[MUSIC] Now, let's go back and finish our calculation of the magnetic field. So we have B sub x and B sub y coming from delta cross a. And if you are now familiar with these two, and add them up, you will be able to understand, this can be simplified into this form. 1 over 4 pi epsilon 0 c squared times p dot + r over c p two dot at the retarded time t- r over c x r over r cube. And let's take a look at the formula. First of all, if we go very far out in r, only the p two dot term counts. The direction of B is given by p cross r, which is at right angles to the radius r and also at right angles to the acceleration. So you see if p two dot terms only dominates and this will be the acceleration. So any vector that is a cross b, is always perpendicular to both of them. So that's why the direction of B, which is given by this one, is at right angles to both the radius vector, r, and the acceleration. So you can see the picture here, where you have acceleration of a dipole along z, and you have resulting B field and E field, which have perpendicular or orthogonal relationship to each other, okay? So then we are now getting closer to what we just discussed before where for the wave or light, where you have wave propagating for a long distance, the electric field is equal to q over 4 pi epsilon 0 c squared r prime, times acceleration of the charge at t- r prime over c, projected at right angles to r prime. So now we are getting closer and closer. Now, let's take a look at what we're not used to at what happens closer in, not far out. According to the law of Blot and Savart, for the magnetic field of an element of current, we found that a current element jdV contributes to magnetic field B the amount dB which is 1 over 4 pi epsilon 0 c squared, times j x r over r cubed dV. So Melody, do you remember this slide? >> Yeah, you teach it out a few lectures ago. >> Exactly, so we covered this, where The Biot-Savart law states the integral, this one, gives B directly in terms of no occurrence. You see we have 1 over 4 pi epsilon 0 c square, integral j of 2 cross e sub 12 over r12 square dV2. And here we didn't mention about any time delay, it was all instantaneous. And as you can see because this is unit vector, if you replace it with r vector, the denominator will be r cubed. So this is equivalent to what we just solved in this lecture. So if we remember that p dot is the current, the law of Biot and Savart looks very much like. And you can see this is the equation that we tackle, and this the law of Biot and Savart. However, there is one difference. For upper equation, the current is to be evaluated at the time (t- r over c), which doesn't appear in the lower equation. So now we have to think about what is happening here, right? So the dB = 1 over 4 pi epsilon 0 c squared j x r over r cubed dV is still very good for small r, so now our approximation kicks in. Because the second term of B, this one, tends to cancel out the effect of the retardation in the first term. This is similar to what we have explored for other equation as well, right? So this will be time delay, and this will be first order approximation to the Taylor series. So altogether, this will make this an instantaneous current. The two terms together in the first equation give a result very near to the second equation when r is small, as we mentioned. And when r is small, (t- r over c) is not very different from t, so we can expand the bracket in a Taylor series. And when we take the sum, the two terms in p two dot cancel, and we are left with the unretarded current p dot and the terms of order (r over c) squared or higher order terms, which will be very small, so we can neglect them. So you can see the first equation here, the first term expanded in the Taylor series, second term, expanded in Taylor series. If you add them up, what is left is really high terms, high order terms, so you can ignore them. And what we are left with is the instantaneous current vector. But for something far out, then 1 over r term dominates. So it's the new kind of effects that we want to mention, right? Then we can have a broader case, and a general law that we can use, right? So we have discussed mostly about the magnetic field of an oscillating dipole, and trying to see if we can get to the point we just started with. Where electric field is -q over 4 pi epsilon 0 c square r prime times the acceleration of the charge at (t- r prime over c) projected at right angles to r prime. So in order to get there, we need to now get the electric field of an oscillating dipole. So for large distances, we can see the answer is going to come out all right. We know that far from the sources, where we have a propagating wave, electric field is perpendicular to magnetic field, and also to r as shown in the figure below here. And the relationship between magnetic field and electric field will be the speed of light. So E is proportional to the acceleration of p two dot, as expected, from the equation above, okay? So we are very near, right, very close. To get the