Possible mistake in an article (rotations and boosts).

[Fredrik]

The physical starting point is that $$
\frac{d^2x'^{\,i}}{dt'^{\,2}} ~=~ 0 ~=~ \frac{d^2x^i}{dt} ~,
$$ i.e., $v'^{\,i}$ and $v^i$ are both (triplets of) constants.
There is no mention of (e.g.,) dx/dy = const in this.

OK, agreed. This is even more natural than the requirement about straight lines. But if this requirement doesn't imply the one about straight lines, I think we're just going to have to introduce other mathematical assumptions inspired by the principle of relativity until we have enough to derive the statement about straight lines.

Anyway, from the rest of your reply I get the feeling we're approaching this with different emphases. (Mine is the physical emphasis of what a local observer can do.)

Yes, I don't care about local stuff at this point. I'm only concerned with theories of physics in which spacetime is a 4-dimensional vector space (or affine space), and the set of functions that change coordinates from one global inertial coordinate system to another is a group that satisfies mathematical statements inspired by the principle of relativity and symmetry ideas (translation, isotropy and maybe parity and/or time reversal).

 

[strangerep]

I think I have now worked out everything in the the 1+1-dimensional case, except how to go from "preserves constant velocity lines" to "preserves straight lines".

In the special case of 1+1D they're the same, aren't they?

 

[Fredrik]

In the special case of 1+1D they're the same, aren't they?

Yes. I started thinking about that an hour after making that comment, and it turned out to be very easy to prove that "preserves lines with zero coordinate acceleration" implies "preserves straight lines that aren't parallel to the x axis". And since we're assuming bijectivity, this means that those lines, i.e. the ones with infinite speed, are preserved as well. For example, the x-axis is sent to the infinite-speed line through $\Lambda(0)$.

So unless I've made a blunder somewhere, I now have a complete proof for the 1+1-dimensional case.

For the 3+1-dimensional case, the only part of the proof that I have completed is the theorem that says that if X is a finite-dimensional vector space such that 2≤dim X<∞, and T:X→X is a bijection that takes straight lines to straight lines, then T is an affine map. I have also done some calculations, like finding a boost for an arbitrary velocity from a boost with a velocity in the 1 direction. But I still haven't decided what exactly I want my assumptions to be.

 

[Fredrik]

I defined $G_p=\{\Lambda\in G|\det\Lambda>0\}$. Now define
$$P=\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix}.$$ For all $\Lambda\in G$, we have $\det(P\Lambda)=-\det(\Lambda)$. If $P\in G$, then this implies that $G-G_p=\{P\Lambda|\Lambda\in G_p\}$, so we have $G=G_p\cup\{P\Lambda|\Lambda\in G_p\}$.

But what if $P\notin G$?
...
...is it possible to prove that $P\notin G$ implies that $G-G_p=\emptyset$?

Edit: ...
Suppose that $P\notin G$. I will prove that $G-G_p$ is empty by deriving a contradiction from the assumption that it's not. So suppose that it's not. Let $\Lambda\in G-G_p$ be arbitrary. Then $\det(\Lambda)=-1$, and $\det(P\Lambda)=1$. So $P\Lambda\in G_p$. Define $\Lambda'=P\Lambda$. We have $P=\Lambda'\Lambda^{-1}\in G$, and this contradicts the assumption that $P\notin G$.

I have found two mistakes in my proof for the 1+1-dimensional case. What I said in the quote above is wrong. If $P\notin G$ and $\Lambda\in G-G_p$, then there's no reason to think that $P\Lambda\in G$.

Maybe I'm missing something obvious, but I'm starting to think that $P\notin G$ doesn't imply that $G=G_p$. Is it possible that $P\notin G$ (the mathematical statement corresponding to the idea that the laws of physics are not invariant under spatial reflections) only implies that $G_p$ doesn't determine $G$? Then the final conclusion won't be that "G+translations" is either the Galilean group or the Poincaré group. It will be that it's a group that has the restricted version of one of those two as a subgroup.

Edit: Nope, the correct final conclusion is that "G+translations" is a subgroup of the Galilean group or the Poincaré group. This is clear when I work with G the whole way instead of restricting my attention to a subgroup. The calculation turned out to be a lot less ugly than I thought it would be, so I will do it this way instead. There's no reason to consider the subgroups as early in the calculation as I did.
The other mistake is in the proof that's supposed to rule out the possibility that K<0. I will have to take another look at that.

 

[Fredrik]

New version of the (main part of) the argument for the 1+1-dimensional case.

We're looking for a non-trivial subgroup of GL(ℝ²), whose members are functions that change coordinates from one global inertial coordinate system to another. We are going to make a small number of assumptions about this group. They will be mathematical statements that can be thought of as making the following ideas precise.

1. Each inertial observer has a finite velocity in the coordinates used by another.
2. For each event x, there are at least two velocities at which an experiment performed at x can test the accuracy of predictions.
3. For each event x, statements that two arbitrary inertial observers at x make about how the other guy's measurements relates to his own coordinate assignments must be symmetrical. (For example, if one of them can say "Your clock is slow by a factor of $\gamma$.", the other one must be able to say the same thing).
4. Positive velocities do not add up to negative velocities unless there's a reflection involved.

It's easy to make the first idea precise, if we first define the velocity of an arbitrary $\Lambda$ in the group. Suppose that S and S' are global inertial coordinate systems, and that $\Lambda$ is the transformation from S to S'. We are going to define the velocity of $\Lambda$ as the "$\Delta x/\Delta t$" of the line that $\Lambda^{-1}$ sends the 0 axis to, because that line is the image in S of the world line of the S' observer. It's also the unique straight line through 0 and $\Lambda^{-1}e_0$. $$\Lambda^{-1}e_0 =\Lambda^{-1}\begin{pmatrix}1\\ 0\end{pmatrix} =\begin{pmatrix}(\Lambda^{-1})_{00}\\ (\Lambda^{-1})_{10}\end{pmatrix}.$$ So we define the velocity of $\Lambda$ as $(\Lambda^{-1})_{10}/(\Lambda^{-1})_{00}$.

