Stanford University. So the goal of today was to show you how a particle like the photon can get a mass from something

called spontaneous symmetry breaking, how it can gain a mass from something called the Higgs phenomenon, two important ingredients that I haven't emphasized tonight, both of them technical,

but also one of them more technical than the other. One of them called spontaneous symmetry breaking, and the other called gauge invariance. We have to go through them if we really want to seriously understand what's going on. They're mathematical ideas, and so we can't avoid

mathematics, as usual. Last time, I think I discussed a little bit an example of spontaneous symmetry breaking, namely a very simple model of magnets, where the magnets, little magnets, individual magnets, can either point up or down,

and where the energy is such that the energy is lower when they're parallel, either this way or this way, and larger if they're anti-parallel. There is no fundamental asymmetry between up and down. Two neighboring up little magnets have the same energy

as two neighboring down magnets. One up and one down has the same energy as one down and one up. So there's no fundamental difference between up, up, and down. But nevertheless, because the spins are,

because it's energetically favorable for them to be parallel, then all you have to do is go very, very far away. Let's imagine now we're talking about the ground state, the state of lowest energy. This magnet has been frozen. All its energy sucked out of it. Which state is it in? You can think of two possibilities.

They could all be up, or they could all be down. One up and one down, anyways, will cost more energy. But which is it? Are they all up, or are they all down? Well, the answer is determined by one spin, very, very far away. You go very, very far away, and you freeze one of them, just one spin very, very far away.

The effect will be that its neighbors, in order to save energy, will want to be parallel to it, all right? So let's say the four are the six neighbors around it, depending on the dimension of space. In three dimensions, there are six neighbors on a square, on a cubic lattice. So it will tell its neighbors, look, you want to save energy?

Point, the same way I point. But then the six neighbors have a bunch of neighbors connected to them, and so forth and so on. And the effect will be that the entire lattice, if we really want to be in the ground state, will all point in one direction.

So there is a symmetry, but the symmetry is spontaneously broken. Now, let's just distinguish that and compare it with what I will call explicit symmetry breaking. Explicit symmetry breaking is different. Imagine that instead, a slightly different system,

perhaps the same system, but one added ingredient. There's a magnetic field, and the magnetic field points up. In that case, just an individual spin, one spin at a time, forget its neighbors, will prefer to point along the direction of the magnetic

field. Prefer, what does that mean? It means the energy will be lower when it's pointing along the magnetic field. This is an explicit breaking. There's something in the system which is single down, up as different than down. The expression for the energy is no longer symmetric.

If all I had was one spin, the energy would be different if it were up and down. And in that case, it really doesn't matter what you do very, very far away. It doesn't make much difference. There's an enormous savings of energy if all of the spins are pointing along the magnetic field.

If you try to take a substantial number of them and flip them in the wrong direction, the energy cost would be enormous. It would be proportional to the total number of spins that you try to change. So one spin far away, holding it in some direction, is not going to have much influence far away.

Why? Because if it did have an influence far away and lines things up with it, the cost and energy would be enormous. So that's explicit symmetry breaking. Does a particle with no charge but spin have a preferred orientation in magnetic field? It has a magnetic moment.

No charge, but magnetic moments. Like the neutron. Neutron has a magnetic moment, but no charge. By spontaneous, you mean they would all flip at the same time, or would they propagate out? Well, that's a good question. I mean, in the real world, they propagate out.

Yeah. Yeah. That's a good question. If you had all the spins lined up in one way, in some way, and then you, spontaneous case, and you came and you flipped one of the spins and held it down, yes, a wave would go out. Just like stack dominoes.

Just like stack dominoes. Yeah. Except in three dimensions. You'd have multiple waves basically moving through a material in a typical case, wouldn't you? You would have multiple waves moving through a large material at any given point from multiple distant

influencers affecting different waves. We're going to talk about waves of reorientation. Yeah, we're going to talk extensively about waves of reorientation. Okay, so question. How would you know?

You find one of these magnetic systems, not a real magnet, it's just a mathematical model with spins which can either be up or down. You find a sample and it's all pointing up. You want to find out whether it's spontaneous symmetry breaking or let's call it explicit symmetry breaking.

What's the difference? Are there any phenomena which are characteristic of one versus the other? The answer is yes. In the case of the simple magnets which can either be up or down, the special phenomenon associated with the spontaneous symmetry breaking is the existence of domain walls. What is a domain wall?

If you had a big sample, there's a big sample here, and you decided to do an experiment, there's a finite-sized sample but very big. So you go out to the boundaries, let's say the left

boundary, and you make all the spins up. You just freeze them. You hold on to them, perhaps by putting the spins on the boundary over here, incidentally this is not time, two space directions just for simplicity, you freeze all of the spins upward over here.

What do you expect to happen? Well, of course, the system will save energy by filling itself up with all up spins, parallel spins, not because there's anything around here which is favoring up spins, but just because each neighbor influences all its neighbors and says, you'll save energy if you're up.

How do you tell the difference between that and explicit symmetry breaking, which is the presence of a magnetic field in the bulk of this sample here, orienting the spins up? Well, one experiment you could do is you could go over to this side over here and place all of the spins down.

Then, what do you expect in the two cases? In the two cases, you expect quite different behavior. If there really was a preference, an explicit symmetry breaking that said, you save energy by being up,

then the effect of what's going on over here would not be very important. It might propagate in a little ways, but it certainly would not flip a large macroscopic percentage of the spins because that just costs energy to turn them relative to the explicit magnetic field.

So if there was a magnetic field pointing up, the effect of turning them down here would only propagate a little ways in and would be too costly in energy to turn over any large macroscopic fraction of them. On the other hand, if up is just as good as down, in

other words, if there is nothing external to the system saying up is preferred, then the effect is quite different. Instead, the favorable energy configuration will be one in which half the sample on one side is down and half the

sample on the other side is up. Now the split might not be at the center. That's not so important, but this would be a perfectly good configuration where half of them would be up, half of them would be down. You cannot find a state with lower energy.

There is no state of lower energy than half being up and half being down. This would be called a domain wall. The separation between the two types of behaviors would be called a domain wall. So for this simple example, the phenomenon that would

characterize spontaneous symmetry breaking would be the existence of domain walls where on either side of the domain wall there would be basically identical things except reflected by 180 degrees. Domain walls would be the existence of domain walls would

be the signature of spontaneous symmetry breakdown for this case. Now we can think of a field theory which has very similar behaviors. Here's a field theory. We have a field.

Let's just call it phi. Is green okay? Does green work all right? Say it again. Black is better. All right. Let's try. What did I do with the black? Here it is. All right.

A field phi. Now for simplicity now, phi is a real valued field, only one component, and it can take on any value, any real value from plus infinity to minus infinity, and there's a dynamics to this field which is controlled by a Lagrangian.

Let's write a Lagrangian for it. It's a relativistic field and the usual expression for the kinetic energy of a relativistic field is to differentiate phi with respect to the direction mu and square that.

Well, you square it with some signs, plus signs and minus signs that are characteristic of relativity and that we describe by putting upper and lower indices, but basically this is just the sums, actually differences of squares of the time derivatives squared, sorry, the difference between the time derivatives squared and the space derivatives

squared. This thing here is one half phi dot squared minus the derivative with respect to x of phi squared, same thing for y and so forth and z, and we'll just represent that by d mu phi, d mu phi. We could even represent it more simply.

