Neutrino energy density vs photon energy density












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$begingroup$


So I'm currently following a course in Cosmology and we're covering the densities of different species in the universe right now. Starting from the photon density $rho_{gamma}$ we need to derive the neutrino density. Apart from a different temperature and a factor 7/8 for the integral of the FD distribution I understand this. But for the photons, we have a degeneracy factor of 2 (two spin states). Then the book lists all characteristics for neutrinos (and the degeneracy factor):




  • spin degree

  • has antiparticles

  • has three generations (flavours)


So I would expect the degeneracy factor of neutrinos to be 2 (spin) x 2 (antiparticle) x 3 (flavour) => g = 12, so another factor of 6 in front of the photon density (g= 12 vs 2). But the book and other books all list a degeneracy factor of 2 x 3 = 6 for the neutrinos. So this results in factor 3 in front of the photon density (together with the other described differences). Why is this? Is the antiparticle part of the neutrinos not taken into account? Can someone please help me.






particle-physics cosmology big-bang neutrinos







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    7












    $begingroup$


    So I'm currently following a course in Cosmology and we're covering the densities of different species in the universe right now. Starting from the photon density $rho_{gamma}$ we need to derive the neutrino density. Apart from a different temperature and a factor 7/8 for the integral of the FD distribution I understand this. But for the photons, we have a degeneracy factor of 2 (two spin states). Then the book lists all characteristics for neutrinos (and the degeneracy factor):




    • spin degree

    • has antiparticles

    • has three generations (flavours)


    So I would expect the degeneracy factor of neutrinos to be 2 (spin) x 2 (antiparticle) x 3 (flavour) => g = 12, so another factor of 6 in front of the photon density (g= 12 vs 2). But the book and other books all list a degeneracy factor of 2 x 3 = 6 for the neutrinos. So this results in factor 3 in front of the photon density (together with the other described differences). Why is this? Is the antiparticle part of the neutrinos not taken into account? Can someone please help me.






    particle-physics cosmology big-bang neutrinos







    share|cite|improve this question











    $endgroup$















      7












      7








      7





      $begingroup$


      So I'm currently following a course in Cosmology and we're covering the densities of different species in the universe right now. Starting from the photon density $rho_{gamma}$ we need to derive the neutrino density. Apart from a different temperature and a factor 7/8 for the integral of the FD distribution I understand this. But for the photons, we have a degeneracy factor of 2 (two spin states). Then the book lists all characteristics for neutrinos (and the degeneracy factor):




      • spin degree

      • has antiparticles

      • has three generations (flavours)


      So I would expect the degeneracy factor of neutrinos to be 2 (spin) x 2 (antiparticle) x 3 (flavour) => g = 12, so another factor of 6 in front of the photon density (g= 12 vs 2). But the book and other books all list a degeneracy factor of 2 x 3 = 6 for the neutrinos. So this results in factor 3 in front of the photon density (together with the other described differences). Why is this? Is the antiparticle part of the neutrinos not taken into account? Can someone please help me.






      particle-physics cosmology big-bang neutrinos







      share|cite|improve this question











      $endgroup$




      So I'm currently following a course in Cosmology and we're covering the densities of different species in the universe right now. Starting from the photon density $rho_{gamma}$ we need to derive the neutrino density. Apart from a different temperature and a factor 7/8 for the integral of the FD distribution I understand this. But for the photons, we have a degeneracy factor of 2 (two spin states). Then the book lists all characteristics for neutrinos (and the degeneracy factor):




      • spin degree

      • has antiparticles

      • has three generations (flavours)


      So I would expect the degeneracy factor of neutrinos to be 2 (spin) x 2 (antiparticle) x 3 (flavour) => g = 12, so another factor of 6 in front of the photon density (g= 12 vs 2). But the book and other books all list a degeneracy factor of 2 x 3 = 6 for the neutrinos. So this results in factor 3 in front of the photon density (together with the other described differences). Why is this? Is the antiparticle part of the neutrinos not taken into account? Can someone please help me.






      particle-physics cosmology big-bang neutrinos




      particle-physics cosmology big-bang neutrinos






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      edited 9 hours ago









      xray0

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      CFRedDemonCFRedDemon

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          In the SM, neutrinos are massless and therefore their helicity corresponds directly to their chirality. Also in the SM we find that neutrinos are always left-handed, right handed neutrinos do not exist. So if all neutrinos are left-handed, all anti-neutrinos must be right handed consequently. For that reason the spin states directly correspond to particle or anti-particle state. This eliminates a factor of 2 in the degeneracy chart.



          Since we know the SM is imcomplete in this regard, this thread discusses the implications to the effective relativistic degrees of freedom if there was a right-handed neutrino: Degrees of freedom of neutrinos






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            In the SM, neutrinos are massless and therefore their helicity corresponds directly to their chirality. Also in the SM we find that neutrinos are always left-handed, right handed neutrinos do not exist. So if all neutrinos are left-handed, all anti-neutrinos must be right handed consequently. For that reason the spin states directly correspond to particle or anti-particle state. This eliminates a factor of 2 in the degeneracy chart.



            Since we know the SM is imcomplete in this regard, this thread discusses the implications to the effective relativistic degrees of freedom if there was a right-handed neutrino: Degrees of freedom of neutrinos






            share|cite|improve this answer









            $endgroup$


















              5












              $begingroup$

              In the SM, neutrinos are massless and therefore their helicity corresponds directly to their chirality. Also in the SM we find that neutrinos are always left-handed, right handed neutrinos do not exist. So if all neutrinos are left-handed, all anti-neutrinos must be right handed consequently. For that reason the spin states directly correspond to particle or anti-particle state. This eliminates a factor of 2 in the degeneracy chart.



              Since we know the SM is imcomplete in this regard, this thread discusses the implications to the effective relativistic degrees of freedom if there was a right-handed neutrino: Degrees of freedom of neutrinos






              share|cite|improve this answer









              $endgroup$
















                5












                5








                5





                $begingroup$

                In the SM, neutrinos are massless and therefore their helicity corresponds directly to their chirality. Also in the SM we find that neutrinos are always left-handed, right handed neutrinos do not exist. So if all neutrinos are left-handed, all anti-neutrinos must be right handed consequently. For that reason the spin states directly correspond to particle or anti-particle state. This eliminates a factor of 2 in the degeneracy chart.



                Since we know the SM is imcomplete in this regard, this thread discusses the implications to the effective relativistic degrees of freedom if there was a right-handed neutrino: Degrees of freedom of neutrinos






                share|cite|improve this answer









                $endgroup$



                In the SM, neutrinos are massless and therefore their helicity corresponds directly to their chirality. Also in the SM we find that neutrinos are always left-handed, right handed neutrinos do not exist. So if all neutrinos are left-handed, all anti-neutrinos must be right handed consequently. For that reason the spin states directly correspond to particle or anti-particle state. This eliminates a factor of 2 in the degeneracy chart.



                Since we know the SM is imcomplete in this regard, this thread discusses the implications to the effective relativistic degrees of freedom if there was a right-handed neutrino: Degrees of freedom of neutrinos







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 16 hours ago









                DomDoeDomDoe

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