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Ether before relativity in quantum theory also

  1. Feb 16, 2007 #1
    I think this all makes sense to me, but I've never heard of this in lectures or in books, so I'll check if I'm getting this right.

    In one dimension Galilean coordinate transformations are [tex]x'=x-ut[/tex] and [tex]t'=t[/tex]. Momentum transforms as [tex]p'=p-mu[/tex], and energy is [tex]E=p^2/(2m)[/tex]. With a quick calculation I get [tex]E't'-p'x'=Et-px-\frac{1}{2}mu^2t-mux[/tex]. So this means, that a wave plane solution [tex]\psi(t,x)=\exp(i(Et-px))[/tex] of the Shrodinger equation is not a solution in other inertial frames? In relativistic theory it goes better, as [tex]p_\mu x^\mu[/tex] is Lorentz invariant, and a solution of Klein-Gordon equation is always a solution also in other inertial frames.

    So if I assume, that a wave function is real in the sense, that there truly is some complex number associated with each space time point, in non-relativistic theory I must assume an ether coordinate set? I think it somehow remarkable, that a Shrodinger equation is not Galilean invariant like the Klein-Gordon or Dirac equations are Lorentz invariant. :uhh:
    Last edited: Feb 16, 2007
  2. jcsd
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