Can gravity accelerate an object past the speed of light?

Question

Imagine we have something very heavy (i.e supermassive black hole) and some object that we can throw with 0.999999 speed of light (i.e proton). We are throwing our particle in the direction of hole. The black hole is so heavy that we can assume that in some moment acceleration of gravity would be say 0.0001 speed of light/s^2. So the question is what will be the speed of proton in few a seconds later, assuming we have such distances that it will not hit the black hole before.

Answer 1

This the classic "hurling a stone into a black hole" problem. It's described in detail in sample problem 3 in chapter 3 of Exploring Black Holes by Edwin F.Taylor and John Archibald Wheeler. Incidentally I strongly recommend this book if you're interested in learning about black holes. It does require some maths, so it's not a book for the general public, but the maths is fairly basic compared to the usual GR textbooks.

The answer to your question is that no-one observes the stone (proton in your example) to move faster than light, no matter how fast you throw it towards the black hole.

I've phrased this carefully because in GR it doesn't make sense to ask questions like "how fast is the stone" moving unless you specify what observer you're talking about. Generally we consider two different types of observer. The Schwarzschild observer sits at infinity (or far enough away to be effectively at infinity) and the shell observer sits at a fixed distance from the event horizon (firing the rockets of his spaceship to stay in place).

These two observers see very different things. For the Schwarzschild observer the stone initially accelerates, but then slows to a stop as it meets the horizon. The Schwarzschild observer will never see the stone cross the event horizon, or not unless they're prepared to wait an infinite time.

The shell observer sees the stone fly past at a velocity less than the speed of light, and the nearer the shell observer gets to the event horizon the faster they see the stone pass. If the shell observer could sit at the event horizon (they can't without an infinitely powerful rocket) they'd see the stone pass at the speed of light.

To calculate the trajectory of a hurled stone you start by calculating the trajectory of a stone falling from rest at infinity. I'm not going to repeat all the details from the Taylor and Wheeler book since they're a bit involved and you can check the book. Instead I'll simply quote the result:

For the Schwarzschild observer:

$$ \frac{dr}{dt} = - \left( 1 - \frac{2M}{r} \right) \left( \frac{2M}{r} \right)^{1/2} $$

For the shell observer:

$$ \frac{dr_{shell}}{dt_{shell}} = - \left( \frac{2M}{r} \right)^{1/2} $$

These equations use geometric units so the speed of light is 1. If you put $r = 2M$ to find the velocities at the event horizon you'll find the Schwarzschild observer gets $v = 0$ and the (hypothetical) shell observer gets $v = 1$ (i.e. $c$).

But this was for a stone that started at rest from infinity. Suppose we give the stone some extra energy by throwing it. This means it corresponds to an object that starts from infinity with a finite velocity $v_\infty$. We'll define $\gamma_\infty$ as the corresponding value of the Lorentz factor. Again I'm only going to give the result, which is:

For the Schwarzschild observer:

$$ \frac{dr}{dt} = - \left( 1 - \frac{2M}{r} \right) \left[ 1 - \frac{1}{\gamma_\infty^2}\left( 1 - \frac{2M}{r} \right) \right]^{1/2} $$

For the shell observer:

$$ \frac{dr_{shell}}{dt_{shell}} = - \left[ 1 - \frac{1}{\gamma_\infty^2}\left( 1 - \frac{2M}{r} \right) \right] ^{1/2} $$

Maybe it's not obvious from a quick glance at the equations that neither $dr/dt$ nor $dr_{shell}/dt_{shell}$ exceeds infinity, but if you increase your stone's initial velocity to near $c$ the value of $\gamma_\infty$ goes to $\infty$ and hence 1/$\gamma^2$ goes to zero. In this limit it's easy to see that the velocity never exceeds $c$.

In his comments Jerry says several times that the velocity exceeds $c$ only after crossing the event horizon. While Jerry knows vaaaaastly more than me about GR I would take him to task for this. It certainly isn't true for the Schwarzschild observer, and you can't even in principle have a shell observer within the event horizon.

Answer 2

Jerry's comment is perfect. Just explaining something I've understood...

I'd advice that normal black-holes are far better than the super-massive ones. Because, they're the largest of 'em and hence effectively less gravitational effect on your proton.

Anyways, The answer is NO because of two things - First of all, Newton's laws (like acceleration of protons) are unusable relative to a black-hole's event horizon. And second, relativity generally restricts a motion faster than light..! Relativity concludes that you'd measure these motions relatively and not absolutely. So, we're using an observer like yourself. General relativity says that gravity influences both space & time to bend out, thereby taking a shorter path which our guys call it - "a geodesic motion"...

Ok. Now, to your question... Let's assume that you're sending something similar to a laser beam of protons. If you're able to see those protons, you'll definitely see a red-shifted beam (becoming dimmer & dimmer with distance) as it approaches the horizon (let's just ignore that it disappears). Now, All paths of the protons turn toward the event horizon of the black-hole where space-time also curves further & further. Even light bends, thereby taking the shortest path (which seems to be accelerated). Instead of mentioning "accelerated", relativity says it gets "curved". Hence, I'd conclude that you'd never cross light speed at any time.

I'm also quoting Jerry's comment that, The protons would appear to travel faster than light relative to you, but you can't observe that in that case because, we can't observe anything inside the black-hole.

Answer 3

I may not have the technical expertise of some here, but I think I can illustrate this idea in a more practical, yet accurate, way.

Imagine you’re floating in a calm lake that connects to a large, deep river flowing toward a massive waterfall. The water represents “space,” and the waterfall represents the black hole. Let’s picture the river as wide and calm on the surface, even though it has a powerful current moving toward the edge.

In water, objects typically have a maximum speed limit due to cavitation—around 67 mph. Although advanced engineering can allow objects to exceed this limit, let’s keep things simple and assume that 67 mph is the absolute speed limit in our illustration, analogous to the “speed of light.”

Now, imagine you’re observing a speedboat in the river, moving at 65 mph toward the waterfall. At first, this seems possible and safe within the speed limits. But as the boat approaches the waterfall, it gets carried by the river’s current, which might add, say, 15 mph to its speed. Suddenly, it appears that the boat is moving at 80 mph, seemingly breaking the speed limit of 67 mph.

However, it’s not actually breaking any limits—the boat is still moving at 65 mph relative to the water around it. The “extra speed” is due to the river itself carrying the boat faster toward the waterfall.

This is much like how a black hole affects objects moving near it. The intense gravity doesn’t actually cause the object to accelerate beyond the speed of light; instead, space itself is “falling” toward the black hole, carrying the object with it. The object’s speed, relative to its surrounding space, remains under the speed of light, just as the boat’s speed remains under the 67 mph limit relative to the water.

To an outside observer, it may look as if the object is breaking the speed limit, but it’s not; rather, it’s the space around it moving faster. The point where this effect takes hold completely is called the event horizon. Beyond this point, space is “falling” into the black hole faster than the speed of light, making it impossible for anything—even light—to escape back out. The object itself remains “stationary” within its local space, but from our perspective, it has been swept beyond reach.