# General Lorenz transformation is not group

1. Sep 13, 2009

### archipatelin

Is corretct, that a general lorenz transformation don't satisfaction axioms of group structure?

Let, GL(A,B) is general Lorenz transformacion from a frame A to frame B, which B is moveing a velocity V with respect to A.
And GL(B,C) is likewise G. L. from B to C. A frame C has a valocity U with respect to B.
Further, Letter W is velocity U observe in frame A.

BUT. Transformation GL(A,C) from frame A to C, where C is moveing just velocity W (to A) is NOT identity with rolling transformacions GL(A,B) (V) "+" GL(B,C)(U)!

I know. This is centrality for new relativistic effect: Thomas precession.

But, I cannot accept, that operation composing of General Lorenez transformations don't make again General Lorenz transformation (a set is not close for this operation).

How, do you explain this?
thx

2. Sep 13, 2009

### JesseM

Not true, GL(A,C) should give the same answer as taking the transformation GL(A,B) and then applying the transformation GL(B,C) to the result. I'm not sure if when you write "general Lorentz transformation" you want to consider arbitrary rotations, but as long as we assume the x-axes of each frame are parallel, then it's not too hard to show this with a little algebra, you just have to keep in mind that according to relativistic velocity addition, the velocity W of C relative to A should be equal to (V + U)/(1 + V*U/c^2).

3. Sep 13, 2009

### archipatelin

NO! This is correct for SPECIAL Lorenz transformation (where velocities are in direct axes X).
BUT I think GENERAL Lorenz transformation, where velicities are general vecor V=(Vx,Vy,Vz). There is this problem for me.

4. Sep 13, 2009

### JesseM

The math would be more complicated for Lorentz transformation that includes rotations...maybe you can show your calculations so others could check them? The Lorentz group does allow for rotations, so if your calculations seem to show it's not a group you must be making an error somewhere.

5. Sep 13, 2009

### archipatelin

Concretely.
GL(A,B): frame B is moveing velocity V=(Vx,0,0) - this is in direct axys X of frame A velocity Vx.
GL(B,C): frame C is movieng velocity U=(0,0,Uz) - in direct axys Z of B (speed Uz)

Velocity W (U with respect A) such is W=(Vx,0,Uz/sqrt(1-Vx^2/c^2)).

Insert it (W) to general form lorentz transformation http://en.wikipedia.org/wiki/Lorentz_transformation" [Broken]
obtained GL(A,C). But this transformation isn't as apply GL(B,C) and than GL(A,B).
This frames are differs in a rotating.

Last edited by a moderator: May 4, 2017
6. Sep 13, 2009

### JesseM

How'd you get that answer for the velocity of C relative to A?

I don't know the general formula for velocity addition with rotated frames, but if the origin of C is moving straight up along B's z'-axis, then at time t' in the B frame, the origin of C is at position x'=0, y'=0, z'=t'*Uz

And if an event happens at x', y', z', t' in B's frame, then in A's frame this should be:

x = gamma*(x' + Vx*t')
y = y'
z = z'
t = gamma*(t' + Vx*x'/c^2)

with gamma = 1/sqrt(1 - Vx^2/c^2)

So the event above on the worldline of C's origin should have the following coordinates in A's frame:

x = gamma*(Vx*t')
y = 0
z = t'*Uz
t = gamma*(t')

Since the origin of C was at x=y=z=0 at time t=0 in A's frame, the x-component of the velocity of the origin of C must be (distance along x-axis)/(time) = gamma*(Vx*t')/gamma*t' = Vx in A's frame, which agrees with what you wrote, while the z-component of the velocity must be (distance along z-axis)/(time) = t'*Uz/gamma*t' = Uz/gamma = Uz*sqrt(1 - Vx^2/c^2), which is different than what you wrote. So I say the velocity W of C in A's frame should be W=(Vx, 0, Uz*sqrt(1-Vx^2/c^2)). Do you disagree?

Last edited: Sep 13, 2009
7. Sep 13, 2009

### archipatelin

Yes. You have right W=(Vx,0,Uz*sqrt(1-Vx^2/c^2). But it was only typing error :)
However, this problem is still here.

8. Sep 13, 2009

### Staff: Mentor

The "general Lorentz transform" is a group, it is the http://en.wikipedia.org/wiki/Poincar%C3%A9_group" [Broken] and essentially consists of translations, rotations, and boosts.

Last edited by a moderator: May 4, 2017
9. Sep 13, 2009

### George Jones

Staff Emeritus
archipatelin, you seem only to be considering boosts, which are Lorentz transformations, but which aren't the most general Lorentz transformations. A general Lorentz transformation can always be expressed as the product of a boost and (as JesseM noted) a rotation The set of all general Lorentz forms a group, but the subset of all boosts does not form a subgroup of the Lorentz group. A product of two non-colinear boosts involves a rotation that is often called a Wigner rotation. In other words,

B_1 B_2 = BR.

