# Calculate a specific boost and rotation

• Frostman

#### Frostman

Homework Statement
In an inertial reference system S, we have a photon with spatial components only along the x axis, its quadrimoment is ##p = E (1, 1, 0, 0)##. There is a special class of Lorentz transformations, called "small group of ##p##", which leaves the components of ##p## unchanged. For example, any rotation around the x axis belongs to this class.
a) Calculate a sequence of boosts and rotations whose product belongs to the small group of p, but which is not a simple rotation around the x axis. To do this, follow the suggestions below.
I) Apply a generic boost in the y-z plane (take advantage of the arbitrariness in deciding the alignment of the y and z axes).
II) Make an appropriate rotation that realigns the photon moment along the abscissa axis.
III) Determine a new boost that brings the photon back to having the initial components.
b) Write explicitly the Lorentz transformation found and verify that it is not a pure rotation.
Relevant Equations
\begin{equation*}
B =
\begin{pmatrix}
\gamma & -\gamma v_x & -\gamma v_y & -\gamma v_z \\
-\gamma v_x & \frac{v_x^2(\gamma - 1)}{|v|^2}+1 & \frac{v_xv_y(\gamma -1)}{|v|^2} & \frac{v_xv_z(\gamma -1)}{|v|^2} \\
-\gamma v_y & \frac{v_xv_y(\gamma -1)}{|v|^2} & \frac{v_y^2(\gamma - 1)}{|v|^2}+1 & \frac{v_yv_z(\gamma -1)}{|v|^2}\\
-\gamma v_z & \frac{v_xv_z(\gamma -1)}{|v|^2} & \frac{v_yv_z(\gamma -1)}{|v|^2} & \frac{v_z^2(\gamma - 1)}{|v|^2}+1
\end{pmatrix}
\end{equation*}
Let's begin with the first point.
a.I) Apply a generic boost in the y-z plane (take advantage of the arbitrariness in deciding the alignment of the y and z axes).
\begin{equation*}
B_{yz} =
\begin{pmatrix}
\gamma & 0 & -\gamma v_y & -\gamma v_z \\
0 & 1 & 0 & 0 \\
-\gamma v_y & 0 & \frac{v_y^2(\gamma - 1)}{|v|^2}+1 & \frac{v_yv_z(\gamma -1)}{|v|^2}\\
-\gamma v_z & 0 & \frac{v_yv_z(\gamma -1)}{|v|^2} & \frac{v_z^2(\gamma - 1)}{|v|^2}+1
\end{pmatrix}
\end{equation*}

I thought that for a generic boost in the yz plane we can consider a boost along the velocity vector ##v=(0, v_y, v_z)##
Applying to our moment vector we have the following result
$$B_{yz} p = E \begin{pmatrix} \gamma \\ 1\\ -\gamma v_y \\ -\gamma v_z \end{pmatrix}$$

a.II) Make an appropriate rotation that realigns the photon moment along the abscissa axis.
Now to realign the moment to the x-axis I have to reset the components ## p_y ## and ## p_z ## with a rotation. But with ##R_x## or ##R_y## or ##R_z## I can't find a right rotation that can make ## p_y = p_z = 0##. How I can find this rotation? Thanks for the help!

Look at the hint about the "arbitrariness" of the boost in the y-z plane. This means you could take the boost to be along the y-axis. That may simplify things.

• Frostman
So I can take a ##B_y## or a ##B_z## satisfying the request that the boost take place in the yz plane? Instead to take a combination of boost along ##y## and ##z##?

So I can take a ##B_y## or a ##B_z## satisfying the request that the boost take place in the yz plane? Instead to take a combination of boost along ##y## and ##z##?
Yes, whatever direction you choose in the y-z plane, I can take that to be my y-axis. We then have the same physical scenario, but my equations will be simpler.

