# Should spacetime transformations make a group?

1. Mar 6, 2009

### lalbatros

In the story below, where would you see possible alternatives, or where would you see a problem?

(01) Let us consider a set of physicists {P0, P1, P2, P3, ...} each at rest in their own inertial frames.
(02) Let us elect one of them (P0) as the boss to manage an experiment.
(03) Let us assume the velocities of each physicist with respect to the boss are {v1, v2, v3, ...}.
(04) The boss asks each physicist to measure in his own frame the spacetime coordinates E(x,t) of a large number of events he has scheduled, something like sparkles in a firework.
(05) All the physicists are using the same measuring instruments.
(06) All physicist "zero" their coordinate systems on the origin of the boss coordinate system.
(07) All the physicists must return their result to the boss for analysis.
(08) The boss is now looking for a relation between all these measurements.
(09) The boss choses for a linear relation between the coordinates measured by all physicists.
(10) The boss assumes this linear relation depends on the velocities of his colaborators.
(11) In summary he assumes: E1=T(v1)E0, E2=T(v2)E0, where T(v) is a 2x2 matrix depending on v
(12) The boss organizes a meeting with his collaborators.
(13) He asks them: "What should be the properties of these matrices T(vi)"?
(14) One of them, called René, writes these properties on the black board:
(15) >> a. T(0) should be the unit matrix such that T(0)Ei=Ei
(16) >> b. Inverse(T(v))=T(-v) to assure forward and backward transformations are similar
(17) >> c. T(vi).T(vj) should also be a transformation matrix, or in other words:
(18) >> d. T(vi).T(vj)=T(vij) without prejudice about how vij should be calculated
(19) >> He concludes saying:
(20) The set of matrices T(v) is a one-parameter matrix group.
(20) The parameter involved is the relative velocity v.

Assume you are one of the other physicists and you are invited to express your opinion:

Q1- What do you think about the overall experiment?
Q2- Is the experimental protocol correctly defined?
Q3- What do you think about the properties suggested by René?
Q4- Do you think all properties should be needed?
Q5- Do you think each property has a physical meaning? And which one?
Q6- Do you agree with each physical meaning?
Q7- Do you think these properties might not all be verified experimentally?
Q8- In summary, according to you, should T(v) be a one-parameter matrix group?

Assuming the conclusion by René is right:

Q9- Do you think there are many different such groups (one-parameter 2x2 matrix group)
Q10- How many could you lists?
Q11- Are some of them already familiar to you?
Q12- To finish the job of the boss, which group would you suggest?
Q13- Write a formula for T(v).
Q14- Would you have ideas for further experiments?

Last edited: Mar 6, 2009
2. Mar 7, 2009

### lalbatros

The assumptions by René are not all necessary I think.
In fact, (18) is not needed.
Assuming a semi-group is enough.

Based on the assumption by René, one of the solution is the Lorentz group based on a certain constant velocity k.
Combining with the constancy of the speed of light leads to k=c, obviously.
Therefore, the first Einstein' postulate should be the same as assuming a semi-group.

Another group is of course the rotation space-group.

Therefore

3. Mar 8, 2009

### Fredrik

Staff Emeritus
If you drop (18), you're saying that if P3 has a constant velocity relative to P2, and P2 has a constant velocity relative to P1, P3 doesn't have a constant velocity relative to P1. Or to put that a different way, you're saying that there's no Lorentz transformation to transform from P1's coordinates to P3's coordinates. That doesn't seem to be consistent with the principle of relativity (the idea that no inertial frame is fundamentally different from any of the others).

4. Mar 8, 2009

### Fredrik

Staff Emeritus
Q9-Q13: Your assumptions 1-18 imply that

$$T(v)=\gamma\begin{pmatrix}1 & -v \\ -v & 1\end{pmatrix}$$

with

$$\gamma=\frac{1}{\sqrt{1-v^2}}$$

(in units such that c=1).

Edit: I just noticed that you didn't assume that the speed of light is the same in all frames. So...

$$T(v)=\gamma\begin{pmatrix}1 & \alpha \\ -v & \beta\end{pmatrix}$$

where $\alpha,\beta,\gamma$ are real numbers, that are not all independent (because of 16-18).

