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A Hadronic era equation of state

  1. Apr 6, 2016 #1
    Hello,

    I was wondering : going back in time up to the time when the universe is hot enough to create nucleons particle antiparticle pairs, when we are not much above this temperature threshold, the nucleons can not be considered ultrarelativistic and if they carry a fraction of the energy density not much smaller than the photons (correct?) and other lighter particles which by contrast are really ultrarelativistic then i would expect that globally the approximation of the ultrarelativistic fluid is not anymore valid (even though it is valid for a colder universe below the energy threshold for pair creation, that's strange!)... so the equation of state would not be rho=3p but somewhere in between rho=3p and p<<rho so that the densities would not decrease as 1/R⁴ but with a law momentarily somewhere in between 1/R³ and 1/R⁴ .. is that correct?, if not what did i miss? thank you in advance.
    I found this strange that i read nowhere before that the equation of state could be significantly modified at such high temperatures ...
     
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  3. Apr 6, 2016 #2

    Chalnoth

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    It's even a bit more complicated than that, as when each fast-interacting particle (e.g. protons, electrons) became non-relativistic, then large numbers of particle/anti-particle pairs annihilated with one another, meaning that the energy that had been in matter was dumped into the radiation fields.

    So yeah, there's quite a lot of fairly complicated physics that goes into calculating precisely what happened back then.
     
  4. Apr 6, 2016 #3

    PeterDonis

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    At the temperature you describe, they don't. Since the temperature is close to the threshold, it is about as easy for a nucleon-antinucleon pair to be destroyed as to be created, i.e., they annihilate each other almost as soon as they are created. That means the average density of nucleons is small compared to the average density of photons. So most of the energy density will be carried by photons.
     
  5. Apr 6, 2016 #4
    OK and then at higher temperatures the fraction of pairs is greater but at the same time these are more energetic ... so it's not easy to figure out at which temperature do we have the most important deviation from a purely radiative rho = 3p equation of state and if the deviation is important. Yet you agree that there is one such deviation each time one is somewhat above a threshold temperature for a given species pair creation...
     
  6. Apr 6, 2016 #5

    PeterDonis

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    In principle, you could model this by using whatever data we have on reaction rates for pair creation and annihilation as a function of temperature to make a prediction for the energy density of nucleons and photons as a function of temperature. But I don't know if anyone has done this.

    There will be a "deviation" in principle whenever any particle species with nonzero rest mass is present, since any such species will have ##\rho > 3p## by some margin, however small. The question is how small a margin, and how much difference it makes to the overall equation of state taking all species into account.
     
  7. Mar 1, 2017 #6
    suppose a kind of scalar theory of gravity : Ricci = 8 pi G (trace of the energy momentum tensor)
    in such theory it's difficult to understand and predict how the source rho-3p will behave at high energy ... even though rho and p are expected to behave as 1/a⁴ (a is the scale factor) i would say that the difference rho-3p will tend to decrease also as the scale factor decreases , hence may be a power low greater than -4 and increasing as the scale factor gets smaller ... this is also the kind of source term for which i expected reaching and somewhat going beyond the threshold for a given specie creation would have the most important effect (rho-3p much more sensitive to it than rho or p alone) ...
    unless there is something i did not correctly figure out ... ? does what i 'm saying now seem reasonable ? i'm not sure because even in standard gravity cosmology, the above scalar equation is as well satisfied and, to my knowledge, nobody has ever been wondering about the specific behaviour of rho-3p as a function of the scale factor
     
    Last edited: Mar 1, 2017
  8. Mar 1, 2017 #7
    Let me be more quantitative to help somebody show me (hopefully) where is my mistake.
    The following (standard cosmology) equation must be satisfied by the scale factor R:
    3(R''/R +R'²/R²) = 4πG (ρ-3p)
    where ' are time derivatives.
    suppose the rhs behaves as 1/Rβ and R itself as tα

    Then the above equation becomes

    3 α (2α-1) .1/t² ∝ 4 π G 1/tαβ

    α=1/2 for the radiative era makes the lhs vanish but the rhs cannot rigourously vanish because ρ-3p =0 is only approximately verified. So α ~ 1/2 rather than = 1/2. But then αβ~2 is required which for α ~ 1/2 implies β~4.

