Welcome back to electrodynamics and its applications. This is lecture 18. My name is Professor Seungbum Hong and to my right side, I have my teaching assistant Melodie Glasser. So, today we're going to start with the solutions of Maxwell's equations in free space. All there is to know about the classical theory of the electric and magnetic fields can be found in the four equations that we have learned for long time. So, you can see here the divergence of electric field is equal to rho over epsilon naught. The curl of electric field is minus round b or round T. The divergence of B field is always zero because we haven't found yet a monopole and C square times del cross B is equal to J over epsilon naught plus around E or around T. So, those are the four equations we have learned and we have deciphered up to now. Now when we put all these equations together, a remarkable new phenomenon occurs. Fields generated by moving charge can leave the sources and travel alone through space. Which means the information of the charge can be transferred into unknown space without deterioration. Isn't that amazing? Yeah, there are probably a lot of scientific measurements we can make from that, right Dr. Hong? Yes, and that's what we do. That's what we do with the telescope or spectroscopy to understand what the stars in the sky are made of, right? So, now we're going to deal with the waves in free space and especially we will think about the plane waves. In the previous lecture, we have thought about the special case where an infinite sheet of current is suddenly turned on in the Y in one dimension. So, we consider that case and if the current has been turned on for the time T, there are uniform electric and magnetic fields extending out the distance CT. C is the speed of light, T is the time considered from the source. So, we can see how fast this information can be spread out from the source. So, at this point let me ask Melodie. Melodie, do you know what the speed of light is? The C? Yeah, I think it's something like three times 10 to the eight meters per second. Yes, it is three times 10 to the eight meters per second. So, in one second, it can travel around the earth more than I think 99 times or eight times. So, that's the speed of light. So you can imagine once this sheet is turned on, the speed at which this information is traveling is very big. So, suppose that the current sheet lies in the YZ plane with a surface current density J going toward positive Y, and the electric field will have only a Y component and the magnetic field only Z component because they are orthogonal to each other. Then the magnitude of the field components is given by, the Y component of E is equal to C times B component of Z. That is equal to minus J over two epsilon nought C. Again you can see the relationship between electric field and magnetic field. As you can see, electric effect is much larger than magnetic effect. As you can see, it is multiplied by the speed of light. That's applicable for the region where the X is between zero and CT for the positive side while for negative side it should be between minus CT and zero. However, for the distance larger than CT, you will see there is no information. The electric field and magnetic field, both of them are zero. Now, consider the following sequence of events. So we turn on a current of unit strength for awhile. So, now we have a motion of charge in one direction. Then suddenly increase the current strengths to three units and hold it constant at this value. What do the fields look like? So, Melodie, can you tell what you think what the fields look like in this case? I think if you increase the strength of the current, there should be an increase in the magnetic field and the electric field. Exactly. We saw that from the equation on the former page. Exactly. What about the timing? So, we turned on four unit strengths, and then a certain time passed, we increased it. So, do you think that will be reflected in our wave that is going out? I think you're going to see like a step function. Yeah, so let's see if Melodie is right. So, we can see what the field will look like in the following way. First, we imagine a current of unit strength that is turned on at T equals zero and left constant forever, and that is depicted here. Next, we ask what would happen if we turned on a steady current of two units at the time T1. So, that is the picture below this one, and you can see because it's in T1, you will see the lengths that is covered as C times T minus T1. When we add these two solutions, you can see you can add this two graphs according to the superposition principle. Then we find that the sum of the two sources is a current of one unit for a time from zero to T1 and a current of three units from times greater than T1 as you can see, you can see it. So this will be what we will see as an answer for the question we ask. Now, let's take a more complicated problem. Consider a current which is turned on to one unit for awhile, then turned up to three units. That's the same as the one which was discussed and later turn off to zero. Then what are the fields for such a current? Melodie. I think it's going to look something like this plot here. That's correct. So, this is the plot. That is the answer to the question here. So we can find the solution as follows: First, we find