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Relativistic Degrees of freedom g(T)

  1. Nov 29, 2014 #1

    ChrisVer

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    Hi, I have one question,
    The rel. dofs are:
    [itex]g_* (T) = \sum_{i=bosons} g_i \Big( \frac{T_i}{T} \Big)^4 + \frac{7}{8} \sum_{j=fermions} g_j \Big( \frac{T_j}{T} \Big)^4 [/itex]
    Before the neutrinos decoupling, the relativistic degrees of freedom are:
    [itex]g_*=g_\gamma + \frac{7}{8} (g_e + g_\nu)= 10.75 [/itex]
    This is clear to me...
    Now what are the relativistic degrees of freedom before and after the electron annihilation?
    I think that before we have thermal equilibrium between photons and electrons, and thus their temperatures are the same... The neutrinos are still relativistic but their temperature is not the same as the rest of the photons, so:
    [itex]g_{*}(T)= g_\gamma + \frac{7}{8} g_e + \frac{7}{8} g_\nu \Big( \frac{T_\nu}{T} \Big)^4 = 5.5 + 5.25 \Big( \frac{T_\nu}{T} \Big)^4[/itex]

    However I haven't seen people using this expression... They instead take the case: neutrino decoupling- after electron decoupling and find the result:
    [itex]g_{*}(T)= g_\gamma + \frac{7}{8} g_\nu \Big( \frac{T_\nu}{T} \Big)^4 = 2 + 5.25 \Big( \frac{T_\nu}{T} \Big)^4 [/itex]

    (Check Equations 14 to 15 : http://www.helsinki.fi/~hkurkisu/cosmology/Cosmo6.pdf ... he may have the entropy but nothing changes between the definitions except for the powers of the temperature)
    Any idea?

    I think my problem lies in when the neutrinos decouple... they say that they decouple at [itex]T \sim 1 ~MeV[/itex] - isn't that the same decoupling temperature for electrons since the annihilation channel is then suppressed by the photons' energy?
     
    Last edited: Nov 29, 2014
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  3. Nov 29, 2014 #2

    Chalnoth

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    Electron decoupling doesn't occur when electrons become non-relativistic. It occurs when they combine with protons to form neutral atoms. Usually called recombination (this was when the CMB was emitted).
     
  4. Nov 29, 2014 #3

    ChrisVer

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    Hm.. sorry I should have written after electron annihilation ....(so electrons can be considered non-relativistic and thus not contributing to g).
    but my problem is the inbetween period, i.e. after neutrino decoupling and before [itex]e^-e^+[/itex] annihilation.
     
  5. Nov 29, 2014 #4

    Chalnoth

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    Okay, I think I see what you're saying.

    The after equation that you have above is not after neutrino decoupling. It's after the electrons become non-relativistic. The argument that they lay out is that once the electrons become non-relativistic, they become a completely negligible component of the entropy.

    Also, your powers appear to be wrong. The temperature should be taken to the third power, not the fourth.
     
  6. Nov 29, 2014 #5

    ChrisVer

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    I am working with [itex]g_*[/itex] of the radiation energy and not [itex]g_{*s}[/itex] of the entropy, and thus the 4th power.

    Let me make a scheme of the periods....

    Before neutrino decoupling ([itex]T_\nu = T_e = T_\gamma [/itex]:
    [itex]g_* = 10.75 [/itex] Eq.1

    After neutrino decoupling and before electron annihilation [itex] T_\nu \ne T_\gamma [/itex]:
    [itex]g_* = 5.5 + 5.25 (\frac{T_\nu}{T})^4[/itex] Eq.2

    After electron annihilation [itex]g_e =0[/itex]:
    [itex]g_* = 2 + 5.25 (\frac{T_\nu}{T})^4[/itex] Eq.3

    Which equation are you refering to?
    I think I found some answer in other literature... They state that even after the neutrino decoupling, the photons and neutrinos still have the same temperature since they scale in the same way , or so I understood...
    http://www.damtp.cam.ac.uk/user/db275/Cosmology/Chapter3.pdf
    (under 3.2.71)
    and they change after electron positron annihilation. So I guess Eq1 and Eq2 are equal.
     
  7. Nov 29, 2014 #6

    Chalnoth

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    Ahh, I see.

    Yes, equation one and equation two are equal (for the most part). I think the point to remember here is that there's some fuzziness about what goes on between neutrino decoupling and electron-positron annihilation. Certainly there is some width to the transition from before to after annihilation, and that transition is going to be pretty complicated. But as long as the electrons are a relativistic gas, yes, equations one and two will be equal.
     
  8. Dec 3, 2014 #7

    ChrisVer

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    I was finally able to solve the problem I had :w just for the record...
    When neutrinos decoupled, they indeed evolved separately from the rest species, but their temperature keeped evolving as [itex]T \sim 1/a[/itex] as the rest of the thermal bath [photons and electrons]. That is a result of their number conservation...
    Then el-pos annihilation transferred heat to the photons [entropy conservation] and not to the decoupled neutrinos and that's why photons and neutrinos don't have equal temperatures after that.
    Of course the width that you mention Chalnoth has some influence on the neutrino temperature as well, since the high energetic neutrinos are still interacting with the electrons at 0.5MeV, something that leads to the effective number of neutrino species not to be taken equal to [itex]N_{eff}=3[/itex] but [itex]N_{eff}=3.046[/itex],

    Do you know any reference where I can find the calculations that give the [itex]N_{eff}=3.046[/itex] ?
     
  9. Dec 3, 2014 #8

    Chronos

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  10. Dec 3, 2014 #9

    ChrisVer

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    he seems to state it rather than proving it...
    From my side I found a book by J.Lesgourgues,G. Mangano, G. Miele and S.Pastor named "Neutrino Cosomology", at around equation 4.21
    I think they are proving that in their book by not using the density on equilibrium of Fermi-Dirac but making an expansion around the equilibrium..I still have to go through their calculations...
     
  11. Dec 3, 2014 #10

    mfb

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    The Planck publications should have useful references.
     
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