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.

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.

share|improve this question
3  
General relativity only restricts the speed of objects so close to each other that the variation in the gravitational field is close to zero. For such observers, there will be no observable gravitational force, and thus your object will still be travelling slightly under the speed of light. Relative to a distant observer, the particle will be moving "faster than light" once it crosses the black hole's horizon. –  Jerry Schirmer Nov 8 '12 at 14:08
    
In this way we also put the laser at same distance and direct it to the black hole. We turn on the lasert and our proton overtakes the ray? –  mirt Nov 8 '12 at 14:21
1  
No. The ray moves in the same background geometry and is always moving faster, locally, than the proton. –  Jerry Schirmer Nov 8 '12 at 14:23
    
Actually i did't understand what a distant observer will see after let's say 3 seconds after start. –  mirt Nov 8 '12 at 14:26
    
The light would outpace the proton, full stop. The actual experiment would depend a lot on the details of the setup. Both would appear to be travelling faster than light relative to the distant observer after they crossed the horizon (let's assume the BH is expanding so that distant observers observe a crossing) –  Jerry Schirmer Nov 8 '12 at 14:28
show 2 more comments

2 Answers 2

up vote 3 down vote accepted

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.

share|improve this answer
    
I'm certainly playing a bit fast and loose with "moving" in this context. You are right to say that things inside the horizon are unobservable, and so there is a bit of a wrongness to the claim. The formal statement would be that a freely falling frame inside the horizon would appear to be moving superlumminally with respect to a freely falling frame outside the horizon--the frames are dragged radially past any static limit--it's basically thinking of the Schwarzschild horizon the same way that you would think of the cosmological horizon. –  Jerry Schirmer Nov 8 '12 at 17:15
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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