electric field completely for all distances, we need to solve for the electrostatic potential. When we computed the current integral to get A(1,t), this one, we made an approximation by disregarding the slight variation of r in the delay terms. You remember that, right? We replace r12 by r, right? So this is what we did, right? We just ignore it. However, this regarding the slight variation r in the delay terms will not work for the elecrostatic potential. You know why? >> Because then we would get the 1 over r time integral of the charge density, which is constant. >> Very good, very good. So it is written here that if we do this approximation, it is a too much of a approximation. As you can see, phi(1,t) = rho 2,t- r12 over c over 4 phi epsilon 0 r12 dV2. And if I replace those two by r, r comes out of this, and this one becomes just q over 4 phi epsilon 0 r, which is not different from static case. So nothing special, nothing new will come out of our approximation. So that's why in this case, we need to think a little bit more, okay? So disregarding the slight variation of r in the delay terms is too rough, as we just discussed. So we need to go to one higher order. So instead of getting involved in that higher-order computation directly, we can do something else. We can determine the scalar of potential from the Lorentz case, right? Delta A is equal to minus 1 over c squared times round feel around T using the vector potential we already found okay, which is good for everything. So Divergence of A is just in our case. It's just around Az over around z, because we only have z component for a and if I put that in here. Right, if I do round or round z, it will be similar to round or round y, what we did. We just replacing y by z, and if I do that, then using the chain rules and everything will get delta A is = to- 1 over 4 pi- c squared. P dot plus r over c p2 dot as t minus r over c dot r over r q. So that's what we will get. >> Looks familiar, right? >> Looks familiar, yes, now, using the Lorentz gauge here and putting these two equations equal will make us round phi over round t is equal to the one that we just solved. And if I integrate them, I will get this one plus a constant. But the constant of integration is static field and for oscillating dipole there's no static field, so we can put that constant to be equal to zero, okay? So now, we get something that is similar in the form with the vector potential, but has different vector operation, okay? So now, we are now able to find the electric field E from E is equal to minus delta V minus dA over dt, right? And if we do that, because we know fee, we know A, we will get this one. Very complicated E of R comma t is equal to minus 1 over 4 pi epsilon. Not r Cube bracket minus 3 times p star dot r or r square minus p star plus 1 over c square bracket p2 dot of t minus r over c cross r bracket cross r, big bracket where p star is p of t minus over c. Plus r of c p dot t minus over c, right? It is very complicated and it takes a lot of papers and time to derive that, but the result can be easily interpreted. The vector p star is the dipole moment retarded and then corrected for the retardation, as you can see here. You see this many times, right? The two terms with p star give just the static dipole field when r is small, as was discussed many times. When r is large determine, p2 dot dominates, and the E field is proportional to the acceleration of the charge, which is this p2 dot. At right angles to r, because this cross r and directed along the projection of p2 dot in a plane perpendicular to r. That is because we have two cross product there, okay? So now, we came up to this point and now, we somehow understand this is related to what we just explained before for general case. So our result above agrees with what we have gotten with the equation below, with some limitations. The equation above is valid only for small motions for which we can take the retardation r/c as constant over source, whereas the equation below works with any motion. So this is more general than this one, but still is very close. And then in order to make it even closer, we will discuss next, how the fields can be obtained for more rapidly moving charge. So you can see even for slowly moving charge it is so difficult and hard to follow. And you can imagine, if we increase the level of complexity, it will take a little more time, right? >> Right. >> So what do you think, Melody? >> I think we should do it. >> Okay, let's go, so we will now take up a calculation of the potentials for a point charge moving in any way with a relativistic theory. Once we have this result, we will have the complete electromagnetism of electric charges. So let's calculate the scalar potential phi of 1 at the point x1, y1, z1, produced by a point charge moving in any manner. So now, we will revisit this equation, phi of (1, t) is equal to 1 over 4 pi epsilon naught integral rho of 2 comma t minus r 12 over c over r 12 dV2, okay? And if I do this for a point charge, then you make