1
$\begingroup$

Varying the electromagnetic action \begin{equation} S=-m c \int d s\left(\dot{z}^{2}\right)^{\frac{1}{2}}-\frac{e}{c} \int d s A_{\mu} \dot{z}^{\mu}-\frac{1}{16 \pi c} \int d^{4} x F_{\mu \nu} F^{\mu \nu} \end{equation} we get the two equations of motion \begin{equation} m \ddot{z}^{\mu}=\frac{e}{c^{2}} F^{\mu \nu} \dot{z}_{\nu} \end{equation}

\begin{equation} \square A^{\mu}-\partial^{\mu}\left(\partial_{\nu} A^{\nu}\right)=\frac{4 \pi}{c} j^{\mu} \end{equation} If we use the Lorenz gauge, the solutions to the second equation are the Liénard–Wiechert potentials.

Is there a general solution to the second equation without fixing a gauge? By solution I mean a closed expression for $A^{\mu}$ in terms of $j^{\mu}$. Or is it possible to proof that such a solution does not exist.

$\endgroup$
3
$\begingroup$

As my2cts said the equation is not invertible. Here's a quick proof. First we go to momentum space by Fourier transforming both sides of the equation

$$ (-k^2 \eta^{\mu\nu}+ k^\mu k^\nu )\tilde A_\nu(k) = \frac{4\pi}{c} \tilde j(k)^\mu$$

Now the solution for $\tilde A(k)$ would be simply given by solving the equation algebrically and then doing an inverse Fourier transform. Thing is, that the tensor operator on the LHS is not invertible.

You can see this easily by noticing that that operator is zero on the vector space spanned by $k^\mu$, since

$$ (-k^2 \eta^{\mu\nu}+ k^\mu k^\nu ) k_\nu = -k^2 k^\mu + k^\mu k^\nu k_\nu = 0$$

The reason why you have to fix the gauge before solving these equations is to get rid of this "null direction". For example imposing $\partial^\mu A_\mu = 0$ is equivalent, in momentum space to $k^\mu A_\mu = 0$. This condition prevents explicitly any configuration that would make the above operator zero and makes it invertible.

$\endgroup$
1
  • $\begingroup$ Thanks a lot. So the argument is, that the operator is not injective, i.e. that it has a nontrivial kernel and thus is not invertible. $\endgroup$
    – NicAG
    Jun 1 at 9:24
1
$\begingroup$

If your question is, as the title suggests, whether the second equation can be inverted, the answer is no. Inversion requires 'gauge fixing'. An operator can be inverted only if it establishes a one to one relation between solution and source. The operator in your second equation assigns many potentials to a single source and hence cannot be inverted.

$\endgroup$
1
  • $\begingroup$ Thanks. Can you maybe provide a reference to a proof that an inversion is not possible? $\endgroup$
    – NicAG
    May 31 at 22:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.