# Einstein postulates and the speed of light

Tags: einstein, light, postulates, speed
Emeritus
PF Gold
P: 9,225
 Quote by Mentz114 I'm about to read what I hope to be the final definitive resolution of the twin paradox.
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.

If what you mean by "the twin paradox" is the question of what's wrong with the calculation that finds that the Earth twin is younger, then there are a number of final definitive resolutions, for example the observation that the time dilation formula doesn't apply since the astronaut twin's world line doesn't coincide with the time axis of any inertial coordinate system. I like to supplement this with a diagram that shows the simultaneity lines of the two inertial coordinate systems that are comoving with the astronaut twin before and after the turnaround.
 HW Helper P: 3,440 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 :)
P: 1,889
 Quote by Fredrik The historical introduction [of Stepanov's paper] was interesting, but then it gets weird. It doesn't include any proofs.
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:
http://arxiv.org/abs/astro-ph/9909311 (which is also in Phys Rev D), and
http://arxiv.org/abs/astro-ph/0111306 (which appears not to have been published in a peer-reviewed journal).
 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?
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.
 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.
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.

Your "open interval around v=0" hypothesis is (essentially) equivalent to the above: there must be a continuous set of physical velocities containing v=0 or the theory is useless.
Emeritus
PF Gold
P: 9,225
 Quote by BruceW 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 :)
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}.

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 too.
Emeritus
PF Gold
P: 9,225
 Quote by strangerep 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.

 Quote by strangerep 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.
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).
P: 1,889
 Quote by Fredrik 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).
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:
http://en.wikipedia.org/wiki/Gyrovector
http://en.wikipedia.org/wiki/Beltram...%93Klein_model
 P: 3,001 Stepanov's paper is really weird, it states as its goal to demonstrate that the second postulate is not necessary only to actually proof in sections 5 and 6 that additional assumptions are needed, that ultimately come down to choose a positive alpha by empirical means. This looks suspiciously similar to the second postulate wich even though it was introduced as a postulate, it's now closer to an empirical fact that leads us to reject Galilean transformations(alpha=0), and alpha<0 is similarly straightforward to discard empirically.
HW Helper
P: 3,440
 Quote by Fredrik 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
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?
Emeritus
PF Gold
P: 9,225
 Quote by BruceW Yeah, negative alpha is not even consistent with 1a.
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.

 Quote by BruceW 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?
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
$$u\oplus v=\frac{u+v}{1-\frac{uv}{c^2}}.$$ This set must be such that the right-hand side of this formula is well-defined for all u,v in the set, i.e. it can't contain u,v such that uv=c2. In particular, it can't contain c or -c. If S is such a set, then
$$\left\{\left.\frac{1}{\sqrt{1+\frac{v^2}{c^2}}} \begin{pmatrix}1 & v/c\\ -v & 1\end{pmatrix}\right|v\in\mathcal S\right\}$$ would be a group that could in principle be used in a theory of physics. But I think any such theory can easily be ruled out by experiments.

 Quote by BruceW Anyway, in the standard group formulation of relativity, negative alpha is not consistent, right?
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.
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P: 3,440
 Quote by Fredrik I don't see a way to completely rule it out without an assumption like 1b.
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.

 Quote by Fredrik 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 $$u\oplus v=\frac{u+v}{1-\frac{uv}{c^2}}.$$
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...
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P: 3,440
 Quote by Mentz114 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.
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.
PF Gold
P: 4,087
 Quote by BruceW 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.
Yes, I think you're correct.

Roughly, the axioms are

1. Defines the domain of the maths. (Looks like the field R4)
2. For any inertial observer, the speed of light is the same everywhere and in every direction, and it is finite. Furthermore, it is possible to send out a light signal in any direction.
3. All inertial observers coordinatize the same set of events.

'Sub-axioms'

4a. Any inertial observer sees himself as standing still at the origin of his coordinates.
4b. Any two inertial observers agree as to the spatial distance between two events if these two events are simultaneous for both of them; furthermore, the speed of light is 1 for all observers.

The first theorem derived from these five axioms is : no inertial observer will 'coordinatize' another inertial observer as travelling at v >= c.

The proof is clunky and inelegant, however ( just my opinion). After this they adopt the Poincare group of transformation as the change of coords between inertial frames. But if the Poincare group is adopted as an axiom, the first theorem (and possibly axiom 4b) is redundant and the theory is more elegant.

So, it doesn't live up to my expectations, but still has some interesting points. I may finish reading it sometime.
P: 1,889
 Quote by BruceW 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?
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.

Edit: In more detail...

Consider the velocity addition formulas in Fredrik's treatment for both cases.

1) First the "negative alpha" case. In units where c=1, the velocity addition formula is
$$u' ~=~ \frac{u + v}{1 - uv} ~=~ \frac{\tan\eta_u + \tan\eta_v}{1 - \tan\eta_u \, \tan\eta_v}$$where ##\eta_u, \eta_v## can take any real value. (In the +ve alpha case, we'd call them "rapidities".)

We now ask: on what domain of ##\eta## is this formula well-defined? Clearly, if ##\eta_v=0##, the denominator is always 1, so that's ok but rather boring since it just means ##u'=u##. But now suppose ##\eta_v\ne 0##. In that case, for any value of ##\eta_v## I can find a value of ##\eta_u## such that the denominator becomes 0. This is because the range of the tan function is ##\pm\infty##. We conclude that this formula is only well-defined on the trivial domain consisting of the single value ##v=0##.

2) Now the "positive alpha" case. Staying with units where c=1, the velocity addition formula is
$$u' ~=~ \frac{u + v}{1 + uv} ~=~ \frac{\tanh\eta_u + \tanh\eta_v}{1 + \tanh\eta_u \, \tanh\eta_v}$$where, as before, ##\eta_u, \eta_v## can take any real value. In this case, we can validly call them "rapidities". In this case, even if ##\tanh\eta_v < 0##, there is no value of ##\eta_u## which makes the denominator 0. This is because the range of the tanh function is bounded by ##\pm 1##, and it approaches these limits only asymptotically. Therefore, on the (open) domain of velocities such that ##|v|<1## the transformation is well-defined.

With closer analysis, one can also show that the limit as ##v\to 1## remains sensible -- in that the result of the "addition" always approaches 1 in that limit. So we can complete this to a closed domain in various situations by taking limits carefully.
Emeritus
PF Gold
P: 9,225
There's a sign error in the "positive alpha" case. The velocity addition formula is
$$u'=\frac{u+v}{1+\alpha uv}$$ in both cases. So the denominator is 1+uv when ##\alpha=1##, and 1-uv when ##\alpha=-1##.
 Quote by strangerep We conclude that this formula is only well-defined on the trivial domain consisting of the single value ##v=0##.
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:
$$(\theta,\eta)\mapsto\begin{cases}\theta+\eta &\text{if }-\pi/2<\theta+\eta<\pi/2\\ \theta+\eta-\pi &\text{if }\theta+\eta>\pi/2\\ \theta+\eta+\pi &\text{if }\theta+\eta<-\pi/2\end{cases}$$ Maybe there's a nicer way to state this definition. Edit: Hey, isn't ##(-\pi/2,\pi/2)\cap\mathbb Q## such a set? Two rational rapidities can't add up to a forbidden value like ##\pi/2## (infinite speed) or ##\pi/4## (speed =c=1), because the forbidden rapidities are all irrational.
P: 1,889
 Quote by Fredrik 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.
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PF Gold
P: 9,225
 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.
P: 1,889
 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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P: 3,440
 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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