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

Please throw light on the above questions.

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

Please throw light on the above questions.

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Just look at the formulacontrolfreak said:

or

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

Please throw light on the above questions.

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.

[add]

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.

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Thanks Pervect. Just a clarification.pervect said: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.

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?

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when x <> ct.

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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.controlfreak said: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 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.

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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.controlfreak said:

when x <> ct.

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Yes? Photons are bosons, you know. Remember Bose-Einstein condensation?pervect said:but even a photon can tell if it's in the same spot as another photon.

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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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