"Start":"06:13.175 ","End":"06:17.030","Text":"Now, it\u0027s not exactly the size because here we\u0027re going to"},{"Start":"06:17.030 ","End":"06:21.155","Text":"have a minus for this term over here."},{"Start":"06:21.155 ","End":"06:23.660","Text":"It would be size if I had a plus over here,"},{"Start":"06:23.660 ","End":"06:26.890","Text":"but because it\u0027s a minus, it\u0027s not exactly the size."},{"Start":"06:26.890 ","End":"06:34.940","Text":"We\u0027re going to get that this is equal to our m_0^2 multiplied by c^2,"},{"Start":"06:34.940 ","End":"06:38.275","Text":"and this is a constant."},{"Start":"06:38.275 ","End":"06:42.510","Text":"This comes from our equation, which we know."},{"Start":"06:42.510 ","End":"06:48.670","Text":"Let\u0027s do it here, it says that our e^2 is equal to"},{"Start":"06:49.010 ","End":"06:57.195","Text":"p^2c^2 plus our m_0^2 multiplied by c^4,"},{"Start":"06:57.195 ","End":"07:01.210","Text":"where our m_0 over here is the mass of our body."},{"Start":"07:01.210 ","End":"07:05.330","Text":"We see that we\u0027ve gotten into the exact same equation."},{"Start":"07:05.330 ","End":"07:07.520","Text":"We can see that this is in fact going to be a"},{"Start":"07:07.520 ","End":"07:09.770","Text":"constant and the exact same thing would have"},{"Start":"07:09.770 ","End":"07:14.980","Text":"happened if I would\u0027ve calculated this in my s tag reference frame."},{"Start":"07:14.980 ","End":"07:18.185","Text":"If we imagine, we added in the tags over here,"},{"Start":"07:18.185 ","End":"07:21.245","Text":"so if we had px tag^2, py tag^2,"},{"Start":"07:21.245 ","End":"07:30.538","Text":"pz tags^2 minus E tag divided by c^2 so we will still get this exact equation over here."},{"Start":"07:30.538 ","End":"07:33.020","Text":"It\u0027s still equal to a constant."},{"Start":"07:33.020 ","End":"07:40.040","Text":"Because this equation is the same not just when we\u0027re in our reference frame for the lab,"},{"Start":"07:40.040 ","End":"07:43.075","Text":"but with every single reference frame."},{"Start":"07:43.075 ","End":"07:46.125","Text":"This is the same equation."},{"Start":"07:46.125 ","End":"07:49.985","Text":"That means that this equation is always going to be a constant and it\u0027s"},{"Start":"07:49.985 ","End":"07:54.385","Text":"always going to be equal to our m^2 multiplied by c^2."},{"Start":"07:54.385 ","End":"07:56.430","Text":"What I said isn\u0027t exactly correct,"},{"Start":"07:56.430 ","End":"08:01.890","Text":"it\u0027s always going to be equal to m^2c^2 if we only have 1 body."},{"Start":"08:01.890 ","End":"08:06.185","Text":"Soon, we\u0027ll speak about the case if we have more bodies but"},{"Start":"08:06.185 ","End":"08:08.540","Text":"the most important thing to realize is that"},{"Start":"08:08.540 ","End":"08:11.736","Text":"this equation is always going to be a constant."},{"Start":"08:11.736 ","End":"08:15.155","Text":"It doesn\u0027t matter in which reference frame I am."},{"Start":"08:15.155 ","End":"08:16.655","Text":"Another important point,"},{"Start":"08:16.655 ","End":"08:22.295","Text":"when I say that it\u0027s always constant then that means between the 2 reference frames."},{"Start":"08:22.295 ","End":"08:26.095","Text":"It could be that my energy and my momentum is changing,"},{"Start":"08:26.095 ","End":"08:29.075","Text":"not to deal with which reference frame."},{"Start":"08:29.075 ","End":"08:33.440","Text":"However, even if we\u0027re in the s reference frame or"},{"Start":"08:33.440 ","End":"08:37.645","Text":"in the s tag reference frame at any given moment in time,"},{"Start":"08:37.645 ","End":"08:41.450","Text":"if the person in the s reference frame were to look at"},{"Start":"08:41.450 ","End":"08:45.672","Text":"the energy and the momentum of the particle."},{"Start":"08:45.672 ","End":"08:49.280","Text":"So the person standing in the S tag reference frame would"},{"Start":"08:49.280 ","End":"08:53.480","Text":"record the exact same value for energy and momentum."},{"Start":"08:53.480 ","End":"08:56.375","Text":"This was for one body."},{"Start":"08:56.375 ","End":"09:00.650","Text":"We\u0027ve understood that this equation remains constant when we have one body."},{"Start":"09:00.650 ","End":"09:06.424","Text":"Now the interesting thing is that even if we have multiple bodies in our system,"},{"Start":"09:06.424 ","End":"09:11.945","Text":"this 4-vector is still going to remain constant."},{"Start":"09:11.945 ","End":"09:16.309","Text":"Let\u0027s imagine for one second that we have two bodies."},{"Start":"09:16.309 ","End":"09:26.740","Text":"Here is our body Number 1 and then let\u0027s say that over here is our second body."},{"Start":"09:26.740 ","End":"09:33.010","Text":"Now what I\u0027m going to do is I\u0027m going to write my 4-vector for the two bodies together."},{"Start":"09:33.010 ","End":"09:35.699","Text":"What that means is that in my vector,"},{"Start":"09:35.699 ","End":"09:38.285","Text":"I\u0027m going to sum each component."},{"Start":"09:38.285 ","End":"09:40.280","Text":"Instead of my P_x,"},{"Start":"09:40.280 ","End":"09:45.695","Text":"I\u0027m going to have my P_x for body Number 1 plus my P_x for body Number 2,"},{"Start":"09:45.695 ","End":"09:50.165","Text":"and so on and so forth for the rest of the components of my 4-vector."},{"Start":"09:50.165 ","End":"09:55.808","Text":"I\u0027ll have my P_1x plus my P_2x,"},{"Start":"09:55.808 ","End":"10:00.815","Text":"then I\u0027ll have my P_1y plus my P_2y."},{"Start":"10:00.815 ","End":"10:08.795","Text":"Then I have my P_1z plus my P_2z,"},{"Start":"10:08.795 ","End":"10:11.090","Text":"and then in my next component,"},{"Start":"10:11.090 ","End":"10:17.180","Text":"I will have E_1 divided by c plus my E_2 divided by"},{"Start":"10:17.180 ","End":"10:24.245","Text":"c. What I have written in black is my 4-vector for my S frame of reference."},{"Start":"10:24.245 ","End":"10:26.975","Text":"Now, what if I\u0027m in my S tag frame of reference?"},{"Start":"10:26.975 ","End":"10:30.515","Text":"Then all I have to do is I\u0027ll have to add in my tags"},{"Start":"10:30.515 ","End":"10:34.520","Text":"over here but it will be the exact same formula,"},{"Start":"10:34.520 ","End":"10:36.710","Text":"just my values will be different."},{"Start":"10:36.710 ","End":"10:40.160","Text":"My P_1x in S frame of reference will be"},{"Start":"10:40.160 ","End":"10:44.675","Text":"different to my P_1x in my S tag frame of reference."},{"Start":"10:44.675 ","End":"10:49.850","Text":"Now what happens is if I work out the size of this 4-vector,"},{"Start":"10:49.850 ","End":"10:52.220","Text":"remember with my minus over here,"},{"Start":"10:52.220 ","End":"11:00.560","Text":"so that will be my P_1x plus P_2x squared,"},{"Start":"11:00.560 ","End":"11:07.580","Text":"plus my P_1y plus P_2y squared,"},{"Start":"11:07.580 ","End":"11:15.470","Text":"plus my P_1z plus P_2z squared, minus,"},{"Start":"11:15.470 ","End":"11:17.810","Text":"remember here there\u0027s a minus,"},{"Start":"11:17.810 ","End":"11:23.600","Text":"E_1 plus E_2 divided by c squared,"},{"Start":"11:23.600 ","End":"11:26.000","Text":"because they have a common denominator."},{"Start":"11:26.000 ","End":"11:32.000","Text":"This is still going to be equal to a constant."},{"Start":"11:32.000 ","End":"11:37.319","Text":"It\u0027s exactly identical to what I had when I had one body."},{"Start":"11:37.319 ","End":"11:39.035","Text":"If I have two bodies,"},{"Start":"11:39.035 ","End":"11:44.420","Text":"it\u0027s the same formula and I come to a constant."},{"Start":"11:44.420 ","End":"11:46.820","Text":"Again, when I say constant,"},{"Start":"11:46.820 ","End":"11:50.060","Text":"then I mean from either frame of reference,"},{"Start":"11:50.060 ","End":"11:56.703","Text":"I\u0027ll have the same value for my constant but that doesn\u0027t mean in time."},{"Start":"11:56.703 ","End":"11:58.820","Text":"This is going to be equal to a constant,"},{"Start":"11:58.820 ","End":"12:01.595","Text":"it\u0027s for a specific moment in time."},{"Start":"12:01.595 ","End":"12:06.710","Text":"If my observer in S and my observer in S tag write down the values"},{"Start":"12:06.710 ","End":"12:12.920","Text":"for my momentum and my energy and then plug them into this formula,"},{"Start":"12:12.920 ","End":"12:18.215","Text":"they\u0027ll both get the same constant for that exact moment of time."},{"Start":"12:18.215 ","End":"12:23.210","Text":"Obviously, when I\u0027m looking from the different reference frames,"},{"Start":"12:23.210 ","End":"12:27.650","Text":"my energy in my reference frame S and my momentum in"},{"Start":"12:27.650 ","End":"12:32.675","Text":"that reference frame will be different to my energy and my momentum in my S tag."},{"Start":"12:32.675 ","End":"12:35.615","Text":"However, in this equation,"},{"Start":"12:35.615 ","End":"12:40.640","Text":"the size of my 4-vector will lead to this same answer."},{"Start":"12:40.640 ","End":"12:45.725","Text":"Now the only difference to note is that even though we do get this constant,"},{"Start":"12:45.725 ","End":"12:53.705","Text":"it\u0027s not going to be equal to our m^2c^2 because our m^2c^2 is specifically for one body."