Difference between revisions of "Scattering Cross Section"

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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.
 
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==
 +
 +
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:
 +
 +
<center><math>dN=\frac{dR}{dt}=\mathcal L \times \sigma_p</math></center>
 +
 +
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.
 +
 +
 +
<center><math>\mathcal L = \Phi \rho \ell \approx i_{beam} \rho \ell</math></center>
 +
 +
where
 +
 +
<center><math>\Phi \equiv </math>flux, or incoming particles per second</center>
 +
 +
 +
<center><math>\rho \equiv </math>target density</center>
 +
 +
 +
<center><math>\ell \equiv </math>length of target</center>
 +
 +
 +
For the differential cross-section:
 +
 +
<center><math>\frac{d\sigma}{d\Omega} =\frac{dN}{\mathcal L d\Omega}=\frac{dN}{\Phi \rho \ell d\Omega}</math></center>
  
 
=Transforming Cross Section Between Frames=
 
=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
 
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
  
Line 22: Line 48:
  
  
This makes the total cross section a Lorentz invariant in that it is not effected by any relativistic transformations
+
This makes the total cross section a Lorentz invariant in that it is not effected by any relativistic transformations.
  
 
<center><math>\therefore\ \sigma_{CM}=\sigma_{Lab}</math></center>
 
<center><math>\therefore\ \sigma_{CM}=\sigma_{Lab}</math></center>
Line 30: Line 56:
  
  
<center><math>\frac{\partial ^2\sigma_{Lab}}{\partial p_{Lab}\,\partial \Omega_{Lab}}dp_{Lab}\,d\Omega_{Lab}=\frac{\partial ^2\sigma_{CM}}{\partial p_{CM}\,\partial \Omega_{CM}}dp_{CM}\, d\Omega_{CM}</math></center>
+
<center><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></center>
  
  
Line 36: Line 62:
  
  
 +
Expressing this in terms of the solid angle components,
  
<center><math>\frac{\partial ^2\sigma_{Lab}}{\partial p_{Lab}\,\partial \Omega_{Lab}}\partial p_{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}</math></center>
+
<center><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></center>
  
  
 
As shown earlier,
 
As shown earlier,
  
<center><math>\phi_{Lab}=\phi_{CM}</math></center>
+
<center><math>\phi=\phi^*</math></center>
 
 
 
 
  
<center><math>\Rightarrow\ d\phi_{Lab}=d\phi_{CM}</math></center>
 
  
 +
Thus,
 +
<center><math>\Rightarrow\ d\phi=d\phi^*</math></center>
  
  
<center><math>\frac{\partial ^2\sigma_{Lab}}{\partial p_{Lab}\,\partial \Omega_{Lab}} dp_{Lab}\,\sin{\theta_{Lab}}\,d\theta_{Lab}=\frac{\partial ^2\sigma_{CM}}{\partial p_{CM}\, \partial \Omega_{CM}}dp_{CM}\, \sin{\theta_{CM}}\,d\theta_{CM}</math></center>
+
Simplify our expression for the cross section gives:
 +
<center><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></center>
  
  
Line 59: Line 86:
  
  
 +
To give
 +
<center><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></center>
  
<center><math>\frac{\partial ^2\sigma_{Lab}}{\partial p_{Lab}\,\partial \Omega_{Lab}} dp_{Lab}\,d(\cos{\theta_{Lab}})=\frac{\partial ^2\sigma_{CM}}{\partial p_{CM}\, \partial \Omega_{CM}}dp_{CM}\, d(\cos{\theta_{CM}})</math></center>
 
  
  
  
 +
<center><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></center>
  
<center><math>\frac{\partial ^2\sigma_{Lab}}{\partial p_{Lab}\,\partial \Omega_{Lab}} =\frac{\partial ^2\sigma_{CM}}{\partial p_{CM}\, \partial \Omega_{CM}} \frac{dp_{CM}\,d(\cos{\theta_{CM}})}{dp_{Lab}\,d(\cos{\theta_{Lab}})}</math></center>
 
 
 
