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[{"Name":"Introduction to Electromagnetic Waves","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Deriving the Wave Equations Using Maxwell","Duration":"19m 58s","ChapterTopicVideoID":21559,"CourseChapterTopicPlaylistID":99477,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/21559.jpeg","UploadDate":"2020-04-21T21:46:07.3800000","DurationForVideoObject":"PT19M58S","Description":null,"MetaTitle":"23-1do Deriving the Wave Equations Using Maxwell: Video + Workbook | Proprep","MetaDescription":"Electromagnetic Waves - Introduction to Electromagnetic Waves. 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-2-electricity-and-magnetism/electromagnetic-waves/introduction-to-electromagnetic-waves/vid22346","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.935","Text":"Hello. In this lesson,"},{"Start":"00:01.935 ","End":"00:07.335","Text":"we\u0027re going to be speaking about the wave equation and its derivation."},{"Start":"00:07.335 ","End":"00:12.240","Text":"The derivation isn\u0027t that useful for solving questions,"},{"Start":"00:12.240 ","End":"00:15.405","Text":"however, the wave equation is imperative."},{"Start":"00:15.405 ","End":"00:18.555","Text":"But we\u0027re going to look at how we derive it."},{"Start":"00:18.555 ","End":"00:21.540","Text":"Now, we\u0027ve seen these 4 equations before."},{"Start":"00:21.540 ","End":"00:28.780","Text":"These are 4 of Maxwell\u0027s equations that we\u0027ve seen in the previous chapters."},{"Start":"00:29.150 ","End":"00:34.275","Text":"The wave equation comes from these 4 equations."},{"Start":"00:34.275 ","End":"00:43.030","Text":"Now, what is important is that the wave equation speaks about waves that are in a vacuum."},{"Start":"00:43.160 ","End":"00:46.050","Text":"This is very important."},{"Start":"00:46.050 ","End":"00:47.345","Text":"What does this mean?"},{"Start":"00:47.345 ","End":"00:48.710","Text":"If it\u0027s in a vacuum,"},{"Start":"00:48.710 ","End":"00:51.650","Text":"that means that the space is completely empty."},{"Start":"00:51.650 ","End":"00:54.365","Text":"There is no other particles."},{"Start":"00:54.365 ","End":"00:58.475","Text":"What does that mean for us with respect to the equation?"},{"Start":"00:58.475 ","End":"01:00.875","Text":"If there\u0027s no other particles,"},{"Start":"01:00.875 ","End":"01:04.385","Text":"then that means that there\u0027s no rho,"},{"Start":"01:04.385 ","End":"01:07.625","Text":"because rho is the charge density."},{"Start":"01:07.625 ","End":"01:10.970","Text":"If there\u0027s no particles,"},{"Start":"01:10.970 ","End":"01:13.490","Text":"that means there\u0027s no charge density,"},{"Start":"01:13.490 ","End":"01:19.850","Text":"so this becomes 0 and then that case this whole equation is equal to 0."},{"Start":"01:19.850 ","End":"01:22.834","Text":"Additionally, if there\u0027s no particles,"},{"Start":"01:22.834 ","End":"01:26.715","Text":"then there can be no current passing through."},{"Start":"01:26.715 ","End":"01:31.169","Text":"We can also see that by the fact that there\u0027s no charge density."},{"Start":"01:31.310 ","End":"01:35.355","Text":"There\u0027s no charges causing the current."},{"Start":"01:35.355 ","End":"01:37.565","Text":"That means that J,"},{"Start":"01:37.565 ","End":"01:39.800","Text":"which is the current density,"},{"Start":"01:39.800 ","End":"01:42.205","Text":"is also equal to 0."},{"Start":"01:42.205 ","End":"01:47.674","Text":"Then we\u0027re just left with this section over here for this equation."},{"Start":"01:47.674 ","End":"01:52.430","Text":"What we can see is that we\u0027re left with 2 variables,"},{"Start":"01:52.430 ","End":"01:56.510","Text":"E for the electric field and B for the magnetic field."},{"Start":"01:56.510 ","End":"01:58.940","Text":"What do we want to do is we want to play around with"},{"Start":"01:58.940 ","End":"02:03.259","Text":"these equations in order to get to 2 different equations,"},{"Start":"02:03.259 ","End":"02:06.545","Text":"1 dependent solely on E and 1"},{"Start":"02:06.545 ","End":"02:11.510","Text":"dependent solely on B and those are called the wave equations."},{"Start":"02:11.510 ","End":"02:14.520","Text":"I\u0027ll write them out over here."},{"Start":"02:14.660 ","End":"02:23.030","Text":"These are the 2 wave equations and we can see that they\u0027re identical."},{"Start":"02:23.030 ","End":"02:28.640","Text":"Just one is with a variable of E and the other one is with a variable of B."},{"Start":"02:28.640 ","End":"02:31.250","Text":"What I\u0027m going to do is I\u0027m going to show how we get to"},{"Start":"02:31.250 ","End":"02:35.990","Text":"the first equation and to get to the second equation,"},{"Start":"02:35.990 ","End":"02:41.670","Text":"we do the exact same thing just with the magnetic field."},{"Start":"02:42.140 ","End":"02:44.790","Text":"Let\u0027s number these equations."},{"Start":"02:44.790 ","End":"02:46.950","Text":"This is 1, 2,"},{"Start":"02:46.950 ","End":"02:49.485","Text":"3, and 4."},{"Start":"02:49.485 ","End":"02:52.700","Text":"The first thing we\u0027re going to do is we\u0027re going to take"},{"Start":"02:52.700 ","End":"02:56.270","Text":"Equation 3. I\u0027ll write it like this."},{"Start":"02:56.270 ","End":"03:00.260","Text":"We\u0027re taking Equation 3 and we\u0027re doing a rotor on both sides."},{"Start":"03:00.260 ","End":"03:05.475","Text":"That means taking the nabla symbol cross all of this to both sides."},{"Start":"03:05.475 ","End":"03:11.720","Text":"What we have is nabla or Del cross and then Equation 3 so then we"},{"Start":"03:11.720 ","End":"03:17.930","Text":"have nabla cross E and this is equal to,"},{"Start":"03:17.930 ","End":"03:20.795","Text":"again, we\u0027re doing the same thing to both sides."},{"Start":"03:20.795 ","End":"03:29.340","Text":"nabla cross and then negative dB by dt."},{"Start":"03:30.500 ","End":"03:34.430","Text":"What we\u0027re going to do is we\u0027re going to look at"},{"Start":"03:34.430 ","End":"03:39.575","Text":"the left side of the equation now to show how we get to"},{"Start":"03:39.575 ","End":"03:43.850","Text":"the left side of the final equation and then we\u0027re going to deal with the right side"},{"Start":"03:43.850 ","End":"03:49.260","Text":"of the equation to see how we get to the right side of the final equation."},{"Start":"03:49.460 ","End":"03:53.645","Text":"Using mathematical laws and operators,"},{"Start":"03:53.645 ","End":"03:56.915","Text":"which you may have learned in one of your maths courses."},{"Start":"03:56.915 ","End":"04:06.815","Text":"We have seen or you may know that the rotor of a vector field"},{"Start":"04:06.815 ","End":"04:14.780","Text":"is equal to nabla.(nabla.the"},{"Start":"04:14.780 ","End":"04:24.840","Text":"vector field) minus nabla^2 of the vector field."},{"Start":"04:25.760 ","End":"04:28.040","Text":"From the rule of operators,"},{"Start":"04:28.040 ","End":"04:32.675","Text":"what we have here is the gradient of the divergence of the E field."},{"Start":"04:32.675 ","End":"04:38.250","Text":"Here we have the Laplacian if you\u0027ve heard of this before."},{"Start":"04:38.290 ","End":"04:42.095","Text":"Now notice these 2 terms are different."},{"Start":"04:42.095 ","End":"04:45.590","Text":"Here first we find the divergence of E,"},{"Start":"04:45.590 ","End":"04:48.740","Text":"and then we do nabla."},{"Start":"04:49.210 ","End":"04:54.135","Text":"Here we take the nabla, the Del,"},{"Start":"04:54.135 ","End":"04:55.990","Text":"and we square it,"},{"Start":"04:55.990 ","End":"05:00.235","Text":"and only then do we apply it to our vector field."},{"Start":"05:00.235 ","End":"05:02.980","Text":"These 2 are different."},{"Start":"05:02.980 ","End":"05:06.130","Text":"If you haven\u0027t learned this or you don\u0027t remember this,"},{"Start":"05:06.130 ","End":"05:12.790","Text":"you can either research it online and look at the mathematical introduction."},{"Start":"05:12.790 ","End":"05:18.768","Text":"But otherwise, it doesn\u0027t really matter as we said at the beginning of the lesson,"},{"Start":"05:18.768 ","End":"05:22.645","Text":"we really just want to know how to use these equations."},{"Start":"05:22.645 ","End":"05:26.830","Text":"Getting there is just also interesting and it\u0027s quite important to know."},{"Start":"05:26.830 ","End":"05:29.200","Text":"But for solving questions,"},{"Start":"05:29.200 ","End":"05:32.035","Text":"you don\u0027t really need to know this."},{"Start":"05:32.035 ","End":"05:35.480","Text":"Don\u0027t panic if you don\u0027t know what\u0027s going on here."},{"Start":"05:35.480 ","End":"05:42.405","Text":"We can see this side is equal to just this."},{"Start":"05:42.405 ","End":"05:47.780","Text":"Now, notice from Equation 1 here we have the divergence of E,"},{"Start":"05:47.780 ","End":"05:52.245","Text":"which from Equation 1 we see is equal to 0."},{"Start":"05:52.245 ","End":"05:56.365","Text":"This then cancels out, this is equal to 0."},{"Start":"05:56.365 ","End":"06:00.410","Text":"What we\u0027re left with is that the whole left side of"},{"Start":"06:00.410 ","End":"06:05.405","Text":"the equation is equal to the negative Laplacian."},{"Start":"06:05.405 ","End":"06:13.680","Text":"We can already see this Laplacian over here on this left side of the final equation."},{"Start":"06:14.650 ","End":"06:17.180","Text":"We\u0027ve gotten to this side."},{"Start":"06:17.180 ","End":"06:23.070","Text":"Now we\u0027re going to look at this section over here."},{"Start":"06:24.200 ","End":"06:26.580","Text":"Let\u0027s begin it over here."},{"Start":"06:26.580 ","End":"06:35.385","Text":"What we have is the rotor of negative dB by dt."},{"Start":"06:35.385 ","End":"06:38.720","Text":"First of all, the negative is like a constant."},{"Start":"06:38.720 ","End":"06:41.410","Text":"It\u0027s equivalent to negative 1."},{"Start":"06:41.410 ","End":"06:47.240","Text":"Because it\u0027s a constant we can just take it out and put it to the side over here."},{"Start":"06:47.240 ","End":"06:49.895","Text":"Also when we\u0027re looking at the rotor,"},{"Start":"06:49.895 ","End":"06:57.890","Text":"this nabla or this Del is taking partial differentials with respect to space."},{"Start":"06:57.890 ","End":"07:02.270","Text":"Coordinates in space, for instance, the x, y,"},{"Start":"07:02.270 ","End":"07:07.415","Text":"z axis or r Theta z or r Theta Phi,"},{"Start":"07:07.415 ","End":"07:10.340","Text":"depending on the coordinate system that we\u0027re using."},{"Start":"07:10.340 ","End":"07:15.140","Text":"Whereas this differential is with respect to time."},{"Start":"07:15.140 ","End":"07:20.330","Text":"Of course, over here we can see that a differential with"},{"Start":"07:20.330 ","End":"07:25.595","Text":"respect to time and with respect to space are independent of one another."},{"Start":"07:25.595 ","End":"07:29.840","Text":"In other words, we can look at this d by"},{"Start":"07:29.840 ","End":"07:35.900","Text":"dt as also being a constant which is independent of x,"},{"Start":"07:35.900 ","End":"07:40.655","Text":"y, z, or coordinate systems relating to space."},{"Start":"07:40.655 ","End":"07:50.570","Text":"Therefore, we can just write this as being equal to negative d by dt of"},{"Start":"07:50.570 ","End":"08:00.855","Text":"the rotor of the magnetic field like so."},{"Start":"08:00.855 ","End":"08:06.500","Text":"Then we can rewrite this as being equal to negative d by"},{"Start":"08:06.500 ","End":"08:13.170","Text":"dt and now the rotor of the magnetic field."},{"Start":"08:13.170 ","End":"08:17.870","Text":"This we can see over here in Maxwell\u0027s Equation number 4,"},{"Start":"08:17.870 ","End":"08:20.480","Text":"we can see the rotor of the magnetic field where"},{"Start":"08:20.480 ","End":"08:24.485","Text":"this term canceled out and we\u0027re left with this term over here,"},{"Start":"08:24.485 ","End":"08:28.670","Text":"which is equal to Mu_naught,"},{"Start":"08:28.670 ","End":"08:34.750","Text":"Epsilon_naught dE by dt."},{"Start":"08:34.750 ","End":"08:38.680","Text":"Then of course we can see that"},{"Start":"08:38.680 ","End":"08:43.180","Text":"Mu_naught and Epsilon_naught I like constants so what we\u0027re left with is"},{"Start":"08:43.180 ","End":"08:46.690","Text":"negative Mu_naught Epsilon_naught because"},{"Start":"08:46.690 ","End":"08:50.260","Text":"we can move them out because they\u0027re like constants d by"},{"Start":"08:50.260 ","End":"08:59.290","Text":"dt of d by dt of the electric field or in other words,"},{"Start":"08:59.290 ","End":"09:03.415","Text":"this is equal to negative Mu_naught Epsilon_naught"},{"Start":"09:03.415 ","End":"09:08.980","Text":"over d^2 E"},{"Start":"09:08.980 ","End":"09:13.930","Text":"by dt^2,"},{"Start":"09:13.930 ","End":"09:16.315","Text":"which is exactly the same as over here."},{"Start":"09:16.315 ","End":"09:22.780","Text":"Now of course, then we can say that here that this is equal"},{"Start":"09:22.780 ","End":"09:31.555","Text":"to this over here so we have negative so here we have this negative over here,"},{"Start":"09:31.555 ","End":"09:36.340","Text":"and here we have this negative over here so we can multiply just both sides."},{"Start":"09:36.340 ","End":"09:41.650","Text":"Remember this is equal to the left side and this is equal to the right side of"},{"Start":"09:41.650 ","End":"09:46.500","Text":"the equation so we can just multiply both sides of the equation"},{"Start":"09:46.500 ","End":"09:52.420","Text":"by negative 1 to get rid of the negative sign and we\u0027re left with this over here."},{"Start":"09:52.560 ","End":"09:59.590","Text":"As we\u0027ve seen, the equations are exactly the same Just one has the variable E and 1 has"},{"Start":"09:59.590 ","End":"10:03.775","Text":"the variable B and we"},{"Start":"10:03.775 ","End":"10:08.680","Text":"derive both of them so also the second equation in the exact same way,"},{"Start":"10:08.680 ","End":"10:12.190","Text":"we do router to both sides,"},{"Start":"10:12.190 ","End":"10:18.220","Text":"and then we substitute in this equation over here and we"},{"Start":"10:18.220 ","End":"10:25.120","Text":"just solve and another explanation which we may speak about later is to show,"},{"Start":"10:25.120 ","End":"10:27.235","Text":"or you might learn it in a different course,"},{"Start":"10:27.235 ","End":"10:30.940","Text":"is that the electric field and the magnetic field are actually the same just"},{"Start":"10:30.940 ","End":"10:34.990","Text":"from different frames of reference gets not important"},{"Start":"10:34.990 ","End":"10:38.800","Text":"to know now just notes and then that\u0027s another explanation of"},{"Start":"10:38.800 ","End":"10:44.395","Text":"why this is a very similar equation."},{"Start":"10:44.395 ","End":"10:46.855","Text":"Now, another thing to note,"},{"Start":"10:46.855 ","End":"10:48.670","Text":"we have this constant over here,"},{"Start":"10:48.670 ","End":"10:52.675","Text":"Mu_naught Epsilon_naught so Mu_naught"},{"Start":"10:52.675 ","End":"11:00.735","Text":"Epsilon_naught multiplied together is also equal to 1 divided by C^2,"},{"Start":"11:00.735 ","End":"11:06.700","Text":"where C is the speed of light."},{"Start":"11:06.700 ","End":"11:10.029","Text":"Now, let\u0027s speak a little bit about the equations."},{"Start":"11:10.029 ","End":"11:13.990","Text":"I\u0027m just going to speak about the equation with the variable E for"},{"Start":"11:13.990 ","End":"11:17.770","Text":"the electric field but everything I say"},{"Start":"11:17.770 ","End":"11:25.945","Text":"about this equation is also correct for the equation relating to the magnetic field."},{"Start":"11:25.945 ","End":"11:28.225","Text":"Let\u0027s just look at the electric field."},{"Start":"11:28.225 ","End":"11:33.190","Text":"The equation for the electric field over here is actually made of"},{"Start":"11:33.190 ","End":"11:38.620","Text":"3 equations because we have over here this nabla,"},{"Start":"11:38.620 ","End":"11:40.540","Text":"so whenever we see the nabla,"},{"Start":"11:40.540 ","End":"11:46.000","Text":"we know that it separates out the vector field into each of its components in space."},{"Start":"11:46.000 ","End":"11:48.429","Text":"Let\u0027s say we\u0027re using Cartesian coordinates."},{"Start":"11:48.429 ","End":"11:52.540","Text":"It\u0027s with respect to the components in the x-direction,"},{"Start":"11:52.540 ","End":"11:55.850","Text":"y-direction, and z-direction."},{"Start":"11:56.370 ","End":"12:02.860","Text":"We can see that this would be nabla^2 Ex,"},{"Start":"12:02.860 ","End":"12:06.850","Text":"which is equal to Mu_naught Epsilon_naught"},{"Start":"12:06.850 ","End":"12:15.760","Text":"d^2 E_x by dt^2 or nabla^2Ey,"},{"Start":"12:15.760 ","End":"12:21.490","Text":"which is equal to Mu_naught Epsilon_naught d^2E_y"},{"Start":"12:21.490 ","End":"12:30.010","Text":"by dt^2 and nabla^2 E_z,"},{"Start":"12:30.010 ","End":"12:38.270","Text":"which is equal to Mu_naught Epsilon_naught d^2 E_z by dt^2."},{"Start":"12:38.790 ","End":"12:41.320","Text":"This is the exact same thing,"},{"Start":"12:41.320 ","End":"12:51.070","Text":"like when we saw Sigma F is equal to ma so that means that we would have Sigma,"},{"Start":"12:51.070 ","End":"12:57.865","Text":"sorry, this is Sigma F_x is equal to ma in the x direction."},{"Start":"12:57.865 ","End":"13:02.380","Text":"The sum of all the forces in the y-direction would equal to the mass multiplied"},{"Start":"13:02.380 ","End":"13:08.180","Text":"by the acceleration in the y-direction and the same for the z-axis."},{"Start":"13:09.000 ","End":"13:12.370","Text":"Now let\u0027s look at each one of these,"},{"Start":"13:12.370 ","End":"13:15.970","Text":"what it means so over here we understand what\u0027s going on,"},{"Start":"13:15.970 ","End":"13:19.360","Text":"on the right side of the equation so these are"},{"Start":"13:19.360 ","End":"13:23.950","Text":"constants multiplied by the E_x component,"},{"Start":"13:23.950 ","End":"13:29.730","Text":"which has been derived twice with respect to t,"},{"Start":"13:29.730 ","End":"13:34.470","Text":"or its derivative has been taken twice with respect to t so if"},{"Start":"13:34.470 ","End":"13:39.990","Text":"our electric field was some constant t^2,"},{"Start":"13:39.990 ","End":"13:42.615","Text":"so we would take its derivative once,"},{"Start":"13:42.615 ","End":"13:46.155","Text":"which would give 2AT,"},{"Start":"13:46.155 ","End":"13:51.250","Text":"and then the second time derivative would just leave us with 2A."