},{"Start":"12:53.705 ","End":"12:57.590","Text":"I\u0027m not going to go into the equation that it will be equal"},{"Start":"12:57.590 ","End":"13:02.330","Text":"to because we have to work it out for two bodies or more bodies,"},{"Start":"13:02.330 ","End":"13:04.265","Text":"but just know it\u0027s not equal to that,"},{"Start":"13:04.265 ","End":"13:08.825","Text":"but it still is equal to some constant number."},{"Start":"13:08.825 ","End":"13:14.960","Text":"Now, let\u0027s take a look at the idea for why this exists."},{"Start":"13:14.960 ","End":"13:17.930","Text":"Now, of course, I can use these equations to work out"},{"Start":"13:17.930 ","End":"13:22.625","Text":"my energy and my momentum for different frames of reference."},{"Start":"13:22.625 ","End":"13:29.308","Text":"It will work out to be a lot of work and sometimes the answers will be very complicated."},{"Start":"13:29.308 ","End":"13:35.995","Text":"Instead what I can do is I can work out the size of my 4-vector in one frame of reference"},{"Start":"13:35.995 ","End":"13:43.520","Text":"and then compare it to the size of my 4-vector in my other frame of reference."},{"Start":"13:43.520 ","End":"13:47.840","Text":"That way my calculations are significantly simplified."},{"Start":"13:47.840 ","End":"13:50.840","Text":"Imagine we\u0027re in the frame of reference of our lab."},{"Start":"13:50.840 ","End":"13:52.490","Text":"Then in my lab,"},{"Start":"13:52.490 ","End":"13:57.200","Text":"I\u0027ve worked out my momentum and all the components,"},{"Start":"13:57.200 ","End":"13:58.730","Text":"my x, y, and z components,"},{"Start":"13:58.730 ","End":"14:01.985","Text":"and my energy for that particle."},{"Start":"14:01.985 ","End":"14:09.530","Text":"Then what I\u0027ll do is I\u0027ll compare that to what I get in my S tag frame of reference."},{"Start":"14:09.530 ","End":"14:13.940","Text":"What I\u0027ll do is I\u0027ll have my P_1x plus my P_2x squared,"},{"Start":"14:13.940 ","End":"14:18.239","Text":"and so on and so forth for my y and z components minus my energies."},{"Start":"14:18.239 ","End":"14:22.985","Text":"That is equal to some constant and this is in my S frame of reference."},{"Start":"14:22.985 ","End":"14:27.440","Text":"Then I\u0027ll do the exact same thing in my S tag frame of reference."},{"Start":"14:27.440 ","End":"14:30.995","Text":"I\u0027m reminding you it\u0027s going to be this exact equation,"},{"Start":"14:30.995 ","End":"14:32.735","Text":"but with tags,"},{"Start":"14:32.735 ","End":"14:35.075","Text":"and then I\u0027ll compare the two."},{"Start":"14:35.075 ","End":"14:39.440","Text":"I\u0027ll say that that is equal to and then I\u0027ll have my P_1x tag"},{"Start":"14:39.440 ","End":"14:44.090","Text":"plus my P_2x tag squared plus."},{"Start":"14:44.090 ","End":"14:48.499","Text":"Then the same for my ys and my zs and my energies."},{"Start":"14:48.499 ","End":"14:51.335","Text":"Then I can get my different values."},{"Start":"14:51.335 ","End":"14:55.670","Text":"Now it\u0027s very important to note that I can only compare my 4-vectors"},{"Start":"14:55.670 ","End":"15:00.560","Text":"between different reference frames when I work out the size of the 4-vectors."},{"Start":"15:00.560 ","End":"15:05.000","Text":"In other words, I can only compare the size of my 4-vectors in"},{"Start":"15:05.000 ","End":"15:10.310","Text":"the reference frames and not the actual vector itself."},{"Start":"15:10.310 ","End":"15:16.880","Text":"The important things to know from this lesson are these equations that we have over"},{"Start":"15:16.880 ","End":"15:20.840","Text":"here for our energy and our momentum when"},{"Start":"15:20.840 ","End":"15:25.190","Text":"we\u0027re doing a transformation from different frames of reference."},{"Start":"15:25.190 ","End":"15:32.120","Text":"The next thing is to know that sometimes these calculations can be long and complicated,"},{"Start":"15:32.120 ","End":"15:34.235","Text":"so you need to know another way,"},{"Start":"15:34.235 ","End":"15:36.110","Text":"which is via the 4-vector."},{"Start":"15:36.110 ","End":"15:37.580","Text":"In the 4-vector,"},{"Start":"15:37.580 ","End":"15:40.775","Text":"we can work out for some frame of reference,"},{"Start":"15:40.775 ","End":"15:42.710","Text":"our momentum in the x,"},{"Start":"15:42.710 ","End":"15:44.045","Text":"y, and z component,"},{"Start":"15:44.045 ","End":"15:47.150","Text":"and our energy of the particle divided by c,"},{"Start":"15:47.150 ","End":"15:49.684","Text":"and that\u0027s what\u0027s called our 4-vector."},{"Start":"15:49.684 ","End":"15:53.405","Text":"Then when we find the size of our 4-vector;"},{"Start":"15:53.405 ","End":"15:58.730","Text":"where it\u0027s not exactly the size because here before our E divided by c squared,"},{"Start":"15:58.730 ","End":"16:01.956","Text":"we have a negative and not a positive;"},{"Start":"16:01.956 ","End":"16:06.800","Text":"so when we take the size of our 4-vector,"},{"Start":"16:06.800 ","End":"16:11.330","Text":"we\u0027re going to come up with some constant."},{"Start":"16:11.330 ","End":"16:15.035","Text":"Then if we take the exact same 4-vector"},{"Start":"16:15.035 ","End":"16:18.620","Text":"for the same particle but in a different frame of reference,"},{"Start":"16:18.620 ","End":"16:21.365","Text":"so from our S tag frame of reference,"},{"Start":"16:21.365 ","End":"16:23.594","Text":"and we write the exact same 4-vector."},{"Start":"16:23.594 ","End":"16:27.320","Text":"We also find the size of that 4-vector at a specific given time,"},{"Start":"16:27.320 ","End":"16:31.610","Text":"we\u0027ll find that the sizes of the two vectors are equal to"},{"Start":"16:31.610 ","End":"16:36.260","Text":"the same constant number and that is to do with one particle."},{"Start":"16:36.260 ","End":"16:40.473","Text":"If, of course, we have two particles in our frames of reference,"},{"Start":"16:40.473 ","End":"16:44.045","Text":"then all we have to do is sum every component."},{"Start":"16:44.045 ","End":"16:53.510","Text":"We\u0027ll have P_1x plus P_2x plus P_3x and so on and so forth until P_nx squared,"},{"Start":"16:53.510 ","End":"16:59.990","Text":"and then the same with y, so P_1y plus P2_y plus etc."},{"Start":"16:59.990 ","End":"17:02.510","Text":"until we have P_ny squared,"},{"Start":"17:02.510 ","End":"17:07.796","Text":"and again for z, and the same thing for our energy is divided by c^2."},{"Start":"17:07.796 ","End":"17:12.500","Text":"Again, we will receive this same constant."},{"Start":"17:12.500 ","End":"17:15.420","Text":"That\u0027s the end of this lesson."}],"ID":9566},{"Watched":false,"Name":"Exercise 10","Duration":"1m 37s","ChapterTopicVideoID":9276,"CourseChapterTopicPlaylistID":9398,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.460","Text":"In this lesson, we\u0027re going to be looking at"},{"Start":"00:02.460 ","End":"00:06.630","Text":"certain particles who have a resting mass of 0."},{"Start":"00:06.630 ","End":"00:10.410","Text":"Such particles are photons and neutrinos."},{"Start":"00:10.410 ","End":"00:14.085","Text":"Our most interesting particle will be our photon."},{"Start":"00:14.085 ","End":"00:16.995","Text":"Our photon is some light particle."},{"Start":"00:16.995 ","End":"00:21.875","Text":"We can consider light as particles and as a wave."},{"Start":"00:21.875 ","End":"00:24.230","Text":"If we\u0027re going to consider light as a particle,"},{"Start":"00:24.230 ","End":"00:27.455","Text":"then we\u0027re speaking about photons and then we can"},{"Start":"00:27.455 ","End":"00:31.335","Text":"use this equation to find out our energy."},{"Start":"00:31.335 ","End":"00:35.000","Text":"Then we will set our mass over here to equal 0."},{"Start":"00:35.000 ","End":"00:38.810","Text":"That is because we said that photons have a resting mass of 0."},{"Start":"00:38.810 ","End":"00:40.925","Text":"Once we substitute that in,"},{"Start":"00:40.925 ","End":"00:47.240","Text":"that mass of the photon is equal to 0 so we\u0027ll be left with this equation over here,"},{"Start":"00:47.240 ","End":"00:51.035","Text":"that our energy is going to be equal to the size of p"},{"Start":"00:51.035 ","End":"00:55.250","Text":"multiplied by c. When we have this equation over here,"},{"Start":"00:55.250 ","End":"00:58.430","Text":"our energy is equal to the size of the momentum multiplied by"},{"Start":"00:58.430 ","End":"01:03.005","Text":"c. This is also equal to h times"},{"Start":"01:03.005 ","End":"01:08.030","Text":"v. So h is our Plank\u0027s constant and it\u0027s"},{"Start":"01:08.030 ","End":"01:13.640","Text":"equal to 6.626 times 10^negative 34js,"},{"Start":"01:13.640 ","End":"01:15.256","Text":"where j is our joules,"},{"Start":"01:15.256 ","End":"01:16.509","Text":"it\u0027s a unit of energy,"},{"Start":"01:16.509 ","End":"01:19.050","Text":"multiplied by s, seconds."},{"Start":"01:19.050 ","End":"01:22.740","Text":"Then this letter over here is our Nu."},{"Start":"01:22.740 ","End":"01:26.955","Text":"Our Nu represents the frequency of the photon."},{"Start":"01:26.955 ","End":"01:31.685","Text":"That\u0027s the end of this brief lesson and soon we\u0027re going to see how we utilize"},{"Start":"01:31.685 ","End":"01:34.370","Text":"these equations in 1 of the exercises"},{"Start":"01:34.370 ","End":"01:38.220","Text":"and it\u0027s important to remember this equation over here."