# Composition of 2 transformations of E

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1. Apr 4, 2015

### Happiness

Suppose frame $S^\prime$ moves in the positive $x$ direction at $v$ with respect to frame $S$, and frame $S^"$ moves in the positive $y$ direction at $v^\prime$ with respect to frame $S^\prime$.

Then,

$E^\prime_x=E_x$
$E^\prime_y=\gamma(E_y - vB_z)$
$E^\prime_z=\gamma(E_z + vB_y)$

and

$E^{\prime\prime}_x=\gamma^\prime(E^\prime_x+v^\prime B^\prime_z)$
$E^{\prime\prime}_y=E^\prime_y$
$E^{\prime\prime}_z=\gamma^\prime(E^\prime_z-v^\prime B^\prime_x)$

Substituting the first set of equations into the second set, we have

$E^{\prime\prime}_x=\gamma^\prime(E_x+v^\prime B^\prime_z)$
$E^{\prime\prime}_y=\gamma(E_y-vB_z)$
$E^{\prime\prime}_z=\gamma^\prime[\gamma(E_z + vB_y)-v^\prime B^\prime_x]$

With respect to frame $S$, frame $S^"$ would not move in the positive $x$ direction but at angle $\theta$ anticlockwise from the positive $x$ direction at a velocity $v_0$.

Let $x_0$ be the axis pointing in the direction of $v_0$.

Then,

$E^"_{x_0}=E^"_x\cos\theta+E^"_y\sin\theta=\gamma^\prime\cos\theta\ (E_x+v^\prime B^\prime_z)+\gamma\sin\theta\ (E_y-vB_z)$

But

$E_{x_0}=E_x\cos\theta+E_y\sin\theta\neq E^"_{x_0}$

2. Apr 4, 2015

### Staff: Mentor

Actually, a pair of boosts is not just a boost, but a boost and a rotation. This may be causing the problem that you are noticing here. The transformations that you have written from S to S' and from S' to S'' are only valid for boosts, and not for rotations. So since the transformation from S to S'' involves a boost and a rotation you would not expect it to remain valid.

3. Apr 4, 2015

### Happiness

Thanks a lot! Really appreciate your answer! Could you recommend some reading materials or textbooks where this is explained in greater details, eg., containing the transformation equations that are valid for both boosts and rotations?

4. Apr 4, 2015

### Staff: Mentor

This is called Thomas precession or Wigner rotation (or some combination of those names and terms).

Here is one reference that seems to present the concepts at various levels:
http://arxiv.org/abs/1102.2001

In particular, it contains a boost matrix formulation that is valid for both boosts and rotations in the "intermediate" section.

You may also want to get started with the relevant Wikipedia information:
http://en.wikipedia.org/wiki/Lorentz_transformation#Composition_of_two_boosts
http://en.wikipedia.org/wiki/Lorentz_group#Restricted_Lorentz_group

5. Apr 7, 2015

### Happiness

Hi DaleSpam, thanks a lot for your recommendations!

Regarding http://arxiv.org/abs/1102.2001, page 9, equation (6),

||$v_{12}\times v_{21}$|| = ||$(v_1+\frac{v_2}{\gamma_1})\times (v_2+\frac{v_1}{\gamma_2})$|| = ||$(v_1\times v_2)+(\frac{1}{\gamma_1\gamma_2}v_2\times v_1)$|| = $(1-\frac{1}{\gamma_1\gamma_2})$ ||$v_1\times v_2$|| = $v_1v_2(1-\frac{1}{\gamma_1\gamma_2})$

The last equality holds only if $v_1$ and $v_2$ are perpendicular. But they are not perpendicular in the mission control's frame, though they are in Alice's frame. Since the Wigner rotation angle $\Omega$ is the angle in the mission control's frame, I would think that ||$v_1\times v_2$|| should be calculated based on the angle between them in the mission control's frame rather than that in Alice's frame.

Also, is there a simple, intuitive reason or explanation for the Wigner rotation angle?