},{"Start":"13:52.290 ","End":"13:57.280","Text":"That\u0027s what we do here, that\u0027s not very new but now let\u0027s look at the left side,"},{"Start":"13:57.280 ","End":"14:06.010","Text":"which is something that is new to us so we have this nabla^2."},{"Start":"14:06.010 ","End":"14:07.810","Text":"What is nabla^2? Nabla,"},{"Start":"14:07.810 ","End":"14:11.755","Text":"we know are the partial derivatives so if it\u0027s squared,"},{"Start":"14:11.755 ","End":"14:22.090","Text":"it\u0027s just the partial derivative squared so we have d2 by dx^2 plus d2"},{"Start":"14:22.090 ","End":"14:30.850","Text":"by dy^2 plus d2 by dz^2 in"},{"Start":"14:30.850 ","End":"14:37.330","Text":"Cartesian coordinates so in actual fact what we\u0027re doing when"},{"Start":"14:37.330 ","End":"14:44.350","Text":"we have this nabla^2E so here we split it up into 3 equations."},{"Start":"14:44.350 ","End":"14:45.910","Text":"In this section,"},{"Start":"14:45.910 ","End":"14:50.710","Text":"we\u0027re taking the x-component of the magnetic field and"},{"Start":"14:50.710 ","End":"14:55.525","Text":"we are taking its partial derivatives with respect to x twice,"},{"Start":"14:55.525 ","End":"14:56.980","Text":"with respect to y twice,"},{"Start":"14:56.980 ","End":"14:58.705","Text":"and with respect to z twice,"},{"Start":"14:58.705 ","End":"15:01.795","Text":"and then adding them up."},{"Start":"15:01.795 ","End":"15:06.400","Text":"Over here we would have dE_x with respect to"},{"Start":"15:06.400 ","End":"15:13.495","Text":"x like so and then we just add them up."},{"Start":"15:13.495 ","End":"15:20.290","Text":"That is all that this means so let\u0027s just go over this again."},{"Start":"15:20.290 ","End":"15:23.200","Text":"Let\u0027s imagine or not imagine our E field is"},{"Start":"15:23.200 ","End":"15:26.365","Text":"a vector field so it could be something like this,"},{"Start":"15:26.365 ","End":"15:30.190","Text":"xy, y^2, z."},{"Start":"15:30.190 ","End":"15:36.550","Text":"Let\u0027s just say that this is our vector field so let\u0027s say I\u0027m doing"},{"Start":"15:36.550 ","End":"15:39.310","Text":"this equation so this side of the equation just for"},{"Start":"15:39.310 ","End":"15:43.615","Text":"the x-component so I\u0027m just looking at this over here."},{"Start":"15:43.615 ","End":"15:50.935","Text":"Let\u0027s underline it so that is what I\u0027m doing."},{"Start":"15:50.935 ","End":"16:01.300","Text":"That means that all I\u0027m doing over here from this nabla^2E_x is I\u0027m taking"},{"Start":"16:01.300 ","End":"16:12.040","Text":"d^2 of the x-component by dx^2 plus d^2 of the x-component of"},{"Start":"16:12.040 ","End":"16:16.465","Text":"the E field by y^2 plus"},{"Start":"16:16.465 ","End":"16:25.180","Text":"d of the x-component by dz^2."},{"Start":"16:25.180 ","End":"16:28.600","Text":"This is our E_x,"},{"Start":"16:28.600 ","End":"16:34.760","Text":"this is the x-component of our E field over here I\u0027ll just change it to x^2,"},{"Start":"16:35.040 ","End":"16:39.700","Text":"y^2 so let\u0027s take the x-component of"},{"Start":"16:39.700 ","End":"16:44.440","Text":"the electric field and take its derivative with respect to x twice."},{"Start":"16:44.440 ","End":"16:50.140","Text":"The first time we\u0027ll be left with 2xy^2 and the second time we\u0027ll just be left"},{"Start":"16:50.140 ","End":"16:56.005","Text":"with 2y^2 and then again the same component,"},{"Start":"16:56.005 ","End":"17:01.030","Text":"but we\u0027re taking the derivative twice with respect to y so the first time we\u0027ll have"},{"Start":"17:01.030 ","End":"17:07.930","Text":"2x^2y and the second derivative will just be 2x^2."},{"Start":"17:07.930 ","End":"17:11.665","Text":"Then we can take it again with respect to z."},{"Start":"17:11.665 ","End":"17:15.040","Text":"We can see that there\u0027s no z variable over here"},{"Start":"17:15.040 ","End":"17:20.325","Text":"so both the first and the second derivatives will give us 0."},{"Start":"17:20.325 ","End":"17:25.570","Text":"Then when we\u0027re just calculating the nabla^2,"},{"Start":"17:25.570 ","End":"17:33.490","Text":"so we\u0027re taking the sum of this so what we\u0027re going to have is 2y^2 plus 2x^2."},{"Start":"17:33.490 ","End":"17:35.600","Text":"That\u0027s all that this means and of course,"},{"Start":"17:35.600 ","End":"17:37.580","Text":"you do it for each component,"},{"Start":"17:37.580 ","End":"17:41.480","Text":"so also for the y-component and also for the z-component,"},{"Start":"17:41.480 ","End":"17:43.950","Text":"the exact same thing."},{"Start":"17:44.880 ","End":"17:48.695","Text":"Then of course this is equal to the right side,"},{"Start":"17:48.695 ","End":"17:50.255","Text":"which is this Mu_naught,"},{"Start":"17:50.255 ","End":"17:52.745","Text":"Epsilon_naught and then we take"},{"Start":"17:52.745 ","End":"17:57.500","Text":"the second derivative of the same component the x-component"},{"Start":"17:57.500 ","End":"18:04.925","Text":"of the E field in this example over here but this time with respect to time."},{"Start":"18:04.925 ","End":"18:10.805","Text":"What we get is some equation in just the x-direction just using,"},{"Start":"18:10.805 ","End":"18:14.870","Text":"or here just using one of the components of the E field."},{"Start":"18:14.870 ","End":"18:16.850","Text":"You can also do it for y and for z,"},{"Start":"18:16.850 ","End":"18:24.140","Text":"that links the second derivative of it with respect to its position in space,"},{"Start":"18:24.140 ","End":"18:28.760","Text":"to the second derivative with respect to"},{"Start":"18:28.760 ","End":"18:37.790","Text":"its change with respect to time so as we can see,"},{"Start":"18:37.790 ","End":"18:42.030","Text":"we\u0027re left with this differential equation that we need to solve somehow."},{"Start":"18:42.030 ","End":"18:47.015","Text":"I\u0027m not going to go over right now how we solve it but in the end,"},{"Start":"18:47.015 ","End":"18:50.270","Text":"if we\u0027re still looking at this x-component of"},{"Start":"18:50.270 ","End":"18:55.350","Text":"our E field we\u0027re meant to get a solution that is E_x"},{"Start":"18:55.350 ","End":"19:00.250","Text":"is equal to a cosine of k.r"},{"Start":"19:00.250 ","End":"19:07.415","Text":"minus Omega t. Now,"},{"Start":"19:07.415 ","End":"19:09.170","Text":"often added to this,"},{"Start":"19:09.170 ","End":"19:13.595","Text":"we\u0027ll have another component over here of Phi,"},{"Start":"19:13.595 ","End":"19:23.060","Text":"where Phi refers to the phase but often this is left out so for simplicity,"},{"Start":"19:23.060 ","End":"19:25.550","Text":"let\u0027s just leave it at this,"},{"Start":"19:25.550 ","End":"19:27.079","Text":"and in later lessons,"},{"Start":"19:27.079 ","End":"19:30.110","Text":"we\u0027ll explain what is going on over here,"},{"Start":"19:30.110 ","End":"19:32.165","Text":"what all of these terms are."},{"Start":"19:32.165 ","End":"19:36.920","Text":"This over here is what is really important to us now, of course,"},{"Start":"19:36.920 ","End":"19:39.365","Text":"there are versions with cosine or"},{"Start":"19:39.365 ","End":"19:42.290","Text":"using sine you can use either right now we\u0027re just looking at"},{"Start":"19:42.290 ","End":"19:47.855","Text":"the cosine version so this is what is important and in the next lessons,"},{"Start":"19:47.855 ","End":"19:50.465","Text":"we\u0027re going to really be looking at this,"},{"Start":"19:50.465 ","End":"19:51.860","Text":"how we use it,"},{"Start":"19:51.860 ","End":"19:55.055","Text":"and what is going on inside here."},{"Start":"19:55.055 ","End":"19:58.470","Text":"That is the end of this lesson."}],"ID":22346},{"Watched":false,"Name":"Wave Equations Solutions and Equations","Duration":"36m 26s","ChapterTopicVideoID":21558,"CourseChapterTopicPlaylistID":99477,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.239","Text":"Hello, this lesson is a continuation of what we learned in the previous lesson."},{"Start":"00:06.239 ","End":"00:07.829","Text":"In the previous lesson,"},{"Start":"00:07.829 ","End":"00:14.399","Text":"we looked at Maxwell\u0027s equations and we saw how to derive the wave equations"},{"Start":"00:14.399 ","End":"00:21.960","Text":"eventually arriving at this solution to this differential equation over here."},{"Start":"00:21.960 ","End":"00:24.255","Text":"In this lesson now,"},{"Start":"00:24.255 ","End":"00:30.520","Text":"we\u0027re going to speak about this solution in slightly more detail."},{"Start":"00:31.460 ","End":"00:35.240","Text":"Let\u0027s take a look at this over here."},{"Start":"00:35.240 ","End":"00:36.680","Text":"First of all, again,"},{"Start":"00:36.680 ","End":"00:40.625","Text":"we\u0027re looking at the x component of the electric field."},{"Start":"00:40.625 ","End":"00:43.469","Text":"As we know, the electric field is a vector field."},{"Start":"00:43.469 ","End":"00:46.024","Text":"If we\u0027re looking at Cartesian coordinates,"},{"Start":"00:46.024 ","End":"00:47.824","Text":"it will have an x component,"},{"Start":"00:47.824 ","End":"00:50.390","Text":"y component, and a z component."},{"Start":"00:50.390 ","End":"00:53.329","Text":"Here, at our example right now we\u0027re just looking at"},{"Start":"00:53.329 ","End":"00:57.870","Text":"the x component but it doesn\u0027t really matter."},{"Start":"00:58.220 ","End":"01:03.199","Text":"This equation over here gives us the magnitude of"},{"Start":"01:03.199 ","End":"01:09.020","Text":"the electric field in the x direction at some given time,"},{"Start":"01:09.020 ","End":"01:11.704","Text":"so as a function of time."},{"Start":"01:11.704 ","End":"01:15.529","Text":"What we can see is that the electric field over here is dependent on"},{"Start":"01:15.529 ","End":"01:19.445","Text":"time and on this r over here,"},{"Start":"01:19.445 ","End":"01:23.405","Text":"where r is a vector representing its position,"},{"Start":"01:23.405 ","End":"01:25.175","Text":"the position of the wave,"},{"Start":"01:25.175 ","End":"01:28.679","Text":"or of the electric field in space."},{"Start":"01:28.880 ","End":"01:39.155","Text":"We can rewrite this as showing that it\u0027s dependent on r and t. What is our r vector?"},{"Start":"01:39.155 ","End":"01:44.254","Text":"Our r vector is equal to in Cartesian coordinates some x,"},{"Start":"01:44.254 ","End":"01:47.199","Text":"y, and z values."},{"Start":"01:47.199 ","End":"01:50.059","Text":"If we\u0027re looking at some axis,"},{"Start":"01:50.059 ","End":"01:52.355","Text":"so this is the x direction."},{"Start":"01:52.355 ","End":"01:55.429","Text":"This is the, let\u0027s do this,"},{"Start":"01:55.429 ","End":"02:01.715","Text":"the z direction, and this is the y direction."},{"Start":"02:01.715 ","End":"02:08.659","Text":"Then we\u0027re looking over here from the origin at some point."},{"Start":"02:08.659 ","End":"02:18.200","Text":"This is going to be our r point,"},{"Start":"02:18.200 ","End":"02:20.344","Text":"some x, y, z value,"},{"Start":"02:20.344 ","End":"02:24.654","Text":"giving us this position over here."},{"Start":"02:24.654 ","End":"02:29.344","Text":"That\u0027s the first thing. Then we also want to look at this k vector."},{"Start":"02:29.344 ","End":"02:31.130","Text":"Now the k vector, we\u0027re going to speak"},{"Start":"02:31.130 ","End":"02:34.340","Text":"about in a little bit more detail later but right now,"},{"Start":"02:34.340 ","End":"02:41.555","Text":"just think of it as just a vector with some constant values."},{"Start":"02:41.555 ","End":"02:45.860","Text":"It also represents something which we\u0027ll speak"},{"Start":"02:45.860 ","End":"02:49.639","Text":"about what it means soon but what it is? It\u0027s just constant."},{"Start":"02:49.639 ","End":"02:52.385","Text":"We have k in the x direction,"},{"Start":"02:52.385 ","End":"02:54.320","Text":"k in the y direction,"},{"Start":"02:54.320 ","End":"02:57.289","Text":"and k in the z direction."},{"Start":"02:57.289 ","End":"02:58.610","Text":"Then if we look at here,"},{"Start":"02:58.610 ","End":"03:02.180","Text":"we have the dot product between k and r,"},{"Start":"03:02.180 ","End":"03:04.940","Text":"which means that we take the x, y,"},{"Start":"03:04.940 ","End":"03:06.709","Text":"and z component of each,"},{"Start":"03:06.709 ","End":"03:09.965","Text":"multiply them together, and then sum this up."},{"Start":"03:09.965 ","End":"03:13.354","Text":"Given the r vector and the k vector,"},{"Start":"03:13.354 ","End":"03:21.665","Text":"we can rewrite this as being equal to A cosine of k_x multiplied by x"},{"Start":"03:21.665 ","End":"03:32.865","Text":"plus k_y multiplied by y plus k_z multiplied by z,"},{"Start":"03:32.865 ","End":"03:40.940","Text":"then minus Omega t. This equation over"},{"Start":"03:40.940 ","End":"03:45.079","Text":"here is correct exactly the same equation"},{"Start":"03:45.079 ","End":"03:50.735","Text":"for all of the other components of the E field so obviously, y and z."},{"Start":"03:50.735 ","End":"03:54.740","Text":"The only difference between each component x, y,"},{"Start":"03:54.740 ","End":"03:57.860","Text":"and z will be this constant A over here,"},{"Start":"03:57.860 ","End":"04:02.510","Text":"which represents the amplitude of the wave in that direction."},{"Start":"04:02.510 ","End":"04:09.160","Text":"The amplitude will be different for x, y, and z."},{"Start":"04:09.530 ","End":"04:15.170","Text":"Obviously, that means that sometimes it will be equal in each direction,"},{"Start":"04:15.170 ","End":"04:20.899","Text":"sometimes the amplitude it could also be 0 in one direction but in general,"},{"Start":"04:20.899 ","End":"04:23.929","Text":"this whole equation is correct for each component,"},{"Start":"04:23.929 ","End":"04:27.650","Text":"just the amplitude is the thing that changes."},{"Start":"04:27.650 ","End":"04:29.450","Text":"The k, the x, y,"},{"Start":"04:29.450 ","End":"04:32.360","Text":"z, and the Omega is different."},{"Start":"04:32.360 ","End":"04:35.044","Text":"The t of course just represents the time."},{"Start":"04:35.044 ","End":"04:38.330","Text":"We can just plug in t is equal to 0,"},{"Start":"04:38.330 ","End":"04:42.539","Text":"t is equal to 2 seconds and calculated."},{"Start":"04:43.790 ","End":"04:49.235","Text":"Equally this exact same equation and the ideas with"},{"Start":"04:49.235 ","End":"04:54.920","Text":"the different components is also correct for the B field, the magnetic field."},{"Start":"04:54.920 ","End":"04:59.144","Text":"Whether we\u0027re looking at the electric field"},{"Start":"04:59.144 ","End":"05:04.050","Text":"in the x direction or the magnetic field in the z direction,"},{"Start":"05:04.050 ","End":"05:06.740","Text":"this equation will be exactly the same,"},{"Start":"05:06.740 ","End":"05:10.645","Text":"just the amplitude will be different."},{"Start":"05:10.645 ","End":"05:14.120","Text":"What we\u0027re going to do now is we\u0027re going to give"},{"Start":"05:14.120 ","End":"05:20.150","Text":"a very easy example just to make this a little bit clearer."},{"Start":"05:20.150 ","End":"05:25.185","Text":"Let\u0027s just draw out a border."},{"Start":"05:25.185 ","End":"05:27.780","Text":"Now we\u0027re doing the example."},{"Start":"05:27.780 ","End":"05:31.910","Text":"Let\u0027s say that we have an electric field that is"},{"Start":"05:31.910 ","End":"05:36.155","Text":"given as a function of its possession and of time."},{"Start":"05:36.155 ","End":"05:38.719","Text":"It\u0027s equal to A,"},{"Start":"05:38.719 ","End":"05:46.725","Text":"the amplitude multiplied by cosine of 2z minus Omega t,"},{"Start":"05:46.725 ","End":"05:48.510","Text":"x in the x direction."},{"Start":"05:48.510 ","End":"05:55.910","Text":"What we can see over here is that our electric field given is only in the x position."},{"Start":"05:55.910 ","End":"06:02.270","Text":"It only has an amplitude in the x direction or in other words,"},{"Start":"06:02.270 ","End":"06:09.455","Text":"its amplitude in the y and in the z directions is equal to 0."},{"Start":"06:09.455 ","End":"06:12.299","Text":"It means the same thing."},{"Start":"06:13.070 ","End":"06:15.975","Text":"That\u0027s with respect to the direction,"},{"Start":"06:15.975 ","End":"06:20.145","Text":"and now what we want to do is we want to find our k vector."},{"Start":"06:20.145 ","End":"06:24.784","Text":"Our k vector, we saw that we\u0027d have the x component of it"},{"Start":"06:24.784 ","End":"06:30.320","Text":"multiplied by the x component of our r vector,"},{"Start":"06:30.320 ","End":"06:31.639","Text":"of our position vector."},{"Start":"06:31.639 ","End":"06:36.260","Text":"The y component multiplied by the y component of"},{"Start":"06:36.260 ","End":"06:41.735","Text":"the r vector and the z component multiplied by the z component of our r vector."},{"Start":"06:41.735 ","End":"06:47.969","Text":"Here we can see that we have 0 multiplied by x."},{"Start":"06:47.969 ","End":"06:52.669","Text":"We can see that our k vector is going to be equal to 0."},{"Start":"06:52.669 ","End":"06:55.594","Text":"We also don\u0027t have any y variable either,"},{"Start":"06:55.594 ","End":"06:59.900","Text":"so it\u0027s going to be 0 multiplied by y and then we can"},{"Start":"06:59.900 ","End":"07:04.895","Text":"see that we have a z variable and its coefficient is 2."},{"Start":"07:04.895 ","End":"07:08.900","Text":"The z component of the k vector is 2."},{"Start":"07:08.900 ","End":"07:13.349","Text":"This is how you would find the k vector."},{"Start":"07:14.390 ","End":"07:21.260","Text":"What we can see is that we have this electric field which is only in the x direction."},{"Start":"07:21.260 ","End":"07:24.964","Text":"However, it is dependent on z."},{"Start":"07:24.964 ","End":"07:27.095","Text":"Don\u0027t get confused between this."},{"Start":"07:27.095 ","End":"07:28.250","Text":"It\u0027s in the x direction,"},{"Start":"07:28.250 ","End":"07:29.944","Text":"but it\u0027s dependent on z."},{"Start":"07:29.944 ","End":"07:33.710","Text":"Let\u0027s draw out an axis."