}],"ID":9567},{"Watched":false,"Name":"Exercise 11","Duration":"12m ","ChapterTopicVideoID":9273,"CourseChapterTopicPlaylistID":9398,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.370","Text":"Hello. In this question,"},{"Start":"00:02.370 ","End":"00:07.440","Text":"we\u0027re being told that a particle of energy E_1 and resting mass m_1"},{"Start":"00:07.440 ","End":"00:13.455","Text":"is moving in the lab in the positive direction of the x axis like so."},{"Start":"00:13.455 ","End":"00:17.100","Text":"Then we\u0027re being told that at some moment the particle breaks down and"},{"Start":"00:17.100 ","End":"00:21.105","Text":"transforms into a photon and particle 2."},{"Start":"00:21.105 ","End":"00:25.550","Text":"This is particle 2 and this is our photon and the photon is moving in"},{"Start":"00:25.550 ","End":"00:31.030","Text":"the positive y direction and has a known energy of E Gamma."},{"Start":"00:31.030 ","End":"00:33.140","Text":"Our first question is,"},{"Start":"00:33.140 ","End":"00:39.100","Text":"what is the momentum of the particle before the decay over here."},{"Start":"00:39.100 ","End":"00:44.285","Text":"Because we know what its energy is and we\u0027re being asked to find the momentum,"},{"Start":"00:44.285 ","End":"00:47.780","Text":"we\u0027re going to use to answer question number 1,"},{"Start":"00:47.780 ","End":"00:51.895","Text":"our equation for connecting energy and momentum."},{"Start":"00:51.895 ","End":"00:59.200","Text":"We\u0027ll have that our E_1^2 is going to be equal to our momentum of this P_1^2"},{"Start":"00:59.200 ","End":"01:08.985","Text":"multiplied by C^2 plus m_1^2 multiplied by C^4."},{"Start":"01:08.985 ","End":"01:12.815","Text":"Now, because what we want to do is we want to find our momentum and"},{"Start":"01:12.815 ","End":"01:18.025","Text":"our E_1,m_1 and c are details that are given to us in the question."},{"Start":"01:18.025 ","End":"01:23.430","Text":"We can isolate out our P_1 and this is a vector."},{"Start":"01:23.430 ","End":"01:26.810","Text":"It\u0027s going to be equal to the square root"},{"Start":"01:26.810 ","End":"01:30.969","Text":"from just doing some algebra of E_1 divided by c^2"},{"Start":"01:30.969 ","End":"01:39.885","Text":"minus m_1^2 C^2 in the x-direction."},{"Start":"01:39.885 ","End":"01:46.235","Text":"What we have with this square root sign is our size of the momentum and then,"},{"Start":"01:46.235 ","End":"01:50.131","Text":"because we\u0027re being told that our particle is moving along the x-axis,"},{"Start":"01:50.131 ","End":"01:52.730","Text":"so we know that its momentum is in the x-direction,"},{"Start":"01:52.730 ","End":"01:54.724","Text":"and then we get a vector quantity."},{"Start":"01:54.724 ","End":"01:56.015","Text":"The next question is,"},{"Start":"01:56.015 ","End":"01:59.135","Text":"at what angle is the momentum of particle 2,"},{"Start":"01:59.135 ","End":"02:00.695","Text":"this particle over here,"},{"Start":"02:00.695 ","End":"02:03.995","Text":"relative to the x-axis?"},{"Start":"02:03.995 ","End":"02:08.210","Text":"What we can do is we can look at our particle number 2 and see that it"},{"Start":"02:08.210 ","End":"02:12.860","Text":"has a y component for momentum,"},{"Start":"02:12.860 ","End":"02:17.149","Text":"and it also has an x-component for momentum."},{"Start":"02:17.149 ","End":"02:23.696","Text":"Then when we use our trig identity for a tan theta,"},{"Start":"02:23.696 ","End":"02:27.830","Text":"we can see that that\u0027s going to be equal to the momentum on"},{"Start":"02:27.830 ","End":"02:33.275","Text":"the y-axis divided by the momentum in the x-axis."},{"Start":"02:33.275 ","End":"02:36.990","Text":"Now, what I can say is that in the x-axis,"},{"Start":"02:36.990 ","End":"02:39.920","Text":"I have conservation of momentum and there\u0027s also,"},{"Start":"02:39.920 ","End":"02:43.490","Text":"of course, conservation of momentum in the y-direction."},{"Start":"02:43.490 ","End":"02:48.665","Text":"In that case, I can say that by momentum of particle 2 in"},{"Start":"02:48.665 ","End":"02:54.440","Text":"the x-direction is going to be equal to the momentum of particle 1 in the x-direction,"},{"Start":"02:54.440 ","End":"02:57.725","Text":"which as we know, is simply the momentum of particle"},{"Start":"02:57.725 ","End":"03:01.100","Text":"1 because it\u0027s only in the x-direction."},{"Start":"03:01.100 ","End":"03:06.200","Text":"Now, we can also see that when our particle m_1 breaks apart,"},{"Start":"03:06.200 ","End":"03:09.350","Text":"it breaks into particle 2 and this photon."},{"Start":"03:09.350 ","End":"03:13.655","Text":"Now, this photon we can see is only moving in the y-direction,"},{"Start":"03:13.655 ","End":"03:17.425","Text":"which means that it has no momentum in the x-direction."},{"Start":"03:17.425 ","End":"03:22.070","Text":"That means that all of the momentum will just go to particle 2 and"},{"Start":"03:22.070 ","End":"03:26.610","Text":"we don\u0027t have to split it between particle 2 and our photon."},{"Start":"03:26.610 ","End":"03:30.360","Text":"That means that our momentum in the x-direction is only going to"},{"Start":"03:30.360 ","End":"03:34.905","Text":"our particle number 2 and it equals to our momentum of particle 1,"},{"Start":"03:34.905 ","End":"03:37.125","Text":"which we found over here."},{"Start":"03:37.125 ","End":"03:40.250","Text":"Now we found that. Now what we have to do is"},{"Start":"03:40.250 ","End":"03:43.190","Text":"we have to find the momentum in the y direction."},{"Start":"03:43.190 ","End":"03:48.710","Text":"Now, over here, we can see that before the particles splits,"},{"Start":"03:48.710 ","End":"03:52.070","Text":"we can see that we only have momentum in the x-direction,"},{"Start":"03:52.070 ","End":"03:55.400","Text":"which means that momentum in the y-direction is equal to 0."},{"Start":"03:55.400 ","End":"03:57.680","Text":"Because of conservation of momentum,"},{"Start":"03:57.680 ","End":"04:04.805","Text":"that means that the momentum in the y-direction over here also has to be equal to 0."},{"Start":"04:04.805 ","End":"04:07.910","Text":"How can that be? That means that our momentum in"},{"Start":"04:07.910 ","End":"04:11.110","Text":"the y-direction of our photon has to be negative"},{"Start":"04:11.110 ","End":"04:17.460","Text":"the momentum in the y-direction of our mass number 2 over here."},{"Start":"04:17.460 ","End":"04:22.220","Text":"Let\u0027s write that out. That means that my P_2 in"},{"Start":"04:22.220 ","End":"04:29.520","Text":"the y direction has to be equal to negative my momentum of the photon,"},{"Start":"04:29.520 ","End":"04:30.755","Text":"or in other words,"},{"Start":"04:30.755 ","End":"04:35.060","Text":"we can also say that the size of the momentum"},{"Start":"04:35.060 ","End":"04:39.440","Text":"in the y-direction for my mass number 2 has to be equal"},{"Start":"04:39.440 ","End":"04:43.640","Text":"to the size of the momentum in"},{"Start":"04:43.640 ","End":"04:48.500","Text":"the y-direction of my photon and we can see that they\u0027re going in opposite directions."},{"Start":"04:48.500 ","End":"04:52.370","Text":"This is the y component of the momentum and y from mass number"},{"Start":"04:52.370 ","End":"04:58.255","Text":"2 and this is the y component for our photon and they\u0027re equal and opposite."},{"Start":"04:58.255 ","End":"05:03.470","Text":"Now what we can do is we can use our equation saying that our energy of"},{"Start":"05:03.470 ","End":"05:08.405","Text":"the photon is equal to the momentum of the photon multiplied by c,"},{"Start":"05:08.405 ","End":"05:12.287","Text":"the speed of light and then because we want to find the momentum of the photon,"},{"Start":"05:12.287 ","End":"05:15.990","Text":"we just isolated out and then we can say, therefore,"},{"Start":"05:15.990 ","End":"05:22.610","Text":"the momentum of the photon is going to be equal to E of the photon divided"},{"Start":"05:22.610 ","End":"05:29.360","Text":"by c and notice this has given to us in the question and that means from this,"},{"Start":"05:29.360 ","End":"05:36.900","Text":"that that also equals to our momentum of mass number 2 in the y-direction."},{"Start":"05:36.900 ","End":"05:40.640","Text":"Now, what I can do from this is I can substitute this"},{"Start":"05:40.640 ","End":"05:45.304","Text":"into this equation because what I\u0027m trying to find is the angle of momentum."},{"Start":"05:45.304 ","End":"05:49.085","Text":"I have that my P_y is equal to this."},{"Start":"05:49.085 ","End":"05:56.450","Text":"I have my E Gamma divided by c and then divide it by my P_x,"},{"Start":"05:56.450 ","End":"05:58.505","Text":"which is this over here."},{"Start":"05:58.505 ","End":"06:03.965","Text":"That\u0027s going to be the square root of E_1 divided by C^2"},{"Start":"06:03.965 ","End":"06:10.320","Text":"minus m_1^2 C^2 and the square root."},{"Start":"06:10.320 ","End":"06:18.440","Text":"Now, all I have to do is put a negative because we said that my momentum of"},{"Start":"06:18.440 ","End":"06:26.719","Text":"particle 2 is in the opposite y direction to that of the photon."},{"Start":"06:26.719 ","End":"06:31.815","Text":"Remember from over here so I place a minus over here."},{"Start":"06:31.815 ","End":"06:35.345","Text":"Now all I have to do in order to find the angle is"},{"Start":"06:35.345 ","End":"06:39.055","Text":"arctan both sides and there we have the answer."