Fundamentals of Physics II

PHYS 201 - Lecture 15 - Maxwell's Equations and Electromagnetic Waves II

Chapter 1: Recap—Solving Maxwell's Equations [00:00:00]

Professor Ramamurti Shankar: Okay, this is the final lecture on electromagnetism. When you come back from the break we will start doing optics. Of course you know by now that's really electromagnetism, but it's going to be somewhat easier on your imagination because it doesn't involve many abstract things, pretty concrete things. Then it'll take a turn for the worse when we do quantum mechanics, but I think the mathematics of that should be quite simple. Before you do that you should make sure you know all about elementary complex numbers like what is eiθ. You should get that completely into your system. It's going to be all about that, then very elementary calculus, just one variable.

So last time I told you that if you want to solve Maxwell's equation in free space you can write down any function you want for E and B that satisfy the four Maxwell equations. Two involve surface integrals, and two involve line integrals, and the answer we got looked like this. This is x, y and z. The electric field is going to point that way, magnetic field is going to point that way, and the functional form E was k (k is unit vector in the z direction) E0sin(ky - ωt), and B was iB0sin(ky - ωt).

When you take this assumed form with this y and this t dependence you will find, as written, they automatically satisfy the surface integral, and the surface integral of these fields on any tiny cube will be 0. When you do line integrals you can take a loop in this plane, and that plane, and the third plane, and you fool around you get some additional constraints which I derived, but I don't want to do that again. But the conditions you derive are that ω should be equal to k times c, and remember the velocity of light emerged at this following combination. Then you will find the magnetic field should be electric field divided by c. These are the two conditions you get from writing the two other. There are three equations for loops. One of them is trivially satisfied 0 = 0, the others give you this. Now I want you to be able to at least visualize these fields. You can write them on a piece of paper, but E is not the letter E. E stands for something physical. So what do these things do? First of all what is k and what is ω? I realize that some of you are not familiar with what k and ω mean.

First of all you know it's a periodic function in y at fixed t, if you like, because sine of anything is periodic. That means at a given time if you take a snapshot of this E it'll be oscillating as you vary y. And at a given point in space, if you study it as a function of time, it'll be also oscillating in time with an angular frequency ω. So let me relate these quantities to more familiar things. If the wave looks like this, and E is going to point in that direction, you know that if you go a full wavelength you come back to where you are. So let's call the wavelength λ. It's clear then whatever happens at 0 happens at λ. At zero, ky is 0, at λ, ky should be 2Π. When you go a distance lambda it should be equal to a 2Π change. So λ, which is more familiar to you as the wavelength, is the reciprocal up to the factor of 2Π of this k. k is called the wave number. The bigger the k the more rapidly the wave oscillates. That is bigger the λ the long wavelength means more slowly the wave oscillates. Similarly you can come to the second guy and you can ask yourself — let me continue. So that's what E is. B, again, is a function of space at a given time if you draw it, but the B vector's pointing this way, so it'll look like this, and you can keep repeating this. So E is going back. So now this is the thing to understand with this graph. You can see this graph, but you should know what it means.

This whole graph describes the situation at one given time only at points on the y-axis. I've drawn it only at points on the y-axis. So this arrow, even though it sticks out here and there, is not talking about a whole region. It's talking about that one point. For example, at this one point the electric field points this way and has that magnitude. At this point the electric field points down and has that magnitude, and the magnetic field points away from the board, here the magnetic field into the board and away from the board, okay? That's the situation at one time.

Then you can ask yourself what happens later. You can do it in two ways. First of all let me tell you the relation between ω and the time period. If you wait a full time period everything should come back to where it was, so the time period T by definition will be this condition. Therefore, the time period that you guys are all familiar with is also related to ω by a similar formula. So one way to write this wave that is in terms of things you're more familiar with would be to say it looks like E0sin(2Πy/λ - 2Πt/T). This is completely equivalent to what I wrote because these are the definitions of ω and k in terms of λ and T. Maybe writing it this way is more transparent: that if you add a λ to y on the top, if you say y goes to y + λ you're adding a 2Π. If you add capital T to small t upstairs you're subtracting the 2Π. That doesn't matter. This shows you the periodicity in space of λ and in time of capital T. You can also write ω as 2Π/T, but 1/T is the frequency. This is what we normally think of as frequency, so many hertz, megahertz or whatever. That's the frequency. The angular frequency omega we use is 2Π times that, because every time the thing completes a cycle it goes by 2Π, and if it does f revolutions per second the angular frequency is 2Πf radians per second. That's the difference between revolutions per second and radians per second. Okay, so this is another way to write it, and B will be the same thing. So you should understand what this wave does. So imagine the plane wave is coming from the blackboard towards you. If it's polarized this way — now I should tell you what polarization is. Polarization is the direction of E. B is understood to be perpendicular to it, and the direction of propagation is perpendicular to both of them. You can see E x B is the direction of propagation of the wave. You do the cross product from E to B and that's the way it advances. So what you will see is that if these rows in this classroom were not curved, but straight, then all the people in the first row will see exactly the same electric field. So maybe it's pointing up. Then people in the second row may see no electric field because they happen to be sitting at that distance. People in the third row may see a negative electric field.

