- #1
eoghan
- 207
- 7
Hi!
I'm trying to derive the cross section of the bhabha scattering:
[tex]\frac{d\sigma}{d\Omega}=\frac{e^4}{32\pi ^2 E^2_{cm}}\left[ \frac{1+cos^4\frac{\theta}{2}}{sin^4\frac{\theta}{2}}-\frac{2cos^4\frac{\theta}{2}}{sin^2\frac{\theta}{2}}+\frac{1+cos^2 \theta}{2}\right] [/tex]
I'm using the "toy theory" where at each vertex of a feynman diagram I associate a factor -ig, for each internal line there corresponds a propagator 1/q^2. The formula I get is:
[tex]\frac{d\sigma}{d\Omega}=\frac{1}{64\pi ^2 E^2_{cm}}\left(\frac{g^2}{E^2_{cm}}\right)^2\left[ \frac{1}{sin^4\frac{\theta}{2}}-\frac{2}{sin^2\frac{\theta}{2}}+1\right] [/tex]
Now, the incorrect angular dependence comes from the fact that in the toy theory I don't consider the spin, I guess. Looking at the correct formula I guess that the spin dependence gives a factor of
[tex]1+cos^2\theta[/tex]
to the annihilation diagram and a factor of
[tex]1+cos^2\frac{\theta}{2}[/tex]
to the exchange diagram.
Is it right?
Then there is the problem of the value of the coupling constant g. The text says that it can't just be "e" because in the toy theory g has the value of a momentum, while in the real world it is dimensionless. So, what is the value of g? It should be
[tex]e^4E^2_{cm}[/tex]
in order to get the right expression?
I'm trying to derive the cross section of the bhabha scattering:
[tex]\frac{d\sigma}{d\Omega}=\frac{e^4}{32\pi ^2 E^2_{cm}}\left[ \frac{1+cos^4\frac{\theta}{2}}{sin^4\frac{\theta}{2}}-\frac{2cos^4\frac{\theta}{2}}{sin^2\frac{\theta}{2}}+\frac{1+cos^2 \theta}{2}\right] [/tex]
I'm using the "toy theory" where at each vertex of a feynman diagram I associate a factor -ig, for each internal line there corresponds a propagator 1/q^2. The formula I get is:
[tex]\frac{d\sigma}{d\Omega}=\frac{1}{64\pi ^2 E^2_{cm}}\left(\frac{g^2}{E^2_{cm}}\right)^2\left[ \frac{1}{sin^4\frac{\theta}{2}}-\frac{2}{sin^2\frac{\theta}{2}}+1\right] [/tex]
Now, the incorrect angular dependence comes from the fact that in the toy theory I don't consider the spin, I guess. Looking at the correct formula I guess that the spin dependence gives a factor of
[tex]1+cos^2\theta[/tex]
to the annihilation diagram and a factor of
[tex]1+cos^2\frac{\theta}{2}[/tex]
to the exchange diagram.
Is it right?
Then there is the problem of the value of the coupling constant g. The text says that it can't just be "e" because in the toy theory g has the value of a momentum, while in the real world it is dimensionless. So, what is the value of g? It should be
[tex]e^4E^2_{cm}[/tex]
in order to get the right expression?