# Do we have to consider decoupled particles in the computation of $g_{*}(T)$?

I am studying the thermal history of the universe and I encountered the definition of effective degrees of freedom $$g_{*}(T)$$ defined as $$g_{*}(T)=\sum_{Bosons}g_{B}(\frac{T_{B}}{T})^{4}+\frac{7}{8}\sum_{Fermions}g_{F}(\frac{T_{F}}{T})^{4}$$ Now, in the computation of this factor, we need to consider just relativistic particles since the NR ones are suppressed from the Boltzmann factor. But what about decoupled particles? Do we have to consider them?

I am making this question because it's always written that neutrino decouples at $$T ≃ 1$$ Mev , and if we want to compute $$g_{*}(T=0.1 Mev)$$ we have to consider just photons and neutrinos, since they are the only relativistic particles left (Notice that we are considering neutrinos that are already decoupled at that temperature).

On the other hand when computing the ratio $$\frac{T_{\nu0}}{T_{CMB}}$$ one has to compute the factor ($$\epsilon<<1$$) $$\frac{T_{\nu0}}{T_{CMB}}=\bigl(\frac{g_{*s}(t_{e}-\epsilon)}{g_{*s}(t_{e}+\epsilon)}\bigr)^{1/3}=(\frac{4}{11})^{1/3}$$ where I've indicated with $$t_{e}$$ the instant at which electrons become non relativistic. The problem is that I don't understand why, in the latter $$g_{*s}(t_{e}-\epsilon)=2+\frac{7}{8}\cdot2\cdot2$$, we consider only photons and electrons, forgetting about neutrinos which are decoupled but still relativistic right?

Those two computations seems to be in contrast in a sense that in one case I consider decoupled neutrinos while on the other I forget them. What do I miss here?

$$g_*$$ is the effective number of relativistic degrees of freedom for the energy density. The idea of $$g_*$$ is that given the temperature $$T$$ of the photons, the total energy density in radiation is $$\rho=\frac{\pi^2}{30}g_* T^4.$$ Decoupled particles must be included in $$g_*$$ because they still contribute to the energy density. The same is true of $$g_{*s}$$, the effective number of relativistic degrees of freedom for the entropy density. The idea of $$g_{*s}$$ is that if the photon temperature is $$T$$, then the total entropy density in radiation is $$s=\frac{2\pi^2}{45}g_{*s}T^3.$$ That also implies all relativistic species should be accounted for.

Thus, it is nonstandard to define a $$g_{*s}$$ that only includes some of the species, as you do when you write

$$g_{*s}(t_{e}-\epsilon)=2+\frac{7}{8}\cdot2\cdot2$$

Nevertheless this approach works, because since the neutrinos are decoupled, the entropy $$\propto sa^3\propto g_{*s} a^3 T^3$$ of the photon-electron bath and that of the neutrinos are separately conserved.

• I have to think about that before accepting the answer, thank you btw. Commented Feb 23 at 17:56
• I still don't really understand, I thought that the effective degrees of freedom $g_{*s}(T)$ by definition take into account every particle at that temperature which is not Boltzmann suppressed, I'd be grateful if you could explain it in more details. Cause $g_{*s}(T)^{1/3}a(T)T=const.$ should consider everything, no? Commented Feb 26 at 15:10
• @Filippo It's nonstandard to define a $g_{*s}$ that only accounts for some species, but if you do so, then the $g_{*s} a^3 T^3$ of each set of coupled species will be separately conserved.
– Sten
Commented Feb 27 at 9:10