de Sitter Temperature and Entropy

De Sitter temperature is defined relative to an observer’s static patch, a normalized horizon generator, and a regular state. In four-dimensional de Sitter space with Hubble parameter $H$, the geodesic observer at the patch center assigns $T_{\rm dS}=H/(2\pi)$ to the Bunch–Davies state and entropy $S_{\rm dS}=A_c/(4G)$. Neither statement turns the cosmological horizon into an observer-independent material surface.

Helpful background. Quantum Fields in Curved Spacetime supplies thermal detector response and regular states; Einstein Equations supplies the de Sitter solution; Black-Hole Geometry and Mechanics supplies Killing horizons and surface gravity; Semiclassical Gravity supplies renormalized entropy; and Causality and Global Structure supplies observer patches.

The static-patch generator

For $H\gt0$, the static patch has $0\le r\lt H^{-1}$ and $t\in\mathbb R$, with line element

$$ds^2=-f(r)\,dt^2+\frac{dr^2}{f(r)}+r^2d\Omega_2^2, \qquad f(r)=1-H^2r^2.$$

The Killing field $\xi=\partial_t$ is normalized so that $\xi^2=-1$ on the central geodesic $r=0$. It becomes null at the cosmological horizon $r_c=H^{-1}$. With this normalization,

$$\kappa_c=\frac12\left|f'(r_c)\right|=H.$$

Rescaling $\xi\mapsto\alpha\xi$ rescales both $\kappa_c$ and Killing energy by $\alpha$. A numerical temperature is therefore meaningful only after the normalization point has been stated. Unlike an asymptotically flat black hole, de Sitter space has no spatial infinity that supplies a canonical unit time.

The patch is observer dependent. Another timelike geodesic has its own static region related by a de Sitter isometry, while a nongeodesic observer at fixed $r$ uses the same Killing flow but a different proper time. The patch is the intersection of the observer’s causal past and future; its future and past null boundaries are distinct branches meeting at a bifurcation sphere in the stationary extension.

Euclidean regularity and temperature

Near $r=r_c$, set $r=r_c-\epsilon$ and continue $t=-i\tau$. To leading order the $(\tau,r)$ plane is polar:

$$ds_E^2\simeq d\rho^2+\kappa_c^2\rho^2d\tau^2+r_c^2d\Omega_2^2.$$

Smoothness at $\rho=0$ requires $\tau$ to have period

$$\beta_c=\frac{2\pi}{\kappa_c}, \qquad T_c=\beta_c^{-1}=\frac{H}{2\pi}.$$

Gibbons and Hawking connect this regular Euclidean periodicity with particle-detector thermality and horizon entropy Gibbons and Hawking 1977, pp. 2739–2741. The statement assumes a field theory admitting the regular Euclidean, or Bunch–Davies, state. It applies, for example, to a conformally coupled massless scalar. The massless minimally coupled scalar is an exception: it admits no de Sitter-invariant Fock vacuum of the ordinary scalar field Allen 1985, § III, pp. 3141–3142. The obstruction is a statement about symmetry and the full two-point function, not simply the existence of a constant classical mode. Neither that exceptional case nor an arbitrary excited state can be handled just by asserting “the Bunch–Davies vacuum.”

Equivalently, restrict the Bunch–Davies two-point function to the central static worldline. Its analytic continuation is periodic in imaginary Killing time by $2\pi/H$ and satisfies the KMS relation for the static Hamiltonian. A stationary two-level detector consequently obeys detailed balance at $T_c$. This is an independent operational check of the Euclidean conical argument. It does not assert that a single global particle-number operator sees a literal gas everywhere: particle notions are observer dependent, whereas the KMS statement refers to the algebra of the chosen static patch and its normalized time flow.

A static observer at radius $r$ measures proper time $d\tau_{\rm prop}=\sqrt{f}\,dt$. The Tolman relation gives

$$T_{\rm loc}(r)=\frac{T_c}{\sqrt{f(r)}}.$$

Its divergence near $r_c$ reflects the unbounded acceleration required to remain static, not a divergent stress tensor in a freely falling frame.