Last edited: Mar 8, 2009
5. Mar 8, 2009

### lalbatros

Fredrik,

1)
When I said that (18,17) can be dropped, I do not mean it is not true.
I simply mean that the Lorentz transformation can be derived without making this assumption.
Having derived the LT in this way, (18) is a consquence.

2)
Indeed, I do not assume the constancy of the speed of light.
Yet the Lorentz transformation results.
The speed of light is simply a definition of the units of time with respect to the units of space.
This is why I also started another https://www.physicsforums.com/showthread.php?t=297931".

Last edited by a moderator: Apr 24, 2017
6. Mar 8, 2009

### lalbatros

Fredrik,

In this formula

$$T(v)=\gamma\begin{pmatrix}1 & \alpha \\ -v & \beta\end{pmatrix}$$

we should have

$$\alpha=\alpha(v) ...$$

And the conditions then restricts the choice of these functions.

In the end, I think, c is only a unit convertion factor.
A subtle problem then, I think, is the Galilean limit.
How can it be understood if c is a unit convertion factor and c->infinity ?

7. Mar 8, 2009

### Fredrik

Staff Emeritus
That sounds weird to me. Have you done this derivation already?

This is probably true (at least if we include all of the other assumptions), but I'd like to see an actual derivation. Maybe I'll try it myself later.

There should be at least one more variable (other than v) that we can't eliminate: the invariant velocity. But the value of variable doesn't label different members of the same group (like v), but instead label the group we're dealing with. I expect that the groups with different but finite invariant velocity are all isomorphic, so we really only get two choices: finte, or infinite. (The Poincaré group or the Galilei group).

That's what I expect, but as I said, I haven't done or seen the calculations.

8. Mar 9, 2009

### lalbatros

Fredrik,

No calculation is needed.
You know the rotation group with its matrix {{cos(q),sin(q)},{-sin(q),cos(q)}}.

The other group is simply based on {{ch(q),sh(q)},{sh(q),ch(q)}}.
Of course, units of space and units of time should be the same.
The relation between q and v is obtained easily.

9. Mar 9, 2009

### lalbatros

I just found this interresting paper:

forums.futura-sciences.com/attachments/physique/8862d1151959759-vitesse-de-lumiere-cette-constante-rr-1-postulat.pdf

The paper is maybe a bit longer than needed.
The linearity of the transformation could have been assumed from the beginning.
Using the speed (v) as the parameter makes the algebra a bit longer.
Actually any one-parameter matrix group is similar to either rotarions or lorentz transformation.

The mere matrix group assumption leads to the conclusion of a limit speed in the general case. Identifying the limit as c is a consequence from the necessary invariance of Maxwell's equations.

10. Mar 9, 2009

### Fredrik

Staff Emeritus
We obviously need to do something to see that those two groups are the only options.

You didn't answer the question about if you have successfully derived (18) from the other assumptions.

11. Mar 9, 2009

### lalbatros

Fredrik,

I think I was wrong claiming that (18) is not useful.
I don't remember exactly why I came to this idea.

The paper ""forums.futura-sciences.com/attachments/physique/8862d1151959759-vitesse-de-lumiere-cette-constante-rr-1-postulat.pdf"[/URL]" derives the Lorentz transformation without using the second postulate "c=constant" (up to the final match with the speed of light).
The group properties are all used in this paper.
The need for a linear transformation is derived from homogeneity.

One question still remains for me now.
As a matter of fact, the velocity (v) used in the transformation is not an additive parameter.
However, the Lorentz group is isomorphic to the addition group, using the parameter q (such that th(q)=v) as a parameter in place of v.
My question is: was there no good reason to assume the existence of this additive parameter right from the beginning?
I guess this is related to the 2x2 dimension of the matrix which is assumed initially.

