Question

3. A particle of rest mass m moving in the a direction at a speed of c/3 abruptly decays electromagnetically, yielding two photons. From the perspective of the home frame, the photon moving in the positive r direction is more energetic than the photon moving in the negative r direction - (a) Determine the energies and frequencies of both photons in the rest frame of the decaying particle. -(b) Using Lorentz transformations, determine the energies and frequencies of both photons emitted in the decay in the home frame. Show that the sum of the energies of the two photons as determined in the home reference frame is γmc2, where γ is calculated or f-1/3. In terms of energy conservation, why is this an expected result?

Answer 1

I'll use c=1 units through out.

Using 4-vector notation, the conservation of 4-momentum is written as

$(m,0,0,0)=(E,-E,0,0)+(E,E,0,0)$

$E=m/2$

So the energy of both the photons is m/2, and the frequency is

$\nu=\frac{m}{2h}$

Summation convention

$p^{\mu}=\Lambda_{\mu'}^{\mu} p^{\mu'}$

So for the photon moving in +x direction

$\begin{bmatrix} \gamma & \beta\gamma & 0 & 0\\ \beta\gamma & \gamma & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 &1 \end{bmatrix} \begin{bmatrix} E\\ E\\ 0\\ 0 \end{bmatrix}$

$p_{+}=E\gamma(1+\beta,1+\beta,0,0)$

Thus, its energy is

$E_{+}=E\gamma(1+\beta)=\sqrt{2}m$

on inserting the values given for speed=1/3.

Similarly for -x

$\begin{bmatrix} \gamma & \beta\gamma & 0 & 0\\ \beta\gamma & \gamma & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 &1 \end{bmatrix} \begin{bmatrix} E\\ -E\\ 0\\ 0 \end{bmatrix}$

$p_{-}=E\gamma(1-\beta,-1+\beta,0,0)$

$E_{-}=\frac{m}{\sqrt{2}}$

Total energy of photons

$E_H=2E\gamma=\gamma m$

This is expected because, a moving mass increases by a factor of $\gamma$ and hence the energy. SO the total energy has to be $\gamma m$