Welcome back to electrodynamics and its applications. This will be lecture 19 in total and the fourth lecture of part four, which is in depth solutions for Maxwell's equations. My name is professors Seungbum Hong and to my right side, I have my teaching assistant Melodie Glasser. So we will discuss about solutions of Maxwell's equations with current density j and charge density rho. So among their solutions, Maxwell's equations have waves of electricity and magnetism. We want to show that Maxwell's equations can form the base for the phenomenon of light. The equation for the electric field produced by a charge, which moves in any arbitrary way is written as follows. These equations we already showed you in the very first lecture, if you remember and we told you they are very complicated, so we're not going to use this equations to understand the electricity and magnetism. We introduced the concept of electric field and magnetic field and use those characters of fields to understand the static and quasi-static world of electricity and magnetism. However, now we are at a point to tackle this complicated equations, which were solved by professor Richard Feynman. He is very smart and genius as you can see While you are taking a look at these equations, we will think about, if the charge moves in an arbitrary way, what would happen? Now, the electric field we would find now, at some point depends only on the position and motion of the charge at an earlier time. This sentence can be applied to our daily lives as well, like what we do today is somehow contributed to by what we did yesterday. Then you may wonder how earlier time it is. It is in an instant which is earlier by the time it would take light to travel the distance r prime from the charge to the field point. So now you can understand this time would be a split second or even smaller than that because most of the distance we're dealing with is much smaller in comparison to the speed of light. Yeah. I'd like to remind students to look at the earlier lecture to see how the electric field changes as a position of function of distance and time as it moves from a charge, to see why this is necessary. I think that's a very kind comment and as Melodie told us, if you are unsure about what we just discussed, we recommend you to revisit some of the earlier chapters. We want the electric field at point one at the time t, we must calculate the location two prime of the charge and its motion at a time t minus r prime over c, where r prime is the distance the point one from the position of the charge two prime at a time 2t minus r prime over c. So you can see the point of interest or observation is close to Melodie and you can see two prime is just a little bit earlier than the point two, and the r prime vector is pointing from the past moment of your charge to the point of interest, point one. So the prime is to remind you that r prime is the so-called retarded distance from the point two prime to the point one. So now let's look at what the equation says and we are going to analyze the equation term by term. Here we're going to focus on the very first term, which is now boxed with an orange color. So the first term is what we would expect for the Coulomb field of the charge at its retarded position, because you can see you have r prime and r prime here and we may call this the retarded Coulomb field. You see here is q over 4pi epsilon nought times e sub r prime over r prime square. We learned from the previous lectures that the static Coulomb field is nothing but q over 4pi epsilon nought times e sub r over r square. You can see it has the same exact inverse square law, with the small difference of this retarded notion. So that's why we will call this the retarded Coulomb field. The electric field depends inversely on the square of the distance and is directed away from the retarded position of the charge, e sub r prime as you can see here. Now, if we didn't know about the other terms and it's just about the first term, we would say all the fields are the same as the static ones but just retarded. But because of the other terms, this is not the case. Now, to the retarded Coulomb field, we must add the other two terms. The second term says there is a correction to the retarded Coulomb field, which is the rate of change of the retarded Coulomb field multiplied by r prime over c the retardation delay. So why don't we take a look at the retardation delay more. Okay. So I think if you look at this, you'll see it's distance over velocity and you also know the equation that velocity is distance divided by time. So if you rearrange this equation, you'll see that this retardation delay actually corresponds to the time and we mentioned that time earlier in the lecture. Great. So as Melodie just wrote the equations here and she's about to erase it but before she does erase it, let me recap what she's just told us. So as you can see the time delay is related to the distance traveled by the light, which is r prime, r prime will be the d here and the velocity in the denominator will be the c. That's why r prime over c will be the time that will take for light to travel from two prime to one. So this term will be the time as you can see and this term on the right side is the derivative of the first term with respect to time. So you know from Taylor series to the first-order approximation, these