Gauge transformation and Occam's razor

[jostpuur]

For electromagnetic field we usually use the Lagrange's density

$$-\frac{1}{4}F_{\mu\nu}F^{\mu\nu},\quad\quad\quad\quad\quad\quad\quad(1)$$

but we could also use a simpler Lagrange's density

$$-\frac{1}{2} (\partial_{\mu} A_{\nu})(\partial^{\mu} A^{\nu}),\quad\quad\quad\quad\quad(2)$$

which gives the same equation of motion if the gauge condition $\partial_{\mu} A^{\mu}=0$ chosen.

Suppose we want to have a theory that explains the Lorentz force and the Maxwell's equations. Why should we use the more complicated Lagrange's density (1) which leads to the gauge invariance, when we could use the simpler one (2) without gauge invariance?

Is it a good thing that there is gauge transformations in theory? I thought that they are usually source of some kind of trouble, but on the other hand people seem to think that the gauge invariance is a sign of some deep properties of the nature.

 

[blechman]

The "trouble" with gauge theories is that you are including unphysical degrees of freedom (such as the longitudinally polarized photon) and you have to make sure that they don't appear in the final calculations, and this complicates matters when you have to calculate things. However, despite the headaches, this can be done.

But the problem with no-gauge symmetry is that the the final result doesn't make sense! Gauge symmetry turns out to be exactly the statement of "charge conservation", so by writing down an operator that breaks the symmetry (as you just did) you will also violate charge-conservation, which is a disaster! This result is due to the famous "Noether's theorem." To see this, include the coupling of the current to the vector potential $$j^\mu A_\mu$$ and you will see that $$\partial_\mu j^\mu \neq 0$$!

So the moral of the story is that even though it looks like the more complicated action is making your life difficult for the moment, trying to make it simpler will only lead to contradictions with experiment.

As a funny aside, it's a theorem in QFT that the ONLY self-consistent theories with spin-1 bosons are gauge theories, and the analogous theorems get even stronger as you go to higher spin. For example, the ONLY self-consistent theory of (massless) spin-2 bosons is Einstein's Gravity! This is called the Coleman-Mandula theorem in case you're interested.

 

[blechman]

Let me make one other observation to your question. You will maintain charge conservation, backing away slightly from what I said above, AS LONG AS you enforce the Lorenz condition. The way to enforce constraints in Lagrangian mechanics is to include a Lagrange multiplier $$\lambda$$, so:
$$\mathcal{L}=-\frac{1}{2} (\partial_{\mu} A_{\nu})(\partial^{\mu} A^{\nu})+\lambda (\partial_\mu A^\mu)^2$$

Now solve for $$\lambda$$, plug back into the Lagrange density and you will end up with your Equation (1). So you see, you haven't actually done anything!

 

[cesiumfrog]

For example, the ONLY self-consistent theory of (massless) spin-2 bosons is Einstein's Gravity! This is called the Coleman-Mandula theorem in case you're interested.

Really? No, that can't be right. There's a good reason why gravity has to be spin-2, isn't there? It would have been much bigger news in the GR crowd if a QFT had done something to verify the (ad hoc and still sometimes disputed form of the) EFE! You must mean something weaker.. and even the wikipedia stub doesn't mention the significance you claim?

 

[blechman]

It is a rather wonderful result (proved by Dick Feynman many years ago, if I remember correctly) that the only way to write down a self-consistent theory of a *massless, spin-2 * object is Einstein's GR. There is simply no other way to do it! The proof is quite non-trivial, but it's a famous result. Of course, there are some important assumptions, such as Lorentz invariance, energy conservation, etc. But given some rather weak assumptions like that, then the only theory that you can write down with massless spin-2 is GR. There is NOTHING at all "ad-hoc" about Einstein's Field Equations (I'm assuming that's what "EFE" stands for). High-spin theories (s>1) are VERY constraining.

People are out to modify GR all the time, and that's okay. But almost always, the way they do it is to ADD something, such as new scalar fields, extra dimensions, etc. But they always end up with Einstein PLUS stuff. You can never get rid of Einstein entirely. Every model I know of that tries always has deep problems!

As to Wikipedia, I have found it to be lacking in many ways - I wouldn't treat them as the final authority for this stuff. I'm certainly not responsible for what they do or do not mention!

CORRECTION:
I see I wrote Coleman-Mandula; that's wrong. That's another result that, roughly speaking, says that you cannot write down a nontrivial theory with s>2 (it says more than that, but that's for another thread). The result I quoted was Feynman's. I apologize for the error in citation.

 

[cesiumfrog]

almost always, the way they [modify GR] is to ADD something, such as new scalar fields, extra dimensions, etc. But they always end up with Einstein PLUS stuff. [...] The result I quoted was Feynman's.

You're saying that adding extra dimensions (like Kaluza-Klein) and fields (like electromagnetism) to GR will produce a theory for which the solutions are a subset of the solutions of standard GR? I guess that's nice to have actually proven (and it seems that relativists often want arguments to eliminate unobserved solutions). But a common modification is not so much to add something, but to alter the actual (historically ad hoc) expression (in the field equation) for $G_{ab}$, usually with the aim of altering the long range effect of gravity (perhaps also zeroing any cosmological constant, but most often to explain dark matter.. not sure whether MOND is also equivalent to doing this), and I'm curious whether you're saying particle physics can also eliminate that. Can you reference a specific publication?

