Difference between revisions of "Scattering Cross Section"

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<center><math>\Phi \equiv flux, or incoming particles per second</math></center>
<center><math>\Phi \equiv </math>flux, or incoming particles per second</center>
<center><math>\rho \equiv target density</math></center>
<center><math>\ell \equiv length of target</math></center>
<center><math>\rho \equiv </math>target density</center>
<center><math>\ell \equiv </math>length of target</center>
=Transforming Cross Section Between Frames=
=Transforming Cross Section Between Frames=

Revision as of 15:08, 18 March 2016

Scattering Cross Section


[math]\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[/math]

[math]where\ d\Omega=\sin{\theta}\,d\theta\,d\phi[/math]

[math]\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[/math]

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.


Luminosity is the quantity that measures the ability of a particle accelerator to produce the required number of interactions and is the proportional to the number of events per second dR/dt and the total cross section σp:

[math]\frac{dR}{dt}=\mathcal L \times \sigma_p[/math]

In order to compute a luminosity for fixed target experiment, it is necessary to take into account the properties of the incoming beam and the stationary target.

[math]\mathcal L = \Phi \rho \ell[/math]


[math]\Phi \equiv [/math]flux, or incoming particles per second

[math]\rho \equiv [/math]target density

[math]\ell \equiv [/math]length of target

Transforming Cross Section Between Frames

Cross Section as a Function of Momentum and Solid Angle

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

[math]\sigma=\frac{N}{\mathcal L}=constant\ number[/math]

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

[math]\therefore\ \sigma_{CM}=\sigma_{Lab}[/math]

This implies that the number of particles going into the solid-angle element d ΩLab and having a momentum between pLab and pLab+dpLabbe the same as the number going into the corresponding solid-angle element CM and having a corresponding momentum between pCM and pCM+dpCM

[math]\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^*[/math]

[math]where\ d\Omega=\sin{\theta}\,d\theta\,d\phi[/math]

Expressing this in terms of the solid angle components,

[math]\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^*[/math]

As shown earlier,



[math]\Rightarrow\ d\phi=d\phi^*[/math]

Simplify our expression for the cross section gives:

[math]\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^*[/math]

We can use the fact that

[math]\sin{\theta}\ d\theta=d(\cos{\theta})[/math]

To give

[math]\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^*})[/math]

[math]\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})}[/math]

[math]\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})}[/math]

Using Chain Rule

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

[math]\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})}[/math]

Starting with the term:

[math]\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^*}[/math]


[math]\frac{\partial (p\, \theta\, \phi)}{\partial (p\, \cos{\theta})}=\frac{1}{\sin{\theta}}[/math]

Using the conversion of cartesian to spherical coordinates we know:

[math]\begin{cases} p_x=p\, \sin{\theta}\, \cos{\phi} \\ p_y=p\, \sin{\theta}\, \sin{\phi} \\ p_z=p\, \cos{\theta} \end{cases}[/math]

and the fact that as was shown earlier, that

[math]\begin{cases} p^*_x=p_x \\ p^*_y=p_y \\ \phi^*=\phi \end{cases}[/math]

This allows us to express the term:

[math]\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}[/math]

[math]\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^*}}[/math]

Again, similarly

[math]\frac{\partial (p_x\, p_y\, p_z)}{\partial (p\, \theta\, \phi)}=p^2\, \sin{\theta}[/math]

To find the middle component in the chain rule expansion,

[math]\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)[/math]

which gives,

[math]\Longrightarrow\begin{cases} E^*=\gamma E-\beta \gamma^* p_z \\ p^*_z=-\beta \gamma\, E+\gamma\, p_z \end{cases}[/math]

[math]\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[/math]

We can use the relativistic definition of the total Energy,


[math]\Rightarrow \frac{\partial E}{\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}[/math]

[math]\frac{\partial p^*_z}{\partial p_z}=-\beta \gamma \frac{\partial E}{\partial p_z}+\gamma=-\beta \gamma \frac{p_z}{E}+\gamma[/math]

Then using the fact that

[math]E^*\equiv \gamma E-\beta \gamma\, p_z [/math]

[math]\frac{\partial p^*_z}{\partial p_z}=-\beta \gamma \frac{p_z}{E}+\frac{\gamma\, E}{E}=\frac{E^*}{E}[/math]

Final Expression

Using the values found above, our expression becomes:

[math]\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})}[/math]

This gives,

[math]\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}[/math]

Cross Section as a Function of Energy, Momentum, and Solid Angle

[math]\frac{\partial ^2\sigma(E,\, \theta ,\, \phi)}{\partial E\,\partial \Omega} dE\,d(\cos{\theta})=\frac{\partial ^2\sigma^*(p^*,\, \theta^* ,\, \phi^*)}{\partial p^*\, \partial \Omega^*}dp^*\, d(\cos{\theta^*})[/math]