So every half-wave length it'll go from 0 to maximum to 0 to minimum back to 0, and so it oscillates at one time. If you wait a little bit the whole pattern will shift towards the back of the class. So what these guys saw now you will saw a little later. How much later? It's the time light takes to go from this row to your row. That's the delay, ωt, in fact I told you, you can write ky - ωt as — let's see, as k times y - ct because ω = kc. And I told you very early on that any function of y - ct is a solution to wave equation, and this tells you there's a wave propagating to the right at speed c, because whatever it does at one time, say at t = 0, at later time it's shifted by an amount ct. Now if I send an electromagnetic wave towards you, first thing you should notice is the electric field is perpendicular to the plane, in which things are happening are normal to the direction of propagation, so the field vector is perpendicular. Both E and B are in this plane and the wave is going that way. For this reason it's called a transverse wave. So transverse means all the action is in a plane perpendicular to the motion.

So one example of transverse waves is if you hold a string; I give it to the people in the last row and I hold it and then I shake it. These ripples will go from me towards the end of the room. The ripples are going like this but the wave is going that way. That's a transverse wave. Or if I shake it this way, same thing, it's going from side to side. The wave is going from me to you. A longitudinal wave is a sound wave. For example, when I talk now some diaphragm in the throat pushes the air out and compresses it, and maybe when it goes the other way decompress it, and the motion of the air is back and forth, namely in the same direction as the propagation. So sound waves are longitudinal, light waves are transverse. Yes?

Student: You said the direction of E x B gives the direction of propagation. Is there a significance to the magnitude of that cross product?

Professor Ramamurti Shankar: Yes. We'll come to that later today. That's correct. So E x B not only tells you the direction it's reasonable to ask "If it is a physical quantity, what does it mean? And we'll find out shortly what it means. Now polarization is the direction of the electric field in the solution, but we've got to realize that not every electric field has to look like this. I can make a new problem where I have solution looking like this. What is this guy doing? Can you tell it's going the other way, because I just changed that to that. So that's the plane wave going to the left. Now what should I do with the B field? Its magnitude should be E/c. I can also change this to going backward, otherwise they don't form a pair. E and B have to form a pair. But I claim this is not enough. If you took this pair it'll, of course, satisfy the wave equation because any function of y + ct will satisfy the wave equation, but I want you to remember there are four Maxwell equations. You jiggle them all you get the wave equation. It's just one consequence of the four. You've got to satisfy all the Maxwell equations, and some of them won't be happy with this choice in which I reverse just the velocity. Can you think about what's the way to fix that or what's wrong with that? I'm just saying that if the wave traveled to the left instead of to the right that's what I've done here. I would just reverse the direction of propagation. That's not a good answer. Yes, you've got an answer?

Student: Does it have to do with the fact that they're a cross product?

Professor Ramamurti Shankar: Yeah, so what's about the cross product?

Student: To make an inverse cross product you have to make it negative.

Professor Ramamurti Shankar: You want the cross product to dictate the direction of motion. So if E and B are like this the cross product comes this way. If you want them to go the other way you've got to reverse the direction of B so it points that way. That is another pair. So this, in other words, you change the velocity here, but you forgot to change it — not you, I mean, I did not change it because I wanted to show you that you just cannot put together any Eand B that you like. You've got to make sure that these form a pair. This is the pair that obeys Maxwell equation. Now I know this will work for another reason. If you took this long wave, okay? I want you to think in your head. Rotate it around this axis by 180 degrees. Can you do that in your head? Take that axis turn the whole pattern and it's going the opposite way, but in the process you can see this B will rotate from pointing this way to pointing that way. And one of the properties of nature is that if something is a solution, then you rotate the whole thing, that's also a solution.

You've got to be careful when you say rotate. For example, if you have a grandfather clock and you rotate only the clock it won't work the same way. That's because the clock is very sensitive to the earth, but if you took the earth and the clock, you rotate both of them it doesn't matter. In fact, that's happening all the time. So all the things relevant should be rotated, but for electromagnetic field in vacuum there's nothing else to rotate that matters so you rotate it. Now another thing you can do instead of rotating around that axis you can rotate it around this axis, around the axis of propagation. You can turn E to that or you can turn B to that. I'm not good at drawing pictures, but in the E-B plane you had them like this, E and B, the thing is coming towards you. I'm saying you can turn this by some angle and turn that by the same angle. That'll also form a good solution. It must. I mean, what's wrong? If you look at a wave this way it should still obey Maxwell equations, right? The titled frame of reference is as good as the original one.

All right, so now we've got a few properties of waves, and one thing maybe worth mentioning is the light from the light bulbs in this room is a chaotic set of waves being sent to you. Each atom emits light at a different polarization not in sync with the other atom, so it doesn't have a definite frequency. It's a big jumble, a big mess, but it's a superposition of elementary plane waves. Actually, I should amend myself. Plane waves are an idealization. Every wave you have, like if you turn on the light bulb the waves go out spherically from the center, but far from the center when this sphere is 10,000 miles in radius if you're a little creature at the end it will look like a plane wave, just like a sphere would look like a sheet to somebody who is very near like the earth does to us. Yep?

Student: Professor, could you just say again what you said about switching the cross product or just ________?

Professor Ramamurti Shankar: Yeah, I was saying that the property of the electromagnetic field that you saw here is that if you take the cross product, E x B...

Student: You get the direction.

Professor Ramamurti Shankar: You get the direction. I then took an opposite direction of propagation, but took the same Eand B. They don't fit. If you reverse the velocity here you've got to change this so that the cross product of E and B vectors is pointing the opposite.

Student: Okay, thank you.

