I have the following process: two ingoing particles, a photon hitting a nucleus, and two outgoing particles, the nucleus and a pion. I have computed $|M|^2$ and the differential cross section in the center of mass frame $\frac{d \sigma}{d \Omega_{CM}}$; I now have to go into the lab frame, where the nucleus is initially at rest, and consider the limit of a infinite massive nucleus $M_N \to \infty$, and compute $\frac{d \sigma}{d \Omega_{lab}}$.

Is there a general procedure to go from the first to the second? I first wrote $\frac{d \sigma}{dt}$ and then multiplied it for a rather complicated expression that I found on a book to obtain $\frac{d \sigma}{d \Omega_{lab}}$. However, taking the infinite massive nucleus limit, the result I get is not what I'm supposed to.

  • $\begingroup$ Hmmm...as the target mass grows the CoM frame gets slower and slower with respect to the lab frame (and should come to a halt in the limit as $M_n \to \infty$), no? $\endgroup$ Apr 3, 2013 at 16:09
  • $\begingroup$ @dmckee so you are saying that just by taking the limit $M_n \to \infty$ in $\frac{d \sigma}{d \Omega_{cm}}$ I would get the right answer for $\frac{d \sigma}{d \Omega_{lab}}$? I didn't think about it. I don't know if it works but I'll try to do it tomorrow $\endgroup$
    – user22710
    Apr 3, 2013 at 16:29
  • $\begingroup$ Well, I haven't sat down and figured it for myself. You might also look to see if you can find a ratio like $M_N^\alpha v_{CM}^\beta$ that stays finite in the limit and write the whole thing in terms of that ratio. But try the simplest possible option first. $\endgroup$ Apr 3, 2013 at 17:02

1 Answer 1


A Lorentz boost can be used to go from the center of mass frame to the lab frame. Mandelstam variables are invariant under a boost. Here is the boost procedure for Compton scattering:

In the center of mass frame, let $p_1$ be the inbound photon, $p_2$ the inbound electron, $p_3$ the scattered photon, $p_4$ the scattered electron.

\begin{equation*} p_1=\begin{pmatrix}\omega\\0\\0\\ \omega\end{pmatrix} \qquad p_2=\begin{pmatrix}E\\0\\0\\-\omega\end{pmatrix} \qquad p_3=\begin{pmatrix} \omega\\ \omega\sin\theta\cos\phi\\ \omega\sin\theta\sin\phi\\ \omega\cos\theta \end{pmatrix} \qquad p_4=\begin{pmatrix} E\\ -\omega\sin\theta\cos\phi\\ -\omega\sin\theta\sin\phi\\ -\omega\cos\theta \end{pmatrix} \end{equation*}

where $E=\sqrt{\omega^2+m^2}$.

It is easy to show that

\begin{equation} \langle|\mathcal{M}|^2\rangle = \frac{e^4}{4} \left( \frac{f_{11}}{(s-m^2)^2} +\frac{f_{12}}{(s-m^2)(u-m^2)} +\frac{f_{12}^*}{(s-m^2)(u-m^2)} +\frac{f_{22}}{(u-m^2)^2} \right) \end{equation}


\begin{equation} \begin{aligned} f_{11}&=-8 s u + 24 s m^2 + 8 u m^2 + 8 m^4 \\ f_{12}&=8 s m^2 + 8 u m^2 + 16 m^4 \\ f_{22}&=-8 s u + 8 s m^2 + 24 u m^2 + 8 m^4 \end{aligned} \end{equation}

for the Mandelstam variables $s=(p_1+p_2)^2$, $t=(p_1-p_3)^2$, $u=(p_1-p_4)^2$.

Next, apply a Lorentz boost to go from the center of mass frame to the lab frame in which the electron is at rest.

\begin{equation*} \Lambda= \begin{pmatrix} E/m & 0 & 0 & \omega/m\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ \omega/m & 0 & 0 & E/m \end{pmatrix}, \qquad \Lambda p_2=\begin{pmatrix}m \\ 0 \\ 0 \\ 0\end{pmatrix} \end{equation*}

The Mandelstam variables are invariant under a boost. \begin{equation} \begin{aligned} s&=(p_1+p_2)^2=(\Lambda p_1+\Lambda p_2)^2 \\ t&=(p_1-p_3)^2=(\Lambda p_1-\Lambda p_3)^2 \\ u&=(p_1-p_4)^2=(\Lambda p_1-\Lambda p_4)^2 \end{aligned} \end{equation}

In the lab frame, let $\omega_L$ be the angular frequency of the incident photon and let $\omega_L'$ be the angular frequency of the scattered photon. \begin{equation} \begin{aligned} \omega_L&=\Lambda p_1\cdot(1,0,0,0)=\frac{\omega^2}{m}+\frac{\omega E}{m} \\ \omega_L'&=\Lambda p_3\cdot(1,0,0,0)=\frac{\omega^2\cos\theta}{m}+\frac{\omega E}{m} \end{aligned} \end{equation}

It follows that \begin{equation} \begin{aligned} s&=(p_1+p_2)^2=2m\omega_L+m^2 \\ t&=(p_1-p_3)^2=2m(\omega_L' - \omega_L) \\ u&=(p_1-p_4)^2=-2 m \omega_L' + m^2 \end{aligned} \end{equation}

Compute $\langle|\mathcal{M}|^2\rangle$ from $s$, $t$, and $u$ that involve $\omega_L$ and $\omega_L'$. \begin{equation*} \langle|\mathcal{M}|^2\rangle= 2e^4\left( \frac{\omega_L}{\omega_L'}+\frac{\omega_L'}{\omega_L} +\left(\frac{m}{\omega_L}-\frac{m}{\omega_L'}+1\right)^2-1 \right) \end{equation*}

From the Compton formula \begin{equation*} \frac{1}{\omega_L'}-\frac{1}{\omega_L}=\frac{1-\cos\theta_L}{m} \end{equation*}

we have \begin{equation*} \cos\theta_L=\frac{m}{\omega_L}-\frac{m}{\omega_L'}+1 \end{equation*}

Hence \begin{equation*} \langle|\mathcal{M}|^2\rangle= 2e^4\left( \frac{\omega_L}{\omega_L'}+\frac{\omega_L'}{\omega_L}+\cos^2\theta_L-1 \right) \end{equation*}

The differential cross section for Compton scattering is \begin{equation*} \frac{d\sigma}{d\Omega}\propto \left(\frac{\omega_L'}{\omega_L}\right)^2\langle|\mathcal{M}|^2\rangle \end{equation*}


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