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Mentz114
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Depends how tall you are. The brevity and unambiguous language is a change from hand-wavey arguments.bcrowell said:There's nothing deep going on here.
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Depends how tall you are. The brevity and unambiguous language is a change from hand-wavey arguments.bcrowell said:There's nothing deep going on here.
If what you mean by "the twin paradox" is the result that the astronaut twin is younger, then the final definitive resolution is a simple calculation of the proper times of the two curves.Mentz114 said:I'm about to read what I hope to be the final definitive resolution of the twin paradox.
You might recall that this is the same author I mentioned in a previous thread that wrestled with how to derive the fractional-linear transformations. You might also recall I advised taking the main body of his papers with a grain of salt. His previous papers, which give more detail if you can interpret the poor English are:Fredrik said:The historical introduction [of Stepanov's paper] was interesting, but then it gets weird. It doesn't include any proofs.
There's quite a lot of literature that works with relativistic velocity space in this way. It's a hyperbolic space (Cayley-Klein-Beltrami projective space). The thing that distinguishes it from an ordinary open ball in ##\mathbb R^3## is the nonlinear velocity addition rule. There's a body of theory going back maybe 100 years or more about how to work with such spaces by tricky mappings.It talks about the curvature of the space of velocities, without even introducing a metric on it. I don't understand this at all. Isn't the space of velocities just an open ball in ##\mathbb R^3##, or ##\mathbb R^3## itself?
It's true (which can be seen more easily in the context of a dynamical approach which I won't elaborate on here), but the devil is in the detail: one finds that the group of transformations is only defined on a slice of velocity phase space where velocity is 0. I.e., all observers must be at rest relative to each other. Off this slice, the transformation is ill-defined. Hence, for physical reasons, we can discard the possibility ##\alpha<0## because it fails to model realworld situations.The paper also claims that the case ##\alpha<0## (where the coordinate transformations are rotations of spacetime) isn't logically inconsistent. I strongly doubt the validity of this claim.
I could drop assumption 1b from the definition of "linear relativistic group", since it's not used in lemma 3, where the formula for Lorentz/Galilei/SO(2) transformations is found. Then K<0 (i.e. ##\alpha<0##) isn't immediately ruled out. However, infinite-speed transformations are still ruled out by assumption 1a, which says that every transformation has a velocity in ℝ, and this makes a lot of velocities "forbidden" in the sense that there's no transformation with that velocity in the group. There is certainly a forbidden velocity in every open interval that contains 0. I haven't thought this through to the end, but I would guess that the set of forbidden velocities is dense in ℝ, but not equal to ℝ-{0}.BruceW said:It is an interesting pdf. It says that negative alpha is inconsistent with the group that they have defined. So I guess that a group-theoretic description allowing negative-alpha would need a different group to the one they define. I wonder if there is an easy adaptation to get such a group. Maybe I'll think about it later when I am more awake :)
Oops, I had completely forgotten.strangerep said:You might recall that this is the same author I mentioned in a previous thread that wrestled with how to derive the fractional-linear transformations. You might also recall I advised taking the main body of his papers with a grain of salt.
OK. Do you know how that space is defined, and how it makes sense to talk about curvature? (If it's a pain to explain it, never mind, I'm just a little bit curious).strangerep said:There's quite a lot of literature that works with relativistic velocity space in this way. It's a hyperbolic space (Cayley-Klein-Beltrami projective space). The thing that makes distinguishes it from an ordinary open ball in ##\mathbb R^3## is the nonlinear velocity addition rule. There's a body of theory going back maybe 100 years or more about how to work with such spaces by tricky mappings.
Well, yes, it's a bit of a pain -- partly (mainly?) because I don't know that subject in much detail. These Wiki pages might be a start, however:Fredrik said:Do you know how that space is defined, and how it makes sense to talk about curvature? (If it's a pain to explain it, never mind, I'm just a little bit curious).