Let's even represent it by an even simpler notation. Let's just call it one half, and the one half is completely conventional, meaning to say it's a convention, but you could change it. It wouldn't make any difference to the physics. A simpler notation would just be to write d phi squared

and it means phi dot squared minus gradient of phi squared, and then you can add another term, which is a potential energy, a potential energy minus v of phi, minus a potential, which depends on phi, potential energy.

It doesn't depend on the gradients of phi, the time derivatives, the space derivatives, it just depends on phi itself, and imagine that v of phi looks something like this, completely symmetric, exactly symmetric with respect to changing plus to minus, all right?

So here would be phi equals zero, right over here. Let's call this point phi equals f, and here phi equals minus f, plus f and minus f, and plus f and minus f are completely symmetric. This function, assume it, is completely symmetric with

respect to flipping about the vertical axis, and under that case, there will once again be two ground states of a field, two states of lowest energy, incidentally the energy has a plus sign for v of phi, the Lagrangian has a minus, the energy has a plus sign.

There will be two states. In one configuration, the field will sit at this point over here. Everywhere is in space, so everywhere in our sample or in space, phi could be minus f. Now why can't it make a jump, obviously the energy is

lowest if phi is at this value or this value, but why can't it jump from here to here without a big cost in energy? Well, supposing the field just went, just a sharp

boundary here, phi equals f, phi equals minus f. Everywhere in this sample, phi is equal to either plus or minus f, and so this potential energy here is minimized, but what about the derivative energies here?

What about the terms in the energy proportional to, let's say, the derivative of phi with respect to x squared? This direction could be x. Then there's a huge jump or a very sudden jump in phi going from here to here.

A sudden jump means a large derivative. In fact, if the jump was sudden, meaning really concentrated, the derivative of phi would be very big, the square of it would be even bigger, and you would pay a large price in energy for a sudden jump.

On the other hand, you pay no price if phi is either equal to f or minus f everywhere. This is very much like the magnets lined up, and again, is a case of spontaneous symmetry breaking.

A different possibility where there would be no spontaneous symmetry breaking would be if the potential energy instead looked like this. Then the ground state would just be phi equals zero.

Well, phi equals zero is the same as phi equals minus zero, and there would be no bias. You would not observe anything in here. You would just have phi equals zero everywheres, and there would be no breaking of the symmetry of left to right

or of plus phi to minus phi. Okay, so that would be the case of a spontaneous, this would be the spontaneously broken case, this would be the unbroken case. In one case, the ground state simply has the field equal to zero. In the other case, there are two ground states, two

identical ground, well, physically identical ground states. And again, in this case here, if you insisted that the field was plus f on this end and minus f on this end by going into the system and holding it down, then there would be a transition somewheres in the middle from minus

f to plus f. It would be costly in energy because of these gradient terms, but still, you have no choice. You said it's plus on this end, minus on this end, and there would be a domain wall down the center. Yeah. So are you assuming that there's no influence from second

nearest neighbors, only the nearest neighbors? No, we're not assuming that. So why wouldn't everything just be random, everything up and down, up and down, everywhere? Because if you look at the secondary nearest neighbors, they would be in the same orientation as the... As long as the second nearest neighbor interactions also

preferred parallelness, it wouldn't make much of a difference. If you had a situation where the first nearest neighbors preferred parallel and the second nearest neighbors anti-parallel, you might get into some kind of competition. Yeah.

No. It refers to the idea that there's really no fundamental asymmetry. There's a question of chance, which way it goes, or if not chance, that you can set off the difference by a very, very tiny perturbation, very tiny perturbation, just one

spin... It is a kind of instability. It's a kind of instability that... What does an instability mean? An instability means that a very, very small perturbation on the system will change it a lot. All right? So small in this case may mean that just one spin far

away is being tampered with. You ordinarily think that's not much of a change in the system. You go far, you have a system which is a million zillion miles by a million million zillion miles, and it's loaded up with a huge number of these spins, and somebody off on Alpha Centauri turns one of them over.

And what's the effect? The effect would be of a propagating wave, which would come in and it would take some time, of course, but which would cause, if you sucked all the energy out of the system, would cause the ground state to be, to realign itself. Is it starting out in one state and then it gets flipped

to the other? Or is it starting out in a bit of unknown? Well, it depends on the particular physical situation. We'll say a little bit more about it. At high temperatures, where there's lots of random fluctuation, it may be, and it typically would be, that the random fluctuation would be such that on the

average, there would be equal amounts of up and down, random fluctuation. Now you start to cool it. You cool it and take energy out of it and take energy out of it. Which way is it going to go? Well, the answer is that it's a sort of random thing.

A patch might form plus over here, minus over here. Another, some patchwork might start to form to lower the energy where they were parallel. And then the patches would connect themselves together and the patches would try to align each other. And eventually, it's a matter of random chance which one will take over the whole system.

Okay? And that's what's called spontaneous symmetry breaking. Now in this case here, you could ask what would correspond to explicit symmetry breaking? Anybody guess?

If the potential had an imbalance so that one was lower than the other, then this would be an explicit symmetry breaking and the ground state would sit here.

So explicit symmetry breaking, that means it's something asymmetric in, if you like, in the Lagrangian itself or in the potential energy. Just an explicit term in the fundamental setup to begin with which favors one side or favors one configuration

instead of the other. That's explicit symmetry breaking. In that case, if you take a very, very big sample, there will not be domain walls. It would be very energetically unfavorable to have domain walls. The domain wall will simply want to minimize the number of wrong-headed, wrong-directed spins or wrong-directed fields

and simply move over and try to wipe out the wrong spins. It might not be able to completely do it, but the domain wall will not be in the center because that will require too many spins to be in the wrong state, too many degrees of freedom, too many points of space to be wrongly aligned.

Okay, so I think we have the basic idea of what a spontaneous symmetry breaking is, but now we're going to talk, and in this case, the two cases that I've discussed, the symmetry which is broken is a discrete symmetry. In the spin case, it was spin up goes to spin down.

In this case, it's left goes to right. It's not a continuous symmetry. A continuous symmetry is one where you can interpolate in between. You can interpolate continuously in between whatever configurations there are.

U1 is a continuous symmetry. We'll discuss it in a moment if I were going to discuss it right now. SU2, SU3, whatever, they're all continuous symmetries where you can interpolate configurations by using the continuous generators of the group to go from one configuration to another.

The magnetic case, real magnets, real magnets often in the real world have continuous symmetries. The continuous symmetry would be not just can you flip a spin, but you can rotate a spin. So, for example, some examples of ferromagnets, real

ferromagnets have continuous symmetries where the spins prefer to be aligned alright, but it's not just up and down, but anywhere in between. In that case, the symmetry group is not just reflection, but rotations and it's a continuous group.

So, we will want, since we're interested in things like SU2 and SU3 and U1, we want to understand what it would mean to spontaneously break a continuous symmetry. I'm going to do one example tonight of both the notion of continuous symmetry breaking, what the idea of a goldstone

boson is, which replaces domain walls, and how this can radically change its nature when the symmetry has to do with gauge bosons. When it has to do with gauge bosons and morphs from

spontaneous symmetry breaking into the Higgs phenomenon, in particular, how Higgs' phenomena gives particles like the Z boson masses, how it provides a mechanism for them to have masses. So, let's try to go through that mathematics.

Alright, first of all, we're talking, as I said, about continuous symmetries and I don't want to do the general case, it's not interesting for us at this point to do the general case. Let's do a very specific case. The specific case involves a field, let's call it phi again, but phi is now a complex field.