The second idea can be made precise by requiring that there's a member of the group that has a non-zero velocity. (This assumption is necessary in the step that determines the relationship between the off-diagonal components).

The third idea can be made precise by requiring that $(\Lambda^{-1}e_\mu)_\mu=(\Lambda e_\mu)_\mu$ for all $\mu\in\{0,1\}$. This is equivalent to saying that $\Lambda^{-1}$ and $\Lambda$ have the same diagonal elements.Theorem: Suppose that $G$ is a subgroup of $\operatorname{GL}(\mathbb R^2)$ such that the following statements are true.

1. There's a $V:G\to\mathbb R$ such that $V(\Lambda)=-\frac{(\Lambda^{-1})_{10}}{(\Lambda^{-1})_{00}}$ for all $\Lambda\in G$.
2. $V(G)\neq\{0\}$.
3. For all $\mu\in\{0,1\}$, we have $(\Lambda^{-1})_{\mu\mu}=\Lambda_{\mu\mu}$.
4. For all $\Lambda',\Lambda''\in G$ with positive velocities and determinants, we have $V(\Lambda'\Lambda'')>0$.

Then there's a $K\geq 0$ such that
$$G\subset\left\{\left.\frac{\sigma}{\sqrt{1-Kv^2}}\begin{pmatrix}1 & -Kv\\ -\rho v & \rho\end{pmatrix}\right|\ \sigma,\rho\in\{-1,1\}, 1-Kv^2>0\right\}.$$

Proof: Let $\Lambda\in G$ be arbitrary. Denote its components by a,b,c,d. We have
$$\Lambda=\begin{pmatrix}a & b\\ c & d\end{pmatrix},\quad \Lambda^{-1}=\frac{1}{ad-bc}\begin{pmatrix}d & -b\\ -c & a\end{pmatrix}.$$ Note that
$$V(\Lambda)=-\frac{(\Lambda^{-1})_{10}}{(\Lambda^{-1})_{00}}=-\frac{c}{d},\qquad V(\Lambda^{-1})=-\frac{\Lambda_{10}}{\Lambda_{00}}=\frac{c}{a}.$$ This means that assumption 1 implies that a≠0, d≠0. Assumption 3 implies that
$$a=\frac{d}{\det\Lambda},\quad d=\frac{a}{\det\Lambda}=\frac{d}{(\det\Lambda)^2}.$$ Since d≠0, this implies that $\det\Lambda=\pm 1$.

Define $\rho=\det\Lambda$, $\gamma=|a|$, $\sigma=\operatorname{sgn}(a)$, $\alpha=b/a$ and $v=-\rho c/a$. Note that since $a=\rho d$, this v is the velocity of $\Lambda$.
$$\Lambda=\begin{pmatrix}a & b\\ c & \rho a\end{pmatrix} =a\begin{pmatrix}1 & b/a\\ c/a & \rho\end{pmatrix} =\sigma\gamma\begin{pmatrix}1 & \alpha\\ -\rho v & \rho \end{pmatrix}.$$ Let $\Lambda',\Lambda''\in G_p$ be arbitrary.
\begin{align}
&G_p\ni \Lambda'\Lambda'' =\sigma'\sigma''\gamma'\gamma''\begin{pmatrix}1 & \alpha'\\ -\rho' v' & \rho'\end{pmatrix}\begin{pmatrix}1 & \alpha''\\ -\rho'' v'' & \rho''\end{pmatrix} =\sigma'\sigma''\gamma'\gamma''\begin{pmatrix}1-\alpha'\rho''v'' & \alpha''+\alpha'\rho''\\ -\rho' v'-\rho''v'' & -\rho' v'\alpha''+\rho'\rho''\end{pmatrix}\\
&\rho'\rho'' =(\det\Lambda')(\det\Lambda'') =\det(\Lambda'\Lambda'')
=\frac{(\Lambda'\Lambda'')_{11}}{(\Lambda'\Lambda'')_{00}} =\frac{-\rho' v'\alpha''+\rho'\rho''}{1-\alpha'\rho''v''}.\end{align} Suppose that $\rho'\rho''=1$. Then we have $\rho'=\rho''$ and
$$1=\frac{-\rho'v'\alpha''+1}{1-\alpha'\rho''v''}.$$ This is equivalent to $\alpha'\rho''v'' =\rho'v'\alpha''$. Since $\rho'=\rho''$, this is equivalent to $\alpha' v''=v'\alpha''$.

Now suppose that $\rho'\rho''=-1$. Then we have $\rho'=-\rho''$ and $$-1=\frac{-\rho'v'\alpha''-1}{1-\alpha'\rho''v''}.$$ This is equivalent to $\alpha'\rho''v''=-\rho'v'\alpha''$. Since $\rho'=-\rho''$, this is equivalent to $\alpha' v''=v'\alpha''$.

So $\alpha'v''=v'\alpha''$ for all $\Lambda',\Lambda''\in G$. If every member of $G$ has velocity 0, then this result tells us nothing. This is why we included assumption 2. It ensures that the result we just obtained implies the following.