two will constitute the original function at the instantaneous time. So these two will compensate the retarded delay. So in fact, it will be the instantaneous field from point where the charge is currently, instead of at the retarded position. So as we just mentioned and discussed, the second term tends to compensate for the retardation in this first term. The first two terms correspond to computing the retarded Coulomb field and then, extrapolating it toward the future by the amount of r prime over c, that is to say right up to the time t. So combining these two, as we mentioned is like having the Taylor series up to the first-order approximation, which will make this the function as a function of the current time, so this will make this field as an instantaneous like this to the person who is observing this field. So that's why we are now talking about instantaneous Coulomb field. If the field is changing slowly, the effect of the retardation is almost completely removed by the correction term and the two terms together, give us an electric field that is instantaneous Coulomb field, that has the Coulomb field of a charge at point the two instead of two prime to a very good approximation. So what we learned before this lecture for static ones becomes valid. Now, we're moving to the third term which is a little bit more difficult to understand and grasp, but this will be the basis for the phenomenon that we just mentioned, the light. So the third term is the second derivative of the unit vector e sub r prime, and for the study of the phenomena of light, we find that far away from the charge the first two terms, which decreases as one over r square becomes very weak in comparison to the last term would decreases as one over r. So that we know for a long distance and when this electric field travel a long way, as a wave or like a light, then we will see this term dominates. So the electric field is, for large distances, proportional to the component of the acceleration of the charge at right angles to the line of sight. So you see, if you do second derivatives with respect to time of e sub r prime, this will be the acceleration. So the physical meaning is now written in this box that electric field is minus q over four Pi Epsilon square r prime, times the acceleration of the charge at t minus r prime over c projected at right angles to r prime. So we are going to deal with this more and more so you can understand better. Okay. So for the moment let's just keep this in your mind. We will diverse your attention to something that we learned before, that is to say spherical waves from a point source, and we know that Maxwell's equations can be solved by letting electric field being the minus Del Phi minus da over dt, and b field is the curl of a vector potential a, where Phi and a must then be the solutions of the equations written below. We know from the Lorentz gauge that these two equations have the same form. So and must also satisfy the condition that I just mentioned the Lorentz gauge which is, Del dot A is equal to minus one over C squared, times round Phi over round t. So let's find a solution Psi of the equation. So we have now two equations and now we're going to use a general form Psi, Del squared Psi which is Laplacian of Psi minus one over C squared times round squared Psi over round t squared is equal to minus S, where S is the source and it is known. So in places where the source is zero, the solution Psi satisfy the three-dimensional wave equation without sources whose mathematical form is as follows. This is the mathematical form when we let S is equal to zero, and we learn that in that case, we have the wave equations, plane waves. So Psi x comma t is equal to function of x minus ct. For spherical waves, when we have a source, it is Psi is equal to f of t minus r over c over r. Just to remind students one of the differences is plane waves don't degrade as a function of diffs and to have spherical waves when they're spreading out will decrease by that one over r term right here. Very good. For the plane waves, we may not have low sources at the origin but for spherical waves, we need to have one. All right, very good. So let's take a look at the equation for a spherical wave, Psi x comma y comma z comma t is equal to f of t minus r over c over r. As Melodie mentioned before, because there's r term in the denominator, this will decrease in amplitude, as we are farther and farther away from the source. We remark that the equation above doesn't represent a wave in free space that there must be charge at the origin to get the outgoing wave started. we may ask, what kind of a source S would give rise to a wave like the equation above? What kind of source? Suppose we have the spherical wave of Psi, f of t minus r over c, over r, and look what is happening for very small r, then, the retardation r over c in f of t minus r over c can be neglected. So provided f is a smooth function, then, Psi becomes f of t over r when we send r to a very small value. Then we're going to see what kind of source can satisfy, the formal equation that we are just dealing with. So Psi is just like a cooling field for a charge at the origin that varies with time, as you can see. Therefore, this function f of t can be like q over four Pi Epsilon naught. Which we know, from the previous slides. So this will be the function that will