I'm considering the process: $$ e^+e^- \longrightarrow \mu^+\mu^-$$ The cross section in the center-of-mass frame: \begin{equation} \left( \frac{d \sigma}{d \Omega} \right)_{CoM}= \frac{\alpha^2}{4 s} \sqrt{1-\frac{4m^2_{\mu}}{s}} \left( 1+\frac{4m^2_{\mu}}{s} + \left( 1-\frac{4m^2_{\mu}}{s} \right)\cos^2\theta\right) \end{equation} Let's fix the parameter $\sqrt{s}=0.212$ GeV, I take $m_e=0$ for semplicity, but I' am not in the case of the ultra-high-energy limit so $m_{\mu} \neq 0$.

I want the probability density function of the observable $\theta$. $$ \Pr [\theta_1 \le \theta \le \theta_2] = \int_{\theta_{1}}^{\theta_{2}} f(x) \, dx $$ $$f(x)= \frac{C}{\sigma_{total}}\left( \frac{d \sigma}{d x} \right)_{CoM} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C \Rightarrow \int_{0}^{\pi}f(x)=1$$ $$f(x)=\frac{2s}{\pi(4m_{\mu}^2+3s)}\left( 1+ \frac{4m^2_{\mu}}{s}+\left( 1-\frac{4m^2_{\mu}}{s} \right)\cos^2x \right)$$ Below are two graphs of the PDF by changing the range on the y axis.

I'm not at all sure that the result makes sense.

In fact it would imply that there are no privileged directions. The variability of the cross section is practically independent of the angle and therefore is at the end constant.

I do not think it could be that way but if it is not so I do not know where my mistake is.

  • 1
    $\begingroup$ You are computing the density function using the CoM, from that perspective all angles will look the same since momentum conservation means $p_{in}=0$, that is no direction is preferred. $\endgroup$ – ohneVal Feb 7 at 9:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.