I am studying the derivation of the black hole temperature by means of the Euclidean approach, i.e. by Wick rotating, compactifying the Euclidean time and identifying the period with the inverse temperature.
Consider the Schwarzschild case as an example. The Euclidean Schwarzschild metric is of course
$ds^2=\left(1-\frac{2M}{r}\right)d\tau^2+\left(1-\frac{2M}{r}\right)^{-1}dr^2$
where $\tau=it$ the Euclidean time. Here $\tau\in[0,\beta]$ with $\beta=T^{-1}$ the inverse temperature, where the points $\tau=0$ and $\tau=\beta$ are equivalent. (I ignore the 2-sphere part of the metric.)
Outside but close to the event horizon $r=2M$, we can (after simple some steps) write this as
$ds^2=\frac{\sigma^2}{16M^2}d\tau^2+d\sigma^2$.
Here $\sigma^2 \equiv 8M(r-2M) $, though this is not really relevant for my question.
Now the next step in all the literature is to require that $\tau/4M$ is periodic with period $2\pi$ to prevent conical singularities. I have trouble understanding this.
- A conical singularity basically means that the point at $\sigma=0$ looks like the tip of a cone, right? So we have a singularity at $\sigma=0$. But don't we still have a singularity there for polar coordinates $(r,\theta)$, since the coordinate chart $\theta$ is not continuous there?
- If so, removing the point $\sigma=0$ would erase the conical singularity, right? Then why do we want to get rid of the conical singularity, why is it any worse than the polar coordinate singularity? Of course polar coordinates just describe flat space without the origin, is this not the case for the conical coordinates?
Clearly I have not understood the concept of a conical singularity…
Last question: suppose I did understand it, and continued the derivation to get the temperature $T=1/8\pi M$. Apparently this is the temperature as measured by an observer at infinity, how can I see this? I know the temperature gets redshifted, like frequency, but I don't see where the derivation identifies $T$ with the one measured at infinity.
Best Answer
Well, the singularity does not concern the differentiable structure: Even around the tip of a cone (including the tip) you can define a smooth differentiable structure (obviously this smooth structure cannot be induced by the natural one in $R^3$ when the cone is viewed as embedded in $R^3$). Here the singularity is metrical however! Consider a $2D$ smooth manifold an a point $p$, suppose that a smooth metric can be defined in a neighborhood of $p$, including $p$ itself. Next consider a curve $\gamma_r$ surrounding $p$ defined as the set of points with constant geodesic distance $r$ from $p$. Let $L(r)$ be the (metric) length of that curve. It is possible to prove that: $$L(r)/(2\pi r) \to 1\quad \mbox{ as $r \to 0$.}\qquad (1)$$ Actually it is quite evident that this result holds. We say that a $2D$ manifold, equipped with a smooth metric in a neighborhood $A-\{p\}$, of $p$ (notice that now $p$ does not belong to the set where the metric is defined), has a conical singularity in $p$ if: $$L(r)/(2\pi r) \to a\quad \mbox{ as $r \to 0$,}$$ with $0<a<1$.
Notice that the class of curves $\gamma_r$ can be defined anyway, even if the metric at $p$ is not defined, since the length of curves and geodesics is however defined (as a limit when an endpoint terminates at $p$). Obviously, if there is a conical singularity in $p$, it is not possible to extend the metric of $A-\{p\}$ to $p$, otherwise (1) would hold true and we know that it is false.
As you can understand, all that is independent from the choice of the coordinates you fix around $p$. Nonetheless, polar coordinates are very convenient to perform computations: The fact that they are not defined exactly at $p$ is irrelevant since we are only interested in what happens around $p$ in computing the limits as above.
Yes, removing the point one would get rid of the singularity, but the fact remains that it is impossible to extend the manifold in order to have a metric defined also in the limit point $p$: the metric on the rest of the manifold remembers of the existence of the conical singularity!
The fact that the Lorentzian manifold has no singularities in the Euclidean section and it is periodic in the Euclidean time coordinate has the following physical interpretation in a manifold with a bifurcate Killing horizon generated by a Killig vecotr field $K$. As soon as you introduce a field theory in the Lorentzian section, the smoothness of the manifold and the periodicity in the Euclidean time, implies that the two-point function of the field, computed with respect to the unique Gaussian state invariant under the Killing time and verifying the so called Hadamard condition (that analytically continued into the Euclidean time to get the Euclidean section) verifies a certain condition said the KMS condition with periodicity $\beta = 8\pi M$.
That condition means that the state is thermal and the period of the imaginary time is the constant $\beta$ of the canonical ensemble described by that state (where also the thermodynamical limit has been taken). So that, the associated "statistical mechanics" temperature is: $$T = 1/\beta = 1/8\pi M\:.$$
However the "thermodynamical temperature" $T(x)$ measured at the event $x$ by a thermometer "at rest with" (i.e. whose world line is tangent to) the Killing time in the Lorentzian section has to be corrected by the known Tolman's factor. It takes into account the fact that the perceived temperature is measured with respect to the proper time of the thermometer, whereas the state of the field is in equilibrium with respect to the Killing time. The ratio of the notions of temperatures is the same as the inverse ratio of the two notions of time, and it is encapsulated in the (square root of the magnitude of the) component $g_{00}$ of the metric $$\frac{T}{T(x)}=\frac{dt_{proper}(x)}{dt_{Killing}(x)} = \sqrt{-g_{00}(x)}\:.$$ In an asymptotically flat spacetime, for $r \to +\infty$, it holds $g_{00} \to -1$ so that the "statistical mechanics" temperature $T$ coincides to that measured by the thermometer $T(r=\infty)$ far away from the black hole horizon. This is an answer to your last question.