be, spherically symmetric. Where Q, the capital Q, is, the integration of rho dv and rho is the charge density. So we know that such a Phi, satisfies the Poisson's equation, so the Laplacian of Phi will be minus rho over Epsilon naught. So following the same mathematics, we would say that the Psi satisfies precisely equal to f over t, over r, and if r goes to zero, this is true, and this will also, satisfy, the Poisson's equations, so it's Laplacian of Psi equal to minus s and where s is related to f, by f, equals capital S over four Pi, where capital S, is the, volume integral of small s, the source. So the only difference is that in the general case s and therefore, the capital S, can be a function of time. Now the important thing, is that if Psi satisfy Laplacian of Psi equal to minus S for small r, it also satisfies Laplacian of Psi minus one over c squared, round, Del squared, Psi over Del, t squared is equal to minus S. So we will now think about what we just mentioned. So if this satisfies for small r, then, this also satisfied for this general equation. So let's take a look what why does the case. As we go very close to the origin, the one over r dependence on Psi causes the space derivative is this one, to become very large in comparison to this one. But the time derivative keeps their same values. So as r goes to zero, this one, this equation, becomes equivalent to the Laplacian of precise equal to minus S. So that's how we can understand why it also satisfy this one. Okay? If the source function S of t of Del squared, Psi minus one over c squared, times Del squared Psi over Del t squared is equal to minus S is localized at the origin, and has the total strength the capital S of t is equal to small s of t dv integral, the solution is, two Psi, x comma y comma z comma t, is equal to one over four pie, capital S of t minus r over c over r. The only effect of the term Del squared Psi over Del t squared is to introduce the retardation t minus r over c in the Coulomb potential and that we can revisit the slides we covered, for the plane wave. Where we saw, this equation leads to, this retardation. All right. So we have found the solution of del square psi which is Laplacian of psi minus 1 over c squared times del squared psi over del t square equals minus s for a point source. The next question is, what is the solution for spread out source? So I am repeating this again. So up to now, what we have done was for a point source and that's why we had 1 over r, which was like Coulomb field from a point charge. But the next question now we are going to ask is, what is the solution for spread out source if we have? So we're now expanding our concept to be applicable to more general case. Now, we can think of any source as x, y, z and t as made up of the sum of many point sources, one for each volume element dV and each with the source strength of s (x, y, z, t) times dV. Using the results of psi of (x, y, z, t) is equal to one over four pi times the capital S of t minus r over c over r. We know that the field d psi at the point )x_1, y_1, z_1) or one for short at the time t from a source element sDV at a point (x_2, y_2, z_2) or two for short is given by d psi (1, t) is equal to S of 2, t minus r_1 2 over C times dV2 over 4pi r_1 2. Where r_1 2 is the distance from two to one. Even though it seems complicated, if you read this out loud as I did, you will be able to follow where and when, and from what. So let's rewrite this equation again. D psi of 1, t is equal to S of 2, t minus r_1 2 over c times dV2 over 4pi r_1 2. Adding the contribution from all the pieces of the source means doing integral over all regions where S is not equal to zero. If S is equal to zero then we don't have to do that because there is no contribution. So if we do that the psi of 1, t is equal to integral of S of 2, t minus r_1 2 over c over 4pi r_1 2 dV2. That is the field at point one at a time t, is the sum of all the spherical waves which leaves a source elements at two at time t minus r_1 2 minus C. So it's like if you throw a stone to a calm pond, you are making spherical waves, that's one point source. If you throw a lot of stones, then those stones will add up, and this is like this. The time here is when you threw the stone, and you are seeing that afterwards. So this is the solution of our wave equation for any set of sources, and we now see how to obtain a general solution for Maxwell equation. So for psi is equal to scalar potential phi, then S is equal to rho over epsilon nought and for psi is equal to vector potential a then S is just equal to j over epsilon nought c squared. So we can use the same equations for the two different potentials. If we know the charge density, rho of x, y, z, t and the current density j of x, y, z, t everywhere, we can immediately write down the solutions of the Laplacian of phi minus 1 over c squared del squared phi over del round T square is equal to minus rho over epsilon nought and Laplacian of A minus 1 over c square, round square A over round t square is equal to minus j over epsilon nought z squared. So it is just technicality that matters here. So if you understand one thing, you can understand the other thing as well. So that's the beauty of math, even though you don't know the details of what's happening down there, you have the same governing equations. If you solve the equations from bottom