the fields for a step current of unit strength. Next we find a fields produced by step current of two units. Finally, we solve 40 fills of step current of minus three units. So, then you can see this weapon type or car type of the field will be the resulting field when we do this kind of current. So, you can now understand the shape of the electric field or the shape of the magnetic field that left the source has the history of all the current variations. So the field is an exact representation of the current. The field distribution in space is a nice graph of the current variation with time only drawn backwards. As time goes on, the whole picture moves outward at speed c, so there is a little blob of field traveling toward positive x, which contains a completely detailed memory of the history of all the current variations. So that's why when we observed the stars that are so, so far away from our galaxy, if we detect the electric field, magnetic field variation by our detector, then we can decipher, what happened in the star so long, long time ago? So that is very, very neat and amazing, isn't it? All right? Right. So if we were to stand miles away, we could tell from the variation of the electric or magnetic field exactly how the current had varied at the source. So that could be a black box of the stars in the galaxy. If we want, we can give a complete mathematical representations of the analysis. We have just done by writing that the electric field at a given place, and a given time is proportional to the current at the source. Only at the same time, but at the earlier time t minus x over c. So it is not at the same time, I just did some mistake here. Not at the same time, but at the earlier time, t minus x over c. So you can see there's a delay. So, here in the equation, you see E sub y of t is equal to minus J of t minus x over c or two Epsilon naught c. So this will be the equation that represent what we have just described here. So let's take a look at the theme of far away from the sources. So now, we want to look in a general way at the behavior of electric and magnetic fields in empty space far away from the sources. That is to say from the currents and charged. Very near the sources, near enough so that during the delay in transmission, the source has not had time to change much. The fields are very much the same as we have found in what we call the electrostatic or magneto static cases, where we have infinite charge and there has an electric field that is not a function of distance. Now, if we go out to distances large enough so that the delays become important, however, the nature of the fields can be radically different from the solutions we have found. In a sense, the fields began to take on a character of their own when they have gone a long way from all the sources. So now they are on their own. So we can begin by discussing the behavior of the fields in a region where there are no currents or charges. So, now we're separating out the wave and the source. Now, let's take a look at electric and magnetic fields far away from the sources. So, what kind of fields can there be in regions where Rho and j are both zero? So let's take a look at the equations that we summarized in terms of electrostatic potential and magnetic vector potential. So this is electric scalar potential. So you see Laplacian of electric potential minus one over c squared times Rho square. Electric potential over Rho t square is equal to minus Rho over Epsilon naught, and the same form is taken by the vector potential. Go ahead. I just wanted to say that reminds students there's three of those for A. For three. Yeah. We have three components for A. So in fact, it seems like two equations, but in fact you need four equations. Correct. Very good. So here, for the right term, Rho and j, because we are far away from sources. We can say they are zeros. So that's why we put zero to the right term, does in free space to scalar potential Phi and each component of the vector potential A all satisfy the same mathematical equation as you can see. So, once you understand solution to one of these equations, you can solve for all of them. Okay. So, universal solution for Maxwell equations is really to solve the one of these equations. Suppose we let Psi stands for any one of the four quantities Phi, A sub x , A sub y, A sub z as melody just mentioned. Then you can see this will be the universal equation and universal solution will be derived from this equation. You can see they're neatly arranged in a linear combination and you can see it's symmetric. Only the time is different because it has different dimensions with the space. So let's take a look at the field as a three-dimensional wave, so we will revisit this equation, the Laplacian of A minus one over c square, Rho square A over Rho t square is equal to zero. Now we are going to use the equivalence of changing the sequence of the operator. So instead of first doing Laplacian we're going to take the kernel and then do Laplacian, and from the fact that B field is the curl of A, we can replace this part by B, so it becomes the Laplacian of B. For the second term here, we will do the similar trick, we'll use a similar trick. So, instead of first doing the curl and then do the time derivatives, we're going to first do the time derivative here, the left side, and