# Scattering Cross Section - At high energies

## 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 $\frac{ e^+ + e^- \rightarrow \mu^+ + \mu^- }{e^+ + e^- \rightarrow e^+ + e^- }$?

## The Attempt at a Solution

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

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

$$\frac{d\sigma}{d\Omega} = \frac{1}{2\nu_e} \Gamma = \frac{1}{2\nu_e} 2\pi |M_{fi}|^2 \frac{dN}{dE_f}$$

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

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

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

Last edited:

Related Advanced Physics Homework Help News on Phys.org
At high energies, $v_\mu \approx c$, so $\frac{dp_\mu}{dE_\mu} = \frac{1}{2} \frac{E_\mu}{p_\mu} = \frac{1}{2c} = \frac{1}{2}$, letting c=1. This gives:

$$d\sigma = \frac{1}{2\nu_e} 2\pi \left( \frac{g_{EM}^2}{4E_e^2} \right)^2 \frac{p^2}{(2\pi)^3} \frac{1}{2} d\Omega$$

$$\sigma = \frac{1}{2\nu_e} 2\pi \left( \frac{g_{EM}^2}{4E_e^2} \right)^2 \frac{p^2}{(2\pi)^3} \frac{1}{2} 4\pi$$

$$= \frac{1}{64} \frac{1}{\nu_e} \frac{1}{\pi} \frac{g_{EM}^2}{E_e^4} p_\mu^2$$

Using $\frac{g_{EM}^2}{4\pi} = \alpha$,

$$\sigma = \frac{\pi}{4} \frac{1}{\nu_e} \frac{p_\mu^2}{E_e^4} \alpha^2$$

For highly relativistic case, $p_\mu \approx E_\mu = E_e$ and $\nu_e \approx c = 1$.

$$\sigma = \frac{\pi}{4} \frac{1}{E^2} = \pi \frac{\alpha^2}{(2E)^2} = \pi \frac{\alpha^2}{s^2}$$
where $s=2E$ is the centre of mass energy.

Last edited:
Since the scattering cross section only depends on:

a) Centre of mass energy
b) $g_{EM}$

Wouldn't both $\rightarrow 1$ at high energies ($p >> m$)?

ChrisVer
Gold Member
Wouldn't both →1\rightarrow 1 at high energies (p>>mp >> m)?
I find the 2nd question a little tricky...
For the first you dealt only for one possible channel : the s-channel ..
For the second case however, the electrons can also "not interact" at all... apart from annihilating and giving again e^+/- pair..

I find the 2nd question a little tricky...
For the first you dealt only for one possible channel : the s-channel ..
For the second case however, the electrons can also "not interact" at all... apart from annihilating and giving again e^+/- pair..
I think I got it.

Part (a)
The cross section is proportional to $|M_{fi}|^2$.

First process is given by:

$$e^+ + e^- \rightarrow \gamma^* \rightarrow \mu^+ + \mu^-$$

The propagator factor for this process is $\frac{1}{P \cdot P - m_\gamma^2} = \frac{1}{P \cdot P}$ since $m_\gamma = 0$.

Second process is given by:

$$e^+ + e^- \rightarrow Z^0 \rightarrow \tau^+ + \tau^-$$

The propagator factor for this process is $\frac{1}{P \cdot P - m_Z^2}$.

At high energies, this propagator factor $\approx \frac{1}{P \cdot P}$

Thus they approach 1 at high energies.

Part (b)

Wouldn't this always be true? Given that the propagator factor for both are equal to $\frac{1}{P \cdot P}$.

ChrisVer
Gold Member
The part a is clear for you I think... Even for photon interactions alone, the ratio you give approaches 1 for high energies...it can also be done for the Z0.

For part b, you don't have to think of propagators alone...
you can have:
$e^- e^+ \rightarrow \gamma^* \rightarrow e^- e^+$ (s-channel)
as well as
$e^- e^+ \rightarrow e^- e^+$ ...by t-channel...they don't have to annihilate to photon and produce the pairs again as you have to do for the case of muon products.

ChrisVer
Gold Member

Their vertex factors are equal, since they the muons and electrons have equal charge.

But what about their propagator factors?

ChrisVer
Gold Member
For high energies, the propagators of the weak and electromagnetic interaction are the same. It goes as $\frac{1}{s}$
The difference comes by the 2nd diagram that is possible for the electron/positron products interaction and doesn't exist for the muons.

For high energies, the propagators of the weak and electromagnetic interaction are the same. It goes as $\frac{1}{s}$
The difference comes by the 2nd diagram that is possible for the electron/positron products interaction.
Which process dominates? Which is the "s-channel" and which is the "t-channel"?

ChrisVer
Gold Member
Which process dominates? Which is the "s-channel" and which is the "t-channel"?
In my figure the diagram with muons and the same one with electrons are the s-channel processes.
The extra diagram for the $e^- \rightarrow e^-$ and $e^+ \rightarrow e^+$ which exists only for the $e^-e^+ \rightarrow e^+ e^-$ process is the t-channel....

I don't understand what you mean by "dominates". I am explainging why:
$\frac{e^+e^- \rightarrow \mu^+\mu^-}{e^+e^- \rightarrow e^+e^-}$ shouldn't be 1.

In my figure the diagram with muons and the same one with electrons are the s-channel processes.
The extra diagram for the $e^- \rightarrow e^-$ and $e^+ \rightarrow e^+$ which exists only for the $e^-e^+ \rightarrow e^+ e^-$ process is the t-channel....

I don't understand what you mean by "dominates". I am explainging why:
$\frac{e^+e^- \rightarrow \mu^+\mu^-}{e^+e^- \rightarrow e^+e^-}$ shouldn't be 1.
All three are in the same picture.

By "dominate" I mean which is most frequent/most likely?

ChrisVer
Gold Member

Got it. For electron-positron scattering, is the s-channel more likely or the t-channel more likely?

ChrisVer
Gold Member
Both are the same (I think), since the interaction is symmetric under the interchange $t \leftrightarrow s$
But nevertheless, how can this "help" you?