 

[blechman]

I think there's a mis-communication. Feynman's result was that the only way to write down a massless spin-2 field is the Einstein-Hilbert action. The fact that we believe the graviton to be a massless spin-2 field is very promising! The equations:

$$G_{\mu\nu}=T_{\mu\nu}$$

are usually altered on the RHS - by a Brans-Dicke field, cosmological constant, quintescence, or whatever you want; this is the "adding stuff" I was alluding to earlier.

Altering the LH side of Einstein's equations is a very nontrivial undertaking. It's funny you mentioned MOND: this theory has some deep problems (ghosts, nonlocalities) that violate precisely the "weak assumptions" I was mentioning earlier. It could very well be that nature DOES violate these assumptions, who knows? But if they do, it must be such that it's consistent with what we've seen so far.

In the end, it all comes down to what you mean by phrases like "self-consistent" and "weak assumptions."

As to references, I'm afraid I cannot provide any direct references to past papers; all I have is my old particle theory class notes. However, according to those notes, the results were worked out by Feynman in the early 60's, and then refined by Stanley Deser and then clinched by the Coleman-Mandula paper; I knew that result came in sooner or later ;-) Before Feynman's result, it was assumed that Einstein was "lucky" in finding his equations mainly by trial and error. But it turns out there was more to it: under the assumptions Einstein made, the LH side of his equation was pretty much fixed!

 

[jostpuur]

The "trouble" with gauge theories is that you are including unphysical degrees of freedom (such as the longitudinally polarized photon) and you have to make sure that they don't appear in the final calculations, and this complicates matters when you have to calculate things. However, despite the headaches, this can be done.

But the problem with no-gauge symmetry is that the the final result doesn't make sense! Gauge symmetry turns out to be exactly the statement of "charge conservation", so by writing down an operator that breaks the symmetry (as you just did) you will also violate charge-conservation, which is a disaster! This result is due to the famous "Noether's theorem." To see this, include the coupling of the current to the vector potential $$j^\mu A_\mu$$ and you will see that $$\partial_\mu j^\mu \neq 0$$!

So the moral of the story is that even though it looks like the more complicated action is making your life difficult for the moment, trying to make it simpler will only lead to contradictions with experiment.

As a funny aside, it's a theorem in QFT that the ONLY self-consistent theories with spin-1 bosons are gauge theories, and the analogous theorems get even stronger as you go to higher spin. For example, the ONLY self-consistent theory of (massless) spin-2 bosons is Einstein's Gravity! This is called the Coleman-Mandula theorem in case you're interested.

I don't understand how gauge symmetry is supposed to be the same thing as the charge conservation. Suppose I define a system with a Lagrange's function

$$L=-\int d^3x'\; \frac{1}{2}\big(\partial_{\mu} A_{\nu}(x')\big)\big(\partial^{\mu} A^{\nu}(x')\big) - \sum_{k=1}^N \Big(q_k A^0(x_k) - q_k v_k\cdot A(x_k) + m_k \sqrt{1 - |v_k|^2}\Big).$$

There's the electromagnetic field, fixed number of relativistic particles, and an interaction term $j_{\mu} A^{\mu}$, where point charges $q_k \delta^3(x'-x_k)(1,v)^{\mu}$ have been substituted. There is no gauge symmetry, but the total charge [tex]\sum_{k=1}^N q_k[/itex] is quite conserved.

Or is there some symmetry here, that I don't know? Anyway, my point was, that the $F_{\mu\nu} F^{\mu\nu}$ still seems unnecessarily complicated.

 

[jostpuur]

Or is there some symmetry here, that I don't know? Anyway, my point was, that the $F_{\mu\nu} F^{\mu\nu}$ still seems unnecessarily complicated.

Oh well, the equations of motion are

$$\partial_{\mu}\partial^{\mu} A^{\nu}(x) = \sum_{k=1}^N q_k\delta^3(x-x_k)(1,v_k)^{\nu}$$
$$F_k = -q_k(\nabla A^0 + \partial_0 A) + q_k v_k\times(\nabla\times A),$$

so there is the gauge symmetry in the sense that the forces, that the charges experience, are invariant. Thus the $A^{\mu}$ cannot be measured uniquely. At least not using charges only.

 

[blechman]

Let's consider the action you just wrote down. The field equations can be read off from Lagrange's equations:

$$\partial^2 A_\mu = j_\mu$$

Now, this field equation is not gauge invariant (just gauge transform the vector potential and you get a new inhomogeneous term). What does this mean for charge conservation?

$$\partial_\mu j^\mu = \partial^2(\partial_\mu A^\mu)$$

The RHS of this equation is not equal to zero UNLESS you enforce the Lorenz condition! But as I mentioned earlier, when you have a constraint in Lagrangian mechanics, you add it with a Lagrange multiplier and this just gives you your gauge-invariant action, and you've done nothing.

Now let's go back and try the gauge-invariant action's equations of motion:

$$(g_{\mu\nu}\partial^2 - \partial_\mu\partial_\nu)A^\nu = j_\mu$$

Now look what happens: the second term that wasn't there before suddenly enforces the conservation of charge, regardless of Lorenz condition or any other gauge choice!

This all comes down to the famous "Noether's Theorem" which says that all (local) conservation laws imply a symmetry, and vice versa. A very powerful result.

 

[jostpuur]

$$\partial_\mu j^\mu = \partial^2(\partial_\mu A^\mu)$$

The RHS of this equation is not equal to zero UNLESS you enforce the Lorenz condition!