Last edited by a moderator: Apr 24, 2017
12. Mar 9, 2009

### Fredrik

Staff Emeritus
I decided to try to work it out on my own. I started with

$$\Lambda=\gamma\begin{pmatrix}1 & \alpha \\ -v & \beta\end{pmatrix}$$

This is the most general form of a 2x2 matrix. I just chose to write it like that so that I could label the velocity v right from the start. (Any 2x2 matrix maps the 0 axis to a line through the origin, and we define v by saying that the slope of this line is -1/v). I have tried to assume as little as possible other than that we're trying to find a group of such matrices and that when we take the inverse, v gets replaced by -v. I didn't make any assumptions about how the greek letters depend on v or on each other.

These assumptions turned out to be sufficient to further constrain the form of $\Lambda$ and its inverse. This is what I got:

$$\Lambda=\gamma\begin{pmatrix}1 & \alpha \\ -v & 1\end{pmatrix}$$

$$\Lambda^{-1}=\frac{1}{\gamma(1+\alpha v)}\begin{pmatrix}1 & -\alpha \\ v & 1\end{pmatrix}$$

But I still don't see how to show that alpha and gamma are uniquely determined by v.

13. Mar 9, 2009

### Fredrik

Staff Emeritus
OK, I got one step further. If $\Lambda$ is a boost to the right, then its inverse is a boost to the left. If "left" isn't fundamentally different from "right", then

$$\Lambda\begin{pmatrix}0 \\ 1\end{pmatrix}$$

should have the same 0 component as

$$\Lambda^{-1}\begin{pmatrix}0 \\ -1\end{pmatrix}$$

and that condition implies

$$\gamma=\pm\frac{1}{\sqrt{1+\alpha v}}$$

So gamma and alpha are not independent. We can choose to think of alpha as an independent parameter and gamma as a function of alpha and v, but it's clear from the explicit expressions for $\Lambda$ and $\Lambda^{-1}$ above that alpha also depends on v. It must be an odd function of v.

We get rid of the minus sign above by adding the assumption that the group we're interested in is path connected (which (I think) it is if and only if it's simply connected, since it's a matrix group). (The Lambda we get when we choose the minus sign doesn't go to an identity transformation when we let v go to 0)

So now we have

$$\Lambda=\gamma(v)\begin{pmatrix}1 & \alpha(v) \\ -v & 1\end{pmatrix}$$

where

$$\gamma(v)=\frac{1}{\sqrt{1+\alpha(v)v}}$$

and alpha is a still undetermined odd function of v. I think we're supposed to get

$$\alpha(v)=-Kv$$

where K is a positive constant, but I still haven't figured out how.

14. Mar 10, 2009

### lalbatros

You are right on the point Fredrik.
If you play the condition (18), you get this extra condition:

$$v1 .\alpha(v2) = v2 .\alpha(v1)$$

(since in the matrix product, the diagonal elements must be the same)

15. Mar 10, 2009

### Fredrik

Staff Emeritus
Thanks. I got that result too (and used it to get my expression for $\Lambda^{-1}$), but I didn't see that it implies $\alpha(v)=-Kv$. Now I see that it does, but it doesn't imply K>0. I haven't had much time to think about this yet, but so far I haven't been able to obtain that result. If K can be negative, the result is very different. For example, the choice K=-1 is just a (Euclidean) rotation in spacetime. We seem to need K>0 to get an invariant velocity.

16. Mar 10, 2009

### Fredrik

Staff Emeritus
I found the answer in the article you linked to above. The condition that I've been writing as

$$\bar\Lambda\Lambda=\tilde\Lambda$$

(on my whiteboard, not in this forum) implies that

$$\tilde\gamma=\bar\gamma\gamma(1+K\bar v v)$$

The gammas are all positive, so if we allow K to be negative, we get a contradiction because it's possible to choose the velocities so that $\tilde\gamma<0$. Therefore, we must have K≥0.

This has been very interesting. I'm going to post the full calculation later, so that I can use this thread as a reference in the future.

17. Mar 10, 2009

### Fredrik

Staff Emeritus
Here's the stuff I was too lazy to LaTeX yesterday.

