Recognitions:

## Possible mistake in an article (rotations and boosts).

 Quote by Fredrik Thank you for those comments. I appreciate them a lot.
Now if only one of the SAs or mentors would even try to answer my questions on the extremely rare occasions when I ask a question.... (sigh).
 1. There is NO proper transformation with velocity c.
Ah, but we do often analyze physical situations in a limit as ##v/c\to 1##. So this opens a new can of worms for you: how to deal more satisfactorily with these limiting situations? Currently, it seems you have no way of handling these usefully. :-)

Mentor
 Quote by strangerep Now if only one of the SAs or mentors would even try to answer my questions on the extremely rare occasions when I ask a question.... (sigh).
I've been very lucky with that sort of thing. Most of my questions are math-oriented these days, and micromass have been answering pretty much all of them within an hour of me asking the question. That guy is awesome.

I don't open the relativity and QM forums to look for new posts as often as I used to, so I'm likely to miss most questions that are being asked. If you ever ask something that you think I might be able to answer, don't hesitate to send me a PM with a link to the thread.

 Quote by strangerep Ah, but we do often analyze physical situations in a limit as ##v/c\to 1##. So this opens a new can of worms for you: how to deal more satisfactorily with these limiting situations? Currently, it seems you have no way of handling these usefully. :-)
I don't see the problem. Do you have an example in mind?

I'm attaching version 2 of the document to this post. I will remove the old version above. The biggest change is to the lemma that rules out K<0. I also split the velocity addition corollary into two very similar corollaries just for clarity, and I made some minor changes here and there.

Edit: I have found some mistakes myself. Just before Lemma 27 (in version 2), I said "we ruled out the possibility K>0". It should of course be K<0. And in Lemma 27 (e) one of the uparrows should be a downarrow. In the unnumbered formula after (52), I have set the velocity to 0 without explaining why. I will have to do something about that.

Recognitions:
 Quote by Fredrik I don't see the problem. Do you have an example in mind?
Oh, never mind for now. If it's a problem that exists anywhere outside my vague imagination, it will re-emerge later. :-)
 I'm attaching version 2 of the document to this post. I will remove the old version above. The biggest change is to the lemma that rules out K<0.
I think that lemma (17) needs a bit more work. Since you're only using a single velocity ##v##, I think you've only proven that rapidities in a certain discrete set ##\{\theta(c)/n\}## are excluded from the allowable group parameter value set. Of course, I'm sure this hole can be plugged by exploiting your original assumption that rational parameter values are dense in an open neighbourhood of 0.

Mentor
 Quote by strangerep I think that lemma (17) needs a bit more work. Since you're only using a single velocity ##v##, I think you've only proven that rapidities in a certain discrete set ##\{\theta(c)/n\}## are excluded from the allowable group parameter value set. Of course, I'm sure this hole can be plugged by exploiting your original assumption that rational parameter values are dense in an open neighbourhood of 0.
I agree that what I'm doing in lemmas 16-17 is shows that those rapidities are excluded (for proper transformations). There are no members of G that have determinant 1 and a rapidity in the set ##\{\theta(c)/n|n\in\mathbb Z^+\}##.

However, assumption 1b says that there's an ε>0 such that the interval (-ε,ε) is a subset of the range of the velocity function V. This means that for each v in that interval, there's a member of G with velocity v. Lemma 15 uses that to show that for each v in that interval, there's a member of G that has velocity v and determinant 1. This implies that for all ##\varphi\in(-\operatorname{arctan}(\varepsilon/c),\operatorname{arctan}(\varepsilon/c))##, there's a member of G that has rapidity ##\varphi## and determinant 1.

These results contradict each other, since for large enough n, we have ##\theta(c)/n\in(-\operatorname{arctan}(\varepsilon/c),\operatorname{arctan}(\varepsilon/c))##. That contradiction is what rules out K<0.