The left side is zero, so whatever happens on the right side, to make it zero, happens automatically.

 

[blechman]

No, jostpuur, you have it the opposite way around! The LHS must be proven to be zero, and this is only true if you enforce the Lorenz condition. Otherwise the RHS is not zero and you are not conserving charge. You can't put the cart before the horse - to prove charge conservation, you show the RHS vanishes, and then you're good.

 

[jostpuur]

No, jostpuur, you have it the opposite way around! The LHS must be proven to be zero, and this is only true if you enforce the Lorenz condition. Otherwise the RHS is not zero and you are not conserving charge. You can't put the cart before the horse - to prove charge conservation, you show the RHS vanishes, and then you're good.

Suppose I define a system with a Lagrange's function

$$L=-\int d^3x'\; \frac{1}{2}\big(\partial_{\mu} A_{\nu}(x')\big)\big(\partial^{\mu} A^{\nu}(x')\big) - \sum_{k=1}^N \Big(q_k A^0(x_k) - q_k v_k\cdot A(x_k) + m_k \sqrt{1 - |v_k|^2}\Big).$$

There's the electromagnetic field, fixed number of relativistic particles, and an interaction term $j_{\mu} A^{\mu}$

If I define a system with the EM-field and fixed number of particles, then the charge is conserved, and the charge current satisfies the continuity equation, very certainly!

(although it can be tricky with those delta functions, perhaps approximating them with some finite peaks makes it clearer)

 

[blechman]

Jostpuur:
Are you trying to tell me that if you take your action, compute the resulting field equations, assume that Lorenz condition does not hold, and yet you still have conservation of charge?! Go back and check your results - you made a mistake!

You can see from my equations that the only way you can get charge conservation from your action is by assuming the Lorenz condition. And this takes you back to your original F^2 Maxwell action, as I already pointed out.

 

[jostpuur]

Jostpuur:
Are you trying to tell me that if you take your action, compute the resulting field equations, assume that Lorenz condition does not hold, and yet you still have conservation of charge?! Go back and check your results - you made a mistake!

I'm telling you, that if my system consists of the EM-field and a fixed number of particles, then the number

$$\sum_{k=1}^N q_k$$

is remaining fixed too.

 

[blechman]

does the total divergence of the current vanish?

If what you're saying is true, then you are able to prove that:

$$\partial^2(\partial_\mu A^\mu$$)=0

vanishes in every gauge. This is simply false.

 

[jostpuur]

does the total divergence of the current vanish?

Perhaps in a sense. Those delta function charge currents are a bit problematic. But if you replace them with some sharp gaussian peaks, then it should vanish. Doesn't it seem obvious physically/intuitively? I haven't gone through the calculation though. Basically each charge would give a contribution of

$$q\Big(\frac{C}{\pi}\Big)^{3/2}e^{-C|x-x(t)|^2}$$

to the charge density, where C is some large constant, and

$$q\dot{x}(t)\Big(\frac{C}{\pi}\Big)^{3/2}e^{-C|x-x(t)|^2}$$

to the three current density.

Charges like this are violating principles of special relativity, though... But I think they can be used in any fixed frame as an approximation. The limit $C\to\infty$ should make the system Lorentz invariant again.

If what you're saying is true, then you are able to prove that:

$$\partial^2(\partial_\mu A^\mu$$)=0

vanishes in every gauge. This is simply false.

If the field satisfies

$$\partial_{\nu}\partial^{\nu} A^{\mu} = j^{\mu}$$

and the right side satisfies

$$\partial_{\mu} j^{\mu} = 0$$

then

$$\partial_{\nu}\partial^{\nu}\partial_{\mu} A^{\mu} = 0$$

is pretty clear. Why would this be wrong?

 

[blechman]

If the field satisfies

$$\partial_{\nu}\partial^{\nu} A^{\mu} = j^{\mu}$$

and the right side satisfies

$$\partial_{\mu} j^{\mu} = 0$$

then

$$\partial_{\nu}\partial^{\nu}\partial_{\mu} A^{\mu} = 0$$

is pretty clear. Why would this be wrong?

Maybe I haven't been saying this properly. The problem is that away from the Lorenz gauge, your equations are no longer Maxwell's equations!

Let me give you a silly example. For your (single) point charge example, you know what the solution of Maxwell's equations are (in the rest frame of the particle):

$$A_0 = \frac{e^2}{r}\qquad\vec{A}=\vec{0}$$

Now this potential satisfies the Lorenz condition, as can be easily seen. However, I can very easily write down a vector potential that gives the same EM fields, and yet does not satisfy Lorenz:

$$A_r=\frac{e^2t}{r^2}$$

and all others zero. Now, this vector potential should give you the same field as the one previous one, but go ahead and compute $\partial^2(\partial_\mu A^\mu)$. Does it vanish? I smell a terrible contradiction in the works!

The source of all this trouble is that outside of the Lorenz gauge, your field equations are not Maxwell's equations! In particular, that is not the equation that describes an electromagnetic field. The only way you can avoid the contradiction is by imposing the Lorenz condition. And as I've been saying all along, imposing that constraint brings you back to the original action.

Putting it a slightly different way: take your field equation, NO Lorenz condtion, and go back and rewrite it in terms of E and B fields. You won't get Maxwell's equations. Go ahead, try it!

