Does Schwarzschild metric in Kruskal-Szekeres coordinates admit asymptotic ($r \to +\infty$) timelike observers?

I thank in advance whoever will answer my question.

Schwarzschild metric in Schwarzschild coordinates in $$\mathbb{R}^{1,3}$$ is [1]: $$ds^2=-\bigg(1-\displaystyle\frac{2M}{r}\bigg)dt^2+\bigg(1-\frac{2M}{r}\bigg)^{-1}dr^2+r^2d\Omega^2.$$

When $$r\to+\infty$$ then $$g_{\mu\nu} \to \eta_{\mu\nu}$$, with $$\eta_{\mu\nu}$$ Minkowski metric (which obviously admits timelike observers e.g. $$(1,0,0,0)$$). By operating the change of coordinate of Kruskal-Szekeres we have [1]:

$$ds^2=\frac{32M^3}{r}e^{-r/2M}(-dT^2+dX^2)+r^2d\Omega^2$$

with $$r=2M(1+W_0[(X^2-T^2)/e])$$ and $$W_0$$ Lambert function. In this case, when $$r \to +\infty$$ we get $$ds^2 \to r^2d\Omega^2>0$$ i.e. any vector becomes spacelike there. How can this discrepancy be explained?

Thanks a lot.

• The “discrepancy” is just an error in taking the limit. At large values of its argument the Lambert W function behaves like logarithm + terms of lower growth rate, so when r is large we recover hyperbolic coordinates of Minkowski space. – A.V.S. Jan 14 at 6:52

You're getting tripped up in two different ways here.

(1) When you reason about the sizes of the terms in the K-S metric, you're ignoring the fact that the expression you're using is written using a mixture of variables from the two coordinate systems. This point is easier to understand with a simpler example. Suppose that $$ds^2=e^{-r}dX^2$$, where $$r=\ln X$$. (As pointed out in a comment by A.V.S., the log is the approximate behavior of the W function.) Implicit differentiation then gives $$dX=e^r dr$$, so that the line element can be rewritten as $$ds^2=e^{r}dr^2$$, and the coefficient clearly doesn't approach zero exponentially.

(Of course every point has a complete light-cone, including the timelike portions, so we knew in advance that there had to be a mistake in your conclusion that all vectors are spacelike in K-S coordinates at large $$r$$.)

(2) If we want a curve to approach spacelike infinity, it's not enough that it have $$r\rightarrow\infty$$. For example, a null geodesic can have $$r\rightarrow\infty$$, but it's going to approach null infinity, not spacelike infinity. The difference between the different idealized boundaries has to do with how fast $$r$$ approaches infinity compared to how fast $$t$$ approaches infinity.

If you look at a Penrose diagram for the Schwarzschild spacetime, it's clear that you can't have a timelike world-line that approaches spacelike infinity.

I don't think there are any $$r\to\infty$$ asymptotic timelike observers. There seems to be a limit of that sort in some coordinates, like Eddington-Finkelstein $$(u,r)$$, but it's an illusion. Conformal spatial infinity is a single point with no time dimension, and its causal past only contains past infinity.

The $$r\to\infty$$ limit of stationary worldlines on the Penrose diagram covers past and future null infinity, with a crossover at spatial infinity. This is the same thing that happens at $$r=2M$$, and the meaning is the same: the observer can't be timelike in the limit, and has to be null instead.

In Minkowski space, you could put the (future) observer at $$(t{-}r)(s)=s$$, with $$t{+}r$$ constant and tending to $$+\infty$$. The "redshift" seen by this observer is the same as that seen by an observer at finite $$r$$, if you take $$s$$ to be its "proper time". In the K-S case, a stationary worldline at finite $$r$$ is given by

$$(T{\pm}X)(t) = \pm R \exp(\pm t/4M)$$

where $$t$$ is the Schwarzschild time and $$R=\sqrt{X^2-T^2}$$, and the observer at infinity can be at

$$(T{-}X)(s) = -R_\infty \exp(-s/4M)$$

where $$R_\infty$$ is any positive value (adjustable by shifting the origin of $$s$$). The "redshift" this observer sees is the limit of redshifts seen by stationary timelike observers.

Thank you very much for your kind and detailed answers.

I had thought about both the problem of the change of coordinates and mixed coordinates; however, what is not clear to me is the reason why $$r = 2M$$ still remains the same region in K-S coordinates while $$r \to +\infty$$ does not. Is this due to the fact that the K-S coordinates are constructed ad-hoc to do that (i.e. to keep the event horizon at $$r = 2M$$)?

Moreover, when I write $$r \to +\infty$$ I mean "in a region very far from the black hole" (i.e. $$r >> 2M\;|\;g_{\mu\nu} \sim \eta_{\mu\nu}$$). This is very useful for calculating time dilation and redshift between a near-horizon observer and a static Earth observer [2].

To "obtain" this region (where $$g_{\mu\nu} \sim \eta_{\mu\nu}$$) in Schwarzschild coordinates it suffices to perform $$r \to +\infty$$; but how can I identify the same region using Kruskal-Szekeres coordinates? How can I go far away from the black hole and put a static timelike observer there?

Thanks again for your help and concern.

References