Gauge fixing in classical electromagnetism

The Classical Solutions

Classical Electromagnetism Lagrangian:

$$\mathcal{L} = - \frac{1}{4}F^{\mu \nu}F_{\mu\nu} + J_{\mu} A^{\mu},$$ where, $F^{\mu \nu} \triangleq \partial^\mu A^\nu - \partial^\nu A^\mu$. The metric is defined as: $$g^{\mu \nu} = \begin{pmatrix} 1 & & & \\ & -1 & & \\ & & -1 & \\ & & & -1 \\ \end{pmatrix}.$$ The theory has an nonphysical gauge symmetry (nonphysical degrees of freedom). But we can still get solutions in classical electromagnetism. The Equation of Motion reads, $$( g_{\mu \nu} \partial^2 - \partial_\mu \partial_\nu ) A^\nu = - J_{\mu}. \qquad (1)$$ The free space solution can be obtained by Fourier Transform: $$( k^2 g^{\mu \nu} - k^\mu k^\nu ) A_\nu = J^\mu.$$ It is convenient to introduce an operator $P^\mu_\nu \equiv \delta^\mu _\nu - \hat{k}^\mu \hat{k}_\nu$. It can be shown, $P\cdot P = P$. Therefore, $P$ is not invertible. We can seek for its pseudo-inverse. Notice that its eigenvalues satisfy $p^2 = p$. Hence $p = 0, 1$. Therefore $P$ is a pseudo-inverse of itself. The full solution can be obtained: $$A^\mu = \frac{g^{\mu \nu} - \frac{k^\mu k^\nu}{k^2}}{k^2} J_\nu + \frac{k^\mu k^\nu} {k^2 } X_\nu$$ where $X_\nu$ is an arbitrary 4-vector. If we choose $X_\nu = \xi J_\nu /k^2$, $A^\mu$ becomes $$A^\mu = \frac{g^{\mu \nu} - (1-\xi) \frac{k^\mu k^\nu}{k^2}}{k^2} J_\nu.$$ The resulted solutions are obviously less than the full solution. Equivalently, we have imposed a constraint on field $A^\mu$. This constraint is called a gauge. The gauge we choose here is called $R_\xi$ gauge, frequently used in Gauge Theory (Quantum Field Theory). $R_\xi$ gauge requires $k\cdot A= 0$ or $\partial_\mu A^\mu = 0$. In fact, in classical electromagnetism, the current is conserved $\partial_\mu J^\mu = 0$ and the general solution is always $$A^\mu = \frac{g^{\mu \nu}}{k^2} J_\nu + \frac{k^\mu k^\nu} {k^2 } X_\nu$$

The Lagrangian with Gauge Fixing

As we mentioned above, gauge fixing introduces a constraint to choose an explicit gauge: $$\mathcal{G}[A] = 0$$
Now we face a system with constraint(s). The normal procedure is to introduce a Lagrange multiplier term.  $$\mathcal{L} = - \frac{1}{4}F^{\mu \nu}F_{\mu\nu} + J_{\mu} A^{\mu} + \lambda \mathcal{G}[A].$$ In this post, we first investigate a co-variant gauge, the Lorenz gauge. By further assume the theory living in an d+1 Minkowski space (hence surfaces terms can be suppressed), the Lagrangian becomes, $$\mathcal{L} = \frac{1}{2} A^\mu ( g_{\mu \nu} \partial^2 - \partial_\mu \partial_\nu ) A^\nu + J_{\mu} A^{\mu} + \frac{\lambda}{2} A^\mu \partial_\mu \partial_\nu A^\nu.$$  The last term, $\frac{1}{2} \lambda(x) A^\mu \partial_\mu \partial_\nu A^\nu$ is a Lagrangian multiplier, that imposes gauge condition $( \partial_\nu A^\nu )^2 = 0$.

Equation of Motion and Green's Function

The resulting Equation of Motion contains two equations:

$( g_{\mu \nu} \partial^2 + (\lambda - 1) \partial_\mu \partial_\nu ) A^\nu = - J_{\mu}. \qquad (1)$
$( \partial_\nu A^\nu )^2 = 0 \qquad (2)$

$(1), (2)$ imply $\partial^2 A^\mu = - J^{\mu}$. The solution in free space is $$A^\mu(x) = \int \mathrm{d}^4 x' J^\mu(x') \int \frac{\mathrm{d}^4 k}{(2\pi)^4} \frac{e^{i k \cdot (x-x')}}{k^2} = \int \mathrm{d}^3 \mathbf{x'} \frac{J^\mu(x'_\pm)}{|\mathbf{x-x'}|}$$ where $x_\pm$ represents retarded and advanced coordinates: $x'_\pm = (t\mp |\mathbf{x-x'}|, \mathbf{x'})$.

Quantum Field Theory with Lagrange Multiplier

It's possible to start from a Lagrangian with Lagrange multiplier field to do QFT. Note that such a QFT converges to classical electromagnetism in classical limit. But whether or not it describes the true physics, is another issue.

references:
[1]: Mark Srednicki, Quantum Field Theory.