thing, hm, if it is a point charge, all 1 and 2 will be the same for every point. So this will be just r12 prime and because r2 prime is a constant, it can come out. If this come out, then it's just integration of the charge density over the source volume. So [SOUND] then it's just the charge quantity, right? So you may think this is correct, where r12 prime is the radius vector from the charge at point 2 to point 1 at the retarded time t minus r12 / c. >> I think that seems a little bit too easy, right? >> It's too easy, what do you think? >> Yeah. >> So do you think this is the correct answer? >> No. >> No, exactly, it's wrong, so what is wrong with this? So let's take a look. So instead of having this simple form, we need to have a correction factor, 1 over 1 minus vr prime over c, where vr prime is the component of the velocity of the charge parallel to r12 prime, to our point 1. So now, we are going to discuss why this is the case with a simple example of a cube chart. To make the argument easier to understand, we will make the calculation first for a point charge that is in the form of a little cube of charge as indicated here. Moving toward the point 1 with the speed of v. Let the length of a side of the cube be a, as you can see here, which is much less than r12, the distance from the center of the charge to the point 1, okay? So with these assumptions, we're going to explain how we end up with this correction term in our equation. So in order to evaluate phi (1, t) = 1 over 4 pi epsilon naught times integration of rho (2, t- r 12 over c) over r12 times dV2, we will return to basic principles. We will write the sum, that is here, sigma i is 1 to capital N, where capital N is number of pieces In this cube, Times rho sub i times delta V sub i over ri, where r sub i is the distance from point 1 to the ith volume element delta V sub i and rho i is the charge density at delta V sub I at the time, t sub i which is = t- ri over c. So let me ask Melody, so you can see from this formula, each piece has it's own time. So I'm asking, will then all the pieces have the same time? >> No, I don't think so, because you can see in this formula, it's ri divided by c, and actually if you look at another slab, maybe this one or this one in the front, the radius will be slightly different. And so, you're actually going to have Different times for every different r that you use in every different slot that you sum up. >> Exactly, so they have their own clocks that takes in the different way. So you will have to think that even though this is one simple cube in space but they have different times, so that may be strange to you. But that's what this involved, that is involving the special relativity. Now, since r sub i is much, much larger than the side length of the cube, which is a, it will be convenient to take our delta V sub i, which is this piece, in the form of thin rectangular slices, perpendicular to r12. Suppose we start by taking the volume elements delta V sub i with some thickness W, which is much less than A, so you can see this piece, you can see it's W. And the individual elements will appear as shown above where we put in more than enough to cover the charts. So this one has bigger space than the size of our cube and you may wonder why, and we will explain that shortly. Now where should we draw the charge? This is the question. For each volume element delta V sub i, we should take rhoi at the time ti which is equal to t- R sub i over c, which Melody explained to us will be different, depending on which piece you were looking at. So if the charge is also moving, it is in a different place for each volume element, delta V sub i. So in other words, we should evaluate rho sub i times delta V sub i over r sub i. At t sub i, which is = t- r sub I/c which is different for different volume element delta V sub i. So it's very complicated, right? So we have to think of different times, also different place because it's moving, okay? Let's say that we begin with the volume element labeled number 1. You can see, this is number 1. In the figure above, chosen so that at the time t1 = t- r 1/c. The back edge of the charge occupies delta v sub 1. Now Melody, among the pieces, right, this one will have the longest length, or distance. >> Right. >> From itself, to point one. Then because it has the longest distance from the perspective of time. Will it have the biggest time or smallest time? >> I think it would have smaller time, because, one divided by c would be larger than perhaps this radius over here. >> Exactly. So, you will see t1, will be the smallest time. Meaning, this will be the earliest time in all of these pieces, okay? So you saw the cube just moved to the right a little bit, while we were taking the second slide, right? So now we want to evaluate rho sub 2 times delta V sub 2, we should use the position of the charge at a slightly later time. As Melody mentioned, t sub one will be the smallest time. Now, it is increasing a bit, because r sub two is smaller than r sub one, when the