 
<center><math>\frac{\partial ^2\sigma_{Lab}}{\partial p_{Lab}\,\partial \Omega_{Lab}} =\frac{\partial ^2\sigma_{CM}}{\partial p_{CM}\, \partial \Omega_{CM}} \frac{\partial (p_{CM}\,\cos{\theta_{CM})}}{\partial (p_{Lab}\,\cos{\theta_{Lab}})}</math></center>
 
  
  
 +
<center><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></center>
  
 +
===Using Chain Rule===
 
We can use the chain rule to find the transformation term on the right hand side:
 
We can use the chain rule to find the transformation term on the right hand side:
  
<center><math>\frac{\partial (p^*\,\cos{\theta^*)}}{\partial (p^*\theta^*\phi^*)} \frac{\partial (p^*\theta^*\phi^*)}{\partial (p^*_xp^*_yp^*_z)} \frac{\partial (p^*_xp^*_yp^*_z)}{\partial (p_xp_yp_z)} \frac{\partial (p_xp_yp_z)}{\partial (p\theta\phi)} \frac{\partial (p\theta\phi)}{\partial (p\,\cos{\theta})}=\frac{\partial (p^*\cos{\theta^*})}{\partial (p\cos{\theta})}</math></center>
+
<center><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></center>
  
  
 
+
Starting with the term:
<center><math>\frac{\partial (p^*\cos{\theta^*})} {\partial (p^*\theta^*\phi^*)}=\frac{\partial p^* \sin{\theta^*} \partial \theta^* \partial \phi^*}{\partial p^*\partial \theta^*\partial \phi^*}=\sin{\theta^*}</math></center>
+
<center><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></center>
  
  
Line 85: Line 111:
  
  
<center><math>\frac{\partial (p\theta\phi)}{\partial (p\,\cos{\theta})}=\frac{1}{\sin{\theta}}</math></center>
+
<center><math>\frac{\partial (p\, \theta\, \phi)}{\partial (p\, \cos{\theta})}=\frac{1}{\sin{\theta}}</math></center>
  
  
Line 92: Line 118:
  
 
<center><math>\begin{cases}
 
<center><math>\begin{cases}
p_x=p\sin{\theta}\cos{\phi} \\
+
p_x=p\, \sin{\theta}\, \cos{\phi} \\
p_y=p\sin{\theta}\sin{\phi} \\
+
p_y=p\, \sin{\theta}\, \sin{\phi} \\
p_z=p\cos{\theta}
+
p_z=p\, \cos{\theta}
 
\end{cases}</math></center>
 
\end{cases}</math></center>
  
Line 111: Line 137:
 
This allows us to express the term:
 
This allows us to express the term:
  
<center><math>\frac{\partial (p^*\theta^*\phi^*)}{\partial (p^*_xp^*_yp^*_z)}=\biggl[\frac{\partial (p^*_xp^*_yp^*_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></center>
+
<center><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></center>
  
  
<center><math>\frac{\partial (p^*\theta^*\phi^*)}{\partial (p^*_xp^*_yp^*_z)}=\biggl[ \frac{\partial p^{*-1}\cos{\theta^{*}}^{-1}} {\partial p\,d\theta^*}\biggr]=\frac{1}{p^{*2}\sin{\theta^*}}</math></center>
+
<center><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></center>
  
  
Line 120: Line 146:
  
  
<center><math>\frac{\partial (p_xp_yp_z)}{\partial (p\theta\phi)}=p^2\sin{\theta}</math></center>
+
<center><math>\frac{\partial (p_x\, p_y\, p_z)}{\partial (p\, \theta\, \phi)}=p^2\, \sin{\theta}</math></center>
  
  
Line 126: Line 152:
 
To find the middle component in the chain rule expansion,
 
To find the middle component in the chain rule expansion,
  
<center><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></center>
+
<center><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></center>
  
 
which gives,
 
which gives,
Line 132: Line 158:
  
 
<center><math>\Longrightarrow\begin{cases}
 
<center><math>\Longrightarrow\begin{cases}
E^*=\gamma^* E-\beta^* \gamma^* p_z \\
+
E^*=\gamma E-\beta \gamma^* p_z \\
p^*_z=-\beta^* \gamma^*E+\gamma^* p_z
+
p^*_z=-\beta \gamma\, E+\gamma\, p_z
 
