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

  1. Feb 26, 2014 #1
    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. jcsd
  3. Feb 26, 2014 #2

    hilbert2

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    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.
     
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