Possible mistake in an article (rotations and boosts).

[Fredrik]

If X and Y are members of the subgroup that consists of all the matrices of the form
$$\begin{pmatrix}A & B\\ 0 & C\end{pmatrix}$$ with det A>0, then there's an R in SO(3) such that $XY^{-1}=U(R)$, where the right-hand side is defined by
$$U(R)=\begin{pmatrix}I & 0\\ 0 & R\end{pmatrix}.$$ This implies that if
$$X=\begin{pmatrix}A & B\\ 0 & C\end{pmatrix},\quad Y=\begin{pmatrix}D & E\\ 0 & F\end{pmatrix},\quad V(X)=V(Y),$$ we have
$$X=U(R)Y=\begin{pmatrix}D & E\\ 0 & RF\end{pmatrix}.$$ So two members of this subgroup with the same velocity differ only in the lower right, and there they differ only by multiplication of a member of SO(2).

Since for all R, $V(U(R)X)=V(X)=V(XU(R))$, this implies that the following statements are true:

1. For all R in SO(2), we have BR=B.
2. For all R in SO(2), there's an R' in SO(2) such that $RC=CR'$.
3. For all R' in SO(2), there's an R in SO(2) such that $RC=CR'$.

The first one implies that B=0. (Choose R to be a rotation by π/2, and the rest is obvious). The results 2 and 3 imply that C is a number times a member of SO(2). I don't see a way to prove that C is diagonal, so I think Gorini has made essentially the same mistake as Giulini. The proof that C is a number times a member of SO(2) is a bit trickier than the corresponding proof in the OP, since now only one of the rotation matrices (R or R') is arbitrary. It's very convenient to choose the arbitrary SO(2) matrix to be a rotation by π/2. To see if the other SO(2) matrix exists at all, we start with the following observations. An SO(2) matrix acting on a 2×2 matrix from the left doesn't change the norm of the columns (viewed as members of ℝ²). An SO(2) matrix acting on a 2×2 matrix from the right doesn't change the norm of the rows. These observations and the results 2-3 above imply that $a^2+c^2=b^2+d^2$ and $a^2+b^2=c^2+d^2$. These results imply that $a^2=d^2$ and $b^2=c^2$. So C is of the form
$$\begin{pmatrix}a & \pm b\\ b & \pm a\end{pmatrix}$$ and it turns out that three of the four possible sign combinations can be ruled out by the observation that an SO(2) matrix acting from the left on a 2×2 matrix doesn't change the inner product of the columns. So the final result is that there exist numbers a,b such that
$$C=\begin{pmatrix}a & -b\\ b & a\end{pmatrix}.$$ The columns (and the rows) are orthogonal and have the same norm. So if we define $k=\sqrt{a^2+b^2}$, there's an R in SO(2) such that C=kR. This means that the X that we started with is of the form
$$\begin{pmatrix}A & 0\\ 0 & kR\end{pmatrix}$$ where k is a real number and R is a member of SO(2). I don't see a way to prove that k=0 right now, but I have only just started to think about it.

It seems impossible to me to prove that the group contains a member with velocity v for each v with |v|<c. If I'm right, it's a pretty big flaw, and the theorem would have be repaired by adding an assumption like my "0 is an interior point of V(G)". Then we would have to go through the same sort of stuff I did in my pdf for the 1+1-dimensional case.

 

[strangerep]

It's a bit difficult for me to follow along properly, since I don't have complete copies of all the papers, and won't be on-campus again for a while. So you might need to include a bit more context in your posts.

As an aside (or maybe a large tangent?) I'll just make the general remark that I think part of the difficulty is that you're still approaching all this as a "geometrical" problem, instead of a "dynamical symmetry" problem. (I had no great difficulty reaching a equivalent point to what these authors reach.)

In the (simplest case of) a "dynamical symmetry" approach, one assumes an independent variable $t$ (time), and a dependent variable $q = q(t)$ (position), and the equation of motion $\Delta \equiv \ddot q = 0$ . In the general theory of symmetries of (systems of) differential equations, one works in a larger space in a larger space, being a Cartesian product of spaces of all variables and the partial derivatives of the dependent variables. The condition $\Delta=0$ then specifies a solution variety within that space.