},{"Start":"07:33.710 ","End":"07:36.470","Text":"Here we have our x-axis,"},{"Start":"07:36.470 ","End":"07:40.549","Text":"here we have our z-axis,"},{"Start":"07:40.549 ","End":"07:44.090","Text":"and here we have our y axis."},{"Start":"07:44.090 ","End":"07:46.760","Text":"Now let\u0027s look at a certain time."},{"Start":"07:46.760 ","End":"07:51.520","Text":"Let\u0027s take a look at the initial moment."},{"Start":"07:51.520 ","End":"07:53.445","Text":"Where t=0."},{"Start":"07:53.445 ","End":"07:55.174","Text":"Here if t=0,"},{"Start":"07:55.174 ","End":"08:03.600","Text":"this all cancels out because of the t. What we\u0027re left with is just A cosine of 2z."},{"Start":"08:03.610 ","End":"08:07.390","Text":"Let\u0027s look over here at the origin."},{"Start":"08:07.390 ","End":"08:10.010","Text":"We can see at the origin,"},{"Start":"08:10.010 ","End":"08:13.250","Text":"we\u0027re at the origin, so z is equal to 0."},{"Start":"08:13.250 ","End":"08:16.535","Text":"Then all will be left with is A cosine of 0."},{"Start":"08:16.535 ","End":"08:18.875","Text":"Cosine of 0 is just 1."},{"Start":"08:18.875 ","End":"08:23.990","Text":"We have just the maximum amplitude in the x direction."},{"Start":"08:23.990 ","End":"08:25.865","Text":"Remember, it\u0027s in the x direction."},{"Start":"08:25.865 ","End":"08:30.230","Text":"The maximum amplitude of our electric field at this point."},{"Start":"08:30.230 ","End":"08:33.080","Text":"Then as z increases,"},{"Start":"08:33.080 ","End":"08:37.565","Text":"we can see just as the cosine function behaves,"},{"Start":"08:37.565 ","End":"08:46.835","Text":"after 0, the cosine function begins decreasing like so as a type of wave."},{"Start":"08:46.835 ","End":"08:49.429","Text":"We get something like this."},{"Start":"08:49.429 ","End":"08:59.150","Text":"Then it moves over to this side until it reaches its maximum amplitude."},{"Start":"08:59.150 ","End":"09:06.049","Text":"Then it begins decreasing again and again on the other side."},{"Start":"09:06.049 ","End":"09:09.929","Text":"This of course continues to infinity."},{"Start":"09:10.120 ","End":"09:14.359","Text":"Then what we can do is we can draw the outline."},{"Start":"09:14.359 ","End":"09:18.440","Text":"I haven\u0027t drawn the arrows in the best way,"},{"Start":"09:18.440 ","End":"09:23.000","Text":"but as you know, it\u0027s meant to look like a cosine wave like so."},{"Start":"09:23.000 ","End":"09:26.639","Text":"Imagine that that is what is drawn here."},{"Start":"09:27.230 ","End":"09:31.115","Text":"Now, let\u0027s take a look at a different moment."},{"Start":"09:31.115 ","End":"09:36.560","Text":"Let\u0027s look 1 second later at t=1."},{"Start":"09:36.560 ","End":"09:39.759","Text":"Let\u0027s see what the wave looks like."},{"Start":"09:39.759 ","End":"09:42.405","Text":"We\u0027ll just draw it in pink."},{"Start":"09:42.405 ","End":"09:43.830","Text":"One second later,"},{"Start":"09:43.830 ","End":"09:47.610","Text":"we can see that here our t will be equal to 1."},{"Start":"09:47.610 ","End":"09:51.360","Text":"We have negative Omega multiplied by 1,"},{"Start":"09:51.360 ","End":"09:53.100","Text":"or just negative Omega."},{"Start":"09:53.100 ","End":"09:55.280","Text":"Because there\u0027s a negative over here,"},{"Start":"09:55.280 ","End":"10:04.510","Text":"that means that this peak will move forward some distance which is dependent on Omega."},{"Start":"10:05.470 ","End":"10:08.915","Text":"Let\u0027s say that it moves up until here."},{"Start":"10:08.915 ","End":"10:13.549","Text":"That means that now our peak is going to be here and"},{"Start":"10:13.549 ","End":"10:19.524","Text":"then slowly it will decrease like so."},{"Start":"10:19.524 ","End":"10:24.727","Text":"Then it just looks like this,"},{"Start":"10:24.727 ","End":"10:27.014","Text":"exactly the same thing."},{"Start":"10:27.014 ","End":"10:31.290","Text":"What we just see is that this peak is moving."},{"Start":"10:31.290 ","End":"10:36.630","Text":"This gives us the direction of the propagation of the wave."},{"Start":"10:36.630 ","End":"10:41.255","Text":"This Omega, and the minus tells us that it\u0027s moving in the rightwards direction."},{"Start":"10:41.255 ","End":"10:42.500","Text":"If there was a plus,"},{"Start":"10:42.500 ","End":"10:46.379","Text":"the peak would move in the leftwards direction."},{"Start":"10:46.430 ","End":"10:51.364","Text":"Then of course, after a certain period of time,"},{"Start":"10:51.364 ","End":"10:56.795","Text":"one of the other peaks in the wave will reach this exact same point again,"},{"Start":"10:56.795 ","End":"11:03.690","Text":"because right now we have a situation like so."},{"Start":"11:03.690 ","End":"11:08.794","Text":"Now, let\u0027s speak about all the components in this equation."},{"Start":"11:08.794 ","End":"11:14.885","Text":"As we can see, Omega is exactly the same as what we saw in circular motion."},{"Start":"11:14.885 ","End":"11:17.794","Text":"It\u0027s equal to 2 Pi f,"},{"Start":"11:17.794 ","End":"11:19.069","Text":"where f is of course,"},{"Start":"11:19.069 ","End":"11:22.430","Text":"the frequency of the wave."},{"Start":"11:22.430 ","End":"11:27.915","Text":"Of course, this is also equal to 2 Pi divided by T,"},{"Start":"11:27.915 ","End":"11:31.080","Text":"where T is the period of the wave."},{"Start":"11:31.080 ","End":"11:38.635","Text":"The time taken for 1 oscillation."},{"Start":"11:38.635 ","End":"11:40.359","Text":"As we were saying before,"},{"Start":"11:40.359 ","End":"11:46.126","Text":"if we started with our peak at the origin,"},{"Start":"11:46.126 ","End":"11:52.524","Text":"the period or T is the time taken for us to,"},{"Start":"11:52.524 ","End":"11:56.877","Text":"again get to a point where there\u0027s a peak at the origin;"},{"Start":"11:56.877 ","End":"12:02.749","Text":"to return to the same image that we had at the beginning."},{"Start":"12:02.749 ","End":"12:06.085","Text":"That is what Omega tells us."},{"Start":"12:06.085 ","End":"12:12.474","Text":"It gives us information on the frequency and on the time period of our wave."},{"Start":"12:12.474 ","End":"12:13.999","Text":"That, in turn,"},{"Start":"12:13.999 ","End":"12:20.932","Text":"gives us information regarding the propagation of the wave;"},{"Start":"12:20.932 ","End":"12:26.350","Text":"how long it takes for our peak to move from some arbitrary point,"},{"Start":"12:26.350 ","End":"12:33.280","Text":"let\u0027s say from being over here at the origin to being a certain distance away,"},{"Start":"12:33.280 ","End":"12:36.295","Text":"so that\u0027s what our Omega gives us."},{"Start":"12:36.295 ","End":"12:41.350","Text":"A larger Omega will mean that in the same timeframe between t is 0,"},{"Start":"12:41.350 ","End":"12:48.085","Text":"when t is 1, our peak could move either less distance or more."},{"Start":"12:48.085 ","End":"12:51.350","Text":"That`s what our Omega gives us."},{"Start":"12:52.020 ","End":"12:56.890","Text":"As said, our Omega tells us the distance moved."},{"Start":"12:56.890 ","End":"12:59.139","Text":"What does our k tell us?"},{"Start":"12:59.139 ","End":"13:04.614","Text":"Our k tells us the direction of this movement."},{"Start":"13:04.614 ","End":"13:07.960","Text":"The distance moved, it\u0027s like I\u0027m walking distance,"},{"Start":"13:07.960 ","End":"13:10.877","Text":"and then the k tells us the direction."},{"Start":"13:10.877 ","End":"13:13.119","Text":"It gives it a size."},{"Start":"13:13.119 ","End":"13:15.475","Text":"This is, if you will, the size,"},{"Start":"13:15.475 ","End":"13:20.869","Text":"and this is the direction of the wave."},{"Start":"13:21.660 ","End":"13:27.564","Text":"Omega tells us how far the peak will move in every time increments."},{"Start":"13:27.564 ","End":"13:31.870","Text":"K will tell us in which direction or in which axis."},{"Start":"13:31.870 ","End":"13:36.205","Text":"Here, specifically, our wave, as we can see,"},{"Start":"13:36.205 ","End":"13:39.453","Text":"it\u0027s not moving in either the x or y direction,"},{"Start":"13:39.453 ","End":"13:41.769","Text":"because we can see 0 in both."},{"Start":"13:41.769 ","End":"13:45.979","Text":"We can see that it\u0027s only moving in the z direction."},{"Start":"13:45.979 ","End":"13:49.329","Text":"This also has a magnitude in the z direction which ties"},{"Start":"13:49.329 ","End":"13:52.809","Text":"in but we can see that it\u0027s only moving in the z direction,"},{"Start":"13:52.809 ","End":"13:55.104","Text":"which is exactly how we drew it."},{"Start":"13:55.104 ","End":"13:57.339","Text":"This is of course an easy example."},{"Start":"13:57.339 ","End":"13:59.170","Text":"A lot of the times,"},{"Start":"13:59.170 ","End":"14:03.610","Text":"we\u0027ll have a k vector that has components also in"},{"Start":"14:03.610 ","End":"14:10.230","Text":"the x and y direction and that just means that it moves a certain magnitude in the x,"},{"Start":"14:10.230 ","End":"14:13.470","Text":"a certain in the y and a certain in the z but here,"},{"Start":"14:13.470 ","End":"14:14.730","Text":"just in this easy example,"},{"Start":"14:14.730 ","End":"14:22.189","Text":"we can see that we are moving with a magnitude of 2 in the z direction each time."},{"Start":"14:22.189 ","End":"14:31.135","Text":"Now another important quality of the k vector is that it\u0027s connected to the wavelength."},{"Start":"14:31.135 ","End":"14:33.190","Text":"When we\u0027re speaking about wavelength,"},{"Start":"14:33.190 ","End":"14:38.154","Text":"we\u0027re speaking about the distance from peak to peak."},{"Start":"14:38.154 ","End":"14:41.049","Text":"This distance over here,"},{"Start":"14:41.049 ","End":"14:43.660","Text":"from peak to peak is Lambda,"},{"Start":"14:43.660 ","End":"14:45.820","Text":"which is the wavelength."},{"Start":"14:45.820 ","End":"14:51.325","Text":"If we take the magnitude of our k vector,"},{"Start":"14:51.325 ","End":"14:59.329","Text":"it is equal to 2 Pi divided by Lambda, the wavelength."},{"Start":"15:00.600 ","End":"15:07.420","Text":"These are 2 very important equations to remember,"},{"Start":"15:07.420 ","End":"15:09.295","Text":"and now there\u0027s another one."},{"Start":"15:09.295 ","End":"15:14.004","Text":"A moment ago, I related the Omega"},{"Start":"15:14.004 ","End":"15:19.885","Text":"to how much the wave moves in a certain direction,"},{"Start":"15:19.885 ","End":"15:21.984","Text":"the distance, if you will,"},{"Start":"15:21.984 ","End":"15:26.680","Text":"and the k to the direction moved but as we can see,"},{"Start":"15:26.680 ","End":"15:29.695","Text":"the k also has a magnitude."},{"Start":"15:29.695 ","End":"15:31.690","Text":"It\u0027s not a unit vector."},{"Start":"15:31.690 ","End":"15:33.760","Text":"For instance, here we can see it\u0027s 2."},{"Start":"15:33.760 ","End":"15:35.199","Text":"If it was a unit vector,"},{"Start":"15:35.199 ","End":"15:38.090","Text":"this value here would just be 1."},{"Start":"15:38.100 ","End":"15:41.935","Text":"K, in actual fact,"},{"Start":"15:41.935 ","End":"15:48.369","Text":"isn\u0027t just the direction that the wave is propagating in,"},{"Start":"15:48.369 ","End":"15:51.124","Text":"but it\u0027s also connected to the distance."},{"Start":"15:51.124 ","End":"15:56.020","Text":"That must mean that k and Omega are somehow related,"},{"Start":"15:56.020 ","End":"16:01.134","Text":"because Omega also gives us something to do with the distance that"},{"Start":"16:01.134 ","End":"16:07.030","Text":"a peak will propagate or any points along the wave will propagate in a certain direction."},{"Start":"16:07.030 ","End":"16:13.330","Text":"We can relate Omega via this equation to k. Omega is equal to c,"},{"Start":"16:13.330 ","End":"16:14.904","Text":"which is the speed of light,"},{"Start":"16:14.904 ","End":"16:21.430","Text":"multiplied by the magnitude of the k vector."},{"Start":"16:21.430 ","End":"16:25.734","Text":"This equation, over here,"},{"Start":"16:25.734 ","End":"16:29.570","Text":"is called the dispersion relationship."},{"Start":"16:29.820 ","End":"16:32.950","Text":"These equations are super important."},{"Start":"16:32.950 ","End":"16:36.174","Text":"Please write them in your equation sheets."},{"Start":"16:36.174 ","End":"16:39.265","Text":"I\u0027ve just labeled that f over here is frequency,"},{"Start":"16:39.265 ","End":"16:41.559","Text":"T over here is period,"},{"Start":"16:41.559 ","End":"16:45.205","Text":"Lambda over here is the wavelength,"},{"Start":"16:45.205 ","End":"16:47.335","Text":"and that this entire equation,"},{"Start":"16:47.335 ","End":"16:51.040","Text":"Omega is equal to c multiplied by the magnitude of"},{"Start":"16:51.040 ","End":"16:55.569","Text":"k. This whole equation is called the dispersion relationship."},{"Start":"16:55.569 ","End":"17:02.725","Text":"As we\u0027ve already spoken about in the past few lessons,"},{"Start":"17:02.725 ","End":"17:05.995","Text":"we\u0027ve said that whenever we have an electric field,"},{"Start":"17:05.995 ","End":"17:08.320","Text":"we also have a magnetic field."},{"Start":"17:08.320 ","End":"17:13.885","Text":"We even said that they are identical just from different reference frames."},{"Start":"17:13.885 ","End":"17:17.320","Text":"We can see that here we have an electric field,"},{"Start":"17:17.320 ","End":"17:20.305","Text":"which in turn is going to create a magnetic field."},{"Start":"17:20.305 ","End":"17:22.929","Text":"If we had a magnetic field,"},{"Start":"17:22.929 ","End":"17:25.525","Text":"it would in turn create an electric field."},{"Start":"17:25.525 ","End":"17:29.785","Text":"Possible question that they could ask is given this electric field,"},{"Start":"17:29.785 ","End":"17:31.930","Text":"find the magnetic field."},{"Start":"17:31.930 ","End":"17:34.569","Text":"Now we\u0027re going to look again at a simple example of"},{"Start":"17:34.569 ","End":"17:38.140","Text":"how to get from the electric field to the magnetic field."},{"Start":"17:38.140 ","End":"17:40.029","Text":"So in order to do this,"},{"Start":"17:40.029 ","End":"17:44.454","Text":"we\u0027re going to again use Maxwell\u0027s equations."},{"Start":"17:44.454 ","End":"17:46.554","Text":"Let\u0027s just go up."},{"Start":"17:46.554 ","End":"17:49.150","Text":"In order to convert between fields,"},{"Start":"17:49.150 ","End":"17:53.334","Text":"we use Maxwell\u0027s 3rd and 4th equations."},{"Start":"17:53.334 ","End":"17:57.230","Text":"This is the 3rd equation;"},{"Start":"17:57.230 ","End":"18:03.055","Text":"all of this, and this is the 4th equation."},{"Start":"18:03.055 ","End":"18:07.419","Text":"All of this, remember that we\u0027re in a vacuum,"},{"Start":"18:07.419 ","End":"18:09.114","Text":"so this is equal to 0."},{"Start":"18:09.114 ","End":"18:15.619","Text":"So we just take into account this section over here."},{"Start":"18:16.350 ","End":"18:20.320","Text":"If we have an electric field as we have here,"},{"Start":"18:20.320 ","End":"18:23.439","Text":"and we want to find the magnetic field,"},{"Start":"18:23.439 ","End":"18:24.919","Text":"we\u0027ll use equation number 3,"},{"Start":"18:24.919 ","End":"18:28.900","Text":"where we\u0027re taking the rotor of the electric field."},{"Start":"18:28.900 ","End":"18:34.450","Text":"Whereas, if we were given a magnetic field and told to calculate the electric field,"},{"Start":"18:34.450 ","End":"18:36.970","Text":"we would use Maxwell\u0027s 4th equation,"},{"Start":"18:36.970 ","End":"18:39.624","Text":"where we take the rotor of the magnetic field,"},{"Start":"18:39.624 ","End":"18:43.465","Text":"and then we will calculate the electric field."},{"Start":"18:43.465 ","End":"18:49.359","Text":"In other words, you take the rotor of the field that you\u0027re given."},{"Start":"18:49.359 ","End":"18:51.399","Text":"If you\u0027re given the electric field,"},{"Start":"18:51.399 ","End":"18:53.275","Text":"you take the rotor of the electric field."},{"Start":"18:53.275 ","End":"18:54.819","Text":"If you\u0027re given the magnetic field,"},{"Start":"18:54.819 ","End":"18:56.770","Text":"you take the rotor of the magnetic field,"},{"Start":"18:56.770 ","End":"19:02.020","Text":"and you use that in order to calculate the other field."},{"Start":"19:02.020 ","End":"19:04.615","Text":"In our question over here,"},{"Start":"19:04.615 ","End":"19:06.919","Text":"we\u0027re given the electric field,"},{"Start":"19:06.919 ","End":"19:08.980","Text":"so we\u0027re going to use Maxwell\u0027s 3rd equation."},{"Start":"19:08.980 ","End":"19:11.320","Text":"We take the rotor of the electric field."},{"Start":"19:11.320 ","End":"19:18.694","Text":"We see that it is equal to the negative time derivative of the magnetic field."},{"Start":"19:18.694 ","End":"19:21.660","Text":"Let\u0027s solve this."},{"Start":"19:21.660 ","End":"19:29.685","Text":"Let\u0027s give us some more space and let\u0027s rewrite out our equation."},{"Start":"19:29.685 ","End":"19:37.649","Text":"We\u0027re using Maxwell\u0027s third equation that says that the rotor of"},{"Start":"19:37.649 ","End":"19:46.440","Text":"the electric field is equal to the negative time derivative of the magnetic field."},{"Start":"19:46.440 ","End":"19:48.765","Text":"Let\u0027s, first of all,"},{"Start":"19:48.765 ","End":"19:55.995","Text":"do the rotor of the electric field."},{"Start":"19:55.995 ","End":"19:59.534","Text":"This is going to give us,"},{"Start":"19:59.534 ","End":"20:02.250","Text":"we\u0027ll just go over how to solve this."},{"Start":"20:02.250 ","End":"20:06.794","Text":"We have our partial derivatives d by dx,"},{"Start":"20:06.794 ","End":"20:14.055","Text":"d by dy, d by dz cross multiplied with our E field."},{"Start":"20:14.055 ","End":"20:17.190","Text":"Our E field is only in the x-direction,"},{"Start":"20:17.190 ","End":"20:20.730","Text":"so it\u0027s equal to A cosine of"},{"Start":"20:20.730 ","End":"20:28.620","Text":"2z minus omega t. Then we have 0 and 0 in the x and y directions."},{"Start":"20:28.620 ","End":"20:31.470","Text":"Now let\u0027s calculate this."},{"Start":"20:31.470 ","End":"20:35.400","Text":"Remember from the rotor we get again a vector."},{"Start":"20:35.400 ","End":"20:40.679","Text":"First, what we do is we cross out over here and then"},{"Start":"20:40.679 ","End":"20:46.064","Text":"we do d by dy of 0 minus d by dz of 0,"},{"Start":"20:46.064 ","End":"20:50.519","Text":"which is equal to 0 in the x-direction."