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 $cos\,(2\theta) = 1-2sin^2\theta = 2\,cos^2 -1$

Thanks!

• Have you writen down what all the 4-momenta are in the CM frame? It would help if you had those on hand May 31, 2018 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, 2018 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 Jun 1, 2018 at 5:00

These identities should work:

\begin{align*} s&=4E^2 \\ t&=-2E^2(1-\cos\theta)=-4E^2\sin^2(\theta/2) \\ u&=-2E^2(1+\cos\theta)=-4E^2\cos^2(\theta/2) \end{align*}

Derivation of probability density for Bhabha scattering:

In a typical collider experiment the momentum vectors are $$\begin{equation*} p_1=\begin{pmatrix}E\\0\\0\\p\end{pmatrix}\qquad p_2=\begin{pmatrix}E\\0\\0\\-p\end{pmatrix}\qquad p_3=\begin{pmatrix} E\\ p\sin\theta\cos\phi\\ p\sin\theta\sin\phi\\ p\cos\theta \end{pmatrix} \qquad p_4=\begin{pmatrix} E\\ -p\sin\theta\cos\phi\\ -p\sin\theta\sin\phi\\ -p\cos\theta \end{pmatrix} \end{equation*}$$

where $$p=\sqrt{E^2-m^2}$$. The spinors are $$\begin{gather*} v_{11}=\begin{pmatrix}p\\0\\E+m\\0\end{pmatrix}\quad u_{21}=\begin{pmatrix}E+m\\0\\-p\\0\end{pmatrix}\quad v_{31}=\begin{pmatrix}p_3^z\\p_3^x+ip_3^y\\E+m\\0\end{pmatrix}\quad u_{41}=\begin{pmatrix}E+m\\0\\p_4^z\\p_4^x+ip_4^y\end{pmatrix} \\ v_{12}=\begin{pmatrix}0\\-p\\0\\E+m\end{pmatrix}\quad u_{22}=\begin{pmatrix}0\\E+m\\0\\p\end{pmatrix}\quad v_{32}=\begin{pmatrix}p_3^x-ip_3^y\\-p_3^z\\0\\E+m\end{pmatrix}\quad u_{42}=\begin{pmatrix}0\\E+m\\p_4^x-ip_4^y\\-p_4^z\end{pmatrix} \end{gather*}$$

The last digit in a spinor subscript is 1 for spin up and 2 for spin down. Note that the spinors are not individually normalized. Instead, a combined spinor normalization constant $$N=(E+m)^4$$ will be used where needed.

This is the probability density for Bhabha scattering. The formula is from Feynman diagrams. $$|\mathcal{M}(s_1,s_2,s_3,s_4)|^2=\frac{e^4}{N} \left| -\frac{1}{t}(\bar{v}_1\gamma^\mu v_3)(\bar{u}_4\gamma_\mu u_2) +\frac{1}{s}(\bar{v}_1\gamma^\nu u_2)(\bar{u}_4\gamma_\nu v_3) \right|^2$$

Symbol $$s_j$$ selects the spin (up or down) of spinor $$j$$. Symbol $$e$$ is electron charge. Symbols $$s$$ and $$t$$ are Mandelstam variables $$s=(p_1+p_2)^2$$ and $$t=(p_1-p_3)^2$$.

Let $$\begin{equation*} a_1=(\bar{v}_1\gamma^\mu v_3)(\bar{u}_4\gamma_\mu u_2) \qquad a_2=(\bar{v}_1\gamma^\nu u_2)(\bar{u}_4\gamma_\nu v_3) \end{equation*}$$

Then \begin{align*} |\mathcal{M}(s_1,s_2,s_3,s_4)|^2 &= \frac{e^4}{N}\left|{-\frac{a_1}{t}} + \frac{a_2}{s}\right|^2\\ &= \frac{e^4}{N}\left(-\frac{a_1}{t} + \frac{a_2}{s}\right)\left(-\frac{a_1}{t} + \frac{a_2}{s}\right)^*\\ &= \frac{e^4}{N} \left( \frac{a_1a_1^*}{t^2} - \frac{a_1a_2^*}{st} - \frac{a_1^*a_2}{st} + \frac{a_2a_2^*}{s^2} \right) \end{align*}

The expected probability density $$\langle|\mathcal{M}|^2\rangle$$ is computed by summing $$|\mathcal{M}|^2$$ over all spin states and dividing by the number of inbound states. There are four inbound states. \begin{align*} \langle|\mathcal{M}|^2\rangle &= \frac{1}{4}\sum_{s_1=1}^2\sum_{s_2=1}^2\sum_{s_3=1}^2\sum_{s_4=1}^2 |\mathcal{M}(s_1,s_2,s_3,s_4)|^2\\ &= \frac{e^4}{4}\sum_{s_1=1}^2\sum_{s_2=1}^2\sum_{s_3=1}^2\sum_{s_4=1}^2 \frac{1}{N} \left( \frac{a_1a_1^*}{t^2} - \frac{a_1a_2^*}{st} - \frac{a_1^*a_2}{st} + \frac{a_2a_2^*}{s^2} \right) \end{align*}