 

[blechman]

I've just read through the thread again, trying to remind myself of how this all started. I see you (jostpuur) were asking specifically about the relationship between gauge invariance and charge conservation. Let me take my last reply and extend it in that direction:

The point is that by breaking the gauge symmetry (imposing the Lorenz constraint does this) you are actually altering Maxwell's equations, as I've alluded to above. This alteration is harmless so long as you stay in the gauge, but if you try to get out of it, you find that the LHS of the field equations is no longer equal to the RHS of the equations (this is the "contradiction in the works" I was hinting at above).

You can interpret this as saying that the current you were coupling to has changed! Going outside your gauge choice has introduced "gauge artifacts" into your conservation equation.

In principle, there is nothing wrong with that. In a strange sense, it's almost more fundamental to do things this way, since by forcing yourself to stay in a certain gauge, you eliminate the unphysical degrees of freedom (what I was saying at the very beginning of the thread). But it turns out to be quite difficult to write down a fully interacting theory of spin-1 particles when you try to fix things this way. By allowing yourself to be in any gauge, you can employ the symmetry to restrict the form of your action. Without that crutch, things are made much harder. This is not such a big deal with such a simple theory as Maxwell's, but in more general Yang-Mills gauge theories, as well as higher spin theories such as gravity, the power of gauge symmetries really shines through.

I was just reading about this recently in the textbook "Gravity and Strings" by Tomas Ortin. It's *very* advanced, but it does a wonderful job in the first few chapters explaining how the gauge symmetry makes enforcing the physical conditions a breeze (basically what I've been trying to say earlier about Lagrange multipliers).

Anyway, I hope that helps.

 

[jostpuur]

You think that the E and B are the fundamental quantities (or the equivalence classes of A), and the potential A is some kind of tool to handle them. I'm asking, that why do we have to deal with the EM-interactions like this, and why couldn't we think that the A is the fundamental quantity. Despite the fact that your answers have been technical, am I right to guess that these technical issues are still derived from the basic assumption, that the E and B are supposed to be postulated as the fundamental ones?

Maybe I haven't been saying this properly. The problem is that away from the Lorenz gauge, your equations are no longer Maxwell's equations!

Let me give you a silly example. For your (single) point charge example, you know what the solution of Maxwell's equations are (in the rest frame of the particle):

$$A_0 = \frac{e^2}{r}\qquad\vec{A}=\vec{0}$$

Now this potential satisfies the Lorenz condition, as can be easily seen. However, I can very easily write down a vector potential that gives the same EM fields, and yet does not satisfy Lorenz:

$$A_r=\frac{e^2t}{r^2}$$

and all others zero. Now, this vector potential should give you the same field as the one previous one, but go ahead and compute $\partial^2(\partial_\mu A^\mu)$. Does it vanish? I smell a terrible contradiction in the works!

This is not a counter example to my claim, because that is not a solution to the equations of motion

$$\partial_{\mu}\partial^{\mu} A^{\nu} = j^{\nu}$$

The source of all this trouble is that outside of the Lorenz gauge, your field equations are not Maxwell's equations! In particular, that is not the equation that describes an electromagnetic field. The only way you can avoid the contradiction is by imposing the Lorenz condition. And as I've been saying all along, imposing that constraint brings you back to the original action.

Putting it a slightly different way: take your field equation, NO Lorenz condtion, and go back and rewrite it in terms of E and B fields. You won't get Maxwell's equations. Go ahead, try it!

Well this was right. The Maxwell's equations are not coming out of the $\partial_{\mu}\partial^{\mu} A^{\nu} = j^{\nu}$ without the additional assumtion $\partial_{\mu} A^{\mu} = 0$.

 

[jostpuur]

The left side is zero, so whatever happens on the right side, to make it zero, happens automatically.

At this point I had forgotten, that in my original post I said

which gives the same equation of motion if the gauge condition $\partial_{\mu}A^{\mu} =0$ chosen.

This caused some confusion. Anyway, the equation

$$\partial_{\nu} \partial_{\mu} \partial^{\mu} A^{\nu} = 0 = \partial_{\nu} j^{\nu}$$

is true in this system (described in the earlier post #8 and #9). I don't think it is relevant to figure out should it be the left or the right side that is supposed to be zero "first".

 

[jostpuur]

No, jostpuur, you have it the opposite way around! The LHS must be proven to be zero, and this is only true if you enforce the Lorenz condition. Otherwise the RHS is not zero and you are not conserving charge. You can't put the cart before the horse - to prove charge conservation, you show the RHS vanishes, and then you're good.

This sounds like that the electric charge is a kind of quantity, that when left alone, could become created out of thin air, and could also vanish into nowhere, and that the equations of motion for the EM-field, or -potential, must have the particular form, that forces the charge to be conserved.

What a strange thought! I don't think it is the purpose of the EM-field to keep charges from violating conservation. It is not that kind of interaction. The fields merely cause forces on the charges.

 

[blechman]

You think that the E and B are the fundamental quantities (or the equivalence classes of A), and the potential A is some kind of tool to handle them. I'm asking, that why do we have to deal with the EM-interactions like this, and why couldn't we think that the A is the fundamental quantity. Despite the fact that your answers have been technical, am I right to guess that these technical issues are still derived from the basic assumption, that the E and B are supposed to be postulated as the fundamental ones?

I don't understand this paragraph at all. I don't know what the word "fundamental" means in the context that you are using it. It is certainly the case that the only observable quantities in nature are by definition gauge-invariant. After all, from the point of view of PHYSICS, we measure forces, and forces are E and B, NOT A! This has nothing to do with "fundamentalism" - one way or the other, we are ultimately interested in these forces, not the potential. What else do you want to calculate?!

