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Schwarzschild metric not stationary inside the horizon?

  1. Aug 21, 2010 #1

    bcrowell

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    The Schwarzschild metric, described in Schwarzschild coordinates, has a Killing vector [itex]\partial_t[/itex]. This vector is timelike outside the horizon, but spacelike inside it. Therefore I would think that a Schwarzschild spacetime should not be considered stationary (which also means it can't be static). This would be despite common usages such as "a static black hole." Below I've described what I've seen in several different textbooks, which leads me to think there may be a difference in definition between different authors.

    Hawking and Ellis's proof of Birkhoff's theorem splits two cases: one where they describe the solution as "static," and the other where they describe it as "spatially homogeneous." This would seem to support the interpretation I gave in the first paragraph above. Their formulation of Birkhoff's theorem is that "any C2 solution of Einstein's empty space equations which is spherically symmetric in an open set V, is locally equivalent to part of the maximally extended Schwarzschild solution in V," which is also consistent with my interpretation.

    Carroll, http://nedwww.ipac.caltech.edu/level5/March01/Carroll3/Carroll7.html , has this: "The fact that the Schwarzschild metric is not just a good solution, but is the unique spherically symmetric vacuum solution, is known as Birkhoff's theorem. It is interesting to note that the result is a static metric." The second sentence seems to me to be false inside the horizon.

    The WP article on Birkhoff's theorem says, "any spherically symmetric solution of the vacuum field equations must be stationary and asymptotically flat." The interior of the Schwarzschild metric seems like it would be a counterexample to this.

    Rindler assumes staticity, derives the Schwarzschild metric, and then says that there's a theorem called Birkhoff's theorem that shows that the assumption of staticity can be dispensed with. This doesn't directly address the issue of the staticity of the continuation of the metric inside the horizon.

    The WP article on stationary spacetimes, http://en.wikipedia.org/wiki/Stationary_spacetime , says: "a spacetime is said to be stationary if it admits a Killing vector that is asymptotically timelike," giving a google books link to a book by Ludvigsen. Ludvigsen doesn't seem to explicitly define "asymptotically timelike," but I assume that the definition would be sort of analogous to the (highly technical) definition of "asymptotically flat."

    Are there two different conventions in terms of terminology? It seems like Ludvigsen and the author of the WP article on Birkhoff's theorem may be defining stationarity in one way, so that the Schwarzschild spacetime inside the horizon is not a counterexample, while Hawking and Ellis may be defining stationarity in another way, which is why they have to split the two cases and distinguish "static" from "spatially homogeneous." Carroll's treatment does not seem self-consistent, though; he defines stationarity by saying that "a metric which possesses a timelike Killing vector is called stationary," but then refers to the Schwarzschild metric as stationary.
     
  2. jcsd
  3. Aug 22, 2010 #2
    Is it really Killing vector below the horizon?
     
  4. Aug 22, 2010 #3

    bcrowell

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    Yes, I think so. Is anything wrong with the following argument? The form of the metric is the same on both sides of the horizon, so its components are still independent of t below the horizon. When a coordinate system exists in which the components of the metric are independent of a particular coordinate q, [itex]\partial_q[/itex] is guaranteed to be a Killing vector.
     
  5. Aug 23, 2010 #4
    I don't know the answer, but in Schwarzschid coordinates the metric has a singulariry at the event horizon. Below the horizon "t" is no more time-like coordinate, and the order of the signature -+++ changes to +-++ if you want to use the same coordinates in the same order.
     
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