1. For all $\Lambda'\in G$, if $v'=0$, then $\alpha'=0$.
2. For all $\Lambda',\Lambda''\in G$ such that $v'\neq 0$ and $v''\neq 0$,
   $$\frac{\alpha''}{v''}=\frac{\alpha'}{v'}.$$

The second result implies that $\alpha'/v'$ has the same value for all $\Lambda'\in G$ such that $v'\neq 0$. Denote this value by -K. The second result implies that if $v\neq 0$, then $\alpha=-Kv$. The first result implies that $\alpha=-Kv$ also when $v=0$.

These results imply that
$$\Lambda=\sigma\gamma\begin{pmatrix}1 & -Kv\\ -\rho v & \rho\end{pmatrix},\quad\Lambda^{-1}=\frac{1}{\sigma\gamma(\rho-\rho Kv^2)}\begin{pmatrix}\rho & Kv\\ \rho v & 1\end{pmatrix} =\frac{\sigma}{\gamma(1-Kv^2)}\begin{pmatrix}1 & \rho Kv\\ v & \rho\end{pmatrix}.$$ Now assumption 3 tells us that
$$\sigma\gamma=\frac{\sigma}{\gamma(1-Kv^2)}.$$ Note that if K>0, this implies that $1-Kv^2>0$ (because $\gamma^2>0$). If we define $c=1/\sqrt{K}$, this is equivalent to $v\in(-c,c)$.

Since $\gamma=|\Lambda_{00}|>0$, the result above implies that
$$\gamma=\frac{1}{\sqrt{1-Kv^2}},$$ and therefore that
$$\Lambda=\frac{\sigma}{\sqrt{1-Kv^2}}
\begin{pmatrix}1 & -Kv\\ -\rho v & \rho\end{pmatrix}.$$ We will prove that K≥0 by deriving a contradiction from the assumption that K<0. So suppose that K<0.

We will prove that there's a $\bar\Lambda\in G$ such that $\det\bar\Lambda>0$ and $v>0$. Let $\Lambda$ be an arbitrary member of G such that v≠0. (Assumption 2 ensures that such a $\Lambda$ exists). If $\det V>0$ and $v>0$, we just define $\bar\Lambda=\Lambda$. If $\det V>0$ and $v<0$, then $\det\Lambda^{-1}>0$, and $V(\Lambda^{-1})=-v>0$. So we can just define $\bar\Lambda=\Lambda^{-1}$. Now suppose that $\det V<0$. For all $\Lambda',\Lambda''\in G$,
$$V(\Lambda'\Lambda'') =-\frac{(\Lambda'\Lambda'')_{10}}{ (\Lambda'\Lambda'')_{00}} =\frac{\rho'v'+\rho''v''}{1-|K|v'v''}.$$ This implies that
$$V(\Lambda^2)=\rho\frac{2v}{1-|K|v^2}\neq 0.$$ We also have $\det(\Lambda^2)=(\det\Lambda)^2=1$, so if $V(\Lambda^2)>0$, we can define $\bar\Lambda=\Lambda^2$. If $V(\Lambda^2)<0$, then we can define $\bar\Lambda=(\Lambda^2)^{-1}$.

Now let $\Lambda$ be an arbitrary member of $G$ such that $\det\Lambda>0$ and $v>0$. Define $\beta=\sqrt{|K|}v$ and $c=1/\sqrt{|K|}$. Let $\theta$ be the unique member of $(-\pi/2,\pi/2)$ such that $\tan\theta=\beta$.\begin{align}
&\cos\theta>0\\
&1=\cos^2\theta+\sin^2\theta=\cos^2\theta(1+\beta^2)\\
&\cos\theta=\frac{1}{\sqrt{1+\beta^2}}\\
&\sin\theta =\frac{\sin\theta}{\cos\theta}\cos\theta =\frac{\beta}{\sqrt{1+\beta^2}}\\
&\Lambda =\frac{\sigma}{\sqrt{1-Kv^2}}
\begin{pmatrix}1 & -Kv\\ -v & 1\end{pmatrix} =\frac{\sigma}{\sqrt{1-Kv^2}}
\begin{pmatrix}1 & -\beta/c\\ -c\beta & 1\end{pmatrix}
=\sigma\begin{pmatrix}\cos\theta & \frac{1}{c}\sin\theta\\ -c\sin\theta & \cos\theta\end{pmatrix}
\end{align} Denote the right-hand side by $T(\theta)$. The above implies that
$$\Lambda^2 =T(\theta)^2 =T(2\theta),$$ and (by induction) that for all $n\in\mathbb Z^+$,
$$\Lambda^n=T(n\theta).$$ Let $n\in\mathbb Z^+$ be such that $n\theta\in(\pi/2,3\pi/2)$.
$$V(\Lambda^n)=V(T(n\theta))=\frac{c\sin(n\theta)}{\cos(n\theta)}<0.$$ This contradicts assumption 4.

 

[Fredrik]

I really like this approach. I especially like that we didn't have to make any assumptions about parameters. For example, we didn't even have to assume that a restricted transformation is fully determined by its velocity.

The requirement of positive determinants in assumption 4 is however pretty ugly. I don't see a way to know that this is the appropriate criterion until we've found the final result for $\Lambda$. But I think I see a solution. Instead of mentioning determinants, we require that G has a path-connected subgroup, and that for any two $\Lambda',\Lambda''$ in it, such that v,v'>0, the velocity of $\Lambda'\Lambda''$ is non-negative. It's especially easy to see how this gets the job done in the case K=-1. Here $\Lambda$ is just a (proper) rotation by an angle θ. If the velocity of $\Lambda'\Lambda''$ is negative, then its rapidity (i.e. the angle of rotation) will be in the interval (π/2,π), and there's no path from 0 to a number in that interval that doesn't go through the values π/2 that corresponds to velocity +∞.

I think this will also make our assumptions strong enough to completely determine the group up to the value of K and the inclusion of parity and/or time reversal. That's a very nice bonus.

