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

Cross Section as a Function of p, θ, Φ

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

$\frac{\partial ^2\sigma(p,\, \theta ,\, \phi)}{\partial p\,\partial \Omega}dp \,d\Omega=\frac{\partial ^2\sigma^*(p^*,\, \theta^* ,\, \phi^*)}{\partial p^* \,\partial \Omega^* }dp^* \, d\Omega^*$

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

Expressing this in terms of the solid angle components,

$\frac{\partial ^2\sigma(p,\, \theta ,\, \phi)}{\partial p\,\partial \Omega}\partial p \,\sin{\theta} \,d\theta \,d\phi=\frac{d^2\sigma^*(p^*,\, \theta^* ,\, \phi^*)}{dp^* \, d\Omega^*}dp^*\, \sin{\theta^*}\,d\theta^* \,d\phi^*$

As shown earlier,

$\phi=\phi^*$

Thus,

$\Rightarrow\ d\phi=d\phi^*$

Simplify our expression for the cross section gives:

$\frac{\partial ^2\sigma(p,\, \theta ,\, \phi)}{\partial p \,\partial \Omega} dp \,\sin{\theta}\,d\theta=\frac{\partial ^2\sigma^*(p^*,\, \theta^* ,\, \phi^*)}{\partial p^*\, \partial \Omega^*}dp^*\, \sin{\theta^*}\,d\theta^*$

We can use the fact that

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

To give

$\frac{\partial ^2\sigma(p,\, \theta ,\, \phi)}{\partial p\,\partial \Omega} dp\,d(\cos{\theta})=\frac{\partial ^2\sigma^*(p^*,\, \theta^* ,\, \phi^*)}{\partial p^*\, \partial \Omega^*}dp^*\, d(\cos{\theta^*})$

$\frac{\partial ^2\sigma(p,\, \theta ,\, \phi)}{\partial p\,\partial \Omega} =\frac{\partial ^2\sigma^*(p^*,\, \theta^* ,\, \phi^*)}{\partial p^*\, \partial \Omega^*} \frac{dp^*\,d(\cos{\theta^*})}{dp\,d(\cos{\theta})}$

$\frac{\partial ^2\sigma(p,\, \theta ,\, \phi)}{\partial p\,\partial \Omega} =\frac{\partial ^2\sigma^*(p^*,\, \theta^* ,\, \phi^*)}{\partial p^*\, \partial \Omega^*} \frac{\partial (p^*\,\cos{\theta^*)}}{\partial (p\,\cos{\theta})}$

Using Chain Rule

We can use the chain rule to find the transformation term on the right hand side:

$\frac{\partial (p^*\, \cos{\theta^*)}}{\partial (p^*\, \theta^*\, \phi^*)} \frac{\partial (p^*\, \theta^*\, \phi^*)}{\partial (p^*_x\, p^*_y\, p^*_z)} \frac{\partial (p^*_x\, p^*_y\, p^*_z)}{\partial (p_x\, p_y\, p_z)} \frac{\partial (p_x\, p_y\, p_z)}{\partial (p\, \theta\, \phi)} \frac{\partial (p\, \theta\, \phi)}{\partial (p\, \cos{\theta})}=\frac{\partial (p^*\, \cos{\theta^*})}{\partial (p\, \cos{\theta})}$

Starting with the term:

$\frac{\partial (p^*\, \cos{\theta^*})} {\partial (p^*\, \theta^*\phi^*)}=\frac{d p^*\, \sin{\theta^*} \, d \theta^* \, d \phi^*}{d p^*\, d \theta^*\, d \phi^*}=\sin{\theta^*}$

Similarly,

$\frac{\partial (p\, \theta\, \phi)}{\partial (p\, \cos{\theta})}=\frac{1}{\sin{\theta}}$

Using the conversion of cartesian to spherical coordinates we know:

$\begin{cases} p_x=p\, \sin{\theta}\, \cos{\phi} \\ p_y=p\, \sin{\theta}\, \sin{\phi} \\ p_z=p\, \cos{\theta} \end{cases}$

and the fact that as was shown earlier, that

$\begin{cases} p^*_x=p_x \\ p^*_y=p_y \\ \phi^*=\phi \end{cases}$

This allows us to express the term:

$\frac{\partial (p^*\, \theta^*\, \phi^*)}{\partial (p^*_x\, p^*_y\, p^*_z)}=\biggl[\frac{\partial (p^*_x\, p^*_y\, p^*_z)}{\partial (p^*\, \theta^*\, \phi^*)}\biggr]^{-1}=\biggl[\frac{\partial (p^*\, \sin{\theta^*}\, \cos{\phi^*}p^*\, \sin{\theta^*}\, \sin{\phi^*}p^*\, \cos{\theta^*})}{\partial p^*\, \partial \theta^*\, \partial \phi^*}\biggr]^{-1}$

$\frac{\partial (p^*\, \theta^*\, \phi^*)}{\partial (p^*_x\, p^*_y\, p^*_z)}=\biggl[ \frac{d (p^{*-1})\, d(\cos{\theta^{*}}^{-1})} {d p^*\, d\theta^*}\biggr]=\frac{1}{p^{*2}\sin{\theta^*}}$

Again, similarly

$\frac{\partial (p_x\, p_y\, p_z)}{\partial (p\, \theta\, \phi)}=p^2\, \sin{\theta}$

To find the middle component in the chain rule expansion,

$\left( \begin{matrix} E^* \\ p^*_x \\ p^*_y \\ p^*_z\end{matrix} \right)=\left(\begin{matrix}\gamma & 0 & 0 & -\beta \gamma\\0 & 1 & 0 & 0 \\ 0 & 0 & 1 &0 \\ -\beta\gamma & 0 & 0 & \gamma \end{matrix} \right) . \left( \begin{matrix}E\\ p_x \\ p_y\\ p_z\end{matrix} \right)$

which gives,

$\Longrightarrow\begin{cases} E^*=\gamma E-\beta \gamma^* p_z \\ p^*_z=-\beta \gamma\, E+\gamma\, p_z \end{cases}$

$\frac{\partial (p^*_x\, p^*_y\, p^*_z)}{\partial (p_x\, p_y\, p_z)}=\frac{\partial p^*_z}{\partial p_z}=\frac{\partial (-\beta \gamma\, E+\gamma\, p_z)}{\partial p_z}=-\beta \gamma \,\frac{\partial E}{\partial p_z}+\gamma$

We can use the relativistic definition of the total Energy,

$E=\sqrt{p^2+m^2}=\sqrt{p_x^2+p_y^2+p_z^2+m^2}$

$\Rightarrow \frac{\partial p^*_z}{\partial p_z}=\frac{\sqrt {p_x^2+p_y^2+p_z^2+m^2}} {\partial p_z}=\frac{p_z}{\sqrt {p_x^2+p_y^2+p_z^2+m^2}}=\frac{p_z}{E}$

$\frac{\partial p^*_z}{\partial p_z}=-\beta \gamma \frac{\partial E}{\partial p_z}+\gamma=-\beta \gamma \frac{p_z}{E}+\gamma$

Then using the fact that

$E^*\equiv \gamma E-\beta \gamma\, p_z$

$\frac{\partial p^*_z}{\partial p_z}=-\beta \gamma \frac{p_z}{E}+\frac{\gamma\, E}{E}=\frac{E^*}{E}$

Final Expression

Using the values found above, our expression becomes:

$\frac{\partial (p^*\, \cos{\theta^*)}}{\partial (p^*\, \theta^*\, \phi^*)} \frac{\partial (p^*\, \theta^*\, \phi^*)}{\partial (p^*_x\, p^*_y\, p^*_z)} \frac{\partial (p^*_x\, p^*_y\, p^*_z)}{\partial (p_x\, p_y\, p_z)} \frac{\partial (p_x\, p_y\, p_z)}{\partial (p\, \theta\, \phi)} \frac{\partial (p\, \theta\, \phi)}{\partial (p\, \cos{\theta})}=\frac{\partial (p^*\, \cos{\theta^*})}{\partial (p\, \cos{\theta})}$

This gives,

$\Longrightarrow \frac{\partial ^2\sigma(p,\, \theta ,\, \phi)}{\partial p\,\partial \Omega} =\frac{\partial ^2\sigma^*(p^*,\, \theta^* ,\, \phi^*)}{\partial p^*\, \partial \Omega^*} \frac{p^2E^*}{p^{*2}E}$