},{"Start":"20:50.519 ","End":"20:56.325","Text":"Then we cross out the middle row and then we flip it over."},{"Start":"20:56.325 ","End":"21:01.379","Text":"We have d by dz of A cosine of"},{"Start":"21:01.379 ","End":"21:06.960","Text":"2z minus omega t minus d by dx of 0, which is 0."},{"Start":"21:06.960 ","End":"21:13.380","Text":"What we have is d by dz of A cosine"},{"Start":"21:13.380 ","End":"21:23.589","Text":"2z minus omega t. Then we cross out the bottom row."},{"Start":"21:23.990 ","End":"21:29.459","Text":"We have d by dy of this,"},{"Start":"21:29.459 ","End":"21:34.619","Text":"so d by dy of A cosine of"},{"Start":"21:34.619 ","End":"21:40.169","Text":"2z minus omega t and then minus d by dx of 0,"},{"Start":"21:40.169 ","End":"21:42.765","Text":"which is just 0."},{"Start":"21:42.765 ","End":"21:45.285","Text":"This is the vector that we have."},{"Start":"21:45.285 ","End":"21:52.200","Text":"Then we can say that this is then equal to 0."},{"Start":"21:52.200 ","End":"22:02.430","Text":"Then we have d by dz of A cosine of 2z minus omega t. This is going to give us negative."},{"Start":"22:02.430 ","End":"22:05.190","Text":"Then we\u0027ll have from the cosine."},{"Start":"22:05.190 ","End":"22:10.920","Text":"We\u0027ll have A so cosine becomes negative sine,"},{"Start":"22:10.920 ","End":"22:13.800","Text":"that\u0027s why negative sine of"},{"Start":"22:13.800 ","End":"22:20.850","Text":"2z minus omega t. Then this is going to be multiplied by the inner derivative."},{"Start":"22:20.850 ","End":"22:22.920","Text":"Here only this term has a z,"},{"Start":"22:22.920 ","End":"22:27.840","Text":"so the derivative of 2z is just 2. So that."},{"Start":"22:27.840 ","End":"22:34.215","Text":"Then here, we have d by dy of the same function and as we can see,"},{"Start":"22:34.215 ","End":"22:36.390","Text":"we have no y variables."},{"Start":"22:36.390 ","End":"22:38.805","Text":"When we take the derivative with respect to y,"},{"Start":"22:38.805 ","End":"22:42.190","Text":"we\u0027re just left with 0."},{"Start":"22:42.230 ","End":"22:44.955","Text":"This is the rotor."},{"Start":"22:44.955 ","End":"22:53.084","Text":"In other words, we can say that our rotor is equal to"},{"Start":"22:53.084 ","End":"23:03.390","Text":"negative 2A sine of 2z minus omega t in the y direction."},{"Start":"23:03.390 ","End":"23:07.335","Text":"We can see it\u0027s the y component of this vector over here."},{"Start":"23:07.335 ","End":"23:08.684","Text":"This is equal to,"},{"Start":"23:08.684 ","End":"23:11.054","Text":"according to Maxwell\u0027s third equation,"},{"Start":"23:11.054 ","End":"23:16.320","Text":"to negative dB by dt."},{"Start":"23:16.320 ","End":"23:21.375","Text":"First of all, we can see that the magnetic field is in the y-direction."},{"Start":"23:21.375 ","End":"23:27.645","Text":"Sometimes we\u0027ll also have other components like x and z but specifically here,"},{"Start":"23:27.645 ","End":"23:31.810","Text":"we can see that the magnetic field is only in the y direction."},{"Start":"23:31.880 ","End":"23:37.905","Text":"Let\u0027s just erase this so that we can play around with both sides."},{"Start":"23:37.905 ","End":"23:43.320","Text":"Both sides have negative signs so we can cancel them out."},{"Start":"23:43.320 ","End":"23:46.065","Text":"They just become pluses."},{"Start":"23:46.065 ","End":"23:51.870","Text":"Now what we\u0027re going to do is we\u0027re going to integrate both sides."},{"Start":"23:51.870 ","End":"24:00.450","Text":"In that case, we\u0027re going to get that the magnetic field is equal to the time"},{"Start":"24:00.450 ","End":"24:06.064","Text":"integral of 2A sine"},{"Start":"24:06.064 ","End":"24:11.750","Text":"of 2z minus omega t dt."},{"Start":"24:11.750 ","End":"24:16.360","Text":"Of course, this is in the y-direction."},{"Start":"24:16.360 ","End":"24:22.889","Text":"We\u0027ve just integrated both sides with respect to t. Of course,"},{"Start":"24:22.889 ","End":"24:26.414","Text":"what we\u0027re going to get over here is in the y direction,"},{"Start":"24:26.414 ","End":"24:30.150","Text":"the integral with respect to t will give"},{"Start":"24:30.150 ","End":"24:36.330","Text":"us negative because the integral of sine is negative cosine."},{"Start":"24:36.330 ","End":"24:40.095","Text":"We have negative cosine of"},{"Start":"24:40.095 ","End":"24:47.235","Text":"2z minus omega t and then multiplied by our constants over here 2A."},{"Start":"24:47.235 ","End":"24:51.389","Text":"Then we divide by the inner derivative with respect to"},{"Start":"24:51.389 ","End":"24:55.109","Text":"t. Only this term over here has a"},{"Start":"24:55.109 ","End":"24:59.550","Text":"t. The derivative of negative omega t is just negative omega."},{"Start":"24:59.550 ","End":"25:05.909","Text":"We divide all of this by negative omega and then again,"},{"Start":"25:05.909 ","End":"25:08.159","Text":"we can see that the minuses here cancel out,"},{"Start":"25:08.159 ","End":"25:11.115","Text":"so they just become positives."},{"Start":"25:11.115 ","End":"25:12.870","Text":"Then, in other words,"},{"Start":"25:12.870 ","End":"25:20.775","Text":"we can see that our magnetic field is equal to 2A divided by omega"},{"Start":"25:20.775 ","End":"25:30.759","Text":"multiplied by cosine of 2z minus omega t in the y direction."},{"Start":"25:31.130 ","End":"25:35.549","Text":"Now of course, we can add a constant"},{"Start":"25:35.549 ","End":"25:39.209","Text":"over here because we had an indefinite integral but usually,"},{"Start":"25:39.209 ","End":"25:41.669","Text":"we say that this constant is equal to 0."},{"Start":"25:41.669 ","End":"25:44.129","Text":"So you can just erase it,"},{"Start":"25:44.129 ","End":"25:45.945","Text":"and this is the magnetic field."},{"Start":"25:45.945 ","End":"25:54.489","Text":"What\u0027s important to note is that the argument or in other words, this over here."},{"Start":"25:54.500 ","End":"26:00.299","Text":"Cosine 2z minus omega t for the magnetic field"},{"Start":"26:00.299 ","End":"26:05.925","Text":"is the exact same thing as we saw for the electric field."},{"Start":"26:05.925 ","End":"26:15.120","Text":"Also, cosine 2z minus omega t. The only difference that there is, is the axis."},{"Start":"26:15.120 ","End":"26:18.960","Text":"So here, the electric field is in the x-direction,"},{"Start":"26:18.960 ","End":"26:20.714","Text":"whereas as we can see here,"},{"Start":"26:20.714 ","End":"26:23.595","Text":"the magnetic field is in the y-direction."},{"Start":"26:23.595 ","End":"26:25.499","Text":"The other difference is,"},{"Start":"26:25.499 ","End":"26:27.645","Text":"as we\u0027ve mentioned before, the amplitude."},{"Start":"26:27.645 ","End":"26:31.290","Text":"Here, the amplitude of the electric field is just A,"},{"Start":"26:31.290 ","End":"26:36.509","Text":"whereas the amplitude of the magnetic field is 2A divided by omega."},{"Start":"26:36.509 ","End":"26:40.320","Text":"However, the direction of the propagation and"},{"Start":"26:40.320 ","End":"26:44.190","Text":"the wavelength of the wave is exactly the same."},{"Start":"26:44.190 ","End":"26:50.205","Text":"If we draw this over here back on our axis,"},{"Start":"26:50.205 ","End":"26:53.925","Text":"so let\u0027s draw it in pink."},{"Start":"26:53.925 ","End":"26:59.670","Text":"What we can see is that our wave is in the y-direction,"},{"Start":"26:59.670 ","End":"27:01.830","Text":"but it\u0027s also dependent on z."},{"Start":"27:01.830 ","End":"27:04.035","Text":"If we start at the origin,"},{"Start":"27:04.035 ","End":"27:07.379","Text":"we\u0027ll again have our peak over"},{"Start":"27:07.379 ","End":"27:12.070","Text":"here at the origin just it\u0027s a different peak because it\u0027s a different amplitude."},{"Start":"27:12.230 ","End":"27:15.479","Text":"It\u0027s in the y direction,"},{"Start":"27:15.479 ","End":"27:17.957","Text":"as we can see, however, it\u0027s dependent on z."},{"Start":"27:17.957 ","End":"27:23.970","Text":"We\u0027re going along the z-axis, like so."},{"Start":"27:23.970 ","End":"27:28.570","Text":"I haven\u0027t actually drawn this very well."},{"Start":"27:28.610 ","End":"27:36.840","Text":"We have the peak over here and then here the peaks correspond."},{"Start":"27:36.840 ","End":"27:39.840","Text":"Then we get to this minimum point,"},{"Start":"27:39.840 ","End":"27:42.930","Text":"and then it flips at the exact same point."},{"Start":"27:42.930 ","End":"27:47.835","Text":"The only difference is that the amplitude is different and"},{"Start":"27:47.835 ","End":"27:54.848","Text":"it\u0027s in the y direction."},{"Start":"27:54.848 ","End":"28:03.040","Text":"We can just see that this is the wave representing the magnetic field."},{"Start":"28:03.040 ","End":"28:04.329","Text":"So both of them,"},{"Start":"28:04.329 ","End":"28:09.879","Text":"the electric field and the magnetic field are propagating in the same direction."},{"Start":"28:09.879 ","End":"28:13.360","Text":"These they\u0027re both propagating in this direction."},{"Start":"28:13.360 ","End":"28:16.509","Text":"They have the exact same wavelength,"},{"Start":"28:16.509 ","End":"28:19.029","Text":"so the same Lambda."},{"Start":"28:19.029 ","End":"28:22.900","Text":"As we can see, it\u0027s the exact same Lambda from peak to peak,"},{"Start":"28:22.900 ","End":"28:24.939","Text":"the same period and everything,"},{"Start":"28:24.939 ","End":"28:30.895","Text":"just their amplitude is different and they\u0027re in the different direction."},{"Start":"28:30.895 ","End":"28:35.995","Text":"The electric field is in the x direction and the magnetic field is in the y direction."},{"Start":"28:35.995 ","End":"28:40.370","Text":"Notice that they\u0027re perpendicular to one another."},{"Start":"28:40.620 ","End":"28:43.179","Text":"The fact that they are perpendicular,"},{"Start":"28:43.179 ","End":"28:46.885","Text":"we can actually write this over here."},{"Start":"28:46.885 ","End":"28:52.449","Text":"The electric field is always going to be perpendicular to"},{"Start":"28:52.449 ","End":"28:58.340","Text":"the magnetic field and perpendicular to the direction of propagation."},{"Start":"28:58.530 ","End":"29:03.490","Text":"For instance here, the direction of propagation is z,"},{"Start":"29:03.490 ","End":"29:08.484","Text":"the direction that the wave is moving in is z."},{"Start":"29:08.484 ","End":"29:11.649","Text":"As we can see, the electric field is in the x-axis,"},{"Start":"29:11.649 ","End":"29:13.314","Text":"which is perpendicular to z,"},{"Start":"29:13.314 ","End":"29:15.115","Text":"and the magnetic field,"},{"Start":"29:15.115 ","End":"29:17.425","Text":"is in the y-axis,"},{"Start":"29:17.425 ","End":"29:20.139","Text":"which is perpendicular to z as well."},{"Start":"29:20.139 ","End":"29:25.910","Text":"Of course, the x and y are also perpendicular to one another."},{"Start":"29:26.550 ","End":"29:32.530","Text":"This is also something important to include in your equation sheets."},{"Start":"29:32.530 ","End":"29:36.325","Text":"Of course, the direction of propagation is given to us by"},{"Start":"29:36.325 ","End":"29:41.590","Text":"k. The amplitude of the electric field will"},{"Start":"29:41.590 ","End":"29:45.054","Text":"be perpendicular to k. The amplitude of"},{"Start":"29:45.054 ","End":"29:50.829","Text":"the magnetic field will also be perpendicular to k. K it\u0027s in the y direction,"},{"Start":"29:50.829 ","End":"29:55.855","Text":"and k represents the direction of propagation."},{"Start":"29:55.855 ","End":"30:00.440","Text":"We can just write this as the K vector."},{"Start":"30:00.840 ","End":"30:03.115","Text":"Now, just a little note,"},{"Start":"30:03.115 ","End":"30:08.890","Text":"we\u0027ve already said this here because the coefficient of Omega was negative 1,"},{"Start":"30:08.890 ","End":"30:11.560","Text":"or in other words, because there\u0027s a negative over here,"},{"Start":"30:11.560 ","End":"30:18.045","Text":"the direction of propagation is as given in our vector K. However,"},{"Start":"30:18.045 ","End":"30:20.535","Text":"if there was a plus here,"},{"Start":"30:20.535 ","End":"30:26.969","Text":"so then we said that the direction of propagation would be leftwards, or in other words,"},{"Start":"30:26.969 ","End":"30:33.189","Text":"in the opposite direction to k. If there\u0027s a minus here,"},{"Start":"30:33.189 ","End":"30:34.495","Text":"it might sound a bit weird."},{"Start":"30:34.495 ","End":"30:36.474","Text":"If there\u0027s a minus over here,"},{"Start":"30:36.474 ","End":"30:40.885","Text":"it\u0027s in the direction of k and if there\u0027s a plus,"},{"Start":"30:40.885 ","End":"30:48.144","Text":"then it\u0027s in the opposite direction to k. Now"},{"Start":"30:48.144 ","End":"30:55.975","Text":"let\u0027s do a little recap of everything we\u0027ve done in this lesson and the previous one."},{"Start":"30:55.975 ","End":"30:57.849","Text":"In the previous lesson,"},{"Start":"30:57.849 ","End":"31:04.480","Text":"we looked at Maxwell\u0027s 4 equations from which we can derive the wave equations."},{"Start":"31:04.480 ","End":"31:07.464","Text":"First of all, we said that the wave equations are"},{"Start":"31:07.464 ","End":"31:10.974","Text":"correct for when we\u0027re looking at waves in a vacuum."},{"Start":"31:10.974 ","End":"31:12.519","Text":"If we\u0027re in a vacuum,"},{"Start":"31:12.519 ","End":"31:15.579","Text":"that means that there\u0027s no free particles or charges,"},{"Start":"31:15.579 ","End":"31:19.240","Text":"which means that Rho is equal to 0 and also that"},{"Start":"31:19.240 ","End":"31:23.058","Text":"current can pass because there\u0027s no charges around."},{"Start":"31:23.058 ","End":"31:28.404","Text":"So J is equal to 0 and then using these equations,"},{"Start":"31:28.404 ","End":"31:31.580","Text":"we got to the wave equations."},{"Start":"31:32.250 ","End":"31:40.090","Text":"It\u0027s not that important to remember how we got to these and then we also explained what"},{"Start":"31:40.090 ","End":"31:43.854","Text":"this Nabla squared means and"},{"Start":"31:43.854 ","End":"31:48.039","Text":"we looked at that and we got this differential equation over here."},{"Start":"31:48.039 ","End":"31:53.799","Text":"So Nabla squared is just each partial derivative taken twice and the sum of"},{"Start":"31:53.799 ","End":"32:00.519","Text":"them on each component of the field be it electric or magnetic."},{"Start":"32:00.519 ","End":"32:03.034","Text":"We got this differential equation where,"},{"Start":"32:03.034 ","End":"32:05.004","Text":"from this differential equation we got"},{"Start":"32:05.004 ","End":"32:08.725","Text":"this solution which is where it becomes interesting."},{"Start":"32:08.725 ","End":"32:11.470","Text":"We saw that we have the r,"},{"Start":"32:11.470 ","End":"32:18.520","Text":"which gives us the position in space and we have the k vector, which represents,"},{"Start":"32:18.520 ","End":"32:21.504","Text":"as we spoke about a little bit later,"},{"Start":"32:21.504 ","End":"32:29.905","Text":"the direction of propagation and also the magnitude of this propagation."},{"Start":"32:29.905 ","End":"32:36.524","Text":"We saw that when we do the dot product between k and r we get this over here."},{"Start":"32:36.524 ","End":"32:40.950","Text":"This type of equation and that we have A over here,"},{"Start":"32:40.950 ","End":"32:42.671","Text":"which gives us the amplitude,"},{"Start":"32:42.671 ","End":"32:47.566","Text":"and Omega which gives us also the direction of propagation."},{"Start":"32:47.566 ","End":"32:54.939","Text":"It also gives us the frequency of the wave and the period of the wave."},{"Start":"32:54.939 ","End":"32:58.420","Text":"Therefore, from that also get"},{"Start":"32:58.420 ","End":"33:04.240","Text":"the wavelength of the wave by playing around with certain things."},{"Start":"33:04.240 ","End":"33:07.660","Text":"From k, we get the wavelength of the wave."},{"Start":"33:07.660 ","End":"33:10.599","Text":"Then we looked at an easy example."},{"Start":"33:10.599 ","End":"33:15.629","Text":"We saw how to find k. We just look for any variables x,"},{"Start":"33:15.629 ","End":"33:19.890","Text":"y, z, and then place their coefficients."},{"Start":"33:19.890 ","End":"33:26.620","Text":"Here we had 0 times x plus 0 times y so it\u0027s is 0,0 plus 2 times z."},{"Start":"33:27.450 ","End":"33:30.310","Text":"This is the direction of propagation."},{"Start":"33:30.310 ","End":"33:34.600","Text":"We can see that the wave is propagating in the z direction and"},{"Start":"33:34.600 ","End":"33:39.670","Text":"the amplitude of the electric field was in the direction given here specifically,"},{"Start":"33:39.670 ","End":"33:46.760","Text":"it\u0027s in the x direction as we had 0 in the y and 0 in the z-direction."},{"Start":"33:46.920 ","End":"33:50.770","Text":"Then we looked at some other important equations."},{"Start":"33:50.770 ","End":"33:53.080","Text":"We have our Omega, which as we said,"},{"Start":"33:53.080 ","End":"33:57.219","Text":"is connected to the frequency of the wave and to the time period."},{"Start":"33:57.219 ","End":"34:03.175","Text":"We saw that the magnitude of k can also give us the wavelength."},{"Start":"34:03.175 ","End":"34:08.034","Text":"We saw that Omega is also connected to k,"},{"Start":"34:08.034 ","End":"34:13.060","Text":"where omega is equal to the speed of light multiplied by the magnitude of"},{"Start":"34:13.060 ","End":"34:20.695","Text":"k. Then we also saw how to convert between the electric field and the magnetic field."},{"Start":"34:20.695 ","End":"34:23.065","Text":"We said that if we\u0027re given the magnetic field,"},{"Start":"34:23.065 ","End":"34:27.399","Text":"we use Maxwell\u0027s 3rd equation and if we have the magnetic fields,"},{"Start":"34:27.399 ","End":"34:30.609","Text":"then we use Maxwell\u0027s 4th equation."},{"Start":"34:30.609 ","End":"34:32.035","Text":"Whichever field we have,"},{"Start":"34:32.035 ","End":"34:37.659","Text":"we do the rotor of it to calculate the other field, the corresponding field."},{"Start":"34:37.659 ","End":"34:41.649","Text":"Remember equations 3 and 4 are not the same."},{"Start":"34:41.649 ","End":"34:47.890","Text":"In equation 4, we also have to take into account Mu naught and Epsilon naught."