Professor Ramamurti Shankar: Now when you buy Polaroid glasses I think you know what they do, right? The polarizers in the glasses will allow light to travel in only one polarization, so even though the stuff coming in can be randomly polarized, what comes out of the other side towards your eye is pointing only one way, so you cut down about fifty percent of the light if you polarize it. In fact, when light reflects off a shiny surface and comes to your eyes it tends to be polarized horizontally. Therefore, if you choose your Polaroid to be vertical in your lens, that's the most effective way to cut the glare. Also, you can take two lenses, Polaroid lenses, and you turn them one relative to the other you can cancel the light completely, because the first one may let it go this way, the second one will only let it go that way. So if you chop it like this first then it won't go through the other way. Anyway, light is a very extensive and interesting subject which probably is more important to you than some of these fields, but we don't have time to go into all the aspects of that. I do want to mention to you that the light that you and I see has a very limited range of the possible wave lengths like 400 nanometers or 4,000 angstroms, namely 400 times 10 to the minus 9 meters.

To roughly double that is all you can see. Stuff on the other side is ultra violet, then you've x-rays, on the other side you've got infrared, you've got radio waves, but they're all electromagnetic waves. All you're doing is varying ω, but nature designed our eyes to respond only to a range of ω's, because that's where our enemies were. If you've got different enemies maybe you will have different eyesight. Maybe if you've got a lot of enemies you'll have eyes all over your head like some animals do, so we got two. Anyway, that's another interesting thing on how nature adapts to properties of light.

Chapter 2: Deriving the Energy and Intensity of an Electromagnetic Wave [00:18:19]

All right, so let me tell you now about the energy contained in this wave. You've got to agree that when you have an electromagnetic wave you have energy that you did not have before. Let me ask you a first question. If I'm sending an electromagnetic wave towards you, and I ask you, which way is it polarized, what will you need to check that, any idea? Look, I've told you. If you cannot measure something or tell me in principle how we'll measure it you don't know what you're saying. Yes? You have an idea? Anybody? Yeah, go ahead.

Student: Was there anything lost maybe?

Professor Ramamurti Shankar: Pardon me?

Student: Was there anything lost?

Professor Ramamurti Shankar: Yeah, but forget the polarizing. Go back to basic definition. How will you know there's an electric field somewhere? You think you'd see a little arrow sticking out in space? Yes?

Student: You take a test charge.

Professor Ramamurti Shankar: Take a charge and you put it there and you see which way it moves. If it moves like this that's the polarization, if it moves like that, that's the polarization. In fact, the antenna on your radio has got a wire and the electric field from the radio station comes and starts moving the charges in the direction of the polarization. So you guys should know this. I'm going to assume that you didn't answer me because you're unusually modest, because you must know how to measure electric field. I could even ask you, how would you find the magnetic field," But I don't want to do that now because we should always know how to measure these things. So anyway, we know that when you have an electric field it has got energy. For example, when you took a capacitor, it took some energy to charge it, to rip the charges from one plate and ram them in the other plate. We saw that as an energy stored between the plates in the form of an electric field, and the energy of the electric field was ε0E2/2. Now you may ask me, you found the energy density in a capacitor, now we're talking about the electric field in vacuum traveling from some radio station, how do I know this formula is good?

And the remarkable thing about the field is that any expression you derive for it is a local expression. It only cares about what the field is at this point. It doesn't care what the origin is. It doesn't matter if this is produced by static charges or maybe it's an electric field produce by changing magnetic field. It does not matter. The answer does not depend on the context. This is the energy density, energy per unit volume. For a magnetic field it looks like B2/2μ0, and you all know by now that whatever ε0 does μ0 does in the other place. If this guy's up that guy's down, or in the field laws μ0/2Π's up and 1/4Πε0 epsilon is down. Therefore, if you have a region where there is, let's say, no electromagnetic field and suddenly a wave goes by that region has now got energy, and how much energy do we have? The electrical energy, you can see, is ε0 over 2E02sin2(ky - ωt). And the magnetic energy is 1 over 2μ0B02sin2(ky - ωt). I will now show you that these energy densities are actually equal.

They're equal because ½μ0 — B0 is E/c, so let's write it as E02/c2. From the definition of the velocity of light and μ0 and ε0 this is simply ε0 over 2E02. Sorry, E02 plus all the sine squares, which I forgot. Therefore, the total energy is equal to E02ε0sin2(ky - ωt). You got a half from this and you got a half from that. Even though the formula looks different by the time you put in the relation between μ0 and ε0 and B0 and E0 it turns out to be equal. So even though the magnetic field is weaker than the electric field by a factor of 1/c you might think it's negligible in terms of energy, but it's got same energy density, and you add them up you get that. Now you can see this energy density is time dependent and space dependent because it's oscillating with time at a given point, and oscillating with space at a given time. So you can ask yourself, let me sit at one place and ask myself, "What does the average energy density average over a full cycle?" If something is up sometime and down sometime what's the average? The average will be ε0E02/2, because the average value of sin2θ over a full cycle is 0 to 2Π divided by 2Π which is ½. This is something we have done before in circuits. You should know that. Average of sine squared is half, average of cosine squared is half, and the check is that sine squared plus cosine squared average is 1 because that's 1 identically. So try to remember this. So this is the average energy density.

Next I want to ask the following question. What is the rate at which energy is coming at me? And that is called intensity, I, and it is equal to the watts per meter squared, just like a flow of liquid except this is the flow of energy. So what I want to do is take a square meter, stand in the way of the beam and ask how many joules cross per second, or if you want, how many watts per square meter. That's easily calculated from the energy density and I'll give you the reason. It's identical to the reasoning for currents traveling in a wire, or fluid flowing in a tube. Suppose you take a region, a cylindrical cross section through which electromagnetic waves are traveling? This is cross section A and I wait 1 second it'll go a distance c. So all the stuff in a cylinder of base A and length c that energy will go past this checkpoint. Can you see that? It's like toothpaste. You squeeze the toothpaste these guys get passed in 1 second. If I want per square meter we'll just call that 1. Therefore the intensity is simply the energy density multiplied by the velocity of light. That's the rate at which the energy flows.

