Bhabha Scattering and its differential cross section

I worked out the differential cross section for Bhabha Scattering in the center of mass frame and I obtained the following:

$$\left(\dfrac{d\sigma}{d\Omega}\right)_{CM} = \dfrac{\alpha^2}{2s} \left[\dfrac{t^2}{s^2} + \dfrac{s^2}{t^2} + u^2 \left(\dfrac{1}{s}+\dfrac{1}{t}\right)^2\right]$$

where s, t and u are the usual Mandelstam variables and $\alpha$ is the fine structure constant.

I need the above equation to be like the following expression:

$$\left(\dfrac{d\sigma}{d\Omega}\right) = \dfrac{\alpha^2}{8E^2} \left[\dfrac{1 + cos^4(\theta/2)}{sin^4(\theta/2)} - \dfrac{2cos^4(\theta/2)}{sin^2(\theta/2)} + \dfrac{1 + cos^2(\theta)}{2}\right]$$

I tried several times but for some reasons I am not able to get it in the desired form. I used the identities $cos\,(2\theta) = 1-2sin^2\theta = 2\,cos^2 -1$

Can someone please help me out to prove this final expression?

Thanks!

• Have you writen down what all the 4-momenta are in the CM frame? It would help if you had those on hand – Triatticus May 31 '18 at 18:20
• I think that's where I am in trouble, I can't seem to get the 4 momenta correct in the CM frame. Could you please help me with that? – SSS Jun 1 '18 at 4:00
• Can you edit in what you got for the momenta and maybe I can point out where the flaw in your logic is, one major thing you want to use is conservation of momentum – Triatticus Jun 1 '18 at 5:00