up, top down, then you will be able to apply the knowledge to different fields as we just discussed before. You can see those differences in these equations here. This equation has the c squared and the rho in the j. Yeah, very good. So you see rho is here, instead here is j and c squared. So with that in mind, we can solve for both phenomena. The fields E and B can then be found by differentiating the potentials that we just solved. So electric field will be minus del phi minus dA over dt and B field will be del cross A. This equation really tells us that when we think about a ring, a metallic ring, and if we put a magnet close to the ring and we shake the magnets so we are inducing an electromotive force, then it is really coming from the change of magnetic potential not the electrostatic potential. So in a ring, it is hard to set up a gradient because the starting point and ending point is the same. But you can set up a vector potential time dependence, so that's how we can understand why we still have electric field around the ring that could make a light bulb lit in this closed loop. So we have solved Maxwell's equations, given the currents and charge in any circumstance, we can find the potentials directly from these integrals and then differentiate and get the fields. So we have finished with the Maxwell theory. All that remains is to take a moving charge, calculate potentials from these integrals and then differentiate to find electric field from the equation we just mentioned. The gradient of the electrostatic potential and the time derivative of vector potential. Then we will get this equation that we just mentioned before and discussed in detail. So Melodie, let me ask you a simple question. Of these three terms, what do these two terms mean? So this first term is the retarded Coulomb field and then this one corrects it into an instantaneous field, and together they're like the first of a Taylor series. Yeah. So as Melodie mentioned, this is the first-order approximation of Taylor series, which makes this retarded Coulomb field to almost instantaneous coulomb field for small distance. So this will be exactly the same form as we learned from statics that we have the Coulomb field. How about the third term? But still account for the acceleration and the relativistic effects. Exactly. So the third term dominates when we travel long distances and is more to do with the light phenomena, wave characteristics. So you can see it depends on the acceleration of the charge rather than the charged quantity and the distance vector of the charge. So that's what we have learned, and this is really the essence of the Maxwell theory. So we have covered the universe of electromagnetism and here's the structure built by Maxwell, complete in all his power and beauty. So in the previous slides, we have dealt with the Maxwell equations in four equations. Introducing the concept of field; an electric field and magnetic field and we have also explained the characters of those fields namely; the flux and circulation. In order to understand flux and circulation of each field, we came up with divergence and curl. So divergence of electric field is here that curl of electric field is there. Divergence of magnetic field is here and curl of magnetic field is there. So with these four equations, we were able to understand most of the static and quasi-static phenomena. Now at this point, we're at the point to solve the general equations in the form of this electric field is the gradient of the electrostatic potential and the time derivative of the vector potential, and magnetic field is curl of magnetic vector potential. With the Lawrence gauge, we were able to come up with this beautiful two sets of equations, where solving these two will give us the most rigorous sets of equations that complete the theory. However, as you can see it's still very difficult and complicated. We will give you some examples that will help you understand those equations in more details. So let's think about the fields of an oscillating dipole as an example where dipole can move up and down. By the way Melodie, what is an electric dipole? So an electric dipole is when you have a set of charges so maybe like positive and negative and there's a distance between them like here. So as Melodie draw here, if you have a set of charges with equal magnitude and opposite polarity, the dipole moment is formed and the dipole moment is defined by the charge times the distance vector here. So you can see the vector is pointing from minus to plus. So here we have to limit ourselves here just to show that in a few examples this phi (1,t) is equal to integral of rho (2,t) minus r_12 over C over 4 pi epsilon nought r_12 times dv2. A (1,t) is integral j (2,t) minus r_12 over c over 4 pi epsilon nought c square, r_12 times dv2 give the same results as the one that we just mentioned again and again and again solved by Richard Feynman. So first, we will show that that e equals q over 4 pi epsilon nought times e_r prime over r prime square plus r prime over c times d over dt, e_r prime over r prime square, plus 1 over c squared d squared over dt square e_r prime and cB is equal to e_r prime cross E gives the correct fields for only restriction that the motion of the charged particles is non relativistic. So if the motion of the charged particle is non-relativistic that means it