So let us now calculate that and you'll get a very interesting result. The intensity is equal to the energy density times c. The energy density was ε0E02 sin2(ky - ωt). This is not the average energy. This is instantaneous. I'll average it in a moment. Are you guys with me now somewhere here, here? Now I'm going to write it as follows, ε0 and E0 then at B0 — I forgot a c here. You realize that? That's another c. Then E0 is B0 times another c times sin2(ky - ωt). And what is that? c2ε0 is 1/μ0. So E0 B0 over μ0 sin2(ky - ωt). But if you now define a vector S, it's called a Poynting vector, which is spelled with a p, to be E x B over μ0. The magnitude of that vector is precisely the intensity. The magnitude of S, which you can either call S or you can call it I, is exactly this.

So E x B except for a factor 1/μ0 not only gives you the direction of the way of propagation it tells you how many watts are going to cross a square meter, or how many joules are going to cross square meter in 1 second. Now these are things you'll find in all the textbooks, so. All I've done is take this u c, put it in the formulas with E and bring B back into the picture so that it's symmetric between E and B, because this gives the impression it's all electric. Remember, this is electric and magnetic. They just happen to be equal. This way you can see the role played by E and B. The Poynting vector tells you the flux of energy.

So here's one example. At the surface of the earth if you take a square meter, and here's the sun, emitting light. You can ask, "What's the intensity of sunlight?" Anybody have an idea how many watts per square meter from the sun.

Student: A thousand?

Professor Ramamurti Shankar: Yes, very close to a thousand. I think there's some other numbers, but for our purposes it's very close to 1,000 watts per meter squared. That's pretty amazing over the entire surface of the earth every second the sun is pumping in 1,000 joules. And you've got to remember the context of the sun. I mean, here is the sun and here we are ninety-three million miles away, and the light energy is going and this is our share. A tiny circle like 7,000 miles in diameter you're intercepting the light, and every square meter of it gets a thousand watts. You can see the stuff coming out of the sun, its a prodigious amount of light. Anyway, that's the electric field. I mean, that's the intensity, so let me write the following thing. This is oscillating rapidly with time, so let's define an average intensity as the average of the sine squared. That's E0B0/2μ0. All these averages are simply half, anything involving in sine squared. I'm not going to worry about the half, but if you took this to be the average intensity you can ask, "How big is the electric field that comes with it," because that light is going to be this electric and magnetic field. That's all light is. So the sunlight produces electromagnetic field, is electromagnetic waves and I'm asking, "How big is the E vector?"

And all you have to do is stick that into this number. If you want you can get rid of B and go back to E because E is just B times c. You'll find a pretty surprising amount. It's roughly 1,000 volts per meter. Remember electric field is measured in volts per meter. What that means is if you took the field, and it's uniform in space, between one place and another place there's a potential difference of 1,000 volts, or it'll take 1,000 joules to shove a coulomb from lower potential to higher potential. That's a pretty strong field, but it's very incoherent in direction. It's doing this for a while and doing that for a while, but if you roughly approximate it and just ask, "What is the average field strength," this is the number you get.

Chapter 3: The Origin of Electromagnetic Waves [00:30:44]

So let's talk about one thing which I've not discussed at all, which is, where are these electromagnetic fields coming from? I said you don't need ρ, you don't need I, you don't need the current, you don't need the charge, these can exist in free space. But what's the origin of the electromagnetic waves? Anybody know? When can you get them? It didn't happen in anything we studied. Yes?

Student: Electron changes energy level?

Professor Ramamurti Shankar: Well, you can say in the atom the electron changes energy level, but before you do the quantum mechanics. In classical theory one can ask, what is the electric field? When does the electric field produce these waves? You take a static charge it has a 1/r2 field which is pinned to the charge. If you go too far from the charge you don't see it, but these guys can go off into space without being anywhere near a charge or current, but what produces them? Yes? You don't know? Yep?

Student: When it oscillates.

Professor Ramamurti Shankar: Pardon me?

Student: When is oscillates.

Professor Ramamurti Shankar: Which oscillates?

Student: The electric field.

Professor Ramamurti Shankar: The electric field is itself oscillating, but what's producing the electric field? What's the cause of the electric field? Electric field fairy? Yes?

Student: When a charge oscillates?

Professor Ramamurti Shankar: Yes. In other words, rather than oscillates the more general answer is whenever a charge accelerates. This is a very important result. Waves are produced by accelerating charges. If they travel at uniform velocity like in a wire they don't produce oscillations, electromagnetic waves, or if they're going in a circuit at constant rate that doesn't produce. But if you have charges which are say oscillating is one example when things are accelerating, right, because you're going back and forth when you radiate light. So every single source of electromagnetic wave is oscillating charges. You can sort of imagine why that will happen. I mean, if you took a capacitor plate, and you connect it to an AC source, let's say, there may be other things to keep it from burning out, then what will happen is charges will be like this for a while then they've got to go back and forth in order to change polarity with the alternating field, therefore the charges are going back and forth, and you have a time dependent electric field here.

When you have a time dependent electric field you'll have a magnetic field going around it because the line integral of B will involve the rate of change of electric flux. And that will also be time dependent, but if that gets time dependent there'll be an electric field going around that. So basically these will curl around each other whenever they're dependent on time, and they can then free themselves loose from the capacitor and take off. All you need is two plates, and an AC source, and you connect them, you will make electromagnetic waves. You'll make them at the frequency of the source, so you won't be able to see it. Your dog won't be able to see, but some gadget will be able to pick it up. That's all you need, oscillating charges. So what happens in a radio station is you could imagine a simple radio station with an LC circuit, the current is oscillating at some rate, and part of the circuit is in the antenna, and the charges are going up and down as the current goes back and forth that sends out the waves, and the waves come to your house.