Instead of just existing on the real axis, instead of having a value on the real axis, it has a complex value. Or another way of saying it is it can take on any value in the plane, the two-dimensional plane. What is a plane? The plane is the plane of the real value of phi and the

imaginary value of phi. And the complex field phi is phi real plus i times phi imaginary. And we'll just call it phi. It's a field, it depends on position, but knowledge of

that field involves knowledge of the two independent components of the field. Now these components are not components in space. This is just a field which is composed out of two fields and grouped together to form something called the complex field phi. And of course, there's also a canonical conjugate, not a

canonical conjugate, a complex conjugate phi star which is just equal to phi real minus i phi imaginary. Alright, so phi is a thing which lives in the plane like

this, meaning to say the values that it takes on are in the plane. And let's suppose that the, in that case, we would write the kinetic energy here not in this form, but as the product of the derivative of the conjugate of phi

times phi itself. This is equivalent to the sum of two terms, one involving phi real and the other involving phi imaginary. You can rewrite this as, if you like, you could rewrite it as d mu phi real squared plus d mu phi imaginary

squared. That's just using the idea that the square of a complex conjugate, a complex number times its own complex conjugate is just the product of the real components or the square of the real components plus the square of the imaginary components. Alright, but let's write it this way.

In fact, the half is normally not put there for complex fields. Now, this Lagrangian has a symmetry and the symmetry is the U1 symmetry, let's write it down, where you replace everywheres phi by phi prime and phi prime is equal to

e to the minus i theta times phi. In other words, we multiply the field at every point, prime does not stand for derivative here, it's just a

redefinition of the field where we multiply it by e to the i theta. Do you know what that means on the complex plane here? Yeah, we rotate it by angle theta. Now, this Lagrangian here does have that symmetry where

at the same time phi prime star is equal to e to the minus i theta times phi prime. Did I write that? Sorry, phi star. Theta is a constant at the moment. Theta is a constant with respect to space.

It could be any phase, but it is not allowed to vary from point to point or time to time at the moment. Okay, why is this invariant under that? Because the e to the i thetas will cancel out.

The derivative, let's write down, the derivative of phi prime, I'm not going to write d by d mu, I'll just write derivative, abstract derivative, of phi prime is just e to the i theta times the derivative of phi. Why? Because theta is constant by assumption. Likewise, the derivative of phi star prime, phi prime star,

is e to the minus i theta times the derivative of phi star. That means the product of d phi prime with d star phi prime is exactly the same.

The e to the i thetas cancel out and so the Lagrangian is unchanged by rotating the field by a phase. That's called asymmetry. That's what asymmetry means, operations that you can do which don't change Lagrangian. Of course, I could write other Lagrangians which don't respect this symmetry, but what else can we add to the

Lagrangian? Let's get rid of these mu's and just call it derivative of phi times derivative of phi star. You can remember that there's some signs in there. You can add a potential. So, let's add a potential. In Lagrangian, that always means subtracting via phi.

But I want via phi to have the same symmetry. In other words, I want it to maintain the symmetry under these phase rotations. How do I do that? And the answer is very simple. You require that the potential is a function only of phi star phi.

In other words, it's only a function of phi real squared plus phi imaginary squared. Say it again. Was there a question? No. Now it's clear that this Lagrangian has the symmetry even with the potential term.

I could break the symmetry. All sorts of ways of breaking the symmetry. If I made the potential a function of some other combination, let's say just phi real by itself without phi imaginary, that would break the symmetry. Okay, what is the symmetry from the point of view of this picture over here? It is the symmetry of rotating the phi plane.

At every point, just take whatever the field is at any point in space and rotate it here. Rotate all the fields simultaneously and that is the symmetry that we're talking about. Okay, let's now focus on two different kinds of

potential energies. One of them spontaneously breaks the symmetry, the other doesn't. The situation is very, very similar to the previous case. First of all, as a simple example, let's take a potential, well, let's draw it.

No, let's write it. Let's just take the potential phi star phi. Phi star phi, where is phi star phi minimized? At zero. It's always positive except when it's zero and so the potential is minimum right over here. That point is obviously a symmetric point with respect to

rotating the plane and so a potential like this, which I'll draw for you in a minute, I'll show you, we'll draw it in a minute, a potential like that is symmetric and there is no either explicit or spontaneous symmetry

breaking associated with it. The ground state is just phi equals zero, completely symmetric. But now let's write another potential. Let's write minus A times this plus B times phi star phi squared.

This potential has the property that near the origin, incidentally, near the origin this is bigger than this. When phi is small, phi to the fourth, this is like phi to the fourth, phi to the fourth is much smaller than phi squared. So near the origin, the potential energy is not minimum

at zero but it's maximum at zero. The field doesn't want to be at zero. The field doesn't want to be zero because it costs more energy for it to be zero than to move away from zero. Potential energy because of the minus sign here favors having phi move away from zero. On the other hand, it doesn't want to move too far from

zero if B is positive. Let's suppose B is positive. If B is positive, for large phi this is bigger than this. So when this is bigger than this, you don't want to have too much field because this will get big and overwhelm this.

What will happen? The competition between these two will take place and there will be a favorable value of phi star phi at some value. We can draw a picture. Let's draw a picture. There's a picture of the potential. Here's the phi plane. I'll make the phi plane horizontal. And vertically is the potential energy.

In the first case, the first case was plus and I didn't even write this here but we could. What does the potential look like? It's a function only of phi star phi. That means it's a function only of the distance from the origin and the potential would look like something like

this. It would be a paraboloid which would have complete rotational symmetry about phi equals zero. Here the potential energy is minimum. That's phi equals zero.

And if we move away from phi equals zero, there's a symmetry which makes sure that the potential energy doesn't depend on angle. That's the character of the first case where we just had plus A phi star phi and in fact we could put in another term. It doesn't matter. The only specific ingredient here is that the potential is

upward and continues to go upward so that the minimum of energy is right at phi equals zero. Under those circumstances, we would say there's no symmetry breaking, no spontaneous symmetry breaking. On the other hand, let's go back to the case with minus here. Then the potential looks like this.

It starts out at the origin and goes negative until the quartic term becomes as large as the quadratic term. And then it turns up.

It's still symmetric about rotation. It's a little bit hard to draw. It's a surface of revolution, meaning to say it's symmetric with respect to rotation about this axis here. And the minimum is not a point, but the minimum lies on a

circle. It lies on a circle in the complex plane. From this point of view over here, the minimum energy state is somewheres along a circle of a radius which is determined by the parameters here.

You simply minimize this function as a function of phi star phi and you find out where that point is. Let's call the radius F. That's a standard notation for it, incidentally. The minimum of the potential is not at a point. It's along a track. It's sort of along a track like this, but a track of radius

F. All right, now, what is the ground state of this system? There are many ground states and they're all of the same energy. They correspond to the field being at any point along the circle here, everywheres.

A way to think about this is to imagine that the field is a bunch of little arrows. Not arrows in space, arrows in this abstract space. They want to be aligned along the same direction. How do I know that?

