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- 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!

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!