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