Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question
2  
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. –  Michael Brown Mar 5 '13 at 1:54
    
@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? –  Super Frog Mar 5 '13 at 16:38
    
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. –  Michael Brown Mar 6 '13 at 2:15
add comment

1 Answer

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.