# Gauge transformation and Occam's razor

1. Oct 19, 2007

### 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.

2. Oct 23, 2007

### 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.

Last edited: Oct 23, 2007
3. Oct 23, 2007

### 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!

4. Oct 23, 2007

### cesiumfrog

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?

5. Oct 24, 2007

### 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.

6. Oct 24, 2007

### cesiumfrog

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?

7. Oct 24, 2007

### 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!

8. Oct 25, 2007

### jostpuur

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 $$\sum_{k=1}^N q_k[/itex] is quite conserved. Or is there some symmetry here, that I don't know? Anyway, my point was, that the $F_{\mu\nu} F^{\mu\nu}$ still seems unnecessarily complicated. Last edited: Oct 25, 2007 9. Oct 25, 2007 ### jostpuur Oh well, the equations of motion are [tex] \partial_{\mu}\partial^{\mu} A^{\nu}(x) = \sum_{k=1}^N q_k\delta^3(x-x_k)(1,v_k)^{\nu}$$
$$F_k = -q_k(\nabla A^0 + \partial_0 A) + q_k v_k\times(\nabla\times A),$$

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.

Last edited: Oct 25, 2007
10. Oct 25, 2007

### 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, irregardless 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.

11. Nov 13, 2007

### jostpuur

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

12. Nov 13, 2007

### 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.

13. Nov 14, 2007

### jostpuur

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)

14. Nov 16, 2007

### 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.

15. Nov 16, 2007

### jostpuur

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.

16. Nov 16, 2007

### 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.

Last edited: Nov 16, 2007
17. Nov 16, 2007

### jostpuur

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 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?

Last edited: Nov 16, 2007
18. Nov 16, 2007

### blechman

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!

19. Nov 16, 2007

### 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.

20. Nov 17, 2007

### 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?

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

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

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 $\partial_{\mu} A^{\mu} = 0$.