Difference between revisions of "Differential Cross-Section"
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+ | <center><math>\therefore E^2\equiv m^2+p^2 </math></center> | ||
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+ | <center><math>\frac{d\sigma}{d\Omega}=\frac{\alpha ^2}{8p^{*2}}\left( \frac{16E^{*4}\sin^4{\frac{\theta}{2}}+16E^{*4}}{16E^{*4}\cos^4{\frac{\theta}{2}}}-\frac{32E^{*4}}{4E^{*2}\sin^2{\frac{\theta}{2}}4E^{*2}\cos^2{\frac{\theta}{2}}}+\frac{16E^{*4}\cos^4{\frac{\theta}{2}}+16E^{*4}}{16E^{*4}\sin^4{\frac{\theta}{2}}}\right )</math></center> | ||
+ | <center><math>\frac{d\sigma}{d\Omega}=\frac{\alpha ^2}{8E^{*2}}\left( \frac{\sin^4{\frac{\theta}{2}}+1}{\cos^4{\frac{\theta}{2}}}-\frac{2}{\sin^2{\frac{\theta}{2}}\cos^2{\frac{\theta}{2}}}+\frac{\cos^4{\frac{\theta}{2}}+1}{\sin^4{\frac{\theta}{2}}}\right )</math></center> | ||
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