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## Homework Statement

Why does ## \frac{ e^+ + e^- \rightarrow \mu^+ + \mu^- }{e^+ + e^- \rightarrow \tau^+ + \tau^- } \rightarrow 1## at high energies?

Would it be the same if it was ## \frac{ e^+ + e^- \rightarrow \mu^+ + \mu^- }{e^+ + e^- \rightarrow e^+ + e^- }##?

## Homework Equations

## The Attempt at a Solution

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For the first reaction, propagator factor for the photon is ##\frac{1}{(2E)^2 - m_\gamma^2} = \frac{1}{4E^2}##. Two vertex factors are ##g_{EM}^2##. Thus ## |M_{fi}|^2 = |\frac{g_{EM}^2}{4E_e^2}|^2##.

Fermi's golden rule is given by:

[tex]\Gamma = 2\pi |M_{fi}|^2 \frac{dN}{dE_f} [/tex]

Differential cross section is number of particles scattered per unit solid angle per unit time divided by flux:

[tex]\frac{d\sigma}{d\Omega} = \frac{1}{2\nu_e} \Gamma = \frac{1}{2\nu_e} 2\pi |M_{fi}|^2 \frac{dN}{dE_f} [/tex]

The density of states is given by: ##dN = \frac{p^2_\mu}{(2\pi)^3} d\Omega dp_\mu## and substituting gives:

[tex] d\sigma = \frac{1}{2\nu_e} 2\pi |M_{fi}|^2 \frac{p^2}{(2\pi)^3} \frac{dp}{dE_0} d\Omega [/tex]

At high energies, ##E_\mu \approx \frac{1}{2} E_0##, so ## \frac{dp_\mu}{dE_0} = \frac{1}{2} \frac{dp_\mu}{dE_\mu} = \frac{E_\mu}{p_\mu}##.

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