# Statistical physics

1. Feb 6, 2007

### Logarythmic

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

$$n_{0 \gamma} = \left( \frac{k_BT_{0r}}{hc} \right)^3 \int_0^{\infty} \frac{8 \pi x^2 dx}{e^x-1} = 2 \frac{\zeta(3)}{\pi^2} \left( \frac{k_BT_{0r}}{hc} \right)^3$$

is obtained for photons by integrating over the Planck distribution appropriate for bosons. In the case of neutrinos (or other fermions), show that the number-density in thermal equilibrium at a temperature $T_{0 \nu}$ is

$$n_{0 \nu} = 3 \frac{\zeta(3)}{2 \pi^2} \left( \frac{k_BT_{0r}}{hc} \right)^3$$

2. Relevant equations
Bose-Einstein distribution:
$$f_{BE}(E) = \frac{1}{e^{E/kT}-1}$$

Fermi-Dirac distribution:
$$f_{FD}(E) = \frac{1}{e^{(E-E_F)/kT}+1}$$

Density of states for bosons:
$$g_{BE} = \frac{8 \pi VE^2}{c^3h^3}$$

Density of states for fermions:
$$g_{FD}(E) = \frac{4 \pi (2m)^{3/2}}{h^3} VE^{1/2}$$

3. The attempt at a solution
First of all, I guess that $T_{0r}$ in the second formula should be $T_{0 \nu}$.
I guess that in the first case, for bosons, the Bose-Einstein distribution is used together with the density of states for bosons to give the correct formula by also using $n_{BE}(E) = g_{BE}(E)f_{BE}(E)$.
But now, in the case of fermions, I have tried to use the Fermi-Dirac distribution together with the density of states for fermions, exactly the same procedure as I used for bosons, but this gives me

$$n_{0 \nu} = \frac{4 \pi (2m)^{3/2}}{h^3} \int_0^{\infty} \frac{(xkT+E_F)^{1/2} kT dx}{e^x + 1}$$

so this cannot be the correct procedure?

Last edited: Feb 7, 2007
2. Feb 7, 2007

### dextercioby

I don't think the integral you've written is correct. Post the derivation.

3. Feb 7, 2007

### Logarythmic

I used

$$\frac{g_{FD}(E)f_{FD}(E)}{V} = \int_0^{\infty} \frac{4 \pi (2m)^{3/2}}{h^3}E^{1/2} \frac{dE}{e^{(E-E_F)/kT}+1}$$

$$\frac{E-E_F}{kT} = x \Rightarrow dE = kTdx , E^{1/2} = (xkT +E_F)^{1/2}$$

so

$$n_{0 \nu} = \frac{4 \pi (2m)^{3/2}}{h^3} \int_0^{\infty} \frac{(xkT+E_F)^{1/2} kT dx}{e^x + 1}$$

But thinking in another way, what's the difference in the states for bosons and fermions? Is it just the spin? Cause then the density of states for fermions should just be 2 times that for bosons, right? But that doesn't solve the problem either...

4. Feb 7, 2007

### mjsd

from this result it appears to me that $$2\pi T_{0r}=T$$

note that to get the given answer for the fermion case, your integral should be in the form
$$\displaystyle{\int_0^{\infty} \frac{x^2}{e^x+1}\; dx}$$

5. Feb 7, 2007

### Logarythmic

Sorry I forgot the index. I know it should be in that form, I just can get it to be in that form. ;)

6. Feb 7, 2007

### mjsd

should check your density of states expression again.. by the way answer doesn't have "m".

7. Feb 7, 2007

### Logarythmic

But $m=E/c^2$, right?

8. Feb 7, 2007

### Logarythmic

and the density of states is written here in my book... Not the same book as the problem is in though.

9. Feb 7, 2007

### dextercioby

I wouldn't have done that substitution, but

$$\frac{E}{kT} = x$$

and the notation

$$-\frac{E_{F}}{kT}\equiv \zeta$$

and tried to solve the resulting integral either by means of special functions or maybe some other tricks.

10. Feb 7, 2007

### mjsd

from your given stuffs, there is no way you get what you want in your formulation

oh... i think your density of states function is the one used for the non-relativistic case... try the ultra-relativistic version

11. Feb 7, 2007

### Logarythmic

And where can I find that one?

12. Feb 7, 2007

### mjsd

the general form is (in my book)
$$\displaymath{\mathcal{D}(E) = 2\frac{4\pi V}{(2\pi\hbar)^3}p^2 \frac{dp}{dE}}$$
where p is momentum, E is energy so ... in ultra-relativistic case E=pc.. you should recover your expression if you use E=p^2/2m

and you will find that the resultant density is proportional to E^2
... mmm... given the answer you can always reverse engineer... hope this works

13. Feb 7, 2007

### Logarythmic

I think we're working too hard. If

$$\int_0^{\infty} \frac{8 \pi x^2 dx}{e^x-1} = 2 \frac{\zeta(3)}{\pi^2}$$

what is

$$I=\int_0^{\infty} \frac{8 \pi x^2 dx}{e^x+1}$$?

14. Feb 7, 2007

### mjsd

firstly,
$$\int_0^{\infty} \frac{8 \pi x^2 dx}{e^x-1} \neq 2 \frac{\zeta(3)}{\pi^2}$$
it should be $$16\pi \zeta(3)=8\pi\times 2\zeta(3)$$
anyway for your
$$I=12\pi \zeta(3) = 8\pi \times \frac{3}{2} \zeta(3)$$

Last edited: Feb 7, 2007
15. Feb 8, 2007

### Logarythmic

I do not understand you.
It's given in my problem that

$$\left( \frac{k_BT_{0r}}{hc} \right)^3 \int_0^{\infty} \frac{8 \pi x^2 dx}{e^x-1} = 2 \frac{\zeta(3)}{\pi^2} \left( \frac{k_BT_{0r}}{hc} \right)^3$$

16. Feb 8, 2007

### dextercioby

I don't think so.

$$\int_{0}^{\infty} \frac{x^{2}}{e^{x}-1}{}dx =2\zeta(3)$$

beyond any doubt.

17. Feb 8, 2007

### Logarythmic

I see that now. The book says that

$$2 \frac{\zeta(3)}{\pi^2} \left( \frac{k_BT_{0r}}{hc} \right)^3 = 420 cm^{-3}$$

which cannot be correct. I'll skip this problem. Thanks for your help.

18. Feb 8, 2007

### mjsd

if it was just an execise of integration .. you may as well leave it... for you would probably use a table anyway...