Difference between revisions of "Scattering Cross Section"
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![Scattering.png](/./images/thumb/4/48/Scattering.png/400px-Scattering.png)
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<center><math>\frac{d\sigma}{d\Omega}=\frac{dN}{\mathcal L\, d\Omega}</math></center> | <center><math>\frac{d\sigma}{d\Omega}=\frac{dN}{\mathcal L\, d\Omega}</math></center> | ||
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<center><math>where\ dN=\ the\ amount\ of\ particles\ per\ unit\ time\ detected,\ also\ called\ the\ event\ rate</math></center> | <center><math>where\ dN=\ the\ amount\ of\ particles\ per\ unit\ time\ detected,\ also\ called\ the\ event\ rate</math></center> |
Revision as of 21:02, 2 February 2016
Scattering Cross Section
![Scattering.png](/./images/thumb/4/48/Scattering.png/400px-Scattering.png)
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.
This is a Lorentz invariant.
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*