• Frostman
Yes, whatever direction you choose in the y-z plane, I can take that to be my y-axis. We then have the same physical scenario, but my equations will be simpler.
For the first point I can use this boost.

\begin{equation*}
B_{y} =
\begin{pmatrix}
\gamma & 0 & -\gamma v_y & 0 \\
0 & 1 & 0 & 0 \\
-\gamma v_y & 0 & \frac{v_y^2(\gamma - 1)}{|v|^2}+1 & 0 \\
0 & 0 & 0 & 1
\end{pmatrix}
\end{equation*}

Which gives me this new quadrimoment ##p=E(\gamma, 1, -\gamma v_y, 0)##.
Now I have to do a rotation to reset ##p_y = p_z = 0## and I can use the only rotation along z-axis, because

\begin{equation*}
R_{x}(\varphi)p =
\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & \cos\varphi & -\sin\varphi \\
0 & 0 & \sin\varphi & \cos\varphi \\
\end{pmatrix}
p=E
\begin{pmatrix}
\gamma \\
1 \\
-\gamma v_y \cos \varphi \\
-\gamma v_y \sin \varphi \\
\end{pmatrix}
\end{equation*}

But ##\cos## and ##\sin## are never ##0## at the same angle.

\begin{equation*}
R_{y}(\varphi)p =
\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & \cos\varphi & 0 & \sin\varphi \\
0 & 0 & 1 & 0 \\
0 & -\sin\varphi & 0 & \cos\varphi \\
\end{pmatrix}
p=E
\begin{pmatrix}
\gamma \\
\cos \varphi \\
-\gamma v_y \\
- \sin \varphi \\
\end{pmatrix}
\end{equation*}

By hypothesis ##-\gamma v_y ≠ 0 ##

In the end we have

\begin{equation*}
R_{z}(\varphi)p =
\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & \cos\varphi & -\sin\varphi & 0 \\
0 & \sin\varphi & \cos\varphi & 0 \\
0 & 0 & 0 & 1 \\
\end{pmatrix}
p=E
\begin{pmatrix}
\gamma \\
\cos \varphi + \gamma v_y \sin \varphi \\
\sin \varphi - \gamma v_y \cos \varphi \\
0 \\
\end{pmatrix}
\end{equation*}

##p_y## is ##0## for ##\sin \varphi - \gamma v_y \cos \varphi = 0## which is true for ##\varphi = \arctan( \gamma v_y)##

• etotheipi
Looks good so far.

• Frostman
Looks good so far.

Calling ## \varphi_z = \arctan( \gamma v_y ) ## and ##\gamma_1## the first boost. We have the last and new quadrimoment:

##p'' = E(\gamma_1, \cos \varphi_z + \gamma_1 v_y \sin \varphi_z, 0, 0) ##
Now let's make the second boost along x-axis:

\begin{equation*}
B_{x} =
\begin{pmatrix}
\gamma_2 & -\gamma_2 v_x & 0 & 0 \\
-\gamma_2 v_x & \frac{v_x^2(\gamma - 1)}{|v|^2}+1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{pmatrix}
\end{equation*}

\begin{equation*}
B_{x}p'' = E
\begin{pmatrix}
\gamma_1 \gamma_2 - \gamma_2 v_x (\cos \varphi_z + \gamma_1 v_y \sin \varphi_z) \\
-\gamma_1 \gamma_2 v_x + (\frac{v_x^2(\gamma_2 - 1)}{|v|^2}+1)(\cos \varphi_z + \gamma_1 v_y \sin \varphi_z) \\
0 \\
0 \\
\end{pmatrix}
= E
\begin{pmatrix}
1\\
1\\
0\\
0\\
\end{pmatrix}
\end{equation*}

I have to solve this 2 equations in 1 variable (##v_x##)
\begin{cases}
\gamma_1 \gamma_2 - \gamma_2 v_x \cos \varphi_z - \gamma_2 v_x \gamma_1 v_y \sin \varphi_z) = 1 \\
-\gamma_1 \gamma_2 v_x + \gamma_2 \cos \varphi_z + \gamma_2 \gamma_1 v_y \sin \varphi_z = 1
\end{cases}