},{"Start":"34:47.890 ","End":"34:54.370","Text":"Here we showed an example of how we do the rotor"},{"Start":"34:54.370 ","End":"35:03.594","Text":"of a vector field and then we got the differential equation so we integrated both sides,"},{"Start":"35:03.594 ","End":"35:08.125","Text":"and therefore, we got to our magnetic field."},{"Start":"35:08.125 ","End":"35:11.139","Text":"We also saw that the argument is going to be the"},{"Start":"35:11.139 ","End":"35:15.010","Text":"same for the magnetic and the electric field."},{"Start":"35:15.010 ","End":"35:16.749","Text":"What changes is that"},{"Start":"35:16.749 ","End":"35:22.240","Text":"the electric and magnetic field will be perpendicular to one another."},{"Start":"35:22.240 ","End":"35:26.139","Text":"Here our electric field was in the x direction,"},{"Start":"35:26.139 ","End":"35:29.141","Text":"and here we can see it\u0027s in the y direction."},{"Start":"35:29.141 ","End":"35:32.019","Text":"Of course, both fields are going to also be"},{"Start":"35:32.019 ","End":"35:35.260","Text":"perpendicular to the direction of propagation."},{"Start":"35:35.260 ","End":"35:38.080","Text":"Here the direction of propagation was z so"},{"Start":"35:38.080 ","End":"35:41.995","Text":"the x and the y-axis are of course, perpendicular to z."},{"Start":"35:41.995 ","End":"35:45.340","Text":"The direction of propagation is of course, as we said,"},{"Start":"35:45.340 ","End":"35:49.285","Text":"given by the vector k. In other words, E,"},{"Start":"35:49.285 ","End":"35:55.495","Text":"B, and K are all going to be perpendicular to one another."},{"Start":"35:55.495 ","End":"35:58.029","Text":"Then the only other difference is"},{"Start":"35:58.029 ","End":"36:02.485","Text":"just the amplitude of the wave but everything else is the same."},{"Start":"36:02.485 ","End":"36:05.065","Text":"Their peaks are at the same points."},{"Start":"36:05.065 ","End":"36:11.919","Text":"Their minimums are at the same points and they cut the axes of their propagation,"},{"Start":"36:11.919 ","End":"36:17.260","Text":"here it\u0027s the z-axis at the same points also."},{"Start":"36:17.260 ","End":"36:21.430","Text":"That is what we saw in the past 2 lessons."},{"Start":"36:21.430 ","End":"36:26.479","Text":"That\u0027s a recap and that is the end of this lesson."}],"ID":22345},{"Watched":false,"Name":"Exercise - Calculate the Corresponding Magnetic Field, New Equation","Duration":"10m 50s","ChapterTopicVideoID":21561,"CourseChapterTopicPlaylistID":99477,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.785","Text":"Hello. In this lesson,"},{"Start":"00:01.785 ","End":"00:04.515","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.515 ","End":"00:09.480","Text":"We\u0027re being asked to calculate the magnetic field given the following electric fields."},{"Start":"00:09.480 ","End":"00:16.875","Text":"We have an electric field which is equal to E_0 with the vector 1, 1, 2,"},{"Start":"00:16.875 ","End":"00:24.090","Text":"and then multiplied by e^i multiplied by 2x minus z,"},{"Start":"00:24.090 ","End":"00:28.335","Text":"minus Omega t. This over here,"},{"Start":"00:28.335 ","End":"00:30.975","Text":"e^i with these brackets,"},{"Start":"00:30.975 ","End":"00:33.780","Text":"is the exact same function as"},{"Start":"00:33.780 ","End":"00:37.890","Text":"cosine of what\u0027s in the brackets or sign of what\u0027s in the brackets."},{"Start":"00:37.890 ","End":"00:42.055","Text":"It\u0027s the exact same thing just written in a different format."},{"Start":"00:42.055 ","End":"00:48.080","Text":"Up until now, we\u0027ve used Maxwell\u0027s equations in order to convert."},{"Start":"00:48.080 ","End":"00:54.390","Text":"Converting between the electric field to the magnetic field, we would do,"},{"Start":"00:54.390 ","End":"01:02.790","Text":"Nabla cross multiplied by the E field is equal to negative dB by dt."},{"Start":"01:03.070 ","End":"01:06.290","Text":"This is how he solved it up until now."},{"Start":"01:06.290 ","End":"01:09.470","Text":"But in this lesson I\u0027m going to show you a new equation,"},{"Start":"01:09.470 ","End":"01:12.500","Text":"which might be a little bit easier."},{"Start":"01:12.500 ","End":"01:18.440","Text":"We calculate the B field by taking our k vector by"},{"Start":"01:18.440 ","End":"01:26.105","Text":"the unit vector cross multiplied by the electric field and divide it by c,"},{"Start":"01:26.105 ","End":"01:28.410","Text":"the speed of light."},{"Start":"01:28.730 ","End":"01:31.295","Text":"This is the equation."},{"Start":"01:31.295 ","End":"01:35.015","Text":"Please put it in your equation sheets."},{"Start":"01:35.015 ","End":"01:37.650","Text":"Now let\u0027s begin calculating."},{"Start":"01:37.650 ","End":"01:44.205","Text":"First of all we want to know what our k vector is so that we can do this."},{"Start":"01:44.205 ","End":"01:45.935","Text":"As we can see over here,"},{"Start":"01:45.935 ","End":"01:49.760","Text":"we have in the x-direction,"},{"Start":"01:49.760 ","End":"01:55.710","Text":"2 and the y is 0 and then the z negative 1."},{"Start":"01:55.710 ","End":"01:59.210","Text":"Then k hat, the unit vector,"},{"Start":"01:59.210 ","End":"02:06.300","Text":"is just equal to our k vector divided by its magnitude."},{"Start":"02:06.300 ","End":"02:10.030","Text":"This is going to be our k vector."},{"Start":"02:10.420 ","End":"02:12.605","Text":"Write it over here."},{"Start":"02:12.605 ","End":"02:15.745","Text":"It\u0027s just 2, 0, negative 1."},{"Start":"02:15.745 ","End":"02:20.780","Text":"Then its magnitude, so 1 divided by its magnitude is"},{"Start":"02:20.780 ","End":"02:27.035","Text":"just the square root of 2^2 plus 0^2 plus negative 1^2,"},{"Start":"02:27.035 ","End":"02:30.530","Text":"so 2^2 is 4 plus 0 plus 1,"},{"Start":"02:30.530 ","End":"02:33.800","Text":"so that\u0027s 5 and the square root of that."},{"Start":"02:33.800 ","End":"02:37.210","Text":"This is just Pythagoras."},{"Start":"02:37.210 ","End":"02:46.815","Text":"Now what we\u0027re going to do is we\u0027re going to cross multiply our k hat with our E. We have"},{"Start":"02:46.815 ","End":"02:51.710","Text":"k hat cross multiplied by E is equal"},{"Start":"02:51.710 ","End":"03:00.060","Text":"to 1 divided by the square root of 5."},{"Start":"03:00.060 ","End":"03:02.535","Text":"Then we have 2,"},{"Start":"03:02.535 ","End":"03:04.995","Text":"0, negative 1."},{"Start":"03:04.995 ","End":"03:08.690","Text":"All of this is cross multiplied by our E field,"},{"Start":"03:08.690 ","End":"03:12.122","Text":"which is E_0 1,1,"},{"Start":"03:12.122 ","End":"03:18.935","Text":"2 multiplied by e^i 2x minus z"},{"Start":"03:18.935 ","End":"03:22.280","Text":"minus Omega t. Now what we\u0027re"},{"Start":"03:22.280 ","End":"03:26.045","Text":"going to do is we\u0027re going to put all the scalars to one side."},{"Start":"03:26.045 ","End":"03:36.290","Text":"This will be simply equal to E_0 as a scalar divided by the square root of 5."},{"Start":"03:36.290 ","End":"03:42.500","Text":"Then also this e^i of 2x minus z"},{"Start":"03:42.500 ","End":"03:50.010","Text":"minus Omega t. Then we have our vector."},{"Start":"03:50.010 ","End":"03:57.660","Text":"We have 2, 0, negative 1 cross multiplied with 1,1, 2."},{"Start":"04:00.230 ","End":"04:04.880","Text":"Let\u0027s just label all of this as in the meantime,"},{"Start":"04:04.880 ","End":"04:09.320","Text":"A, so that we don\u0027t have to rewrite everything."},{"Start":"04:09.320 ","End":"04:12.050","Text":"We can cross out the first line."},{"Start":"04:12.050 ","End":"04:14.485","Text":"We have 0 times 2,"},{"Start":"04:14.485 ","End":"04:22.605","Text":"which is 0 minus negative 1 times 1,"},{"Start":"04:22.605 ","End":"04:24.044","Text":"which is just negative 1,"},{"Start":"04:24.044 ","End":"04:26.000","Text":"so this becomes plus 1."},{"Start":"04:26.000 ","End":"04:29.045","Text":"Then we cross out the middle line."},{"Start":"04:29.045 ","End":"04:30.635","Text":"We switch the order."},{"Start":"04:30.635 ","End":"04:35.270","Text":"We have negative 1 times 1 is negative 1,"},{"Start":"04:35.270 ","End":"04:40.345","Text":"and then minus 2 times 2, which is 4."},{"Start":"04:40.345 ","End":"04:48.845","Text":"Then we cross out the final line."},{"Start":"04:48.845 ","End":"04:58.920","Text":"We have 2 times 1 is 2 minus 0 times 1 is plus 0."},{"Start":"04:59.810 ","End":"05:06.260","Text":"Then what we get is A multiplied by the vector 1,"},{"Start":"05:06.260 ","End":"05:08.870","Text":"negative 5, 2."},{"Start":"05:12.230 ","End":"05:17.230","Text":"This is our k cross E, therefore,"},{"Start":"05:17.230 ","End":"05:23.800","Text":"let\u0027s write out our magnetic field is equal to our k hat,"},{"Start":"05:23.800 ","End":"05:27.875","Text":"cross E, which is what we calculated over here,"},{"Start":"05:27.875 ","End":"05:37.810","Text":"divided by c. We just take all of this and divide it by c. We have E_0 divided by root 5"},{"Start":"05:37.810 ","End":"05:46.480","Text":"c multiplied by e^i of 2x minus z minus"},{"Start":"05:46.480 ","End":"05:56.590","Text":"Omega t. Then multiplied by the vector 1 negative 5, 2."},{"Start":"05:58.520 ","End":"06:01.310","Text":"This is the answer to the question."},{"Start":"06:01.310 ","End":"06:09.750","Text":"Notice over here, root 5 is the magnitude of the k vector."},{"Start":"06:10.070 ","End":"06:13.850","Text":"It\u0027s multiplied by c. As we\u0027ve seen before,"},{"Start":"06:13.850 ","End":"06:17.340","Text":"that is just equal to Omega."},{"Start":"06:17.870 ","End":"06:19.940","Text":"Instead of this over here,"},{"Start":"06:19.940 ","End":"06:21.440","Text":"we could have just written Omega."},{"Start":"06:21.440 ","End":"06:24.905","Text":"We can see now how we get here,"},{"Start":"06:24.905 ","End":"06:27.830","Text":"and that is the answer."},{"Start":"06:27.830 ","End":"06:30.635","Text":"Now, another thing that we can check is to make sure"},{"Start":"06:30.635 ","End":"06:35.600","Text":"that our magnetic field is perpendicular to the electric field."},{"Start":"06:35.600 ","End":"06:38.240","Text":"As we know, they always have to be perpendicular."},{"Start":"06:38.240 ","End":"06:45.080","Text":"In order to check that the electric field is perpendicular to the magnetic fields,"},{"Start":"06:45.080 ","End":"06:50.810","Text":"we do the dot product between the two."},{"Start":"06:50.810 ","End":"06:56.189","Text":"What we\u0027ll have is this is in the 1, 1,"},{"Start":"06:56.189 ","End":"07:01.130","Text":"2 direction, dot product with the B field,"},{"Start":"07:01.130 ","End":"07:05.030","Text":"which is in the 1 negative 5, 2 direction."},{"Start":"07:05.030 ","End":"07:12.345","Text":"What we have is 1 times 1 is 1 plus 1 times minus 5."},{"Start":"07:12.345 ","End":"07:19.620","Text":"Negative 5 plus 2 times 2 is 4,"},{"Start":"07:19.620 ","End":"07:23.655","Text":"so 1 plus 4 is 5 minus 5 is 0."},{"Start":"07:23.655 ","End":"07:31.390","Text":"Therefore, E and B are perpendicular."},{"Start":"07:31.430 ","End":"07:35.000","Text":"In addition to this equation here, of course,"},{"Start":"07:35.000 ","End":"07:38.345","Text":"we can see that if we have the electric field,"},{"Start":"07:38.345 ","End":"07:42.650","Text":"we can calculate the corresponding magnetic field as we did in this lesson."},{"Start":"07:42.650 ","End":"07:46.520","Text":"But what if we have the magnetic field and"},{"Start":"07:46.520 ","End":"07:51.035","Text":"we want to calculate the electric fields using this equation."},{"Start":"07:51.035 ","End":"07:56.405","Text":"The electric field is equal to C,"},{"Start":"07:56.405 ","End":"08:03.463","Text":"which is the speed of light multiplied by B cross our k hat,"},{"Start":"08:03.463 ","End":"08:07.500","Text":"so the unit vector for our k vector."},{"Start":"08:07.790 ","End":"08:11.485","Text":"Write this into your equation sheets as well."},{"Start":"08:11.485 ","End":"08:16.420","Text":"Remember this is if you\u0027re finding B and this is if you\u0027re finding E,"},{"Start":"08:16.420 ","End":"08:20.455","Text":"which is the opposite of what we saw in Maxwell\u0027s equations."},{"Start":"08:20.455 ","End":"08:24.685","Text":"Where here on the left side we put in the fields that we had,"},{"Start":"08:24.685 ","End":"08:27.715","Text":"not the field that we were trying to find."},{"Start":"08:27.715 ","End":"08:30.530","Text":"Just remember that."},{"Start":"08:32.270 ","End":"08:36.150","Text":"We\u0027ve seen that these are perpendicular."},{"Start":"08:36.150 ","End":"08:38.710","Text":"We can see that the argument over here,"},{"Start":"08:38.710 ","End":"08:41.365","Text":"e^i of what\u0027s in the brackets here,"},{"Start":"08:41.365 ","End":"08:47.095","Text":"is, as usual, the exact same as in the electric field."},{"Start":"08:47.095 ","End":"08:49.790","Text":"The only difference is that they\u0027re perpendicular to one"},{"Start":"08:49.790 ","End":"08:54.750","Text":"another and that the amplitude is different."},{"Start":"08:54.980 ","End":"09:00.150","Text":"A short little notes on these new equations of why they work."},{"Start":"09:00.150 ","End":"09:05.885","Text":"We can imagine them as being in an axis."},{"Start":"09:05.885 ","End":"09:10.580","Text":"As we know, our k vector, k,"},{"Start":"09:10.580 ","End":"09:13.400","Text":"the direction of propagation is always"},{"Start":"09:13.400 ","End":"09:18.320","Text":"perpendicular to both the electric field and to the magnetic field."},{"Start":"09:18.320 ","End":"09:20.735","Text":"We know that the electric field and"},{"Start":"09:20.735 ","End":"09:23.930","Text":"the magnetic fields are always perpendicular to one another."},{"Start":"09:23.930 ","End":"09:26.540","Text":"We can think of this as being here,"},{"Start":"09:26.540 ","End":"09:29.060","Text":"the magnetic field, here,"},{"Start":"09:29.060 ","End":"09:32.855","Text":"our k vector, the direction of propagation,"},{"Start":"09:32.855 ","End":"09:35.965","Text":"and here the electric field."},{"Start":"09:35.965 ","End":"09:40.130","Text":"As we can see they\u0027re always perpendicular to one another."},{"Start":"09:40.130 ","End":"09:41.660","Text":"We can liken this to this being,"},{"Start":"09:41.660 ","End":"09:43.280","Text":"let\u0027s say the x,"},{"Start":"09:43.280 ","End":"09:46.535","Text":"this the y, and this the z-axis."},{"Start":"09:46.535 ","End":"09:50.105","Text":"Just like with our Cartesian coordinates, like so,"},{"Start":"09:50.105 ","End":"09:54.955","Text":"if we have x cross y, we get z."},{"Start":"09:54.955 ","End":"09:59.685","Text":"If we have y cross z, we get x."},{"Start":"09:59.685 ","End":"10:03.845","Text":"Similarly we can get all these different values."},{"Start":"10:03.845 ","End":"10:08.435","Text":"In the same way if we take B cross k,"},{"Start":"10:08.435 ","End":"10:14.640","Text":"we get E. If we do k cross E,"},{"Start":"10:14.640 ","End":"10:19.170","Text":"we get B. K cross E, we get B."},{"Start":"10:19.170 ","End":"10:26.090","Text":"Exactly as we would if we were using these coordinates of x, y, z."},{"Start":"10:26.090 ","End":"10:29.120","Text":"The only thing to remember is that we have"},{"Start":"10:29.120 ","End":"10:33.230","Text":"this value c for the speed of light, which is inside."},{"Start":"10:33.230 ","End":"10:35.645","Text":"When we\u0027re doing k cross E,"},{"Start":"10:35.645 ","End":"10:36.910","Text":"we divide by C,"},{"Start":"10:36.910 ","End":"10:39.380","Text":"and when we\u0027re doing B cross k,"},{"Start":"10:39.380 ","End":"10:47.465","Text":"we multiply it by c. I hope that\u0027s an explanation for why these equations work."},{"Start":"10:47.465 ","End":"10:50.910","Text":"That is the end of this lesson."}],"ID":22348},{"Watched":false,"Name":"Exercise - Wave Equations","Duration":"19m 13s","ChapterTopicVideoID":21562,"CourseChapterTopicPlaylistID":99477,"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":"00:05.415","Text":"we\u0027re going to be answering the following question."},{"Start":"00:05.415 ","End":"00:12.600","Text":"We have here the general wave equation for all waves where we have Nabla squared of Phi."},{"Start":"00:12.600 ","End":"00:16.320","Text":"The Laplacian of Phi is equal to 1 divided"},{"Start":"00:16.320 ","End":"00:23.460","Text":"by v^2 multiplied by the second derivative with respect to time of Phi."},{"Start":"00:23.460 ","End":"00:27.203","Text":"Phi is, of course, the wave function,"},{"Start":"00:27.203 ","End":"00:29.607","Text":"and v is the velocity of the wave."},{"Start":"00:29.607 ","End":"00:34.130","Text":"In general, the velocity of the wave is equal to Omega divided"},{"Start":"00:34.130 ","End":"00:39.440","Text":"by k. In this specific case of electromagnetic waves,"},{"Start":"00:39.440 ","End":"00:43.100","Text":"Phi is the function of the electric or magnetic field,"},{"Start":"00:43.100 ","End":"00:45.785","Text":"and v, the velocity of the wave,"},{"Start":"00:45.785 ","End":"00:49.230","Text":"is equal to c, the speed of light."},{"Start":"00:49.240 ","End":"00:53.030","Text":"What we saw in previous lessons that instead of Phi,"},{"Start":"00:53.030 ","End":"00:57.934","Text":"we wrote here e, the x component,"},{"Start":"00:57.934 ","End":"01:01.970","Text":"and then also here so this is what we\u0027ve seen up until now,"},{"Start":"01:01.970 ","End":"01:07.800","Text":"but generally speaking, we\u0027ll write it out as Phi."},{"Start":"01:08.060 ","End":"01:12.120","Text":"Let\u0027s answer question number 1,"},{"Start":"01:12.120 ","End":"01:15.950","Text":"which is to show that Phi as a function of x and"},{"Start":"01:15.950 ","End":"01:21.425","Text":"t is equal to A cosine of kx multiplied by sine of Omega t,"},{"Start":"01:21.425 ","End":"01:24.268","Text":"fulfills the wave equation,"},{"Start":"01:24.268 ","End":"01:30.520","Text":"and that it is a possible solution to the equation."},{"Start":"01:30.860 ","End":"01:34.670","Text":"The first thing that we\u0027re going to do is we\u0027re going to write"},{"Start":"01:34.670 ","End":"01:39.380","Text":"out the Laplacian of our Phi."},{"Start":"01:39.380 ","End":"01:45.395","Text":"The Laplacian is just the second derivative"},{"Start":"01:45.395 ","End":"01:51.875","Text":"with respect to x of our function Phi,"},{"Start":"01:51.875 ","End":"02:01.040","Text":"plus the second derivative with respect to y of our function Phi,"},{"Start":"02:01.040 ","End":"02:08.650","Text":"plus the second derivative with respect to z of our function Phi."