},{"Start":"06:39.055 ","End":"06:41.210","Text":"Now, quickly, another explanation for this"},{"Start":"06:41.210 ","End":"06:44.105","Text":"minus before we move on to question number 3,"},{"Start":"06:44.105 ","End":"06:48.125","Text":"is that we can see that our angle theta is opening"},{"Start":"06:48.125 ","End":"06:53.135","Text":"up in this direction towards the negative y."},{"Start":"06:53.135 ","End":"07:01.250","Text":"That\u0027s how we know that our angle is going to have to be some 180 plus this theta."},{"Start":"07:01.250 ","End":"07:07.460","Text":"Because we\u0027re going down in the y direction so we also have a negative over here,"},{"Start":"07:07.460 ","End":"07:11.240","Text":"because our theta isn\u0027t located over here."},{"Start":"07:11.240 ","End":"07:12.950","Text":"It\u0027s located down here,"},{"Start":"07:12.950 ","End":"07:16.410","Text":"which means that we\u0027ll have a y over there."},{"Start":"07:16.410 ","End":"07:21.260","Text":"Question 3 is asking me to find the new frame of reference S tag,"},{"Start":"07:21.260 ","End":"07:26.470","Text":"where the photon will be ejected in the opposite direction to that of particle 2."},{"Start":"07:26.470 ","End":"07:29.300","Text":"Then we\u0027re being asked what is the velocity of"},{"Start":"07:29.300 ","End":"07:33.640","Text":"S tag compared to the labs frame of reference."},{"Start":"07:33.640 ","End":"07:35.030","Text":"In questions like this,"},{"Start":"07:35.030 ","End":"07:37.940","Text":"the first thing that we want to do is we want to"},{"Start":"07:37.940 ","End":"07:41.765","Text":"move to the frame of reference of our center of mass."},{"Start":"07:41.765 ","End":"07:45.125","Text":"When we\u0027re working with a frame of reference of our center of mass,"},{"Start":"07:45.125 ","End":"07:51.020","Text":"that means that the total momentum is going to be equal to 0."},{"Start":"07:51.020 ","End":"07:54.565","Text":"That means that in this type of system,"},{"Start":"07:54.565 ","End":"08:02.030","Text":"we can look at it like this and if we have our 1 particle that\u0027s moving before the split,"},{"Start":"08:02.030 ","End":"08:05.480","Text":"with a total momentum equal to 0."},{"Start":"08:05.480 ","End":"08:07.250","Text":"That means after the split,"},{"Start":"08:07.250 ","End":"08:10.340","Text":"the only possible way that we can draw"},{"Start":"08:10.340 ","End":"08:15.350","Text":"this diagram where the total momentum is still going to be equal to 0 is if"},{"Start":"08:15.350 ","End":"08:21.920","Text":"both particles have equal and opposite momentums and that way they\u0027ll cancel each other"},{"Start":"08:21.920 ","End":"08:25.610","Text":"out and we\u0027ll remain with our total momentum being"},{"Start":"08:25.610 ","End":"08:29.770","Text":"equal to 0 because of course we have conservation of momentum."},{"Start":"08:29.770 ","End":"08:34.055","Text":"Whenever we\u0027re using our frame of reference for our center of mass,"},{"Start":"08:34.055 ","End":"08:37.250","Text":"the total momentum is always going to be equal to 0."},{"Start":"08:37.250 ","End":"08:41.000","Text":"It\u0027s always conserved and that means that all of our bodies are"},{"Start":"08:41.000 ","End":"08:46.950","Text":"moving on the same line just in opposite directions."},{"Start":"08:46.950 ","End":"08:51.320","Text":"When we\u0027re working with our center of mass frame,"},{"Start":"08:51.320 ","End":"08:55.915","Text":"it\u0027s going to be the same before and after our split."},{"Start":"08:55.915 ","End":"08:59.930","Text":"What I\u0027m going to work out is the velocity of mass number"},{"Start":"08:59.930 ","End":"09:06.760","Text":"1 before the split and then I\u0027m going to use what\u0027s given to me in the question."},{"Start":"09:06.760 ","End":"09:12.574","Text":"What we\u0027re going to do is we\u0027re going to use our equation for our energy"},{"Start":"09:12.574 ","End":"09:18.090","Text":"is equal to our mass times C^2 times our Gamma."},{"Start":"09:18.090 ","End":"09:23.330","Text":"What we\u0027re trying to find is our velocity inside our Gamma over here."},{"Start":"09:23.330 ","End":"09:28.760","Text":"Now, it\u0027s obvious that when we\u0027re dealing with assisting with only 1 body,"},{"Start":"09:28.760 ","End":"09:33.115","Text":"that means the center of mass is going to be that body itself."},{"Start":"09:33.115 ","End":"09:38.210","Text":"Now, we\u0027re looking at a system where our mass over here is going to be"},{"Start":"09:38.210 ","End":"09:43.205","Text":"stationary because our frame of reference is from its position."},{"Start":"09:43.205 ","End":"09:46.980","Text":"As in someone standing over here and"},{"Start":"09:46.980 ","End":"09:50.825","Text":"they see everything moving around them but they feel that they are stationary."},{"Start":"09:50.825 ","End":"09:54.530","Text":"We\u0027re in the frame of reference of the mass itself."},{"Start":"09:54.530 ","End":"10:00.746","Text":"Now, obviously, if in this frame of reference now our mass has 0 velocity,"},{"Start":"10:00.746 ","End":"10:04.345","Text":"it means that the momentum is obviously going to be equal to 0."},{"Start":"10:04.345 ","End":"10:08.885","Text":"Now, we can say that our Gamma of our particle is"},{"Start":"10:08.885 ","End":"10:14.280","Text":"equal to the Gamma of the system that we\u0027re trying to find."},{"Start":"10:14.480 ","End":"10:19.080","Text":"Now we can say that our Gamma 0 is equal to,"},{"Start":"10:19.080 ","End":"10:20.730","Text":"and now we know this equation."},{"Start":"10:20.730 ","End":"10:26.330","Text":"It\u0027s 1 divided by the square root of 1 minus v_0 divided by"},{"Start":"10:26.330 ","End":"10:33.000","Text":"c^2 and this is always going to be the equation."},{"Start":"10:33.000 ","End":"10:37.220","Text":"Now, because I want to find out what my v_0 is,"},{"Start":"10:37.220 ","End":"10:39.405","Text":"I can isolate this out."},{"Start":"10:39.405 ","End":"10:41.610","Text":"By doing some simple algebra,"},{"Start":"10:41.610 ","End":"10:45.830","Text":"I can get that my v_0 divided by c is going to be equal to"},{"Start":"10:45.830 ","End":"10:52.595","Text":"the square root of 1 minus 1 divided by Gamma 0^2."},{"Start":"10:52.595 ","End":"10:56.510","Text":"Now, instead of substituting in my Gamma 0^2,"},{"Start":"10:56.510 ","End":"10:59.750","Text":"I know that my Gamma 0 is equal to my Gamma"},{"Start":"10:59.750 ","End":"11:05.070","Text":"1 and I know that my Gamma 1 is equal to this."},{"Start":"11:05.070 ","End":"11:06.435","Text":"Let\u0027s see what that is."},{"Start":"11:06.435 ","End":"11:09.495","Text":"My Gamma 1, which is equal to my Gamma 0,"},{"Start":"11:09.495 ","End":"11:15.120","Text":"is equal to my E_1 divided by m_1c^2"},{"Start":"11:15.120 ","End":"11:20.975","Text":"because this is the information that I\u0027m given in my question."},{"Start":"11:20.975 ","End":"11:24.595","Text":"I know what my E_1 and my m_1 and my c is equal to."},{"Start":"11:24.595 ","End":"11:27.450","Text":"I\u0027m going to write out my Gamma in terms of what I"},{"Start":"11:27.450 ","End":"11:30.755","Text":"know and then I\u0027m just going to substitute that in."},{"Start":"11:30.755 ","End":"11:35.150","Text":"Therefore, we\u0027ll get that my V_0 is going to be equal"},{"Start":"11:35.150 ","End":"11:40.650","Text":"to the square root of 1 minus 1 divided by Gamma which is simply"},{"Start":"11:40.650 ","End":"11:50.795","Text":"going to be m_1 multiplied by c^2 divided by E_1 and all of this is going to be squared."},{"Start":"11:50.795 ","End":"11:55.760","Text":"Then we\u0027ll take the square root of this and multiply it by c. There we have it."},{"Start":"11:55.760 ","End":"11:58.370","Text":"That is our final answer."},{"Start":"11:58.370 ","End":"12:01.469","Text":"That\u0027s the end of this lesson."}],"ID":9568},{"Watched":false,"Name":"Minimum Energy To Form Particles","Duration":"7m 5s","ChapterTopicVideoID":9270,"CourseChapterTopicPlaylistID":9398,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:05.090 ","End":"00:07.860","Text":"Hello. In this lesson,"},{"Start":"00:07.860 ","End":"00:13.815","Text":"I\u0027m going to explain how to find the minimum energy required to form a particle."},{"Start":"00:13.815 ","End":"00:18.585","Text":"What we\u0027re going to do is we\u0027re going to start from the end."},{"Start":"00:18.585 ","End":"00:21.090","Text":"I\u0027m already going to tell you what the condition is."},{"Start":"00:21.090 ","End":"00:25.874","Text":"The condition required for finding minimum energy needed for the process,"},{"Start":"00:25.874 ","End":"00:28.874","Text":"the process of forming 2 new particles,"},{"Start":"00:28.874 ","End":"00:30.915","Text":"or however many new particles,"},{"Start":"00:30.915 ","End":"00:38.805","Text":"is that the particles after the collision will be at rest in the center of mass system."},{"Start":"00:38.805 ","End":"00:42.120","Text":"If we\u0027re going with a frame of reference,"},{"Start":"00:42.120 ","End":"00:44.911","Text":"which is our center of mass,"},{"Start":"00:44.911 ","End":"00:50.345","Text":"so the particles in that frame of reference must be stationary."},{"Start":"00:50.345 ","End":"00:52.370","Text":"That is the condition."},{"Start":"00:52.370 ","End":"00:56.315","Text":"Now, let\u0027s try and understand why that is."},{"Start":"00:56.315 ","End":"00:58.555","Text":"Now let\u0027s go back to the beginning."