So here's the picture. Here's the radio station, and the waves are emitted in big circles, and this is your house, and here's your little antenna. It's a piece of wire, and the electric field, if it's polarized this way, will move the charges up and down, and charges can be part of an LC circuit. So here are the antennas, if you like; part of the circuit. As the charge goes up and down the AC current will try to flow here. And if you tune this capacitor so that it resonates with the frequency you'll get a hefty signal from the radio station. So in the end it's all charges. Charges produce the field. Charges respond to the field. That was true in the static case. That's true in the time dependent case. Yes?

Student: When you say that the field if it's loose from the capacitor, what do you mean free from the capacitor?

Professor Ramamurti Shankar: It means it can go thousands of miles from the capacitor. And if you simply look at the Coulomb force due to the charges, the plus and minus charges, they die like 1/r2. So if you calculate 1/r2 you will get a negligible number compared to the actual electric field. So it's really like you and your parents. I mean, at some point you are free from your parents. You are able to manage on your own, but you had parents somewhere sometime, right? That's what it is. The electromagnetic waves can go on their own, but they are not produced on their own. They're produced by charges. It is just that unlike the static fields, which are very near the currents and charges that produce them, the time dependent fields propagate on their own. E keeps B alive and B keeps E alive. If E tries to die there's a dE/dT that produces a B. If the B tries to go down it produces at dB/dT that produces an E, so they go back and forth. It's really like oscillations in which you have kinetic to potential transfer. You can have energy transfer. So the fields cannot die. They are self-sustaining, but to get all of that physics you had to put the term that Mr. Maxwell put in. Without that term you don't have this phenomenon. You don't get it from statics.

So one typical problem you have is this radio station is, let's say, 100 kilowatts and you're sitting here at some distance r from the radio station. Then the intensity in your house will be 100 kilowatts spread over a sphere of radius R. That will be your intensity. Then you can go from the intensity and translate it to an electric field and say, "The electric field produced by the radio station oscillates with the following amplitude. I've got to build a circuit that's smart enough to pick up that tiny field."

Chapter 4: Relativity and Maxwell's Equations [00:37:07]

Okay, so I want to switch now to my favorite theme. The remarkable thing about electromagnetism is that you can ask what happened when physics went through the Einstein's revolution with special relativity. We know everything changed after Einstein, and all the Newtonian mechanics had to be modified. And so far I never mentioned the word relativity, so you can ask yourself, "How are these modified by Einstein's work?" So first let me tell you, remind you, now you guys did relativity last term, right? Is there anybody who's never seen it before?

Okay, well you don't have to know a whole lot, but let me just say the following. In the Newtonian world there was a principle of relativity according to which the equations like F = ma you can ask, "Who is allowed to use this equation?" It says by definition an inertial observer can use this equation. And you say, "Who is an inertial observer?" You say, "Inertial observer is anybody who can use this equation." Sort of seems to be meaningless tautology. What makes it meaningful is the following. There are some people, at least, for whom this equation works. So there are at least some inertial observers. For example, we are an inertial observer because if you want to test this equation you can say, "I leave a piece of chalk here. I don't apply force. Does it have an acceleration?" It doesn't. Okay, so I'm obeying at least the first of the three laws of Newton. On the other hand, if I leave my iPod in Grand Central Station and I come back it's gone, that's not a violation of Newton's Laws. That just means I'm stupid because there are other forces acting on that iPod, and those forces moved it, so I can understand that. So I won't be that traumatized by the loss of the iPod because I feel as I know there's an explanation. But if you go to a plane and the plane is about to take off you leave stuff on the floor it will slide to the rear end of the plane. You have no F, but you have an a.

That means an accelerating plane is a reference frame in which people cannot use F = ma. Stuff will accelerate for no apparent reason, so not everybody's inertial. So you can ask, "If I have at least one inertial observer in the universe does it imply others?," and the answer is yes. If I'm inertial and you move relative to me at constant velocity you're also inertial. There's a large number of people in the universe all allowed to use Newton's Laws if you've got one of them, and they differ by velocity, constant velocity, and we can understand that from Newton's Laws. Here is a mass and spring system. Newton's Law takes the form -kx = ma, let's say, or let's write mdv/dt. Now if you go to a train or you carry this on a train and you go at constant velocity I know it will obey this equation because I've not done anything. It's not my fault you're going on a train and this is riding with you on the train. It will obey this equation. By the way, this x is really x - x0, where x0 is the rest length of the spring and x - x0 is the deviation from that. That's the equation. Now you and I differ by what?

You and I differ by a constant velocity. The constant velocity just means x prime is equal to x - ut. Our origins are differing by an amount ut after time T, so if an event occurs here for me, you have moved a distance ut and it occurs at a distance x' which is x - ut. Then you can see the laws are going to work for you also because even though we disagree on velocity we don't disagree on the rate of change of velocity. Your velocity and mine differ by a constant, which has no derivative, so they have the same acceleration. As far as the spring is concerned if I think it's stretched by two inches you will also think it's stretched by two inches. I mean, your x is zooming to the right according to me and your x0 is also zooming to the right, but the extension of the spring is the same. So F = ma doesn't have to be modified by going to a moving frame. And that is the relativity of Newtonian mechanics. If these are the laws of motion we can understand why if you're inside a train, which is completely closed, you cannot look outside and it's going at uniform velocity with respect to the ground, you cannot tell. You cannot tell because nothing you do will be different. Because everything you observe in the world is controlled by Newton's Laws, and Newton's Laws are unaffected by adding a constant velocity to everything. So when you wake up and I say, "Is the train moving," you cannot tell.

