Difference between revisions of "Differential Cross-Section"
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<center><math>\left(10\right)\qquad \qquad \frac{-2us}{t^2}=\frac{4p^{*2}\left(1+\cos{\theta}\right)4E^{*2}}{4p^{*2}\left(1-\cos{\theta}\right)^2}=\frac{E^{*2}\left(1+\cos{\theta}\right)\csc^4{\frac{\theta}{2}}}{p^{*2}}=\frac{E^{*2}p^{*2}\left(1+\cos{\theta}\right)}{p^{*4}\left(1-\cos{\theta}\right)^2}=\frac{E^{*2}p^{*2}\left(1+\cos{\theta}\right)\left(1+\cos{\theta}\right)^2}{p^{*4}\left(1-\cos{\theta}\right)^2\left(1+\cos{\theta}\right)^2}=\frac{E^{*2}p^{*2}\left(\cos^3{\theta}+3\cos^2{\theta}+3\cos{\theta}+1\right)}{p^{*4}\sin^4{\theta}}</math></center> | <center><math>\left(10\right)\qquad \qquad \frac{-2us}{t^2}=\frac{4p^{*2}\left(1+\cos{\theta}\right)4E^{*2}}{4p^{*2}\left(1-\cos{\theta}\right)^2}=\frac{E^{*2}\left(1+\cos{\theta}\right)\csc^4{\frac{\theta}{2}}}{p^{*2}}=\frac{E^{*2}p^{*2}\left(1+\cos{\theta}\right)}{p^{*4}\left(1-\cos{\theta}\right)^2}=\frac{E^{*2}p^{*2}\left(1+\cos{\theta}\right)\left(1+\cos{\theta}\right)^2}{p^{*4}\left(1-\cos{\theta}\right)^2\left(1+\cos{\theta}\right)^2}=\frac{E^{*2}p^{*2}\left(\cos^3{\theta}+3\cos^2{\theta}+3\cos{\theta}+1\right)}{p^{*4}\sin^4{\theta}}</math></center> | ||
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| + | Reducing by the common denominator, and combing like terms in the numerator | ||
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| + | <center><math> \left(1\right)\rightarrow \qquad \qquad 2p^{*4}-4p^{*4}\cos^2{\theta}+2p^{*4}\cos^4{\theta}</math></center> | ||
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| + | <center><math> \left(2 \right) \rightarrow \qquad \qquad 8E^{*4}-8E^{*4}\cos^2{\theta}</math></center> | ||
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| + | <center><math> \left(3/4\right) \rightarrow \qquad \qquad 2p^{*4}\cos^4{\theta}+12p^{*4}\cos^2{\theta}+2p^{*4}</math></center> | ||
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| + | <center><math> \left(5/6\right) \rightarrow \qquad \qquad 8E^{*4}\cos^2{\theta}+*E^{*4}</math></center> | ||
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| + | <center><math> \left(7/8\right) \rightarrow \qquad \qquad 8E^{*2}p^{*2}-8E^{*2}p^{*2}\cos^2{\theta}</math></center> | ||
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| + | <center><math> \left(9/10\right) \rightarrow \qquad \qquad 6E^{*2}p^{*2}\cos^2{\theta}+2E^{*2}p^{*2}</math></center> | ||
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Revision as of 17:19, 1 January 2019
Differential Cross-Section
Working in the center of mass frame
Determining the scattering amplitude in the center of mass frame
Using the fine structure constant ()
In the center of mass frame the Mandelstam variables are given by:
Calculating the parts to have common denominators:
Reducing by the common denominator, and combing like terms in the numerator