Because if there's variation in the field from one point to another, then this term has a lot of energy in it. Okay, the gradient term. So, the field wants to stay parallel in this internal abstract space there. Everywheres in space, the field will want to lie at the

same angle, but there's an ambiguity. Which direction? The energy is the same along any direction as long as the field has magnitude F. In other words, as long as it's stuck in this track here

or equivalently in this track over here, then the energy is minimum. Can you say anything beyond that? No. That's it. The energy will be minimum somewheres, everywheres along here, but it will cost energy to cause it to vary from

point to point. Even if it stays in the track here, there will be a cost in energy of having it change from point to point. So, that's the ground state. The ground state is any one of these possible positions for the field, but always sort of locked together. Again, spontaneous symmetry breaking, but this time the

symmetry is a symmetry of rotation about here. Alright, now, are there domain walls? Interesting question. In this case, are there domain walls? So, let's go back to the Gedanken experiment that we might imagine to check whether there were domain walls. We take a big sample and one place in the sample we put

the field at this point over here. I don't know. Let's just call it A. And at the other end, we put it at B.

So, along here, the field lies in the A direction. Along here, it lies in the B direction. What do you think the ground state, the state of lowest energy, subject to this constraint is? Is it that the field is uniform until it gets to some

point and then makes a rapid transition to the other value? The answer is no. That's not the preferred energy configuration. And the reason is that cost a good deal of energy because making the transition from A to B suddenly has the effect of making the gradient terms in the energy large.

A much cheaper thing to do is to have the field gradually vary from here to here. This was not an option in the case where the symmetry was just a reflection symmetry. But now we have the option of going smoothly from A to B. So, the field at one end of space is at A and gradually

as we move across space, it varies staying in the track here. Staying in the track so that it never costs us much potential energy. There are two possible ways to go.

That's right. There are two possible ways. So, let's resolve that by not putting this at point B. Let's put it over here to resolve that ambiguity. But you're right. There is an ambiguity about which way you go from one point to another. If this point is over here, not at the extreme diametrically opposite point, it will pay to go this way

and not go around the long way because the field gradient will be less. Yeah, if it was over here, you could get it trapped the other way. But given a pair of points A and B, unless you have this very singular situation where they're exactly diametrically opposed, there will be a particular best state.

But it will be a state with smooth variation from one end to the other. In other words, effectively, there's a field in the system for which it does not cost much energy to vary smoothly

from one place to another. The energy cost in having a gradient, that's all there is. Only the energy cost in having a gradient and nothing else. A field with that character, not phi itself, but we'll

discuss what field it is. A field of that character where there's only gradient energy is massless. It corresponds to a massless particle.

Okay, so it is a theorem. This you can prove that the way to go from A to B in the minimum energy is to have this angle very linearly from

one side to another. That's not hard to prove. And not going away from the ground state, just change the angle linearly. That's a theorem you can prove, not too hard to prove. So that is, and how do you prove it?

You prove it by using some calculus of variation type minimization and you prove the answer is linear. At this point, we don't have any, it's minimum all the way around.

Yeah, we have not tilted the potential. You could imagine an explicit symmetry breaking which would correspond to tilting the potential a little bit. If you tilted it, then one point would be preferred relative to the other and it would be the ground state. Okay, let's discuss from the point of view of wave

equations. I just want to remind you that from the point of view of wave equations, a field with only gradient terms in its energy is massless. What does it mean that it's massless?

It means that the quanta associated with it are massless, but it also means if you imagine a wave, a wave of wave number k, in other words, a wave which looked like e to the ikx going from one point to another, just a single

field now. Let's forget this complex field for a minute. Just a field which goes like e to the ikx, then we would say that it has a momentum k. It has a momentum k, k is related to the wave number, it has a momentum k, and if the energy in that field goes to

zero as k goes to zero, that means that the quanta of that field have zero energy when the momentum goes to zero. K is like momentum. If the energy stored in a field with k equals zero

vanishes, then that says there's no energy for a particle of very, very low wavelength, a very, very long wavelength, excuse me, very low wave number. So, and that would correspond to saying a particle at rest would have no energy. If a particle at rest has no energy, it's called a massless particle.

So, if the energy of a wave gets lower and lower, as the wave gets longer and longer wavelength, in other words, then you call that a massless particle. That's the case of a massless particle.

Okay, so let's look at various potentials. Let's start first of all with a potential which looks like this. M squared over two, phi squared or phi star phi, it doesn't matter.

It's clear that if you created a wave that varied from place to place, no matter how long that wavelength was, as long as there was variation in the field away from phi equals zero, there's going to be an appreciable cost of energy.

Why? Because any place where the field is not zero, in particular if it's a wave like this, there's going to be a lot of places where it's not zero and those places are going to cost energy. That is the case of a massive particle.

That's the case of a particle which has energy no matter how long the wavelength of the particle is. If this is absent, then a very, very long wavelength has very small energy. The only terms in the energy come from the gradient

terms and very, very long wavelengths have small gradients. The gradients are very, very small when the wavelength gets small and that's the case of a massless particle. A term like M squared phi squared is called a mass term and it represents the mass of a particle. We've discussed this before so I won't go through it in

detail again, but let me just show you how you can think about decomposing. In this case here where the field likes to lie on this circle here, let's go through a little bit of mathematics and see if we can identify why there's a

massless particle and what that massless particle is. The massless particle incidentally just corresponds to a wave where the field stays in the lowest energy here. If it stays in the lowest energy, then the long wavelength excitations, that just means the field varies

gradually from place to place, only an angle and never creeps out of this track here that will have very low energy and will correspond to something massless. But let's see if we can see some mathematics for it.

So let's take the potential to be derivative of, let's take the Lagrangian to be derivative of phi times the star phi. And now let's represent the field in a new way, not by

the real and imaginary part of the field, but let's write phi is equal to rho times e to the i alpha. What is alpha? Alpha is the angle on the complex plane, rho is the

radial value of the field, in other words the magnitude of the field and alpha is the angle. Rho and alpha are also fields. Rho and alpha can vary from place to place and let's rewrite the Lagrangian in terms of rho and alpha.

Now notice, alpha, this is alpha not phi, notice moving around on the track here corresponds to keeping rho constant and varying alpha. So you can think of alpha as a field whose value is on this circle here if you like or at least it has to do

with an angle. Rho is a field which when you change it, changes the potential energy and pushes you away from the minimum energy. Okay, let's try to write down what the Lagrangian is in terms of rho and alpha, very easy. The derivative of phi with respect to any direction in

space-time, the derivative of phi is equal to the derivative of rho times e to the i alpha, that's differentiating this factor leaving this one alone and then the other term is plus rho times the derivative

of e to the i alpha. What's that? Did I leave out, sorry, I should have put rho here. Sorry, what am I writing, derivative of rho, thank you. Derivative of rho times e to the i alpha plus rho times

the derivative of e to the i alpha and that's i rho times the derivative of alpha all times e to the i alpha. Let's factor out e to the i alpha and put it on the outside.

So that's the derivative of phi. What about the derivative of phi star? Let's multiply, let's work that out. Derivative of phi star, that's exactly the same thing except with a minus sign over here, derivative of rho minus i rho derivative of alpha times e to the minus i

alpha. Now what happens when we multiply these two? When we multiply these two, the cross term goes away and we just get two terms. One of them is just a derivative of rho squared.

The e to the i alpha cancels the e to the minus i alpha here. So that you could just dispense with from the beginning. e to the i alpha goes away and what do we get? We get the derivative of rho squared and then we get

plus rho squared times the derivative of alpha squared. And then finally, v of, what is phi star phi in terms of rho and alpha?