Yeah, negative alpha is not even consistent with 1a. I think Stepanov wanted to allow these infinite relative velocities, and if this is the case, then negative alpha is not consistent with the "linear relativistic group". If I am following things correctly, am I right in saying that negative alpha is not necessarily a stupid idea, but until someone comes up with a nice group of transformations for it, it is pretty much a useless idea? I wonder if Frank and Rothe (which Stepanov mentions) thought about a group for these transforms... Maybe they were too early in the formulation of SR to be thinking about groups. Anyway, in the standard group formulation of relativity, negative alpha is not consistent, right?Fredrik said:However, infinite-speed transformations are still ruled out by assumption 1a, which says that every transformation has a velocity in ℝ, and this makes a lot of velocities "forbidden" in the sense that there's no transformation with that velocity in the group... (I skip some of it here)
So if we want to allow K<0 for some reason, we either have to take the set of allowed velocities to be full of holes, or we allow transformations with infinite speed, i.e. ##\Lambda\in G## such that ##(\Lambda^{-1})_{00}=0##. Buuuut...here's something I learned very recently: That would imply that the zero-velocity subgroup of the proper and orthochronous subgroup is not the group of all rotations of space. That sounds undesirable to
I don't see a way to completely rule it out without an assumption like 1b. But of course, the results that most velocities are "forbidden", and that positive velocities can add up to negative ones, are both pretty unappealing.BruceW said:Yeah, negative alpha is not even consistent with 1a.
We would have to do the work to find a subset of ℝ that contains 0, and is closed under the "addition" operation ##\oplus## defined byBruceW said:If I am following things correctly, am I right in saying that negative alpha is not necessarily a stupid idea, but until someone comes up with a nice group of transformations for it, it is pretty much a useless idea?
Special relativity is a theory with a positive alpha, so I'm not sure I understand the question. In my approach, I included assumption 1b (which holds in SR), specifically to rule out negative alphas. It was the simplest assumption I could find that got the job done.BruceW said:Anyway, in the standard group formulation of relativity, negative alpha is not consistent, right?
Ah, yeah. I was using the equations derived by Stepanov, then comparing them to 1a and seeing that they are not consistent. I didn't make that clear, sorry.Fredrik said:I don't see a way to completely rule it out without an assumption like 1b.
I was thinking of some kind of group which allows the infinite relative velocity transformations. I'm guessing it could be very different to the "linear relativistic group"... Like "who even knows what that would look like" kind of different...Fredrik said:We would have to do the work to find a subset of ℝ that contains 0, and is closed under the "addition" operation ##\oplus## defined by
$$u\oplus v=\frac{u+v}{1-\frac{uv}{c^2}}.$$
Ah, thanks man. To be honest, it was too much maths for me, I skipped over most of it. Am I right in thinking that one of the first theorems they state is that the speed of light is invariant? So I guess this axiomatization would be very far from the idea of having a choice of values for an invariant speed.Mentz114 said:Thanks for finding this. It's blown me away. I never thought my training in formal systems would be useful in physics.
Having axiomatized SR, they extend it to include accelerated observers by adding some axioms. I'm about to read what I hope to be the final definitive resolution of the twin paradox.
Yes, I think you're correct.BruceW said:Ah, thanks man. To be honest, it was too much maths for me, I skipped over most of it. Am I right in thinking that one of the first theorems they state is that the speed of light is invariant? So I guess this axiomatization would be very far from the idea of having a choice of values for an invariant speed.
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.BruceW said:am I right in saying that negative alpha is not necessarily a stupid idea, but until someone comes up with a nice group of transformations for it, it is pretty much a useless idea?
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)##? In terms of rapidities, is {0} really the only subset of ##(-\pi/2,\pi/2)## that's closed under the corresponding operation? I think the operation can be defined like this:strangerep said:We conclude that this formula is only well-defined on the trivial domain consisting of the single value ##v=0##.
Corrected, thanks.Fredrik said:There's a sign error in the "positive alpha" case.
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.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)##?
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.strangerep said: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.
Actually, I was just using the standard notion of "Lie group". No need for anything more elaborate.Fredrik said:By this I assume you mean [...]
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.1. This assumption implies my 1b, and is much stronger than my 1b.
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##.2. This assumption is not one of the statements that turns the principle of relativity into a mathematically precise statement.
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## .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.
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.strangerep said: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.
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.strangerep said:Actually, I was just using the standard notion of "Lie group". No need for anything more elaborate.
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 said: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##.
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. :-)Fredrik said:But it doesn't seem to imply that we need a Lie group.
PAllen said: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.