put curl here. So, in fact, I told you to put the first here, but in fact the sequences that you do the time derivative first and then do curl here. But in this case, you do curl first and then do time derivative. Now, from the same definition, we can see that we can replace this del cross A by B, so it becomes one over c square Rho square B over Rho t squared. So if we do this transformation we can understand that the Laplacian of B minus one over c squared times Rho squared B or Rho t square is equal to zero. So you can see they have the same form and so each component of the magnetic field B satisfies the three-dimensional wave equation. Now likewise, let's take a look at electric field as a three-dimensional wave as well. You can probably think oh, because we know they are coupled and they have orthogonal relationship that should be something that we expect to see. But the mathematical derivation might be a little bit more complicated. So let's take a look here. Look at this equation. So, electric potential, Laplacian minus one over c squared times Rho squared Phi over Rho t squared, we can use the fact that the electric field is the combination of the electrostatic potential gradient and the time derivative of magnetostatic potential. Using the fact, we can now expand the curl of E, and if I took this here, the curl be E becomes minus del cross del Phi minus del cross d over dt. We know from our earlier lecture, gradient of a scalar function has no curl. There's no rotation, so this should be zero. So what is left is only the latter part. Now this latter part we can do the same trick to change a sequence of the operations. So instead of del cross d over dt, we can do the d over dt del cross A, and del cross A again is B. All right. So then you can see it's minus Rho B over Rho T. In fact, this is the second equation of Maxwell equations. Then we can also use the fact del cross, del cross E from the backup rule is del, del.E minus Laplacian of E field, and the divergence of E is the charge density at the point of interest divided by permittivity. But because we are free of source, this should be zero. So, what is left is only this Laplacian of E field. Laplacian E field from here as you can see, this one goes into here. So, it is equal to minus del cross Rho B over Rho t. We can change the sequence again, which becomes minus Rho over Rho t del cross B. Del cross B we know from the equation, that it is minus one over c squared times Rho squared E over Rho t because the source J is zero. So, putting this one and putting this one equal we come up with the equation here, which has the same form as three-dimensional waves. So, again, it follows that the electric field E in free space also satisfies the three-dimensional wave equation. So then I know magnetic potential that the potential electric potential, scalar potential, magnetic field, electric field they all follow the wave equation. So, now, it is the time to solve for the wave equation. So all of our electromagnet fields satisfy the same wave equation. That is to say. The Laplacian of Psi minus one over c square, Rho square Psi over Rho t square is equal to zero. So, what is the most general solution to this equation? So Melody, do you have any clue what the form of the solution will be for wave equation? I'm a little bit rusty on my physics I didn't memorize that. Okay. That's okay. So here's the thing. If you don't memorize this as our teaching assistants does, there's no worry to do. You can try to follow us and repeat it until it becomes second-hand nature. So for here, we will see the general solution will take a form of a sinusoidal or cosine function, and inside the cosine or sinusoidal function, you will have an argument that is a function of the position and time, and then you will have defined the wave vector and the angular velocity of the wave. But that is okay. So rather than tackling this difficult question right away, we will look first at what can be said general about the solution where nothing varies in y and z. So we'll go step-by-step. So let's suppose that the magnitudes of the fields depend only upon x, that there are no variations of field with y and z. So we are considering plain waves again. So why do we do it all over again? We didn't show the waves, we found what most general solutions for plane waves. We found the fields only from a very particular current source, which was all of a sudden a plain current was turned on. That was a particular example, but now we want to expand our horizon. So, what is the most general one-dimensional wave that can be in free space? We cannot find that by seeing what happens for this or that particular source, but must work with greater generality. We're going to work this time with differential equations instead of with integral forms. Although, we will get the same results, it's a way of practicing back and forth to show that it doesn't make any difference which way you go. We did it before for many cases. You should know how to do things every which way, because when you get a hard problem, you will often find that only one of the various ways is tractable. So you need to have a lot of weapons on your hand, so to choose which one suits you best. Now, we want to start right from the beginning with Maxwell equations in free space, so that we can see their close relationship