is moving very slow. Its motion is very slow. So we consider a situation where we have a blob of charge that is moving about in some way in small region and we will find the fields far away. So from a distance, this blob of charge may look like a point charge, To put it another way, we're finding the field at any distance from a point charge that is shaking up and down in very small motion. Since light is usually emitted from neutral objects such as atoms, we'll consider that our wiggling charge q is located near an equal and opposite charges at rest. So you can imagine you have protons which is very heavy, fixed, an electron cloud moving up and down, that will make your dipole moving up and down. If the separation between the centers of the charges d as we mentioned before, the charge will have a dipole moment p which is equal to q times d which we take to be a function of time. So at close distance. So you may remember these slides. Now we should expect that if we look at the fields close to the charge, we won't have to worry about the delay because it's so close. The electric field will be exactly the same as the one we have calculated earlier for an electrostatic dipole using the instantaneous dipole moment p of t. So you can remember the field line coming out of this dipole moment where E_z is P over 4 pi epsilon nought times 3 cosine squared theta minus 1 over r cube. E in-plane direction of electric field is the square root of E_x squared plus e_y squared, due to the Pythagorean theorem, is equal to p over 4 pi epsilon nought times 3 cosine theta, sin theta over r cubed. You know that if theta is zero, the E_z only remains. But if it is 90 degrees, this also have only z but in the opposite direction and other than those two angles, we will have in plain field as well. Time part comes in here. Yes. So the time pulse comes with p. The p will change as a function of time because now we are moving this up and down. But for far distance, let's say we go very far out we ought to find a term in the field that goes as one over r like in the case we just mentioned before, and depends on the acceleration of the charge perpendicular to the line of sight. So now, you see this was from a single charge, then how about from a dipole? So in order to understand how this term, the third term changes for dipole moment, we will detour, we will go a different route. So let's begin by calculating the vector potential a using this one. A (1,t) is equal to integral j (2,t) minus r_12 over z over 4 pi epsilon nought c squared r_12 times dV2. Suppose that our moving charge is in a small blob as depicted here, whose charged density is given by rho( x, y, z) and the whole thing is moving at any instant with the velocity v. So this blob is moving upward. So we are aligning our Cartesian coordinates so that the z-axis is in parallel with the v vector. The current density j (x, y, z ) will be equal to v times rho (x, y, z). This is what we learned before. The current density is the number of charge passing through a unit area per unit time. So if you understand that rho is the number of charge given volume and v is the velocity vector, then you can understand these two are equivalent. Now it will be convenient to take our coordinate system so that z-axis is in the direction of v as we just mentioned to make the problem easier to solve. So we want the integral, the j of two comma t minus r12 over c over R12 times tb2 and now we want to simplify this. So now if the size of the chart blob is really very small compared with r12 which is the distance from the blob to the point of interest or point of observation, we can set the r12 term the denominator equal to r the distance to the center of the blob and take r outside the integral. So if we simplify this to this one, then r can come out of this integral because it is no more a function of the blob. Next we will set r12 equals r in the numerator as well which is not quite right. It is not right because we should take j at the top of the blob at a slightly different time than we use for j at the bottom of the blob because the time here will be different from top to bottom. But if we ignore that, that's why we are using non-relativistic calculation then we can set this r12 to r. So when we set r12 is equal r and j of t minus r2 over c, we're taking the current density for the whole blob at the same time t minus r over c. This is an approximation that will be good only if the velocity v of the charge is much less than c. As I told you if the motion is slow. So we're making a non-relativistic calculation. In the later slides, we will show you how to take this into account for relativistic calculation. So here, we're going to replace j by rho and v and as you can see the equation j of 2 comma t minus r over c or r dV2 will be replaced by one over r times this integration because r is constant and is not affected by the source. Then we're going to change the j into v times rho. Since all the charge has the same velocity, the integral above is just v over r times the total charge q. As you can see these constants, so is escaping out of the integral, it becomes v over r and this rho of 2 comma t minus r over c dV2 is really the charge density which is uniform at a time of t minus r over c. So this will be the total charge q at time t minus r over c. So q you can see it is v over r times q of t minus r