# Spacelike slicing of Schwarzschild geometry

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?

• 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

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.
Your hypersurfaces do not meet. Here is a spacetime diagram in Kruskal-Szekeres coordinates, taking the black hole mass to be $M=1$, and suppressing the $\theta$ and $\phi$ coordinates. The straight blue line is the hypersurface $t=-1$, for all $r>2M$. The curved blue line is the hypersurface $r=1.75$ for all $t$. Actually this is clear from thinking in Schwarzschild coordinates: a $r=\textrm{const}<2M$ surface cannot intersect a $2M<r<\infty$ surface (i.e. the $t=\textrm{const}$ surface). 