},{"Start":"02:09.260 ","End":"02:14.180","Text":"Let\u0027s take the second derivative with respect to x."},{"Start":"02:14.180 ","End":"02:17.970","Text":"Only here we have our variable x."},{"Start":"02:18.380 ","End":"02:21.435","Text":"When we take the first derivative,"},{"Start":"02:21.435 ","End":"02:23.490","Text":"we\u0027ll have A,"},{"Start":"02:23.490 ","End":"02:28.270","Text":"and the first derivative is negative sine."},{"Start":"02:28.270 ","End":"02:33.695","Text":"Let\u0027s actually write this in blue over here that it\u0027s the first derivative."},{"Start":"02:33.695 ","End":"02:36.585","Text":"We\u0027ll have A of cosine,"},{"Start":"02:36.585 ","End":"02:38.625","Text":"the first derivative is negative,"},{"Start":"02:38.625 ","End":"02:45.720","Text":"sine of kx multiplied by sine of Omega t,"},{"Start":"02:45.720 ","End":"02:50.660","Text":"and this is multiplied by the inner derivative over here with respect to x,"},{"Start":"02:50.660 ","End":"02:57.210","Text":"which is k. We can write it over here, negative kA."},{"Start":"02:58.670 ","End":"03:01.050","Text":"This was the first derivative."},{"Start":"03:01.050 ","End":"03:03.480","Text":"Now, let\u0027s take the second derivative,"},{"Start":"03:03.480 ","End":"03:05.760","Text":"again, with respect to x."},{"Start":"03:05.760 ","End":"03:10.318","Text":"We have negative kA,"},{"Start":"03:10.318 ","End":"03:12.830","Text":"and then with respect to x again,"},{"Start":"03:12.830 ","End":"03:19.325","Text":"so the derivative of sine is going to be cosine."},{"Start":"03:19.325 ","End":"03:28.325","Text":"We have cosine of kx sine of Omega t. Then again,"},{"Start":"03:28.325 ","End":"03:31.580","Text":"we have to multiply by the anti-derivative,"},{"Start":"03:31.580 ","End":"03:37.175","Text":"which over here again is k. We have k multiplied by k, which is k^2."},{"Start":"03:37.175 ","End":"03:41.950","Text":"Now, I\u0027ll rub this out."},{"Start":"03:42.500 ","End":"03:46.700","Text":"Now, let\u0027s add in the second derivative with respect to y."},{"Start":"03:46.700 ","End":"03:48.290","Text":"If we look at the equation,"},{"Start":"03:48.290 ","End":"03:52.490","Text":"we have no y-variable and similarly,"},{"Start":"03:52.490 ","End":"03:55.625","Text":"with respect to z, we have no z variable."},{"Start":"03:55.625 ","End":"03:59.550","Text":"This is just what we are left with."},{"Start":"03:59.780 ","End":"04:09.360","Text":"Now, let\u0027s work out the second derivative of our wave function with respect to time."},{"Start":"04:11.030 ","End":"04:15.195","Text":"That will be the second derivative of this."},{"Start":"04:15.195 ","End":"04:21.260","Text":"Again, only over here do we have this t variable so what"},{"Start":"04:21.260 ","End":"04:29.175","Text":"we\u0027ll have is A cosine of kx,"},{"Start":"04:29.175 ","End":"04:35.005","Text":"and then, if we take the first derivative of sine Omega t,"},{"Start":"04:35.005 ","End":"04:36.815","Text":"so we get cosine,"},{"Start":"04:36.815 ","End":"04:40.980","Text":"so we\u0027ll be left here with cosine of"},{"Start":"04:40.980 ","End":"04:46.830","Text":"Omega t multiplied by the anti-derivative, which is Omega."},{"Start":"04:46.830 ","End":"04:51.290","Text":"Then when we take the second derivative of this,"},{"Start":"04:51.310 ","End":"04:54.725","Text":"the cosine becomes negative sine,"},{"Start":"04:54.725 ","End":"04:57.515","Text":"so we just put here a negative,"},{"Start":"04:57.515 ","End":"05:00.920","Text":"and we just change from cosine to"},{"Start":"05:00.920 ","End":"05:05.840","Text":"sine and then we again multiply by the anti-derivative, which is Omega."},{"Start":"05:05.840 ","End":"05:08.300","Text":"We\u0027re left with negative Omega^2,"},{"Start":"05:08.300 ","End":"05:11.160","Text":"A cosine of kx,"},{"Start":"05:11.160 ","End":"05:14.040","Text":"sine of Omega t. Now,"},{"Start":"05:14.040 ","End":"05:21.495","Text":"what we can see is A cosine sine is our original function,"},{"Start":"05:21.495 ","End":"05:24.450","Text":"Phi, and also over here,"},{"Start":"05:24.450 ","End":"05:28.539","Text":"A cosine sine is our original function."},{"Start":"05:29.030 ","End":"05:34.610","Text":"In that case, we can write this as negative k^2 of"},{"Start":"05:34.610 ","End":"05:41.280","Text":"Phi and this as being equal to negative Omega^2 of Phi."},{"Start":"05:41.280 ","End":"05:45.255","Text":"Now, we want to make them equal."},{"Start":"05:45.255 ","End":"05:48.450","Text":"What we have is we say that,"},{"Start":"05:48.450 ","End":"05:57.870","Text":"negative k^2 of Phi is equal to 1 divided by v^2 of this over here,"},{"Start":"05:57.870 ","End":"06:02.700","Text":"which is negative Omega^2 of Phi."},{"Start":"06:02.700 ","End":"06:06.485","Text":"We can cross out the Omega from each side and"},{"Start":"06:06.485 ","End":"06:13.460","Text":"the minus sign and then we\u0027re left with k^2 or let\u0027s take the square root of both sides."},{"Start":"06:13.460 ","End":"06:19.610","Text":"We have k is equal to Omega divided by v. Now, if we look here,"},{"Start":"06:19.610 ","End":"06:26.105","Text":"we\u0027re given that v is equal to Omega k. If we multiply both sides by v and divide by k,"},{"Start":"06:26.105 ","End":"06:31.370","Text":"we get over here that v is equal to Omega divided by k,"},{"Start":"06:31.370 ","End":"06:36.760","Text":"which we know is meant to be correct from what is given."},{"Start":"06:36.760 ","End":"06:42.305","Text":"That\u0027s how we show that this equation for Phi over here fulfills"},{"Start":"06:42.305 ","End":"06:48.730","Text":"the wave equation and that therefore it is a possible solution to the equation."},{"Start":"06:48.730 ","End":"06:52.305","Text":"Now, let\u0027s answer question number 2."},{"Start":"06:52.305 ","End":"06:59.113","Text":"Question number 2 says that d\u0027Alembert solution to the wave equation says that"},{"Start":"06:59.113 ","End":"07:07.245","Text":"every solution must be of the form f(x) minus vt plus g(x) plus vt,"},{"Start":"07:07.245 ","End":"07:09.720","Text":"where f and g are functions."},{"Start":"07:09.720 ","End":"07:13.555","Text":"We\u0027re being asked to show that the function from 1,"},{"Start":"07:13.555 ","End":"07:19.385","Text":"the function over here is also a solution according to d\u0027Alembert."},{"Start":"07:19.385 ","End":"07:25.225","Text":"The hint for us to show this is to use trigonometric identities."},{"Start":"07:25.225 ","End":"07:30.061","Text":"For those of you that remember trigonometric identities very well,"},{"Start":"07:30.061 ","End":"07:32.270","Text":"that is very useful, but otherwise,"},{"Start":"07:32.270 ","End":"07:37.730","Text":"2 very useful identities to remember are the"},{"Start":"07:37.730 ","End":"07:43.740","Text":"following and I also suggest you write them in your equation sheets."},{"Start":"07:43.740 ","End":"07:53.685","Text":"The first is sine of Alpha plus sine of Beta"},{"Start":"07:53.685 ","End":"08:01.390","Text":"is equal to 2 sine of Alpha plus Beta over"},{"Start":"08:01.390 ","End":"08:10.270","Text":"2 multiplied by cosine of Alpha minus Beta over 2."},{"Start":"08:10.270 ","End":"08:14.935","Text":"This is the first one and the second one is sine of Alpha"},{"Start":"08:14.935 ","End":"08:22.330","Text":"minus sine of Beta is equal to 2 sine,"},{"Start":"08:22.330 ","End":"08:24.610","Text":"and then the signs switch over here."},{"Start":"08:24.610 ","End":"08:30.145","Text":"Here it\u0027s Alpha minus Beta over 2 multiplied by"},{"Start":"08:30.145 ","End":"08:36.925","Text":"cosine of Alpha plus Beta over 2."},{"Start":"08:36.925 ","End":"08:39.595","Text":"Here if we have a positive,"},{"Start":"08:39.595 ","End":"08:41.320","Text":"it\u0027s like so plus the minus,"},{"Start":"08:41.320 ","End":"08:42.370","Text":"and if there\u0027s a negative,"},{"Start":"08:42.370 ","End":"08:45.500","Text":"it\u0027s negative then plus."},{"Start":"08:45.840 ","End":"08:49.525","Text":"Now let\u0027s write out our function."},{"Start":"08:49.525 ","End":"09:00.135","Text":"We have Phi is equal to A cosine of kx"},{"Start":"09:00.135 ","End":"09:05.450","Text":"multiplied by sine of Omega t."},{"Start":"09:05.610 ","End":"09:14.180","Text":"Now what we want to do is we want to write this out in one of these formats over here."},{"Start":"09:14.580 ","End":"09:18.610","Text":"Taking into account this 2 over here."},{"Start":"09:18.610 ","End":"09:25.345","Text":"We can say that this is equal to A and then we\u0027ll divide and multiply by 2."},{"Start":"09:25.345 ","End":"09:28.270","Text":"The 2\u0027s will cancel out."},{"Start":"09:28.270 ","End":"09:33.610","Text":"But what this does is this gives us this constant in relation to the 2 over here,"},{"Start":"09:33.610 ","End":"09:37.345","Text":"so we can equate it to this format over here."},{"Start":"09:37.345 ","End":"09:46.870","Text":"Then we have cosine of kx multiplied by"},{"Start":"09:46.870 ","End":"09:56.440","Text":"sine of Omega t. We can just show this."},{"Start":"09:56.440 ","End":"10:01.160","Text":"Now we can see that we\u0027re just beginning to get this format."},{"Start":"10:05.490 ","End":"10:08.110","Text":"Which format are we going to use here?"},{"Start":"10:08.110 ","End":"10:16.240","Text":"We see that we want to have f of this plus g. We\u0027ll look at this format over here,"},{"Start":"10:16.240 ","End":"10:18.800","Text":"where we have a plus."},{"Start":"10:19.080 ","End":"10:21.910","Text":"We see that the cosine,"},{"Start":"10:21.910 ","End":"10:27.730","Text":"where we have kx has to be"},{"Start":"10:27.730 ","End":"10:34.765","Text":"equal to Alpha minus Beta divided by 2."},{"Start":"10:34.765 ","End":"10:39.835","Text":"Omega t has to be equal to the sine component,"},{"Start":"10:39.835 ","End":"10:44.935","Text":"which is Alpha plus Beta divided by 2."},{"Start":"10:44.935 ","End":"10:50.365","Text":"Let\u0027s call this a and let\u0027s call this B,"},{"Start":"10:50.365 ","End":"10:52.540","Text":"so these 2 equations."},{"Start":"10:52.540 ","End":"10:55.960","Text":"Now, in order to find out what Alpha and Beta are,"},{"Start":"10:55.960 ","End":"11:00.295","Text":"so first we can do a plus b."},{"Start":"11:00.295 ","End":"11:02.200","Text":"We add these 2 equations,"},{"Start":"11:02.200 ","End":"11:06.490","Text":"so then we\u0027ll have Alpha over 2 plus Alpha over 2 is just Alpha."},{"Start":"11:06.490 ","End":"11:10.420","Text":"Then we have negative Beta over 2 plus Beta over 2,"},{"Start":"11:10.420 ","End":"11:12.520","Text":"so it cancels out."},{"Start":"11:12.520 ","End":"11:15.940","Text":"Alpha is equal to a plus b,"},{"Start":"11:15.940 ","End":"11:25.015","Text":"which is just kx plus Omega t. Then if we take a minus b,"},{"Start":"11:25.015 ","End":"11:30.550","Text":"so we get 1/2 Alpha minus 1/2 Alpha is 0."},{"Start":"11:30.550 ","End":"11:38.477","Text":"Then we have negative 1/2 Beta which is a,"},{"Start":"11:38.477 ","End":"11:44.470","Text":"negative b, so negative 1/2 Beta which is negative Beta."},{"Start":"11:44.470 ","End":"11:49.135","Text":"We get that negative Beta is equal to a minus b,"},{"Start":"11:49.135 ","End":"11:53.545","Text":"so it\u0027s equal to kx minus b,"},{"Start":"11:53.545 ","End":"11:56.980","Text":"which is Omega t. Of course,"},{"Start":"11:56.980 ","End":"11:58.660","Text":"we can also flip these around,"},{"Start":"11:58.660 ","End":"12:00.970","Text":"it doesn\u0027t really make a difference."},{"Start":"12:00.970 ","End":"12:07.845","Text":"In other words, we can say that Beta is equal to Omega t minus kx."},{"Start":"12:07.845 ","End":"12:12.250","Text":"We multiply both sides by negative 1."},{"Start":"12:12.720 ","End":"12:17.230","Text":"In that case we can say that our Phi,"},{"Start":"12:17.230 ","End":"12:19.405","Text":"back to this equation over here,"},{"Start":"12:19.405 ","End":"12:24.130","Text":"is equal to A divided by 2 multiplied by,"},{"Start":"12:24.130 ","End":"12:28.085","Text":"so now we\u0027re just going to write it out and solve in this format over here,"},{"Start":"12:28.085 ","End":"12:30.045","Text":"which is what we have written over here,"},{"Start":"12:30.045 ","End":"12:33.100","Text":"we\u0027ll write it in this format here."},{"Start":"12:33.530 ","End":"12:38.800","Text":"Multiplied by sine of Alpha."},{"Start":"12:39.570 ","End":"12:50.390","Text":"Alpha as we said was equal to kx plus Omega t plus sine of Beta,"},{"Start":"12:50.390 ","End":"13:00.500","Text":"where Beta is equal to Omega t minus kx."},{"Start":"13:03.480 ","End":"13:08.200","Text":"As we saw, we had here 2 cosine of something,"},{"Start":"13:08.200 ","End":"13:10.645","Text":"sine of something, which is like this."},{"Start":"13:10.645 ","End":"13:13.405","Text":"2 cosine of something, sine of something,"},{"Start":"13:13.405 ","End":"13:19.330","Text":"so we wrote it in this format so that we can convert from this side to this side."},{"Start":"13:19.330 ","End":"13:23.665","Text":"That\u0027s why this 2 over here is important to us."},{"Start":"13:23.665 ","End":"13:27.355","Text":"Now we\u0027re writing that it\u0027s equal to this over here."},{"Start":"13:27.355 ","End":"13:30.100","Text":"Now, if we look back into our question,"},{"Start":"13:30.100 ","End":"13:33.850","Text":"we have f(x) minus something as"},{"Start":"13:33.850 ","End":"13:37.690","Text":"a function of t. We have this x plus something as a function of t,"},{"Start":"13:37.690 ","End":"13:38.710","Text":"but that\u0027s okay,"},{"Start":"13:38.710 ","End":"13:42.655","Text":"it just means this something is just minus of this."},{"Start":"13:42.655 ","End":"13:50.540","Text":"That\u0027s fine, and then plus g(x) plus."},{"Start":"13:50.540 ","End":"13:54.640","Text":"Here, what we can do is we can use"},{"Start":"13:54.640 ","End":"14:01.120","Text":"the identity saying that sine of something,"},{"Start":"14:01.120 ","End":"14:02.845","Text":"so let\u0027s say we\u0027ll write it out here."},{"Start":"14:02.845 ","End":"14:09.145","Text":"Omega t minus kx is equal to negative sine"},{"Start":"14:09.145 ","End":"14:16.285","Text":"of kx minus Omega t. Therefore,"},{"Start":"14:16.285 ","End":"14:22.330","Text":"we can say that Phi is"},{"Start":"14:22.330 ","End":"14:28.840","Text":"equal to A over 2 of sine,"},{"Start":"14:28.840 ","End":"14:32.095","Text":"so then we can write this over here."},{"Start":"14:32.095 ","End":"14:39.955","Text":"Sine of kx plus Omega t plus or rather minus over here,"},{"Start":"14:39.955 ","End":"14:46.000","Text":"minus sine of kx"},{"Start":"14:46.000 ","End":"14:52.250","Text":"minus Omega t from this identity over here."},{"Start":"14:56.490 ","End":"15:04.765","Text":"We\u0027re almost done. What we can see in d\u0027Alembert is our x variable is alone,"},{"Start":"15:04.765 ","End":"15:10.030","Text":"so it doesn\u0027t have a coefficient."},{"Start":"15:10.030 ","End":"15:16.480","Text":"Here we have the coefficient k. We\u0027re going to take the k out."},{"Start":"15:16.480 ","End":"15:24.985","Text":"We\u0027re just going to rewrite this as A over 2 of sine,"},{"Start":"15:24.985 ","End":"15:26.350","Text":"and then we have k,"},{"Start":"15:26.350 ","End":"15:30.800","Text":"and then we\u0027ll have x plus Omega over kt."},{"Start":"15:34.080 ","End":"15:40.510","Text":"Minus sine, so I\u0027m just taking again k out as a factor,"},{"Start":"15:40.510 ","End":"15:45.710","Text":"so kx minus Omega over kt."},{"Start":"15:48.660 ","End":"15:51.145","Text":"As we saw before,"},{"Start":"15:51.145 ","End":"15:54.940","Text":"Omega over k=v over here,"},{"Start":"15:54.940 ","End":"15:58.415","Text":"and we also saw it in question number 1 at the end."},{"Start":"15:58.415 ","End":"16:00.665","Text":"V is equal to Omega over k,"},{"Start":"16:00.665 ","End":"16:06.965","Text":"so we can just erase this and write this as v,"},{"Start":"16:06.965 ","End":"16:10.160","Text":"which is exactly what we wanted here,"},{"Start":"16:10.160 ","End":"16:14.100","Text":"vt. Now we have vt."},{"Start":"16:14.160 ","End":"16:20.530","Text":"If we just multiply both by A divided by 2,"},{"Start":"16:20.530 ","End":"16:26.620","Text":"so we have A2 multiplied by sine of kx"},{"Start":"16:26.620 ","End":"16:33.250","Text":"plus vt minus or rather plus and then let\u0027s write it minus,"},{"Start":"16:33.250 ","End":"16:34.745","Text":"I\u0027ll explain this in a second."},{"Start":"16:34.745 ","End":"16:42.610","Text":"A divided by 2 of sine of kx minus vt."},{"Start":"16:42.610 ","End":"16:51.320","Text":"In that case, we can see that this term over here is because we have the plus over here."},{"Start":"16:51.320 ","End":"16:57.859","Text":"This is our g as a function of x plus vt."},{"Start":"16:57.859 ","End":"17:01.880","Text":"This term over here with the minus,"},{"Start":"17:01.880 ","End":"17:08.450","Text":"because here we have a minus is our f. This is a function of f,"},{"Start":"17:08.450 ","End":"17:17.480","Text":"which is x minus vt. Now we can see that we got from our original equation,"},{"Start":"17:17.480 ","End":"17:20.070","Text":"this one over here."},{"Start":"17:20.070 ","End":"17:22.565","Text":"Our original wave equation,"},{"Start":"17:22.565 ","End":"17:27.410","Text":"we use d\u0027Alembert theory in order to"},{"Start":"17:27.410 ","End":"17:30.665","Text":"manipulate it using trigonometric identities"},{"Start":"17:30.665 ","End":"17:33.545","Text":"in order to get it into this format over here,"},{"Start":"17:33.545 ","End":"17:37.740","Text":"where it is the sum of 2 functions."},{"Start":"17:38.730 ","End":"17:41.470","Text":"We\u0027ve shown how to do this,"},{"Start":"17:41.470 ","End":"17:43.530","Text":"so this is the end of question number 2."},{"Start":"17:43.530 ","End":"17:47.090","Text":"I just want to speak about what type of wave this is, because at the end of the day,"},{"Start":"17:47.090 ","End":"17:51.860","Text":"we started off with a wave function that we just manipulated into a different format,"},{"Start":"17:51.860 ","End":"17:54.785","Text":"but it\u0027s still a wave function."