One then considers a Lie group $G$ acting on both the dependent and independent variables, and the so-called higher prolongation(s) [*ref 1] of $G$ acting in a space thus augmented by Cartesian product with spaces of various derivatives of the dependent variable(s). The idea is to find the most general transformation of the larger space such that the variety $\Delta=0$ is mapped into itself. There are reasonably straightforward formulas for the 1st and 2nd prolongations of the Lie algebra of $G$, and the symmetries can thus be found in a couple of pages. (I now know that this is actually easier than all the previous ways we've discussed about obtaining the straight-line-preserving maps.)

My point here is that velocity, i.e., $\dot q$, is an integral part of this whole approach, rather than an afterthought, and one can also apply the 1st prolongation to find out how velocity is involved in the transformations. Since the basic variables are continuous and differentiable, so is the velocity, at least piece-wise.

Anyway, maybe this was indeed just me running off on a tangent.

*Ref 1: P. J. Olver, "Applications of Lie Groups to Differential Equations", 2nd Ed.

 

[Fredrik]

I don't know... The approach you're describing sounds a lot less appealing to me. One of the reasons is that it involves so many technical terms that even if I would find it easier to learn, I still wouldn't be able to explain it to a physics student who has studied linear algebra and special relativity, or even to some physics PhD's. But I'm still curious enough that I might take a look at that approach when I'm done with this one. (If nothing else, I'll at least find out what equations of motion have to do with this).

If it really is possible to do what Gorini claims to be able to do (I will know when I've worked through the rest of the proof), then it seems to be the ultimate theorem of this type. Gorini's theorem says that if G is a subgroup of GL(ℝ⁴) such that the subgroup of G that takes the 0 axis to itself is equal to
$$\left\{\begin{pmatrix}1 & 0^T\\ 0 & R\end{pmatrix}\bigg| \,R\in\operatorname{SO}(3)\right\}$$ (where 0 is a 3×1 matrix), then G is either the group of Galilean boosts, the group of Lorentz boosts with some invariant speed $c\in(0,\infty)$, or the group of rotations that I just mentioned. (Unfortunately he says this in a rather complicated way).

It's understandable that you're having difficulties following my notes on lemmas 1-3 above. They're not as well thought out or as detailed as the stuff in my pdf. They are more like a "version 0.1" of a new document. I need to make notes of what I find somewhere, that I can later develop into a "version 1", and I figured I might as well make them here. If you find them useful, that's a bonus. If not, I'm not offended or anything. This also applies to what I'm saying in this post. If you too are interested in understanding Gorini's proof, I'll be happy to answer questions on the parts of it I understand so far.

Here's a summary of my thoughts on lemmas 1-3. First a comment about velocities. For all $\Lambda\in G$ such that $(\Lambda^{-1})_{00}\neq 0$, I will call the 3×1 matrix with components $(\Lambda^{-1})_{i0}/(\Lambda^{-1})_{00}$ the velocity of $\Lambda$. (If $\Lambda$ changes coordinates from a system S to a system S', then this is the velocity of S' in S). For all $\Lambda\in G$ such that $\Lambda_{00}\neq 0$, I will call the 3×1 matrix with components $\Lambda_{i0}/\Lambda_{00}$ the reciprocal velocity of $\Lambda$. (This is the velocity of S in S').

