Electromagnetic gauge invariance with boundary conditions

[Xezlec]

Hello. I'm trying to wrap my head around how Lagrangians work in classical field theory.

I have a book that is talking about the gauge invariance of the Lagrangian: \mathscr{L} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}-J^\mu A_\mu. It shows that we can replace A^\mu with A^\mu+\partial^\mu\chi for some arbitrary field \chi and this adds an extra term to the action: \Delta S=- \int J_\mu\,\partial^\mu\chi\,d^4x. Ok, clear enough.

Now here's the trouble: from this they derive that \Delta S=\int \partial^\mu J_\mu\,\chi\,d^4x (and that's how we get charge conservation), by integrating by parts and assuming the boundary terms (J_\mu\chi I guess) vanish. But why should the boundary terms vanish, especially in the time dimension? If I consider a region of space between two time instants such that some opposite charges separate, and remain separated at the end time, shouldn't the behavior of the field still minimize the action within that boundary? And shouldn't charge still be conserved? So why do I wind up with an action that differs depending on the gauge? It looks like that J_\mu\chi is now going to have some interesting value at the boundary.

I hope I said that clearly.

 

[samalkhaiat]

I hope I said that clearly.

Well, it was not clear enough to me. However, if you are asking why \int d^4x \partial_{\mu}(\chi J^{\mu}) = 0, have look at post #7 in
https://www.physicsforums.com/threads/gauge-invariance-of-interaction-lagrangian.775672/

 

[Xezlec]

Well, it was not clear enough to me.

Sorry, can you elaborate any? What part was unclear?

Do you understand the derivation that I'm referring to? I can paste the book's exact wording if that would help. I'm just trying to understand why the boundary terms can be neglected.

 

[samalkhaiat]

Sorry,
I'm just trying to understand why the boundary terms can be neglected.

Did you read post #7 in the link I gave you?

 

[Xezlec]

Did you read post #7 in the link I gave you?

Maybe I'm just missing something really obvious, but I can't seem to follow that discussion. Specifically, I don't understand the first equality in post #1, nor what \Lambda represents. This seems to be a different presentation of gauge symmetry than the one in my book.

The book I'm reading illustrates a gauge transformation as the addition of an arbitrary curlfree vector field to A_\mu. Such a curlfree vector field is conveniently formed by taking the gradient of some arbitrary scalar field \chi. They then look at the extra term that this transformation adds into the resulting action, and require that this extra term be identically zero. At no point in the derivation does a derivative of a product occur.

 

[samalkhaiat]

Maybe I'm just missing something really obvious, but I can't seem to follow that discussion. Specifically, I don't understand the first equality in post #1, nor what \Lambda represents.

\Lambda in there denotes the gauge function which you call \chi

This seems to be a different presentation of gauge symmetry than the one in my book.

There is no difference. So, you have no problem with \Delta S = - \int d^{4}x \ J_{\mu} \partial^{\mu}\chi? Now, use
- J_{\mu} \partial^{\mu}\chi = \chi \partial^{\mu}J_{\mu} - \partial^{\mu}(\chi J_{\mu}) .
You know that the current is conserved, i.e., \partial^{\mu}J_{\mu} = 0. So
\Delta S = - \int d^{4}x \ \partial^{\mu}(\chi J_{\mu}) .
Now, use the divergence theorem to go to 3D integral over the boundary "surface" M
\Delta S = - \int_{M} d^{3}x \ n^{\mu} \chi J_{\mu} ,
where, n^{\mu} is the outer pointing unit vector on the boundary surface M. Now, go to that post and try to understand why this integral vanishes.

 

[Xezlec]

Now, use
- J_{\mu} \partial^{\mu}\chi = \chi \partial^{\mu}J_{\mu} - \partial^{\mu}(\chi J_{\mu}) .

Whoops! Duh. Thanks. Not sure how I confused myself on that.

Now I can rephrase my problem in terms of post #7 on that other thread: I now fully agree that if we restrict \Lambda to be constant on \partial D_1 and \partial D_2, everything works out properly. But why should that restriction be true? \Lambda is an entirely arbitrary scalar function, right?