We're looking for a simply connected group of 2×2 matrices that satisfies the following conditions:

1. If

$$\Lambda=\gamma\begin{pmatrix}1 & \alpha\\ -v & \beta\end{pmatrix}$$​

and

$$\Lambda^{-1}=\gamma'\begin{pmatrix}1 & \alpha'\\ -v' & \beta'\end{pmatrix}$$​

then $v'=-v$.

2. The 0 component (the "upper" component) of

$$\Lambda\begin{pmatrix}0 \\ 1\end{pmatrix}$$​

is equal to the 0 component of

$$\Lambda^{-1}\begin{pmatrix}0 \\ -1\end{pmatrix}$$​

The requirement that these matrices form a group implies that the product of two such matrices is another such matrix. Let's write this as

$$\bar\Lambda\Lambda=\tilde\Lambda$$​

We get

$$\tilde\gamma\begin{pmatrix}1 & \tilde\alpha\\ -\tilde v & \tilde\beta\end{pmatrix}=\tilde\Lambda=\bar\Lambda\Lambda=\bar\gamma\gamma\begin{pmatrix}1 & \bar\alpha\\ -\bar v & \bar\beta\end{pmatrix}\begin{pmatrix}1 & \alpha\\ -v & \beta\end{pmatrix} =\bar\gamma\gamma\begin{pmatrix}1-\bar\alpha v & \alpha+\bar\alpha\beta\\ -\bar v-\bar\beta v & -\bar v\alpha+\bar\beta\beta\end{pmatrix}$$​

$$=\bar\gamma\gamma(1-\bar\alpha v)\begin{pmatrix}1 & \frac{\alpha+\bar\alpha\beta}{1-\bar\alpha v}\\ \frac{-\bar v-\bar\beta v}{1-\bar\alpha v} & \frac{-\bar v\alpha+\bar\beta\beta}{1-\bar\alpha v}\end{pmatrix}$$​

Now let's choose $\bar\Lambda=\Lambda^{-1}$. This choice reduces the left-hand side to the identity matrix, so we get

$$1=\tilde\gamma=\bar\gamma\gamma(1-\bar\alpha v) \implies \bar\gamma\gamma=\frac{1}{1-\bar\alpha v}$$

$$0=\tilde v\implies 0=-\bar v-\bar\beta v=v-\bar\beta v\implies\bar\beta=1$$

We got this from $\bar\Lambda\Lambda=1$, and since we must also have $\Lambda\bar\Lambda=1$, we can also conclude that $\beta=1$.

$$1=\tilde\gamma\tilde\beta=\bar\gamma\gamma(-\bar v\alpha+\bar\beta\beta)=\bar\gamma\gamma(-\bar v\alpha+1)\implies \bar\gamma\gamma=\frac{1}{1-\alpha\bar v}$$

Compare this to the other expression we got for $\bar\gamma\gamma$, and we see that $\bar\alpha v=\alpha\bar v$.

$$0=\tilde\alpha\implies 0=\alpha+\bar\alpha\beta=\alpha+\bar\alpha\implies\bar\alpha=-\alpha$$

These results tell us that $\Lambda$ and its inverse can be expressed as in #12. Now #13 tells us how to find $\gamma$ explicitly, and the results $\bar\alpha=-\alpha$ and $\bar\alpha v=\alpha\bar v$ together imply that $\alpha=-Kv$. The argument from #16 now implies that K≥0.

So the final result is

$$\Lambda=\gamma(v)\begin{pmatrix}1 & -Kv \\ -v & 1\end{pmatrix}$$

where

$$\gamma(v)=\frac{1}{\sqrt{1-Kv^2}}$$

and K is a non-negative real number. The choice K=0 gives us the restricted Galilei group. K=1 gives us the restricted Lorentz group, and all other choices give us a group that's isomorphic to the restricted Lorentz group.

18. Mar 10, 2009

### Fredrik

Staff Emeritus
Note also that the 10 component of the matrix equation $$\tilde\Lambda=\bar\Lambda\Lambda$$ is the velocity addition rule, and that if we use the other results we obtained, we can express it in a form that looks a lot like what we're used to seeing:

$$\tilde v=-\frac{-\bar v-\bar\beta v}{1-\bar\alpha v}=\frac{\bar v+v}{1+K\bar v v}$$

Last edited: Mar 10, 2009
19. Mar 12, 2009

### Fredrik

Staff Emeritus
Adding a few more details for completeness...

