# Scattering Cross Section - At high energies

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1. Feb 24, 2015

### unscientific

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. The attempt at a solution

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: Feb 24, 2015
2. Feb 24, 2015

### unscientific

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: Feb 24, 2015
3. Feb 24, 2015

### unscientific

Since the scattering cross section only depends on:

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

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

4. Feb 28, 2015

### ChrisVer

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..

5. Mar 16, 2015

### unscientific

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 $\frac{1}{P \cdot P}$.

6. Mar 16, 2015

### ChrisVer

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.

7. Mar 16, 2015

### ChrisVer

8. Mar 16, 2015

### unscientific

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

But what about their propagator factors?

9. Mar 16, 2015

### ChrisVer

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.

10. Mar 16, 2015

### unscientific

Which process dominates? Which is the "s-channel" and which is the "t-channel"?

11. Mar 16, 2015

### ChrisVer

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.

12. Mar 16, 2015

### unscientific

All three are in the same picture.

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

13. Mar 16, 2015

### ChrisVer

14. Mar 16, 2015

### unscientific

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

15. Mar 16, 2015

### ChrisVer

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