Last edited: Apr 7, 2015
6. Apr 7, 2015

### Staff: Mentor

Yes, I agree that this is confusing, but this type of thing is actually pretty common in relativity. You usually want to be able to determine what something is like in one frame, given a description in a different frame. So here the question is, given that the velocities are perpendicular in Alice's frame, how does the direction transform to mission control's frame.

Not that I know of. To me, it remains one of the weirder things about special relativity.

7. Apr 7, 2015

### Staff: Mentor

It depends on what you consider "simple" and "intuitive". The underlying reason is that Lorentz boosts in different directions do not commute; that is, the result you get depends on the order in which you do the boosts. So, for example, a boost in the $x$ direction, then a boost in the $y$ direction, does not give you the same result as a boost in the $y$ direction, then a boost in the $x$ direction. The Wigner rotation angle is related to the commutator of the two boosts, i.e., the difference between doing them in one order vs. the other.

(The non-commutativity of Lorentz boosts is similar to the non-commutativity of ordinary rotations in 3-space about different axes; for example, a rotation about the $x$ axis, then a rotation about the $y$ axis, gives a different result than a rotation about the $y$ axis, then a rotation about the $x$ axis. That is easier to visualize--try it with a 6-sided die, for example. It works the same with boosts except that it's hyperbolic rotations and the "axis" points partly in the time direction--or, alternately, you can think of boosts as hyperbolic rotations in the $t-x$, $t-y$, or $t-z$ planes, while ordinary rotations are in the $x-y$, $x-z$, or $y-z$ planes.)

8. Apr 7, 2015

### Happiness

Hi DaleSpam, thanks for your reply! I figured out why I was so confused. The $v_2$ in the diagram of page 8 exists in Alice's frame and so is $v_2$ j', where j' is the unit vector pointing in the y' direction; while those $v_2$ in page 9 actually meant $v_2$ j, where $v_2$ is just a scalar for which no transformation is needed in going from one frame to another.

9. Apr 8, 2015

### Happiness

Hi PeterDonis, thanks for your answer! I would consider your explanation more "technical" or "mathematical". With that explanation, one would continue to ask why Lorentz boosts do not commute. The explanation about the non-commutativity of rotation is all well and good. But perhaps what I was looking for would be something that links Wigner rotation angle to the relativity of simultaneity or length contraction or the conservation of angular momentum or something of that sort or something we can relate to easily. Or maybe there just isn't such a connection.

10. Apr 8, 2015

### Staff: Mentor

Do you understand why ordinary spatial rotations do not commute? If you do, boosts are just the spacetime version of the same thing. If you don't, then it might help to work through that case first. Again, I recommend actually trying it with something like a 6-sided die.

The paper DaleSpam linked to gives another way of looking at it: it's the angle between the two "velocity sum" vectors you get when you add a pair of non-collinear velocity vectors, using the relativistic formula, in opposite order.

11. Apr 9, 2015

### Happiness

And why don't ordinary spatial rotations commute? I rotated an object and verified that, indeed, rotations don't commute. I'm 100% convinced that they don't commute. But still, that conviction doesn't stop me from asking why? I don't see any explanation other than "this is just maths" or "this is just the way the world is".

A more intuitive explanation to me would be as follows: the composition of two Lorentz boosts is not exactly analogous to the addition of two vectors (using the parallelogram law). It is the misconception that it is so that leads to the counter-intuitive feeling we get from the Wigner rotation angle or the composition of Lorentz boosts. Composing "going right" and then "going up" is not the same as composing "going up" and then "going right"; in the former case, the time dilation is "in the rightward direction" and there is a "double (or composed) time dilation" "in the upward direction", and this is different in the latter case. The two cases are not identical; they are only mirror images along the "diagonal". It is the equating (or identification) of a symmetry along the "diagonal" with "identity" that causes the problem.

To me, that's just a theorem rather than an explanation; it's just a different perspective, if you like.