[MUSIC] You will see that not only is any function of x-ct a solution, but any function of x+ct is also a solution to the wave equation. You can go through the process that we just covered and solve it, right? Since the wave equation only contains c squared, changing the sign of c makes no difference. In fact, the most general solution of the one-dimensional wave equation is the sum of two arbitrary functions, one of (x-ct) and the other of (x+ct). And this is superposition rule, so we can say psi is equal to function of (x-ct) + g(x+ct). The first term represents a wave traveling toward positive x, as we saw before, and the second term an arbitrary wave traveling toward negative x. The general solution is the superposition of two such waves, both existing at the same time. Now, here is a homework question. Psi cosine kx times cosine kct. The above equation is not in the form of a function of (x-ct) or of (x+ct), is it? >> No. >> No. Yet, you can easily show that this function is a solution of the wave function, wave equation below. Now, how can we then say that the general solution is the form of the equation below here? Mallory, can you give some hint to our students? >> I think it has to do with some simplifications you can make with trigonometry, if I were to guess. >> Exactly, exactly, the addition or subtraction of two trigonometric functions can result in this form. So with that hint, I guess you will be able to answer the homework question, okay? So in our solution, we chose to consider the special case where the electric field has only a y-component, and there's clearly another solution for waves traveling plus or minus x-direction within electric field which has only a z-component. Since Maxwell's equations are linear, the general solution for one-dimensional waves propagating in the x-direction is the sum waves of E sub y, and waves of E sub z. And this general solution is summarized in the following equations, as you can see here, E = (0, Ey, Ez). And you see the Ey and Ez follows the superposition of those two wave equations, wave solutions, and B = ( 0, By, Bz). Again, the component has cBz and cBy with the two wave functions that are superimposed together, right? And such electromagnetic waves have an E-vector whose direction is not constant but which gyrates around in some arbitrary way in these yz-plane. And at every point, the magnetic field is always perpendicular to the electric field and to the direction of propagation. So if there are only waves traveling in one direction, for example, the positive x-direction, there is a simple rule that tells the relative orientation of E and B fields. And that rule is that the cross product E cross B. E cross B Is pointing in the direction where the waves is traveling, okay? So that will be what we have learned up to now. Now, since we have covered one-dimensional solution, now we are going to step up and think about three-dimensional waves. We have already seen that the vector E satisfies the wave equation, and it is also easy to arrive at the same conclusion by arguing directly from Maxwell's equations. So let's take a look at the second equation, where del cross E is = to minus round B over round t. We are taking one more curl here, and we are changing the sequence of the operation. And we can see from this side, where we have the Rule to replace here. And here we are using the Maxwell's force equation. Insert it and replace it with something in terms of E. When we do that, because this part is 0. So why is this 0, Melody? >> Because, sorry. >> Because the diversions of E field is the first law, right? >> Yeah. >> Is equal to the density, charged density over epsilon naught. Now, if we know that, why is this 0? >> Because we don't have any charged density, we've removed the source. >> Exactly. We removed the source, so this should be 0. So we only have of electric field and here we have this time derivative, right? So then, we just spelled it out, and this is the equation we have in three dimensions, right? So how shall we find the general wave equation? The answer is that all the solutions of the three-dimensional solutions we have already found. Since we have written out our equations in vector form, the three-dimensional wave equation can have solutions that are plane waves moving in any direction at all. Again, since the equations are linear, we may have simultaneously as many plane waves as we wish, traveling in as many different directions. So if we revisit the plane wave, plane wave has a characteristic that the vector of E field or B field is 0 in the traveling direction, right? The most general solution of the three-dimensional wave equation is a superposition of all sorts of plane waves moving in all sorts of directions. And the other thing is, once you have defined the plane within the plane you don't have any variation, okay? That's another character. Now, le's talk about scientific imagination, okay? Let's try to imagine what the waves look like in our mind, and that's pretty challenging, in fact. You may have some wiggling picture. You may saw this kind of wiggling picture like, if you have three-dimension axis here, E field is wiggling like this. And B field is wiggling in perpendicular to this. And this is, let's say x, y, and z, right? You may have seen some pictures like this. But does this really represent how the waves are propagating? And we have already discussed, from the relativity point of view, depending