Well this was right. The Maxwell's equations are not coming out of the $\partial_{\mu}\partial^{\mu} A^{\nu} = j^{\nu}$ without the additional assumtion $\partial_{\mu} A^{\mu} = 0$.

Anyway, the equation

$$\partial_{\nu} \partial_{\mu} \partial^{\mu} A^{\nu} = 0 = \partial_{\nu} j^{\nu}$$

is true in this system (described in the earlier post #8 and #9). I don't think it is relevant to figure out should it be the left or the right side that is supposed to be zero "first".

There's some sort of lack of communication going on here. Noether's theorem works both ways: a conservation law implies a symmetry, and vice versa. Make up your mind which one you want: If you have a gauge symmetry, the current is automatically divergenceless, as I've been saying all along. If you would rather never talk about gauge invariance but you still write down a conserved current, then you can show that the action must have a gauge symmetry. If you start with your action, notice that the interaction term $A_\mu j^\mu$ is already gauge invariant if the current is conserved. If you perform a gauge transformation with gauge function $\lambda$ on your "kinetic" term, notice that (after some integration by parts) you get a term in the action:

$$\Delta S=\int d^4x\lambda(2\partial_\mu A^\mu)$$

Then in order for this to be consistent with Noether's theorem, you derive the Lorenz condition as a constraint equation.

This sounds like that the electric charge is a kind of quantity, that when left alone, could become created out of thin air, and could also vanish into nowhere, and that the equations of motion for the EM-field, or -potential, must have the particular form, that forces the charge to be conserved.

I think you're getting a little confused over what it means for local conservation of charge to be violated:

$$\partial_\mu j^\mu = 0 \Rightarrow Q_{\rm in} = Q_{\rm out}$$

So if the current isn't divergenceless, it doesn't mean that charge comes out of nowhere (although that is one way it could happen), but it means that charge can accumulate without any outlet. It was this kind of flagrant contradiction with observation that led Maxwell to propose his displacement current that augments Ampere's Law and completes Maxwell's equations (and makes sure that conservation of charge is consistent with the field equations).

From the point of view of the quantum field theory: there is a theorem that says that the only way to write down a self-consistent interacting theory of particles with spin >= 1 is to couple it to a conserved current. This is what I was talking about long ago. So conservation of charge in the quantum theory is intimately related to the fact that the photon is a spin 1 field. I don't think there's a classical analogy to this, though. Just empirical fact.

What a strange thought! I don't think it is the purpose of the EM-field to keep charges from violating conservation. It is not that kind of interaction. The fields merely cause forces on the charges.

What does this paragraph even mean?! Any equations that describe E&M must be consistent with a charge conservation law. This is motivated by empirical fact, as well as mathematical consistency.


***********************

Your original question, jostpuur, was if you can write down the Maxwell action in a simpler form than the usual F^2. I have said all along that the answer is that you can, but only if you fix the gauge (in your choice, the Lorenz gauge). If you agree to fix that gauge, then your action is fine. However, if you choose to go to another gauge, then your action is wrong! You seem to agree with me, at least on that.

Now the next step is to understand the connection between gauge invariance and charge conservation. The connection is Noether's theorem, which says that these two things are the same thing.

So what does it mean when you break the gauge invariance by fixing a gauge, from the point of view of charge conservation? The answer is that you still have charge conservation, but you *also* no longer have complete gauge freedom. In particular, Maxwell's equations are still gauge invariant, but your gauge-fixed equations for the potential are not. Putting it more explicitly, your field equation $\partial^2A^\mu = j^\mu$ must be supplemented with the Lorenz condition $\partial_\mu A^\mu=0$, or else your theory is not Maxwell anymore.

You prefer to talk in terms of conserved currents. Fine: then your constraint equation is $\partial_\mu j^\mu=0$, and the Lorenz condition follows. But you still have a field equation and a constraint, and nothing has changed!

You tried to pull a fast one by explicitly writing down a current that you stole from Jackson! But where did this current really come from? Why did you chose it, as opposed to ANY other 4-vector? The answer is that you used the conservation law explicitly to construct the current, and then, of course, your (choice of) field equations reproduce the Lorenz condition. If you chose another gauge, your (different) field equations would reproduce the (different) gauge-fixing condition. Or, as is typically done, you can not chose a gauge, leave the field equations in gauge-invariant form, and then derive the charge-conservation condition. Each of these approaches is acceptable, and depending on what it is that you are doing, anyone of these choices might be more convenient than the others. But at the end of the day, they're all equivalent.

*******************************

I think this finally answers your question. Looking back, I think that my explanations might have been more confusing than helpful, but I hope that this last post cleans things up a bit.

In summary: you can treat spin-1 theories in one of three ways:

1) Construct a gauge-invariant theory (which automatically enforces charge conservation).

2) Write down the theory in a specific gauge and invoke the gauge-fixing condition as a constraint equation. This is equivalent to (1) after the Lagrange multiplier trick I mentioned earlier.

3) Write down the theory in a specific gauge and invoke a charge-conservation constraint. This is equivalent to (2) as you have shown here.

From a practical point of view, Option (1) is the most-often-used choice of field theorists, since you don't have to worry about any explicit constraint equations (they're already built in), and the resulting calculations are just a lot cleaner.

