Bhabha Scattering Cross Section Derivation

[eoghan]

Hi!
I'm trying to derive the cross section of the bhabha scattering:
\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]

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:

\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]

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
1+cos^2\theta
to the annihilation diagram and a factor of
1+cos^2\frac{\theta}{2}
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
e^4E^2_{cm}
in order to get the right expression?

 

[member 553083]

Yes, you are correct about the spin dependence giving the factors 1+cos^2θ for the annihilation diagram and 1+cos^2θ/2 for the exchange diagram.The value of g is indeed e^4 E^2_{cm}. This is because the momentum in the toy theory has the same dimension as the energy, so we need to convert the energy to a momentum by multiplying it with the speed of light c. Thus, the coupling constant is g=e^4 E^2_{cm}/c^2.