2
$\begingroup$

It is well known that nothing can escape a black hole, including gravitational radiation. Many questions have been asked here about this topic, such as:

and they all have the same answer. Gravitational radiation cannot propagate faster than the speed of light.

On the other hand, it also seems to be generally agreed that if an Alcubierre drive is physically possible, one can theoretically use one to escape a black hole. (Could a ship equipped with Alcubierre drive theoretically escape from a black hole?)

Now an Alcubierre drive is essentially just a particular spacetime metric. You don't even need negative energy density to make one. In particular, it was recently shown that one can construct a superluminal soliton solution in general relativity using only positive energy densities.

Connecting these ideas, it seems that it should be possible, in principle, to fall through the event horizon of a black hole; construct an Alcubierre spacetime by manipulating a collection of very massive objects; and escape the event horizon with information from inside the black hole. Then my question is: where is the line between gravitational waves and Alcubierre spacetimes? Could there be other Alcubierre-like metrics that do not contain a full warp bubble, but still allow information to escape from a black hole?

$\endgroup$
3
  • $\begingroup$ " construct an Alcubierre spacetime by manipulating a collection of very massive objects", inside the horizon? before spaggetification?en.wikipedia.org/wiki/Spaghettification $\endgroup$
    – anna v
    Jun 25, 2021 at 19:19
  • 1
    $\begingroup$ @annav Let us suppose that spaghettification is not a concern. Either our spacecraft is made of very strong materials, or we are dealing with a very large black hole. As Wikipedia points out, an astronaut may cross the event horizon of a supermassive black hole without noticing any squashing and pulling. $\endgroup$
    – Thorondor
    Jun 25, 2021 at 19:20
  • 2
    $\begingroup$ Any external observer would not see you crossing the horizon in the eternity of time in this universe. And you cannot return before you cross. Thus you cannot return to this universe while it still exists. $\endgroup$
    – safesphere
    Jun 25, 2021 at 22:27

2 Answers 2

2
$\begingroup$

There is a fundamental problem with this scheme: constructing a warp drive spacetime. The papers exploring such spacetimes all work by (1) assuming some negative energy density exotic matter to hold together the bubble, and (2) that the bubble already exists. (1) is problematic since we have never seen such exotic matter, but (2) is more fundamental: there are to my knowledge no warp drive spacetimes that start locally as flattish spacetime, become a spacetime with a bubble, and then stop having a bubble.

Indeed, there are theorems in general relativity that state that if you have a Cauchy surface (a "flat" initial state at some coordinate time) then you will always retain that topology at later times. However, these theorems are broken by the exotic matter assumption, so how much they can be trusted here is debatable. Similarly the topological censorship theorem fails here. The issue here is that by allowing arbitrary exotic fields we can make spacetime manifolds with almost any crazy topology... but these fields are not guaranteed to exist or even be possible.

But if you can make warp drive spacetimes out of positive matter densities then the second difficulty gets you instead. Now those theorems apply, and you cannot get the nontrivial topology of spacetime trajectories you want from a warp drive. Even disregarding the issue of what the black hole would do to them.

$\endgroup$
0
$\begingroup$

The reality is that the question you are asking is just too hard to answer at the moment.

The state of the art can barely describe the properties of isolated warp drives solutions in flat space time, while this would be a warp drive falling in the curved geometry of a black hole. As the boundary conditions and assumptions of the most common warp solutions (e.g. Alcubierre) are invalidated, you would need to perform a numerical analysis to get accurate predictions.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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