Okay, so that's by doing physics experiments. You cannot say, "It says Amtrak so I know it's not moving." That kind of argument is based on sociological axioms. I'm just saying can you — after all it's possible, theoretically, Amtrak trains can move, so you must admit the possibility.

Okay, now you come to Maxwell's equations and electromagnetic theory. Let me write down what we have. The first thing we have is q equals — I'm sorry, q times E + v x B is the force. Then let me write down one other consequence, d2E/dx2 − (1/c2)d2E/dt2 is 0. These are some of the results we got from electromagnetic theory. Now comes the important question. This is the velocity of a particle according to whom? That question came up earlier in the class. Who is supposed to use it? I may assume it worked for me, but how do I know then when you see it it's got a different velocity will you get the same physical world that I get. Now let's look at this equation, this also came from Maxwell's equations, and compare the equation for a string. Let me call ψ as the displacement of the string rather than y, d2ψ/dx2 is (1/v2)d2ψ/dt2. I just wrote the equation shifting to the other side. They look very similar. Here v is the velocity of the waves according to a person for whom the string is at rest. Okay, according to whom the string is at rest. You understand that? This velocity, because the waves are traveling in the string, a at speed v. So this equation is to be used in its present form only by a person for whom the string is at rest. If you want to see the string from a moving frame then x' is x - ut, and in classical mechanics t' is equal to t. You can do the change of variables with these partial derivatives. I don't want to do that, but you can say d/dx of ψ is equal to d/dx' of ψ times dx' over dx, then dψ/dt', then dt' over dx. Now t' is the same as t, but formally we can change variables, and we can take this equation and rewrite it in a frame moving to the right at speed u, and I promise you it won't look like this. It will look very different. More importantly at least understand conceptually that the velocity u of the moving observer will appear in the final equations, because these derivatives, x' over x and so on contain the velocity u. Can you see that, dx'/dt has the velocity u? So when you make all these changes and put them in you'll get a new equation involving x' and t' in which the velocity u will appear. So if you are that person you have to decide what's the velocity to use for you, and the answer is unique. It's your velocity relative to the string. The string is anchored in the lab. If you happen to have a speed u relative to that that's the speed you should put in. Therefore, the equation is not the same for everybody. There is a special observer for whom the string is at rest, namely in the laboratory frame for whom this equation works, and v is the velocity for that person. Now we come to this equation. It has no reference to the velocity of the observer. It's got a velocity of light, and you can ask who is supposed to use it. Whereas in the string we know the privileged frame of reference is where the string is nailed down, but the light is traveling in vacuum.

There is no frame of reference. People thought maybe even the vacuum contains a medium called ether. Because everything needs a medium to travel they said there is an ether. Then, of course, this is to be used only by people who live in that ether at rest, but we are moving relative to the ether because we are on the earth which is going around the sun. You may say, "Well, maybe today I just happen to be at rest relative to the ether. That's possible, but then tomorrow I cannot be because I'm going around the sun. Six months from now I'm going the opposite way around the sun at a huge speed, but what I find is every single day of the year I'm able to use these equations. That means they apply to me no matter what my velocity is. So these equations, it turns out, are valid for any observer who is inertial, namely one for whom at low velocities Newton's Laws apply. Now what people were worried about in the old days they said, "Look, let's take x'x - ut and t' equal to t and put it into this equation." Then they found the equation changed their form because it's just like this wave equation here. Then they said, "We've got to change the equation because it depends on our speed u, but there is no valid choice for what our u is. What is our speed relative to this magical ether?"

That was what they were worried about until Einstein came and said, "There is no either and this is the wrong set of transformations." If you use x' equal to x - ut divided by this 1 − u2 over c2 and t' is t - ux over c2 divided by the same square root, if you change coordinates this way you'll find amazingly, if you change all the d/dx's to d/dx' and did the whole partial derivatives and chain rule and so on you will find that in the new frame of reference you'll find the equation will look like this. It will look the same for anybody moving relative to me at any speed. That's why it's not clear that I'm the privileged user. All people at uniform relative motion can use the very same equation with the very same number c entering. So this was the great triumph of the Maxwell theory. It was that it was already consist with the relativity. In fact it is what led to Einstein's revolution because this equation said no matter who you are a light pulse is going to travel at a speed c for you, no matter who you are. That's very strange because every signal we know has a property that if you move along the signal its speed is reduced, right?

If you've got a bullet going at 700 miles per second if you travel at 400 you will think it is going at 300, but if it's a beam of light it's supposed to have the same velocity for everybody, even those moving in the same direction and the opposite direction. It doesn't matter. Therefore, something had to change with our definition of space, and time, and velocities, and Einstein replaced it with these new equations. And one of the consequences of that equation is that if I have an object that is going at a speed v and they're moving the same direction at the speed u in the old days you will subtract your speed in the same direction, but the correct answer is — this is 1 - uv over c2. This is the relativistic equation about how to change velocities. v is the speed of an object according to me. You are moving in the same direction at speed u. You will measure the speed w given by this. If u and v are much smaller than c you can forget this and it looks like the good old Newtonian days, but if u and v are comparable to c then the denominator is one less than, one minus something, so the whole thing will be a little bigger than what you thought because you're dividing by 1 minus something.