 

[samalkhaiat]

\Lambda is an entirely arbitrary scalar function, right?

That is correct. However, \Lambda must be well-behaved so that the physical quantities do not depends on it. To see this, let us define the following “current”
J^{\mu}_{\Lambda} \equiv J^{\mu}_{em}\Lambda - \frac{\partial \mathcal{L}}{\partial(\partial_{\mu}A_{\nu})} \partial_{\nu}\Lambda , \ \ \ \ \ (1)
where,
J^{\mu}_{em}= \frac{\partial \mathcal{L}}{\partial A_{\mu}},
is the conserved electromagnetic current of the matter fields. Notice that J^{\mu}_{\Lambda} \propto J^{\mu}_{em}, when \Lambda is constant.
Using the equation of motion for A_{\mu}, we can rewrite (1) as
<br /> \begin{align*}<br /> J^{\mu}_{\Lambda} &amp;= \frac{\partial \mathcal{L}}{\partial A_{\mu}} \Lambda - \frac{\partial \mathcal{L}}{\partial(\partial_{\mu}A_{\nu})} \partial_{\nu}\Lambda \\<br /> &amp;= \partial_{\nu} \left(\frac{\partial \mathcal{L}}{\partial(\partial_{\nu}A_{\mu})} \right) \Lambda + \frac{\partial \mathcal{L}}{\partial(\partial_{\nu}A_{\mu})} \partial_{\nu} \Lambda \\<br /> &amp;= \partial_{\nu} \left(\frac{\partial \mathcal{L}}{\partial(\partial_{\nu}A_{\mu})} \Lambda \right) .<br /> \end{align*}<br />
Clearly, this current is identically conserved, \partial_{\mu}J^{\mu}_{\Lambda}=0, because of the antisymmetry in (\mu\nu). The charge Q_{\Lambda}, associated with this current, is obtained by integrating J^{0}_{\Lambda} over the whole 3D volume,
Q_{\Lambda} = \int d^{3}x \ J^{0}_{\Lambda}(x) .
Now
\frac{d}{dt}Q_{\Lambda} = \int d^{3}x \ \partial_{0}J^{0}_{\Lambda}(x) = \int d^{3}x \left( \partial_{\mu}J^{\mu}_{\Lambda} - \vec{\nabla} \cdot \vec{J}_{\Lambda} \right) .
In the last integral, the first term vanishes by current conservation, and the second can be converted to a surface integral at infinity:
\frac{d}{dt}Q_{\Lambda} = - \oint_{\infty} d\vec{S} \cdot \vec{J}_{\Lambda} .
Thus, the charge is conserved, dQ_{\Lambda}/dt = 0, provided \vec{J}_{\Lambda} falls off sufficiently fast as |\vec{x}| \to \infty. Thus, in order for Q_{\Lambda} to be conserved, \Lambda (\infty) must be bounded.
Let’s look at the explicit form of Q_{\Lambda}:
Q_{\Lambda} = \int d^{3}x \partial_{j}\left( \frac{\partial \mathcal{L}}{\partial(\partial_{j}A_{0})} \Lambda (x) \right) = \oint_{\infty} d\vec{S} \cdot \vec{E} \ \Lambda (x) .
It thus appears that we have an infinite class of conserved charges! But, there is only one physical charge, which is the electromagnetic charge
Q_{em} = \oint d\vec{S} \cdot \vec{E} .
Notice though E \sim \frac{1}{r^{2}}. So, if \Lambda approaches an angle-independent limit at infinity, where the integral is evaluated, we then obtain
Q_{\Lambda(\infty)} = \Lambda(\infty) \ Q_{em} ,
and we really have just one physical charge.

 

[Xezlec]

OK, I think I figured it out: since \Lambda is arbitrary, we ought to be able to choose a \Lambda that makes the boundary term go away, and the Lagrangian should be invariant. When we do that, we find that charge must be conserved. I guess I was thinking of the idea as being that gauge invariance and charge conservation are equivalent, but really the simple argument in my book only shows that the former implies the latter.