},{"Start":"00:58.555 ","End":"01:00.930","Text":"If we\u0027re going to look at 2 particles,"},{"Start":"01:00.930 ","End":"01:03.390","Text":"so we have Mass number 1 and Mass number 2."},{"Start":"01:03.390 ","End":"01:06.800","Text":"Mass number 1 is moving with a velocity of v_1,"},{"Start":"01:06.800 ","End":"01:10.655","Text":"and Mass number 2, let\u0027s just say that it\u0027s at rest."},{"Start":"01:10.655 ","End":"01:15.160","Text":"This is what it looks like before the collision."},{"Start":"01:15.350 ","End":"01:18.080","Text":"Then these 2 particles collide,"},{"Start":"01:18.080 ","End":"01:19.535","Text":"and then after the collision,"},{"Start":"01:19.535 ","End":"01:21.425","Text":"we have 2 new particles."},{"Start":"01:21.425 ","End":"01:22.730","Text":"We have particle 3,"},{"Start":"01:22.730 ","End":"01:24.425","Text":"and particle 4,"},{"Start":"01:24.425 ","End":"01:28.520","Text":"and each 1 has its own velocity, v_3 and v_4."},{"Start":"01:28.520 ","End":"01:30.920","Text":"We don\u0027t know what the size of these velocities are,"},{"Start":"01:30.920 ","End":"01:32.705","Text":"and we don\u0027t know their direction."},{"Start":"01:32.705 ","End":"01:38.900","Text":"Now, what I want to find is the minimum energy required to form these particles."},{"Start":"01:38.900 ","End":"01:42.245","Text":"Now, usually in order to find the minimum energy,"},{"Start":"01:42.245 ","End":"01:46.310","Text":"we\u0027d go about finding the general energy that we have."},{"Start":"01:46.310 ","End":"01:50.345","Text":"I\u0027ll have v_3^2 and our v_4^2."},{"Start":"01:50.345 ","End":"01:52.235","Text":"But as we can see in this case,"},{"Start":"01:52.235 ","End":"01:56.480","Text":"because I don\u0027t know the size or direction of either of the velocities,"},{"Start":"01:56.480 ","End":"02:00.230","Text":"so how am I going to find the minimum energy?"},{"Start":"02:00.230 ","End":"02:04.344","Text":"Because the options that I have are endless."},{"Start":"02:04.344 ","End":"02:10.460","Text":"What we have to do is we have to find a slightly easier way to find our minimum energy."},{"Start":"02:10.460 ","End":"02:13.535","Text":"Now, of course, we know that our energy will be at a minimum"},{"Start":"02:13.535 ","End":"02:16.805","Text":"if these velocities are equal to 0."},{"Start":"02:16.805 ","End":"02:21.335","Text":"However, because during this collision we have conservation of momentum,"},{"Start":"02:21.335 ","End":"02:24.380","Text":"we know therefore that after the collisions,"},{"Start":"02:24.380 ","End":"02:29.545","Text":"our particles must have some velocity that does not equal to 0."},{"Start":"02:29.545 ","End":"02:33.815","Text":"Because otherwise, that means that momentum won\u0027t be conserved,"},{"Start":"02:33.815 ","End":"02:37.015","Text":"which we know is not the case. What\u0027s written here?"},{"Start":"02:37.015 ","End":"02:39.905","Text":"Momentum is conserved during the collision,"},{"Start":"02:39.905 ","End":"02:41.890","Text":"and therefore our bodies,"},{"Start":"02:41.890 ","End":"02:46.310","Text":"mass 3 and 4, can never be at rest after the collision."},{"Start":"02:46.310 ","End":"02:50.059","Text":"Now what we have to do is we have to somehow balance"},{"Start":"02:50.059 ","End":"02:54.800","Text":"that we need to use the information that momentum is conserved and therefore,"},{"Start":"02:54.800 ","End":"02:57.545","Text":"we\u0027re going to have velocity v_3 and v_4."},{"Start":"02:57.545 ","End":"03:03.530","Text":"However, we also want our v_3 and v_4 to be as small as possible,"},{"Start":"03:03.530 ","End":"03:08.420","Text":"such that our energy is going to be at a minimum value."},{"Start":"03:08.420 ","End":"03:10.505","Text":"Of course, as you know,"},{"Start":"03:10.505 ","End":"03:15.844","Text":"not only the size of the velocities over here come into the equation,"},{"Start":"03:15.844 ","End":"03:20.300","Text":"but also their directions and the size of each mass."},{"Start":"03:20.300 ","End":"03:24.695","Text":"That makes this calculation very difficult."},{"Start":"03:24.695 ","End":"03:27.265","Text":"Now comes our trick."},{"Start":"03:27.265 ","End":"03:31.099","Text":"If we move into a system of the center of mass,"},{"Start":"03:31.099 ","End":"03:32.300","Text":"or a frame of reference,"},{"Start":"03:32.300 ","End":"03:33.980","Text":"which is the center of mass,"},{"Start":"03:33.980 ","End":"03:38.330","Text":"then we know that all the time when we\u0027re in the system of center of mass,"},{"Start":"03:38.330 ","End":"03:42.355","Text":"our total momentum is going to be equal to 0."},{"Start":"03:42.355 ","End":"03:45.170","Text":"Then again, if we\u0027re in this frame of reference,"},{"Start":"03:45.170 ","End":"03:46.475","Text":"so our center of mass,"},{"Start":"03:46.475 ","End":"03:49.985","Text":"that because the total momentum is equal to 0,"},{"Start":"03:49.985 ","End":"03:52.130","Text":"then we know that in this frame of reference,"},{"Start":"03:52.130 ","End":"03:57.140","Text":"we\u0027re going to have particle 1 and particle 2 moving in a way such that they collide,"},{"Start":"03:57.140 ","End":"04:01.489","Text":"and that the momentum p_1 tag is equal to negative"},{"Start":"04:01.489 ","End":"04:06.720","Text":"the momentum of p_2 tag of Mass number 2 tag."},{"Start":"04:07.810 ","End":"04:12.695","Text":"Then we can see that they\u0027re of equal size but opposite direction."},{"Start":"04:12.695 ","End":"04:14.870","Text":"Then they will cancel each other out."},{"Start":"04:14.870 ","End":"04:16.970","Text":"We will see that before our collision,"},{"Start":"04:16.970 ","End":"04:21.740","Text":"our total momentum in this case is going to be equal to 0."},{"Start":"04:21.740 ","End":"04:26.065","Text":"Then that means that after our collision over here,"},{"Start":"04:26.065 ","End":"04:32.015","Text":"so we know that both of our masses that are a result from this collision,"},{"Start":"04:32.015 ","End":"04:36.770","Text":"are going to be having to move also again in the opposite direction,"},{"Start":"04:36.770 ","End":"04:40.205","Text":"and such that the momentum of body 3,"},{"Start":"04:40.205 ","End":"04:43.985","Text":"so p_3 tag has to be of an equal size,"},{"Start":"04:43.985 ","End":"04:48.325","Text":"but in opposite direction to the momentum of body 4."},{"Start":"04:48.325 ","End":"04:52.205","Text":"Then that means that our total momentum"},{"Start":"04:52.205 ","End":"04:56.940","Text":"again after the collision is going to be equal to 0."},{"Start":"04:57.220 ","End":"04:59.915","Text":"Now, when we\u0027re looking at this,"},{"Start":"04:59.915 ","End":"05:02.600","Text":"it\u0027s very easy for us to see which type of"},{"Start":"05:02.600 ","End":"05:07.275","Text":"condition we\u0027re going to require in order to get minimum energy."},{"Start":"05:07.275 ","End":"05:11.550","Text":"That is simply that our v_3 and our v_4,"},{"Start":"05:11.550 ","End":"05:14.600","Text":"when added together, are equal to 0."},{"Start":"05:14.600 ","End":"05:20.606","Text":"Notice that that also works out when we apply that into our equation for momentum,"},{"Start":"05:20.606 ","End":"05:23.510","Text":"because momentum is mass times velocity,"},{"Start":"05:23.510 ","End":"05:26.995","Text":"and we know that our total momentum is equal to 0."},{"Start":"05:26.995 ","End":"05:32.630","Text":"That means that if we have equal sized velocity has been in opposite direction,"},{"Start":"05:32.630 ","End":"05:36.020","Text":"then once we apply that into our momentum equations,"},{"Start":"05:36.020 ","End":"05:41.290","Text":"we\u0027ll see that the total momentum of this entire system is equal to 0."},{"Start":"05:41.290 ","End":"05:43.760","Text":"When we\u0027re in the system of center of mass,"},{"Start":"05:43.760 ","End":"05:47.008","Text":"it\u0027s very easy to look at what the collision will look like,"},{"Start":"05:47.008 ","End":"05:50.935","Text":"and then we can see what we need to find our minimum energy."},{"Start":"05:50.935 ","End":"05:55.475","Text":"We\u0027ve already seen that in order to get our P total,"},{"Start":"05:55.475 ","End":"05:58.610","Text":"or our total momentum to be equal to 0,"},{"Start":"05:58.610 ","End":"06:03.965","Text":"then our velocities are going to have to be of equal size and in opposite direction."},{"Start":"06:03.965 ","End":"06:08.074","Text":"That gives us some energy in order to form particles."},{"Start":"06:08.074 ","End":"06:10.490","Text":"However, because we want our minimum energy,"},{"Start":"06:10.490 ","End":"06:11.990","Text":"all we really need,"},{"Start":"06:11.990 ","End":"06:13.670","Text":"which is what I said at the beginning,"},{"Start":"06:13.670 ","End":"06:17.645","Text":"is that our total momentum is going to be equal to 0,"},{"Start":"06:17.645 ","End":"06:20.330","Text":"which can also be done if both of"},{"Start":"06:20.330 ","End":"06:25.700","Text":"our velocities are equal to 0 in our frame of reference for the center of mass."},{"Start":"06:25.700 ","End":"06:29.615","Text":"That means that any velocity,"},{"Start":"06:29.615 ","End":"06:30.980","Text":"v_3 or v_4,"},{"Start":"06:30.980 ","End":"06:32.465","Text":"which is above 0,"},{"Start":"06:32.465 ","End":"06:34.820","Text":"will give us energy, which is correct."