Statistical physics

  • #1
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Homework Statement


The result

[tex]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[/tex]

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 [itex]T_{0 \nu}[/itex] is

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


Homework Equations


Bose-Einstein distribution:
[tex]f_{BE}(E) = \frac{1}{e^{E/kT}-1}[/tex]

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

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

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



The Attempt at a Solution


First of all, I guess that [itex]T_{0r}[/itex] in the second formula should be [itex]T_{0 \nu}[/itex].
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 [itex]n_{BE}(E) = g_{BE}(E)f_{BE}(E)[/itex].
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

[tex]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}[/tex]

so this cannot be the correct procedure?
 
Last edited:

Answers and Replies

  • #2
dextercioby
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I don't think the integral you've written is correct. Post the derivation.
 
  • #3
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I used

[tex]\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}[/tex]

and made a substitution

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

so

[tex]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}[/tex]

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
mjsd
Homework Helper
726
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Homework Statement


The result

[tex]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[/tex]
from this result it appears to me that [tex]2\pi T_{0r}=T[/tex]

note that to get the given answer for the fermion case, your integral should be in the form
[tex]\displaystyle{\int_0^{\infty} \frac{x^2}{e^x+1}\; dx}[/tex]
 
  • #5
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Sorry I forgot the index. I know it should be in that form, I just can get it to be in that form. ;)
 
  • #6
mjsd
Homework Helper
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should check your density of states expression again.. by the way answer doesn't have "m".
 
  • #7
281
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But [itex]m=E/c^2[/itex], right?
 
  • #8
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and the density of states is written here in my book... Not the same book as the problem is in though.
 
  • #9
dextercioby
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I used

[tex]\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}[/tex]

and made a substitution

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

so

[tex]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}[/tex]

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

[tex] \frac{E}{kT} = x [/tex]

and the notation

[tex] -\frac{E_{F}}{kT}\equiv \zeta [/tex]

and tried to solve the resulting integral either by means of special functions or maybe some other tricks.
 
  • #10
mjsd
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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 :smile:
 
  • #11
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And where can I find that one?
 
  • #12
mjsd
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the general form is (in my book)
[tex]\displaymath{\mathcal{D}(E) = 2\frac{4\pi V}{(2\pi\hbar)^3}p^2 \frac{dp}{dE}}[/tex]
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 :smile:
 
  • #13
281
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I think we're working too hard. If

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

what is

[tex]I=\int_0^{\infty} \frac{8 \pi x^2 dx}{e^x+1}[/tex]?
 
  • #14
mjsd
Homework Helper
726
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firstly,
[tex]\int_0^{\infty} \frac{8 \pi x^2 dx}{e^x-1} \neq 2 \frac{\zeta(3)}{\pi^2} [/tex]
it should be [tex]16\pi \zeta(3)=8\pi\times 2\zeta(3)[/tex]
anyway for your
[tex]I=12\pi \zeta(3) = 8\pi \times \frac{3}{2} \zeta(3)[/tex]
 
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  • #15
281
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I do not understand you.
It's given in my problem that

[tex]\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[/tex]
 
  • #16
dextercioby
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I do not understand you.
It's given in my problem that

[tex]\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[/tex]
I don't think so.

[tex] \int_{0}^{\infty} \frac{x^{2}}{e^{x}-1}{}dx =2\zeta(3) [/tex]

beyond any doubt.
 
  • #17
281
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I see that now. The book says that

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

which cannot be correct. I'll skip this problem. Thanks for your help.
 
  • #18
mjsd
Homework Helper
726
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if it was just an execise of integration .. you may as well leave it... for you would probably use a table anyway... :smile:
 

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