# Electromagnetic gauge invariance with boundary conditions

1. Dec 28, 2015

### 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.

2. Dec 28, 2015

### samalkhaiat

3. Dec 29, 2015

### Xezlec

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.

4. Dec 29, 2015

### samalkhaiat

5. Dec 29, 2015

### Xezlec

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.

6. Dec 29, 2015

### samalkhaiat

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

7. Dec 30, 2015

### Xezlec

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?

8. Dec 31, 2015

### samalkhaiat

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
\begin{align*} J^{\mu}_{\Lambda} &= \frac{\partial \mathcal{L}}{\partial A_{\mu}} \Lambda - \frac{\partial \mathcal{L}}{\partial(\partial_{\mu}A_{\nu})} \partial_{\nu}\Lambda \\ &= \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 \\ &= \partial_{\nu} \left(\frac{\partial \mathcal{L}}{\partial(\partial_{\nu}A_{\mu})} \Lambda \right) . \end{align*}
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.

9. Dec 31, 2015

### 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.