Example 2: Photon emission

If the particle 2 is photon, $T_2 = p_2 = E_2 = E_{\gamma }$, then

$$\begin{eqnarray*}E_{\rm d} &=& \frac{F}{\sqrt{m_1^2 + F} + m_1} \\ F &=& 2 m_1... ...a } \gamma _1 \left( 1 - \beta _1 \cos \theta _{12} \right) , \end{eqnarray*}$$

where the recoil effect by emitting photon is included. If one neglect this, i.e. $F/m_1^2 \rightarrow 0$,

$$\begin{eqnarray*}E_{\rm d} &\rightarrow & E_{\gamma } \gamma _1 \left( 1 - \beta _1 \cos \theta _{12} \right) , \end{eqnarray*}$$

which is well-known formula for the Doppler shift.