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Einstein postulates and the speed of light

 Quote by strangerep Corrected, thanks. If the transformations form a group which is continuous and differentiable in a real parameter (i.e., ##\eta## here), we must allow ##\eta## to takes arbitrary real values.
By this I assume you mean that there's a function that takes each ##\Lambda## in the group to the rapidity of ##\Lambda##, and that we require this function to be continuous in some sense, and differentiable in some sense. We could e.g. use the Hilbert-Schmidt norm on the set of matrices (the norm obtained from the inner product ##\langle A,B\rangle=\mathrm{Tr}A^TB##) to define a topology on the group. An (equivalent?) alternative is to instead consider the function that takes the 4-tuple of components of the 2×2 matrix ##\Lambda## to the rapidity of ##\Lambda##. For this function, we can use the standard definitions of continuity and differentiability from calculus.

There is of course nothing wrong with such assumptions, but I'd like to point out two things. 1. This assumption implies my 1b, and is much stronger than my 1b. 2. This assumption is not one of the statements that turns the principle of relativity into a mathematically precise statement. Rather, this assumption should be thought of as making "boost invariance" mathematically precise. This is when we are talking specifically about rapidities. If the parameter had been a position or an Euler angle, it would have been part of making the principles of "translation invariance" (="spatial homogeneity") or "rotation invariance" (="spatial isotropy") precise.

So I think we have to consider ##\alpha<0## with a non-trivial set of allowed rapidities to be consistent with the principle of relativity, but not consistent with these other principles.

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 Quote by Fredrik By this I assume you mean [...]
Actually, I was just using the standard notion of "Lie group". No need for anything more elaborate.

 1. This assumption implies my 1b, and is much stronger than my 1b.
Certainly one needs the transformation to be well-defined in an open neighbourhood of the identity, else taking derivatives is a problem. So one could alternatively regard what I wrote above as saying that the 1st case is incompatible with such an assumption.

 2. This assumption is not one of the statements that turns the principle of relativity into a mathematically precise statement.
That depends on whether one's starting point is dynamical or geometric. If the former, then notions of continuity and differentiability are already there, gratis, since we start with ##d^2 x/dt^2 = 0##.
 So I think we have to consider ##\alpha<0## with a non-trivial set of allowed rapidities to be consistent with the principle of relativity, but not consistent with these other principles.
Well, in the dynamical approach with Lie symmetries it is not consistent with invariance of the equation ##d^2 x/dt^2 = 0## on a nontrivial range of values of ##dx/dt## .

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 Quote by strangerep As I tried to explain briefly in a previous post, there is already a valid transformation group for negative alpha, but it is only well-defined on v=0. I.e., it only makes mathematical sense in a situation where all observers are at rest relative to each other. Thus, it is indeed "pretty much useless" for physics.
Right. Yeah, that is definitely not what we want. So negative alpha is pretty much useless until someone comes up with a transformation group that can give us all velocities.

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 Quote by strangerep Actually, I was just using the standard notion of "Lie group". No need for anything more elaborate.
OK. But if we're assuming that our group is a Lie group, then we don't need to talk about derivatives to rule out velocity sets like the one I mentioned in the negative alpha case. We only need to use that a manifold is locally homeomorphic to ##\mathbb R^n##. And this also rules out the velocity set {0}, because a Lie group can't be a singleton set.

 Quote by strangerep That depends on whether one's starting point is dynamical or geometric. If the former, then notions of continuity and differentiability are already there, gratis, since we start with ##d^2 x/dt^2 = 0##.
I don't follow this argument. This implies that the world line of a non-accelerating object is a differentiable curve. But it doesn't seem to imply that we need a Lie group.

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