The escape velocity of a black hole at the event horizon is the speed of light, this means an object that is dropped from the top of the black hole's gravitational well will reach the speed of light (just like an object dropped from the top of earth's gravitational well will reach 15km/s when it hits earth's surface, ard 15km/s is earth's escape velocity at earth's surface). So what if the ball is thrown towards the ground? In the earth example, it will reach 15.01km/s, and in the black hole example it will reach 300,000.01km/s. Does this mean the speed of light in not a speed limit?
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To make this question meaningful, you would have to say what this velocity is to be measured relative to. Furthermore, it has to be measured relative to something local, not something distant. See How do frames of reference work in general relativity, and are they described by coordinate systems?
Relative to some other material object that is also at the horizon, the ball can have any speed you like, as long as it's less than $c$. It depends on the ball's object relative to the other object.
You can say that you want the other object to be one that's at rest relative to the black hole, but that won't work. There are no static, timelike world-lines at or within the event horizon.
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the velocity of the ball would be measured relative to whatever the escape velocity is relative to. thought that was obvious... – Hierarchist Nov 24 '19 at 14:54
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@Hierarchist: When we talk about an escape velocity, we're talking about a velocity relative to a static observer. As explained in the answer, there are no static, timelike world-lines at or within the event horizon. – Nov 25 '19 at 16:24
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So there is no escape velocity at the event horizon? LOL – Hierarchist Nov 25 '19 at 16:33
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I'm not talking about the event horizon, I'm talking about one plank distance away from the event horizon. – Hierarchist Nov 26 '19 at 19:42
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@safesphere, if the escape velocity at some point in space is some velocity, then that is the velocity of the falling object. – Hierarchist Nov 27 '19 at 22:11
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@safesphere, escape velocity is relative, yes. – Hierarchist Nov 28 '19 at 16:41
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You are not saying it, but I assume you are talking about an object with rest mass. An object with rest mass can never reach the speed of light as per SR.
The velocity of the dropped particle is hiding in the $\gamma$. After some algebra we can find: $$ v = c \sqrt{1 - \left[ \frac{1}{\frac{GM}{c^2} \left( \frac{1}{r_\mathrm{final}} - \frac{1}{r_\mathrm{initial}} \right) + 1} \right]^2}$$ You can check and see that the square root term is always less than 1, so the dropped particle does not go faster than the speed of light!
You are correct, the escape velocity of the black hole is c at the event horizon.
The escape velocity from the surface (i.e., the event horizon) of a Black Hole is exactly c, the speed of light.
What is the escape velocity of a Black Hole?
Though, as per SR, no object with rest mass can reach the speed of light, and even if you drop in your example an object towards the event horizon, it will never reach the speed of light.
Árpád Szendrei
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1"You are correct, the escape velocity of the black hole is c at the event horizon.... Though, as per SR, no object with rest mass can reach the speed of light, and even if you drop in your example an object towards the event horizon, it will never reach the speed of light." That's a contradiction. – Hierarchist Nov 24 '19 at 14:45
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@Hierarchist There is no contradiction in the quoted statement. Objects falling to a black hole never reach the speed of light. See section 3 here for a detailed explanation: https://arxiv.org/abs/0804.3619 – safesphere Nov 24 '19 at 18:02
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@safesphere, if the escape velocity at the event horizon is the speed of light, then so too is the speed of an object falling from the top of the gravitational well to the event horizon. Let me know if you need help understanding this. – Hierarchist Nov 24 '19 at 19:39
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I am so delighted to see this equation! I would just point out that if you plug in the limits of an initial distance of infinity and a final distance of the schwarzschild radius, you end up with a fall velocity of 0.75c. I came up with the same result using a highly simplistic approach setting kinetic energy equal to the gravitational potential energy. As to what the velocity is relative to, that would be to the same reference frame in which we measure the schwarzschild radius as having length 2GM/c^2. – Ralph Berger Dec 05 '20 at 17:18
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2@Ralph Bear in mind that the above equation (quoted from Paul T.'s answer) is a semi-relativistic hybrid. It uses Special Relativity to calculate kinetic energy and Newtonian gravity to calculate potential energy. And it totally ignores the issue of coordinate systems. It's an ok approximation in (relatively) flat spacetime, but it is not accurate in the highly curved spacetime in the vicinity of a black hole. You really need General Relativity to do this properly. – PM 2Ring Dec 06 '20 at 10:59