Rho squared. Notice that the potential only depends on rho and not alpha. That's just the symmetry. Well now let's say that the very, very low energy excitations, the very, very low energy excitations, the

very, very long wavelength configurations where the field varies slowly from point to point, where it stays down in the track here, number one. And number two, where it varies slowly. If we only have, we don't have much energy, we're just trying to make a very low energy excitation, it's not going to pay us to change rho very much.

We want to keep rho right on the track here. So in some approximation where we don't ever excite any motion up and down or away from the track here, we can just think of rho as a constant.

Rho doesn't vary from point to point and therefore we can say that rho is, what do we call rho? We call it the rho at the minimum, f. So we can just replace this by f squared, derivative of alpha squared. And then the rest of it has only to do with rho, has to do

with variations of rho from place to place, climbing up and down this well for example, and also potential energies that have to do with the value of rho. The potential energy is going to say, hey, don't get far from this point f, but still it's there.

This term, what does that term look like? That term just looks like the Lagrangian of a field with no potential at all. What about this f? What do we do with this f? It's just a number. Remember, it's just a number. It's fixed. It doesn't vary from place to place. We can absorb it into the redefinition of alpha.

We can say let f times alpha be a new field. What can we call it? We call it beta. And this would just become the derivative of beta squared. That would be characteristic of a field with no mass. So the f here, that's unimportant.

As I said, it can be absorbed in here and just change the definition from alpha to f times alpha since it's a constant. And this just becomes the Lagrangian for a massless field with no potential energy at all. The meaning of that is if all you're allowed to do is

vary the angle here, then excitations of that type behave like massless fields, behave like fields whose quanta have no masses. Now, on the other hand, what about rho? Rho is an independent field. It's an independent field. It has a Lagrangian derivative of rho squared and a term

v of rho. Supposing we choose the ground state to be over here at this point. Remember, we could choose the ground state to be in any direction. Then we can move in two directions. We can move around the circle. That's just exciting alpha. We've already taken care of that.

That's just a massless excitation that costs no energy except for the gradient terms. But we could also move radially outward. What would happen if we started the field over here, radially outward? And then what?

Let's not give it any rotation. Yeah. It would just oscillate about this point. It would oscillate about this point in very much the same way as the field would oscillate about the origin if there was a five star phi term. That corresponds to a mass.

Oscillations, if you take the field everywheres and displace it, it will just oscillate back and forth. That corresponds to a mass. In other words, this V of rho here, which could contain, for example, a rho squared or it could contain a term which oscillates about here, that does correspond to a

mass. What a mass is, is it's an energy that's associated with zero momentum. Zero momentum means that the field is homogeneous and doesn't vary from point to point. Zero momentum means no variation at all. So if there's energy stored in a way, not a way, it's

not hardly a way, if there's energy stored in a configuration where you move everybody simultaneously, then it would, and it just rocks back and forth, that frequency, the frequency of its motion is the mass. Alright, so if you took the field everywheres and you displaced it up onto the side of the potential, it

would start to oscillate everywheres, it would correspond to a mass. On the other hand, if you displaced it along the track here, nothing much happens. If you displaced it everywheres along the track, there would be no restoring force. The restoring force, which gives a periodic oscillation motion, that is what a massive field does.

It oscillates if you excite it homogeneously and okay. Whenever you have a symmetry, and the symmetry is, whenever you have a continuous symmetry, in this case the continuous symmetry is a rotation symmetry around

here, and that symmetry is spontaneously broken, there is always the possibility of changing the orientation of things very gradually from place to place if you have a continuous symmetry, and in that case you have massless excitations, massless particles.

Those particles are called Goldstone bosons. Goldstone boson is a particle or a field, a Goldstone field, is a field which is massless, which has no restoring force by virtue of a symmetry, by virtue of a

the fact that you're moving the field in a direction that costs no energy. So whenever you have a spontaneously broken symmetry, there are goldstone bosons. There are goldstone bosons. Well, there are goldstone bosons. Let's leave it at that. In this case, the goldstone boson is the quanta of the field alpha, which is just

motion along this track here. That's a very important concept in physics. It's very important in particle physics. It gives rise to massless particles, which we'll see

are really unwanted in a sense. And it's also a very important concept in condensed matter physics. In condensed matter physics, for example, ferromagnets, which like to line up, can be a ferromagnet where you can orient the magnets in any direction. But there's a symmetry under rotation.

There are spin waves. Spin waves are gradual variations of the direction of spin from point to point, gradual. And for the same reason that I've described here, a very gradual variation of the direction of spin from place to place is not costly in energy. It costs almost no energy.

And those spin waves behave like massless particles. So the phenomena or the idea of Goldstone bosons is a really important concept in field theory and condensed matter physics. Where else does it appear?

All kinds of places. Is there any physics which isn't either field theory or condensed matter physics? In general relativity, it also appears, but it won't tonight.

A long elastic string is an example with the motions, the waves on a long string. Let's just take it, not a string of string theory, just a long rubber string from one place to the next. That's an example of a Goldstone boson or a Goldstone field.

What would be the symmetry associated with it? Can you guess? I'll set up an example for you. Forget gravity. Gravity is not important here. Let's suppose there's a pole over here, another pole over here, a ring around this pole with a string connected to it, and a ring around here.

The reason I've set it up in this way is so that you can move the whole thing up and down like that. OK. We know that if we make this very, very, very long, there are massless waves which propagate up and down it. Massless means they cost very little energy,

only gradient energy. Can you guess what the symmetry in this case is? Translation up and down. Right. As I said, imagine no gravitational field. It costs no energy to move this whole thing up or down.

So it's a symmetry. It's translation symmetry, just translation symmetry up and down. So first of all, there's a symmetry. Next, you can imagine changing the orientation. And not orientation, but position gradually as you go from point to point. This Goldstone-Boson idea where

you imagine a little symmetry transformation, but a gradual change from one point to another, what would that look like? That would just look like a very slow variation of the vertical position. So an overall motion up and down costs no energy.

And a very small variation across here will also cost very little energy. And the waves, the long wavelengths which propagate up and down a very, very long string are also Goldstone-Bosons. They're Goldstone-Bosons of the symmetry of translation.

Spin waves are Goldstone-Bosons of rotational symmetry. And this field is the Goldstone-Boson of a U1 symmetry. Of a U1 symmetry here.

Well, if the thing was standing still, it has no momentum. Oh, you want to give it some oscillation energy?

Be careful. Be careful what you mean about momentum. You're speaking about vertical momentum. Vertical. Yeah, we could certainly make it have zero vertical momentum. But what about horizontal momentum?

Yeah. Well, no, there's waves. There's waves with wave numbers. So it's actually, if you had the standing wave, it would be a linear superposition of plus momentum and minus momentum. So what I would mean by momentum in this case would not be the vertical momentum, but the momentum along the wave, along the wave axis.

The vertical direction here is just to stand in for some internal symmetry. Good. Another example. Another example. A slinky.

Let's say you pin down the slinky at two ends. Does everybody know what a slinky is? Yeah, you know what a slinky is. You can move the whole slinky horizontally. Doesn't cost any energy.

So there's a symmetry of horizontal. Just think of it as being infinitely long for all practical purposes. That if you move it horizontally, costs no energy. So there's a symmetry. In this case, it's a horizontal motion. On the other hand, you can imagine slowly varying from place to place how much you translated it.

And in that case, you would make a wave. Tighter over here, less tight over here. But if you do it slowly enough, then the wave will not cost you much energy. You do it slowly enough from place to place.

