Question

a) A photon with energy Eph in a reference frame S moves at an angle 0 with respect to the y axis, in the x-y plane (i.e., its propagation vector is  = (sin 0, cos 0, 0)). Write out the photon's momentum four-vector pph in frame S. b) Write out the 4x4 matrix that describes a Lorentz boost into a frame (S') that moves with velocity v (corresponding to Lorentz factor v) in the y direction of frame S. c) Use the transformation matrix from part b) to carry out the Lorentz transforma- tion of the photon four momentum pon from frame S into frame S' (+ p'ph). [6]

Answer 1

The 4 momentum components of the photon are

$p^t_{ph}=E_{ph}/c,\quad p^x_{ph}=E_{ph}/c \times (\hat k_x)=E_{ph}\sin\theta /c,\\ \\ p^y_{ph}=E_{ph}/c \times (\hat k_y)=E_{ph}\cos\theta /c,\quad p^z_{ph}=E_{ph}/c \times (\hat k_z)=0$

In cloumn vector notation

$p^\mu_{ph}=\begin{pmatrix} p^t_{ph}\\ p^x_{ph}\\ p^y_{ph}\\ p^z_{ph} \end{pmatrix}=\begin{pmatrix} E_{ph}/c\\ E_{ph}\sin\theta /c\\ E_{ph}\cos\theta /c\\ 0 \end{pmatrix}$

b) The Lorentz transformation matrix for transformation along Y axis

$\Lambda^{\mu'}_{\,\,\,\,\mu}=\begin{pmatrix} \gamma & 0 & -v\gamma/c & 0 \\ 0 & 1 & 0 & 0 \\ -v\gamma/c & 0 & \gamma & 0 \\ 0 &0 & 0 & 1 \end{pmatrix}$

c) Therefore the momentum in S' frame is

$p^{\mu'}_{ph}=\Lambda^{\mu'}_{\,\,\,\,\mu}p^\mu_{ph}=\begin{pmatrix} \gamma & 0 & -v\gamma/c & 0 \\ 0 & 1 & 0 & 0 \\ -v\gamma/c & 0 & \gamma & 0 \\ 0 &0 & 0 & 1 \end{pmatrix} \begin{pmatrix} E_{ph}/c\\ E_{ph}\sin\theta /c\\ E_{ph}\cos\theta /c\\ 0 \end{pmatrix}=\left( \begin{array}{c} \gamma E_{ph}/c -\gamma v E_{ph}\cos\theta/c^2 \\ E_{ph}\sin\theta/c \\ \gamma E_{ph}\cos\theta/c-\gamma E_{ph}v/c^2 \\ 0 \\ \end{array} \right)$