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What is the Limits of Lorentz Transformation when (v/c->1)?.

  1. Feb 27, 2005 #1
    Is the limit for lorentz transformation when v/c -> 1 known?


    Is there a proof which says that such a limit doesnt exist?

    Please throw light on the above questions.
  2. jcsd
  3. Feb 27, 2005 #2
    First may be we should look at the mathematical possibility, then later ponder upon the physics significance and applicability.
  4. Feb 27, 2005 #3


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    Just look at the formula

    x' = (x-vt)/sqrt(1-v^2)

    Obviously there is no general limit as v-> 1. A limit can only exist if the limit as v->1 of x-vt is zero.

    Let v=1-eps. Then we have

    (x-vt) / sqrt(1-(1-eps)^2) =
    (x-vt) / sqrt(eps)sqrt(2-eps)

    Taking the limit as eps goes to zero we have

    (x-vt) / (sqrt(2)(sqrt(eps)))

    So there will be no limit if x-vt is proportional to eps, because we would have
    k*eps/sqrt(eps), which goes to infinity.

    So it takes a highly unusual set of circumstances for the limit to exist. Similar concerns apply for the other part of the Loerntz transform.
    Last edited: Feb 27, 2005
  5. Feb 27, 2005 #4
    Thanks Pervect. Just a clarification.

    I suppose you mean the formula is:

    x' = (x-vt)/sqrt(1-(v/c)^2)

    and What you intend to say is when v/c->1, a limit will eixts only is x-vt is zero. Right?

    So limits for lorentz transformation might hold good for x coordinates where x-vt=0. And as we are taking the case where v->c. So in a way we can extend the statement to say that the limit will exist for coordinates where x-ct=0 or where x=ct and not for any point where x>ct or x<ct.

    If so what is that limit for x' when x=ct? Is it 0?

    Infact the same is true for the transformation of time coordinates, the limits for t' will exists only when x=ct. That is when the numerator is zero like the denominator.

    If so what is that limit for t' when x=ct? Is it 0?
  6. Feb 27, 2005 #5
    What is the physical significance of the lorentz transformation limit (when v/c->1) existing when x=ct (if it exists?) and the significance of the limit not existing
    when x <> ct.
  7. Feb 27, 2005 #6


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    Basically, yes. The first thing I did was to assume c=1, which I usually do, but I should have mentioned that! Otherwise my post is just too unclear.

    So I'll mention that I'm assuming c=1 now, belatedly.

    The question is not only does x-vt go to zero in the limit as v/c->1 (or, giving my assumption that c=1, the limit as v->1), but how fast x-vt goes to zero. x-vt is a function of v, after all. If x and t are not functions of v, then by setting v = 1-eps

    (x-vt) will be (x-(1-eps)*t) = (x-t) + eps*t

    So if x and t are not a function of v (which is likely), the only case where we will have a limit is if x-t = 0. In that case the limit IS defined (I screwed up), but equal to zero. The limit of (eps/sqrt(eps)) exists as eps->0, it's sqrt(eps), which is zero. Note that physically we are really interested in the limit as eps-> 0+, i.e. epsilon approaches zero while remaninig positive.
  8. Feb 28, 2005 #7


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    I don't see a lot of physical significance - one might say something vague like "photons don't have a rest frame (you can't do a lorentz transform with v=c, nor can you take the limit as v->c in general), but even a photon can tell if it's in the same spot as another photon.
  9. Feb 28, 2005 #8


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    Yes? Photons are bosons, you know. Remember Bose-Einstein condensation?
  10. Feb 28, 2005 #9


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    Bose-Einstein condensates seems to be quite a leap from what we were talking about, but it brings up an interesting question. Are all particles which travel at 'c' bosons?

    The neutrino used to be a good candidate for a fermion that travelled at 'c', but people now think it has mass, so it travels a hair less than 'c'.
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