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.

If space is warped by objects in space, and black holes are made of infinitely heavy objects, can space be torn by black holes?

share|improve this question
1  
I'm a bit surprised at the downvotes. While this is a naive question it seems to me to be a fair one. –  John Rennie Feb 17 at 9:11
1  
see arxiv.org/abs/hep-th/9109030 for a related paper from the arxiv; extending general relativity to degenerate metrics, some apparently violent processes sometimes can be realized as smooth deformations; also, I believe string theory has some things to say about the subject as well... –  Christoph Feb 17 at 9:45
    
At least it can be "twisted" –  Theraot Feb 17 at 13:47
add comment

3 Answers 3

That's a surprisingly subtle question, and it depends on what you mean by torn.

You've probably seen the rubber sheet analogy for spacetime, and it's tempting to think that because a rubber sheet will snap if you stretch it too far maybe the same thing will happen to spacetime. However this is taking the analogy too far. Spacetime isn't an object, it's a manifold. The term manifold has a precise mathematical definition that we don't need to go into - you can think of it simply as something you can move around in.

Anyhow, spacetime can be deformed as much as you want. Indeed, at the singularity in a black hole the deformation becomes infinite, though most of us think some theory of quantum gravity will become important at very small distances and will prevent the curvature becoming infinite. So even at the centre of a black hole space isn't torn.

You could argue that creating a hole in spacetime is similar to tearing it, and you could also argue that this is what a wormhole does. If you could get hold of enough exotic matter this could be used to create a wormhole. I discuss this in my answer to Negative Energy and Wormholes. Whether this constitutes a tear is a matter of debate.

Actually a spinning or charged black hole can act as a sort of wormhole because it can join two causally disconnected areas of spacetime. See my answer to Entering a black hole, jumping into another universe---with questions for more on this. Again, whether this constitutes a tear is debatable.

share|improve this answer
    
Thanks, your answer helps a lot. Or comes close. Doesn't quantum mechanics theorizes the possibility of multiple universes close to one another? If so is it possible to have space warped/torn enough to touch the other universe. And what would be the results. Would the warped space return to a normal state once the matter is released. Isn't it true that matter is neither created nor destroyed? I know it can be transformed but Could it be transferred? –  user40752 Feb 20 at 20:42
    
@user40752: There are speculative, and I emphasise speculative, theories based on eternal inflation. In these theories spacetime is still continuous, but you get universes of normal spacetime embedded in an inflating spacetime. It is possible for these universes to overlap. Nothing in this theory requires spacetime to be torn. Actually it's quite an interesting area and there are several questions about it on this site. –  John Rennie Feb 21 at 7:56
add comment

Black holes are not "infinitely heavy"; all black holes have a finite mass. Nevertheless, this mass is indeed concentrated, as far as we can tell, at a single point, so that a static black hole can be said to be infinitely dense. Within the framework of classical General Relativity, black holes contain a singularity which can be said to be a "tear" in spacetime, but that is very informal language, is not used by any serious physicist, and should be avoided because it clouds the exact nature of the singularity.

I must also mention that no serious physicist expects General Relativity to be valid all the way down to those scales. We expect generally that some form of quantum gravity, yet to be discovered, will take over before that, and that it will not contain either infinite densities nor singularities.

share|improve this answer
add comment

First, black holes are not infinitely heavy. They tend to have well-defined masses, even locally.

http://science.nasa.gov/astrophysics/focus-areas/black-holes/

By locally, I mean that the Newtonian limit tends to work, and you don't really need the ADM definitions.

http://en.wikipedia.org/wiki/Mass_in_general_relativity

Perhaps you mean the singularity, which doesn't have infinite mass, and whose exact definition is hard to pin down, but the best is due to Hawking and Ellis:

Given the generalized affine parameter $u$ on $\lambda(t)$, a singly-differentiable curve through the point $p$ on the manifold $M$ equipped with a tangent bundle containing the tangent vectors $V=(\frac{\partial}{\partial_t})_{\lambda_t}$:

$$u=\int_p \sqrt{\sum_i V^i V^i}dt$$

Then the Manifold $M$ with the metric $g$ is bundle complete if there is an endpoint for every $C^1$ curve of finite length as measured by $u$. (Recall that $C^n$ means n-times differentiable.)

A singularity lies on a curve which is not bundle complete. (Hawking and Ellis, Chapter 8).

Now, there exists a class of solutions to GR called the Pp-wave spacetime:

http://en.wikipedia.org/wiki/Pp-wave_spacetime

For which Sir Penrose has found a particular solution, infamously nicknamed the thunderbolt or "wave of death".

http://www.ias.ac.in/jarch/jaa/20/233-248.pdf

This is a non scalar null curvature singularity which propagates through spacetime, destroying it as it goes. Any paths containing the thunderbolt are geodesically incomplete, essentially making the thunderbolt track a rip in spacetime.

References:

Hawking, Stephen W., G. F. R. Ellis, P. V. Landshoff, D. R. Nelson, D. W. Sciama, and S. Weinberg. The Large Scale Structure of Space-Time. Cambridge University Press, 1975.

share|improve this answer
add comment

protected by Qmechanic 2 days ago

Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?

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