So I disagree that there's a hole in the proof, but I still consider this very valuable input, because I think this means that I need to explain the overall plan for lemmas 15-17 somewhere. I think this is what I'll do: Right after the second version of the velocity addition rule (corollary 12 in version 2), I add a comment about how it looks like we may have a division by 0 problem when K<0. (When K=0, there's clearly no problem, and we have already ruled out velocities v such that |v|≥c for the case K>0). Then I explain that I'm going to use this observation to rule out K<0, and describe the strategy for lemmas 15-17.
 Mentor I found a serious mistake as I was thinking about what to say after the velocity addition rule. Lemma 9 (The range of ##\Lambda_K## is closed under matrix multiplication) is wrong. When K<0, it's simply not true that the range is closed under matrix multiplication. I'm sure that the problem is fixable, but it requires a substantial rewrite.

Recognitions:
 Quote by Fredrik I found a serious mistake as I was thinking about what to say after the velocity addition rule. Lemma 9 (The range of ##\Lambda_K## is closed under matrix multiplication) is wrong. When K<0, it's simply not true that the range is closed under matrix multiplication.
 Recognitions: Science Advisor Unfortunately, I've not the time to dulge into this interesting thread. What I liked most as a "derivation" of the Lorentz transform is the following paper. Perhaps, you find it interesting too: V. Berzi, V. Gorini, Reciprocity Principle and the Lorentz Transformations, Jour. Math. Phys. 10, 1518 (1969) http://dx.doi.org/10.1063/1.1665000

Mentor
 Quote by vanhees71 Unfortunately, I've not the time to dulge into this interesting thread. What I liked most as a "derivation" of the Lorentz transform is the following paper. Perhaps, you find it interesting too: V. Berzi, V. Gorini, Reciprocity Principle and the Lorentz Transformations, Jour. Math. Phys. 10, 1518 (1969) http://dx.doi.org/10.1063/1.1665000
Thanks for the tip. I haven't been able to access that paper (I searched for it a few weeks ago), but the paper by Giulini that I linked to in the OP claims to be doing essentially the same thing as Berzi and Gorini. There are a few things I don't like about that approach. In particular, I think it's a bit ugly to assume that the domain of the function that takes velocities to boosts is an open ball of radius c, where c is a non-negative real number or +∞. I want the possibility of a "speed limit" to be a derived result, not one of the assumptions. Giulini also assumes that this velocity function is continuous, and uses that to make a fairly sophisticated argument based on analysis in one step. He also claims that Berzi & Gorini made an additional assumption of continuity that he didn't need to make.

I think I can avoid all of that by starting with a set of assumptions that makes better use of the principle of relativity. You could say my mathematical assumptions that are based on "principles" are stronger, and as a result, (I think) I can avoid technical assumptions and arguments based on analysis. But there are still mistakes in my pdf, so I guess I can't say that for sure yet. I'm trying to fix them now.

My pdf is about the 1+1-dimensional case, but I think that once I've gotten that right, the step to 3+1 dimensions will be much easier than the full proof of that 1+1-dimensional case. I have a pretty good idea about how to do it.

Another issue I have is with Giulini's approach is that he doesn't rigorously prove that Euclidean rotations of spacetime can be ruled out as an option. Instead of showing that they contradict his assumptions, he argues that they contradict physical common sense. To make his version of that part of the proof rigorous, we would have to make another assumption that makes that common sense precise. I think I can do this part much better.

Also, the first step in Giulini's article is incorrect. This is what we discussed on page 1. I don't know if he inherited that mistake from Berzi & Gorini or if it's one of the things he did differently.

Recognitions:
 Quote by vanhees71 V. Berzi, V. Gorini, Reciprocity Principle and the Lorentz Transformations, Jour. Math. Phys. 10, 1518 (1969) http://dx.doi.org/10.1063/1.1665000
For those who have trouble accessing behind the paywall, some related material is here:

 Mentor I'm still working on the rewrite of my pdf. That mistake I made has caused an avalanche of changes. It's super annoying. It will probably take another day or two. In the mean time, I want to mention that I have some concerns about my assumption 2 (which says that ##\Lambda## and ##\Lambda^{-1}## have the same diagonal elements). The concern is that it may not make sense to interpret it as a mathematically precise statement of an aspect of the principle of relativity alone. In that case, it's probably a precise statement of an aspect of the combination of the principle of relativity and the idea of reflection invariance. The problem with that is that I'm defining $$P=\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix}$$ and want to interpret the statements ##P\in G## and ##P\notin G## respectively as "space is reflection invariant" and "space is not reflection invariant". This won't make sense if we have already made a mathematical assumption inspired by the principle of reflection invariance. I got concerned about this when I read a comment in Berzi & Gorini (I have obtained a copy of the article) that I had already read in Giulini, but not given enough thought. What they say is this: If v is the velocity of S' in S, and v' is the velocity of S in S', then the principle of relativity doesn't justify the assumption ##v'=-v##. If the function that takes v to v' is denoted by ##\varphi##, the principle of relativity does however justify the assumptions ##\varphi(v)=v'## and ##\varphi(v')=v##, which imply that ##\varphi\circ\varphi## is the identity map. But that's it. So now they have to make some continuity assumption and use analysis to prove that the continuity assumption and the result ##\phi\circ\phi=\operatorname{id}## together imply that ##\phi(v)=-v## for all v. I tried to think of a physical argument for why we should have v'=-v, but they all start with something like "consider two identical guns pointing in opposite directions, both fired at the same event, while moving such that the bullet fired from gun A will end up comoving with gun B". This is definitely something I will have to think about some more. If my assumption 2 has the same problem as the assumption v'=-v (it probably does), then maybe I can still avoid reflection invariance by stating the assumptions in the context of 3+1 dimensions and using rotation invariance.

Recognitions:
 Quote by Fredrik So now they have to make some continuity assumption and use analysis to prove that the continuity assumption and the result ##\phi\circ\phi=\operatorname{id}## together imply that ##\phi(v)=-v## for all v. [...] If my assumption 2 has the same problem as the assumption v'=-v (it probably does), then maybe I can still avoid reflection invariance by stating the assumptions in the context of 3+1 dimensions and using rotation invariance.
In my 1+3D derivation (i.e., my rework of Manida's derivation), I started out with such a reciprocity assumption, just like Manida. But then I found I was able to use spatial isotropy (i.e., invariance of the transformation equations under rotation around the boost axis) to derive the desired condition. I.e., that the parameter for the inverse transformation corresponds to ##-v##.

Levy-Leblond does a similar trick (a bit less obviously) in the paper I cited earlier.

In your 1+1D derivation, I don't think you have any choice but to rely on parity invariance. But when you graduate up to 1+3D, that part of the proof can indeed be changed to use rotational invariance. I wouldn't waste too much time worrying about it in the 1+1D case.

Mentor
 Quote by strangerep In my 1+3D derivation (i.e., my rework of Manida's derivation), I started out with such a reciprocity assumption, just like Manida. But then I found I was able to use spatial isotropy (i.e., invariance of the transformation equations under rotation around the boost axis) to derive the desired condition. I.e., that the parameter for the inverse transformation corresponds to ##-v##. Levy-Leblond does a similar trick (a bit less obviously) in the paper I cited earlier. In your 1+1D derivation, I don't think you have any choice but to rely on parity invariance. But when you graduate up to 1+3D, that part of the proof can indeed be changed to use rotational invariance. I wouldn't waste too much time worrying about it in the 1+1D case.
That sounds good. Makes me a bit less worried.

For anyone who's interested, here's version 3 of the pdf that proves the theorem that was posted (incorrectly) in post #42 and (correctly) in post #48.

If this post doesn't have an attachment, look for a newer version in my posts below.
Attached Files
 nbr.pdf (94.1 KB, 5 views)

Mentor
 Quote by strangerep For those who have trouble accessing behind the paywall, some related material is here: http://books.google.com.au/books?id=...ariance%22&lr=
Hey, this is a great link. Thanks for finding it and posting it. I can't see all the pages, but I can see the statement of the theorem, and he makes exactly the kind of assumptions that I'm OK with. There are no weird technical assumptions about continuity, about the group being a connected Lie group, or anything like that. There's no assumption about some function that takes velocities to boosts, or anything like that. He just sets out to find all groups ##G\subset\operatorname{GL}(\mathbb R^4)## such that the subgroup ##\{\Lambda\in G|V(\Lambda^{-1})=0\}## is the set of all matrices
$$\begin{pmatrix}1 & 0 & 0 & 0\\ 0 & & &\\ 0 & & R &\\ 0 & & &\end{pmatrix},$$ with ##R\in\operatorname{SO}(3)##. His notation and statement of the theorem is kind of ugly, but that's a ******* beautiful theorem. It's a far more awesome theorem than I thought would exist, after I had read Giulini. I'm going to have to get a copy of that book somehow.
 Mentor Gorini's theorem looks so awesome that it really frustrates me that the library isn't open today. He's really making the absolute minimum of assumptions.