<DEEP BREATH!>

 

[jostpuur]

I don't understand this paragraph at all. I don't know what the word "fundamental" means in the context that you are using it. It is certainly the case that the only observable quantities in nature are by definition gauge-invariant. After all, from the point of view of PHYSICS, we measure forces, and forces are E and B, NOT A! This has nothing to do with "fundamentalism" - one way or the other, we are ultimately interested in these forces, not the potential. What else do you want to calculate?!

We want to have a theory, that explains the EM-interactions. If we can explain them by assuming that there exists some specific potential A, what would be a problem with that?

Is the problem, that the A is not supposed to be assumed fundamental, because it cannot be uniquely measured?

There's some sort of lack of communication going on here. Noether's theorem works both ways: a conservation law implies a symmetry, and vice versa.

But wasn't the gauge fixing Lagrangian I wrote, an example of a system where there is charge conservation without gauge symmetry related to it?

What does this paragraph even mean?!

The equations of motion for the EM-fields (the Maxwell's equations) give the time evolution of the EM-fields. When you want to find out the time evolution of the EM-fields, the equations of motion for them are the answer.

The equations of motion for the charges (the Lorentz force) give the time evolution of the charge distribution. When you want to find out the time evolution of the charges, the equations of motion for them are the answer.

In particular, the equations of motion for the EM-fields, are not the ones that tell what happens to the charges!

That is what it meant, but...

Any equations that describe E&M must be consistent with a charge conservation law. This is motivated by empirical fact, as well as mathematical consistency.

...yeah, I agree with this. Consistency is preferable. It would be unpleasant to have $\neq$ in some equation due to wrong gauge.

This starts to look like understanding. I must point out that your tone here:

Your original question, jostpuur, was if you can write down the Maxwell action in a simpler form than the usual F^2. I have said all along that the answer is that you can, but only if you fix the gauge (in your choice, the Lorenz gauge). If you agree to fix that gauge, then your action is fine. However, if you choose to go to another gauge, then your action is wrong! You seem to agree with me, at least on that.

is different from what it was here in the first response:

But the problem with no-gauge symmetry is that the the final result doesn't make sense! Gauge symmetry turns out to be exactly the statement of "charge conservation", so by writing down an operator that breaks the symmetry (as you just did) you will also violate charge-conservation, which is a disaster! This result is due to the famous "Noether's theorem." To see this, include the coupling of the current to the vector potential $j^{\mu}A_{\mu}$ and you will see that $\partial_{\mu}j^{\mu} \neq 0$!

You tried to pull a fast one by explicitly writing down a current that you stole from Jackson!

This must have been accidental. Are you talking about the Gauss peaks I used for charge distributions?

 

[jostpuur]

The motivation btw stems from the QFT. If we define the system by using the Lagrangian with $(\partial_{\mu} A_{\nu})(\partial^{\mu} A^{\nu})$, then this is simply a four component Klein-Gordon field, and it can be quantized easily. If we use $F_{\mu\nu}F^{\mu\nu}$ instead, I don't really know how the quantization occurs. So far all texts I have encountered have been very confusing.

 

[jostpuur]

Is it clear that the gauge symmetry and the charge conservation are related according to the Noether's theorem? If I coupled a complex Klein-Gordon field to the EM-field, wouldn't it still be the internal symmetry of the KG-field that corresponds to the conserved KG-current?

 

[blechman]

We want to have a theory, that explains the EM-interactions. If we can explain them by assuming that there exists some specific potential A, what would be a problem with that?

Is the problem, that the A is not supposed to be assumed fundamental, because it cannot be uniquely measured?

What, precisely, does the word "explain" mean in this context? We need to be able to compute EM forces (as I believe you said earlier). These forces are given by fields, not potentials. Therefore, the FIELDS are the right thing to be interested in. After all, an infinite number of potentials give the same physics (fields), as you know.

But wasn't the gauge fixing Lagrangian I wrote, an example of a system where there is charge conservation without gauge symmetry related to it?

I don't think so - what you wrote down was a a system with a conserved current coupled to a vector field, and this re-created the gauge symmetry, thus keeping in line with Noether's theorem. That's what I was trying to say earlier: by writing down a conserved current coupling to a vector field, you AUTOMATICALLY get a gauge symmetry (in your case, in the form coming from the Lorenz gauge (see below).

Remember: your original action in the first post did NOT have a current. Since you did not already specify that the current was conserved (which, once again, is the SAME THING as demanding gauge invariance), that's why I was saying that you were in trouble. Later, when you wrote down the current, I should have let up a little, but...

This must have been accidental. Are you talking about the Gauss peaks I used for charge distributions?

All I meant by the obnoxious "pulling a fast one" comment is that you have no right (at the outset) to decide that the current you wrote down was in fact the correct one - what you did was choose the correct current (which of course is divergenceless) and then found that it was divergenceless! But from the point of view of Field Theory, you generally do not know what the current is before you start - after all, that's part of the point: you want to be able to "compute" the current from first principles. That's all I was trying to say.

This starts to look like understanding. I must point out that your tone here...
is different from what it was here in the first response...

I think I was just being stupid back then! What I should have said is that your action can only be the correct action if you impose the Lorenz condition (back when you didn't have a current); or equivalently, when you wrote down your current, you had to make sure it was conserved (which then enforced the Lorenz condition automatically, as you pointed out).

