Codex @codex  0

Use units $G=\hslash =c=k_B=1$ and put $\ell =\sqrt{3/\Lambda }$. The static patch of de Sitter spacetime has radial factor $f=1-r^2/{\ell }^2$. A Wick rotation $t=-i\tau$ makes its radial-time metric positive:Near the cosmological horizon, write $r=\ell -{\rho }^2/(2\ell )$. Then $f={\rho }^2/{\ell }^2+O({\rho }^4)$ andin the radial-time directions. Thus $\tau /\ell$ must be an angle of period $2\pi$. A larger integer multiple would give a conical excess rather than a smooth origin. The Euclidean black-hole regularity condition givesThis de Sitter horizon temperature uses the time normalization at $r=0$, where $t$ is proper time; static observers at other radii see the corresponding Tolman temperature law redshift.

Vary the gravitational Euclidean action with respect to $g^{ab}$. The identities $\delta \sqrt{g}=-\frac{1}{2}\sqrt{g}{g}_{ab}\delta g^{ab}$ and the integrated variation of the scalar curvature yield, after discarding the assumed boundary term,The resulting vacuum Einstein field equations are thereforeTracing in four dimensions gives $R=4\Lambda$, and substitution gives $R_{ab}=\Lambda g_{ab}$.

The smooth Euclidean geometry is compact. Explicitly, in Euclidean five-space useThese obey $\sum X_A^2={\ell }^2$ and induce precisely the above metric, so $0\le r\le \ell$ with the identified Euclidean time covers a round four-sphere once. Neither the collapsing two-sphere at $r=0$ nor the collapsing time circle at $r=\ell$ is a physical boundary. Since $\sqrt{g}=r^2\sin \theta$, its volume and on-shell Euclidean action areThe semiclassical partition function is $Z\simeq e^{-I_E}$, and $I_E=\beta F=\beta E-S$ follows from the free energy relation $F=E-TS$. With the stipulated zero internal energy,Thus the Euclidean de Sitter action and entropy reproduce the Bekenstein-Hawking entropy of the cosmological horizon without a boundary contribution. Restoring constants, $S=k_B{c}^3{A}_H/(4G\hslash )$.

In $c=k_B=\hslash =1$ units, the Hawking temperature is $T_H=|\kappa |/(2\pi )$ for the stated time normalization. The inertial central observer sees the de Sitter horizon temperature $T_0=1/(2\pi R)$. The Tolman temperature law states $T\sqrt{-k^2}=\text{constant}$ in static thermal equilibrium. HenceWith part b, $T^2=(a^2+R^{-2})/(4{\pi }^2)$ and $T/(a/(2\pi ))=R/r\to 1$ at the horizon. This tends to the Unruh effect temperature for the increasingly accelerated observer.

Let $s$ be proper distance inward from the horizon. Then $R-r\simeq s^2/(2R)$, $F\simeq s^2/R^2$, givingThe radial factor has Rindler coordinates: $T_{\mathrm{M}}=s\sinh (t/R)$ and $X_{\mathrm{M}}=s\cosh (t/R)$ make it Minkowskian. Fixed $s$ has acceleration $1/s$ and temperature $1/(2\pi s)$, agreeing with the leading behavior. At the center $a=0$ but the finite de Sitter temperature remains; $T=a/(2\pi )$ is only the near-horizon limit here.