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Relativity: what's wrong with this logic?

  1. Feb 19, 2012 #1
    Okay, suppose we know that the laws of physics are invariant with respect to a shift in position, and invariant wrt a shift in time (ie the transformations [itex]\mathbf{r}\ \rightarrow\ \mathbf{r}+\delta\mathbf{r}\ \mathrm{and}\ t\ \rightarrow\ t+\delta t[/itex] preserve the laws of physics). Then wouldn't that imply that the laws are *also* invariant wrt a change in inertial reference frame because a change in velocity amounts to continuously alternating between the transformation [itex]\mathbf{r}\ \rightarrow\ \mathbf{r}+\delta\mathbf{r}\ \mathrm{and}\ t\ \rightarrow\ t+\delta t[/itex]? Or is there something wrong with this logic?

    Incidentally, I posted this in another site, and someone replied that the logic is fine. But I'm skeptical: I've always viewed the principle of relativity as an experimental fact which is not deducible from other, simpler symmetries.
     
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  3. Feb 19, 2012 #2

    bcrowell

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    As a counterexample, take a theory involving two particles at positions [itex]\mathbf{r}_1[/itex] and [itex]\mathbf{r}_2[/itex]. This theory has one law of physics, which is that [itex]d\mathbf{r}_1/dt=-d\mathbf{r}_2/dt=k(\mathbf{r}_1-\mathbf{r}_2)[/itex]. This predicts that the particles scoot directly away from each other at an exponentially increasing speed. The theory is clearly not invariant with respect to Galilean relativity or Lorentz transformations. For example, in a frame that happens to be co-moving with particle 1 at a certain time, particle 1 disobeys the law of motion. But the law of motion is form-invariant with respect to translation in time and space.
     
  4. Feb 19, 2012 #3
    Perhaps there exists another coordinate-transformation law which preserves your law of physics.

    One could have said something similar about Maxwell's equations: they are symmetric wrt temporal and spacial translations, but are not invariant wrt Galilean transformations. But they contain within themselves the seeds of the Lorentz transformations, which do preserve Maxwell's equations.
     
  5. Feb 19, 2012 #4

    PeterDonis

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    No, it doesn't. You can translate in space and time without changing velocity, and you can change velocity without translating in space and time. They are independent transformations.

    For example, the Lorentz transformation:

    [tex]x' = \gamma \left( x - vt \right)[/tex]

    [tex]t' = \gamma \left( t - vx \right)[/tex]

    boosts the velocity but does not translate in either space or time. And of course a translation like

    [tex]x' = x + X_{0}[/tex]

    [tex]t' = t + T_{0}[/tex]

    translates in space and time but doesn't change velocity.
     
  6. Feb 20, 2012 #5

    Mentz114

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    The laws of physics need to be locally invariant under translation. What you've described is global invariance. Local invariance is expressed by
    [tex]
    x^\mu \rightarrow x^{\mu '}=x^\mu + \epsilon^\mu(x^\mu)
    [/tex]
    which is saying that the translations depend on position. Global invariance entails action-at-a-distance which is not relativistic.

    Interestingly, the translation group is associated with energy and momentum, which are the sources of gravity, and it's possible to make gravity appear in the same way as the Lorentz force, by demanding local translation invariance.
     
  7. Feb 20, 2012 #6

    bcrowell

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    You're missing the point. The logic in your #1 can't be correct, because if it were correct, it would apply to this theory, but the contradiction it leads to when applied to this theory is false.
     
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