on which frame of reference, sometimes E field may appear, disappear, B field can appear or disappear. So you will easily find that imagining the shape of the wave, it's not a trivial one. It's not a trivial one. So with that being said, let's try, still, to discuss the scientific imagination of the waves. So even if we cannot see beauty, in particular, the measured results, we can already claim to see a certain beauty in the equation which describes general physics law, right? Even though describing or depicting the shape of the wave is challenging, but still, looking at the equation that describe the wave, we can discuss the beauty of this. So to some students, it may seem dull, but to me, it seems beautiful. How about you Melody? >> Yeah, I like it. >> Good, so for example in the wave equation, there is something nice about the regularity of the appearance of the x, of the y, the z, and the t. This is this nice symmetry in appearance of the x, y, z, and t suggests to the mind still a greater beauty which has to do with the four dimensions, the possibility that space has four-dimensional symmetry. The possibility of analyzing this four-dimensional symmetry, and the developments of the special theory of relativity. So there is plenty of intellectual beauty associated with the equations. But we will just stop here for the discussion and leave our students to think more about this, okay? Now, let's change our topic to spherical waves. So plane waves is something, you have tidal waves that has very long parallel lines like in a vast ocean, right? Spherical waves is something you see when you drop a rock in a pond or a lake and you see the waves goes out in a circular fashion. Have you done that before? >> Yeah. >> And do you like the way the waves propagate? >> Yeah, I think it's really cool. >> Okay, if that's cool, let's think about how we can describe it in mathematical way. >> Great. >> Have you done that? >> I have not, but I'm excited to. >> Okay, so we have seen there are solutions of the wave equation which corresponds to plane waves and that any electromagnetic wave can be described as a superposition of many plain waves. In certain special cases, however, it is more convenient to describe the wave field in a different mathematical form. Now, we'd like to discuss the theory of spherical waves, waves which corresponds to spherical surfaces that are spreading out from some center. Suppose we have a function psi of (r) that is spherically symmetric, where by r we mean r = square root of x squared + y squared + z squared. This is Pythagorean Theorem, the radial distance from the origin, okay? In order to find out what functions psi of r satisfy the wave equation, we will need an expression for the Laplacian of psi. So round psi of r over round x is = to round prime of r, round r over round x. This is chain rule as we just discussed. So with this in mind, if I do the second derivative, you see you can get this result. And bear in mind that round squared r over round x squared can be found from round over round x, which is coming from the fact that r is equal to square root of x squared plus y squared plus z squared. And if you do this, you will be able to approve that is x over r. Once you know that, you can do the second derivative as well. Once you do that, and insert it into this equation, you will find round squared psi over round x squared is equal to x squared over r squared psi two prime plus 1 over r times 1 minus x squared over r squared psi prime. That is mathematical manipulation, okay? So with that in mind, the spherical waves will be described with the equations in the box which is written here. And knowing the fact that x squared + y squared + z squared is r squared, you will be able to say, by adding them up, that the Laplacian of phi of r is equal to psi 2 prime r because all the nominators add up to r squared, and then it annihilates with the denominator r squared plus 2 over r psi prime r. Because here, you edit this up. This is 3 over r, and this is minus 1 over r, so that's why you have 12 over r, all right? So all in all, we can show that the Laplacian of psi is equal to 1 over r, d squared Over dr squared of r phi, okay? This is something you can prove by yourself. So if we wish to consider spherical symmetric fields which can propagate as spherical waves, our field quantity must be a function of both r and t. Suppose we ask what functions psi(r,t) are solutions of the three-dimensional wave equation? So we just insert it here. And using the fact that the Laplacian of the psy function is 1 over r, d squared over dr squared r psi, we just insert this here, what we get one slide before. And once we do that, we got this equation. And from here, we do some mathematical trick. Can you imagine what mathematical trick we do here, Melanie? >> Well, you can move the r around. >> Yeah, we are moving the r around by multiplying the r to both sides. So here, there's no effect, right? Here, the r and here, r can go into here. So you can make a package for r psi for the function we do the derivatives. So you can make a neat equation like this, round squared over round r squared, r psi minus 1 over c-squared, round squared over round t squared r psi is equal to 0. So r psi becomes the member of wave equation, okay? >> Yes. >> r psi being a member of wave equation has a meaning here. So this equation tells us that the functions r psi satisfy the one-dimensional wave equation in the