Derivation of Lorentz invariant

  • Thread starter ralqs
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  • #1
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Some time ago, I came across a nice justification (by Einstein IIRC) for the formula [tex]x'^2 + y'^2 + z'^2 - c^2t'^2 = x^2 + y^2 + z^2 - c^2t^2[/tex].

The argument went something like this:
(1) [tex]x'^2 + y'^2 + z'^2 - c^2t'^2 = x^2 + y^2 + z^2 - c^2t^2 = 0[/tex] for light.
(2) *reasoning I forget*, therefore [tex]x'^2 + y'^2 + z'^2 - c^2t'^2 = \sigma(x^2 + y^2 + z^2 - c^2t^2)[/tex]
(3) [tex]y'^2 = y^2 [/tex], so [tex]\sigma = 1[/tex].

I can't remember or deduce the argument for step 2. I'm guessing it came down to some argument about linearity...does anyone have any ideas? Thanks.
 

Answers and Replies

  • #2
Matterwave
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The argument is (ds')^2=f(p)ds^2+C

Where f(p) is some function of space, velocity, and time. The reason we can expect this form is that we can expect ds to at least be an infinitesimal of the same order when we make a coordinate change. You can get rid of C by your point (1). You can further say that f(p) cannot depend on space and time because of homogeneity and isotropy of space and homogeneity of time. You can go another step and say that f(p) can then only depend on the absolute value of the relative velocity. For the last step, you can transform to and back from ds' to ds to see that (f(p))^2=1. You conclude therefore that f(p)=1 (I don't recall the argument for dropping the negative 1 option).
 
  • #3
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Hmmm, that only proves that ds'2 = ds2, which doesn't necessarily mean that (delta)s' = (delta)s.

At any rate, I figured out another way of proving it based on your idea, so thanks anyways!
 
  • #4
Matterwave
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To go to the finite case, you can simply integrate.
 

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