# Statistical Physics Homework: Neutrinos in Thermal Equilibrium

• Logarythmic
In summary, the result for the number-density of photons in thermal equilibrium at a temperature T_{0 \gamma} is obtained by integrating over the Planck distribution appropriate for bosons, while the result for the number-density of neutrinos (or other fermions) is obtained by integrating over the Fermi-Dirac distribution appropriate for fermions. The density of states for bosons is derived as g_{BE} = \frac{8 \pi VE^2}{c^3h^3}, while the density of states for fermions is derived as g_{FD}(E) = \frac{4 \pi (2m)^{3/2}}{h^3} VE^{1/2}. The integral for the

## Homework Statement

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

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

## Homework 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:
$$g_{FD}(E) = \frac{4 \pi (2m)^{3/2}}{h^3} VE^{1/2}$$

## 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?

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

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...

Logarythmic said:

## Homework Statement

The result

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

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
$$\displaystyle{\int_0^{\infty} \frac{x^2}{e^x+1}\; dx}$$

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

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

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

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

Logarythmic said:
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...

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.

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

And where can I find that one?

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

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

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

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
$$I=12\pi \zeta(3) = 8\pi \times \frac{3}{2} \zeta(3)$$

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

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

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

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

I don't think so.

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

beyond any doubt.

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.

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

## 1. What is statistical physics and how does it relate to neutrinos in thermal equilibrium?

Statistical physics is the branch of physics that studies the behavior of large systems of particles. In the context of neutrinos in thermal equilibrium, statistical physics is used to understand the statistical distribution of neutrinos at different energies and temperatures, as well as their interactions with other particles in the system.

## 2. What is thermal equilibrium and why is it important in studying neutrinos?

Thermal equilibrium is a state where the temperature of a system is uniform and there is no net flow of heat. In the context of neutrinos, this means that the neutrinos are in equilibrium with their surroundings, and their energy distribution is determined by the temperature of the system. This is important because it allows us to make predictions about the behavior of neutrinos in various environments, such as the early universe or a supernova.

## 3. What is the role of the neutrino's mass in statistical physics?

The mass of the neutrino is an important factor in statistical physics because it affects the way neutrinos interact with other particles in the system. Neutrinos with different masses have different probabilities of interacting with other particles, and this can have significant implications for the overall behavior of the system.

## 4. How do you calculate the equilibrium distribution of neutrinos in a system?

The equilibrium distribution of neutrinos can be calculated using statistical mechanics, specifically the Fermi-Dirac distribution. This distribution takes into account the temperature and the mass of the neutrino to determine the probability of finding a neutrino at a given energy level in the system.

## 5. What are some applications of studying neutrinos in thermal equilibrium?

Studying neutrinos in thermal equilibrium has many applications in astrophysics, cosmology, and particle physics. For example, it can help us understand the properties of the early universe and the formation of structures such as galaxies. It can also provide insights into the nature of dark matter and the behavior of neutrinos in extreme environments, such as supernovae and black holes.