},{"Start":"06:34.820 ","End":"06:36.710","Text":"However, it won\u0027t be the minimum."},{"Start":"06:36.710 ","End":"06:40.220","Text":"It will be something higher than our E_min required."},{"Start":"06:40.220 ","End":"06:44.465","Text":"Now we got to our condition for finding the minimum energy"},{"Start":"06:44.465 ","End":"06:49.079","Text":"needed for the process to form new particles after collision,"},{"Start":"06:49.079 ","End":"06:55.420","Text":"and that is that the particles will be at rest in the center of mass system."},{"Start":"06:55.420 ","End":"06:57.210","Text":"That\u0027s the end of this lesson,"},{"Start":"06:57.210 ","End":"07:00.800","Text":"and now let\u0027s go and do a practice example in order to"},{"Start":"07:00.800 ","End":"07:05.700","Text":"see how we can use this condition to our advantage."}],"ID":9569},{"Watched":false,"Name":"Exercise 12","Duration":"13m 17s","ChapterTopicVideoID":9275,"CourseChapterTopicPlaylistID":9398,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.430","Text":"Hello. In this question,"},{"Start":"00:02.430 ","End":"00:07.545","Text":"we have a proton which is at rest in the lab and it\u0027s being hit by a photon."},{"Start":"00:07.545 ","End":"00:12.450","Text":"We\u0027re being asked what is the minimum energy required from the photon in"},{"Start":"00:12.450 ","End":"00:17.385","Text":"order for a pion and a new proton to be formed after the collision."},{"Start":"00:17.385 ","End":"00:20.700","Text":"We\u0027re being told that the masses of the proton and pion,"},{"Start":"00:20.700 ","End":"00:23.205","Text":"m_Pi and m_p given,"},{"Start":"00:23.205 ","End":"00:27.225","Text":"and of course also the mass of the photon is given."},{"Start":"00:27.225 ","End":"00:31.230","Text":"However, its mass is equal to 0."},{"Start":"00:31.230 ","End":"00:34.745","Text":"How are we going to find this minimum energy?"},{"Start":"00:34.745 ","End":"00:38.630","Text":"Now, what we\u0027re going to do is we\u0027re going to use the notion of conservation"},{"Start":"00:38.630 ","End":"00:42.695","Text":"of energy and conservation of momentum during the collision."},{"Start":"00:42.695 ","End":"00:47.735","Text":"That means that we can work out the energy and momentum before the collision,"},{"Start":"00:47.735 ","End":"00:50.160","Text":"the energy and momentum after the collision,"},{"Start":"00:50.160 ","End":"00:56.595","Text":"and we can use all of that in order to find out what our minimum energy is."},{"Start":"00:56.595 ","End":"01:02.420","Text":"Let\u0027s start by writing our equations for before the collision."},{"Start":"01:02.420 ","End":"01:05.495","Text":"What we\u0027re going to do is we\u0027re going to start by writing out"},{"Start":"01:05.495 ","End":"01:08.835","Text":"our momentum for our proton,"},{"Start":"01:08.835 ","End":"01:12.830","Text":"and we\u0027re going to write this as a 4-vector."},{"Start":"01:12.830 ","End":"01:17.690","Text":"We\u0027re going to have our momentum in the x-direction,"},{"Start":"01:17.690 ","End":"01:19.760","Text":"our momentum in the y-direction,"},{"Start":"01:19.760 ","End":"01:21.740","Text":"our momentum in the z-direction,"},{"Start":"01:21.740 ","End":"01:28.325","Text":"and then the energy of the proton divided by c. Let\u0027s see what this is."},{"Start":"01:28.325 ","End":"01:33.545","Text":"Now we\u0027re being told that the proton before the collision is at rest."},{"Start":"01:33.545 ","End":"01:37.790","Text":"Because momentum is mass times velocity and its velocity is 0,"},{"Start":"01:37.790 ","End":"01:40.970","Text":"so our momentum in the x, y,"},{"Start":"01:40.970 ","End":"01:44.045","Text":"and z-direction is going to be equal to 0,"},{"Start":"01:44.045 ","End":"01:49.550","Text":"and our E divided by c is simply going to be the mass of the proton,"},{"Start":"01:49.550 ","End":"01:52.520","Text":"which has given to us in the question multiplied by c,"},{"Start":"01:52.520 ","End":"01:54.230","Text":"the speed of light."},{"Start":"01:54.230 ","End":"01:56.540","Text":"Just to remind you what this is,"},{"Start":"01:56.540 ","End":"02:01.965","Text":"remember that we learned about a very famous equation by Einstein,"},{"Start":"02:01.965 ","End":"02:04.536","Text":"and that\u0027s the resting mass energy."},{"Start":"02:04.536 ","End":"02:07.700","Text":"Which is equal to the mass of the particle,"},{"Start":"02:07.700 ","End":"02:09.485","Text":"here specifically it\u0027s a proton,"},{"Start":"02:09.485 ","End":"02:12.865","Text":"multiplied by the speed of light squared."},{"Start":"02:12.865 ","End":"02:20.105","Text":"Over here, I have to substitute my E divided by c so when I have here divided by c,"},{"Start":"02:20.105 ","End":"02:22.495","Text":"I\u0027ll have mc over here."},{"Start":"02:22.495 ","End":"02:26.225","Text":"Now we\u0027re going to write the exact same thing,"},{"Start":"02:26.225 ","End":"02:30.195","Text":"but for our photon, our P Gamma."},{"Start":"02:30.195 ","End":"02:32.164","Text":"The momentum of our photon,"},{"Start":"02:32.164 ","End":"02:35.675","Text":"so let\u0027s see, its momentum in the x-direction."},{"Start":"02:35.675 ","End":"02:37.805","Text":"Let\u0027s see how we work this out."},{"Start":"02:37.805 ","End":"02:39.560","Text":"We know that it\u0027s going to have momentum in"},{"Start":"02:39.560 ","End":"02:41.975","Text":"the x-direction because it\u0027s moving in the x-direction."},{"Start":"02:41.975 ","End":"02:45.110","Text":"However, we know that our momentum is mass times velocity,"},{"Start":"02:45.110 ","End":"02:49.010","Text":"and we just said that the mass of a photon is equal to 0."},{"Start":"02:49.010 ","End":"02:54.770","Text":"What we\u0027re going to use is our equation from relativity,"},{"Start":"02:54.770 ","End":"02:58.565","Text":"which says that our energy is equal to the size"},{"Start":"02:58.565 ","End":"03:03.370","Text":"of our momentum multiplied by the speed of light."},{"Start":"03:03.370 ","End":"03:07.895","Text":"That means that we can rearrange this equation to say that"},{"Start":"03:07.895 ","End":"03:14.335","Text":"our momentum P is equal to our energy divided by the speed of light."},{"Start":"03:14.335 ","End":"03:19.820","Text":"Now here we know that the energy that a photon has is called E Gamma."},{"Start":"03:19.820 ","End":"03:22.010","Text":"This is our unknown,"},{"Start":"03:22.010 ","End":"03:23.810","Text":"which is what we\u0027re trying to find,"},{"Start":"03:23.810 ","End":"03:27.830","Text":"divided by c, which is this over here."},{"Start":"03:27.830 ","End":"03:31.460","Text":"Then we can see that our momentum is only in the x-direction,"},{"Start":"03:31.460 ","End":"03:36.530","Text":"which means that we have no components for momentum in the y or z-direction."},{"Start":"03:36.530 ","End":"03:38.870","Text":"Then it\u0027s energy,"},{"Start":"03:38.870 ","End":"03:43.590","Text":"our fourth component in our 4-vector is E divided by c,"},{"Start":"03:43.590 ","End":"03:44.880","Text":"which again, as we know,"},{"Start":"03:44.880 ","End":"03:50.050","Text":"is our E Gamma and again divided by c, our speed of light."},{"Start":"03:50.050 ","End":"03:52.505","Text":"It\u0027s this E Gamma,"},{"Start":"03:52.505 ","End":"03:55.410","Text":"which we\u0027re trying to figure out."},{"Start":"03:56.120 ","End":"04:01.055","Text":"Now we\u0027re going to do is we\u0027re going to write a 4-vector for"},{"Start":"04:01.055 ","End":"04:04.610","Text":"the total momentum of everything before"},{"Start":"04:04.610 ","End":"04:08.160","Text":"the collision of the photon and the proton together."},{"Start":"04:08.160 ","End":"04:11.233","Text":"How we do that is we\u0027re just going to add components,"},{"Start":"04:11.233 ","End":"04:15.245","Text":"so our x component or our first component from here and from here."},{"Start":"04:15.245 ","End":"04:18.550","Text":"That\u0027s simply going to be E Gamma divided by c,"},{"Start":"04:18.550 ","End":"04:20.730","Text":"0 plus 0 is 0,"},{"Start":"04:20.730 ","End":"04:22.550","Text":"0 plus 0 is 0,"},{"Start":"04:22.550 ","End":"04:28.665","Text":"and then we\u0027re going to have m_p c plus E Gamma"},{"Start":"04:28.665 ","End":"04:35.390","Text":"divided by c. These are all of the equations that we need for before the collision,"},{"Start":"04:35.390 ","End":"04:39.860","Text":"and now we\u0027re going to deal with the equations for after the collision."},{"Start":"04:39.860 ","End":"04:42.860","Text":"Now, the important thing to remember at"},{"Start":"04:42.860 ","End":"04:45.870","Text":"this stage is what we learned in the previous video."},{"Start":"04:45.870 ","End":"04:53.615","Text":"That is, that in order for us to find the minimum energy required after some collision."},{"Start":"04:53.615 ","End":"04:55.535","Text":"In order to find that,"},{"Start":"04:55.535 ","End":"04:59.150","Text":"we remember that our particles in our frame"},{"Start":"04:59.150 ","End":"05:03.175","Text":"of reference of our center of mass must be stationary,"},{"Start":"05:03.175 ","End":"05:04.845","Text":"so that\u0027s Number 1."},{"Start":"05:04.845 ","End":"05:10.565","Text":"Number 2 is that when we\u0027re using a center of mass frame of reference,"},{"Start":"05:10.565 ","End":"05:18.960","Text":"that means that the total momentum of the system must be equal to 0."