I'm considering the $4D$ de-Sitter spacetime, in static coordinates (I'm using $c = 1$ and $k_{\text{B}} = 1$): \begin{equation}\tag{1} ds^2 = (1 - \frac{\Lambda}{3} \, r^2) \, dt^2 - \frac{1}{1 - \frac{\Lambda}{3} \, r^2} \, dr^2 - r^2 \, d\Omega^2, \end{equation} where $\Lambda > 0$ is the cosmological constant. This spacetime has an horizon around any static observer, at $r = \ell_{\Lambda} \equiv \sqrt{3 / \Lambda}$. The whole space volume inside that horizon is easily calculated from the metric above (it is not $4 \pi \ell_{\Lambda}^3 / 3$): \begin{equation}\tag{2} \mathcal{V} = \pi^2 \ell_{\Lambda}^3, \end{equation} and the horizon area is $\mathcal{A} = 4 \pi \ell_{\Lambda}^2$. The vacuum has an energy density and pressure: \begin{align}\tag{3} \rho &= \frac{\Lambda}{8 \pi G}, & p &= -\, \rho. \end{align} Thus, the vacuum energy inside the whole volume of the observable de-Sitter universe is \begin{equation}\tag{4} E = \rho \, \mathcal{V} = \frac{3 \pi \ell_{\Lambda}}{8 G}. \end{equation} Note that the enthalpy is trivially 0 (what does that mean?): \begin{equation} H = E + p \mathcal{V} = 0. \end{equation}

I'm now considering the thermodynamic first law, comparing various de-Sitter universes which have slightly different $\Lambda$ (or $\ell_{\Lambda}$): \begin{equation}\tag{5} dE = T \, dS - p \, d\mathcal{V} = T \, dS + \rho \, d\mathcal{V}. \end{equation} Inserting (2) and (4) give the following: \begin{equation}\tag{6} T \, dS = -\, \frac{3 \pi}{4 G} \, d\ell_{\Lambda}. \end{equation} If $d\ell_{\Lambda} > 0$ and $dS > 0$, this implies a negative temperature! If I use the entropy $S = \mathcal{A}/ 4 G$ (take note that this entropy formula is very controversial for $\Lambda > 0$), then $dS = 2 \pi \ell_{\Lambda} \, d\ell_{\Lambda} / G$ and \begin{equation}\tag{7} T = -\, \frac{3}{8 \, \ell_{\Lambda}}. \end{equation} This result is puzzling!

I'm now wondering if the $T \, dS$ term would better be replaced with the work done by the surface tension on the horizon, instead: $T \, dS \; \Rightarrow \; -\, \tau \, d\mathcal{A}$ (I'm not sure of the proper sign in front of $\tau$). In this case, I get the tension of the horizon (I don't know if this makes any sense!): \begin{equation}\tag{8} \tau = \frac{3}{32 G \ell_{\Lambda}}. \end{equation} So is the reasoning above buggy? What is wrong with all this? Any reference that confirms that the de-Sitter Horizon's temperature could be negative, or that the entropy is really undefined there (or that $S = \mathcal{A} / 4 G$ is wrong in this case)? Or should the entropy term $T \, dS$ actually be interpreted as the tension work $-\, \tau \, d\mathcal{A}$ on the horizon instead?

In (4) and (5), is it legit to use the energy inside the horizon only, excluding the exterior part?

EDIT: The energy (4) is the energy of vacuum inside the horizon. It doesn't take into account the gravitational energy. I now believe that it's the Komar energy in the same volume that should be considered. The integration gives the following Komar energy inside the volume (2): \begin{equation}\tag{9} E_K = -\, \frac{\ell_{\Lambda}}{G}. \end{equation} But then, the trouble with the temperature is still the same: temperature is negative if $d\ell_{\Lambda} > 0$ (which is the same as $d\Lambda < 0$) and assume $dS > 0$ (or $S = \mathcal{A}/ 4 G$, which may be false for the de-Sitter spacetime).

