Einstein postulates and the speed of light

  • Thread starter MHD93
  • Start date

strangerep

Science Advisor
3,001
797
There's a sign error in the "positive alpha" case.
Corrected, thanks.
I'm not convinced that this is accurate (without an assumption like my 1b). Is {0} really the only subset of ℝ that's closed under the operation ##(u,v)\mapsto (u+v)/(1-uv)##?
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.
Consider:
$$
e^{\eta K} e^{\eta K} ~=~ e^{2\eta K}
$$ so by composing transformations I can get arbitrarily large rapidities. Any restriction on the values of ##\eta## must also be compatible with arbitrarily many such compositions, else we don't have a group. The only such valid restriction (afaict) is the restriction to ##\eta=0##, i.e., a group consisting trivially of the identity and nothing else.
 

Fredrik

Staff Emeritus
Science Advisor
Gold Member
10,730
406
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.
 

strangerep

Science Advisor
3,001
797
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## .
 

BruceW

Homework Helper
3,609
119
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.
 

Fredrik

Staff Emeritus
Science Advisor
Gold Member
10,730
406
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.

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.
 

strangerep

Science Advisor
3,001
797
But it doesn't seem to imply that we need a Lie group.
If we're considering a family of (zero-acceleration-preserving) mappings of the ambient ##(t,x)## space such that ##(t',x') = F(t,x,\xi)## is continuous (where ##\xi## parameterizes the family), that suggests Lie groups quite readily. :-)

Of course, there may also be discrete groups as well.
 

Nugatory

Mentor
12,385
4,867
Maxwell's equations include wave propagation. All prior experience with waves suggested a propagation medium. It is easy from our modern standpoint to laugh at aether, but I think it was quite natural for many physicists in the 1800s to suppose all waves must have a material medium to propagate in. Once you assume aether is material, however exotic, it is not necessary to assume that Maxwell's equations, in the 'standard' form, hold only in the aether frame, is a violation of POR.
I agree with this; it's part of the historical perspective that I believe is essential to understanding why Einstein developed the second postulate in the way that he did.
 

Related Threads for: Einstein postulates and the speed of light

Replies
2
Views
4K
Replies
15
Views
1K
Replies
6
Views
1K
Replies
6
Views
3K
  • Last Post
Replies
17
Views
3K
Replies
6
Views
2K
Replies
28
Views
3K
Top