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An object cannot escape the event horizon by catapulting it outside the black hole. However, what if instead of relying on escape velocity, the object was tethered to a ship orbiting well outside the event horizon? The object needs not to be pulled out at speeds higher than $c$, but rather can be pulled slowly.

Assume the gravity gradient between the ship's orbit and the tether's other end is manageable, the mass of the object is small enough to be pulled without too much burden on the ship's engines, and the tether is strong enough. (Or it's just a loose end). Would such an object be pulled-out of the event horizon?

Qmechanic
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Christmas Snow
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3 Answers3

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In General Relativity, no amount of force, exerted through a tether or in any other way, can extract an object from the interior of a black hole. There are no “tricks” to get around this fact, any more than there are tricks to make a perpetual motion machine possible.

All future-directed timelike worldlines within the interior lead to the singularity, not just ones for freely falling objects. This is a consequence of the black hole’s geometry.

The gravity gradient is irrelevant. The mass of the object is irrelevant. The strength of the tether is irrelevant. All that matters is the spacetime geometry and the possible worldlines that it allows.

G. Smith
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    It should be added that there is no way the tether would keep the object at a constant radius beyond the event horizon, either. It will break or (if "indestructible") pull the spaceship in. – The Vee Aug 28 '19 at 08:46
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    @TheVee Well, in a world with indestructible tethers, black holes are impossible and vice versa, so you don't really need to consider them :) – Luaan Aug 28 '19 at 10:11
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The problem here is that while the escape speed is indeed $c$ at the horizon, the classical (Newtonian) way of conceiving of the significance of that speed is much different in general relativity versus in Newtonian gravity - or, actually, it can be thought of in the same way in both, but it really becomes important when you are considering general relativity.

You should not think of the escape speed as simply being a speed which only applies to ballistic escapes, but rather as a speed that signifies how hard it is to escape the gravity well of whatever the thing in question is, by any method. Namely, it is "as hard as accelerating your whatsit up to that speed", whether you actually achieve that acceleration or not: remember that when considering a "slow" climb out of a gravity well even in Newtonian mechanics, your rockets have to be firing continuously and they will use at least (and likely much more) as much energy and propellant as reaching that escape speed requires despite the slow climb.

Hence, when in general relativity you see escape velocity $c$, what it really means is "it is as hard to get away from here as it is to travel at exactly the speed of light". In other words, right at the horizon, getting away is equivalent to sending your massive spaceship at the speed of light: something you already should know from special relativity is infinitely hard. Going below the horizon, it becomes "beyond infinite" - so hard that it is described with an imaginary number, which in relativityese actually means "as hard as going faster than light" and thus "as hard as making a time machine", and therefore, you really really can't get away.

Likewise the same applies to tethers: you will have to pull on it with infinite muscle to get it just up from being exactly at the horizon, and no rope can be infinitely strong, so when lowering something, all ropes must break before the suspended objects reach the horizon.

(In a sense, you could say "gravity becomes infinitely strong" at the horizon, not the singularity, but it's better to say "gravity becomes irresistible", or that the hovering force becomes infinite, because the "strength of gravity" has other definitions that are more appropriate to the general-relativistic setting. In particular, the gravitational field must be described by a tensor, not a vector, in general relativity, and this tensor does not become infinite at the horizon, but the function mapping from this tensor field to the needed hovering force does become infinite there.)

The_Sympathizer
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  • Is that allusion to imaginary (complex) numbers ever representing any "numbers beyond infinity", er, real mathematical apparatus of GR, or was it just a non-mathematical metaphor? The idea sounds to me a bit surprising as I normally don't see complex numbers used as linearly ordered in any way. – Jirka Hanika Aug 28 '19 at 16:53
  • @Jirka Hanika : Yeah, it's more of a metaphor. – The_Sympathizer Aug 28 '19 at 20:56
  • Although, I suppose, you could try to make it more mathematically literal by considering the codomain of the hovering-strength function to be a linearly ordered set consisting of the union of the positive reals and positive imaginaries where the latter are ordered to be after the former. It will still be at least order-consistent with $+$, though not $*$ by non-real numbers. – The_Sympathizer Aug 28 '19 at 20:57
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The object needs not to be pulled out at speeds higher than c, but rather can be pulled slowly.

For simplicity, consider the Schwarzschild black hole solution. Inside the horizon, the world-lines of 'outwardly' directed light not only remain within the horizon, they end on the singularity (the singularity is in the future of all world lines within the horizon).

The world line of a massive object remains within its future light cone (since speed must be less than $c$) and so must also end on the singularity if within the horizon.

To 'pull the object out' would require that the object's world-line cross out of its future light cone which is as impossible as is having a speed greater than $c$.

Hal Hollis
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