Gauge Invariance in Electrodynamics

Question

I am studying Electrodynamics and I have been introduced to the concept of Gauge Invariance.

This was introduced by noting that $E$ and $B$ amount to 6 six degrees of freedom and the Maxwell equations amount of 3 degrees of freedom. On the other hand, if we write $$E = - \nabla \phi - \frac{\partial A}{\partial t}, \qquad B = \nabla \times A$$

this contains 4 degrees of freedom.

The extra degree of freedom forms part of this Gauge invariance.

My lecture notes go on to talk about the Neumann gauge and the Lorenz gauge and how these are both 'natural' choices for a Gauge.

I have come to Stack Exchange because I am fairly confused. I am not sure what a 'gauge' even is and what their point is. It's not obvious from what I've put above...

Furthermore, I read on in my lecture notes and it says that in the Lorenz gauge, $A$ and $\phi$ satisfy wave equations. Again, I don't see how this is useful, but maybe a user on here can shed some light on gauges and this will make sense.

Answer 1

Answering the reason it is useful:

In the Lorenz gauge, the wave equations are

$$\mu_{0} \epsilon_{0}\frac{\partial^2\Phi}{\partial t^2}-\vec{\nabla}^2\Phi = \frac{\rho}{\epsilon_{0}},$$

$$\mu_{0} \epsilon_{0}\frac{\partial^2\vec{A}}{\partial t^2}-\vec{\nabla}^2\vec{A} = \mu_{0}\vec{J}$$.

We can combine $\Phi$ and $\vec{A}$ into a 4-vector, called the 4-potential, and combine the wave equations into a single expression:

$$A^{\mu} = \left(\Phi, \vec{A}\right),$$

$$\partial^{\mu}\partial_{\mu}A^{\alpha} = \frac{4\pi}{c}J^{\alpha}.$$

With the 4-potential now defined, we can go back to the fields and write them in terms of the 4-potential component partials:

$$E_{x} = -\left(\partial^{0}A^{1} + \partial^{1}A^{0}\right), \textrm{ etc.}$$

$$B_{x} = -\left(\partial^{2}A^{3} + \partial^{3}A^{2}\right), \textrm{ etc.}$$

(For the $\vec{B}$ components, the indexes are permuted according to the cross product.) We want the fields to be Lorentz invariant, but it becomes clear at this point they cannot be replaced by an ordinary 4-vector. Instead, we can replace them with a rank-2 antisymmetric field tensor:

$$F^{\alpha\beta} = \partial^{\alpha}A^{\beta} - \partial^{\beta}A^{\alpha}.$$

The non-zero components of $F^{\alpha\beta}$ are just the components of $\vec{E}$ and $\vec{B}$, with either positive or negative signs depending on the specific component. The diagonal elements of $F^{\alpha\beta}$ all vanish.

From here, we would proceed to verify that $F^{\alpha\beta}$ is Lorentz invariant, and eventually we derive the Lorentz transformation equations for $\vec{E}$ and $\vec{B}$. The choice of gauge at the beginning and identifying the 4-potential $A^{\mu}$ makes all this possible.

All of this is a choice of gauge, i.e., it's a choice of a particular $\vec{A}$ and $\Phi$. The Lorenz gauge is useful because it is Lorentz invariant. See What is a gauge in a gauge theory?.

Answer 2

My edited answer: The concept of electromagnetic gauge invariance is useless. It is quite possible to formulate the theory of electromagnetism without this invariance. See my paper, of which I am proud, on https://arxiv.org/abs/physics/0106078 , published as Eur. Phys. J. D, vol. 8, p 9-12 (2000).