The motivation btw stems from the QFT. If we define the system by using the Lagrangian with $(\partial_{\mu} A_{\nu})(\partial^{\mu} A^{\nu})$, then this is simply a four component Klein-Gordon field, and it can be quantized easily. If we use $F_{\mu\nu}F^{\mu\nu}$ instead, I don't really know how the quantization occurs. So far all texts I have encountered have been very confusing.

I'm going to change tactics with you on this point. I hope you don't mind.

You started by trying to quote Occam's Razor (something I've always taken issue with, between you and me, but that's besides the point). Let me take this opportunity to turn your question on its head. As a general rule, you are required to write down every operator that cannot be forbidden by some principle (symmetry) - since all such operators must be allowed to contribute. To write down a theory of a vector field, you need two derivatives (to make it a dynamical field) and two fields in the leading action. There are *two* independent ways to do this:

$$a(\partial_\mu A_\nu)(\partial^\mu A^\nu)+b(\partial_\mu A_\nu)(\partial^\nu A^\mu)$$

where a and b are some constants. In what you did, you took it upon yourself to set $b=0$. Can I now go back and ask: why would you ever do that? What right do you have? I say: to hell with Occam's Razor. QM says you must include everything, since if you don't, such operators will always be generated when you start computing quantum corrections. So why would you drop this second operator?

As to quantization: whether you have this term or not doesn't change quantization issues. For example: in your KG instance, you still seem to have FOUR fields oscillating. How do you make sure the ghost-like $A^0$ KG operator doesn't come along and violate unitarity? You still need the gauge symmetry to show that this can never happen! So I don't think setting $b=0$ above really did anything for you.

No matter what gauge you pick, you have to do work to quantize vector fields (there's a reason it took 10-20 years longer to figure out than scalar fields!) In the canonical formalism, you use the Gupta-Bleuler formalism; or in Path Integrals, you can use the Fadeev-Popov formalism (which is MUCH! easier to generalize to the non-abelian case, although G-B is good for a QFT student to have to suffer through once ). But in short answer to your question, I don't think your fix of going to Lorenz gauge (or Feynman-'tHooft gauge if you're talking about QED) will help you. Your problems await you in any gauge, I'm afraid.

Is it clear that the gauge symmetry and the charge conservation are related according to the Noether's theorem? If I coupled a complex Klein-Gordon field to the EM-field, wouldn't it still be the internal symmetry of the KG-field that corresponds to the conserved KG-current?

Yes, it is clear. I'm a little confused by this paragraph. Are you agreeing with me? The gauge symmetry applies to a complex KG field as well:

$$\phi(x)\rightarrow e^{i\lambda(x)}\phi(x)\qquad A_\mu(x)\rightarrow A_\mu(x)+\partial_\mu\lambda(x)$$

is still how the gauge symmetry acts, just like the fermions. So you are agreeing with me that the gauge symmetry and the charge conservation law are the same thing, yes? Or am I missing something?

 

[jostpuur]

Yes, it is clear. I'm a little confused by this paragraph. Are you agreeing with me? The gauge symmetry applies to a complex KG field as well:

$$\phi(x)\rightarrow e^{i\lambda(x)}\phi(x)\qquad A_\mu(x)\rightarrow A_\mu(x)+\partial_\mu\lambda(x)$$

is still how the gauge symmetry acts, just like the fermions. So you are agreeing with me that the gauge symmetry and the charge conservation law are the same thing, yes? Or am I missing something?

Oh! Yes! I forgot about this symmetry. I was thinking about the global complex phase rotation, where $\lambda$ did not depend on the x. Now I'll have to work with some details before returning to this. I'll be back...

 

[blechman]

Now I'll have to work with some details before returning to this. I'll be back...

I'll be waiting...

 

[jostpuur]

What am I doing wrong here?

$$\mathcal{L}(x) = -\frac{1}{4}F_{\mu\nu} F^{\mu\nu}\; +\; \frac{1}{2}(\partial_{\mu}\phi^*)(\partial^{\mu}\phi)\; - \;\frac{1}{2}m^2\phi^*\phi\; -\; gA_{\mu}\textrm{Im}(\phi^* \partial^{\mu}\phi)$$

This is a Lagrangian of an electromagnetic field and a complex scalar Klein-Gordon field, and the current

$$j^{\mu} = g\textrm{Im}(\phi^*\partial^{\mu}\phi) = -\frac{ig}{2}\big(\phi^*\partial^{\mu}\phi - (\partial^{\mu}\phi^*)\phi\big)$$

where g is some constant, being coupled to the four potential as usual.

The gauge transformation should be

$$A^{\mu}(x)\mapsto A^{\mu}(x) + \partial^{\mu}\Lambda(x) \quad\quad\quad \phi(x)\mapsto e^{i\alpha\Lambda(x)}\phi(x)$$

where alpha is some constant to be fixed later. In this transformation the following expressions transform as follows

$$F^{\mu\nu}\mapsto F^{\mu\nu}$$

$$(\partial_{\mu}\phi^*)(\partial^{\mu}\phi)\mapsto (\partial_{\mu}\phi^*)(\partial^{\mu}\phi)\; + \;2\alpha(\partial_{\mu}\Lambda)\textrm{Im}(\phi^*\partial^{\mu}\phi)\; + \;\alpha^2(\partial_{\mu}\Lambda)(\partial^{\mu}\Lambda)\phi^*\phi$$

$$\phi^*\phi\mapsto\phi^*\phi$$

$$A_{\mu}\textrm{Im}(\phi^*\partial^{\mu}\phi)\mapsto A_{\mu}\textrm{Im}(\phi^*\partial^{\mu}\phi)\; + \;(\partial_{\mu}\Lambda)\textrm{Im}(\phi^*\partial^{\mu}\phi)\; +\; \alpha A_{\mu}(\partial^{\mu}\Lambda)\phi^*\phi\; +\; \alpha(\partial_{\mu}\Lambda)(\partial^{\mu}\Lambda)\phi^*\phi$$