},{"Start":"17:54.785 ","End":"17:57.800","Text":"What we can see, which is what we\u0027ve seen in"},{"Start":"17:57.800 ","End":"18:01.205","Text":"previous videos is that because there\u0027s a plus over here,"},{"Start":"18:01.205 ","End":"18:05.690","Text":"that means that this wave is traveling in this negative direction."},{"Start":"18:05.690 ","End":"18:07.925","Text":"Because there\u0027s a minus over here,"},{"Start":"18:07.925 ","End":"18:11.715","Text":"that means that this wave is traveling in"},{"Start":"18:11.715 ","End":"18:19.120","Text":"this positive direction along a certain axis."},{"Start":"18:19.120 ","End":"18:21.245","Text":"This is what we can see,"},{"Start":"18:21.245 ","End":"18:24.390","Text":"that we have 2 waves."},{"Start":"18:24.390 ","End":"18:27.290","Text":"They\u0027re both equal in magnitude,"},{"Start":"18:27.290 ","End":"18:29.960","Text":"but they\u0027re traveling in different directions."},{"Start":"18:29.960 ","End":"18:35.305","Text":"What we get in a case like this is a standing wave."},{"Start":"18:35.305 ","End":"18:41.105","Text":"I\u0027m not going to go into too much detail with what a standing wave is right now,"},{"Start":"18:41.105 ","End":"18:45.890","Text":"you\u0027ll learn it in later chapters and in your later studies."},{"Start":"18:45.890 ","End":"18:50.090","Text":"But this is just to give you a sense of what this means."},{"Start":"18:50.090 ","End":"18:54.410","Text":"We have one wave that has the exact same magnitude,"},{"Start":"18:54.410 ","End":"18:56.780","Text":"where one is traveling in one direction and the"},{"Start":"18:56.780 ","End":"19:00.449","Text":"other is traveling in the opposite direction."},{"Start":"19:00.660 ","End":"19:02.935","Text":"This is a standing wave,"},{"Start":"19:02.935 ","End":"19:09.380","Text":"this might give you a little bit more intuition later in more advanced wave courses."},{"Start":"19:09.380 ","End":"19:13.559","Text":"That is the end of this lesson."}],"ID":22349},{"Watched":false,"Name":"Deriving the Equation for the Dispersion Relationship","Duration":"11m 47s","ChapterTopicVideoID":21362,"CourseChapterTopicPlaylistID":99477,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.385","Text":"Hello. In the previous lesson,"},{"Start":"00:02.385 ","End":"00:05.175","Text":"we learned about the dispersion relationship,"},{"Start":"00:05.175 ","End":"00:11.520","Text":"which was the equation linking Omega 2k in the wave equation."},{"Start":"00:11.520 ","End":"00:16.185","Text":"We saw that Omega is equal to c multiplied by the magnitude of k,"},{"Start":"00:16.185 ","End":"00:17.580","Text":"where c is, of course,"},{"Start":"00:17.580 ","End":"00:18.990","Text":"the speed of light."},{"Start":"00:18.990 ","End":"00:24.730","Text":"In this lesson, we\u0027re going to see how we arrive at this relationship."},{"Start":"00:25.460 ","End":"00:28.390","Text":"Let\u0027s show how we do this."},{"Start":"00:28.390 ","End":"00:29.505","Text":"In the previous lesson,"},{"Start":"00:29.505 ","End":"00:32.655","Text":"we saw Maxwell\u0027s 4 equations,"},{"Start":"00:32.655 ","End":"00:40.940","Text":"which by using them we get to the wave equations which describe waves in a vacuum."},{"Start":"00:40.940 ","End":"00:47.095","Text":"Then we saw what this Nabla^2 was equal to."},{"Start":"00:47.095 ","End":"00:50.360","Text":"It\u0027s the partial derivative or"},{"Start":"00:50.360 ","End":"00:52.715","Text":"the second partial derivative of"},{"Start":"00:52.715 ","End":"00:56.525","Text":"a certain component of the electric field according to x,"},{"Start":"00:56.525 ","End":"00:57.740","Text":"y, and z,"},{"Start":"00:57.740 ","End":"01:00.260","Text":"and then their summation."},{"Start":"01:00.260 ","End":"01:07.010","Text":"Also, we saw that Mu naught Epsilon naught was equal to 1 divided by c^2,"},{"Start":"01:07.010 ","End":"01:11.495","Text":"where of course, c is the speed of light."},{"Start":"01:11.495 ","End":"01:16.470","Text":"Now, let\u0027s just write this out again."},{"Start":"01:16.470 ","End":"01:18.950","Text":"Of course, we can do this exact same thing as we"},{"Start":"01:18.950 ","End":"01:22.130","Text":"saw here for the other components of the E field."},{"Start":"01:22.130 ","End":"01:25.369","Text":"The y-component and also the z-component,"},{"Start":"01:25.369 ","End":"01:26.735","Text":"it\u0027s the exact same thing."},{"Start":"01:26.735 ","End":"01:31.910","Text":"We\u0027re just showing the example over here with the x-component."},{"Start":"01:31.910 ","End":"01:38.870","Text":"We can just rewrite this as being equal to 1 divided by c^2 for"},{"Start":"01:38.870 ","End":"01:41.883","Text":"the Mu naught Epsilon naught"},{"Start":"01:41.883 ","End":"01:50.620","Text":"of d^2E_x by dtx^2."},{"Start":"01:50.620 ","End":"01:54.215","Text":"Now, what we\u0027re going to do is we\u0027re going to plug"},{"Start":"01:54.215 ","End":"01:58.940","Text":"in the equation for the x-component of the E field."},{"Start":"01:58.940 ","End":"02:00.695","Text":"As we\u0027ve seen before,"},{"Start":"02:00.695 ","End":"02:02.870","Text":"we can write it as E_x,"},{"Start":"02:02.870 ","End":"02:05.540","Text":"which is as a function of r,"},{"Start":"02:05.540 ","End":"02:12.920","Text":"its position and t time is equal to A cosine of"},{"Start":"02:12.920 ","End":"02:18.035","Text":"k vector dot product with r vector minus"},{"Start":"02:18.035 ","End":"02:24.500","Text":"Omega t. This is our x-component and,"},{"Start":"02:24.500 ","End":"02:29.270","Text":"of course, we can write the same thing for the y and z-components,"},{"Start":"02:29.270 ","End":"02:32.960","Text":"just, they might have different amplitudes."},{"Start":"02:32.960 ","End":"02:38.435","Text":"We can again write this as A cosine of,"},{"Start":"02:38.435 ","End":"02:48.475","Text":"and then we remember that the r vector is equal to some position x, y, z."},{"Start":"02:48.475 ","End":"02:51.125","Text":"When we do the dot product between k and r,"},{"Start":"02:51.125 ","End":"02:57.490","Text":"we get k_xx plus k_yy plus"},{"Start":"02:57.490 ","End":"03:04.170","Text":"k_zz minus Omega t."},{"Start":"03:04.220 ","End":"03:08.160","Text":"This is our x-component of the E field."},{"Start":"03:08.160 ","End":"03:17.550","Text":"Now, what we\u0027re going to do is we\u0027re going to plug this in to this equation over here."},{"Start":"03:18.290 ","End":"03:27.210","Text":"Let\u0027s first of all look at this d^2E_x by dx^2."},{"Start":"03:27.210 ","End":"03:28.740","Text":"We\u0027re just plugging this in,"},{"Start":"03:28.740 ","End":"03:34.120","Text":"so let\u0027s just write this d^2E_x by dx^2."},{"Start":"03:34.130 ","End":"03:41.850","Text":"That will be d^2 by dx^2 of this so of"},{"Start":"03:41.850 ","End":"03:48.990","Text":"A cosine of k_xx plus k_yy"},{"Start":"03:48.990 ","End":"03:58.350","Text":"plus k_zz minus Omega t. Now,"},{"Start":"03:58.350 ","End":"04:03.355","Text":"we\u0027re taking the derivative twice with respect to x."},{"Start":"04:03.355 ","End":"04:06.485","Text":"If we take the first derivative,"},{"Start":"04:06.485 ","End":"04:11.015","Text":"so let\u0027s write the first derivative in blue at the bottom."},{"Start":"04:11.015 ","End":"04:14.060","Text":"The derivative of cosine is negative sine,"},{"Start":"04:14.060 ","End":"04:21.380","Text":"so we\u0027ll have negative A sine of whatever is in the brackets k_xx plus"},{"Start":"04:21.380 ","End":"04:29.175","Text":"k_yy plus k_zz minus Omega t,"},{"Start":"04:29.175 ","End":"04:36.240","Text":"and then we\u0027re going to multiply by the inner derivative with respect to x."},{"Start":"04:36.240 ","End":"04:41.035","Text":"This is over here. It will just be multiplied by k_x."},{"Start":"04:41.035 ","End":"04:45.465","Text":"Remember, k_x is the x-component of the k vector,"},{"Start":"04:45.465 ","End":"04:48.230","Text":"k_y is the y-component of the k vector,"},{"Start":"04:48.230 ","End":"04:50.900","Text":"and z is the z-component of the k vector,"},{"Start":"04:50.900 ","End":"04:52.850","Text":"just in case you forgot that."},{"Start":"04:52.850 ","End":"04:57.715","Text":"Now we\u0027ll write out the second derivative over here."},{"Start":"04:57.715 ","End":"05:00.535","Text":"We\u0027re taking this derivative again."},{"Start":"05:00.535 ","End":"05:11.210","Text":"The derivative of sine is just cosine so we still have negative Ak_x"},{"Start":"05:11.210 ","End":"05:19.850","Text":"and then cosine of k_xx plus k_yy plus"},{"Start":"05:19.850 ","End":"05:25.055","Text":"k_zz minus Omega t and then"},{"Start":"05:25.055 ","End":"05:30.605","Text":"we multiply this again by the derivative of our x values."},{"Start":"05:30.605 ","End":"05:34.865","Text":"Again, only here inside the brackets do we have this."},{"Start":"05:34.865 ","End":"05:42.330","Text":"Again, we\u0027re multiplying by just k_x so we can write k_x^2 over here."},{"Start":"05:43.040 ","End":"05:45.210","Text":"As we\u0027ll notice here,"},{"Start":"05:45.210 ","End":"05:50.175","Text":"we have some coefficients, these are constants."},{"Start":"05:50.175 ","End":"05:57.590","Text":"Over here we have cosine of k_xx, k_yy,"},{"Start":"05:57.590 ","End":"06:03.925","Text":"k_zz minus Omega t. In other words,"},{"Start":"06:03.925 ","End":"06:14.440","Text":"if we take into account also the amplitude and this together we have just E_x,"},{"Start":"06:14.440 ","End":"06:20.220","Text":"the x-component of our electric field."},{"Start":"06:20.900 ","End":"06:25.910","Text":"In other words, we can just write this as being equal to"},{"Start":"06:25.910 ","End":"06:30.440","Text":"negative k_x^2"},{"Start":"06:30.440 ","End":"06:38.320","Text":"multiplied by E_x."},{"Start":"06:38.320 ","End":"06:47.700","Text":"Now, let\u0027s scroll down and it\u0027s of course the exact same thing for d^2E_x by dy^2."},{"Start":"06:48.440 ","End":"06:51.680","Text":"We\u0027re going to have the exact same thing except"},{"Start":"06:51.680 ","End":"06:54.755","Text":"instead of taking the second derivative with respect to x,"},{"Start":"06:54.755 ","End":"06:59.420","Text":"we\u0027re taking the 2 derivatives with respect to y."},{"Start":"06:59.420 ","End":"07:05.300","Text":"Instead of having k_x^2, we\u0027ll have k_y^2."},{"Start":"07:05.300 ","End":"07:07.835","Text":"We do the exact same steps,"},{"Start":"07:07.835 ","End":"07:12.290","Text":"but we have k_y^2 because we\u0027re taking the derivative with respect to y."},{"Start":"07:12.290 ","End":"07:22.010","Text":"What we\u0027ll have is that this is equal to simply negative k_y^2E_x again and of course,"},{"Start":"07:22.010 ","End":"07:30.515","Text":"d^2E_x by dz^2 is the same thing but taking the derivative with respect to z."},{"Start":"07:30.515 ","End":"07:40.405","Text":"We\u0027ll have negative k_z^2E_x."},{"Start":"07:40.405 ","End":"07:44.405","Text":"Here we have all of these equations and of course,"},{"Start":"07:44.405 ","End":"07:47.150","Text":"this is equal to this over here."},{"Start":"07:47.150 ","End":"07:53.840","Text":"1 divided by c^2 of d^2E_x by dt^2."},{"Start":"07:53.840 ","End":"08:03.750","Text":"Now, let\u0027s write out d^2E_x by dt^2."},{"Start":"08:03.750 ","End":"08:11.655","Text":"Now, again, we\u0027re taking the second derivative of this but with respect to t, to time."},{"Start":"08:11.655 ","End":"08:17.255","Text":"Just like before, we\u0027re going to get that this is equal to the negative,"},{"Start":"08:17.255 ","End":"08:21.545","Text":"which comes from taking the first derivative of cosine,"},{"Start":"08:21.545 ","End":"08:28.740","Text":"and then it\u0027s just going to be the coefficient of the variable squared."},{"Start":"08:28.740 ","End":"08:35.080","Text":"The coefficient of t is negative Omega^2 but a negative squared is a positive"},{"Start":"08:35.080 ","End":"08:42.034","Text":"so we\u0027re just going to have Omega^2 multiplied by E_x."},{"Start":"08:42.034 ","End":"08:43.700","Text":"Just like we\u0027ve seen before,"},{"Start":"08:43.700 ","End":"08:47.730","Text":"we\u0027ve done the exact same as what we\u0027ve done here."},{"Start":"08:48.680 ","End":"08:56.295","Text":"Now, we take all of this so let\u0027s just show this in pink."},{"Start":"08:56.295 ","End":"08:59.794","Text":"We take this value, this value,"},{"Start":"08:59.794 ","End":"09:05.180","Text":"this value, and this value and we just plug it into this equation."},{"Start":"09:05.180 ","End":"09:11.550","Text":"Therefore, what we get is d^2E_x by dx^2 so we have"},{"Start":"09:11.550 ","End":"09:18.375","Text":"negative k_x^2E_x plus d^2E_x by dy^2,"},{"Start":"09:18.375 ","End":"09:28.170","Text":"so plus negative so we can just write negative k_y^2E_x plus negative."},{"Start":"09:28.170 ","End":"09:32.440","Text":"Just negative k_z^2E_x."},{"Start":"09:32.450 ","End":"09:41.315","Text":"All of this is equal to 1 divided by c^2d^2E_x by dt^2,"},{"Start":"09:41.315 ","End":"09:44.660","Text":"which is just this negative Omega^2E_x."},{"Start":"09:44.660 ","End":"09:50.960","Text":"We can just write the negative over here, Omega^2 E_x."},{"Start":"09:51.400 ","End":"09:56.990","Text":"Now, what we can see is we can divide both sides by E_x so this cancels out,"},{"Start":"09:56.990 ","End":"09:58.340","Text":"this cancels out, this cancels out,"},{"Start":"09:58.340 ","End":"10:06.360","Text":"this cancels out and by minus 1 so all of these become positives."},{"Start":"10:06.360 ","End":"10:08.310","Text":"Now, let\u0027s just rewrite this."},{"Start":"10:08.310 ","End":"10:15.815","Text":"We have k_x^2 plus k_y^2 plus k_z^2,"},{"Start":"10:15.815 ","End":"10:21.560","Text":"which is equal to Omega^2 divided by c^2."},{"Start":"10:21.560 ","End":"10:31.070","Text":"Now, what we can say is that what we have here is we just have our k vector squared."},{"Start":"10:31.070 ","End":"10:33.420","Text":"If our k vector was k_x, k_y,"},{"Start":"10:33.420 ","End":"10:35.865","Text":"k_z so if we square it,"},{"Start":"10:35.865 ","End":"10:38.835","Text":"we have k_x^2, k_y^2, k_z^2."},{"Start":"10:38.835 ","End":"10:42.499","Text":"What we have is our k vector squared,"},{"Start":"10:42.499 ","End":"10:47.620","Text":"which is equal to Omega^2 divided by c^2."},{"Start":"10:47.620 ","End":"10:52.475","Text":"Here, this is the same as just taking the magnitude of the k vector."},{"Start":"10:52.475 ","End":"10:54.170","Text":"I\u0027ve just written it like this."},{"Start":"10:54.170 ","End":"10:56.780","Text":"Now, if we take the square root of both sides,"},{"Start":"10:56.780 ","End":"11:01.910","Text":"so we\u0027re still left with the magnitude of our k vector so I\u0027ll leave it"},{"Start":"11:01.910 ","End":"11:07.370","Text":"written like this as being equal to Omega divided by c. Now,"},{"Start":"11:07.370 ","End":"11:09.575","Text":"if we multiply both sides by c,"},{"Start":"11:09.575 ","End":"11:13.835","Text":"we\u0027re left with the relationship of Omega is equal to c"},{"Start":"11:13.835 ","End":"11:20.440","Text":"multiplied by the magnitude of our k vector."},{"Start":"11:22.010 ","End":"11:25.700","Text":"Now, we\u0027ve reached this,"},{"Start":"11:25.700 ","End":"11:29.270","Text":"which is exactly the dispersion relationship,"},{"Start":"11:29.270 ","End":"11:30.990","Text":"which we wanted to show."},{"Start":"11:30.990 ","End":"11:33.635","Text":"This is how we get from"},{"Start":"11:33.635 ","End":"11:40.130","Text":"Maxwell\u0027s equations to the wave equation and from the wave equations,"},{"Start":"11:40.130 ","End":"11:44.020","Text":"we get to the dispersion relationship."},{"Start":"11:44.020 ","End":"11:47.830","Text":"That is the end of this lesson."}],"ID":21442},{"Watched":false,"Name":"Exercise","Duration":"17m 43s","ChapterTopicVideoID":21560,"CourseChapterTopicPlaylistID":99477,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.845","Text":"Hello. In this lesson,"},{"Start":"00:01.845 ","End":"00:04.590","Text":"we\u0027re going to be answering the following questions."},{"Start":"00:04.590 ","End":"00:09.936","Text":"We\u0027re given a magnetic field that is equal to B naught multiplied by cosine of"},{"Start":"00:09.936 ","End":"00:16.575","Text":"Ax minus 2Ay minus Omega t in the z direction."},{"Start":"00:16.575 ","End":"00:21.480","Text":"The first thing that we\u0027re being asked to do is to find this field\u0027s wave vector."},{"Start":"00:21.480 ","End":"00:22.935","Text":"What is wave vector?"},{"Start":"00:22.935 ","End":"00:26.500","Text":"The wave vector is just our k vector."},{"Start":"00:26.630 ","End":"00:31.604","Text":"Our k vector is any coefficient of x,"},{"Start":"00:31.604 ","End":"00:33.600","Text":"which over here is A,"},{"Start":"00:33.600 ","End":"00:35.700","Text":"any coefficient of y,"},{"Start":"00:35.700 ","End":"00:38.805","Text":"which over here is negative 2A,"},{"Start":"00:38.805 ","End":"00:46.050","Text":"and any coefficient of z which we can see we have no variable here of z, so it\u0027s 0."},{"Start":"00:46.490 ","End":"00:51.795","Text":"Super easy, this is the answer to question number 1."},{"Start":"00:51.795 ","End":"00:56.580","Text":"Question number 2 is, what is this wave\u0027s frequency?"},{"Start":"00:56.580 ","End":"00:59.930","Text":"When we\u0027re asking in this type of question about frequency,"},{"Start":"00:59.930 ","End":"01:02.075","Text":"we\u0027re looking for Omega."},{"Start":"01:02.075 ","End":"01:09.995","Text":"We\u0027re going to use the equation linking Omega to our k vector or to our wave vector,"},{"Start":"01:09.995 ","End":"01:11.660","Text":"which we\u0027ve already found."},{"Start":"01:11.660 ","End":"01:15.835","Text":"We know that this is the dispersion relationship."},{"Start":"01:15.835 ","End":"01:20.660","Text":"We know that this is the equation Omega is equal to the speed of"},{"Start":"01:20.660 ","End":"01:25.345","Text":"light multiplied by the magnitude of our k vector."},{"Start":"01:25.345 ","End":"01:27.780","Text":"We have C multiplied by,"},{"Start":"01:27.780 ","End":"01:31.245","Text":"the magnitude of our k vector is just Pythagoras,"},{"Start":"01:31.245 ","End":"01:37.530","Text":"we have our x component squared plus our y component squared,"},{"Start":"01:37.530 ","End":"01:40.480","Text":"that\u0027s plus negative 2A squared,"},{"Start":"01:40.480 ","End":"01:46.630","Text":"which is just the same as 2A squared and the square root of that,"},{"Start":"01:46.630 ","End":"01:50.935","Text":"which is just equal to C multiplied by,"},{"Start":"01:50.935 ","End":"01:55.555","Text":"here we have just 5A squared."