Now some thoughts about the 3+1-dimensional case. It would be pretty cool if we could just modify the assumptions to say essentially the same things about a group that's a subgroup of GL(ℝ⁴) instead of a subgroup of GL(ℝ²), and then add an assumption about rotations. This is probably doable in principle, but it looks very hard, because the formula for $\Lambda^{-1}$ is now much more complicated.

Giulini assumes that there's a subgroup of boosts, and a subgroup of rotations. He assumes that a boost is fully determined by a velocity. This is a bit ugly in my opinion, because we didn't have to make any assumptions at all about parameters in the 1+1-dimensional case. If we are willing to make assumptions that strong, then why not choose our assumptions so that we can use as much as possible of what we did in the 1+1-dimensional case? So I suggest something like this:

Assumption 1. The set of all $\Lambda\in G$ of the form
$$\Lambda=\begin{pmatrix}B & 0\\ 0 & I\end{pmatrix}$$ where B,0 are 2×2 matrices and I is the 2×2 identity matrix, is a subgroup.

Assumption 2. The set of all matrices
$$\Lambda=\begin{pmatrix}1 & 0 & 0 & 0\\ 0 & & & \\ 0 & & R & \\ 0 & & &\end{pmatrix}$$ where $R\in SO(3)$, is a subgroup of G.

Assumption 3. The assumptions from the 1+1-dimensional case, with some obvious modifications, apply to the subgroup mentioned in assumption 1 here.

Assumption 4: Something like Giulini's $B(Rv)=R(D)B(v)R(D^{-1})$.

Asumption 5: Every member of the G is a boost times a rotation. (This one is ugly, but how else can we know that G is the Lorentz group when K=1, rather than "has the Lorentz group as a subgroup"?)

 

[strangerep]

I'm not quite sure where/how you've banished the dilation subgroup. I guess it's lurking somewhere in your assumption 3?

Now some thoughts about the 3+1-dimensional case.

This is where you'll have to pin down what spatial isotropy means (physically, at least). It can be done by demanding that an observer cannot pick out a distinguished spatial direction by any means. As a consequence, spatial angles must be preserved, else axes which are squeezed together can be distinguished from those which are pushed apart. Finding just the subgroup that preserves spatial angles gives you O(3) -- since one also assumes a locally-Euclidean spatial metric. Then you need to have an assumption about homogeneity to dispense with dilations, and hence get down to SO(3).

Then you can pick an arbitrary spatial direction denoted by a unit vector $\widehat u$ and consider velocity boosts along this direction only. (Do an orthogonal decomposition of other vectors wrt $\widehat u$. Then you can re-use some of the 1+1D stuff.)

Then, changing to a different velocity direction is now easy since you've already got SO(3).

You'll still need an assumption about spatiotemporal homogeneity to banish the dilation subgroup though. In physics textbooks, this usually comes in the form of assumptions about standardized clocks and rulers...

 

[Fredrik]

I'm not quite sure where/how you've banished the dilation subgroup. I guess it's lurking somewhere in your assumption 3?

I'm not sure what exactly a dilation is. Is it a number times the identity matrix? In that case, the requirement $(\Lambda^{-1})_{00}=\Lambda_{00}$ is sufficient to rule it out.

If it's an arbitrary diagonal matrix, then the result $d=(\det\Lambda)a$ (which follows from assumption 3) takes care of it.

Is it supposed to dilate spacetime or just space? Hm, a $\Lambda$ that stretches vectors in the 1 direction without changing their direction in spacetime has b=0, i.e. $\alpha=0$. Ouch. This is probably what you're talking about, because I see a problem. The result $\alpha'v''=\alpha''v'$ doesn't tell us anything when the alphas are 0. So it seems possible that $\alpha/v$ has the same value for all members of the group with both $\alpha$ and v non-zero, but a different value (0 to be specific) for a bunch of members of the group with $\alpha=0$.

Edit: On second thought, I don't think there is a problem. My proof is correct, and when K≠0, the possibility that I was concerned with leads to a contradiction.

This is where you'll have to pin down what spatial isotropy means (physically, at least).

I'm just thinking that a measuring device should find the theory's predictions equally accurate regardless of which way we point it. (If someone is concerned about changing the properties of the device as we physically rotate it, we can instead imagine a new device being built to point in another direction, from individual atoms if necessary, according to identical specifications). It seems acceptable to me to just say that we're making this idea mathematically precise with my assumption 2 from my previous post.

Of course, the reason I find it acceptable is that I've done proofs like this. I should probably make that part of the explanation of why I think of the assumption as making the idea of isotropy mathematically precise.

It can be done by demanding that an observer cannot pick out a distinguished spatial direction by any means. As a consequence, spatial angles must be preserved, else axes which are squeezed together can be distinguished from those which are pushed apart. Finding just the subgroup that preserves spatial angles gives you O(3) -- since one also assumes a locally-Euclidean spatial metric. Then you need to have an assumption about homogeneity to dispense with dilations, and hence get down to SO(3).

Interesting ideas. I will have to think about how much I want to include. (If I do include it, I think it will be in the "explanation" part that precedes the actual theorem).

 

[strangerep]

I'm not sure what exactly a dilation is. Is it a number times the identity matrix? [...] Is it supposed to dilate spacetime or just space?

"Yes" and "spacetime" -- since I was talking about uniform dilations. Sometimes one speaks of a nonuniform dilation, but a stretch of one spatial axis compared to another conflicts with spatial isotropy.

I'm just thinking that a measuring device should find the theory's predictions equally accurate regardless of which way we point it. [...] It seems acceptable to me to just say that we're making this idea mathematically precise with my assumption 2 from my previous post.

Your current assumption 2 could refer to 2 velocities in the same direction, hence might need enhancement.

