In ordinary text books the vacuum energy density $u_{\small\Lambda}$ associated with the cosmological constant $\Lambda$ is given by $u_{\small\Lambda}=\Lambda/\kappa$, where $\kappa$ is Einstein's gravitational constant. Given that there is no other source of radiation, is it valid to apply the Stefan-Boltzmann law $ u_{\small\Lambda}=\sigma T^4$ to obtain the corresponding temperature?
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Another possible formula to use would be $E=k_bT$ – NicAG Apr 11 '20 at 16:47
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@NicAG For what reason? – Apr 11 '20 at 17:46
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It's the most natural link between energy and temperature. – NicAG Apr 11 '20 at 19:48
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Probably from kinetic gas theory. – Apr 11 '20 at 19:51
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It is important to contrast the two answers already given: the one referring to the vacuum energy density, $\Lambda$, and the other referring to the energy density associated with the de Sitter horizon in spacetimes which possess a nonzero $\Lambda$.
The former is generally thought to arise from the vacuum fluctuations of quantum fields, albeit the observed value of $\Lambda$ does not agree with the naive summation of fluctuations arising in the Standard Model. This energy density does not have a temperature, as indeed it has no thermal degrees of freedom.
A universe with a $\Lambda$ that dominates other forms of matter/energy undergoes accelerated expansion; in the limit that the universe contains nothing but $\Lambda$ this is called de Sitter expansion. Due to the accelerated expansion, a cosmological event horizon develops which possesses a temperature, much akin to Hawking radiation. This temperature is due to the excitation of quantum fluctuations in the rapidly expanding spacetime, a process described by L. Parker in 1968 (see "Particle Creation in Expanding Universes"). The de Sitter temperature is given by $T = 1/2\pi r$, where $r$ is the horizon radius. Often this is written $T = H/2\pi$ where $H = 1/r$ is the Hubble parameter.
bapowell
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1As far as I know, the thermalization in de Sitter space is due to the euclideanization in quantum field theory which is (imho) somewhat vague, - such as if one wanted to get this result. On the other hand, why shouldn't excitations of ordinary vacuum fluctuations not be any sort of a statistical ensemble associated with an ordinary Gibbs entropy? (+1) – Apr 17 '20 at 05:25
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1The de Sitter temperature, or really, the growth of classical perturbations during inflation, can be studied using regular old quantum field theory in curved spacetime (though, yes, Gibbons and Hawking derived it also using Euclidean methods). The temperature is related to the vacuum autocorrelation function of the fields, which vanishes trivially in non-dynamical spacetimes. I believe this is why vacuum fluctuations in Minkowski space don't have a temperature. – bapowell Apr 17 '20 at 13:28
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In a paper of Lima and Santos (1995) there is a derivation of a generalized Stefan-Boltzmann law to general relativity. Therein, thermodynamics arguments have been employed to derive how the energy density $u$ depends on the temperature $T$, for a fluid whose pressure $p$ obeys the equation of state $p = (\gamma-1)\, u$, where $\gamma$ is a constant. This leads to the expression
$$u = \eta\, T^{\frac{\gamma}{\gamma-1}}$$
where $\eta$ is a constant.
Applying this to the vacuum equation of state $p=-u_\Lambda$ implies $\gamma=0$. Accordingly, the answer of the question is no.
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Assuming you're referring to this vacuum energy, then (to my knowledge) it doesn't really have a temperature. It's a consequence of quantum fluctuations, not thermal fluctuations.
According to wikipedia, Lorenz invariance requires the vacuum energy to be distributed like $f^3$, so not like blackbody radiation.
The cosmic microwave background, which is blackbody radiation, does have a well defined temperature of 2.726 Kelvin.
Disclaimer: I am not a cosmologist.
taciteloquence
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Do you know a source where it is further explained that quantum fluctuations are not thermal? – Apr 17 '20 at 08:20
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An asymptotic cosmological constant yields de Sitter spacetime as late time cosmological evolution. This spacetime has a horizon. There should be a temperature related to this horizon as it is for black holes. However, I'm not sure whether there is the same kind of Hawking radiation associated with this horizon an temperature. However, it seems more natural for me than the Stefan-Boltzmann law or the relation from kinetic gas theory.
p6majo
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This implies that the energy density of my post would by proportional to the square of this temperature like $u_{\small\Lambda}\propto T_{\small H}^2$. A rather peculiar notion. – Apr 13 '20 at 11:33