Indeed the magnitude of the static observer’s proper acceleration is $a_{\rm obs}=H^2r/\sqrt{1-H^2r^2}$. Substitution gives an independent check,

$$T_{\rm loc}^2=\frac{H^2+a_{\rm obs}^2}{4\pi^2}.$$

The central observer has $a_{\rm obs}=0$ but still detects the de Sitter temperature. Near the horizon, acceleration dominates. This also explains why the thermometer reading depends on the observer even though the horizon area in the chosen patch is fixed.

Horizon entropy and its sign

The cosmological-horizon area and leading gravitational entropy are

$$A_c=4\pi H^{-2}, \qquad S_c=\frac{A_c}{4G}=\frac{\pi}{GH^2}.$$

Higher-curvature actions replace $A/(4G)$ by the corresponding Wald entropy plus quantum corrections. The area formula is thus the leading Einstein-gravity term, evaluated with renormalized $G$ in a controlled semiclassical expansion.

The sign of a patch first law deserves care. Add a small central mass to obtain Schwarzschild–de Sitter,

$$f(r)=1-\frac{2GM}{r}-H^2r^2.$$

At fixed $H$ and to first order, the cosmological root shifts from $H^{-1}$ to $r_c=H^{-1}-GM$. Consequently

$$\delta S_c=\frac{2\pi r_c}{G}\,\delta r_c \simeq-\frac{2\pi}{H}\delta M, \qquad \delta M=-T_c\,\delta S_c.$$

Positive energy inside the patch shrinks its cosmological horizon. Using the black-hole sign $+\!T\delta S$ without tracking which boundary and energy are varied would give the wrong conclusion. Broader de Sitter thermodynamic interpretations and their observer dependence are reviewed in Spradlin, Strominger, and Volovich 2001, §§ 3.2–3.3, PDF.

Checks and limitations

Dimensions provide an immediate check: $H$ and $T$ have inverse-length units, while $A/G$ is dimensionless. In the flat limit $H\to0$, the horizon recedes, $T\to0$, and $S\to\infty$ because the static patch becomes unbounded; the divergent entropy is not a finite Minkowski thermal entropy. The Euclidean derivation assumes an exactly stationary geometry. In a slowly evolving cosmology one may define approximate horizons and temperatures, but corrections depend on the chosen horizon and adiabatic hierarchy.

Pure de Sitter also describes one temperature. Spacetimes with both black-hole and cosmological horizons generally have unequal surface gravities, so a single smooth Euclidean period does not regularize both. That nonequilibrium multi-horizon problem cannot be solved by assigning the pure de Sitter formula to each boundary independently.

Exercise

For a static observer at $r=\sqrt{3}/(2H)$, compute the locally measured temperature. Then explain why this value does not change the central observer’s entropy $A_c/(4G)$.

Solution

At that radius $f=1-H^2r^2=1/4$, so

$$T_{\rm loc}=\frac{H/(2\pi)}{\sqrt{1/4}}=\frac{H}{\pi}.$$

The factor of two is the gravitational redshift between the observer’s proper time and the normalized central Killing time. The horizon cross-section and its area are unchanged; entropy is a geometric charge for the specified horizon, not a local thermometer reading. Rescaling time changes energy and temperature together but not $A_c/(4G)$.

References

• Allen, B. (1985). “Vacuum States in de Sitter Space.” Physical Review D 32, 3136–3149. DOI.
• Gibbons, G. W., and Hawking, S. W. (1977). “Cosmological Event Horizons, Thermodynamics, and Particle Creation.” Physical Review D 15, 2738–2751. DOI.
• Spradlin, M., Strominger, A., and Volovich, A. (2002). “De Sitter Space.” In Unity from Duality: Gravity, Gauge Theory and Strings, pp. 423–453. DOI. The 2001 preprint is titled “Les Houches Lectures on de Sitter Space”; section locators above refer to that version. Open PDF.