And finally, if what I was looking at was a light pulse, that means v is equal to c, I get c - u divided by 1 - uc over c2. And if you fiddle with that you'll find it's c. So this law of transformation of velocity has the amazing property that if I'm observing a light pulse that's got a speed c you will also get a speed c. So it's a very beautiful way in which the mystery was resolved. So Maxwell gave these equations, did not give a preferred frame of reference. Then you've got to ask yourself, "What coordinate transformation should exist so that the equation has the same form for everybody," because it doesn't tell you who supposed to use it. Then you get this equation. You can get to this equation simply by demanding that two observers looking at a light pulse somehow get the same speed. If you fiddle with that and the symmetry between the two observers you will get this. Anyway, I just wanted to tell you that there are many things you have to change, but you don't have to change any of electromagnetic theory. The equations I wrote down are correct.

Chapter 5: Deducing the Presence of Magnetism [00:51:46]

Okay, the last thing I want to do is something I promised long ago, which is the following thing. If you believe in relativity, namely if you believe that the laws of physics should have the same form for people in uniform relative motion, you can deduce the presence of magnetism given just the presence of electrostatics. In other words, suppose you never heard of magnetism. I can show you that it must exist. magnetic forces must exist, and I show that as follows. Before I show that you need a couple of results that you guys may not remember all the time. First result is if you've got a wire and it's got n is the number of carriers, charge carriers per unit volume, and A is the cross section of the wire, and e is the charge of the carrier, and v is the velocity of the carrier, then the current is equal to n, sorry, nAve. Let me tell you why that is true. Look at the wire. I'm going to stretch it a little bit so it looks like a thing with a finite cross section. It's just like what I did earlier on. If you wait 1 second the carriers in that cylinder will have crossed the checkpoint, and the volume of that region is A times v. This is the number of carriers per unit volume, and each one carries charge e, that many coulombs would have gone past this point and that's the meaning of current, first thing to know.

Second thing to know, if I took a rod of length L and I put some charges on it, and they had a certain density n0, if the rod moves at the velocity v you know it will shrink, therefore these plus signs will be compressed, and to a person seeing the moving rod the density n will be n0 (the density at rest) divided by 1 − v2 over c2. This is a relativistic effect: that the number of charges get squeezed because the rod itself gets squeezed. So a moving rod which is charged will appear to have a higher charge density. And the final result I'm going to invoke is the following. What is the charge per unit length? If the charge per unit volume is n then I claim the answer is n times a. That's also easy to understand. If you took a unit length of this wire, unit length, the volume of that is just A times 1 and that times the density is the amount of charge that's there. So if you have a very thin wire you may like to think about charge per unit length rather than charge per unit volume, and this is the way to go from one to the other. Now we are ready.

Now I'm ready to show you how just by thinking you can deduce the presence of magnetism. And I like this argument because quite often this is how people make, theorists make discoveries. They will take something that's known. They'll appeal to a principle like symmetry, or relativity or whatever. Then they will say, "This implies that there is a new force, and I'm going to tell you what the new force is."

So here's what I want you to imagine. There's a very long infinite line of charge and right on top of it there's another infinite line of negative charge. If they're just sitting there there's nothing interesting. Now what I want to do is I want to have the upper thing going at a speed v. Now n0+is charge density of plus charges in the rest frame of the rod, of the plus rod, and likewise minus. Are you with me? Each rod has a certain density when it's at rest. I'm going to call that with a subscript zero. Now I want to arrange this wire to be electrically neutral. I want it to be neutral, and I'm producing the current by moving the plus charges to the right. I'm dragging them. The minus charges are not doing anything.

So what is the current? The current is going to be the density of plus charges, times the velocity, times area, times the value of each charge. But for the wire to be neutral — this is where I want you to follow me closely. The two of them cannot have the same densities at rest because if the plus charge had the same density at rest as the minus charge when it starts moving and it starts compressing the plus density will exceed the minus density and the wire will be neutral. So I cook it up so that n0+ with this factor is the density of minus charges at rest. Are you with me? I take a rod with positive charge somewhat less than the density that the negative charges are going to have, but by moving it at the suitable speed I bring these two densities to the same value. So I want you to follow this in some detail because it's not so difficult. This is the neutrality condition for the wire. I want the wire to be neutral. See normally when you don't think in relativist terms you'll say, "Take a plus charge rod and a minus charge rod and make sure the charges are equal per length, and drag one of them, or you've got a current." That works in a non-relativistic limit, but if you take into account length contraction you won't get neutrality unless the plus charge had a slightly smaller density in its own rest frame that got boosted by this factor to equal the density of negative charge.

Okay, now imagine a particle at rest here, and you know all about electrostatics. What do you think it will do? You've never heard of magnetism. You've heard of electrostatics. What will this charge do? Yep?

Student: It is positive. It'd probably be attracted.

Professor Ramamurti Shankar: Why would it be attracted?

Student: Because the negative wire has more density and...

Professor Ramamurti Shankar: No, no. That's what I said. No. I arranged it so that the positive wire, the moving positive wire has the same density as the static negative wire. You understand? I cooked it up so that the wire is neutral. All I'm telling you is, if one set of charges are sliding to the right at a certain speed that rod must have a density that's somewhat lower in its own rest frame, but by the time you translate it to the laboratory frame it's electrically neutral. So what'll this charge do is my question.

