Astroparticle Physics

194 9 The Early Universe9.2.1 Energy and Number Densitiesinternal degrees of freedomphoton spin statesStefan–Boltzmann lawnumber densitiesIn the limit where the particles are relativistic, i.e., T ≫ m, 1the energy density for a given particle type is⎧π⎪⎨230 gT 4 for bosons,ϱ =⎪⎩ 7 π 28 30 gT 4 for fermions.(9.10)Here g is the number of internal degrees of freedom forthe particle. For a particle of spin J , for example, one has2J + 1 spin degrees of freedom. In addition, the factor gincludes different colour states for quarks and gluons. Forexample, for electrons one has g e − = 2 or, for electronsand positrons considered together, g e = 4. The spin-1/2 uquark together with the ū has g u = 12, i.e., 2 from spin,3 from colour, and 2 for considering particle and antiparticletogether. Note that the photon has J = 1 but only twospin states, which correspond to the two transverse polarizationstates. The longitudinal polarization is absent as a consequenceof these particles having zero mass. For photonsthis gives ϱ γ = (π 2 /15)T 4 , the well-known formula for theenergy density of blackbody radiation (Stefan–Boltzmannlaw). Indeed, one sees that all relativistic particles have asimilar behaviour with ϱ ∼ T 4 , the only differences arisingfrom the number of degrees of freedom and from the factor7/8 if one considers fermions.In a similar manner one can show that the number densityn is given by⎧ζ(3)⎪⎨πn =2 gT 3 for bosons,(9.11)3 ζ(3) ⎪⎩4 π 2 gT 3 for fermions.Here ζ is the Riemann zeta function and ζ(3) ≈ 1.202 06 ....Notice that in particle physics units the number density hasdimension of energy cubed. To convert this to a normalnumber per unit volume, one has to divide by (¯hc) 3 ≈(0.2GeVfm) 3 .From the number and energy densities one can obtainthe average energy per particle, 〈E〉 =ϱ/n.ForT ≫ m onefinds1 Note that Boltzmann’s constant k has also been set equal to one.