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Spacelike slicing of Schwarzschild geometry

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I am having trouble understanding how to obtain a spacelike slicing of the Schwarchild black hole. I understand there is not a globally well defined timelike killing vector, so we can define t=cte slices outside the horizon and r=cte slices inside the horizon.

In the literature people define connector slices that join these two spacelike surfaces.

What is the formal definition of a spacelike connector slice? What is the most practical way to go about finding its mathematical expression?

This post imported from StackExchange Physics at 2014-03-22 16:57 (UCT), posted by SE-user Super Frog
asked Mar 5, 2013 in Theoretical Physics by Super Frog (5 points) [ no revision ]
Is there a reason you're not using Kruskal coordinates? In those coordinates there is no horizon singularity, and the "radial" coordinate stays spacelike everywhere.

This post imported from StackExchange Physics at 2014-03-22 16:57 (UCT), posted by SE-user Michael Brown
@MichaelBrown Thanks for the answer. I assume once I have my slice in Kruskal coordinates I can go back to any other coordinate system and I will see how the connector looks like?

This post imported from StackExchange Physics at 2014-03-22 16:57 (UCT), posted by SE-user Super Frog
I'm not exactly sure what a "connector" is - but yes, you can certainly go to any other coordinate system from Kruskal. If there is a coordinate singularity at the horizon in the new coordinate system then the coordinate transformation will have a singularity that you need to be careful about, but there is no reason in principle that it shouldn't work.

This post imported from StackExchange Physics at 2014-03-22 16:57 (UCT), posted by SE-user Michael Brown

1 Answer

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When defining a foliation by spacelike slices given by a function $t$=const, there is no need to require $\frac{\partial}{\partial t}$ to be a Killing vector. For example you can foliate a Schwarzschild spacetime by using $t$=const slices in the Painleve Gullstrand form of the metric $$ ds^2 = -(1-\frac{2M}{r})dt^2 + 2\sqrt{\frac{2M}{r}}dtdr + dr^2+r^2d\Omega^2$$ Or, as Michael Brown said in the comment, Kruskal coordinates is another choice. Both these choices are nonsingular across the horizon.

This post imported from StackExchange Physics at 2014-03-22 16:57 (UCT), posted by SE-user twistor59
answered Mar 5, 2013 by twistor59 (2,490 points) [ no revision ]

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