Let $\Lambda\in G$ be arbitrary and consider a transformation
$$\Lambda\mapsto U(R)\Lambda U(R')$$ where $U:\operatorname{SO}(3)\to \operatorname{GL}(\mathbb R^4)$ is defined by
$$U(R)=\begin{pmatrix}1 & 0^T\\ 0 & R\end{pmatrix}$$ for all R in SO(3). (This 0 is a 3×1 matrix). Denote $U(R)\Lambda U(R')$ by $\Lambda'$. It turns out that there's a clever choice of R and R' that ensures that $\Lambda'$ is of the form
$$\begin{pmatrix}A & B\\ 0 & C\end{pmatrix},$$ where A,B,C are 2×2 matrices such that $\det A>0$. When we prove this, we also see that $\Lambda_{00}\neq 0$ and $\Lambda'_{11}\neq 0$. Since the inverse of the matrix above is
$$\begin{pmatrix}A^{-1} & A^{-1}B C^{-1}\\ 0 & C^{-1}\end{pmatrix}$$
and we have
$$A^{-1}=\frac{1}{\det A}\begin{pmatrix}A_{11} & -A_{01}\\ -A_{10} & A_{00}\end{pmatrix},$$
the velocity of $\Lambda'$ is
$$\begin{pmatrix}(\Lambda'^{-1})_{10}/(\Lambda'^{-1})_{00}\\ (\Lambda'^{-1})_{20}/(\Lambda'^{-1})_{00}\\ (\Lambda'^{-1})_{30}/(\Lambda'^{-1})_{00}\end{pmatrix} =\begin{pmatrix}-\Lambda'_{10}/\Lambda'_{11}\\ 0 \\ 0\end{pmatrix}.$$
So every member of G has a well-defined reciprocal velocity, and every member of G that has the special form above has a well-defined velocity, and it's given by a simple formula: $V(\Lambda)=-\Lambda_{10}/\Lambda_{11}.$.

Now let X,Y be two arbitrary members of G that have the special form above, and also have the same velocity. Then we can prove that $XY^{-1}$ has velocity 0, and must therefore (by the only assumption we made about G) be equal to U(R) for some R in SO(3). Further, we can show that $(XY^{-1})_{11}=1$. This implies that the R is a rotation around the 1 axis. So if we define $T:SO(2)\to\operatorname{GL}(\mathbb R^4)$ by
$$T(R)=\begin{pmatrix}I & 0\\ 0 & R\end{pmatrix}$$ for all R in SO(2), then we have $XY^{-1}=T(R)$ for some R in SO(2). This implies that X=T(R)Y, and this tells us that X and Y can only differ in the lower right 2×2 corner. In that corner they differ at most by a factor of R. This means that the top two rows of $\Lambda'$ are completely determined by the velocity $v=-\Lambda_{10}/\Lambda_{11}$.

Since $T(R)\Lambda'$ and $\Lambda' T(R)$ have the same velocity as $\Lambda'$ for all R in SO(2), these three matrices can only differ in the lower right. We can use this to show that the upper right 2×2 corner of $\Lambda'$ is 0.

Now Gorini claims that these results imply that there's a R in SO(2) such that $\Lambda' T(R)$ is of the form
$$\begin{pmatrix}d(v) & c(v) & 0 & 0\\ -va(v) & a(v) & 0 & 0\\ 0 & 0 & e(v) & 0\\ 0 & 0 & 0 & f(v)\end{pmatrix}$$ where e(v) is positive and f(v)=±e(v). He doesn't prove this in the book. He just claims that reference 12 (the article about isotropy) proves it.

I actually got the result that e(v)=f(v). I will have to check my calculations to see if this is a mistake. Edit: I did the calculation again. This time my result agrees with Gorini's. So in spite of the suspicions I've had, I now think that there are no mistakes in lemmas 1-3.

Now at this point, it would be incredibly awesome if we could somehow prove a result like that a transformation of this form has the same diagonal elements as its inverse. Then we can proceed as I did in the 1+1-dimensional case. But Gorini doesn't do anything like that. Instead he starts a long and tedious-looking argument that I still haven't studied. I'm sort of hoping that it can be avoided, but even if it can, I probably won't see how to avoid it until I've studied the argument. So I will have to do that. But maybe not today.