to the electromagnetic waves. So we start with Maxwell's equation setting the charge and currents equal to zero. So let's revisit these four equations, and put Rho zero J zero. Then, you can see that we write the first equation out in components, where you can see here, and because we told you that E sub y and E sub z will not vary as a function of Y or Z, we will see these will be all out, because there are no variations with y and z, so these are zero. Then, in order to satisfy this round E sub x and over round x should be zero as well. So then you may argue, if this is also zero, does it mean there is no variation with x as well? Not necessarily, right? Yes it has. So, in space, there is no variation for sure. But in time, there may be variations. So that's something you have to be careful. These two has no variations in space and time whatsoever. But this one, maybe in space is constant but in time it may vary. So, the static field in one-dimension round E_x over round x is equal to zero. Its solution is that, E sub x, the component of electric field in x direction is constant in space. Supposing no B variation in y and z either, we can see that E sub x is also constant in time. Now, with this one, we couldn't tell whether the variation in time was constant or not. But with this second restriction, because B is not varying in space in y and z, meaning the x component of y should also not vary in time. Then, we can see it's constant. Such a field could be the steady DC field from some charged place along distance way as we have solved before. That for infinite sheet of charge, and when with density of Sigma, we feel this Sigma over E epsilon nought, no matter where you are. Is no change in time or space, and this is not interesting. So this is not the case we will cover here. Okay. So, we are interested in dynamically varying fields in one dimension. So we are at the moment interested only in dynamically varying fields. For dynamic fields, E sub x should be zero, and that's peculiar. You may think "Oh, isn't that the same as the former case where we had E sub x constant?" But bear with us. So we then have the important result that for the propagation of plane waves in any direction, electric field must be at right angles to the direction of propagation. It can still vary in a complicated way with the coordinate x. So, as we mentioned, the time change of electric field should be zero because the space change in y and z was always constant, there were no change. So if you look at the former equation here, this should always be zero. Meaning the X-component should always be zero. So the only way we can think of that without static field is to put it zero. So there is no field at all. Then you can now start to think Oh, then in the other case where we discussed before, y and z component should now vary in time at least. Otherwise, you'd have nothing. All of them will vanish. So, now the transfers E field can always be resolved into two components say the y and z components. Let's work out a case where E field has only one transfers components. So, we're choosing one of them, because then out of the symmetry, we can solve the other one, and then we can do superposition to any arbitrary angle. So, we will take first an E field that is always in the y direction with zero. Z component, evidently, if we solve this problem we can also solve for the case where the E field is always in the z direction, by just changing subscript from y to z. The general solution can always be expressed as a superposition of two such fields. So that's why we're now going to simplify that to E sub y only. So, here is the thing. So, E sub x and E sub z is zero, and E sub y is non-zero, and E sub y is a function of x and t. Del cross E sub x is equal to round E sub z over round y minus round E sub y over round z, and you know this one E is zero. E sub Y is only function of X. So, it has no z component. So this will be zero. Here the second equation E sub x is zero, E sub z is zero. So it's zero. Now, only that are surviving one is the last one, the z component of the Dell cross E, because E sub y is function of x. If you do partial derivative with respect to x, you will have something. Here E sub x is zero. So the remaining one is round E sub y over round x, and round B sub x over round t is zero because the x component is zero, round B sub y over round t is zero because the z-component is only the one that we have for magnetic field. So round B sub z that is only the surviving one over t is equal to minus round E sub y over round x. So since the x and y components of the B field have both zero time derivatives, these two components are just constant field. We will ignore these constant field and set B sub x and B sub y equal to zero. So we simplify everything. Okay. So, Melodie, from this simplification. Can you see that we only have z-component in B and y component E, they are perpendicular to each other, and the traveling direction is x. So now you see in the traveling direction, we do not have any component? Right. Okay, good. So let's cover again the dynamically varying E sub y field in one-dimension, we just solved the equation round B sub x over round t is equal to zero. Round B sub y over round t is equal to zero. Round B sub z over round t is equal to minus round E sub y over round x. In plane