[...] making the idea of isotropy mathematically precise.

Thinking it over, it's probably simpler just to exploit the spatial part of the metric being $\delta_{ij}$, since one assumes that every inertial observer perceives locally Euclidean geometry.

 

[Fredrik]

Your current assumption 2 could refer to 2 velocities in the same direction, hence might need enhancement.

Are you referring to item 2 on the first list in post #35? (That's the only place associated with the number 2 that your comment seems to apply to). This is the statement that's supposed to provide some motivation for assumption 2. The actual assumption 2 is V(G)≠{0}.

Do I need to say "two distinct velocities" instead of "two velocities"? Maybe it would be more clear that way. But I'm not sure it's logically necessary. For example, wouldn't we consider the following statement false? "There two positive real numbers in the set {0,1}".

I'm leaning towards replacing assumption 2 with the following: 0 is an interior point of V(G). This has some pretty wonderful consequences, if I'm not mistaken. I think it allows me to drop assumption 4, and also to obtain the stronger result that G is equal to one of the desired groups, rather than just a subset of one of them. I will probably post another comment about that when I have checked the details.

 

[strangerep]

Are you referring to item 2 on the first list in post #35?

I was. But whether my comment remains relevant or not depends on how you proceed, so I'll leave it at that for now.

 

[Fredrik]

I still haven't worked out all the details, but I think I can prove the following:

Theorem? Suppose that G is a subgroup of $\operatorname{GL}(\mathbb R^2)$ such that

1. There's a $V:G\to\mathbb R$ such that
   • For all $\Lambda\in G$, $V(\Lambda)=-\frac{(\Lambda^{-1})_{10}}{(\Lambda^{-1})_{00}}$.
   • 0 is an interior point of V(G).
2. For all $\mu\in\{0,1\}$, we have $(\Lambda^{-1})_{\mu\mu}=\Lambda_{\mu\mu}$.

Then there's a $K\geq 0$ such that
$$G=\left\{\left.\frac{\sigma}{\sqrt{1-Kv^2}}\begin{pmatrix}1 & -Kv\\ -\rho v & \rho\end{pmatrix}\right|\,v\in\mathbb R,\,1-Kv^2>0,\,(\sigma,\rho)\in S\right\},$$ where S is one of the sets
\begin{align}
&S_f=\{-1,1\}\times\{-1,1\}, && S_p=\{(-1,1),(1,1)\},\\
&S_o=\{(1,-1),(1,1)\}, && S_r=\{(1,1)\}.
\end{align}
My strategy to rule out K<0 is as follows. Suppose that K<0. The result for $\gamma$ in my previous proof (obtained from assumption 1a and 2, which were part of my old assumptions as well) implies that c (defined as $1/\sqrt{|K|}$) is not a member of V(G). So c is a "forbidden" velocity, and this implies that $\pi/2$ is a forbidden rapidity. Let ε>0 be such that $(-\varepsilon,\varepsilon)\subset V(G)$. Let $\theta_\varepsilon$ be the rapidity corresponding to velocity ε, and choose an integer n such that $\pi/(2n)<\varepsilon$. Define $\theta_0=(2\pi)/n$, and let $v_0$ be the corresponding velocity. Then I show that there's a $\Lambda\in G$ such that the velocity of $\Lambda^2$ is $v_0$. (If I just choose a $\Lambda$ with velocity $v_0$, its determinant may be <0, and I don't think the whole thing with rapidities work with negative determinants). Such a $\Lambda$ satisfies $V(\Lambda^{2n})=c$, which implies that $\Lambda\notin V(G)$, and we have a contradiction.

I still need to make sure that these assumptions can rule out all pathological possibilities like G contains all the proper transformations with rational velocities and all the improper ones with irrational velocities.

Assuming that this works, I think my assumptions for the 3+1-dimensional case will be something like this:

1. There's a $V:G\to\mathbb R^3$ such that
   • For all $\Lambda\in G$ and all $i\in\{1,2,3\}$, $V^i(\Lambda)=-\frac{(\Lambda^{-1})_{i0}}{(\Lambda^{-1})_{00}}$.
   • 0 is an interior point of V(G).
2. The set of all $\Lambda\in G$ of the form
   $$\Lambda=\begin{pmatrix}B & 0\\ 0 & I\end{pmatrix}$$ where B,0 are 2×2 matrices and I is the 2×2 identity matrix, is a subgroup.
3. For all $\Lambda$ in that subgroup and all $\mu\in\{0,1\}$, we have $(\Lambda^{-1})_{\mu\mu}=\Lambda_{\mu\mu}$.
4. The set of all matrices
   $$\Lambda=\begin{pmatrix}1 & 0 & 0 & 0\\ 0 & & & \\ 0 & & R & \\ 0 & & &\end{pmatrix}$$ where $R\in SO(3)$, is a subgroup of G.
5. I haven't decided how I want to state this assumption, but it will ensure that all boosts can be obtained from the ones in the 1 direction, using a formula like Giulini's $B(Rv)=R(D)B(v)R(D^{-1})$.
6. I haven't decided how I want to state this assumption, but it will say something like "every member of the group is a product of a member of the first subgroup with a member of the second subgroup", and it may also mention parity and time reversal in some way (but it probably won't need to).

Assumptions 4-5 are both aspects of isotropy, so I maybe I should merge those two assumptions into one.

 

[Fredrik]

I'm still not sure if my approach works. There's a problem even with the K=0 case. One of the things I want to show is that for each real number v, there's a proper and orthochronous transformation in G with velocity v.