 

[Fredrik]

I started looking at the main body of the proof today. It's really beautiful and really ugly at the same time. The beauty is in the following idea. Denote the last 4×4 matrix in my previous post by N(v). Its velocity is in the 1 direction. There's no way to get a transformation with a velocity that has a magnitude different from |v| just by applying rotation operators to N(v), but if Q is a rotation around the 3 axis, then $N(v)^{-1}QN(v)$ is a transformation with a velocity that's not in the 1 direction. Now we can find two rotations U and U' such that the lower left corner of $UN(v)^{-1}QN(v)U'$ are all zeroes. Then the upper right corner is automatically all zeroes. Finally, we can multiply this from the right with a rotation around the 1 axis, to get a result that we can denote by N(w), where N is the same function as before, but w is a velocity in the 1 direction that may be different from v.

The really ugly part is that this is a crazy exercise in matrix multiplication. Having to calculate $N(v)^{-1}QN(v)$ is annoying enough (and the result is ugly), but then we have to find the appropriate U and U' (both ugly), and compute $UN(v)^{-1}QN(v)U'$.

This way Gorini finds the result $N(w)^{-1}=N(-w)$, as well as a lot of other details about the components of N(v). I'm still hoping that I can find some sort of shortcut, but I'm perhaps being naive.

 

[Fredrik]

I'm still working on this. I've been trying to find ways to simplify the proof, but I have so far failed miserably at that. I can make some statements clearer, but that's it.

We want to find all groups $G\subset\mathrm{GL}(\mathbb R^4)$ such that the set of all $\Lambda\in G$ that take points on the 0 axis to points on the 0 axis is the rotation subgroup
$$\left\{\left.\begin{pmatrix}1 & 0^T\\ 0 & R\end{pmatrix}\right|R\in\mathrm{SO}(3)\right\},$$ where 0 denotes the 3×1 matrix with all components zero.

It's useful to define the following notations:
$$\begin{align}U(R) &=\begin{pmatrix}1 & 0^T\\ 0 & R\end{pmatrix}\\
T(Q)&=\begin{pmatrix}I & 0\\ 0 & Q\end{pmatrix}\\
F(t)&=\begin{pmatrix}1 & 0 & 0 & 0\\ 0 & \cos t & -\sin t & 0\\ 0 & \sin t & \cos t & 0\\ 0 & 0 & 0 & 1\end{pmatrix},\end{align}$$ where I and Q (and the zeroes next to them) are 2×2 matrices. I denotes the identity matrix.

Let's write an arbitrary $\Lambda\in G$ as
$$\Lambda=\begin{pmatrix}a & b^T\\ c & D\end{pmatrix},$$ where b,c are 3×1 matrices and D is a 3×3 matrix. Let $R,R'\in\mathrm{SO}(3)$. Define $\Lambda'=U(R)\Lambda U(R')$. We have
$$\Lambda'=\begin{pmatrix}a & b^T R'\\ Rc & RDR'\end{pmatrix}.$$ If we choose the first row of R parallel to c, and the first column of R' orthogonal to the second and third rows of RD, then the 20,30,21,31 components of $\Lambda'$ will all be zero. This result is lemma 1.

The other lemmas show that if $\Lambda$ is such that the 20,30,21,31 components (the lower left) are all zero, then so are the 02,03,12,13 components (the upper right). Further, the upper left is fully determined by the velocity, and the lower right is determined up to multiplication by an O(2) matrix. We can use this to show that there's a unique k>0, a unique 2×2 matrix A, and a unique O(2) matrix Q, all determined by the velocity of $\Lambda$, such that
$$\Lambda=\begin{pmatrix}A & 0\\ 0 & kQ\end{pmatrix}.$$ Now, if we multiply this from the right by T(R), where R is either equal to Q (if det Q=1) or equal to Q with the sign of the second row flipped (if det Q=-1), then we get a matrix with an even simpler form, like the N I will mention below.

The strategy in the main body of the proof is this: If there's a $\Lambda\in G$ that isn't a rotation, then there's also an $N\in G$ of the form
$$N=\begin{pmatrix}d & c & 0 & 0\\ -va & a & 0 & 0\\ 0 & 0 & e & 0\\ 0 & 0 & 0 & f\end{pmatrix},$$ where v≠0, e>0, f=±e, and a,d,c,e,f are all fully determined by v. (The point of the lemmas is to show this. I'm sure I understand this part well enough).