charge is in the position above. And we will keep on doing this, and so on for delta V3. You see it move to the right, and you have the third slice. And also for delta W sub 4. So if we do that, step by step, we can evaluate the sum, which is rho sub i delta V sub i over r sub i from i=1 to capital N. So you saw that you have covered all the pieces that is from one to capital N. And we test larger lengths than the side lengths of a cube, okay? So, you can now understand why we had a larger space reserved for our charges. Now since the thickness of each delta V sub i is w, its volume is w a squared, you can see in this three dimensional perspective, right, w is the width of this, aa, so it's w a squared. Each delta V sub i that overlaps the charge distribution contains the amount of charge, which is volume times the density. So that's w a squared times rho. When the distance from the charge to point 1 is large, we'll make a negligible error by setting all the r sub i's in the denominators, equal to some average value, say the retarded position r prime on the center of the chart. If this is so small in comparison to this distance, then we can make that kind of approximation. And if we do that, we can change this into sigma i is 1 to Capital N, rho w a squared over r prime and you know that this rho wa squared sigma N is simply capital N times rho w a squared over r prime. And if we rearrange it a little bit, we make a mathematical trick. We multiply a and divided by a then, you can see the n capital N times W is b. You have n pieces here, each quiz this W. So if you multiply that, that will be length speed, so if you do that, it will be equal to q over r prime because, row eight cube is cube, the charge in this cube, times b over a, is that clear? >> Yes. >> Okay. So now, you will see that we have solved this equation. Phi of 1,t is equal to 1 over 4 pi epsilon knot, times integral rho of 2,t- r12 over c over r12 times dV2. Or this one was replaced by out summation, and with that we know this q over r prime times b over a will be the answer. So if we add that, it will be q over 4 pi epsilon our prime, times b over a. So what is b, Melody. >> As you can see right here, b is the length of the cube of the charge increased by the distance moved by the charge between t1 and tn. >> Exactly. >> So, the diagram is right here. >> Yeah, so, the time for the first piece and the last piece, the difference between that. And this is the distance the charge moves in the time, where's delta t is define by b or c. Meaning, the time a light can travel from the first piece to the last piece, while it is moving at the speed of V. So that's the theoretical meaning of B. And that's the correction factor to what we just intuitively guessed in the first place. So, again delta t will be b over c. B is the length of this one, c is the speed of light. Since the speed of charge is v, the distance moved is v x delta t, which will be b over c. If you do simple arithmetic, b =a + b delta t, and this is equal to a to vb/c. So b = a over 1- v over c. This factor we have already discussed at the first place. So v means the velocity at the retarded time t prime, which is equal to t- r prime over c. And we can indicate it by writing this formula in the form which is written here, q over 4 pi epsilon naught r prime times 1 over bracket 1- v over c bracket at retarded time. Which means t prime, okay? So now you can see the potential, phi, from the static point of view, which is q over 4 pi epsilon naught r will be increased by amount 1 over 1- v over c at the retarded time. But if v is small or even zero, then you will see it will be exactly the same as the static equation. Now, at this point, let me ask Melody, what happens if v becomes very fast? Let's say v approaches c. >> Well, just mathematically speaking, this would go to 0, and then the potential would kind of be infinite, or something. >> Yes, it will diverge to infinity. So you can see that the elective potential will soar as we increase the speed of charge. So you can see now, it's kind of a total recall, that we mentioned that we need this correction term 1 over 1- v sub r prime over c to evaluate the pi of 1, t. And if you compare these two, they're exactly the same. Where r1 2 prime is the radius vector from the charge at point 2 to point 1 at the retarded time t- r1 over c. And v sub r prime is the component of the velocity of the charge parallel to r prime sub 1,2 toward 1. So this is how we get to this correct answer through a simple illustration of a cube charge with uniform density. So there is a correction term, as we just mentioned, which comes about because the charge is moving as our integral sweeps over the charge. We just saw the animation. As we evaluate each piece the cube was moving to the right. When the charge is moving toward the point 1, its contribution to the integral is increased by a ratio of b over a, which we just discussed. For a static case it's almost one, for speed approaching the speed of light it is going to infinity. So you