The obvious way to start an attempted proof goes like this: Let $v\in\mathbb R$ be arbitrary. Let $\varepsilon>0$ be such that $(-\varepsilon,\varepsilon)\subset V(G)$. Let n be a positive integer such that $v/n\in(-\varepsilon,\varepsilon)$. Let $\Lambda\in G$ be such that $V(\Lambda)=v/n$. If $\Lambda$ is proper and orthochronous, then so is $\Lambda^n$, and we have $V(\Lambda^n)=nV(\Lambda)=v$. If $\Lambda$ isn't proper and orthochronous, we have to try something else.

The fact that $\Lambda$ may not be orthochronous is not much of an issue. We could have defined $\Lambda$ as the transformation with velocity v/(2n) instead. If this $\Lambda$ is proper but not orthochronous, then $\Lambda^2$ is proper and orthochronous. So $\Lambda^{2n}$ is proper and orthochronous, and $V(\Lambda^{2n})=v$. Edit: (Removed a ${}^{-1}$ from this sentence that didn't belong there).

The real problem is that this $\Lambda$ may not be proper. If that's the case, we have $V(\Lambda^2)=0$ instead of $V(\Lambda^2)=2 V(\Lambda)\det\Lambda$. Of course, that's not a problem if the transformation with velocity v/(4n) is proper, or if the one with velocity v/(8n) is. In fact, we only have a problem if the entire sequence $\langle v/(2^kn)\rangle_{k=1^\infty}$ are velocities of improper transformations only.

So the only thing that can prevent the restricted subgroup of G from being "filled" with transformations of all velocities (in the K=0 case at least), is that there are lots of improper transformations with small velocities. Since the product of any odd number of improper transformations is improper, there are a lot more improper transformations than the ones in that sequence. I would guess that the set that contains all these improper velocities is dense in (-ε,ε).

If there's an entire interval (-a,a) of velocities of improper transformations, then I think that since the product of an even number of improper transformations is proper, that interval and therefore the entire real line (in the K=0 case) will be filled with velocities of proper transformations as well.

But if the set of velocities of improper transformations only, is a dense proper subset of (-ε,ε), then maybe we have a real problem. But the thing is, the complement of that set in (-ε,ε) consists of velocities of proper transformations, and the composition of a proper and an improper transformation is improper, so it seems that we should be able to fill that interval of improper velocities after all.

This is pretty annoying. It seems like it should work, but I haven't been able to turn this into a proof yet. It doesn't help that the general velocity addition rule is pretty complicated:
$$V(\Lambda\Lambda')=\frac{\rho v+\rho\rho'v}{1+Kvv'\rho'}.$$

 

[strangerep]

I can't say I follow your thinking in detail, so I'll just mention some thoughts that occurred to me as I was reading your previous posts...

You seem not to be exploiting the notion of continuity sufficiently. For the improper transformations, can you show that every such improper transformation can be decomposed into a proper transformation and an ordinary $v$-independent reflection? If you can thus factor out the improper transformations, then you can work with the identity-continuous part of the group, hence the infinitesimal generators only, hence simplifying much of the calculation.

IOW, you still seem to be trying to do everything in one mega-calculation, whereas "decompose-and-conquer" is usually the key to such group analysis problems.

But, as usual, I'm unsure whether these thoughts are relevant...

 

[Fredrik]

I can't say I follow your thinking in detail,

That's understandable. I didn't put as much effort into this explanation as I usually do since I didn't have a proof and was only trying to point out some of the issues.

For the improper transformations, can you show that every such improper transformation can be decomposed into a proper transformation and an ordinary $v$-independent reflection?

I haven't been able to do that yet. It's complicated because my assumptions do not immediately tell me if $P\in G$. (I defined P earlier as the diagonal matrix with 1 and -1 on the diagonal). Suppose that $P\in G$. If $\Lambda\in G$ is improper, then $P\Lambda$ is proper and $P\Lambda\in G$. So we can just write $\Lambda=P^2\Lambda=P(P\Lambda)$.

I believe that my assumptions are strong enough to ensure that for all $v\in\mathbb R$ such that $1-Kv^2>0$, there's a proper and orthochronous transformation in G with velocity v. (This is the part I'm trying to prove). If I'm right, then the following argument proves that either none of the improper transformations is in G, or all of the improper orthochronous ones are. Edit: The original version of this post only had an incomplete proof of this statement. I have now completed the proof and fixed a couple of typos.

Let P be the parity transformation (i.e. the diagonal matrix with 1 and -1 on the diagonal). Either $P\in G$ or $P\notin G$. Suppose that $P\in G$. Let $\Lambda$ be an arbitrary improper orthochronous transformation. Let $\Lambda'$ be the proper orthochronous transformation with velocity $-V(\Lambda)$. Since P and $\Lambda'$ are in G, $P\Lambda'$ is in G. $P\Lambda'$ is improper and orthochronous, and has velocity $-V(\Lambda')=V(\Lambda)$. So $\Lambda=P\Lambda'\in G$.

Now suppose that $P\notin G$. I will prove that there's no improper transformation in G by deriving a contradiction from the assumption that there is. So suppose that $\Lambda\in G$ is improper. Let $\Lambda'$ be the proper and orthochronous transformation with velocity $-V(\Lambda)$. $\Lambda\Lambda'$ is improper, and we have
$$V(\Lambda\Lambda')=\frac{\rho v+\rho\rho' v'}{1+Kvv'\rho'}=\frac{-v-v'}{1+Kvv'}=0.$$ This implies that $\Lambda\Lambda'=\sigma P$. If $\Lambda$ is orthochronous, this means that $P\in G$, and we have a contradiction. If $\Lambda$ is non-orthochronous, then $-P\in G$. But if we let $\Lambda''$ be the proper and orthochronous transformation such that $V(\Lambda'')=-V(\Lambda)$, we have $\Lambda\Lambda''=\sigma I=-I$. So both -I and -P are in G, and that means that $P=(-I)(-P)\in G$, contradicting the assumption that $P\notin G$.