  • $\begingroup$ @A.V.S., the Komar mass cannot be defined in a de-Sitter spacetime: it isn't asymptotically flat. And $\xi^{\mu} \, \xi_{\mu} = 1 - \frac{\Lambda}{3} \, r^2$. You would have to integrate for $r > \ell_{\Lambda}$ where $\xi^{\mu}$ is spacelike. $\endgroup$
    – Cham
    Commented Oct 11, 2019 at 10:07
  • $\begingroup$ @A.V.S., you still have a space like vector on the exterior side of the horizon. How do you calculate de Komar mass in this case? And even if you restrict the integration to the interior of the horizon, it would probably give energy (4) above (with possibly a numerical factor, since pressure is equal to minus the energy density). $\endgroup$
    – Cham
    Commented Oct 11, 2019 at 14:43
  • $\begingroup$ Indeed, if I restrict the integration to the interior of the horizon, the Komar mass is the same as (4), up to a factor of $\frac{1}{4}$ which isn't clear yet. $\endgroup$
    – Cham
    Commented Oct 11, 2019 at 15:04
  • $\begingroup$ @A.V.S., the numerical factor $\frac{1}{4}$ makes things worst: I still get a negative temperature! Also, if the Komar mass of the de-Sitter spacetime was defined, it would already be well known by now! I don't find it anywhere. $\endgroup$
    – Cham
    Commented Oct 11, 2019 at 15:14
  • $\begingroup$ About the Komar mass, I asked a question there: physics.stackexchange.com/questions/507638/… $\endgroup$
    – Cham
    Commented Oct 11, 2019 at 15:56

1 Answer 1


The future cosmic Event horizon is the source of de Sitter (aka cosmic Hawking) radiation, also characterised by a specific temperature, the de Sitter temperature $T$ (as per the OP). It is the minimum possible temperature of the universe.

To an observer in our universe, a de Sitter Universe is in their infinite future, i.e. when the Hubble sphere and Event horizon are coincident. Now, we can assign the de Sitter minimum length as $l_Λ=2$ and de Sitter $Λ=3/4$ in natural units. If you don’t like this, no matter, just stick with the symbolic equations.

Unlike a Schwarzschild black hole solution, the de Sitter solution has a non-zero pressure. So, the following by the OP are correct:

  • having the PV term in equation (5)
  • the entropy expression, i.e. $S=A/4G=π.l_Λ^2=4π$
  • energy density and pressure in (3)

However, because (4) is an expression of the horizon energy $E_H$ the relevant volume is not (2) rather it is the so-called areal volume (page 6) which is $V=4πl_Λ^3/3$. Then, the energy is: $$E_H=U= ρV=(l_Λ^3/6).Λ= (4/3).Λ=1 (Eqn.4)$$ The energy of the horizon equals the energy in the bulk, as per the holographic principle so: $$TS= ρV=1 (Eqn.4b)$$ $$T.4π= (l_Λ^3/6).Λ$$ $$T= (l_Λ^3/24π).Λ=1/4π=1/(2π.l_Λ )$$

Giving the de Sitter temperature $T$ as expected (Page 3, i.e. Gibbons and Hawking, 1977). Or equivalently: $$T= (1/2π).√(Λ/3)= H_o/2π$$ The thermodynamic first law: $$TS-E=pV (Eqn.5)$$ $$E= TS- pV$$ $$E=2TS=2$$ This is the maximum mass-energy of the de Sitter observable universe, and we have also found the universal relation $E=2TS$ as per Padmanabhan (page 42). This result also corresponds with Boehmer & Harko (page 3) mass-energy of an observable universe (natural units): $$m_P.E.c^2=(c^4/G) √(3/Λ)=E=2 (Eqn.5b)$$ Finally, yes, the enthalpy $H$ is indeed zero for a de Sitter universe. This means de Sitter space is unstable, as is known, and so spontaneously (no magician needed) created a rabbit (our Universe). Free energy $G=H-TS= -TS=-1$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.