},{"Start":"05:20.180 ","End":"05:23.570","Text":"That means that if both of our particles are at"},{"Start":"05:23.570 ","End":"05:27.380","Text":"rest in our center of mass frame of reference."},{"Start":"05:27.380 ","End":"05:30.750","Text":"That means that the momentum of"},{"Start":"05:30.750 ","End":"05:36.710","Text":"our pion is going to be equal to the momentum of our new proton,"},{"Start":"05:36.710 ","End":"05:42.050","Text":"and those are both going to be equal to 0."},{"Start":"05:42.050 ","End":"05:47.120","Text":"Because we know through our equations that momentum is mass times velocity,"},{"Start":"05:47.120 ","End":"05:52.060","Text":"and velocity must be equal to 0 in order to find a minimum energy."},{"Start":"05:52.060 ","End":"05:54.095","Text":"Now, what we\u0027re going to do,"},{"Start":"05:54.095 ","End":"05:57.785","Text":"is we\u0027re going to write our equations for energy for each particle."},{"Start":"05:57.785 ","End":"06:01.490","Text":"We\u0027ll have that our energy for our pion is going to be"},{"Start":"06:01.490 ","End":"06:05.540","Text":"equal to what we wrote in this equation over here,"},{"Start":"06:05.540 ","End":"06:07.175","Text":"our rest mass energy."},{"Start":"06:07.175 ","End":"06:10.430","Text":"It\u0027s going to be the mass of the pion which is given to"},{"Start":"06:10.430 ","End":"06:14.045","Text":"us in the question multiplied by c^2."},{"Start":"06:14.045 ","End":"06:15.980","Text":"I remind you we\u0027re working out"},{"Start":"06:15.980 ","End":"06:20.690","Text":"our rest mass energy because we know that both of our particles after"},{"Start":"06:20.690 ","End":"06:23.480","Text":"the collision in our center of mass frame of"},{"Start":"06:23.480 ","End":"06:27.640","Text":"reference are going to be stationary, they\u0027re both resting."},{"Start":"06:27.640 ","End":"06:31.795","Text":"Then we also have our rest mass for our new proton,"},{"Start":"06:31.795 ","End":"06:35.630","Text":"which is again going to be the mass of the proton which is given to us in"},{"Start":"06:35.630 ","End":"06:40.385","Text":"the question and multiplied by the speed of light squared."},{"Start":"06:40.385 ","End":"06:46.040","Text":"Now that we know our energies or our rest mass energies,"},{"Start":"06:46.040 ","End":"06:49.565","Text":"what we can do is we can write 4-vector for"},{"Start":"06:49.565 ","End":"06:55.280","Text":"the total momentum of the system after our collision."},{"Start":"06:55.280 ","End":"06:58.400","Text":"This is before and this is after."},{"Start":"06:58.400 ","End":"07:01.610","Text":"We know that our momentum in the x, y,"},{"Start":"07:01.610 ","End":"07:07.830","Text":"and z-direction is going to be equal to 0 because we\u0027re at rest."},{"Start":"07:07.830 ","End":"07:10.670","Text":"Then our energy is going to be equal to"},{"Start":"07:10.670 ","End":"07:17.340","Text":"the rest mass energy of our pion plus the rest mass energy of our proton."},{"Start":"07:17.420 ","End":"07:23.245","Text":"Remember that our equation for this section over here is E over c,"},{"Start":"07:23.245 ","End":"07:29.115","Text":"so E over c over here will be m_Pi multiplied by c. Over here,"},{"Start":"07:29.115 ","End":"07:35.785","Text":"it will be m_p multiplied by c. We\u0027ll just going to add them all together."},{"Start":"07:35.785 ","End":"07:46.500","Text":"Plus m_p multiplied by c. Now we have all of our equations,"},{"Start":"07:46.500 ","End":"07:51.360","Text":"but the question is, why am I using this 4-vector?"},{"Start":"07:51.880 ","End":"07:54.290","Text":"What I could have done instead,"},{"Start":"07:54.290 ","End":"07:58.745","Text":"is take these equations for my energy and then transform them"},{"Start":"07:58.745 ","End":"08:03.845","Text":"back into equations that work in my lab\u0027s frame of reference."},{"Start":"08:03.845 ","End":"08:06.870","Text":"Because over here they\u0027re in the frame of reference of the center of mass,"},{"Start":"08:06.870 ","End":"08:11.750","Text":"so I have to transform them back into my lab\u0027s frame of reference."},{"Start":"08:11.750 ","End":"08:16.085","Text":"Do the same thing also for my momentum equations."},{"Start":"08:16.085 ","End":"08:21.785","Text":"Then I would use my conservation of energy and momentum in order to"},{"Start":"08:21.785 ","End":"08:27.995","Text":"figure this question out but as you can see from all of the steps that I just mentioned,"},{"Start":"08:27.995 ","End":"08:32.645","Text":"it\u0027s going to be pretty difficult to finicky and a bit of a headache in order to do that."},{"Start":"08:32.645 ","End":"08:36.060","Text":"What I can do instead is this."},{"Start":"08:36.110 ","End":"08:40.545","Text":"If I take the size of my 4-vector."},{"Start":"08:40.545 ","End":"08:45.280","Text":"Now, I\u0027m just going to remind you what the size of a 4-vector is."},{"Start":"08:45.280 ","End":"08:49.540","Text":"It\u0027s going to be my P_x ^2 plus"},{"Start":"08:49.540 ","End":"08:57.455","Text":"my P_y ^2 plus my P_z ^2 minus my fourth component,"},{"Start":"08:57.455 ","End":"09:01.620","Text":"which is my E over c and that squared,"},{"Start":"09:01.620 ","End":"09:04.540","Text":"so this is the size of my 4-vector."},{"Start":"09:04.540 ","End":"09:07.080","Text":"Remember here there\u0027s a minus over here,"},{"Start":"09:07.080 ","End":"09:08.510","Text":"so that\u0027s very important."},{"Start":"09:08.510 ","End":"09:13.685","Text":"Now, the interesting and very important thing about this is that"},{"Start":"09:13.685 ","End":"09:18.625","Text":"it\u0027s invariant 2 frames of reference. What does that mean?"},{"Start":"09:18.625 ","End":"09:21.930","Text":"That means that the size of my 4-vector"},{"Start":"09:21.930 ","End":"09:25.640","Text":"in my center of mass frame of reference is going to"},{"Start":"09:25.640 ","End":"09:33.145","Text":"be equal to the size of my 4-vector from my lab\u0027s frame of reference over here."},{"Start":"09:33.145 ","End":"09:37.550","Text":"The size will be exactly the same and it doesn\u0027t matter which frame of"},{"Start":"09:37.550 ","End":"09:42.755","Text":"reference I\u0027m looking at. Let\u0027s see."},{"Start":"09:42.755 ","End":"09:46.865","Text":"Let\u0027s write the size of our 4-vector before our collision."},{"Start":"09:46.865 ","End":"09:49.535","Text":"We\u0027re going to have our P_x ^2,"},{"Start":"09:49.535 ","End":"09:53.735","Text":"so that\u0027s going to be our E Gamma divided by c squared,"},{"Start":"09:53.735 ","End":"09:58.575","Text":"and then plus our P_y ^2 squared plus our P_z ^2, which is 0."},{"Start":"09:58.575 ","End":"10:02.280","Text":"Then minus our fourth component squared,"},{"Start":"10:02.280 ","End":"10:10.195","Text":"so we\u0027ll have m_p c plus E Gamma divided by c squared."},{"Start":"10:10.195 ","End":"10:17.695","Text":"Then this is going to be equal to my 4-vector after the collision,"},{"Start":"10:17.695 ","End":"10:19.740","Text":"so the size of that."},{"Start":"10:19.740 ","End":"10:21.090","Text":"I have my P_x,"},{"Start":"10:21.090 ","End":"10:22.605","Text":"P_y, and P_z squared,"},{"Start":"10:22.605 ","End":"10:28.251","Text":"so that\u0027s 0^2 plus 0^2 plus 0^2 plus my fourth component squared."},{"Start":"10:28.251 ","End":"10:37.190","Text":"That\u0027s going to be m_Pi plus m_p squared, c^2."},{"Start":"10:37.190 ","End":"10:40.059","Text":"Now when I open up my brackets,"},{"Start":"10:40.059 ","End":"10:49.914","Text":"I\u0027ll have E Gamma divided by c^2 minus E Gamma^2"},{"Start":"10:49.914 ","End":"10:54.175","Text":"divided by c^2 minus"},{"Start":"10:54.175 ","End":"11:02.030","Text":"2E Gamma m_p minus m_p^2c^2."},{"Start":"11:04.100 ","End":"11:07.290","Text":"I just opened up all of these brackets,"},{"Start":"11:07.290 ","End":"11:11.025","Text":"and that\u0027s going to be equal to this over here."},{"Start":"11:11.025 ","End":"11:15.975","Text":"That\u0027s going to be m_Pi^2c^2"},{"Start":"11:15.975 ","End":"11:24.203","Text":"minus 2m_Pi m_p"},{"Start":"11:24.203 ","End":"11:31.320","Text":"c^2 minus m_p^2c^2."},{"Start":"11:31.320 ","End":"11:34.860","Text":"Now we can see that these two terms will cross off."},{"Start":"11:34.860 ","End":"11:39.075","Text":"Then this term and this term will cross off."},{"Start":"11:39.075 ","End":"11:43.160","Text":"Then all I have to do is I have to just do some algebra to"},{"Start":"11:43.160 ","End":"11:48.490","Text":"rearrange over here in order to isolate out my E Gamma."},{"Start":"11:48.490 ","End":"11:50.540","Text":"Then in the end, once I\u0027ve done that,"},{"Start":"11:50.540 ","End":"11:59.320","Text":"I\u0027ll get them our E Gamma is equal to 1 divided by 2m_p multiplied by"},{"Start":"11:59.320 ","End":"12:04.790","Text":"m_Pi^2 plus"},{"Start":"12:04.790 ","End":"12:13.110","Text":"2m_Pi m_p multiplied by c^2."},{"Start":"12:14.510 ","End":"12:21.770","Text":"This is the energy required by the photon in order for this reaction to occur,"},{"Start":"12:21.770 ","End":"12:26.270","Text":"in order for a pion and another proton to be fond."},{"Start":"12:26.270 ","End":"12:27.845","Text":"Now also a little note,"},{"Start":"12:27.845 ","End":"12:30.655","Text":"I forgot to add here a minus."},{"Start":"12:30.655 ","End":"12:35.700","Text":"Because remember that it\u0027s minus our fourth component squared."