If there are no big gradients, it won't cost you much energy. So in this case, the symmetry would be translation horizontally. And the waves on a slinky are also sound waves, are also Goldstone bosons. They also correspond to translation

of all the molecules. If you have a sample, a system, with an infinite system with lots of molecules, and you translate all the molecules, it costs you no energy. But if you translate them a little differently over here than over here than over here, that creates a very long wavelength pressure wave or density wave.

Those are also Goldstone bosons. And they all have the character that they cost no energy when the wavelength goes to infinity or when the momentum goes to zero. And we call them massless. So we've seen some connections now between symmetries and massless bosons called gauge bosons.

Goldstone bosons. Goldstone bosons. I'm sorry that I was apologizing to my friend Goldstone, who was a friend of mine. I called him gauge by mistake.

Sorry, Jeffrey. Now anybody instantly fails this course if they say gouage. Now let's talk about a totally different subject, also connected with U1 symmetry like this.

So it's not totally disconnected. Let's talk about gauge invariance. I'm not sure we'll put the two concepts together tonight. It may take until next week. Another concept called gauge invariance, also associated with this U1 symmetry.

Remember that U1 symmetry, symmetry under this kind of operation, is what we identified earlier in past lectures with electric charge conservation. It doesn't have to be electric charge. But let's think about electric charge. We're going to wind up with gauge bosons with mass.

Well, the real photon doesn't have a mass. But we could be thinking about some other gauge boson, like the z boson, which does have a mass. So for the moment, I'll just call the charge electric charge. But it might not be electric charge. It might be something of a different character.

And we're thinking about something which is not really electric charge. But we did identify the symmetry of this type with the conservation of the electric charge carried by the quanta of the phi field. Charges of various kinds are always

carried by fields which are complex. And the conservation law is associated with the symmetry. The symmetry is associated with the U1 transformation, namely multiplication by a phase. We've talked about this at length previous times. So we won't go into it again.

But let's erase the squiggles here. I'll leave this up because we'll use it again. And this is the symmetry that we're interested in. Phi goes to phi prime.

Now, question. Must we require, in order for it to be a symmetry, must we require that theta is constant over space? Are we allowed to make transformations where theta can vary from place to place?

Not theta, yeah, theta. Are we allowed to imagine transformations where the symmetry parameter, theta, theta's not a field now. It's just a symmetry parameter where it is allowed to vary from place to place. In other words, where we rotate

the phase of a field differently in some places than others. Is that a symmetry? So let's find out. Let's find out by doing such an operation and seeing whether it changes the Lagrangian or not. That's the test. If you think something might be a symmetry, try it out by applying it to the fields

and then stick them back into the Lagrangian and see if the Lagrangian changes. OK, so here's our transformation. Phi prime is equal to e to the i theta of x, where x could be space or time, space and time in general, space time. Running out of ink again.

Let's try blue.

Oh, good. OK, so here's our symmetry operation. Let's calculate the Lagrangian. Let's first calculate. Remember what goes into the Lagrangian. What goes into the Lagrangian is the squares of some derivatives and maybe some potential energy.

So let's start with derivatives. The derivative, arbitrary direction. It doesn't matter what direction. You can pick a direction, space or time, and calculate the derivative of phi prime. Here's phi prime. Well, first of all, it will contain e to the i theta

times the derivative of phi, right? The unmodified, before the transformation takes place. And then it'll contain a second term. And that second term will be phi i phi derivative of theta with respect to x times e to the i theta.

In fact, let's factor e to the i theta out. It's an overall multiplicative factor. Notice there are two terms. One of them, did I write this correctly? I think I did.

Yeah. One of them doesn't care one way or the other whether theta varied with respect to x or not. That was this term here. The other term actually involves the variation from point to point of theta. OK, now let's write the Lagrangian, as we might think of it, in terms of phi prime.

It contains, we know, phi prime, we're interested in derivative of phi prime, derivative of phi star prime. Oh, let's write the derivative of the conjugate. The derivative of phi prime star.

That's just derivative of phi star minus i phi star derivative of theta with respect to x times e to the minus i theta. Now what we're going to do is we're going to multiply these two and see if it's the same as just multiplying

the original derivatives of phi and phi star. Well, obviously the answer is no, because there are some extra terms here. But let's see what they are. Let's just check. First of all, the e to the i thetas will cancel out. So when I multiply the derivative of phi prime

times the derivative of phi prime star, at least these factors will cancel out. In other words, the factors will cancel out if theta doesn't depend on position. If theta didn't depend on position, then these would cancel out. So let's just erase them.

Well, I'm not going to erase them. I don't want to write the wrong equation, so I'll leave them up there. When I multiply these together, I'm going to get, well actually this is all I was hoping to get. I was hoping to get the original Lagrangian back and show that the Lagrangian is invariant.

But it's not. There's a couple of extra terms. First of all, there's a term i. Let's see what there is. i phi d phi star minus phi star d phi.

This is not zero. This is phi, the derivative of phi star. This is phi star times the derivative of phi. This is not zero. Times d theta by dx. That's new. I didn't want that there. And then finally, there's the squared term, which is phi star phi times derivative of theta

with respect to x squared. Well, that's a disaster. I mean, it's a disaster if we were hoping that this would be a symmetry. It's not a symmetry. The new Lagrangian when expressed in terms of phi prime

is not the same as the Lagrangian when expressed in terms of phi. What about the potential energy? If the potential energy is a function only of phi star phi, then phi star phi will not change

when we do one of these symmetry transformations, even if it depends on space. So v, the potential energy, that's symmetric. But the kinetic energy, the derivative terms in the energy, are not symmetric. And if we trace it down, we see what the problem is. The problem is simply associated with the fact

that the derivative of phi prime is not the same as the derivative of phi. Well, let's go to the upper equation. And even if we ignore the c to the i theta, that does cancel. Even if we ignore that, the derivative of phi prime contains this extra term. So we just don't have a symmetry.

Well, can we do anything to restore the symmetry? Is there anything we can do Lagrangian to restore? Well, yes, of course. We could just throw away these terms in Lagrangian. That would not be a very interesting Lagrangian with just the potential. Has no interesting dynamics whatsoever.

The answer is yes. But to do so, you have to introduce more fields. And the additional fields are the gauge boson fields. In this case, if we were really talking about electromagnetism, it would be the electromagnetic field. So let's introduce the electromagnetic field. Incidentally, a transformation like this,

which depends on position, just a symmetry, hoped for symmetry transformation, which depends on position like this, is called a gauge symmetry. Actually, I do know where the term came from. But it's a historical glitch.

It doesn't have to do with tire gauges or anything like that. It's not important where it came from, the term. But that's called a gauge symmetry. Gauge symmetry implies that the symmetry parameter

can be chosen to vary from place to place. Now remember, theta is not itself a field. It's just a generalized symmetry parameter, which is now allowed to vary from place to place. What's actually happening here is when we make the phase of the field

vary from place to place, we're picking up a little bit of gradient energy. The gradients are changing a little bit on us. And so we're picking up some gradient energy that wasn't there before. That's all. OK, let's, uh, let's, uh, no, no, no, it's just a gradient.

It's just if we make a configuration where the field, where the phase of the field varies from place to place, it costs energy. That's all. Yeah, that's all. That's all that's happening here.