The invariant velocity corresponding to a value of K is found by looking for simultaneous eigenvectors that of all the $\Lambda(v)$ (i.e. vectors that are eigenvectors of $\Lambda(v)$ no matter what v is). An eigenvalue $\lambda$ must satisfy

$$0=\begin{vmatrix}1-\lambda & -Kv\\ -v & 1-\lambda\end{vmatrix}=(1-\lambda)^2-Kv^2$$

So we get

$$\lambda=1\pm\sqrt Kv$$

If K=0, a simultaneous eigenvector (a,b) of all the $\Lambda(v)$ must satisfy

$$\begin{pmatrix}a\\ b\end{pmatrix}=\begin{pmatrix}1 & 0\\ -v & 1\end{pmatrix}\begin{pmatrix}a\\ b\end{pmatrix}=\begin{pmatrix}a\\ -va+b\end{pmatrix}$$

for all v. This is the case if and only if a=0, so the only simultaneous eigenvector is (0,1). This is what we expect from a Galilei transformation. The x axis is mapped to itself, and the only invariant velocity is infinite.

If K>0, a simultaneous eigenvector (a,b) of all the $\Lambda(v)$ must satisfy

$$(1\pm\sqrt Kv)\begin{pmatrix}a\\ b\end{pmatrix}=\begin{pmatrix}1 & -Kv\\ -v & 1\end{pmatrix}\begin{pmatrix}a\\ b\end{pmatrix}=\begin{pmatrix}a-Kvb\\ -va+b\end{pmatrix}$$

for all v. This is the case if and only if $b=\mp a/\sqrt K$. If we define $c=1/\sqrt K$, we can write that as $b=\mp ac$, and the simultaneous eigenvectors as $(1,\mp c)$. Note that straight lines through the origin in those directions represent motion with velocities $\mp c$.

Last edited: Mar 12, 2009
20. Mar 12, 2009

### Fredrik

Staff Emeritus
I claimed earlier that if we define $G_K$ to be the group of matrices of the form

$$\Lambda(v)=\frac{1}{\sqrt{1-Kv^2}}\begin{pmatrix}1 & -Kv\\ -v & 1\end{pmatrix}$$

(with K fixed and v variable), $G_K$ is isomorphic to $G_{K'}$ for all K,K'>0. I'm going to prove that, by showing that the function $f:G_K\rightarrow G_{K'}$ defined by

$$f(\Lambda(v))=\Lambda'\Big(\sqrt{\frac{K}{K'}}v\Big)$$

is a group isomorphism. The $\Lambda'$ that appears in the equation above is of course defined by

$$\Lambda'(v)=\frac{1}{\sqrt{1-K'v^2}}\begin{pmatrix}1 & -K'v\\ -v & 1\end{pmatrix}$$

Using these definitions, and the velocity addition law obtained earlier, we see that

$$f(\Lambda(\bar v)\Lambda(v))=f\Big(\Lambda\Big(\frac{\bar v+v}{1+K\bar v v}\Big)\Big)=\Lambda'\Big(\sqrt{\frac{K}{K'}}\frac{\bar v+v}{1+K\bar v v}\Big)=\Lambda'\Bigg(\frac{\sqrt\frac{K}{K'}\bar v+\sqrt\frac{K}{K'}v}{1+K'\Big(\sqrt\frac{K}{K'}\bar v\Big) \Big(\sqrt\frac{K}{K'}v\Big)}\Bigg)$$

$$=\Lambda'\Big(\sqrt\frac{K}{K'}\bar v\Big)\Lambda'\Big(\sqrt\frac{K}{K'}v\Big)=f(\Lambda(\bar v))f(\Lambda(v))$$

The reason why this doesn't also prove that $G_K$ with K>0 is isomorphic to $G_0$ (i.e. that the Lorentz group is isomorphic to the Galilei group) is of course that with K'=0, the calculation includes division by zero in a few places.

Last edited: Mar 12, 2009