    Of course it is well known that both rho and p behave as 1/R4 but now this is also required for ρ-3p to satisfy the cosmological equation. But how can this be in a fluid that is not only made of photons.
    for massive particles of mass m in a fluid element locally inertial coordinate system
    ρ=T00= n E
    ρ0= n m
    3p= n P²/E

    where n is the particle number density, E is the mean energy per particle and P the mean momentum per particle
    thus rho-3p = n m²/E = (nm)²/nE =ρ0²/ρ for ultrarelativistic particles. SO where did i fail because i can hardly believe that the last expression would behave as 1/R4 unless so does ρ0 which is a rest mass density!?!?
     
    Last edited: Mar 1, 2017
  9. Mar 1, 2017 #8

    PeterDonis

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    In the general case, it doesn't, because there will be a mixture of species with different ##\beta##. So you can't just model it as one value for ##\beta##. You have to take into account all of the species present.

    This will also not be true in the general case, because different species will lead to different values for ##\alpha## (and in fact at least one species, dark energy, has ##R## going like ##e^t##, which corresponds to an infinite sum of terms of the form ##t^\alpha##).

    By starting with two assumptions that are invalid if there is more than one species present. See above.
     
  10. Mar 1, 2017 #9
    the assumption here are not meant to reproduce reality which as you say is a mix ...
    The question is that even if there was one unique specie i 'm disturbed by the fact that ρ0 should vary like ρ as 1/R⁴
    while i was expecting it to vary as 1/R³ naively because it's a rest mass density.

    Probably it's this expectation that after all is naive ... it must be wrong since if ρ0 ~ 1/R³ even for ultrarelativistic particles that have ρ ~1/R⁴ then rho-3p = ρ0²/ρ would vary as 1/R² , and the scale factor evolution would be R(t)~t instead of the well known R(t)~t1/2 in the radiative era...but R(t)~t is strongly excluded by BBN and CMB data ...

    dark energy is also negligible in the radiative era, moreover even if there is a mix of many species , provided these are all ultrarelativistic and someof them at least are not massless , i would expect the above result to be true eg R(t)~t following the above too naive argument. Probably the ρ0 i isolated in my computation cannot be considered to vary in the same way in the radiative era as in the cold era ...
     
    Last edited: Mar 1, 2017
  11. Mar 1, 2017 #10

    Chalnoth

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    ##\rho - 3p = 0## identically for radiation.

    If you are suggesting a universe with mostly radiation, but some other stuff as well, then the ##R = t^\alpha## assumption doesn't work. I don't think there is an analytical solution for ##R(t)## for the case with multiple types of matter/energy.
     
  12. Mar 1, 2017 #11
    but the radiative era is known to be a R(t) ~ t1/2 evolution era yet it is by no mean a purely radiation (photons or massless particles only) era. There are plenty of other ultrarelativistic massive species which even if these are a minority make rho - 3p non zero. If my above argument was correct those particles alone would drive the evolution of R(t) to be ~ t , rather than t¹/2...this is what i meant ... right ?
     
  13. Mar 1, 2017 #12

    PeterDonis

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    ##1/R^3## for rest mass assumes that the matter is cold, i.e., zero pressure, i.e., non-relativistic. It does not apply to ultrarelativistic matter; for that you would have ##\approx 1/R^4##, as you derived. In other words, ultrarelativistic matter behaves like radiation, at least to a good approximation.
     
  14. Mar 1, 2017 #13

    Chalnoth

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    That's why it's only approximately ##R(t) \propto t^{1/2}##. During this era, the energy density in radiation was so vastly higher than that in non-relativistic matter that this approximation worked pretty well.
     
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