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Unnecessary step in deriving the Lorentz transformations?

  1. Jan 30, 2016 #1


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    I'm going through Ray D'Iverno's "Introducing Einstein's Relativity", and there is a step he makes in deriving the Lorentz transformations that doesn't seem necessary to me. So I'm not sure what I'm missing. He derives them from Einsteins postulates of relativity. From the postulate that the speed of light is the same in all reference frames, we take a rest frame S and a frame moving with respect to S, S'. When their origins meet, a pulse of light moves out from their origins as a sphere. The events constituting this sphere satisfy the equations I=x2+y2+z2-c2t2=0 for the rest frame and I'=x'2+y'2+z'2-c2t'2=0 for the moving frame. He sets these equations equal and derives the Lorentz transformations, which is pretty straight forward. The step that I'm not sure about is before setting I=I', he says, "..it follows that under a transformation connecting S and S', I=0 ⇔ I'=0, and since the transformation is linear.." (by the first postulate) "we may conclude I=nI'." He then goes on to show how we can reverse the roles of S and S' giving I'=nI and combining the equations gives, n2=1 ⇒ n=±1, and in the limit as v→0 the two frames coincide so I' → I so we must take n=1. It's at this point that he sets I=I'. That's fine with me, I just don't understand why all of that was necessary in showing I=I'. Why couldn't we have just concluded that I=I' when we said earlier that I=0 and I'=0? I'm suspicious that there is some necessary mathematical rigor that I'm missing, in this step.
  2. jcsd
  3. Jan 30, 2016 #2


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    Because just knowing that l=l' when both are zero does not guarantee that l=l' when both are not zero.
  4. Jan 30, 2016 #3


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    What other reason would you have had for concluding that I'=I? Given that I=0 iff I'=0, you could have, for example, I=-3I'.
  5. Jan 30, 2016 #4


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    Ah of course, thanks.
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