# Electromagnetic energy is not Gauge invariant?

1. Feb 8, 2009

### JustinLevy

I assume I am making a mistake here. Can you please help me learn how to fix them?

In electrodynamics, the gauge transformations are:
$$\vec{A} \rightarrow \vec{A} + \vec{\nabla}\lambda$$
$$V \rightarrow V - \frac{\partial}{\partial t}\lambda$$

These leave the electric and magnetic fields unchanged. The electromagnetic energy density is proportional to:
$$u \propto (E^2 + B^2)$$
So the energy density should be gauge invariant. A guage transformation shouldn't even shift it by a constant.

However, the energy density can also be written in terms of the potentials and charge distributions as:
$$u = \frac{1}{2}(\rho V + \vec{j} \cdot \vec{A})$$

If I let $\lambda$ = e^(-r^2), then V is unchanged, and A is just changed by a radial field which vanishes at infinity. Here, the energy density IS changed, and not by a constant amount either ... it changes by an amount depending on j. What gives?

Even weirder, is some textbooks start with the potentials form to derive the fields form. I don't see any place they fix any gauge in doing such derivation. Please help.

2. Feb 8, 2009

3. Feb 8, 2009

### JustinLevy

I wanted to talk about the electromagnetic energy density (proportional to $(E^2+B^2)$), but you have instead talked about a relativistic scalar density (proportional to $(E^2-B^2)$). Therefore I am not sure how to relate your response back to the original question. Are you saying $A_{\mu}J^{\mu}$ is gauge invariant but the electromagnetic energy density is NOT?

I understand that
$$\partial_{\mu} J^{\mu}=0$$
because it is a statement of conservation of charge. But I don't understand why the following is zero for every possible J or lambda
$$\int d^{3} x \ \delta (A_{\mu}J^{\mu}) = \int d^{3} x \ J^{\mu} \partial_{\mu} \lambda = 0$$.

Let's try just a simple circulating current, and a lambda polynomial in x and y:
$$\vec{j} = y\hat{x} + x\hat{y}$$
$$\lambda = xy$$
noting that for this choice
$$\partial_{\mu} \lambda = y\hat{x} + x\hat{y}$$

Looking at the result:
$$\int d^{3} x \ J^{\mu} \partial_{\mu} \lambda = \int d^{3} x \ (y\hat{x} + x\hat{y}) \cdot (y\hat{x} + x\hat{y}) \neq 0$$

So that doesn't seem to be gauge invariant either!?

4. Feb 11, 2009

### samalkhaiat

Last edited: Feb 11, 2009
5. Feb 14, 2009

### JustinLevy

Ah, okay. That is why my "counter-example" fails.
The rest of what you wrote makes sense as well.