over c. Now, q times velocity is just round p over round t because the dipole moment p is equal to q times d and if q is constant then the time derivative will be round p round t which is velocity. Therefore, the rate of change of the dipole moment to be evaluated at the retarded time t minus r over c will be equivalent to this qv. So therefore, we can change this equation into one over r round p of t minus r over c round t which we can neatly denote as one over r, p dot of t minus r over c. Where dot represents the time derivative. I'd like to remind students unfamiliar with this that we covered in an earlier lecture so they can go back and see it if they want to. Yes. So as Melodie mentioned, you can go back to dipole moment and polarization if you feel unsure about what we just discussed right now. So we get for the vector potential a of 1 comma t is equal to 1 over four Pi Epsilon 0 c squared times p.of t minus r over c over r. Our result says that the current in a varying dipole produces a vector potential in the form of spherical waves whose source strength is p dot over 4 Pi Epsilon 0 c squared. So not too far away from now, we discussed about spherical waves and that the spherical waves Psi has the form of f of t minus r over c over r and if you compare these two equations they have the exact same form. Such a function represents a general serve spherical wave traveling outward from the origin at the speed c. Now, we can get the magnetic field from the fact that magnetic field is equal to curl of magnetic electric potential. Since p is totally in the z direction, a the magnetic potential a has only a z component and there are only two non-zero derivatives in the curl. If you remember the formula for curl, then you will understand the z component will vanish because it only has z component a. So bx will be equal to round a sub z over around y and b sub y will be equal to minus round a sub z over round x. If I solve the visa x that's equal to round a sub z round y, I'm going to take a derivative of this dipole moment function with respect to y. Remembering that r is equal to square root of x squared plus y squared plus z squared, we will be able to transform this into a simpler form. As you can know from the law of derivatives, we can split this into two by having one over r first being being acted upon and then the denominator being acted upon by round of round y operator. So remembering that round r round y is equal to minus y over r. Then you can change the first term into this form which is minus 1 over four Pi Epsilon 0 c squared times y times p dot of t minus r over c over r cubed. Now the first term drops off as one over r squared like the fields of a static dipole that we learned before as you can see, and because y over r is constant for a given direction this is falling off as 1 over r squared instead of 1 over r cube. So it's exactly the same aesthetic dipole potential. The second term gives us the new effects. It is responsible for radiation and if we do the algebraic manipulation as we did before, you can see that this one over four Pi Epsilon 0 c squared times 1 over r round over around y p dot of t minus r over c will become like the one in the red box through this operation. We have been using the chain rule two times to get to this point. The p2 dot will represent the second derivative of p with respect to t. Now let's examine more in detail how this radiation term comes about. The A of 1 comma t is equal to 1 over 4 Pi Epilson 0 c squared p dot of t minus over c over r. The expression above has a one over r dependence and is therefore like a Coulomb potential except for a delay in the numerator. Now, why is it then that we don't just get a one of r-square filled with the corresponding time delays when we differentiate with respect to space coordinates to get the fields? So Melodie do you have the answer for this? No, but I think it's on the next page hopefully. Exactly. So we will cover this question with a little bit simpler examples. So to show you that we will get there but it is not that simple. So suppose that we let our dipole oscillates up and down in a sinusoidal motion, so this will be a very simple example. Then the p, the dipole moment is equal to p sub z which is equal to p 0 sine Omega t. If we put that into our equation a sub z is equal to 1 over 4 Pi Epsilon 0 c squared times omega p 0, cosine Omega t minus r over c. If you plot this equation in any graph or or plotter, you will see that you will have an oscillating function that is oscillating between two envelopes that are function of one over r. The peak amplitude decreases as one over r but there is an oscillation in space bounded by one over r envelope as we mentioned. When we take the spatial derivatives, they will be proportional to the slope of the curve, and from the figure, we see that there are slopes much steeper than the slope of the one over r curve itself and you can see here the hills. For a given frequency, the peak slopes are proportional to the amplitude of the wave which varies as one over r and this explains the dropout rate of the radiation term. It all comes about because the variations with time at the source, this dipole moment, are translated into variations in space as you can see in the graph as the waves are propagated outward. So the magnetic fields depend on the spatial derivatives of the potential.