But I've been very explicit about it. And I've identified this as the villain in the problem here, this term here, right. So the question is, can we find some clever way to cancel that out? Now, why should we want to do that? Ultimately, it's an experimental fact that all of the interactions of nature

are associated with gauge symmetries like this. Not all the interactions, but all the gauge symmetries, all the gauge interactions. Photons, z bosons, w bosons, all are associated with symmetries, not just of the kind where you can rigidly everywheres do the symmetry transformation,

but symmetries where you graduate, well, it doesn't even have to be gradual, where you're an arbitrary variation of the symmetry parameters from place to place. OK, so let's say, what's the ingredient that you have to add? You have to add a new collection of fields. In this case, just one new field,

or I should say four new fields, one for each direction of space. You must add a four vector of fields. And that four vector is, if this were electrodynamics, would just be the vector potential. The vector potential, A mu, four vector potential,

out of which you build the electromagnetic field. I'll just remind you quickly what the connection is. The electric field, well, first of all, the time component of A is the electrostatic potential.

It's gradient in space is the electric field. The space components of A form a three vector, and the curl of that three vector is the magnetic field. I won't bother writing it down. This is the standard vector potential out of which you build the electromagnetic field tensor

composed of electric and magnetic fields. We'll come back to that. But let's add in to the symmetry operation. We now ask, when you add a new field, and you're interested in the symmetry, you always have to address the question, how does this new field transform

when you do the symmetry operation? In other words, if we do the symmetry operation that takes us from phi to phi prime, is there also something that we have to do to A to take us from A to A prime? And the answer is yes, because the justification for this

is the result in the end, that you can keep gauge invariance. But what you do to A is you just add to it the derivative mu of theta. In other words, the transformation property of A is not some multiplication by a phase of A,

but just adding to A the gradient of theta. A is a four vector. The derivative of theta is also a four vector. That's the operation that will allow us to have a symmetry. Next, we have to change the definition of the Lagrangian.

And the definition of the Lagrangian is the new definition of the Lagrangian will be to replace ordinary derivatives by a new kind of derivative that's called the covariant derivative. It's not the covariant derivative

of general relativity. It's the covariant derivative of gauge theory. Let me define it for you. Define d mu on phi, the covariant derivative of the charged scalar field.

Or let's say, yes, the covariant derivative is defined to be the ordinary derivative. We can take away these mu's if we want. It wouldn't go, but that's right. Plus, let me get the sign straight.

Plus i a mu times phi. Definition. That's the definition of the covariant derivative. If you like, you can just think of it as a replacement

of derivative by derivative plus i times a. Now, that really doesn't mean anything by itself. It's something which acts on a charged field. This is definition of the covariant derivative.

Same thing, or the analogous thing, for d mu phi star. It's equal to d mu phi star minus i a mu phi star. Keep in mind, this wouldn't make any sense

if phi was a real valued field. Why not? Because if phi was a real valued field, we wouldn't want to add something imaginary. But phi is itself a complex field, and so adding something additional complex, that's fine, no problem. OK, this is the definition of the covariant derivative.

And the kinetic terms in the Lagrangian, forget the potential terms for a minute. The kinetic terms in the Lagrangian are just replaced by the square, or d mu phi, times d mu phi star. And let me show you why that is gauge invariant. Why that now has the invariance that was lost over here.

How this extra little trick of inventing an extra field, giving it a transformation property, now takes a Lagrangian which did not have this gauge invariance and makes it gauge invariant.

We don't even have to work out the Lagrangian. It's enough just to look at the covariant derivatives. Let's prove, let's find the connection. Here's phi and phi prime, where is phi and phi prime? Here's phi and phi prime.

Let's calculate the covariant derivative of phi prime. Let's calculate the covariant derivative of phi prime. And what we want to show is that the covariant derivative of phi prime is essentially the same as the covariant derivative of phi.

That would then establish that the new Lagrangian, which involved the square of these, was really invariant. So let's check that out. Let's calculate, let's recalculate, a covariant derivative of phi prime.

That's equal to, first of all, the first term is just the ordinary derivative of phi prime. And there it is up there. Derivative of phi plus i phi derivative

of theta with respect to x. I wonder if I have a sign wrong someplace. I may have a sign wrong here. Let's see, I think we can fix the sign by changing the sign over, I think, over here.

If it comes out wrong, we'll change it back again. OK, now that's the first term in the covariant derivative

of phi prime, just the derivative of phi prime. But then we have to add plus i something times phi prime. What do I want to put over here? A?

A prime. A prime, because I'm calculating the covariant derivative of the primed fields. So I want to put there a prime. So let's put a prime in. What is a prime? A prime is a mu minus d mu theta.

D mu theta times, sorry, this is phi prime a mu minus d mu

theta all times phi prime. Is that right? I think that's right. Phi.

And I must have left them out in an overall phase. Yeah, both of them have a factor of e to the i theta.

Now you look at it. Here was this extra nasty term which destroyed the invariance of this expression. But here is exactly the same term, but with an opposite sign. One term coming from the transformation of the derivatives of theta, and the other term coming

from the transformation of a. Notice they're exactly the same, I hope. i phi, here we have i phi e to the i theta and d theta by dx. That's d theta. Let's get rid of all the mu's.

Yeah. So this term and this term, this derivative of theta, this term and this term are exactly the same. They cancel. We succeeded in canceling out or playing off the transformation properties of the derivative

with the transformation properties of a and causing the extra term to cancel. So what do we find? We find that a covariant derivative of phi prime is exactly the same as the covariant derivative of phi, except for this factor e to the i theta here.

We have to put that back. What about the covariant derivative of the complex conjugate, phi prime star? That's equal to the covariant derivative of phi star times

e to the minus i theta. And guess what? When we multiply this by this, the e to the i thetas cancel out, and the Lagrangian is gauge invariant. So this a field here was an invention, a device, an invention to sort of forcibly make the Lagrangian

be gauge invariant. But we had to invent a transformation property of the gauge field itself, of the vector potential itself. Now, next.

We're missing something in the Lagrangian. What are we missing? Oh, yes, of course, we are missing the potential. Let's add that in. We can multiply this by this and get d mu phi Lagrangian is d mu phi, d mu phi star minus v of phi star phi.

This was gauge invariant to begin with. We don't have to modify that. But we are missing something rather important than a gauge theory.

This wouldn't be a very interesting gauge theory without some dynamics for the vector potential itself. Now, the vector potential is really the thing out of which you create the electric and magnetic fields.

Anybody know what the Lagrangian for the electromagnetic field is? Anybody know what the energy for the electromagnetic field is? e squared plus b squared. How about the Lagrangian? It's e squared minus b squared. e squared minus b squared. But it can also be represented in terms of the field tensor.

So let's write it in terms of the field tensor. The field tensor is called f mu nu. And I'll just remind you what it is. f mu nu is an anti-symmetric object. It's the derivative in the mu-th direction of a nu

minus the derivative in the nu-th direction of a mu. So it's a kind of four-dimensional curl. The diagonal components are 0. f 0, 0 is 0. Why is that? Because it's d0 a0 times d0 a0 minus d0 a0.

All of the diagonal elements are 0. It's anti-symmetric on the mu nu interchange. And it's components with one space and one time. What are they? The component of f mu nu with one space and one time

is the electric field. It has one space component. And that determines the direction of the electromagnetic field. And what about the terms with space-space components? The magnetic field. For example, f xy is equal to b z,

where b is the magnetic field, and so forth, and cycle it through. So this is the electromagnetic field tensor. And the standard dynamics, the Maxwell, ordinary Maxwell dynamics, is determined

by a Lagrangian. And the Lagrangian is just f squared. It's just f mu nu, f mu nu, raising the indices. There's just a couple of sign changes here and there. The Lagrangian is e squared minus b squared. The energy is e squared plus b squared. But the important thing is that the Lagrangian

for the electromagnetic field involves nothing but f mu nu itself. How about f mu nu? Is f mu nu gauge invariant? Let's check it. To find out, we have to calculate f mu nu prime.