},{"Start":"12:35.700 ","End":"12:40.560","Text":"I forgot to put a minus over here for our fourth component squared."},{"Start":"12:41.360 ","End":"12:43.850","Text":"That\u0027s the end of this question."},{"Start":"12:43.850 ","End":"12:46.250","Text":"Just to have a little recap of what we have"},{"Start":"12:46.250 ","End":"12:48.755","Text":"to do in order to answer these types of questions."},{"Start":"12:48.755 ","End":"12:53.645","Text":"All we have to do is write out a 4-vector for it before the collision,"},{"Start":"12:53.645 ","End":"12:56.750","Text":"using and remembering these equations,"},{"Start":"12:56.750 ","End":"12:58.790","Text":"so the total 4-vector,"},{"Start":"12:58.790 ","End":"13:01.895","Text":"then write the total 4-vector for after the collision."},{"Start":"13:01.895 ","End":"13:05.680","Text":"Find the size of the two 4-vectors,"},{"Start":"13:05.680 ","End":"13:07.835","Text":"remembering that here there\u0027s a minus,"},{"Start":"13:07.835 ","End":"13:10.340","Text":"and then setting them equal to one another."},{"Start":"13:10.340 ","End":"13:14.345","Text":"Then that\u0027s it we just solve in order to find our unknown."},{"Start":"13:14.345 ","End":"13:18.150","Text":"That\u0027s the end of this lesson."}],"ID":9570},{"Watched":false,"Name":"Compton Scattering","Duration":"5m 36s","ChapterTopicVideoID":9264,"CourseChapterTopicPlaylistID":9398,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.815","Text":"Hello. In this lesson,"},{"Start":"00:01.815 ","End":"00:04.725","Text":"we\u0027re going to be speaking about Compton scattering."},{"Start":"00:04.725 ","End":"00:09.720","Text":"Now what happens in Compton scattering is that we have an electron,"},{"Start":"00:09.720 ","End":"00:12.660","Text":"and then we have a photon,"},{"Start":"00:12.660 ","End":"00:15.435","Text":"which I\u0027m going to symbolize by Gamma,"},{"Start":"00:15.435 ","End":"00:17.940","Text":"which travels towards it and hits it."},{"Start":"00:17.940 ","End":"00:24.770","Text":"Now what happens after the collision is that our photon will fly off in this direction,"},{"Start":"00:24.770 ","End":"00:30.110","Text":"which according to the x-direction is going to be at an angle of Theta,"},{"Start":"00:30.110 ","End":"00:35.120","Text":"and our electron will fly off in this angle over here,"},{"Start":"00:35.120 ","End":"00:37.310","Text":"which we\u0027ll call Phi."},{"Start":"00:37.310 ","End":"00:39.664","Text":"Now, before the collision,"},{"Start":"00:39.664 ","End":"00:43.775","Text":"we\u0027re given the energy of our photon which is E Gamma,"},{"Start":"00:43.775 ","End":"00:47.210","Text":"and of course, then we also know our momentum,"},{"Start":"00:47.210 ","End":"00:54.080","Text":"which is simply going to be equal to E Gamma divided by c. Now after the collision,"},{"Start":"00:54.080 ","End":"00:58.355","Text":"but still in the same frame of reference just after the collision."},{"Start":"00:58.355 ","End":"01:01.190","Text":"We can label that our energy"},{"Start":"01:01.190 ","End":"01:04.250","Text":"after the collision in the same frame of reference as E Gamma,"},{"Start":"01:04.250 ","End":"01:06.815","Text":"and that our momentum after the collision,"},{"Start":"01:06.815 ","End":"01:08.660","Text":"also in the same frame of reference,"},{"Start":"01:08.660 ","End":"01:14.705","Text":"is going to be equal to our energy after the collision divided by c. Now,"},{"Start":"01:14.705 ","End":"01:18.770","Text":"we can also say that after the collision that our electron is going to"},{"Start":"01:18.770 ","End":"01:23.540","Text":"have some momentum p and its energy is going to be"},{"Start":"01:23.540 ","End":"01:33.190","Text":"equal to the square root of pc^2 plus the m,"},{"Start":"01:33.190 ","End":"01:36.620","Text":"mass of the electron multiplied by c^2."},{"Start":"01:36.620 ","End":"01:38.600","Text":"All of that squared,"},{"Start":"01:38.600 ","End":"01:41.065","Text":"and the square root of all of this."},{"Start":"01:41.065 ","End":"01:45.020","Text":"All we have to do over here is we have to"},{"Start":"01:45.020 ","End":"01:49.190","Text":"use the idea of conservation of momentum and conservation of energy."},{"Start":"01:49.190 ","End":"01:52.070","Text":"Let\u0027s write out our equations now."},{"Start":"01:52.070 ","End":"01:58.340","Text":"We can say that our initial energy before the collision is going to be equal"},{"Start":"01:58.340 ","End":"02:04.640","Text":"to the energy of our photon plus the rest mass energy of our electron,"},{"Start":"02:04.640 ","End":"02:06.845","Text":"which as we know, is going to be equal to the mass of"},{"Start":"02:06.845 ","End":"02:11.010","Text":"the electron multiplied by the speed of light squared."},{"Start":"02:11.090 ","End":"02:15.565","Text":"This was the initial energy and then from conservation of energy,"},{"Start":"02:15.565 ","End":"02:20.124","Text":"this has to be equal to the energy after the collision,"},{"Start":"02:20.124 ","End":"02:23.430","Text":"which is simply going to be equal to over here,"},{"Start":"02:23.430 ","End":"02:32.100","Text":"so I have the square root of pc^2 plus our m_ec^2 squared,"},{"Start":"02:32.100 ","End":"02:34.380","Text":"squared and all of that is square"},{"Start":"02:34.380 ","End":"02:40.495","Text":"rooted plus our energy of the photon after the collision,"},{"Start":"02:40.495 ","End":"02:43.325","Text":"which is E Gamma tag."},{"Start":"02:43.325 ","End":"02:46.210","Text":"That\u0027s the equation for conservation of energy."},{"Start":"02:46.210 ","End":"02:50.200","Text":"Now let\u0027s write out our equation for conservation of momentum."},{"Start":"02:50.200 ","End":"02:55.520","Text":"We\u0027ll write that our momentum in the x-direction."},{"Start":"02:55.560 ","End":"03:03.770","Text":"Our Px initial is going to be equal to the momentum of the photon,"},{"Start":"03:03.770 ","End":"03:05.690","Text":"which is only in the x-direction,"},{"Start":"03:05.690 ","End":"03:08.660","Text":"so we know that that\u0027s going to be equal to E Gamma"},{"Start":"03:08.660 ","End":"03:11.870","Text":"divided by c plus the momentum of the electron,"},{"Start":"03:11.870 ","End":"03:14.645","Text":"which is stationary so that\u0027s going to be equal to 0,"},{"Start":"03:14.645 ","End":"03:19.175","Text":"and then that equals to the momentum in the x-direction after."},{"Start":"03:19.175 ","End":"03:24.485","Text":"As we can see, we\u0027re going to have the momentum of the x-direction and then,"},{"Start":"03:24.485 ","End":"03:28.175","Text":"sorry, the momentum of the electron in the x-direction,"},{"Start":"03:28.175 ","End":"03:35.795","Text":"so that\u0027s going to be P cosine Phi plus the momentum of the photon in the x-direction,"},{"Start":"03:35.795 ","End":"03:39.150","Text":"which is going to be equal to"},{"Start":"03:39.680 ","End":"03:47.050","Text":"E Gamma tag divided by c multiplied by cosine of Theta."},{"Start":"03:47.050 ","End":"03:52.895","Text":"Now for conservation of momentum in the y-direction."},{"Start":"03:52.895 ","End":"03:57.035","Text":"We\u0027ll have our initial momentum in the y-direction,"},{"Start":"03:57.035 ","End":"04:00.515","Text":"which as we can see, the photon has no momentum in the y-direction,"},{"Start":"04:00.515 ","End":"04:05.340","Text":"and our electron also has no momentum in the y-direction,"},{"Start":"04:05.340 ","End":"04:12.060","Text":"and then that\u0027s going to be equal to the momentum of the electron in the y-direction,"},{"Start":"04:12.060 ","End":"04:19.715","Text":"which is going to be P sine of Phi plus the momentum of the photon in the y-direction,"},{"Start":"04:19.715 ","End":"04:26.470","Text":"which is going to be E Gamma tag divided by c sine of Theta."},{"Start":"04:26.470 ","End":"04:30.020","Text":"Now what we have to do is we have to isolate out"},{"Start":"04:30.020 ","End":"04:33.620","Text":"some things and do some algebra and substitution,"},{"Start":"04:33.620 ","End":"04:39.655","Text":"and what we\u0027ll get from these 3 equations is one final equation afterwards,"},{"Start":"04:39.655 ","End":"04:41.180","Text":"I\u0027ll write it in red,"},{"Start":"04:41.180 ","End":"04:42.950","Text":"which is going to be useful to us,"},{"Start":"04:42.950 ","End":"04:52.700","Text":"which is equal to 1 divided by E Gamma tag minus 1 divided by E Gamma is"},{"Start":"04:52.700 ","End":"04:57.395","Text":"going to be equal to 1 divided by m_ec^2"},{"Start":"04:57.395 ","End":"05:05.500","Text":"multiplied by 1 minus cosine of Theta."},{"Start":"05:05.500 ","End":"05:10.160","Text":"What this equation does is it gives us a relationship between"},{"Start":"05:10.160 ","End":"05:16.265","Text":"the initial energy of the photon to the final energy of the photon,"},{"Start":"05:16.265 ","End":"05:21.275","Text":"and it also shows the relationship between these values for energy and"},{"Start":"05:21.275 ","End":"05:26.735","Text":"the angle that the photon bounces out at or is scattered out."},{"Start":"05:26.735 ","End":"05:31.655","Text":"So this is the equation that you want to remember and that\u0027s it."},{"Start":"05:31.655 ","End":"05:34.205","Text":"That\u0027s the story with Compton scattering."},{"Start":"05:34.205 ","End":"05:36.390","Text":"That\u0027s the end of this lesson."}],"ID":9571}],"Thumbnail":null,"ID":9398}]

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