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I have been dealing with the following issue related to the Schwarzschild geometry recently. When expressed as:

$$ ds^{2}=-\left(1-\frac{2GM}{r}\right)dt^{2}+\frac{1}{1-\frac{2GM}{r}}dr^{2}+d\Omega_{2}^{2}$$

one can find a Killing vector $\xi=\partial_{t}$, since there are no components of the metric depending on $t$. This Killing vector is timelike for $r>2GM$, but spacelike for $r<2GM$ (since $\xi^{\mu}\xi_{\mu}=-\left(1-\frac{2GM}{r}\right)$). My question is:

  1. Can we find any timelike vector for the region $r<2GM$?
  2. If not, this would imply that the Schwarzschild solution is not stationary for $r<2GM$. But it is usually referred to as a "static spacetime". This wouldn't be true for the region $r<2GM$. So is this an abuse of language?
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3 Answers 3

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The are only four Killing vectors of Schwarzschild. They are $\partial_t$ and the three rotational Killing vectors. No linear combination of these is globally timelike within the horizon, so there is no global timelike Killing vector.

I suppose whether or not Schwarzschild is static depends on one's definiton of "static." If you define it to mean that there is a global timelike Killing vector, then yes, Schwarzschild is not static. However I think the word is implicitly used to only refer to patches of spacetimes. So the region outside the horizon could indeed be called "static." This is also the case in de Sitter, where one often talks about the "static patch."

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  • $\begingroup$ The accepted answer by user1379857 says that this is just defined on a patch, but that's silly. What patch would be the correct patch to use? The term "stationary" is often taken to mean "asymptotically stationary." Some authors asymptotical stationarity as their definition of "stationary." Others, such as Carroll, use context to disambiguate the definition. $\endgroup$
    – user321137
    Dec 5, 2021 at 19:32
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    $\begingroup$ Yes, I believe the standard usage is that only the exterior of a Schwartzchild black hole is static, not the interior. $\endgroup$
    – tparker
    Dec 5, 2021 at 19:48
  • $\begingroup$ Almost true. The interior Schwarzschild metric with critical compactness parameter $\alpha=r_{S}/R $ lower as $8/9$, is static, too. The same is valid for any other interior solution but in that case the critical $\alpha$ is lower as $8/9$. $\endgroup$ Dec 6, 2021 at 11:21
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Suppose $\xi$ is a Killing field. Then its flow is a local isometry, so for any scalar $K$ we have that the derivative of $K$ in the direction of $\xi$ is zero i.e. $dK(\xi)=0$. Take the Kretschmann scalar for $K$, this implies that $dr(\xi)=0$. Therefore inside the horizon you have that $\xi^\mu\xi_\mu>0$, because all the terms are positive and the $dr$ term is zero, hence it cannot be timelike.

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A time-line coordinate flips only if one assumes that Schwarzschild's vacuum solution is valid in the interior of a black hole, too. In my view, it is not correct. A solution of Einstein field equations spans the whole spacetime. In case of a constant energy density in some region of spacetime (step function) the solution (metric) can be cast into two parts: an interior and an exterior one. It is admissible to work with only one of them, and glue it to each other, but it is not admissible to extend them over their validity domain. For example the Schwarzschild interior solution remains static, although maybe not stable, even above the critical compactness parameter $\alpha=8/9$. This view gave rise to Pawel O. Mazur and Emil Mottola idea of gravastar, https://en.wikipedia.org/wiki/Gravastar . Their paper you can read here: https://arxiv.org/abs/1501.03806. If you like see also: https://physics.stackexchange.com/a/679431/281096 and https://physics.stackexchange.com/a/674311/281096 .

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