Scattering Cross Section

Scattering Cross Section

$d\sigma = \frac{\left(\frac{number\ of\ particles\ scattered}{sec}\right) into\ d\Omega\ at\ \theta , \phi}{\left(\frac{\left(\frac{number\ of\ particles}{cm^2}}{sec}\right) in\ incoming\ beam}\right) = I(\theta,\phi)d\Omega$

$d\sigma = \frac{number\ of\ particles\ scattered\ into\ d\Omega\ at\ \theta , \phi}{number\ of\ particles\ /cm^2\ in\ incoming\ beam} = I(\theta,\phi)d\Omega$

$d\sigma=\frac{dN}{\mathcal L}$

$where\ \frac{d\sigma}{d\Omega}\equiv differential\ scattering\ cross\ section$

$and\ \sigma=\int\limits_{\theta=0}^{\pi} \int\limits_{\phi=0}^{2\pi} \left(\frac{d\sigma}{d\Omega}\right)\ d\Omega \equiv total\ scattering\ cross\ section$

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

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.

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

This is a Lorentz invariant.

$d\sigma=I_{lab}(\theta_{lab},\ \phi_{lab})\, d\Omega_{lab}=I_{CM}(\theta_{CM},\ \phi_{CM})\, d\Omega_{CM}$

Since the number of particles per second going into the detector is the same for both frames. (Only the z component of the momentum and the Energy are Lorentz transformed)

This is that the number of particles going into the solid-angle element d\Omega and having a moentum between p and p+dp be the same as the number going into the correspoiding solid-angle element d\Omega^* and having a corresponding momentum between p^* and p*+dp*

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