can have some feel for what this correction means for the electrostatic potential. Now, if the velocity of the charge is not directed towards the observation point 1, you can see that what matters is the component of its velocity toward point 1. And that was what we learned from special relativity, right? So that will be this one, v sub r, okay? So that's the component toward the 0.1. And we can also rewrite that using vector operation, the dot product. So we can change r prime into r- v.r over c and bracket all of them at the retarded time, where R is the vector from the charge to 0.1. Where phi being evaluated, and all the quantities in the bracket are to have their values at t prime, which is equal to t minus r prime or c. >> So does that mean if the velocity is heading towards 1, then the potential will be the greatest? >> Can you repeat the question please? >> So if the charge is heading directly, its velocity is only towards 0.1, then the potential will be the greatest or? >> Yes, yes, that is correct. So if the velocity is aligned with the position vector toward 0.1, you see this is maximized, meaning your denominator will be minimized, and that will increase your potential to the maximum. So you can imagine the minimum will be when the velocity is against this position vector that is going away. Then this will be negative, which will increase the denominator and therefore decrease the potential. >> Well, thank you. >> Okay, the analysis we have made goes exactly the same way for a charged distribution of any shape. It's not only for this cubic, because in our equation there is no information about the shape of our charge. Or also, since the size of the charge a doesn't enter into the final result, the same result holds when we shrink the charge to any size, even to a point. And that way, you can now use it for a point charge as well. So now we are approaching to Lienard-Wiechert potentials, which you can find on the right side of these two equations. The potentials for a point charge were first deduced in the forms above by Lienard and Wiechert, and are called Lienard-Wiechert potentials. And you can see, we can derive those two by using the simple cubic analysis. And once we derive these potentials based on the relationship we have explored, where electric field is minus del phi- dA over dt. And magnetic field is the curl of a, we can get electric field and magnetic field from those potentials. And this will much look like the one that we have just mentioned, and which were derived by Richard Feynman. So let's take a look at these Lienard-Wiechert potentials, that is potentials for a charge moving with constant v. This is a special case, and here we assume that the velocity of the charge is uniform. It's not changing with time and it is in a straight line. So there you can apply special relativity. We already know what the potentials are when we are standing in the rest frame of a charge, we have just discussed that. And when the charge is moving, we can figure everything out by a relativistic transformation from one system to the other. And this has been nicely presented by Albert Einstein in 1905, and you can find his paper using this source, and relativity had its origin in the theory of electricity and magnetism. So as you can see from the title, it is on the electrodynamics of moving bodies. And Albert Einstein was tackling then, dilemma of a moving body electrodynamics, okay? Suppose we have a charge moving along the x axis with speed v, and we want the potential at the point p, which is next to Melody. Now, I'm going to ask Melody then what is happening here in this frame? So Melody, please. >> Okay, so as you can see, we have a charge right here. When t was 0, we kind of put the origin right here. And then the charge travels out this way and it's travelling at some velocity. So we know the change in distance would be equal to that velocity times the time it took to get here. And then there is no movement in the y or z-axis when z = 0. >> Exactly. >> However, there's some sort of time delay, right? So even though the charge has gone all the way to this point here, it's still sitting from the perspective of point p at this retarded time which is equal to t minus r prime divided by c. >> Exactly, so if I add a little bit more details to what Mallory just explained, you see even though at time t, which is present time, the charge is in this position. When we measure anything that is coming from the charge, that is coming from a little bit earlier past which is t-prime, okay? So you need to always have in mind the current time t and retarded time t-prime and at time t, when I gather any information from charge, it is coming not from the present time of our charge but a little bit earlier, which is the past or our charge. So at the earlier time t-prime, the charge was at x equal v t-prime. So at this point, it was giving off the information, which reached p to let us know what is the potential. And r-prime therefore is square root of x- v t-prime squared, + y squared, + z