Student: [inaudible]

Professor Ramamurti Shankar: Pardon me? It should not experience any electrical force, you understand that? Our prediction is that it will just stay there because it's neutral and all we know is electrostatics. It can have any velocity, but I want it to have a velocity v. Electrostatics doesn't care what the velocity of the charge is. It's still supposed to go in a straight line because the wire is neutral. But now I say I'm allowed to view the physics from any frame of reference I want. Let me go to the frame of reference of this charge. I move to the right. So in general, if you want to analyze the problem carefully, you can have this wire moving at one speed and this charge moving at a different speed, but I want to do the easier case where they're both the same. It makes life easier. So go to the rest frame of charge. You can already see what's going to happen in the rest frame of the charge. If you go to the frame moving with the particle the plus wire comes to rest, so the density of the plus charges will just become n0+, the rest density of the plus charge, but the minus charges now they will be boosted by this factor because they're going the opposite way, and it doesn't matter which way they're moving. Things depend on v squared. The minus charges will appear compressed.

So if previously they're balanced now they won't balance. In fact, n- is n0- over this, which is n0+ divided by 1 - v2/c2. Therefore the wire will appear to be negatively charged because the negative charges are that value, the positive charges are that value, so net charge on the wire, net charge density will be equal to -n0+over 1 - v2/c2 + n0+. If you like, this is the plus charge and that's the minus charge times e, if you like. And I'm going to make an approximation. I'm going to approximate this is n0+times 1 + v2/c2. There are more terms, but in the binomial expansion I'm going to stop with the first term. So this is the result that's very good in the limit of small v/c, but v/c is not set equal to 0. It is set equal to a finite but small amount. So if you compare those two you will find the net charge is - n0+ v2/c2. That's going to be the charge on the wire. So to the person moving with this charge the wire looks charged. In fact it looks negatively charged. So you know, knowing just electrostatics, what the charge will do. The charge will be drawn to the wire, start moving towards the wire. That's a fact.

Well, if it's moving towards the wire for me it better be moving towards the wire for the original person also because even if you and I move horizontally the fact that your moving transverse to a velocity is going to be a true statement in both frames. From that you will conclude that even in the original laboratory frame this moving charge would have been attracted to that wire, and it's attracted by virtue of its velocity. So that means there is a new force in which a charge is attracted to the current in the same direction, right? So let's ask, "What is the force of attraction?" In Newtonian mechanics — I mean, at this point once you've got this length contraction you can just think in terms of Newtonian formulas because the errors you make involve higher powers of v2/c2. The leading term is this. So look at that. You have a net charge of n+0 v2 /c2. You have a charge per unit length which is the area of the wire times that. I sorry, that's charge e. That's the density. That's the actual charge. That's the λ. So what's the electric field of a wire with charge per unit λ? I remind you it's λ/2Πε0 r.

So for this particular lambda it's n+0 v2 over c2 eA times ½Πε0r. In other words in its particle rest frame there'll be an electrical field because there's a net charge, negative charge, of that strength on the wire. Once you've got a charge per unit length λ/2Πε0r is the field attracting it toward the center, so that's the force. So I'm going to write it as follows. I'm going to write it as n+0veA1 over 2Πε0 c2 times another e and another v. This is the force, this e times E. In Newtonian approximation the force perpendicular to the motion is the same for all observers, therefore I ask you — and this force must be present even in the lab, but look at this force. This guy is the current, nevA is the current. This guy 1/ε0c2 is μ0/2Π. This is ev. You see, this is the exactly the magnetic force we got, μ0I/2Πr. I forgot a 1/r here. That is the azimuthal magnetic field around the wire that times ev is the force. Because you can tell that no matter where the charge was, whether it was here, or whether it was there or anywhere around the wire it'll have the same attraction.

So you can deduce in the laboratory frame there is an attractive force towards the wire, and that agrees with the exact formula you got for the magnetism. If you don't ignore the higher powers of v2/c2 you will find that the force in the lab frame and the force in the moving frame are slightly different. That's because in relativistic theories the force is not the same for everybody. The force for me and the force for you do not actually agree because time is not the same for everybody. But the leading order in v2/c2 you can see this. There's only one thing I did which is a little fudge, which you guys may not have noticed, is that the density here is n+0whereas the actual current in the lab is n+and the two of them differ by this factor. However, if you have a v2/c2 in front of your expression and you've ignored v4 or c4 there is no point in keeping this guy here because that makes an error of order v4/c4. If I'm going to keep that I should go back right here and keep such terms. So the leading order you don't have to worry about the difference in density. This is a very subtle calculation because sometimes you worry about the difference and sometimes you don't. So the rule always is if you're trying to do calculations to a certain order, namely v/c the whole thing squared then things that make corrections of order v/c to the fourth can be dropped.

Anyway, I'm not that worried about the details, but I want you to understand at least what the logic is. That's more important. I know about electrostatics. I cleared the neutral current carrying wire. It's neutral because the negative charges are at rest at some density. The positive charges in the moving rod are also at the same density, but the rest density is somewhat lower. Then I predicted the particle won't move because it's got no electric attraction. Then I go to the rest frame of the particle, then I find that the positive charges have come to rest, and therefore to a lower density. Negative charges are moving the other way, therefore at the higher density. They no longer cancel.

The wire is charged. I expect this charge to be attracted to the wire. That means going back to the lab I expect the moving charge also to be attracted to the wire because when something goes towards the wire that goes towards the wire according to all people. But you can go beyond and actually compute the force and equate the force and deduce that that is a new force now. Whenever you have a current carrying wire with current I, and that's a particle of charge e moving at speed v it will be attracted to the wire by this amount.

Okay, so I'll see you guys after the break, and my suggestion to you on what to do with the break is to try to carry your textbook with you, and the textbook has got lots of problems, some of which are a lot simpler than the one I gave in the early days, but roughly the same level of difficulty as the midterm. And do as many problems as you can. Look at the worked examples and try to solve them.

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