|
|
| Line 50: |
Line 50: |
| | <center><math>s \equiv 4E^{*2}</math></center> | | <center><math>s \equiv 4E^{*2}</math></center> |
| | | | |
| | + | <center><math>t \equiv -2E^{*2}(1-\cos{\theta})=-2E^{*2}\left (1-2\cos^2{\frac{\theta}{2}}+1 \right )=-4E^{*2} \left (1-2\cos^2{\frac{\theta}{2}} \right )=-4E^{*2}\sin^2{\frac{\theta}{2}}</math></center> |
| | + | |
| | + | |
| | + | <center><math>u \equiv -2E^{*2}(1+\cos{\theta})=-2E^{*2}\left (1+2\cos^2{\frac{\theta}{2}}-1 \right )=-4E^{*2}\cos^2{\frac{\theta}{2}}</math></center> |
| | | | |
| | Using the relationship | | Using the relationship |
| Line 70: |
Line 74: |
| | | | |
| | | | |
| − | <center><math>t \equiv -2E^{*2}(1-\cos{\theta})=-2E^{*2}\left (1-2\cos^2{\frac{\theta}{2}}+1 \right )=-4E^{*2} \left (1-2\cos^2{\frac{\theta}{2}} \right )=-4E^{*2}\sin^2{\frac{\theta}{2}}</math></center>
| |
| | | | |
| − |
| |
| − | <center><math>u \equiv -2E^{*2}(1+\cos{\theta})=-2E^{*2}\left (1+2\cos^2{\frac{\theta}{2}}-1 \right )=-4E^{*2}\cos^2{\frac{\theta}{2}}</math></center>
| |
| | | | |
| | | | |
Revision as of 22:03, 29 December 2018
[math]\underline{\textbf{Navigation}}[/math]
[math]\vartriangleleft [/math]
[math]\triangle [/math]
[math]\vartriangleright [/math]
Differential Cross-Section
[math]\frac{d\sigma}{d\Omega}=\frac{1}{64\pi ^2 s}\frac{\mathbf p_{final}}{\mathbf p_{initial}} |\mathfrak{M} |^2[/math]
Working in the center of mass frame
[math]\mathbf p_{final}=\mathbf p_{initial}[/math]
Determining the scattering amplitude in the center of mass frame
[math]\mathfrak{M}=e^2 \left ( \frac{u-s}{t}+\frac{t-s}{u} \right )[/math]
[math]\mathfrak{M}^2=e^4 \left ( \frac{u-s}{t}+\frac{t-s}{u} \right )\left ( \frac{u-s}{t}+\frac{t-s}{u} \right )[/math]
[math]\mathfrak{M}^2=e^4 \left ( \frac{(u-s)^2}{t^2}+\frac{(t-s)^2}{u^2} +2\frac{(u-s)}{t}\frac{(t-s)}{u}\right )[/math]
[math]\mathfrak{M}^2=e^4 \left ( \frac{(u^2-2us+s^2)}{t^2}+\frac{(t^2-2ts+s^2)}{u^2} +2\frac{(ut-st+s^2-us)}{tu}\right )[/math]
[math]\mathfrak{M}^2=e^4 \left ( \frac{(t^2+s^2)}{u^2}-\frac{2s^2}{tu}+\frac{(u^2+s^2)}{t^2}\right )[/math]
Using the fine structure constant ([math]with\ c=\hbar=\epsilon_0=1[/math])
[math]\alpha \equiv \frac{e^2}{4\pi}[/math]
[math]\frac{d\sigma}{d\Omega}=\frac{\alpha ^2}{2s}\left ( \frac{(t^2+s^2)}{u^2}-\frac{2s^2}{tu}+\frac{(u^2+s^2)}{t^2}\right )[/math]
In the center of mass frame the Mandelstam variables are given by:
[math]s \equiv 4E^{*2}[/math]
[math]t \equiv -2E^{*2}(1-\cos{\theta})=-2E^{*2}\left (1-2\cos^2{\frac{\theta}{2}}+1 \right )=-4E^{*2} \left (1-2\cos^2{\frac{\theta}{2}} \right )=-4E^{*2}\sin^2{\frac{\theta}{2}}[/math]
[math]u \equiv -2E^{*2}(1+\cos{\theta})=-2E^{*2}\left (1+2\cos^2{\frac{\theta}{2}}-1 \right )=-4E^{*2}\cos^2{\frac{\theta}{2}}[/math]
Using the relationship
[math]\cos{\theta}=-1+\cos{\frac{\theta}{2}}[/math]
In the ultra-relativistic limit, the electron mass is small enough compared to the energy such that it can be neglected when compared to the momentum
[math]m \lll p[/math]
[math]\therefore E^2\equiv m^2+p^2 \rightarrow E^2 \approx p^2[/math]
[math]\frac{d\sigma}{d\Omega}=\frac{\alpha ^2}{8E^{*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]
[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]
[math]\underline{\textbf{Navigation}}[/math]
[math]\vartriangleleft [/math]
[math]\triangle [/math]
[math]\vartriangleright [/math]
