Calculate a specific boost and rotation

[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!

 

[PeroK]

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 $z$?

 

[PeroK]

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 $\varphi = \arctan( \gamma v_y)$

Can I consider this second point correct? Or I miss something about this rotation?

 

[PeroK]

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.

 

[PeroK]

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 already begin by thanking you for the help you are giving me.

 

[PeroK]

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 already begin by thanking you for the help you are giving me.

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?

 

[PeroK]

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 $\gamma$.

 

[PeroK]

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 $(-1, 1, 1, 1)$.

 

[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 ($R^TR≠\mathbb{1}$)?

 

[PeroK]

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!

 

[PeroK]

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:
$$\gamma = \cosh \phi, \ \gamma v = \sinh \phi \ \ \text{and} \ \ v = \tanh \phi$$