electromagnetic waves, the B-field, as well as the E-field, must be directed at right angles to the direction of propagation. If I draw the relationship here, we have say this is y-direction and say this is x-direction, then this will be z-direction. So the wave direction is along x and I can see the electric field is here and magnetic field is here. So you can see the orthogonal relationship between three of them here. In addition, if the E-field has only a y-component, the B-field will have only z-component. So E and B are at right angles to each other. So we can confer the special case we dealt with last time here. So you can see the infinite space sheet of charge is going up, in that case, you have B-field here, E-field here and the wave is propagating along x-axis. So it was an exception to this rule. Now we're ready to use the last of Maxwell's equations for free-space. Writing out the components, we have three equations as described here. Of the six derivatives of the components of B, only the term round Bz over round x is not equal to zero. I will leave the reasons to the students to think about. Once you believe that's the case and everything else is zero, then you will see that why we have minus c squared times round B sub c over round x is equal to round E sub y over round t. So the result of all our work is that only one component each of the E and B fields is not zero and that these components must satisfy the equations below. So round B sub c over round t is equal to minus E sub y over round x. Minus c squared times round B sub z over round x is equal to round E sub y over round t. If we rearrange that, we can understand that we have this form of equation, which is the wave equation. As you can see with the wave equation we dealt with was three-dimensional wave. So this is the wave equation for one dimensional waves. If any quantity Psi satisfies the one-dimensional wave equation which is here. So now we're getting back to the question that I asked to Melody, then one possible solution is a function of Psi x of t of the form. Psi f sub t is equal to function of x minus ct. That is, some function of the single variable x minus ct. X is the position in one-dimension. Ct, C is the speed of light, T is the time. The function f of x minus ct represents a rigid pattern and x will travels toward positive x at the speed c. So let's take a look if this is the case. so for example, if the function f has a maximum, we have maximum here, when its argument x minus ct is zero, then for t equals zero, the maximum for Psi will occur at x equals zero. Because if this is zero and t is zero, x should be zero. So in x-axis, you should have maximum at x equals zero. Now, at some time later or some later time say t equals 10, Psi will have its maximum at x equal 10c. Why? Because if you insert 10 then it x minus 10c but still this should be zero to have maximum. So X should be 10c. So here you see a 10c. The maximum should be moving. So that's why this wave should move to the right. As time goes on, the next one moves toward positive x at speed of c. You can see ct will be the rate at which this peak is moving along the x-axis. Now you can see why the argument x minus ct in the function makes sense to work as a solution to the wave equation. Now, let's show that f of x minus ct is indeed a solution of wave equation. Let's insert it. Since it is a function only of one variable, the variable x minus ct, we will let f prime represent the derivative of f with respect to its variable and if f two primes represent the second derivative of f. Psi of x,t is equal to f of x minus ct. So if I do the derivation with respect to x then this becomes f prime. If I do two times it becomes two primes. Fair enough. Taking derivatives of Psi with respect to t, time t, we find the first derivative. Because now we are doing this one, so now you may be a little bit confused. So for your reference, what I'm going to do is take a look at this. I'm going to change the parameter to u, this one, x minus ct. If I change that then it becomes f of u. But then you want to, in fact, know round f over round t. If I change the variable to u, what you have to do is round f over round u times round u over round t. This is the chain rule. This one will be f prime x minus ct. Then this one, as you can see round u over round t is equal to minus c, will come here. That's why you will have this one. Then you may ask, "What about round derivative with respect to x?" It's the same, but in that case, round u over round x is equal to, here, one. so that's why you don't have to care about that. You can do the same for the second derivative. Now you can understand why you have to do minus c again because you are doing the same thing again and why we have c squared instead of minus c and you have second derivative. So with this, you see this one will be f two prime x minus ct because we are dividing this function by c square. So c square is annihilated and here you have f two prime of x minus ct. So you see exactly this and this they are the same. If you subtract something there that are same you have zero. Now you see this function satisfy the equation. So it can be the solution. So we see that Psi does indeed satisfy the one-dimensional wave equation. Now, we answer the question. So you see what type of function we will have for wave equation.