# Wave eqn. Greens func. depends only on rel. distance proof.

1. Feb 26, 2014

### center o bass

The Greens function for the wave equation satisfies the differential equation

$$\square G(x,x') = \delta^{(4)}(x-x')$$

where $\square= \partial_\nu \partial^\nu$ is the wave operator and $x,x'$ are four vectors labeling events in spacetime. If we require $G(x,x') \to 0$ as $x \to \infty$ it is often stated that $G(x,x')$ must depend only on the relative distance due to the symmetry of the problem. But I wonder, how can one prove this rigorously?

2. Feb 26, 2014

### hilbert2

It follows from the translation invariance of the problem: if the function $f(x)$ satisfies the wave equation $\partial_{\nu}\partial^{\nu}f(x)=0$, then also the shifted function, $f(x+c)$, where $c$ is any constant four-vector, satisfies the same equation.

Try forming a "shifted" Green's function $G(x+c,x'+c)$ and show that if it's not equal to $G(x,x')$, you get a contradiction with the translation invariance.