Now we note that for all t, the matrix $N^{-1}F(t)N$ can be brought to this pretty form by a transformation $X\mapsto U(R)XU(R')$. So let's write $N'=U(R)N^{-1}F(t)NU(R')$, where $R,R'\in\mathrm{SO}(3)$ are chosen only to ensure that we get the simple form above. Somewhere in this calculation of N', a miracle occurs, and we end up with an even simpler form.
$$N'=\begin{pmatrix}a' & c' & 0 & 0\\ -wa' & a' & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\end{pmatrix}.$$ The value of w depends on the angle t, and Gorini claims that the possible values of w form a closed interval. (This is why he doesn't need an assumption like my 1b). Now he starts deriving a bunch of results about these especially simple matrices. v may not be in that closed interval, but there's an $N'$ with velocity w in that interval such that $(N')^n=N$ for some positive integer n. At least I think that's what he's saying. I'm still not clear on all the details at the end.

I've been thinking that if I can understand why that "miracle" occurs (i.e. why N' has an even simpler form than N), then maybe I can use that insight to simplify the proof considerably. But I still don't see what's really going on there.

I have typed up my version of the lemmas. I guess I'll start typing up the main proof as well.

 

[Fredrik]

Any thoughts on what "rotational" invariance really means in this approach? The title of Gorini's article is "Derivation of the Lorentz and Galilei groups from rotational invariance". But the assumption in his theorem is that the zero-velocity subgroup is equal the rotation subgroup. That's hardly the condition that best matches our intuition about what rotation invariance of space means. It may however be the best match to our intuition about "rotation invariance, and no kind of reflection invariance".

I'm thinking that a statement that we choose to think of as a mathematically precise statement of rotation invariance should imply this:

(1) The set
$$\left\{\left.\begin{pmatrix}1 & 0^T\\
0 & R\end{pmatrix}\right|R\in\mathrm{SO}(3)\right\},$$ where 0 denotes a 3×1 matrix whose components are all zeroes, is a subgroup of G.

This is a weaker statement than Gorini's. I think it's too weak to imply anything like his result. But I also think that there's more to rotation invariance than this. The velocity of a boost singles out a direction in space, but a zero-velocity transformation doesn't do anything like that. So a mathematically precise statement of rotation invariance should also imply things like this:

(2) For all $\Lambda\in G$ with zero velocity, the projection of $\Lambda x$ onto the 0 axis should have the same value for all $x\in\mathbb R^4$ such that $x_0=0$ and $x_1^2+x_2^2+x_3^2=1$.

If we assume (2), then all transformations of the form
\begin{pmatrix}* & * & * & *\\ 0 & * & * & *\\ 0 & * & * & *\\ 0 & * & * & *\end{pmatrix} are actually of the form \begin{pmatrix}* & 0 & 0 & 0\\ 0 & * & * & *\\ 0 & * & * & *\\ 0 & * & * & *\end{pmatrix} This takes us a long way toward Gorini's assumption. I think Gorini took things a bit too far, but I think that we should be able to conclude that transformations of the form above are actually of the form
\begin{pmatrix}a & 0^T\\ 0 & Q\end{pmatrix} where $a\neq 0$ and $Q\in\mathrm{O}(3)$. Unfortunately I don't see how to get there with an assumption like (2), which everyone would agree is an aspect of rotation invariance.


Edit: It makes sense to require that an arbitrary $\Lambda\in G$ with zero velocity takes the unit sphere to a sphere. (If it takes the unit sphere to some other shape, then some directions in space are different from others). But why does it have to have the same radius? I don't see why the inclusion of a transformation of the form
\begin{pmatrix}a & 0^T\\ 0 & bQ\end{pmatrix} in the group should be thought of as inconsistent with rotation invariance. $a,b\neq 0, Q\in\mathrm{O}(3)$.

