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Homework Help: Statistical mechanics of neutrinos

  1. Aug 27, 2011 #1
    1. The problem statement, all variables and given/known data

    Neutrinos are massless spin-1/2 particles (ignore their tiny finite masses).
    There are 6 types of neutrinos (3 flavours of neutrinos and 3 of anti-neutrinos),
    and each has just one possible polarization state. In the early universe neutrinos
    and antineutrinos were in thermal equilibrium with zero chemical potential,
    and filled the universe as a kind of background radiation. Derive a formula for
    the total energy density (energy per unit volume) of the neutrino-antineutrino
    background radiation when it was at temperature T . Leave your answer as a
    constant times a definite integral over a dimensionless variable. Don’t forget to
    determine the value of this constant.

    2. Relevant equations

    Fermi-Dirac Distribution with zero chemical potential.

    [tex] n_{r}(\epsilon) = \frac{1}{e^{\beta \epsilon_{r}}+1} [/tex]

    3. The attempt at a solution

    Just a quick issues about statistical mechanis in general. If I had a gas of one sort of neutrino at a temperature T each particle would occupy a different one of the ppssible states and the distribution of this would be given by the Fermi-Dirac Distribution shown above.

    A gas of a different neutrino flavour should (I think) have a slightly different array of possible states and (though still described by FD stats) the distribution over these states should would be different than the one above.

    However, if I mix these gases or 6 of them I'm not sure how I should combine them. Does it count as one distribution function with the energy being a sum of the energies of the components

    [tex]\epsilon= \epsilon_{1} + \epsilon_{2} + .....[/tex] etc.

    Or am I just thinking about the whole thing wrong.

    Thanks for your help.
  2. jcsd
  3. Aug 28, 2011 #2
    I asked a similar question about how energy was divided in the early universe between photons and electrons,


    Near the end of the post there is a factor of 7/8 that might come into play? Your answer should be of order an 3*8/7 times the energy density of photons?
  4. Aug 28, 2011 #3
    Thanks for that. I had hoped to get away with treating the particles non-relativistically for a 1st order approx and from the nature of the test. Maybe I should rethink that.

    However, I still need to seal up a weakness in my statistical mechanics. That is, how does the Fermi-dirac distribution work for a gas of different types of Fermions. Most textbook deal with the usual example of a one-component Fermi gas e.g electrons. What if it was a gas of electrons and neutrons or this one of different neutrinos?

    In a similar vein, if I wanted to count the number of states by integrating over the Fermi-sphere would I need to do it for each neutrino (which would have the same answer) and then add them?

    Or maybe I'm overcomplicating this. I have than tendency :)
  5. Aug 28, 2011 #4


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    The exclusion principle doesn't operate between different species of fermions, nor between fermions of opposite polarization. Sure, add 6 identical Fermi spheres.
  6. Aug 28, 2011 #5
    So integrating over the Fermi sphere for one type of neutrino the number of states N1.

    [tex] N_{1} = \int^{p}_{0} (\frac{L}{2\pi\hbar})^{3} 4\pi p^{2}.dp = \frac{V}{6\pi^{2}\hbar^{3}} p^{3}[/tex]

    I multiply by six for the six different neutrinos and sub in [tex] \frac{p^{2}}{2m} = \epsilon [/tex]

    Where (is this correct?)

    [tex] \epsilon = \epsilon_{1} + \epsilon_{2} + \epsilon_{3} + \epsilon_{4} + \epsilon_{5} + \epsilon_{6} [/tex]

    We should, after some hopefully correct tidying end up with:

    [tex] N = \frac{V}{\pi^{2}\hbar^{3}} (2m\epsilon)^{3/2} [/tex]

    Is this right?

    If so I can just differentiate with respect to E to get the density of states and work from there to get the energy density.
  7. Aug 29, 2011 #6


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    It was really sloppy of me to echo your 'Fermi sphere' terminology. Fermi spheres are for degenerate cases, like white dwarfs. It's a fermi gas, like a black body of photons. Calculate it like that. You just need to change the exp(E/T-1) to exp(E/T+1). There's a brief discussion of issues like this in Kolb and Turner's The Early Universe. Section 3.3.
  8. Aug 29, 2011 #7
    I'm sorry but could you elaborate a little on what you mean by "Fermi spheres are for degenerate cases." I'm not sure I understand.

