# Why differential scattering cross sectional area is same for Lab frame and centre of mass frame

Differential scattering cross section is denoted by:

$$\frac{{d\sigma}}{{d\Omega}}$$

Here, "$$d\sigma$$" represents the infinitesimal differential scattering cross-sectional area, which is not a physical area but a mathematical concept defined as:

$$d\sigma=\dfrac{Number\; of\; particles\; scattered\; per\; unit\; time\; into\; the\; solid\; angle\; d\Omega}{flux\; of\; incident\; particles}$$

and "$$d\Omega$$" represents the solid angle into which particles are scattered during a collision or scattering event.

In Laboratory frame and Centre of mass frame we write differential scattering cross sections as $$\frac{{d\sigma}}{{d\Omega_{lab}}}$$ and $$\frac{{d\sigma}}{{d\Omega_{cm}}}$$

I understand that $$d\Omega = sin\theta d\theta d\phi$$ must be different for both frames as scattering angles are different in both frames.

My question is that how come $$d\sigma$$ remains same in both frames.

As $$d\sigma$$ is about the number of particles scattered into some infinitesimal solid angle so it should be different according to the solid angles $${d\Omega_{lab}}$$ or $${d\Omega_{cm}}$$, into which particles will scatter in lab or cm frame.

Please help me understand why this quantity $$d\sigma$$ remains same in both frames.

It is true that $$\langle|\mathcal{M}|^2\rangle_{COM} =\langle|\mathcal{M}|^2\rangle_{LAB}$$ because Mandelstam variables are invariant. However, the angular dependence of $$d\sigma$$ does change in the lab frame.

For example, consider Compton scattering. For Compton scattering in the center of mass frame we have the following momentum vectors and spinors. \begin{align*} p_1&=\underset{\text{inbound \gamma}} {\begin{pmatrix}\omega\\0\\0\\ \omega\end{pmatrix}} \\[1ex] p_2&=\underset{\text{inbound e^-}} {\begin{pmatrix}E\\0\\0\\-\omega\end{pmatrix}} & u_{21}&=\underset{\substack{\text{inbound e^-}\\\text{spin up}}} {\begin{pmatrix}E+m\\0\\-\omega\\0\end{pmatrix}} & u_{22}&=\underset{\substack{\text{inbound e^-}\\\text{spin down}}} {\begin{pmatrix}0\\E+m\\0\\\omega\end{pmatrix}} \\[1ex] p_3&=\underset{\text{outbound \gamma}} {\begin{pmatrix}\omega\\\omega\sin\theta\cos\phi\\\omega\sin\theta\sin\phi\\\omega\cos\theta\end{pmatrix}} \\[1ex] p_4&=\underset{\text{outbound e^-}} {\begin{pmatrix}E\\-\omega\sin\theta\cos\phi\\-\omega\sin\theta\sin\phi\\-\omega\cos\theta\end{pmatrix}} & u_{41}&=\underset{\substack{\text{outbound e^-}\\\text{spin up}}} {\begin{pmatrix}E+m\\0\\p_{4z}\\p_{4x}+ip_{4y}\end{pmatrix}} & u_{42}&=\underset{\substack{\text{outbound e^-}\\\text{spin down}}} {\begin{pmatrix}0\\E+m\\p_{4x}-ip_{4y}\\-p_{4z}\end{pmatrix}} \end{align*}

The expected probability density is $$\begin{equation*} \langle|\mathcal{M}|^2\rangle_{COM} = \frac{e^4}{4} \left( \frac{f_{11}}{(s-m^2)^2} +\frac{2f_{12}}{(s-m^2)(u-m^2)} %+\frac{f_{12}^*}{(s-m^2)(u-m^2)} +\frac{f_{22}}{(u-m^2)^2} \right) \tag{1} \end{equation*}$$ where \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} \tag{2} \end{equation*}

Recall the Mandelstam variables are \begin{align*} s&=(p_1+p_2)^2 \\ t&=(p_1-p_3)^2 \\ u&=(p_1-p_4)^2 \end{align*}

Compton scattering experiments are typically done in the lab frame where the electron is at rest. Define Lorentz boost $$\Lambda$$ for transforming momentum vectors to the lab frame. $$\begin{equation*} \Lambda= \begin{pmatrix} E/m & 0 & 0 & \omega/m\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ \omega/m & 0 & 0 & E/m \end{pmatrix} \end{equation*}$$

The electron is at rest in the lab frame. $$\begin{equation*} \Lambda p_2=\begin{pmatrix}m \\ 0 \\ 0 \\ 0\end{pmatrix} \end{equation*}$$

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 \begin{pmatrix}1\\0\\0\\0\end{pmatrix} =\frac{\omega^2}{m}+\frac{\omega E}{m} \\[1ex] \omega_L'&=\Lambda p_3\cdot \begin{pmatrix}1\\0\\0\\0\end{pmatrix} =\frac{\omega^2\cos\theta}{m}+\frac{\omega E}{m} \end{aligned} \tag{3} \end{equation*}

It can be shown that \begin{equation*} \begin{aligned} s&=m^2+2m\omega_L \\ t&=2m(\omega_L' - \omega_L) \\ u&=m^2-2 m \omega_L' \end{aligned} \tag{4} \end{equation*}

Then by (1), (2), and (4) we have $$\begin{equation*} \langle|\mathcal{M}|^2\rangle_{COM}= 2e^4\left( \frac{\omega_L}{\omega_L'}+\frac{\omega_L'}{\omega_L} +\left(\frac{m}{\omega_L}-\frac{m}{\omega_L'}+1\right)^2-1 \right) \end{equation*}$$

Hence $$\begin{equation*} \langle|\mathcal{M}|^2\rangle_{COM} =\langle|\mathcal{M}|^2\rangle_{LAB} \end{equation*}$$

Lab scattering angle $$\theta_L$$ is given by the Compton formula. $$\begin{equation*} \cos\theta_L=\frac{m}{\omega_L}-\frac{m}{\omega_L'}+1 \tag{5} \end{equation*}$$

It follows that \begin{align*} \langle|\mathcal{M}|^2\rangle_{LAB} &=2e^4\left( \frac{\omega_L}{\omega_L'}+\frac{\omega_L'}{\omega_L}+\cos^2\theta_L-1 \right) \\ &=2e^4\left( \frac{\omega_L}{\omega_L'}+\frac{\omega_L'}{\omega_L}-\sin^2\theta_L \right) \end{align*}

To show that $$\theta_L\ne\theta$$, substitute (3) into (5) to obtain $$\begin{equation*} \cos\theta_L=\frac{m^2}{\omega^2+\omega E} -\frac{m^2}{\omega^2\cos\theta+\omega E}+1 \end{equation*}$$

Hence $$\theta_L\ne\theta$$.