Speed at singularity

[Ender]

I was thinking about escape velocities and I ended up thinking about black holes and what would happen

if someone or something fell into one. In particular, I asked myself: if something was to start still at the

event horizon and then, drawn by the gravitational force, was start "falling" toward the center,

at what speed would it reach the singularity? Does it even make sense to ask a question like that?

 

I have some calculus that may make no sense, but I wanted to show it to you to know what you thought:

Probably you all know how to calculate the escape velocity:

E® is the initial kinetic and gravitational potential energy, on the surface of the planet

E(infinity) is the energy when you reach an infinite distance from the planet

 

E®=1/2*m*v^2-G*M*m/r

E(infinity)=0

 

From E®=E(infinity) you find v.

But what if you thought about it in another way?

 

E®=mgr

E(0)=1/2*m*v^2

 

In this case it's like if you are starting from the surface of the planet and you're falling at the center of it.. Does that make sense?

If you use this last formula to find the escape velocity from Earth you'll find that it's the same (because g=G*M/r^2)

but this time you're falling in, not out, so it's the velocity you have when you arrive at the center.

Now let's apply it to black holes: on the event horizon the escape velocity is c.. does that mean that if you start on the

event horizon and fall down the black hole, when you arrive at the singularity your velocity is c?

 

Thank you very much for your time.

 

I'm sorry, all the ( r ) became ®

[xyzt]

I was thinking about escape velocities and I ended up thinking about black holes and what would happen

if someone or something fell into one. In particular, I asked myself: if something was to start still at the

event horizon and then, drawn by the gravitational force, was start "falling" toward the center,

at what speed would it reach the singularity? Does it even make sense to ask a question like that?

 

I have some calculus that may make no sense, but I wanted to show it to you to know what you thought:

Probably you all know how to calculate the escape velocity:

E® is the initial kinetic and gravitational potential energy, on the surface of the planet

E(infinity) is the energy when you reach an infinite distance from the planet

 

E®=1/2*m*v^2-G*M*m/r

E(infinity)=0

 

From E®=E(infinity) you find v.

But what if you thought about it in another way?

 

E®=mgr

E(0)=1/2*m*v^2

 

In this case it's like if you are starting from the surface of the planet and you're falling at the center of it.. Does that make sense?

If you use this last formula to find the escape velocity from Earth you'll find that it's the same (because g=G*M/r^2)

but this time you're falling in, not out, so it's the velocity you have when you arrive at the center.

Now let's apply it to black holes: on the event horizon the escape velocity is c.. does that mean that if you start on the

event horizon and fall down the black hole, when you arrive at the singularity your velocity is c?

 

Thank you very much for your time.

 

I'm sorry, all the ( r ) became ®

Your calculations are based on Newtonian mechanics, as such they are invalid since Newtonian mechanics cannot deal with the issue correctly. In order to get the correct answer you need to apply GR. When you do that you find that the speed is [math]c\sqrt{\frac{r_s}{r}}[/math]. At the singularity , [math]r=0[/math] so....

On a different note, at the even horizon [math]r=r_s[/math], so the infalling particle attains light speed already there.

Edited June 2, 2014 by xyzt

[Ender]

Ok but what is r_s? Is it a distance? Because if it is, the formula doesn't give the dimension of a speed..

Thank you for the response by the way

[xyzt]

Ok but what is r_s? Is it a distance? Because if it is, the formula doesn't give the dimension of a speed..

Thank you for the response by the way

[math]r_s[/math] is the Schwarzschild radius.

[Ender]

Ok thank you. But when the infalling object is, for example, a person, he cannot reach the speed of light. And also, maybe

there's a spaceship bringing him right near the Schwarzschild radius and giving him an initial speed that's almost zero, how

can he reach the speed of light when at the Schwarzschild radius?

[xyzt]

But when the infalling object is, for example, a person, he cannot reach the speed of light.

GR says that it can. Not only that it can, but, upon crossing the event horizon, it will exceed the speed of light.

For a rigorous derivation, see here.