So the main issue is the problem of finding a proper and orthochronous transformation for each velocity.

If you can thus factor out the improper transformations, then you can work with the identity-continuous part of the group, hence the infinitesimal generators only, hence simplifying much of the calculation.

The thing is, I'm deliberately trying to avoid technical statements in the assumptions, like "G is a connected Lie group", because these are less intuitive and harder to motivate, and also because I'd like to be able to explain this to people who don't know Lie groups.

IOW, you still seem to be trying to do everything in one mega-calculation, whereas "decompose-and-conquer" is usually the key to such group analysis problems.

Yes, it might be better to break this up into smaller theorems and then combine them into the theorem I want.

 

[Fredrik]

I think I can explain the issue a bit better. (If nothing else, this will make me a bit less confused). Let G be a group that satisfies my assumptions (1a, 1b and 2 in the "Theorem?" in post #42). Suppose also that for this G, K=0. (This is just so that I won't have to deal with a complicated velocity addition rule in this post). I will call a real number r ugly if all $\Lambda\in G$ such that v=r are improper. Let S be the set of ugly real numbers.

I would like to prove that for each $r\in\mathbb R$, there's a proper and orthochronous $\Lambda\in G$ such that $v=r$.

I have made some major edits of the stuff below.

Let $r\in\mathbb R$ be arbitrary. Let ε>0 be such that $(-\varepsilon,\varepsilon)\subset V(G)$. Let $n\in\mathbb Z^+$ be such that $r/n\in(-\varepsilon,\varepsilon)$. There's a $\Lambda\in G$ with v=r/(2n), but is there a proper $\Lambda\in G$ with that velocity? If there is, then $\Lambda^2$ is proper and orthochronous, and this implies that $\Lambda^{2n}$ is proper and orthochronous. Since $V(\Lambda^{2n})=2nV(\Lambda)=r$, this means that $\Lambda^{2n}$ is a proper and orthochronous transformation with velocity r.

If there's no proper $\Lambda$ with v=r/(2n), then there is an improper one. But there's no easy way to use it to prove that there's a proper and orthochronous transformation with velocity r. All we can say right now is that r/(2n) is an ugly velocity.

But I think I can show that there are no ugly velocities. (This would imply that there is a proper $\Lambda$ such that $v=r/(2n)$, and fix the proof above). For all ugly v', v'/2 is ugly. (If it's not, then there's a proper $\Lambda'$ with velocity v'/2, but then $\Lambda'^2$ is proper and has velocity v', and this implies that v' is not ugly). For all ugly v' and v'', the difference v''-v is not ugly. The reason is this: Let $\Lambda',\Lambda''$ be arbitrary members of G with velocities v' and v'' respectively. Since v' and v'' are ugly, $\Lambda'$ and $\Lambda''$ must be improper. This means that $\Lambda'\Lambda''$ is proper, so the velocity of $\Lambda'\Lambda''$ is not ugly. Since $V(\Lambda'\Lambda'')=\rho'v'+\rho'\rho''v''=v''-v'$, this means that v''-v is not ugly.

Now suppose that there's an ugly velocity v'. One of the results above tells us that v'/2 is ugly, and another one that v'/2=v'-v'/2 is not ugly. This is a contradiction, so we are forced to conclude that there are no ugly velocities. In particular, the number r/(2n) is not ugly.

This seems to solve the problem in the K=0 case. I haven't thought about the other cases yet. (I will probably have to do this even for the K<0 case, and use the result when I prove that K≥0). Instead of "half the velocity", I expect to use "half the rapidity".

 

[Fredrik]

It looks like the proof that there are no "ugly" velocities in the K=0 case is easy enough to modify for the K>0 and K<0 cases. Where I considered "half the velocity" in the K=0 case, I can just consider "half the rapidity". So I think I am able to complete the proof of the theorem stated in post #42.

The proof of that theorem is pretty long, so I think I'm going to break it up into a number of smaller theorems and proofs.

 

[Fredrik]

It took me a long time, but I was finally able to prove the following theorem for the 1+1-dimensional case.

Theorem: Suppose that G is a subgroup of GL(ℝ²) such that

1. There's a $V:G\to\mathbb R$ such that
   • For all $\Lambda\in G$, $V(\Lambda)=\frac{(\Lambda^{-1})_{10}}{(\Lambda^{-1})_{00}}$.
   • 0 is an interior point of V(G).
2. For all $\mu\in\{0,1\}$, we have $(\Lambda^{-1})_{\mu\mu}=\Lambda_{\mu\mu}$.

Then there's a $K\geq 0$ such that
$$G=\left\{\left.\frac{\sigma}{\sqrt{1-Kv^2}}\begin{pmatrix}1 & -Kv\\ -\rho v & \rho\end{pmatrix}\right|\,v\in\mathbb R,\,1-Kv^2>0,\,(\sigma,\rho)\in S\right\},$$ where S is one of the sets
\begin{align}
S_r &=\{(1,1)\}\\
S_p &=\{(1,1),(-1,1)\}\\
S_o &=\{(1,1),(1,-1)\}\\
S_n &=\{(1,1),(-1,-1)\}\\
S_f &=\{-1,1\}\times\{-1,1\}.
\end{align}

The proof is about 14 pages long. I broke it up into a bunch of smaller definitions, theorems and proofs. I have more or less finished typing it up, but I want to go through it again before I upload a pdf. So it will probably be tomorrow.

