Scattering Cross Section

Scattering Cross Section

$\frac{d\sigma}{d\Omega} = \frac{\left(\frac{number\ of\ particles\ scattered/second}{d\Omega}\right)}{\left(\frac{number\ of\ incoming\ particles/second}{cm^2}\right)}=\frac{dN}{\mathcal L\, d\Omega} =differential\ scattering\ cross\ section$

$where\ d\Omega=\sin{\theta}\,d\theta\,d\phi$

$\Rightarrow \sigma=\int\limits_{\theta=0}^{\pi} \int\limits_{\phi=0}^{2\pi} \left(\frac{d\sigma}{d\Omega}\right)\ \sin{\theta}\,d\theta\,d\phi =\frac{N}{\mathcal L}\equiv total\ scattering\ cross\ section$

Since this is just a ratio of detected particles to total particles, this gives the cross section as a relative probablity of a scattering, or reaction, to occur.

Transforming Cross Section Between Frames

Transforming the cross section between two different frames of reference has the condition that the quantity must be equal in both frames. This is due to the fact that

$\sigma=\frac{N}{\mathcal L}=constant\ number$

This makes the total cross section a Lorentz invariant in that it is not effected by any relativistic transformations

$\therefore\ \sigma_{CM}=\sigma_{Lab}$

This implies that the number of particles going into the solid-angle element d Ω_Lab and having a momentum between p_Lab and p_Lab+dp_Labbe the same as the number going into the corresponding solid-angle element dΩ_CM and having a corresponding momentum between p_CM and p_CM+dp_CM

$\frac{d^2\sigma_{Lab}}{dp_{Lab}\,d\Omega_{Lab}}dp_{Lab}\,d\Omega_{Lab}=\frac{d^2\sigma_{CM}}{dp_{CM}\, d\Omega_{CM}}dp_{CM}\, d\Omega_{CM}$

$where\ d\Omega=\sin{\theta}\,d\theta\,d\phi$

$\frac{d^2\sigma_{Lab}}{dp_{Lab}\,d\Omega_{Lab}}dp_{Lab}\,\sin{\theta_{Lab}}\,d\theta_{Lab}\,d\phi_{Lab}=\frac{d^2\sigma_{CM}}{dp_{CM}\, d\Omega_{CM}}dp_{CM}\, \sin{\theta_{CM}}\,d\theta_{CM}\,d\phi_{CM}$

As shown earlier,

$\phi_{Lab}=\phi_{CM}$

$\Rightarrow\ d\phi_{Lab}=d\phi_{CM}$

$\frac{d^2\sigma_{Lab}}{dp_{Lab}\,d\Omega_{Lab}} dp_{Lab}\,\sin{\theta_{Lab}}\,d\theta_{Lab}=\frac{d^2\sigma_{CM}}{dp_{CM}\, d\Omega_{CM}}dp_{CM}\, \sin{\theta_{CM}}\,d\theta_{CM}$

We can use the fact that

$\sin{\theta}\ d\theta=d(\cos{\theta})$

$\frac{d^2\sigma_{Lab}}{dp_{Lab}\,d\Omega_{Lab}} dp_{Lab}\,d(\cos{\theta_{Lab}})=\frac{d^2\sigma_{CM}}{dp_{CM}\, d\Omega_{CM}}dp_{CM}\, d(\cos{\theta_{CM}})$

$\frac{d^2\sigma_{Lab}}{dp_{Lab}\,d\Omega_{Lab}} =\frac{d^2\sigma_{CM}}{dp_{CM}\, d\Omega_{CM}} \frac{dp_{CM}\,d(\cos{\theta_{CM}})}{dp_{Lab}\,d(\cos{\theta_{Lab}})}$

$\frac{d^2\sigma_{Lab}}{dp_{Lab}\,d\Omega_{Lab}} =\frac{d^2\sigma_{CM}}{dp_{CM}\, d\Omega_{CM}} \frac{d(p_{CM}\,\cos{\theta_{CM})}}{d(p_{Lab}\,\cos{\theta_{Lab}})}$

We can use the Jacobian to find the transformation term on the right

$J=\frac{\partial p_{CM}\partial \cos{\theta_{CM})}}{\partial p_{Lab}\partial \cos{\theta_{Lab})}}$

= $$\begin{vmatrix} \frac{\partial p_{CM}}{\partial p_{Lab}} & \frac{\partial p_{CM}}{\partial \cos{\theta_{Lab})}} \\ z & v \end{vmatrix}$$

$\left( \begin{matrix} E_1^*+E_2^* \\ p^*_{1(x)}+p^*_{2(x)} \\ p^*_{1(y)}+p^*_{2(y)} \\ p^*_{1(z)}+p^*_{2(z)}\end{matrix} \right)=\left(\begin{matrix}\gamma^* & 0 & 0 & -\beta^* \gamma^*\\0 & 1 & 0 & 0 \\ 0 & 0 & 1 &0 \\ -\beta^*\gamma^* & 0 & 0 & \gamma^* \end{matrix} \right) . \left( \begin{matrix}E_{1}+E_{2}\\ p_{1(x)}+p_{2(x)} \\ p_{1(y)}+p_{2(y)} \\ p_{1(z)}+p_{2(z)}\end{matrix} \right)$