\end{cases}</math></center>
 
\end{cases}</math></center>
  
  
<center><math>\frac{\partial (p^*_xp^*_yp^*_z)}{\partial (p_xp_yp_z)}=\frac{\partial p^*_z}{\partial p_z}=\frac{\partial (-\beta^* \gamma^*E+\gamma^* p_z)}{\partial p_z}=-\beta^* \gamma^*\frac{\partial E}{\partialp_z}+\gamma^*</math></center>
+
<center><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></center>
 +
 
 +
 
 +
We can use the relativistic definition of the total Energy,
 +
 
 +
 
 +
<center><math>E=\sqrt{p^2+m^2}=\sqrt{p_x^2+p_y^2+p_z^2+m^2}</math></center>
 +
 
 +
 
 +
 
 +
<center><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></center>
 +
 
 +
 
 +
 
 +
<center><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></center>
 +
 
 +
 
 +
Then using the fact that
 +
 
 +
<center><math>E^*\equiv \gamma E-\beta \gamma\, p_z </math></center>
 +
 
 +
 
 +
 
 +
<center><math>\frac{\partial p^*_z}{\partial p_z}=-\beta \gamma \frac{p_z}{E}+\frac{\gamma\, E}{E}=\frac{E^*}{E}</math></center>
 +
 
 +
===Final Expression===
 +
Using the values found above, our expression becomes:
 +
<center><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></center>
 +
 
 +
 
 +
 
 +
 
 +
This gives,
 +
<center><math></math></center>
 +
{| class="wikitable" align="center"
 +
| style="background: gray"      | <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==
 +
<center><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></center>
 +
 
 +
 
 +
 
 +
 
 +
<center><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></center>
 +
 
 +
 
 +
 
 +
<center><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></center>
 +
 
 +
 
 +
We can use the chain rule to find the transformation term on the right hand side:
 +
 
 +
<center><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></center>
 +
 
 +
 
 +
Using the expression found earlier,
 +
<center><math>\frac{\partial (p^*\, \cos{\theta^*)}}{\partial (p\, \cos{\theta})}=\frac{p^2E^*}{p^{*2}E}</math></center>
 +
 
 +
 
 +
This gives,
 +
<center><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></center>
 +
 
 +
 
 +
where it was already shown that since p<sub>z</sub> is the only cartesian component affected by the Lorentz shift
 +
<center><math>\frac{dE}{dp}=\frac{dE}{dp_z}=\frac{p_z}{E}=\frac{p}{E}\Rightarrow \frac{dp}{dE}=\frac{E}{p}</math></center>
 +
 
 +
 
 +
<center><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></center>
 +
 
 +
Our final expression
 +
 
 +
{| class="wikitable" align="center"
 +
| style="background: gray"      | <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,
 +
{| class="wikitable" align="center"
 +
| style="background: gray"      | <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==
 +
<center><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></center>
 +
 
 +
 
 +
<center><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></center>
 +
 
 +
 
 +
<center><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></center>
 +
 
 +
 
 +
 
 +
Using the chain rule
 +
 
 +
<center><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></center>
 +
 
 +
 
 +
Using previous results
 +
 
 +
<center><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></center>
 +
 
 +
 
 +
{| class="wikitable" align="center"
 +
| style="background: gray"      | <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>
 +
|}

Latest revision as of 00:44, 18 December 2016

Scattering Cross Section

Scattering.png


[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

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]dN=\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 \approx i_{beam} \rho \ell[/math]

where

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


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


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


For the differential cross-section:

[math]\frac{d\sigma}{d\Omega} =\frac{dN}{\mathcal L d\Omega}=\frac{dN}{\Phi \rho \ell d\Omega}[/math]

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


Thus,

[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]


Similarly,


[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]E=\sqrt{p^2+m^2}=\sqrt{p_x^2+p_y^2+p_z^2+m^2}[/math]


[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][/math]
[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]