Showing posts with label Green's function. Show all posts
Showing posts with label Green's function. Show all posts

Feb 29, 2012

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.