    My plane was to follow the black body method, but the derivation I'm familiar with involves an integral over the density of states which is what I was hoping to drive from the Fermi spheres. However, maybe I should stick to an approach based purely on the distribution or partition functions?
  9. Aug 29, 2011 #8


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    A Fermi sphere is where the density of states is 1 inside the sphere and 0 outside. It's a cold (hence degenerate) gas. Follow the black body method. Just substitute Fermi statistics for Bose statistics. That's where the 7/8 factor comes from.
  10. Aug 29, 2011 #9
    So in general an integral over the Fermi sphere will only work in cases were things are cold enough (i.e. temperatures corresponding to below the Fermi energy). Otherwise just work out the average energy using the distribution function.
  11. Aug 29, 2011 #10


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    Yes. I think so. Just integrate over phase space using the appropriate statistics.
  12. Aug 30, 2011 #11
    Okay, still abit confused. So I read throughthe derivation of the black body a couple from a couple of places to refresh it in my head. Every version seems to involve an integral with the density of states (or an integral in k-space to find the number of modes/states).

    However, the only way I can think to find the density of states/number of modes is to do an integral over k-space as I described. If I do this between the limit of 0 and infinity (or an arbitrary wavevector k )(rather than the Fermi wavevector [itex]k_F[/itex]). I'm not technically integrating within the Fermi sphere but a continuous range of k-states. That would apply to this situation correct?
  13. Aug 30, 2011 #12


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    Yes, you don't cut off the momentum integration at some energy. It goes to infinity.
  14. Aug 30, 2011 #13
    Doesn't that leave a divergent integral?

    [tex] N_{1} = \int^{\infty}_{0} (\frac{L}{2\pi\hbar})^{3} 4\pi p^{2}.dp [/tex]
  15. Aug 30, 2011 #14


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    Sure, that's divergent. Are you still doing the Fermi sphere thing? That's wrong. I thought you were going to do a blackbody type calculation. Where's the temperature T?
  16. Aug 30, 2011 #15
    Most of the blackbody calculation methods I'm familar with require a (photon) density of states. I thought by integrating over the momenta I would get the number of states which I could differentiate with respect to energy to get the density of states then combine that with the Fermi-Dirac distribution expression for the energy.


    [tex] N = \int^{p}_{0} (\frac{L}{2\pi\hbar})^{3} 4\pi p^{2}.dp;[/tex]

    [tex]\frac{dN}{d\epsilon} = g(\epsilon)[/tex]

    [tex]E = \int g(\epsilon) \frac{\epsilon}{e^{\frac{\epsilon}{kT}}+1} .d\epsilon [/tex]

    Though now that I think about it. Perhaps if I sub in the expression for the energy of massless particles [itex] \epsilon = pc [/itex] into the first expression to give an integral over energy I can find [itex] g(\epsilon) [/itex] by equation to

    [tex] \int g(\epsilon).d\epsilon [/tex]

    Apologies if I was unclear before. I checked through some textbook derivations of the blackbody and they often used the photon density of states or derived the number of modes in the cavity by use of a k-space integral.

    Thank you very much for your patience in helping me out, Dick.
  17. Aug 31, 2011 #16


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    You are making this too complicated, I think. Try starting from (3.47) in Kolb and Turner:

    [tex]\rho = \frac{g}{(2 \pi)^3} \int E(\vec{p}) f(\vec{p}) d^3 p[/tex]

    rho is the energy density, g is the number of degrees of freedom (6 for you). E(p) is energy as a function of p and f(p) is the phase space occupancy (Bose or Fermi). That's straightforward, isn't it?
    Last edited: Aug 31, 2011
  18. Aug 31, 2011 #17
    I see it now.

    It's just a matter of turning

    [itex] E = \sum \epsilon(p)f(p) [/itex]


    [itex] E = \int \epsilon(p)f(p).dp [/itex]

    where we've made the realisation that that these energies momenta can be treated as continuous.

    Yes, I'm afraid making things too complicated is a worrying habit of mine. Thanks so much for all your help and sorry for the confusion.
  19. Aug 31, 2011 #18


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    Exactly! You're welcome!
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