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)