# Double Lorentz transformation

1. Jan 13, 2014

### Libra82

1. The problem statement, all variables and given/known data
Question as stated: In special relativity consider the following coordinate transformation between inertial frames: first make a velocity boost $v_x$ in the x-direction, then make a velocity boost $v_y$ in the y-direction. 1) Is this a Lorentz transformation? 2) Find the matrix of this transformation. 3) Consider the boosts in inverse order - is it the same transformation?

2. Relevant equations
$\beta_i = \frac{v_i}{c}$
$\gamma_i = \frac{1}{\sqrt{1-\frac{v_i^2}{c^2}}}$

3. The attempt at a solution
I use the $c = 1$ convention.

I wrote down the two transformations as:

x-direction:
$\begin{pmatrix} t' \\ x'\\ y' \\ z' \end{pmatrix} = \begin{pmatrix} \gamma_x & -\beta_x \gamma_x & 0 & 0 \\ -\beta_x \gamma_x & \gamma_x & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} t\\ x\\ y\\ z \end{pmatrix}$
and for the y-direction:
$\begin{pmatrix} t'' \\ x''\\ y'' \\ z'' \end{pmatrix} = \begin{pmatrix} \gamma_y & 0 & -\beta_y \gamma_y & 0 \\ 0 & 1 & 0 & 0 \\ -\beta_y \gamma_y & 0 & \gamma_y & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} t' \\ x' \\ y' \\ z' \end{pmatrix}$
and combined these to get
$\begin{pmatrix} t'' \\ x''\\ y'' \\ z'' \end{pmatrix} = \begin{pmatrix} \gamma_y & 0 & -\beta_y \gamma_y & 0 \\ 0 & 1 & 0 & 0 \\ -\beta_y \gamma_y & 0 & \gamma_y & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} \gamma_x & -\beta_x \gamma_x & 0 & 0 \\ -\beta_x \gamma_x & \gamma_x & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} t\\ x\\ y\\ z \end{pmatrix} = \begin{pmatrix} \gamma_x \gamma_y & -\beta_x \gamma_x \gamma_y & -\beta_y \gamma_y & 0 \\ -\beta_x \gamma_x & \gamma_x & 0 & 0 \\ -\beta_y \gamma_x \gamma_y & \beta_x \beta_y \gamma_x \gamma_y & \gamma_y & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} t\\ x\\ y\\ z \end{pmatrix}$
This should be the answer to question 2).

If I inverse the order of the boosts I notice that the resulting transformation matrix is the transpose of the above matrix.

As the transformation matrix for the two cases in question are not equal the two transformations are not the same? (question 3)

I am uncertain on how to explain whether or not the resulting transformations are Lorentz transformations.

why do the question headlines automatically appear each time I preview my post?

2. Jan 14, 2014

### maajdl

First of all, I am uncomfortable with the vy part of the question.
What is vy?
Is that the velocity of the K'' frame origin with respect to the initial K frame?
Or is that the velocity of the K'' frame with respect to the K' frame?
Clarifying this may help.

The resulting transformations are each Lorentz transformations.
There are many ways to prove that, depending on which starting point is assumed.
If the Lorentz transform is defined as a linear transformation that keeps the ds² invariant, then the proof is obvious.

A more algebraic would try to indentify the resulting transformation as a boost along a certain resultant velocity.

Finally, note also the relation Lt(v) = L-1(-v) .
The transformations are different, but nevertheless closely related.

Last edited: Jan 14, 2014
3. Jan 14, 2014

### Libra82

I am, however, unsure of the interpretation of the vy part of the assignment. The exercise is written as in the OP. In my solution I have assumed vy was K'' with respect to K'.

Of course I could check whether or not ds² is invariant with respect to the double transformation! Thanks for the hint (and now I'm sad that I didn't figure this out for myself).

The final relation between the two is the same one I noticed while carrying out the calculations. Different transformations though as one matrix entry switches place.

4. Jan 14, 2014

### maajdl

You made a choice for the meaning of vy.
You simply need to keep that in mind.
Of course if you reverse the order of the boost, you need to switch the meanings too.

5. Jan 14, 2014

### maajdl

If you learned about velocity composition in SR, then you could make use of of this knowledge.