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Neutrinos are sometimes considered to contribute to dark matter, see e.g. E.Siegel. Why not for dark energy?

There is a similar scale in energy density involved. If you use the current upper limit for mass <~ 0.1 eV and the corresponding Compton wavelength as formal length scale (I know neutrinos are point like but that does not necessary imply that they can not be related to some volume) you get an upper limit of energy density <~ 0.1 J/m^3 compared to the vacuum energy of ~ 1E-9 J/m^3 (Wikipedia). Since energy density goes ~Energy^4 that’s a mere 2 orders of magnitude apart. In addition, according to PDG total neutrino average number density today is: n(ν) = 339.5 /cm^3 which in turn with a value of 0.1 eV for neutrino energy gives an energy density ~ 1E-11J/m^3, again close to vacuum energy.

As for expansion/acceleration: There is a steady source of additional neutrinos from fusion in stars.

Is there a reason to exclude neutrinos to contribute to DE?

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Dark energy has a characteristic equation of state relating the density to the pressure. Typically in cosmology we assume that the relationship between density $\rho$ and pressure $p$ is given by an equation of state of the form \begin{equation} p = w \rho \end{equation} where $w$ is the equation of state parameter.

The equation of state parameter is important because it determines the expansion rate of the Universe. For a single component Universe, the expansion rate (or Hubble parameter) $H$ grows with the scale factor $a$ as \begin{equation} H^2 \propto a^{-3(1+w)} \end{equation} When $w\approx -1$, the expansion rate is nearly constant, which is characteristic of exponential growth.

A cosmological constant has an equation of state parameter $w=-1$. "Dark energy", broadly defined, has an equation of state "near" $-1$. Observationally, there are fairly tight bounds on how far $w$ can be from $-1$ to explain the acceleration of the Universe.

Relativistic matter has an equation of state parameter $w\approx 1/3$, while non-relativistic matter has an equation of state parameter $w\approx 0$. We would expect neutrinos in a cosmological neutrino background to be traveling relativistically, so have an equation of state parameter around $1/3$. This would not drive exponential expansion of the Universe, and would not explain observational evidence of a component of the Universe with an equation of state parameter near $-1$.

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  • $\begingroup$ excellent! thanks for this explanation, I had not seen it before. -NN $\endgroup$ Commented Dec 17, 2022 at 20:12
  • $\begingroup$ Andrew said: we assume that the relationship between density ρ and pressure p is given by an equation of state of the form p=wρ .Why is this assumed? $\endgroup$
    – user353451
    Commented Dec 17, 2022 at 23:36
  • $\begingroup$ @DaveTheWave Essentially it follows from the symmetry of FRW spacetimes (homogeneity and isotropy) that the stress energy tensor has the form $T^{i}_{\ \ j} = {\rm diag}(-\rho, p, p, p)$, so there's only really one relationship needed in the equation of state. You can also calculate the stress energy explicitly for special cases, like pressureless dust, the Maxwell stress-energy tensor for the electromagnetic field, or for the cosmological constant, and $\rho=wp$ falls out naturally in those cases. $\endgroup$
    – Andrew
    Commented Dec 18, 2022 at 2:19
  • $\begingroup$ w will give me something to think about. However, the equation of state does not take into account possible source terms, does it? Could it change the game? $\endgroup$
    – qatch
    Commented Dec 18, 2022 at 10:44
  • $\begingroup$ @qatch I'm not sure what you mean. You can calculate the equation of state for a relativistic gas of particles and you find $p=\frac{1}{3}\rho$. $\endgroup$
    – Andrew
    Commented Dec 18, 2022 at 12:31

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