Temperature of the vacuum energy density

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.