},{"Start":"01:55.555 ","End":"02:04.130","Text":"C multiplied by A multiplied by the square root of 5."},{"Start":"02:05.520 ","End":"02:08.005","Text":"That\u0027s the answer to Question 2."},{"Start":"02:08.005 ","End":"02:09.940","Text":"Again very, very easy."},{"Start":"02:09.940 ","End":"02:11.750","Text":"This is 5."},{"Start":"02:11.750 ","End":"02:14.645","Text":"Now let\u0027s take a look at Question 3,"},{"Start":"02:14.645 ","End":"02:17.620","Text":"calculate the corresponding electric field."},{"Start":"02:17.620 ","End":"02:24.715","Text":"We remember that if we have the magnetic field and we want to find the electric field,"},{"Start":"02:24.715 ","End":"02:29.730","Text":"we use Maxwell\u0027s 4th equation."},{"Start":"02:29.730 ","End":"02:34.050","Text":"I\u0027ll just write Maxwell\u0027s 4th equation."},{"Start":"02:34.050 ","End":"02:37.070","Text":"The equation is like so,"},{"Start":"02:37.070 ","End":"02:42.580","Text":"it says that the rotor of our magnetic field is equal"},{"Start":"02:42.580 ","End":"02:50.080","Text":"to Mu naught Epsilon naught multiplied by dE by dt,"},{"Start":"02:50.080 ","End":"02:53.720","Text":"the time derivative of our electric field."},{"Start":"02:53.720 ","End":"02:57.880","Text":"First of all, let\u0027s find the rotor of the magnetic field."},{"Start":"02:57.880 ","End":"03:02.300","Text":"The rotor, as we know is just partial derivative."},{"Start":"03:02.300 ","End":"03:04.010","Text":"So d by dx,"},{"Start":"03:04.010 ","End":"03:08.815","Text":"d by dy, d by dz."},{"Start":"03:08.815 ","End":"03:13.355","Text":"We know that we\u0027re going to get a vector back."},{"Start":"03:13.355 ","End":"03:19.220","Text":"Now our B field we can see is only in the z direction."},{"Start":"03:19.220 ","End":"03:21.625","Text":"It only has a z component,"},{"Start":"03:21.625 ","End":"03:24.780","Text":"so x and y components are 0."},{"Start":"03:24.780 ","End":"03:30.360","Text":"Then we just have B naught cosine of Ax minus"},{"Start":"03:30.360 ","End":"03:40.415","Text":"2Ay minus Omega t. Now we cross out the first,"},{"Start":"03:40.415 ","End":"03:43.930","Text":"and then we have d by dy of this."},{"Start":"03:43.930 ","End":"03:46.230","Text":"Again, we get a vector."},{"Start":"03:46.230 ","End":"03:55.190","Text":"First we have d by dy of B naught multiplied by cosine of Ax"},{"Start":"03:55.190 ","End":"04:04.395","Text":"minus 2Ay minus Omega t minus d by dz of 0,"},{"Start":"04:04.395 ","End":"04:09.045","Text":"so minus 0, or we can just leave it as this."},{"Start":"04:09.045 ","End":"04:12.185","Text":"Then we cross out the middle,"},{"Start":"04:12.185 ","End":"04:14.495","Text":"and then we flip it."},{"Start":"04:14.495 ","End":"04:21.510","Text":"We do d by dz of 0 is 0 minus d by dx."},{"Start":"04:21.510 ","End":"04:29.005","Text":"We have negative d by dx of B naught cosine of"},{"Start":"04:29.005 ","End":"04:36.755","Text":"Ax minus 2Ay minus Omega t. Then finally,"},{"Start":"04:36.755 ","End":"04:38.255","Text":"we cross out here."},{"Start":"04:38.255 ","End":"04:43.615","Text":"D by dx of 0 is 0 minus d by dy of 0 is 0."},{"Start":"04:43.615 ","End":"04:46.540","Text":"We\u0027re just left with this."},{"Start":"04:46.540 ","End":"04:50.905","Text":"Now let us do these partial derivatives."},{"Start":"04:50.905 ","End":"04:56.340","Text":"Here, in the x components,"},{"Start":"04:56.340 ","End":"05:05.235","Text":"we\u0027ll have negative B naught sine of Ax"},{"Start":"05:05.235 ","End":"05:15.215","Text":"minus 2Ay minus Omega t multiplied by the inner derivative with respect to y,"},{"Start":"05:15.215 ","End":"05:17.570","Text":"which is just negative 2A."},{"Start":"05:17.570 ","End":"05:19.745","Text":"This becomes positive,"},{"Start":"05:19.745 ","End":"05:25.140","Text":"so we can just delete it, 2A."},{"Start":"05:25.600 ","End":"05:30.660","Text":"Then here we have the partial derivative with respect to x."},{"Start":"05:30.660 ","End":"05:33.570","Text":"We have negative already over here."},{"Start":"05:33.570 ","End":"05:37.695","Text":"Then we have a negative from when cosine becomes sine."},{"Start":"05:37.695 ","End":"05:39.060","Text":"This just becomes a positive,"},{"Start":"05:39.060 ","End":"05:41.160","Text":"a negative and negative."},{"Start":"05:41.160 ","End":"05:44.040","Text":"We\u0027ll just erase it."},{"Start":"05:44.040 ","End":"05:51.285","Text":"We have again B naught multiplied by sine of"},{"Start":"05:51.285 ","End":"05:58.520","Text":"Ax minus 2Ay minus Omega t multiplied by the inner derivative,"},{"Start":"05:58.520 ","End":"06:01.145","Text":"but again with respect to x over here."},{"Start":"06:01.145 ","End":"06:03.470","Text":"We\u0027re taking the derivative with respect to x."},{"Start":"06:03.470 ","End":"06:07.425","Text":"The coefficient of x over here is just A."},{"Start":"06:07.425 ","End":"06:10.455","Text":"Here of course we have a 0."},{"Start":"06:10.455 ","End":"06:15.470","Text":"What we can do is we can just rewrite this as"},{"Start":"06:15.470 ","End":"06:24.960","Text":"being equal to 2AB naught."},{"Start":"06:24.960 ","End":"06:27.765","Text":"Actually we will write it with like terms."},{"Start":"06:27.765 ","End":"06:34.860","Text":"Everything has AB naught sine of Ax minus"},{"Start":"06:34.860 ","End":"06:40.200","Text":"2Ay minus Omega t. Then this is"},{"Start":"06:40.200 ","End":"06:46.523","Text":"multiplied by 2 in the x direction,"},{"Start":"06:46.523 ","End":"06:51.460","Text":"and 1 in the y direction."},{"Start":"06:56.120 ","End":"06:59.130","Text":"Now what we have to do,"},{"Start":"06:59.130 ","End":"07:00.920","Text":"this is our rotor,"},{"Start":"07:00.920 ","End":"07:08.220","Text":"and now we have to say that this is equal to Mu naught Epsilon dE by dt."},{"Start":"07:09.530 ","End":"07:17.205","Text":"All of this is equal to Mu naught Epsilon naught,"},{"Start":"07:17.205 ","End":"07:27.010","Text":"which is also the same as 1 divided by C squared of dE by dt."},{"Start":"07:27.140 ","End":"07:31.395","Text":"Let\u0027s multiply both sides by C squared."},{"Start":"07:31.395 ","End":"07:41.445","Text":"What we have over here is C squared AB naught sine of"},{"Start":"07:41.445 ","End":"07:47.225","Text":"Ax minus 2Ay minus Omega t"},{"Start":"07:47.225 ","End":"07:53.795","Text":"multiplied by 2 in the x direction plus 1 in the y direction,"},{"Start":"07:53.795 ","End":"07:59.075","Text":"which is equal to dE by dt."},{"Start":"07:59.075 ","End":"08:02.640","Text":"Now we\u0027re left with this differential equation."},{"Start":"08:02.640 ","End":"08:07.625","Text":"What we have to do is we have to integrate both sides with respect to"},{"Start":"08:07.625 ","End":"08:14.490","Text":"t. Then what will be left with is the E field over here on this side."},{"Start":"08:14.490 ","End":"08:18.045","Text":"Now we just have to integrate here."},{"Start":"08:18.045 ","End":"08:23.615","Text":"What we\u0027ll have is the integral of sine is negative cosine."},{"Start":"08:23.615 ","End":"08:25.085","Text":"We have negative,"},{"Start":"08:25.085 ","End":"08:30.830","Text":"and then we have all of our C squared AB naught are constants."},{"Start":"08:30.830 ","End":"08:37.835","Text":"Then we have cosine of Ax minus 2Ay minus"},{"Start":"08:37.835 ","End":"08:46.115","Text":"Omega t. Then these are just the directions,"},{"Start":"08:46.115 ","End":"08:49.385","Text":"and the axes that we\u0027re going along,"},{"Start":"08:49.385 ","End":"08:51.440","Text":"and they\u0027re constants, 2 and 1,"},{"Start":"08:51.440 ","End":"08:58.260","Text":"so this is still 2 in the x plus 1 in the y direction."},{"Start":"08:58.260 ","End":"09:03.140","Text":"Then of course, we have to divide by the inner derivative"},{"Start":"09:03.140 ","End":"09:07.995","Text":"with respect to our variable which is t. The inner derivative,"},{"Start":"09:07.995 ","End":"09:10.505","Text":"we only have the variable t over here."},{"Start":"09:10.505 ","End":"09:14.613","Text":"The derivative of this is negative Omega;"},{"Start":"09:14.613 ","End":"09:18.710","Text":"the derivative of negative Omega t is negative Omega."},{"Start":"09:18.710 ","End":"09:24.050","Text":"All of this is divided by negative Omega,"},{"Start":"09:24.050 ","End":"09:27.770","Text":"so the minuses cancel out to make positives."},{"Start":"09:27.770 ","End":"09:31.288","Text":"What we\u0027re left with is an E field which is equal to C"},{"Start":"09:31.288 ","End":"09:37.260","Text":"squared AB naught divided by Omega multiplied by"},{"Start":"09:37.260 ","End":"09:43.050","Text":"cosine of Ax minus 2Ay minus"},{"Start":"09:43.050 ","End":"09:49.730","Text":"Omega t. Then we have 2 in the x direction,"},{"Start":"09:49.730 ","End":"09:54.690","Text":"and 1 in the y direction."},{"Start":"09:55.410 ","End":"09:59.605","Text":"This is our electric field,"},{"Start":"09:59.605 ","End":"10:03.910","Text":"and I\u0027m just going to copy it and put it next to our magnetic fields so that we can"},{"Start":"10:03.910 ","End":"10:09.200","Text":"just go over some things that are similar."},{"Start":"10:10.200 ","End":"10:18.235","Text":"What we can see is that our electric field has the same argument."},{"Start":"10:18.235 ","End":"10:21.400","Text":"Also, just like the magnetic field,"},{"Start":"10:21.400 ","End":"10:24.250","Text":"Cosine Ax minus 2A minus Omega t,"},{"Start":"10:24.250 ","End":"10:27.220","Text":"the exact same thing over here,"},{"Start":"10:27.220 ","End":"10:30.520","Text":"and the major difference is just the amplitude,"},{"Start":"10:30.520 ","End":"10:35.455","Text":"so here the amplitude of the magnetic field wave is B_0,"},{"Start":"10:35.455 ","End":"10:40.480","Text":"and here it\u0027s C^2 AB_0 divided by Omega,"},{"Start":"10:40.480 ","End":"10:42.955","Text":"and of course the direction."},{"Start":"10:42.955 ","End":"10:45.535","Text":"Here we can see that the magnetic field,"},{"Start":"10:45.535 ","End":"10:48.640","Text":"the amplitude is only in the z direction,"},{"Start":"10:48.640 ","End":"10:53.935","Text":"and here we can see that it\u0027s in the x and the y-direction."},{"Start":"10:53.935 ","End":"10:58.330","Text":"We can see that both waves are propagating in the same direction according to"},{"Start":"10:58.330 ","End":"11:04.570","Text":"the k vector A and negative 2A."},{"Start":"11:04.570 ","End":"11:08.140","Text":"We can also see that the 2 fields are perpendicular."},{"Start":"11:08.140 ","End":"11:13.000","Text":"The magnetic field only has a component of amplitude,"},{"Start":"11:13.000 ","End":"11:14.545","Text":"in the z direction,"},{"Start":"11:14.545 ","End":"11:21.865","Text":"whereas the electric field has amplitude either in the x or the y or in both directions."},{"Start":"11:21.865 ","End":"11:25.240","Text":"Both x and y are perpendicular to z."},{"Start":"11:25.240 ","End":"11:32.425","Text":"We can also check that this is correct by saying that we have no z component over here."},{"Start":"11:32.425 ","End":"11:36.160","Text":"This was our answer to question number 3."},{"Start":"11:36.160 ","End":"11:38.470","Text":"Now let\u0027s answer question number 4."},{"Start":"11:38.470 ","End":"11:41.245","Text":"Calculate the force acting on a charge Q,"},{"Start":"11:41.245 ","End":"11:46.390","Text":"that is located at time t= 0 at the origin,"},{"Start":"11:46.390 ","End":"11:52.285","Text":"and the charge has a velocity of v_0 in the x-direction."},{"Start":"11:52.285 ","End":"11:57.280","Text":"In order to calculate the force acting on this particle or on this charge,"},{"Start":"11:57.280 ","End":"11:59.680","Text":"we\u0027re using Lorentz law,"},{"Start":"11:59.680 ","End":"12:01.675","Text":"which is equal to q,"},{"Start":"12:01.675 ","End":"12:04.990","Text":"the charge multiplied by the electric field,"},{"Start":"12:04.990 ","End":"12:14.240","Text":"plus q multiplied by the velocity cross multiplied with the magnetic field."},{"Start":"12:14.510 ","End":"12:19.095","Text":"We know our charge, we\u0027re given the charge,"},{"Start":"12:19.095 ","End":"12:21.465","Text":"the velocity and t=0,"},{"Start":"12:21.465 ","End":"12:24.835","Text":"and we\u0027re told that it is at the origin."},{"Start":"12:24.835 ","End":"12:30.010","Text":"First of all, we want to find out what the electric field is at that point."},{"Start":"12:30.010 ","End":"12:32.080","Text":"Here we have the electric field."},{"Start":"12:32.080 ","End":"12:33.640","Text":"Now if we\u0027re at the origin,"},{"Start":"12:33.640 ","End":"12:37.445","Text":"that means x, y and z = 0."},{"Start":"12:37.445 ","End":"12:40.365","Text":"Here we can see we have x and y,"},{"Start":"12:40.365 ","End":"12:45.780","Text":"so if they are = 0 and of course our t = 0."},{"Start":"12:45.780 ","End":"12:48.930","Text":"That means that we have Cosine(0),"},{"Start":"12:48.930 ","End":"12:54.480","Text":"which is just 1, which means that the E field is just the amplitude."},{"Start":"12:54.480 ","End":"12:59.445","Text":"The E field is going to be ="},{"Start":"12:59.445 ","End":"13:07.390","Text":"2C^2AB_0 divided by Omega in the x direction,"},{"Start":"13:07.390 ","End":"13:08.950","Text":"plus 1 times this,"},{"Start":"13:08.950 ","End":"13:17.095","Text":"so just C^2 AB_0 divided by Omega in the y direction."},{"Start":"13:17.095 ","End":"13:22.645","Text":"Then we know that our B field,"},{"Start":"13:22.645 ","End":"13:24.715","Text":"so our magnetic field,"},{"Start":"13:24.715 ","End":"13:28.030","Text":"we are again using the equation where x, y,"},{"Start":"13:28.030 ","End":"13:32.755","Text":"and z is =0 and where t=0."},{"Start":"13:32.755 ","End":"13:35.080","Text":"Our B field is this over here so."},{"Start":"13:35.080 ","End":"13:38.050","Text":"Again, all of what\u0027s in the brackets here is equal to 0,"},{"Start":"13:38.050 ","End":"13:39.685","Text":"so we have Cosine(0),"},{"Start":"13:39.685 ","End":"13:43.675","Text":"which is 1 multiplied by B_0."},{"Start":"13:43.675 ","End":"13:49.060","Text":"We\u0027re just left with B_0 in the z direction."},{"Start":"13:49.060 ","End":"13:58.855","Text":"Now what we want to do is we want to calculate what V cross B is equal to."},{"Start":"13:58.855 ","End":"14:01.150","Text":"V is only in the x direction,"},{"Start":"14:01.150 ","End":"14:04.180","Text":"so we have V_0 0,"},{"Start":"14:04.180 ","End":"14:08.320","Text":"0 cross multiplied with our B field, which is 0,"},{"Start":"14:08.320 ","End":"14:10.660","Text":"0, B_0,"},{"Start":"14:10.660 ","End":"14:14.350","Text":"because it\u0027s only in the z direction."},{"Start":"14:14.350 ","End":"14:19.705","Text":"This we can already see is just going to be equal to"},{"Start":"14:19.705 ","End":"14:28.820","Text":"V_0 B_0 in the negative y direction."},{"Start":"14:30.030 ","End":"14:34.660","Text":"Now all we have to do is to just plug this into the equation."},{"Start":"14:34.660 ","End":"14:41.335","Text":"We have that F is equal to the charge Q multiplied by the E field,"},{"Start":"14:41.335 ","End":"14:51.280","Text":"so that is just going to be equal to 2C^2AB_0 divided by"},{"Start":"14:51.280 ","End":"14:54.760","Text":"Omega in the x direction"},{"Start":"14:54.760 ","End":"15:02.280","Text":"plus C^2AB_0 divided by Omega in the y direction."},{"Start":"15:02.280 ","End":"15:06.410","Text":"This is plus our charge,"},{"Start":"15:06.410 ","End":"15:09.850","Text":"so a charge Q multiplied by V cross B,"},{"Start":"15:09.850 ","End":"15:12.370","Text":"which is just V_0,"},{"Start":"15:12.370 ","End":"15:15.700","Text":"B_0 in the negative y- direction,"},{"Start":"15:15.700 ","End":"15:22.640","Text":"so we can just have negative over here and y hat."},{"Start":"15:23.970 ","End":"15:27.895","Text":"Now that we\u0027ve seen the answer to question 4,"},{"Start":"15:27.895 ","End":"15:29.980","Text":"let\u0027s do question 5."},{"Start":"15:29.980 ","End":"15:32.590","Text":"Calculate the Poynting vector,"},{"Start":"15:32.590 ","End":"15:37.960","Text":"so the Poynting vector S is equal to 1 divided by"},{"Start":"15:37.960 ","End":"15:46.280","Text":"Mu_0 of the electric field cross multiplied with the magnetic field."},{"Start":"15:46.320 ","End":"15:54.160","Text":"This is just going to be equal to 1 divided by Mu_0 of the electric field,"},{"Start":"15:54.160 ","End":"15:57.175","Text":"which we saw over here."},{"Start":"15:57.175 ","End":"16:04.375","Text":"We can take out all the constants C^2AB_0 divided by Omega."},{"Start":"16:04.375 ","End":"16:12.295","Text":"Then we have Cosine(Ax"},{"Start":"16:12.295 ","End":"16:17.530","Text":"minus 2Ay minus Omega t),"},{"Start":"16:17.530 ","End":"16:21.580","Text":"and this we have in the x direction,"},{"Start":"16:21.580 ","End":"16:25.190","Text":"and in the y direction."},{"Start":"16:25.560 ","End":"16:30.519","Text":"This is cross multiplied with our B field,"},{"Start":"16:30.519 ","End":"16:36.355","Text":"which is equal to B_0."},{"Start":"16:36.355 ","End":"16:42.250","Text":"Let\u0027s just put these together so we have B_0 over here, and again,"},{"Start":"16:42.250 ","End":"16:49.525","Text":"Cosine of the exact same thing so we can have Cosine^2,"},{"Start":"16:49.525 ","End":"16:56.050","Text":"however, this time we\u0027re cross-multiplying this in the z direction."},{"Start":"16:56.050 ","End":"17:04.630","Text":"What we\u0027re going to get is C^2AB_0^2,"},{"Start":"17:04.630 ","End":"17:09.595","Text":"divided by Mu_0 Omega multiplied by"},{"Start":"17:09.595 ","End":"17:17.710","Text":"Cosine^2 (Ax minus 2Ay minus Omega t)."},{"Start":"17:17.710 ","End":"17:24.055","Text":"Then here when we cross multiply 2x hat with z hat,"},{"Start":"17:24.055 ","End":"17:29.590","Text":"we\u0027re going to have negative 2y hat and when we cross"},{"Start":"17:29.590 ","End":"17:37.255","Text":"multiply y with z we\u0027ll have plus x hat."},{"Start":"17:37.255 ","End":"17:41.140","Text":"This is the answer to question number 5,"},{"Start":"17:41.140 ","End":"17:44.240","Text":"and that is the end of this lesson."}],"ID":22347}],"Thumbnail":null,"ID":99477}]

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[22346,22345,22348,22349,21442,22347];

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