Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Staff Emeritus
    Science Advisor

    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


    User Avatar
    Staff Emeritus
    Science Advisor

    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


    User Avatar
    Staff Emeritus
    Science Advisor

    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


    User Avatar
    Staff Emeritus
    Gold Member
    Dearly Missed

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


    User Avatar
    Staff Emeritus
    Science Advisor

    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'.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

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