variable r, but is more than that. Using the general principle that the same equations always have the same solutions, we know that the spherical waves must have the form of f(r-ct), right? And in fact, we can change it to (t- r over c). This one, you can see obvious relationship. And then if you move r to the right side, then this wave function, the wave form becomes one over r function. What is the character of a point over r, Melody? >> It degrades at- >> Exactly, it decrease as a function of distance. So now, you're losing information as you go far and far away, right? >> We saw that with the point charge, [INAUDIBLE]. >> Yeah, we saw that with a point charge, but for a plain waive, we didn't see that. >> Mm-hm. >> For the plain chart, we didn't see that. So that's a big, big difference, right? >> Right. >> Such a function represents a general spherical wave travelling outward from the origin to the speed c. The factor r in the denominator says that the amplitude of the wave decreases in proportion to 1 over r as the wave propagates. Unlike a plane wave where the amplitude remains constant as the wave runs along, in a spherical wave the amplitude steadily decreases. So you can see this trend, right? So as you move the function in time or space, it has to decrease. Now, we know that the energy density in a wave depends on the square of the wave amplitude, is a square. As the wave spreads, its energy spread over larger and larger areas proportional to the radial distance squared. If the total energy is conserved, the energy density must fall as 1 over r squared. And you probably remember the Gaussian law where we had this light bulb and you have photons spreading out in a radial fashion, and if the number of photons are conserved, we know that the photon density will decrease as you move far away from the source, right? Likewise, if the total energy is conserved, the energy density must fall as 1 over r squared, and the amplitude of the wave must decrease as 1 over r. So this is a reasonable function that we just derived for the spherical wave. Now, we're going to make a special assumption that the waves generated by the source are only the waves which go outward. You can also go inward in mathematics, but that doesn't make sense in physics. Since we know that waves are caused the motion of charge, we want to think that the waves proceed outward from the charge. It would be really strange to imagine that before charge were set in motions, freak waves started out from infinity and arrive at the charge just at a time they began to move. So that's like a Avengers movie where you turn the time backward, and everything goes in a backward fashion. And that's a possible solution in mathematics, but experience shows that when charges are accelerated, the waves travel outward from the charge. We must mention another important point. The function psi is infinite at the origin. Can you see that? If you put r equals 0, this diverges. It becomes infinite, right? We would like to have a wave solution which is smooth everywhere, otherwise, we cannot do our mathematical manipulation. Our solution must represent physically a situation where there is some source at the origin, otherwise this is not explicable. We have inadvertently made a mistake. We haven't solved for the free wave equation everywhere, we have solved the equation with zero on the right everywhere except at the origin. Our mistake crept in because some of the steps in our derivation are not legal when r is equal to 0, and that applies to spherical coordinates, and you will see that. So similar mistake can be made in electrostatics. We didn't cover that before, but now we are going to cover it. So let's show that it's easy to make the same kind of mistake in a electrostatic problem, mathematically. Suppose we want a solution of this equation for an electrostatic potential in free space. The Laplacian is equal to zero, because it's free space. We don't have any charge in the space and you will see what about a spherically symmetric solution to this equation? So Laplacian equation here. Laplacian of sin is equal to zero. And if it is Cartesian coordinate, we know that is zero everywhere. But in spherical coordinate, you can see this is 1 over r, d-squared or dr-squared, r-psi equal to zero. So in fact, this should be zero, right? This should be zero. And if you do this as a zero, then you know from mathematics the solution to this equation is r psi = ar + b, right? Otherwise, we cannot solve it. So if you solve like this, you see psi = a + b over r. Something is evidently wrong here. In the region where there are no electric charge, we know the solution for electrostatic potential. If there is no charge, the potential should be constant. Simple, it's constant. But here it's not constant. That corresponds the first term here, a. But we also have the second term which says that there is a contribution to the potential that varies as 1 over r. And that potential is coming from point charge, right? So where did we bring this point charge? So although we thought we were solving for the potential in free space, our solution also gives the field for a point source at the origin. If there were really no charge or currents at the origin, there wouldn't be spherical outgoing waves. So here, you should be very careful, okay? Okay, with that I'd like to wrap up our lecture here and we're going to meet you again in the next lecture. Have a nice day. Bye, bye.