Now I have to remember that
##\gamma_1 = \frac{1}{\sqrt{1-v_y^2}} ##
##\gamma_2 = \frac{1}{\sqrt{1-v_x^2}} ##
##\cos(\arctan(\gamma_1 v_y)) = \frac{1}{\sqrt{1+\gamma_1^2 v_y^2}} = \sqrt{1-v_y^2} = \gamma_1^{-1}##
##\sin(\arctan(\gamma_1 v_y)) = \frac{\gamma_1 v_y}{\sqrt{1+\gamma_1^2 v_y^2}}=v_y ##

And simplify these equations.
\begin{cases}
\gamma_1 \gamma_2 (1-v_x)=1 \\
\gamma_1 \gamma_2 (1-v_x)=1 \\
\end{cases}

\begin{cases}
\frac{ 1 }{ \sqrt{1-v_y^2}} \frac{1}{\sqrt{1-v_x^2}} (1-v_x)=1 \\
\frac{ 1 }{ \sqrt{1-v_y^2}} \frac{1}{\sqrt{1-v_x^2}} (1-v_x)=1 \\
\end{cases}

Solving them I find out:

$$v_{x_{1,2}} = \frac{1 \pm \sqrt{1 - (2-v_y^2) v_y^2}}{ 2 - v_y^2}$$

That gives me a condition another condition for ##v_y##: ## v_y < -1 \vee v_y > 1 ##
Can you confirm?

And this gives me the sequence of the Lorentz's transformation that I should apply:
$$B_xR_z(\varphi_z)B_yp=p$$

So, if is it all correct ##B_xR_z(\varphi_z)B_y=\mathbb{1}##.
It seemed like a very long exercise full of calculus. Isn't it that I did something wrong or is there a much faster way? It is an exam topic and it is 1 of the 4 exercises to do.

If ##v## is the first boost in the y direction, with gamma factor ##\gamma##, then try ##u = \frac{\gamma^2 -1}{\gamma^2 + 1}## for the boost in the x-direction.

There are a number of ways to simplify what you've done. If I have time, I'll post this later.

• Frostman
I don't understand why did you choose this boost ##u = \frac{\gamma^2-1}{\gamma^2+1}##.
I'm interested and I want to try understand on this ways to simplify my solution.

I don't understand why did you choose this boost ##u = \frac{\gamma^2-1}{\gamma^2+1}##.
I'm interested and I want to try understand on this ways to simplify my solution.
After the second transformation, the problem reduces to one spatial dimension. You want:
$$\begin{pmatrix} \gamma_u &-\gamma_u u \\ -\gamma_u u & \gamma_u \\ \end{pmatrix} \begin{pmatrix} \gamma \\ \gamma \\ \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ \end{pmatrix}$$
Solving that using ##\gamma_u^2 = \frac 1 {1-u^2}## gives you that equation for ##u## in terms of ##\gamma##.

• Frostman
Oh... I should have notice this before...

##\cos(\arctan(\gamma_1 v_y)) = \frac{1}{\sqrt{1+\gamma_1^2 v_y^2}} = \sqrt{1-v_y^2} = \gamma_1^{-1}##
##\sin(\arctan(\gamma_1 v_y)) = \frac{\gamma_1 v_y}{\sqrt{1+\gamma_1^2 v_y^2}}=v_y ##

With this last step is the most simplified solution or did I miss something again?

Oh... I should have notice this before...

##\cos(\arctan(\gamma_1 v_y)) = \frac{1}{\sqrt{1+\gamma_1^2 v_y^2}} = \sqrt{1-v_y^2} = \gamma_1^{-1}##
##\sin(\arctan(\gamma_1 v_y)) = \frac{\gamma_1 v_y}{\sqrt{1+\gamma_1^2 v_y^2}}=v_y ##

With this last step is the most simplified solution or did I miss something again?
That's messy. Just let ##\cos \theta = \frac 1 \gamma## and ##\sin \theta = -v##.

