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 | ||
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− | 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> | ||
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− | <center><math>\frac{ | + | <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> |
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+ | Expressing this in terms of the solid angle components, | ||
− | <center><math>\frac{ | + | <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>\ | + | <center><math>\phi=\phi^*</math></center> |
+ | |||
+ | |||
+ | Thus, | ||
+ | <center><math>\Rightarrow\ d\phi=d\phi^*</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> | ||
+ | |||
+ | |||
+ | |||
+ | We can use the fact that | ||
+ | |||
+ | <center><math>\sin{\theta}\ d\theta=d(\cos{\theta})</math></center> | ||
+ | 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{ | + | <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> |
− | We can use the fact that | + | <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: | ||
+ | |||
+ | <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{d p^*\, \sin{\theta^*} \, d \theta^* \, d \phi^*}{d p^*\, d \theta^*\, d \phi^*}=\sin{\theta^*}</math></center> | ||
+ | |||
+ | |||
+ | Similarly, | ||
+ | |||
+ | |||
+ | <center><math>\frac{\partial (p\, \theta\, \phi)}{\partial (p\, \cos{\theta})}=\frac{1}{\sin{\theta}}</math></center> | ||
+ | |||
+ | |||
+ | |||
+ | Using the conversion of cartesian to spherical coordinates we know: | ||
+ | |||
+ | <center><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></center> | ||
+ | |||
+ | |||
+ | and the fact that as was shown earlier, that | ||
+ | |||
+ | |||
+ | <center><math>\begin{cases} | ||
+ | p^*_x=p_x \\ | ||
+ | p^*_y=p_y \\ | ||
+ | \phi^*=\phi | ||
+ | \end{cases}</math></center> | ||
+ | |||
+ | |||
+ | |||
+ | This allows us to express the term: | ||
+ | |||
+ | <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^*_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> | ||
+ | |||
+ | |||
+ | Again, similarly | ||
+ | |||
+ | |||
+ | <center><math>\frac{\partial (p_x\, p_y\, p_z)}{\partial (p\, \theta\, \phi)}=p^2\, \sin{\theta}</math></center> | ||
+ | |||
+ | |||
+ | |||
+ | 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> | ||
+ | |||
+ | which gives, | ||
+ | |||
+ | |||
+ | <center><math>\Longrightarrow\begin{cases} | ||
+ | E^*=\gamma E-\beta \gamma^* p_z \\ | ||
+ | p^*_z=-\beta \gamma\, E+\gamma\, p_z | ||
+ | \end{cases}</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> | ||
− | |||
− | <center><math>\frac{ | + | 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: | We can use the chain rule to find the transformation term on the right hand side: | ||
− | <center><math>\frac{ | + | <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{ | + | <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](/./images/thumb/4/48/Scattering.png/400px-Scattering.png)
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:
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.
where
For the differential cross-section:
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
This makes the total cross section a Lorentz invariant in that it is not effected by any relativistic transformations.
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 dΩCM and having a corresponding momentum between pCM and pCM+dpCM
Expressing this in terms of the solid angle components,
As shown earlier,
Thus,
Simplify our expression for the cross section gives:
We can use the fact that
To give
Using Chain Rule
We can use the chain rule to find the transformation term on the right hand side:
Starting with the term:
Similarly,
Using the conversion of cartesian to spherical coordinates we know:
and the fact that as was shown earlier, that
This allows us to express the term:
Again, similarly
To find the middle component in the chain rule expansion,
which gives,
We can use the relativistic definition of the total Energy,
Then using the fact that
Final Expression
Using the values found above, our expression becomes:
This gives,
Cross Section as a Function of Energy, Momentum, and Solid Angle
We can use the chain rule to find the transformation term on the right hand side:
Using the expression found earlier,
This gives,
where it was already shown that since pz is the only cartesian component affected by the Lorentz shift
Our final expression
By symmetry,
Cross Section as a Function of Energy, and Solid Angle
Using the chain rule
Using previous results