When I substitute these into the Lagrangian, I get

$$\mathcal{L}(x)\mapsto\mathcal{L}(x)\; +\; (\alpha - g)(\partial_{\mu}\Lambda)\textrm{Im}(\phi^*\partial^{\mu}\phi)\; + \;(\frac{1}{2}\alpha - g)\alpha (\partial_{\mu}\Lambda)(\partial^{\mu}\Lambda)\phi^*\phi\; -\; g\alpha A_{\mu}(\partial^{\mu}\Lambda)\phi^*\phi$$

The first additional term would vanish with choise $\alpha=g$, but the second term would vanish with $\alpha=2g$, and the third vanishes with $\alpha=0$. So I cannot get the Lagrangian invariant!

I remember I have calculated a similar exercise with the Maxwell+Dirac Lagrangian on some courses without problems, and it seems I've done some silly mistake here now, but I'm more interested in finding the mistake than hiding it... so that's why the post.

Oh yes! The situation was different with the Dirac current, because there is the gamma matrix instead of a derivative operator, in the current $\overline{\psi}\gamma^{\mu}\psi$. This is invariant in the local phase rotation, unlike the $\phi^*\partial^{\mu}\phi$. So... is there a mistake anywhere, after all? The Maxwell+KG system doesn't have the analogous symmetry to the Maxwell+Dirac system?

 

[blechman]

You're missing a term!

$$\Delta\mathcal{L}=g^2 A_\mu A^\mu \phi^{*}\phi$$

To see where this term comes from, better to use the covariant derivative:

$$\mathcal{L}_{\rm scalar QED}=|D_\mu\phi|^2-m^2|\phi|^2-\frac{1}{4}F^2$$

where

$$D_\mu\phi=(\partial_\mu+igA_\mu)\phi$$

The cross terms give you the current, but then there is that 2 photon - 2 scalar interaction that is not there in QED, and is vital for maintaining gauge invariance, as you so correctly just showed!

BTW: in case you're wondering, this is how the Higgs mechanism works: when $\phi$ gets a vacuum expectation value (goes to a constant if you don't know what that means), then this term generates a mass term for the photon! This is how the W and Z bosons get masses in particle physics. Also (if you're interested) it's how to derive the Meisner effect in (Landau-Ginzberg) superconductivity, where you can think of the photon as acquiring a mass inside the superconducting material. This sometimes goes under the name of "Anderson-Higgs" mechanism. That's the problem with spontaneous symmetry breaking: there are too many big names that contributed to it!

 

[jostpuur]

Distant memories from the particle physics course are coming back. I didn't understand anything on that course, and wasn't following very carefully anyway.

Does this mean that the conserving current is also

$$\textrm{Im}(\phi^* D^{\mu} \phi),$$

and not

$$\textrm{Im}(\phi^*\partial^{\mu}\phi)?$$

 

[blechman]

Distant memories from the particle physics course are coming back. I didn't understand anything on that course, and wasn't following very carefully anyway.

Does this mean that the conserving current is also

$$\textrm{Im}(\phi^* D^{\mu} \phi),$$

and not

$$\textrm{Im}(\phi^*\partial^{\mu}\phi)?$$

That's right, since your first current is not even gauge invariant!

 

[jostpuur]

blechman, you claimed I was doing something wrong, when I said that total charge conserves trivially when the system consists of discrete particles, even without gauge symmetries of fields. What do you think about the mass conservation discussion https://www.physicsforums.com/showthread.php?t=209929? Can you tell what is the symmetry behind the mass conservation, or is it the mass instead something that is trivially conserved with discrete particles, even without the corresponding symmetry?

If you say, that charge cannot be conserved without corresponding symmetry, but mass can, then I'm curious to hear explanations for this. After all the only difference between

$$\sum_{k=1}^N q_k$$

and

$$\sum_{k=1}^N m_k$$

is the used symbol!

 

[blechman]

sigh... that thread makes my head spin!

First of all, mass is not TRULY conserved - or else there would be no nuclear bombs! Rather, mass is a form of energy, and it is the energy that is conserved when time is homogeneous, according to Noether's theorem. This is manifest directly in the relativistic formulation.

Noether's theorem specifically states that if there is a symmetry, then there is a conserved charge. The symmetry that enforces electromagnetic charge conservation is the gauge symmetry of E&M. The symmetry that enforces ENERGY conservation is homogeneity of time.

If a particle is losing mass, then time is not homogeneous: as you move forward or backward in time, the particle has changed (lost or gained mass). So the symmetry is broken and so is the conservation law!

As to formulating this with a Lagrangian, I don't know what those people on that thread were thinking! If mass is now a DYNAMICAL quantity, you must include its dynamics! That means including some Lagrangian density that describes the flow of mass. It's exactly the same as E&M! If you have a static field, you can just write the Lagrangian as a particle moving in a background field. But if your field is allowed to evolve in time as well, then you must include you $\int d^4x (-1/4)F_{\mu\nu}F^{\mu\nu}$ term as well to take that into account! I'll post this on that thread as well.