In general, you don't need to solve all these intermediate equations. You can simply use the relationship between variables.

Also, using both ##v## and ##\gamma## is useful. You can express some things in terms of ##v## and some in terms of ##\gamma##.

• Frostman
Here's a simple approach. First, we have a boost in the y-direction:
$$B_y(v) = \begin{pmatrix} \gamma & 0 & -\gamma v & 0 \\ 0 & 1 & 0 & 0 \\ -\gamma v & 0 & \gamma & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix}$$
Then, we have a rotation about the z-axis:
$$R_z(\theta) = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & \cos \theta & -\sin \theta & 0 \\ 0 & \sin \theta & \cos \theta & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix}$$
And, if we have ##\sin \theta = v## and ##\cos \theta = \frac 1 \gamma## (check that this is valid), then we get:
$$R_z(\theta)B_y(v) \begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \\ \end{pmatrix} = \begin{pmatrix} \gamma \\ \gamma \\ 0 \\ 0 \\ \end{pmatrix}$$
Then we have a boost in the x-direction, with ##u = \frac{\gamma^2 -1}{\gamma^2 + 1}##.

I got a simplified form for the final Lorentz transformation:
$$L = B_x(u)R_z(\theta)B_y(v) = \begin{pmatrix} 1 + \frac{v^2}{2} & -\frac{v^2}{2} & -v & 0 \\ \frac{v^2}{2} & 1 - \frac{v^2}{2} & -v & 0 \\ -v & v & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix}$$
That gives you something to shoot for. You can see that it leaves the original vector components unchanged. You could also check that ##L## is a Lorentz Transformation:
$$L^{T}\eta L = \eta$$
Where ##\eta## is the diagonal matrix ##(1, -1, -1, -1)## or ##(-1, 1, 1, 1)##.

• etotheipi and Frostman
Really appreciate. To check if is a Lorentz Transformation I agree, I studied this demonstration. The equalities ##\sin \theta = v ## and ##\cos \theta = \frac{1}{\gamma}## come from the observation that I did before about the relation between ##\cos \theta## and ##\sin \theta## with ##\arctan \theta##? If I want to check that ##L## it's not a pure rotation I can use the fact that ##L^TL≠\mathbb{1}## like a pure rotation (##R^TR≠\mathbb{1}##)?

Really appreciate. To check if is a Lorentz Transformation I agree, I studied this demonstration. The equalities ##\sin \theta = v ## and ##\cos \theta = \frac{1}{\gamma}## come from the observation that I did before about the relation between ##\cos \theta## and ##\sin \theta## with ##\arctan \theta##? If I want to check that ##L## it's not a pure rotation I can use the fact that ##L^TL≠\mathbb{1}## like a pure rotation (##R^TR≠\mathbb{1}##)?
The matrix representing a rotation about the x-axis has a specific form. In particular for the time components. You can see "by inspection" that it's not a pure rotation, for ##v \ne 0##.

• Frostman
I understand, that's a good point of view without make any calculus.

Thank you so much!

• berkeman
The equalities ##\sin \theta = v ## and ##\cos \theta = \frac{1}{\gamma}## come from the observation that I did before about the relation between ##\cos \theta## and ##\sin \theta## with ##\arctan \theta##?
Interestingly:
$$\sin^2 \theta + \cos^2 \theta = v^2 + \frac 1 {\gamma^2} = v^2 + (1 - v^2) = 1$$
You can compare that with the rapidity ##\phi##, where:
$$\gamma = \cosh \phi, \ \gamma v = \sinh \phi \ \ \text{and} \ \ v = \tanh \phi$$

• Frostman