Recognitions:
 Mentor Some of my early thoughts on the proof, after studying only the first two lemmas in Gorini's chapter of the book... I will use lowercase letters for numbers and 3×1 matrices, and uppercase letters for square matrices (2×2 or bigger). (See e.g. my notation for an arbitrary ##\Lambda## below). I'm still numbering my rows and columns from 0 to 3. Let G be a subgroup of GL(ℝ4) such that $$\big\{\Lambda\in G\,|\, \Lambda_{10}=\Lambda_{20} =\Lambda_{30}=0\big\} =\left\{\begin{pmatrix}1 & 0^T\\ 0 & R\end{pmatrix}\bigg|R\in\operatorname{SO}(3)\right\}.$$ The goal is to show, without any other assumptions, that G is the restricted Lorentz group, the group of Galilean rotations and boosts, or SO(4). Here's the gist of the first two lemmas. Let ##\Lambda\in G## be arbitrary. I will write it as $$\Lambda=\begin{pmatrix}a & b^T\\ c & D\end{pmatrix}.$$ Let U, U' be such that $$U=\begin{pmatrix}1 & 0^T\\ 0 & R\end{pmatrix},\quad U'=\begin{pmatrix}1 & 0^T\\ 0 & R'\end{pmatrix},$$ where ##R,R'\in SO(3)##. Choose R such that ##Rc## is parallel to the standard basis vector ##e_1##. Let s be the real number such that ##Rc=se_1##. Choose ##R'## such that the first column of R' is parallel to the first row of RD. (This makes the other two columns of R' orthogonal to the first row of RD). Let ##\Lambda'=U\Lambda U'##, ##D'=RDR'## and ##b'=b^TR'##. We have $$\Lambda' =U\Lambda U'=\begin{pmatrix}a & b^TR'\\ Rc & RDR'\end{pmatrix} =\begin{pmatrix}a & b_1' & b_2' & b_3 '\\ s & D'_{11} & D'_{12} & D'_{13}\\ 0 & 0 & D'_{22} & D'_{23}\\ 0 & 0 & D'_{32} & D'_{33}\end{pmatrix}.$$ So now we know that there's a member of G that has only zeros in the lower left quarter. It's easy to see that $$0\neq \det\Lambda'=\begin{vmatrix}a & b_1'\\ s & D'_{11}\end{vmatrix}\begin{vmatrix}D'_{22} & D'_{23}\\ D'_{32} & D'_{33}\end{vmatrix}.$$ Now we want to prove that ##a\neq 0## and ##D'_{11}\neq 0##. I don't understand what Gorini is doing there. It looks wrong to me. But I think I see another way to obtain a contradiction from the assumption that one of these two variables is 0. So hopefully I have either just misunderstood something simple, or I have a way around the problem. This is why I think what he's doing is wrong. Define $$P=\begin{pmatrix}1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & 1\end{pmatrix}.$$ Note that ##P\Lambda'^{-1}## has the same components as ##\Lambda'^{-1}##, except that the middle two rows have the opposite sign. This implies that ##\Lambda' P\Lambda'^{-1}## can differ from ##\Lambda'\Lambda'^{-1}## only in the middle two columns. (We can make a similar case for why they can only differ in the middle two rows). So the 0 column of ##\Lambda' P\Lambda'^{-1}## is the same as the 0 column of ##\Lambda'\Lambda'^{-1}=I##. In particular, ##(\Lambda'P\Lambda'^{-1})_{00}=1##. But my translation of what Gorini is saying into my notation, is that ##D'_{11}=0## implies that ##(\Lambda'P\Lambda'^{-1})_{00}=-1##. I'm still not sure about this, but I think that one way or another, it is possible to prove that those two variables are non-zero. And I think that's very cool. When I proved my theorem for 1+1 dimensions, I had to assume that the 00 component is non-zero. (This is part of my assumption 1a). Here we seem to have the weakest possible assumptions, and we are already recovering my most basic assumption.