[math]\frac{\partial ^2\sigma(E,\, \theta ,\, \phi)}{\partial E\,\partial \Omega} =\frac{\partial ^2\sigma^*(p^*,\, \theta^* ,\, \phi^*)}{\partial p^*\, \partial \Omega^*} \frac{dp^*\,d(\cos{\theta^*})}{dE\,d(\cos{\theta})}[/math]

[math]\frac{\partial ^2\sigma(E,\, \theta ,\, \phi)}{\partial E\,\partial \Omega} =\frac{\partial ^2\sigma^*(p^*,\, \theta^* ,\, \phi^*)}{\partial p^*\, \partial \Omega^*} \frac{\partial (p^*\,\cos{\theta^*)}}{\partial (E\,\cos{\theta})}[/math]

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

[math]\frac{\partial (p^*\, \cos{\theta^*)}}{\partial (p\, \cos{\theta})} \frac{\partial (p\, \cos{\theta})}{\partial (E\, \cos{\theta})}=\frac{\partial (p^*\, \cos{\theta^*})}{\partial (E\, \cos{\theta})}[/math]

Using the expression found earlier,

[math]\frac{\partial (p^*\, \cos{\theta^*)}}{\partial (p\, \cos{\theta})}=\frac{p^2E^*}{p^{*2}E}[/math]

This gives,

[math]\frac{p^2E^*}{p^{*2}E} \frac{\partial (p\, \cos{\theta})}{\partial (E\, \cos{\theta})}=\frac{p^2E^*}{p^{*2}E} \frac{dp\, d(\cos{\theta})}{dE\, d(\cos{\theta})}=\frac{p^2E^*}{p^{*2}E} \frac{dp}{dE}[/math]

where it was already shown that since pz is the only cartesian component affected by the Lorentz shift

[math]\frac{dE}{dp}=\frac{dE}{dp_z}=\frac{p_z}{E}=\frac{p}{E}\Rightarrow \frac{dp}{dE}=\frac{E}{p}[/math]

[math]\frac{p^2E^*}{p^{*2}E} \frac{\partial (p\, \cos{\theta})}{\partial (E\, \cos{\theta})}=\frac{p^2E^*}{p^{*2}E} \frac{E}{p}=\frac{pE^*}{p^{*2}}[/math]

Our final expression

[math]\frac{\partial ^2\sigma(E,\, \theta ,\, \phi)}{\partial E\,\partial \Omega} =\frac{\partial ^2\sigma^*(p^*,\, \theta^* ,\, \phi^*)}{\partial p^*\, \partial \Omega^*} \frac{pE^*}{p^{*2}}[/math]

By symmetry,

[math]\frac{\partial ^2\sigma(p,\, \theta ,\, \phi)}{\partial p\, \partial \Omega^*} = \frac{\partial ^2\sigma^*(E^*,\, \theta^* ,\, \phi^*)}{\partial E^*\,\partial \Omega^*}\frac{p^2}{p^{*}E}[/math]

Cross Section as a Function of Energy, and Solid Angle

[math]\frac{\partial ^2\sigma(E,\, \theta ,\, \phi)}{\partial E\,\partial \Omega} dE\,d(\cos{\theta})=\frac{\partial ^2\sigma^*(E^*,\, \theta^* ,\, \phi^*)}{\partial E^*\, \partial \Omega^*}dE^*\, d(\cos{\theta^*})[/math]

[math]\frac{\partial ^2\sigma(E,\, \theta ,\, \phi)}{\partial E\,\partial \Omega} =\frac{\partial ^2\sigma^*(E^*,\, \theta^* ,\, \phi^*)}{\partial E^*\, \partial \Omega^*}\frac{dE^*\, d(\cos{\theta^*})}{dE\,d(\cos{\theta})}[/math]

[math]\frac{\partial ^2\sigma(E,\, \theta ,\, \phi)}{\partial E\,\partial \Omega} =\frac{\partial ^2\sigma^*(E^*,\, \theta^* ,\, \phi^*)}{\partial E^*\, \partial \Omega^*}\frac{\partial (E^*\, \cos{\theta^*})}{\partial (E\,\cos{\theta})}[/math]

Using the chain rule

[math]\frac{\partial (E^*\, \cos{\theta^*})}{\partial (E\,\cos{\theta})}=\frac{\partial (E^*\, \cos{\theta^*})}{\partial (p^*\, \cos{\theta^*})} \frac{\partial (p^*\, \cos{\theta^*})}{\partial (p\, \cos{\theta})} \frac{\partial (p\, \cos{\theta})}{\partial (E\,\cos{\theta})}[/math]

Using previous results

[math]\frac{\partial (E^*\, \cos{\theta^*})}{\partial (E\,\cos{\theta})}=\frac{p^*}{E^*} \frac{p^{2}E^*}{p^{*2}E} \frac{E}{p}=\frac{p}{p^*}[/math]

[math]\frac{\partial ^2\sigma(E,\, \theta ,\, \phi)}{\partial E\, \partial \Omega} = \frac{\partial ^2\sigma^*(E^*,\, \theta^* ,\, \phi^*)}{\partial E^*\,\partial \Omega^*}\frac{p}{p^{*}}[/math]