squared. And to find our prime or t-prime we have to combine this equation with t-prime = t- r-prime over c. And if we do a simple mathematical manipulation, we can understand c squared times ( t- t-prime) squared = x- vt-prime square + y squared + z squared. And if I arrange that neatly, then it will have this form, and this is quadratic equation in terms of t-prime. You have the second order here, first order and zeroth order there. So Melody, do you remember if I have the second order type of equations, the formula to get the solution? >> I don't remember. >> Yes, among our students, there maybe some who remember that, but if you don't remember you can look up the solution and if you can't take this coefficients a and b and c, you will have the formula to get two solutions for the equations. >> Yeah. >> And solving for t-prime using those solutions, you will get this one t-prime = xv- c squared t +,- square root of xv- c squared t squared- v squared- c squared times (x squared + y squared + c squared- ct squared over v squared- c squared. Maybe this equation will make you fall asleep, but anyway, this is the equation you will have. Okay, so further solving for t-prime, as you can see, if you change the order of this denominator because obviously c is larger than v, so this is negative. And if you switch it like this then, you will have to switch the plus minus to minus plus and this order will be change, and you can rearrange that into this form, okay? So going further, then you can see that we have divided the numerator and denominator by c, in order to get this familiar form 1- v2 over c2, right? And we still have two solutions here. And one is small and one is large. And then we have to find phi of this, using what we have learned before, which is phi (x, y, z, t) = q over 4 pi epsilon naught, bracket r- v.r over c, at retarded time, right? And you can rewrite it using r prime here which is written here. And because this is v is equal to constant, that's why we could use this, okay? Then, what we have now is, This t prime is equal to t- r-prime over c, which we can rearrange to have this one, and v.r-prime, which is this one, if you look at this red triangle = vr-prime cosine theta, which is equal to v times x- vt-prime. So x- vt-prime is this length of this triangle. Get it, right? So phi (x, y, z, t) = q over 4 pi epsilon naught times 1 over c times t- t-prime- v over c times x minus vt-prime, okay? And then we further arrange it with c outside of this, right? As you can see in the denominator. And then, we are going to use t-prime that we have. And remember, we had minus and plus, right? However, from the form we had before, t-prime cannot be larger that t, therefore, we have to adapt minus. And if we do adapt minus and put it here, you can see this is annihilated. And if you put in there, you will have the simple form where phi is equal to q over 4 pie epsilon naught times 1 over square root x- vt square + 1- v-square over c square times y square + z square. So no we're really ready to find phi from the from the form we just derived where we can also plot 1 more factor which is 1 over square root 1 minus v square over c square, which happens to be a factor that you often find in special relativity. And you can now recognize this transformation which is exactly the same as Lorenz transformation. So finally we get this phi = q over 4 pi epsilon naught, times 1 over square root 1- v square over c square times 1 over square root x- vt over square root 1- v square over c square square + y square + z square. Now with this knowledge Melody, can we also get what it is for magnet vector potential A? >> Yeah, as you can see right here, you just need to add an additional v over C squared to get your A. >> Exactly, so you came along with us, a long way, to get the equation for phi, but for A, you can just add V over C squared. So now you have this complicated formula ready for magnetic vector potential as well, okay? So here is the beginning of the Lorentz transformation. If the charge were at the origin in its own rest frame, its potential would be phi ( x, y, z, t) which is = q over 4 pi epsilon naught, times 1 over square root x square, + y square + z square. This is just 1 over r, right? We are seeing it in a moving coordinate system and it appears that the coordinates should be transformed by this transformation. Where you have x is replaced by x- vt over square root 1 over 1- v squared over c squared y goes to y z goes to z and rho goes to rho 0 over square root 1- v squared over c squared. And you'd probably remember this when we have lengths change of the wire that the density also changed. The charge itself doesn't change. >> Correct. >> Yeah, so you can see if we do this transformation to this original form, we will get the formula that we just derived for so many pages, it's just simple, right? So using Lorentz Transformation, you can solve this with less time. So with that, I'd like to wrap up our lecture here, and I hope to see you again in the next lecture. See you.