Edit 2: Nevermind. I figured out the answers. The statements (1), (2), and the requirement that zero-velocity transformations take spheres to spheres, can together be thought of as "rotation invariance". Transformations like
\begin{pmatrix}a & 0^T\\ 0 & bQ\end{pmatrix} with Q in O(2) and a,b not both equal to 1, aren't ruled out by rotation invariance. They are ruled out by "dilation non-invariance", something that's even more intuitive than rotation invariance; it's obvious that if a transformation e.g. changes the length of a meter stick, length measurements will not have the same results as before.

Gorini doesn't derive these groups "from rotation invariance". He derives them from rotation invariance, dilation non-invariance, spatial reflection non-invariance and time reversal non-invariance. I think the proof would work with only minor modifications if we drop the last two assumptions, but we're going to have to keep the dilation non-invariance.

 

[strangerep]

Dunno whether you've already got this earlier Gorini paper (attached hopefully). Still can't find the "Isotropy of Space" paper though.

 

[Fredrik]

Thank you. I already had it though. I found a pdf version of it at this URL a couple of weeks ago: http://physics.sharif.ir/~sperel/paper1.pdf. I haven't had a chance to look at the isotropy paper yet. I didn't find a pdf of that one.

 

[Fredrik]

I've been trying to generalize Gorini's theorem, and I'm stuck on a silly-looking detail, so figured I might as well ask if you see something I don't.

Gorini's assumption is that the zero-velocity subgroup is equal to the rotation subgroup. Since this only gives us the proper and orthochronous groups, I want to weaken the assumption. I'm not 100% sure what the appropriate weaker assumption should be, but I suspect that it's this one: The zero-velocity subgroup has the rotations as a subgroup, and is itself a subgroup of the group
$$\left\{\begin{pmatrix}\sigma & 0^T\\ 0 & Q\end{pmatrix}\,\bigg|\,\sigma\in\{-1,1\}, Q\in\mathrm{O}(3)\right\}$$ (where 0 denotes a 3×1 matrix).

Because of this weakening, I quickly run into a problem. Let $\Lambda$ be an arbitrary member of the group, with only zeroes in the lower left corner (i.e. the 20,30,21,31 components). Gorini proved that the 00 and 11 components of such a $\Lambda$ must be non-zero, so this is what I would like to do. Let F be a rotation by $\pi$ around the 3 axis:
$$F=\begin{pmatrix}1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & 1\end{pmatrix}.$$ Now I can show e.g. that if $\Lambda_{11}=0$, then $\Lambda F\Lambda^{-1}$ is a zero-velocity transformation with its 00 component equal to -1. With Gorini's assumption, this is clearly a contradiction that proves that $\Lambda_{11}\neq 0$ (because every zero-velcocity transformation is a rotation and the 00 component of a rotation is 1). But with my weaker assumption, I don't see how $(\Lambda F\Lambda^{-1})_{00}=-1$ contradicts anything.

I may have to modify the assumption to ensure that this results contradicts it. But I want the assumptions to be mathematical statements that can be thought of as making aspects of the concept of "rotation invariance" mathematically precise. So the question is, what aspect of rotation invariance is violated by this result?

Note that $\Lambda$ having all zeroes in the lower left corner means that it's going to turn out to be a boost in the 1-direction. So in physical terms, what this result is saying is that if I stand on a treadmill, speed it up so I have to run, then turn around and run backwards for a while, and then stop the treadmill, my clock is know running backwards! But to make this argument, I have to anticipate the result of the theorem. At the point where I'm making the calculation, I'm not sure we know enough about the group to even define what an orthochronous transformation is.

 

[strangerep]

Have you been able to obtain a copy of Gorini's "Isotropy of Space" paper? I.e., J. Math. Phys. 11, 2226 (1970) ?

 

[Fredrik]

No, I haven't. I guess I should make another trip to the library to get that. I didn't need it to understand the part of the proof that he said could be found in the isotropy article, so I wasn't very motivated to go back to the library just for that. Now that want to generalize the theorem, I seem to need a deeper understanding of what rotation invariance entails than I currently have. Maybe I can get that from the isotropy article, maybe not.