Edited June 2, 2014 by xyzt

[MigL]

The spaceship that's bringing him arbitrarily close to the event horizon is, by necessity, moving at very close to the speed of light. And so is he.

 

Does that help clarify the situation, Ender ?

[md65536]

The horizon is a light-like surface, and it doesn't have a rest frame, so it doesn't make sense to compare an object's speed relative to it. Similarly, it doesn't make sense to say that we're all traveling at c relative to a beam of light. It makes no sense conceptually or mathematically, and trying to imagine it will confuse.

 

Just as a pulse of light passes an object at c, the horizon passes an in-falling object at c, but in neither case is the object traveling at c relative to anything.

 

 

If it doesn't make sense, consider that how things look near the horizon is very different depending on if you're inertial or trying to escape. At the horizon, escaping is equivalent to trying to outrun light (conceptually and mathematically impossible), and approaching that approaches infinite length contraction, severely distorting space relative to the inertial in-falling observer.

[Ender]

The spaceship that's bringing him arbitrarily close to the event horizon is, by necessity, moving at very close to the speed of light. And so is he.

 

Does that help clarify the situation, Ender ?

Ok, maybe I understand what you're saying. But, let me know if

I'm wrong, couldn't it be that the spaceship is just accelerating

away from the horizon with an acceleration equal to the one that's

pulling it in (since you're not in the black hole you should be able to do that)?

And in that case it's like if you were still and when you let go of the

person, he will be pulled inside starting from almost stil.. Does that make sense?

 

The horizon is a light-like surface, and it doesn't have a rest frame, so it doesn't make sense to compare an object's speed relative to it. Similarly, it doesn't make sense to say that we're all traveling at c relative to a beam of light. It makes no sense conceptually or mathematically, and trying to imagine it will confuse.Just as a pulse of light passes an object at c, the horizon passes an in-falling object at c, but in neither case is the object traveling at c relative to anything.If it doesn't make sense, consider that how things look near the horizon is very different depending on if you're inertial or trying to escape. At the horizon, escaping is equivalent to trying to outrun light (conceptually and mathematically impossible), and approaching that approaches infinite length contraction, severely distorting space relative to the inertial in-falling observer.

I don't think I understood what you said. You are saying that the horizon is moving at

the velocity c? And the infalling observer would experience lenght contraction at the

horizon? But I've always read that due to GR the horizon must be a place like

everywhere else in the Universe right? So I thought it would be nothing fancy, passing through the horizon..

[md65536]

I don't think I understood what you said. You are saying that the horizon is moving at

the velocity c? And the infalling observer would experience lenght contraction at the

horizon? But I've always read that due to GR the horizon must be a place like

everywhere else in the Universe right? So I thought it would be nothing fancy, passing through the horizon..

Yes, the horizon is moving at the speed of light. Locally this is c. From far away, it has a coordinate speed of 0 (it is "frozen"). I think that it means that whatever speed you're going when you fall in (eg. if you start from infinity and are close to c relative to the singularity, or if you're trying to escape and fall in slower), the horizon will pass by you at c.

 

Yes I think you're right that passing the horizon would be nothing fancy. It is only the non-inertial observer, who is trying to hover near the horizon, that sees things as weird. I think it measures extreme length contraction. Then "a lot of light" is contracted near the horizon (whereas the in-falling inertial observer doesn't experience this, and sees no special burst of light as the horizon passes it). Also the rest of the universe should appear faster, brighter, and bluer, but I think only to the hovering observer?

 

It sounds right that the horizon must be a definite place, but it *is* a light-like surface. I think to resolve this it must be that the horizon is stationary at a definite location because it is "frozen" due to infinite time dilation, and this applies to any stationary outside observer. But locally (ie. when falling in) it is not frozen, and light-like things move at a coordinate speed of c.

 

 

("Frozen" here means only in the sense of it being a "frozen star", with a coordinate speed of 0 due to infinite time dilation.)

Edited June 5, 2014 by md65536

[Strange]

This is a bit clearer in Gullstrand-Painlevé coordinates:

http://jila.colorado.edu/~ajsh/insidebh/waterfall.html