# Calculating Cross Section

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1. Jul 25, 2016

### Muh. Fauzi M.

1. The problem statement, all variables and given/known data
I have a problem in calculating cross-section in elektron-positron -> muon-antimuon scattering.

2. Relevant equations
In the relativistic limit, we find the differential cross-section of {e}+{e^-} -> {μ}+{μ^-} is

\frac{d\sigma}{d\Omega}=\frac{\alpha^2}{4s}*(1+cos^2{\theta}).

When I integrating over \theta, i get an answer:

\sigma=\frac{3*\pi*\alpha^2}{8*s},

but that is not the correct answer. Due to my reference (Halzen and Martin), the answer should be

\sigma=\frac{4*\pi*\alpha^2}{3*s}.

3. The attempt at a solution
Can someone help me through this problem? My intuition said I have mistaken the integration. But I've tried to re-integral-ing it, still not come up with correct solution.

2. Jul 25, 2016

### BvU

Hello,

Is $d\Omega$ just $d\theta$ or has $\phi$ something to do with it too ?

Anyway, show your steps in detail, please. That makaes it easier to follow and point out where things may go wrong...

3. Jul 26, 2016

### Muh. Fauzi M.

Hello Mr., thank you for your respond.

In my first reference (Halzen and Martin), the instructor is to integrate over $\theta$ and $\phi$. Here is my doodling:

$$\sigma = \frac{\alpha^2}{4s}\int{(1+cos^2{\theta})}d\theta d\phi$$

Substituting the trigonometry identity,

$$\sigma = \frac{\alpha^2}{4s} \int {\frac{(3+cos{2\theta})}{2}} d\theta d\phi$$.

Dividing the integration,

$$\int 3 d\theta d\phi , \quad \int {cos{2\theta}}d\theta d\phi$$

and integrating from $0$ to $\pi$ for $\theta$, $0$ to $2\pi$ fpr $\phi$. The second term is vanish, so I have

$$\int 3 d\theta d\phi=3(\pi)(2\pi)$$.

Thus, when inserting to $\sigma$, I have

$$\sigma = \frac{3\pi^2\alpha^2}{4s}$$.

Using my second ref (Schwartz), the integration just applied over $\theta$ and I get

$$\sigma = \frac{3\pi\alpha^2}{8s}$$.

Which step do I miss?

4. Jul 26, 2016

### BvU

If I remember well, $d\Omega \ne d\phi d\theta$ because then you would end up with $2\pi^2$ where you expect $4\pi$ . Wasn't it $d\Omega = \sin\theta \;d\theta d\phi$ ?

5. Jul 26, 2016

### Muh. Fauzi M.

Speechless. That's true Mr. Thank you very much. (shy)