Weak interaction cross-section

[Orbital]

Homework Statement

Cross-sections for weak interactions at an energy E increase with E as $E^2$. Show that the rate of weak interactions in the early universe depends on the temperature T as $\sigma_{wk} \propto T^5$2. The attempt at a solution
The only formula I can find is
$$\sigma_{wk} = g_{wk}^2 \left[ \frac{k_B T}{(\bar{h} c)^2} \right]^2$$
where $g_{wk} \approx 1.4 \times 10^{49}$ erg cm^3 is the weak interaction coupling constant.
I have no clue on how to proceed.

 

[Dick]

The 'rate' depends upon more than the cross section. Find a formula for rate.

 

[Orbital]

I cannot find any such formula...

 

[Dick]

Ok. Then we'll have to make one up. What other factors besides cross section would affect the interaction rate?

 

[Orbital]

The number density and the speed of the particles i guess.

 

[Dick]

Hey, you're pretty good at this! Would you say proportional to both and the cross section (at least roughly)?

 

[Orbital]

I'm not sure... I found a formula

$$\tau_{coll} = \frac{1}{n \sigma_{wk} c}$$

where

$$n = 0.2 \left( \frac{k_BT}{\bar{h} c} \right)^3$$

is the number density. This would give

$$\tau_{coll} \propto \frac{1}{T^5}$$

but this is not it, is it?

 

[Dick]

That is it! tau is the inverse of the rate, right? Under what conditions can you use that formula and are they compatible with the description 'early universe'? What happened to our velocity dependence?

 

[Orbital]

Maybe the particles have the speed of light? But what about the energy in the problem statement?

 

[Dick]

Yes, it assumes the particles are relativistic. I'll throw the other question back to you. What's the relation between T and E?

 

[Orbital]

They are essentially the same?

 

[Dick]

Up to a constant (Boltzmann's to be specific), sure. If you understand why number density is proportional to T^3 then I think you have the whole thing (hint: it also assumes the universe is radiation dominated).

 

[Dick]

You know, I think number density is proportional to T^3 regardless of domination.

 

[Orbital]

Uhm, has it got something to do with 3d space?

 

[Dick]

It also has to do with red-shift. E is proportional to 1/a, where a is the scale factor.