I think this is a pretty cool result. The assumptions are very weak, but they're strong enough to rule out the possibility K<0 and all but five groups for each K≥0. And all of those groups satisfy the assumptions 1a, 1b and 2. The proof also makes it clear that G is completely determined by its intersection with {-I,P,T}, where
$$P=\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix},\qquad T=\begin{pmatrix}-1 & 0\\ 0 & 1\end{pmatrix},\qquad -I=\begin{pmatrix}-1 & 0\\ 0 & -1\end{pmatrix}$$ (Since PT=-I, T(-I)=P and -IP=T, there are at most five subsets of {-I,P,T} that can be equal to $G\cap \{-I,P,T\}).$

This means that if we want to narrow it down to only one group, we can do it by adding assumptions like $P\in G$, $T\notin G$, $-I\notin G$. Such assumptions have natural interpretations as symmetry principles. (By a "principle", I mean an idea stated in non-mathematical terms). For example, $P\in G$ can be interpreted as saying that space is invariant under reflections.

 

[Fredrik]

I'm attaching a pdf that contains my proof. Apparently it's 16 pages in this font.

To anyone who's interested in seeing the proof: If this isn't the last post in the thread, scroll down to see if you can find comments about an updated version of the document. If you find a mistake in the latest version of the document, I would appreciate if you let me know, even if it's been a long time since this thread was active. Bump the thread if it hasn't been auto-locked. Send me a PM if it's locked.

Edit: What the...did I do something wrong when I attached the pdf? Or do attachments need to be approved? I'll just try to attach it again. OK, now it's been attached correctly.

 

[strangerep]

1) Minor typo: Just after eq(12) on p3:

[...], the inequality $1-Kv^2$ is [...]

I guess you meant $1-Kv^2 > 0$ ?

2) In the proof of Lemma 16 ("The relativity is non-negative")...

For each $v\in R$, define a corresponding angle $\theta\in(-\pi,\pi)$ by $\theta=\arctan(v/c)$. The angle that corresponds to $c$ is $\pi/2$. [...]

What do you mean by "the angle that corresponds to $c$" ? If you mean "...corresponds to $v=c$" (as suggested by the last line of the proof), then the statement seems incorrect, unless you meant $\pi/4$ or I've misunderstood something else. Or maybe you meant "...corresponds to $c=0$"? But, hmm, no -- then I don't understand eq(35).

3) In the proof of Lemma 13, part (b), you mentioned "Corollary 9".
Did you mean "Lemma 9"?

4) Back to Lemma 16 again... I don't think you've proved $K\ge 0$, but only that $c = 1/\sqrt{K}$ is not an allowable value for $v$ if $K<0$. But that was already apparent from the velocity addition formula, which blows up, unless I'm mistaken. Maybe you intended to show that this means any nontrivial velocity interval $(-\epsilon,\epsilon)$ leads to a blowup, and this is in conflict with the "interior point" assumption?

5) Wouldn't hurt to put a version number and date near the start of the document.

 

[Fredrik]

Thank you for those comments. I appreciate them a lot.

1. Fixed.

2. I meant the angle that corresponds to the velocity c, i.e. the $\theta$ such that $c\tan\theta=c$. You're right that this is $\pi/4$, not $\pi/2$.

3. Yes, I've been changing my mind over and over about when to use the words "theorem", "lemma", "corollary", so I ended up with lots of mistakes like this. I had already fixed most of them when I uploaded the pdf, but apparently not all. I found this specific mistake in four places. I will make another sweep for similar mistakes.

4. The velocity addition formula is
$$V(\Lambda\Lambda')=\frac{\rho'v+v'}{1+Kvv'\rho'}.$$ When K<0 and $\Lambda'=\Lambda$, this turns into
$$V(\Lambda^2)=\frac{\rho v+v}{1-|K|v^2\rho}.$$ This blows up at v=c (and v=-c) only for proper transformations. But that's actually all I need, so I should definitely use this.

My strategy to rule out K<0 is to prove the following:

• There is NO proper transformation with velocity c.
• If ε>0 is such that (-ε,ε) is in the range of the velocity function V (such an ε exists because of my assumption 1b), then for each v in that interval, there's a proper transformation with velocity v.
• For each proper $\Lambda$, the angle corresponding to $\Lambda^n$ is n times the angle corresponding to $\Lambda$.
• Find the angle $\theta_c$ corresponding to the velocity c (which is forbidden for proper transformations), then chose an integer n such that $\theta_c/n<\varepsilon$. Let $\Lambda$ be a proper transformation with velocity $\arctan(\theta_c/n)$. Then $\Lambda^n$ is proper and the angle corresponding to $\Lambda^n$ is $n\theta_c/n=\theta_c$, so the velocity of $\Lambda^n$ is $c\tan\theta_c$. Since $\theta_c=\pi/4$ and $\tan(\pi/4)=1$, this means that the velocity of $\Lambda^n$ is c, and we have a contradiction.

I see that I need to make some rewrites to make this clearer. I will upload a new version when I've fixed these problems. I will add a comment about it to this post when I'm done.

5. Good idea. I will do this.

Edit: The inequality $\theta_c/n<\varepsilon$ above doesn't make much sense. The right-hand side should be the angle that corresponds to the velocity $\varepsilon$. See the lemma titled "The relativity is non-negative" in the latest version of the document. (Scroll down). In version 2, this is lemma 17.

 

[strangerep]

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. :-)

 

[Fredrik]

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.

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.

 

[strangerep]

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.

 

[Fredrik]

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.

 

[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. I'm sure that the problem is fixable, but it requires a substantial rewrite.

 

[strangerep]

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.

Yeah, I had wondered about something similar, but I hadn't gotten around to thinking about it carefully...

 

[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

 

[Fredrik]

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.

 

[strangerep]

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:

http://books.google.com.au/books?id...8T1KCDS&dq=gorini+"rotational+invariance"&lr=