 

[strangerep]

Yeah -- I just figured it would be nice to see what Gorini actually says. (I haven't been on-campus for a while either.)

As for what rotation invariance (or spatial isotropy) means, I think of it as follows.

Suppose you have an expression $F(t,x)$ (which may involve derivatives). Then the equation $$F(t,x) ~=~ 0$$ (in general, on some domain) is said to be rotationally invariant if $$F(t, Rx) ~=~ 0$$ on the same domain (where $R$ is an arbitrary spatial rotation).

 

[strangerep]

Maybe you should just try to contact him direct:
http://www.uninsubria.eu/research/physmath/cv_Gorini.htm

Perhaps he'll send you a copy... :-)

 

[Fredrik]

The thing is, I need to translate "rotation invariance" into a set of conditions on the group G that we're trying to find. An obvious condition is that the group of spatial rotations is a subgroup of G. This corresponds to saying that no matter which way we rotate the laboratory, there's an inertial coordinate system with its spatial axes aligned with the laboratory walls.

A less obvious condition is that zero-velocity transformations must preserve simultaneity. The reason is, a linear transformation either preserves simultaneity or "tilts" the simultaneity hyperplanes. Such a "tilt" would make one direction in space "special". (It's easy to visualize this in a 2+1-dimensional spacetime diagram). This is OK when we're dealing with a transformation with a non-zero velocity in that special direction. But when we're dealing with a zero-velocity transformation, the transformation doesn't single out a direction, so if space itself doesn't have a preferred direction, the simultaneity plane can't be tilted. (In what direction would it tilt?)

The consequence of this is that a zero-velocity transformation (which by definition has three zeroes in column 0) must have three zeroes in row 0. This is easy to see. Just note that the "t=0" hyperplane is preserved by $\Lambda$ only if the 0 component of
$$\Lambda\begin{pmatrix}0\\ x\\ y\\ z\end{pmatrix}$$ is 0 for all $x,y,z\in\mathbb R$. This implies that $\Lambda_{01}=\Lambda_{02}=\Lambda_{03}=0$. Because of this, we choose the theorem's assumptions such that every $\Lambda\in G$ with $\Lambda_{10}=\Lambda_{20}=\Lambda_{30}=0$ also satisfies $\Lambda_{01}=\Lambda_{02}=\Lambda_{03}=0$.

I suspect that I will just have to think of more arguments of this sort, until I find one that takes care of the problem with time reversal that I described.

 

[strangerep]

A less obvious condition is that zero-velocity transformations must preserve simultaneity.

But how do you know that a priori if you're starting from the relativity principle alone, and trying to derive the relativity group(s)? You can't appeal to geometric intuitions from Minkowski spacetime since the latter is really only an afterthought -- a homogeneous space constructed from a given relativity group.

(BTW, your term "zero-velocity transformation" is a bit misleading. I think "velocity-preserving transformation" is clearer.)

 

[Fredrik]

But how do you know that a priori if you're starting from the relativity principle alone, and trying to derive the relativity group(s)? You can't appeal to geometric intuitions from Minkowski spacetime since the latter is really only an afterthought -- a homogeneous space constructed from a given relativity group.

I think I explained how I know that, but feel free to ask about the details if I need to clarify something. Note that I didn't use the Minkowski metric. I just used what I know about linear transformations. There isn't a whole lot of things that a linear transformation can do to a simultaneity hyperplane. It can preserve it, or it can tilt it. If a transformation tilts a simultaneity hyperplane, that always favors a direction in space: the direction of the tilt.

A linear transformation can also stretch or rotate a simultaneity hyperplane, but the preservation of simultaneity depends only on whether the hyperplane gets tilted or not.

(BTW, your term "zero-velocity transformation" is a bit misleading. I think "velocity-preserving transformation" is clearer.)

I define the velocity of a transformation $\Lambda$ as the vector with components $(\Lambda^{-1})_{i0}/(\Lambda^{-1})_{00}$. I think this terminology is appropriate. The velocity of the transformation is the velocity of the second observer in the coordinates of the first.