This does satisfy closure of the Lorentz group, since rotations are also Lorentz transformations.

Also, be careful with notation: GL is often used to denote the general linear group.
Actually, the Lorentz group, which is generated by boosts and rotations (no translations) is a proper subgroup of the Poincare group.

Last edited by a moderator: May 4, 2017
10. Sep 13, 2009

### archipatelin

Yes, I know general lorentz group is Poincare group. But I think this: composition two "general" lorentz transformations (=lorentz transformation for arbitrary vector of valocity) are not "general" lorentz transformation (lorentz transformation for transformated valocity vector). But it is composition of this and specific rotations. Existing this nontrivial rotation is problem for me.

Last edited by a moderator: May 4, 2017
11. Sep 13, 2009

### Staff: Mentor

Good point. I always jump right to the Poincare group since the laws of physics seem to be translation invariant also, but you are of course correct.

12. Sep 13, 2009

### Staff: Mentor

As both JesseM and George Jones have pointed out, that is well known and is why the Lorentz group includes boosts and rotations. You are correct, the boosts do not form a group. That is well known. The boosts and rotations form a group (Lorentz group) as do the boosts, rotations, and translations (Poincare group).

13. Sep 13, 2009

### archipatelin

I agree, boost with rotation is group structure.My problem is this: Im siting on frame A and observe friend who sit on frame B. This frame (B) is moveing general vector of valocity with respect to me (A). He something measure there (on B) and I mesure this thing here (on A). Transformation between our values is "general" lorentz transformations, true?
Let now exist third observer and his frame C. He also measure this same thing.
And I say he got a measured value, but this value is consisten with his frame C and so do "general" Lorentz transformation to me frame A. But this value in my frame isn't identiy with what I measure (missing this rotation). This is my crucial problem.

14. Sep 13, 2009

### Staff: Mentor

Yes, because the boosts do not form a group. Again, this is well known!

The key point that I think you are missing is that "friend who sit on frame B" (which I assume means that your friend is at rest in reference frame B) does not uniquely define reference frame B. In fact, your friend is at rest in an infinite number of inertial reference frames. These reference frames are related to one another via translations and rotations. This is fundamentally why the Poincare group must include translations boosts and rotations and not just boosts.

15. Sep 13, 2009

### archipatelin

You say: if frame A and B are parallel and simultaneously B and C are parallel then A and C needn't be parallel (relation of be parallel isn't transitiv)?

16. Sep 13, 2009

### Staff: Mentor

That is correct. (Not what I said, but still correct)

17. Sep 13, 2009

### Fredrik

Staff Emeritus
Not sure if it will help, but the general velocity addition law is

$$\vec u\oplus\vec v=\frac{1}{1+\vec u\cdot\vec v}\bigg(\vec u+\vec v+\frac{\gamma_{\vec u}}{1+\gamma_{\vec u}}\vec u\times(\vec u\times\vec v)\bigg)$$

There is a way to express this that at least looks simpler. If we write a Lorentz transformation as

$$\Lambda(v)=\gamma\begin{pmatrix}1 & -v^T\beta \\ -v & \beta\end{pmatrix},\quad\gamma=\frac{1}{\sqrt{1-v^Tv}}, \quad\beta\beta^T=\frac{I}{\gamma^2}+vv^T$$

where v is a 3x1 matrix and $\beta$ is a 3x3 matrix, the velocity addition law can be expressed as

$$u\oplus v=\frac{u+\beta_u v}{1+u^Tv}$$

To see that these two are the same, you would have to use the fact that a pure boost has a $\beta$ that satisfies

$$\beta x=x_\parallel+\frac 1 \gamma x_\perp$$

for arbitrary 3x1 matrices x, where $x_\parallel$ is the projection of x onto the direction of v, and $x_\perp$ is the projection of x onto the plane that's perpendicular to v. We can use this to show that the $\beta$ of a pure boost is

$$\beta=\frac{I}{\gamma}+\left(1-\frac{1}{\gamma}\right)\frac{vv^T}{v^Tv}$$

and then use that to prove that the two forms of the velocity addition law are equivalent.

18. Sep 15, 2009

### archipatelin

Thank you everybody for help!
I see. I thought intuitively (corse wrongly), that they frames are mutually in parallel relation. For special lorentz transformation (movement is on shared axis X), where is contracting just axis X and Y,Z are no changes, relation of parallel is true. But for general direct of valocity (general lorentz transformation) is occurs a contraction of each axis (X,Y,Z) in direct of velocity.Therefore cannot speak about parallel, and exist a appended rotating tranformation is needed.
Thus to understand the new.

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