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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.

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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.

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1 Answer 1

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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 $\theta_L\ne\theta$.

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