Well, that's just equal to d mu a prime nu minus d nu a prime mu. But now let's remember, did we erase what a prime is? We did. a prime is just equal to a, I think it was minus the derivative of theta.

So let's just plug it in. This is going to be derivative mu of a nu minus the derivative of mu of the derivative of nu of theta. I've just plugged in for a prime a and then

the derivative with respect to nu of theta. What's going to happen when I put this term in? It's going to subtract derivative nu of a mu. And then it's going to add plus d nu d mu of theta.

But d nu d nu theta is the same as d nu d mu theta. It doesn't matter which order you differentiate in. This is just the fact that the order of partial differentiation is immaterial. So this cancels. And what do we find? We find that f prime is the same as f. This is f.

So the electromagnetic field tensor is gauge invariant. That means that we can construct a completely gauge invariant dynamics of the electromagnetic field coupled to a field whose quanta are charged particles. This is the field whose quanta are charged particles.

What is it? It's d mu. This is a mistake here. This is d mu. Only these covariant derivatives, not ordinary derivatives. A potential and add plus f squared.

I'll just call it f squared, meaning f mu nu, the appropriate thing, e squared minus b squared. There's some numerical factors which are conventional. There's usually a quarter in front of this. But that's not important. The important thing is all terms here are gauge invariant. So we've discovered the fact that we

can promote a symmetry like U1 here to a local symmetry. A local symmetry means that we can locally vary the phase, or we can locally do the operation, whatever it is, from point to point, and still have the symmetry. But it's always at the cost of an extra field,

an extra vector potential field. And that vector potential field can also be assembled into a Maxwell Lagrangian that would determine Maxwell's equations. Maxwell's equations coupled to charge carrying matter.

That's what this is. OK, what can we not add? Let me tell you what we can't add, and still be gauge invariant. First of all, we can't add potentials which are functions other than functions of phi star phi. If they were functions of just the real part of phi,

that would kill the symmetry. But as far as modifying the electromagnetic field, what we cannot add in is a term like, let's call it m squared over 2 times a mu a mu.

Remember, the mass term for a boson field is typically an m squared times the square of a field. A mass term for a photon would be a term like this. If we added this term to the Maxwell theory here,

we would be adding a mass to the photon. However, it's forbidden if we want the theory to be gauge invariant. This is not gauge invariant. This is not the same, not equal to m squared over 2 a prime mu a prime mu.

So this is forbidden by gauge invariance. There is no way by hand to give the photon a mass. That's what this means. The photon is necessarily massless. If you believe in gauge invariance, and every physicist believes in gauge invariance for a wide variety of reasons, including

mathematical consistency, the theory really breaks down badly if you try to do this. So the photon can't have a mass, or can it? Or can it? Is there no device here by which taking this Lagrangian,

just as it is, is there any trick that we can do to give the photon a mass? I'm going to tell you right now what it is, and we'll do it next week. If we spontaneously break the U1 symmetry by making the potential here have one of these shapes

like this, so that the field likes to sit over here, that will mean that the U1 symmetry is spontaneously broken, not explicitly. And spontaneous symmetry breaking, this Lagrangian is gauge invariant.

The spontaneous symmetry breaking will have the effect of giving the photon a mass. And it will also miraculously remove the Goldstone boson. So the effect of spontaneous symmetry breaking is twofold. It removes the Goldstone boson that was associated with slow variations of phi

from one point of phase of phi on the one hand, and at the same time gives a mass to the vector potential, so it gives a mass to the photon. Now, the real photon doesn't have a mass, but the Z boson does. This is called, we're going to work this out next time. We'll work it through next time and just see how spontaneous symmetry breaking gives rise

to a mass for the gauge boson. That's called the Higgs phenomenon. And this would be an example of how the Higgs phenomenon gives mass to a particle which mathematically you're not allowed to just put in a mass by hand. All right? That's the signature or the basic idea

of the Higgs phenomenon, that there may be deep mathematical reasons why you cannot just by hand give a particle a mass, but spontaneous symmetry breaking can do it anyway. Yeah? So the real electron involves the deep derivatives, not the covariant derivatives.

Say it again. The Lagrangian that we started with involves the real derivatives, not the- Not the covariant derivatives. So why should we accept the fact that only for covariant derivatives, it's invariant? What I'm telling you is that if you have some first statement, you can

create a Lagrangian which is gauge invariant. Second statement, the coupling between charge-carrying fields and the electromagnetic potential would become inconsistent if it weren't gauge invariant.

I'm not going to try to demonstrate that. It really does become mathematically inconsistent unless it's gauge invariant. So why should you believe- What is the question? The question is- No, the answer is that you can construct

a gauge invariant Lagrangian. The result is a nice consistent mathematical structure, a gauge theory. If you try to do anything else, you will wind up with a theory which in quantum mechanics which will have some terrible things, it'll have negative probabilities in it, even worse things, but I'm not going to go through that.

Gauge invariance is an extremely fundamental principle in theoretical physics, whether you like it or not, and for our purposes, I think we just have to accept it. We can also accept it as an empirical fact that all of the particles, namely photons, Z bosons, W plus and W minus

bosons, nature seems to have chosen their Lagrangians to be exactly gauge invariant. It's a good thing because we don't want negative probabilities in a theory. You justified it by saying there's an additional field. Yeah.

And if you start with just a plain thing and keep it that way, you add this new field and you describe it as being the- We have to add a new field in order to implement the symmetry. Right. And that's described using the covariate derivative. Yeah. That's part of the explanation. So that's why I understand that it's okay to do that

with the new field. Right. Because in general. Yeah. That's right. But we had to add a new field to do it. Without the new field, we could not keep- Now, there's much deeper mathematics here that I haven't discussed, fiber bundles, all kinds of interesting mathematics. A mu is technically what a mathematician would call a

connection. And there's all sorts of interesting mathematics, which we are not going to get into. All right. So let me see if I understand what you're saying. The gauge invariance leads us to this modification, which

gives rise to the Goldstone boson. The Goldstone boson was a feature of the theory without vector potentials and without gauge invariance.

What the Goldstone boson was, the energy of the Goldstone boson, was exactly the energy of making these variable phases. But now we find that there's a way of making the variable phases without a cost in energy. Why? Because the Lagrangian is invariant under those gradual

changes of phase. What does that mean? That means there's no cost in energy in this gradual change of phase, no Goldstone boson. But we'll do that next time. We'll do it next time. We'll see that this Lagrangian does not have Goldstone bosons, and it does have a mass for the photon.

Does that have anything to do with why rivers meander? Why what meander? Rivers. I don't think, not that well. Why rivers don't flow in a straight line on a relatively labeled plane? I don't think it is so.

And the length of the increase is equal to 5, the distance. Not that I know of, but I would not be at all surprised if somebody has found the gauge theory of rivers. I don't know of any connection.

I do know that there are connections with the motion and shapes of amoeba. Well, I wouldn't be totally surprised if there was some connection, but none that I've thought of.

For more, please visit us at stanford.edu.