# Is the Schwarzschild singularity stretched in space as a straight line?

I am trying to visualize the Schwarzschild geometry and would appreciate a help of the experts. The geometrized radial ($$\theta=\phi=0$$) Schwarzschild metric outside the horizon is

$$d\tau^2 = \left(1-\dfrac{r_s}{r}\right)\,dt^2\tag{1}-\left(1- \frac{r_s}{r}\right)^{-1} \,dr^2$$

Inside the horizon, the metric becomes

$$d\tau^2 = \left(\frac{r_s}{r}-1\right)^{-1} \,dr^2 - \left(\frac{r_s}{r}-1 \right)\,dt^2\tag{2}$$

which is the very same equation just differently rearranged for clarity. The radial coordinate $$r$$ is spacelike outside, but timelike inside. Similarly, the $$t$$ coordinate is timelike outside, but spacelike inside. Using symmetry, we can plot this space in a reduced number of dimensions as shown below (where obviously the plotting grid does not represent actual intervals since they are dilated, etc.)

If this logic and the plot are correct, $$t$$ inside the horizon represents a spatial coordinate that does not point to the singularity thus stretching the singularity in space into a line along this coordinate. Geodesics of free falling objects (solid curves below) and light rays (dotted curves below) terminate at different points of this line (the vertical axis below)

Is this interpretation correct? Otherwise, where is the logical error and what is the correct interpretation?

I realize that geodesics are undefined at $$r=0$$, so the singularity is not an ordinary spacelike interval. However, this question is simply whether the singularity is "stretched in space along $$t$$" or "focused to a point in all dimensions" (as many believe).

All coordinates are in the Schwarzschild frame of reference of a distant observer. This question is about geometry of spacetime. Any issues related to matter or its density in the singularity are out of scope. I would appreciate an answer rather than a comment, even if brief. Thanks for your help!

EDIT: Based on comments, this question requires a more precise definition, so here it is:

At $$r\ll r_s$$, is the hypersurface $$r=const$$ spacelike and infinitely long?

Or is it spatially compact rather than stretched along one dimension?

• This is maybe more suited for Physics.SE or MathOverflow. – Giuseppe Negro Sep 25 '18 at 7:33

## 2 Answers

Given your update, the $$r=const$$ slices are uniform in $$t$$:

$$\mathrm{d}\tau = \sqrt{\left(1- \frac{r_s}{r}\right)^{-1}} \,\mathrm{d}t = (\mathrm{const}) \cdot \mathrm{d}t$$

So you get an ordinary Euclidean line.

It's not an easy task to define topology and even harder to define geometry of your singularity. You can consider your singularity to be the set of time-like geodesics ending in finite time. You would probably want to identify some of those geodesics with some others, if they become too "close" to each other. And then somehow put a topology on this set. But how would you do that?

Mathematics starts with definitions, unless you can provide a definition that you want to use, your question is too vague for a mathematician. In fact, one can give definitions to get both answers: a point and a $$3$$-dimensional surface.

The intuition for the latter is similar to yours and comes from the Penrose diagram (which can be drawn nicely using Krushkal coordinates). I can't speak for everyone, but my intuition for the point comes from the fact that the event-horizon is a time-like sphere of constant finite size. While the universe I don't imagine as a single point at $$t=0$$. But all of this is imprecise.