Use the Casimir trick to replace sums over spins with matrix products. \begin{align*} f_{11}&=\frac{1}{N}\sum_\text{spins}a_1a_1^*= \mathop{\rm Tr}\left( (\not p_1-m)\gamma^\mu(\not p_3-m)\gamma^\nu \right) \mathop{\rm Tr}\left( (\not p_4+m)\gamma_\mu(\not p_2+m)\gamma_\nu \right) \\ f_{12}&=\frac{1}{N}\sum_{\rm spins}a_1a_2^*= \mathop{\rm Tr}\left( (\not p_1-m)\gamma^\mu(\not p_2+m)\gamma^\nu (\not p_4+m)\gamma_\mu(\not p_3-m)\gamma_\nu \right) \\ f_{22}&=\frac{1}{N}\sum_\text{spins}a_2a_2^*= \mathop{\rm Tr}\left( (\not p_1-m)\gamma^\mu(\not p_2+m)\gamma^\nu \right) \mathop{\rm Tr}\left( (\not p_4+m)\gamma_\mu(\not p_3-m)\gamma_\nu \right) \end{align*}

Hence $$\begin{equation*} \langle|\mathcal{M}|^2\rangle = \frac{e^4}{4} \left( \frac{f_{11}}{t^2} - \frac{f_{12}}{st} - \frac{f_{12}^*}{st} + \frac{f_{22}}{s^2} \right) \end{equation*}$$

These formulas compute probability densities directly from momentum vectors. \begin{align*} f_{11}&= 32(p_1\cdot p_2)(p_3\cdot p_4)+32(p_1\cdot p_4)(p_2\cdot p_3) -32 m^2(p_1\cdot p_3)-32 m^2(p_2\cdot p_4)+64 m^4 \\ f_{12}&= -32 (p_1\cdot p_4) (p_2\cdot p_3) -16 m^2 (p_1\cdot p_2) + 16 m^2 (p_1\cdot p_3) - 16 m^2 (p_1\cdot p_4)\\ &\phantom{=}\qquad {}- 16 m^2 (p_2\cdot p_3) + 16 m^2 (p_2\cdot p_4) - 16 m^2 (p_3\cdot p_4) - 32 m^4 \\ f_{22}&= 32(p_1\cdot p_3)(p_2\cdot p_4)+32(p_1\cdot p_4)(p_2\cdot p_3) +32 m^2(p_1\cdot p_2)+32 m^2(p_3\cdot p_4)+64 m^4 \end{align*}

In Mandelstam variables $$s=(p_1+p_2)^2$$, $$t=(p_1-p_3)^2$$, $$u=(p_1-p_4)^2$$ the formulas are \begin{align*} f_{11} &= 8 s^2 + 8 u^2 - 64 s m^2 - 64 u m^2 + 192 m^4 \\ f_{12} &= -8 u^2 + 64 u m^2 - 96 m^4 \\ f_{22} &= 8 t^2 + 8 u^2 - 64 t m^2 - 64 u m^2 + 192 m^4 \end{align*}

When $$E\gg m$$ a useful approximation is to set $$m=0$$ and obtain \begin{align*} f_{11}&= 8 s^2 + 8 u^2\\ f_{12}&= -8 u^2\\ f_{22}&= 8 t^2 + 8 u^2 \end{align*}

For $$m=0$$ the Mandelstam variables are \begin{align*} s&=4E^2 \\ t&=-2E^2(1-\cos\theta)=-4E^2\sin^2(\theta/2) \\ u&=-2E^2(1+\cos\theta)=-4E^2\cos^2(\theta/2) \end{align*}

The corresponding expected probability density is \begin{align*} \langle|\mathcal{M}|^2\rangle &= \frac{e^4}{4} \left( \frac{8s^2+8u^2}{t^2}+\frac{16u^2}{st}+\frac{8t^2+8u^2}{s^2} \right) \\ &=2e^4\left(\frac{s^2+u^2}{t^2}+\frac{2u^2}{st}+\frac{t^2+u^2}{s^2}\right) \\ &=2e^4 \left( \frac{1+\cos^4(\theta/2)}{\sin^4(\theta/2)} -\frac{2\cos^4(\theta/2)}{\sin^2(\theta/2)} +\frac{1+\cos^2\theta}{2} \right) \end{align*}