Acceleration At The Event Horizon

1. Apr 22, 2005

mrjeffy321

For an object exactly at the event horizon of a black hole, is the acceleration infinite or does the object accelerate at the speed of light?
I have read that the force on the object causing it to accelerate when it is at the event horizon is infitite (so the acceleration would be too), but the reason I am thinking that it would not be is that for a couple reasons,
-wouldnt your acceleration be infinite right when you reach the singularity and not the event horizon?
-you are still a certain distance away from the mass of the black hole, so
r != 0, in the equation for the universal law of gravitation
--but you cant use that here because it is a black hole?
-if a beam of light were pointed staight outward from the black hole, wouldnt it be frozen there, not falling back in and not going out?

2. Apr 22, 2005

Labguy

For a large BH, the acceleration would not be anywhere near c except for photons already at c. No object with mass would approach c. Acceleration at the EH of a smaller BH would be faster, but still not at c. Infinity is not possible at all, even for photons.

3. Apr 23, 2005

mrjeffy321

that was a poor choice of words I used above, "acceleration infinite or does the object accelerate at the speed of light?".
what I was trying to do by finding the acceleration at given points was to find the objects, "weight".
for example, if I was 1 km above the Event Horizon (EH from now on), 1 m above, ..., right on it.

4. Apr 23, 2005

pervect

Staff Emeritus
If you try to hover near the event horizon of a Schwarzschild black hole the required proper acceleration (as measured by the seat of your pants) will approach infinity as you approach the event horizon (when you try to hover, i.e. maintain a constant r coordinate in the Schwarzschild geometry).

Wald gives the proper acceleration at a distance r away from the black hole as being proportional to

$$1/\sqrt{r-2M}$$

when one is close to the black hole.

(note this is in geometric units, so that G=c=1, and the Schwarzschild radius is at r=2M)

The length unit here used to measure the acceleration is a meter stick carried by the observer (pointed in the radial direction), and the time unit used to measure the acceleration is a clock carried by the observer. The r coordinate, though, is the Schwarzschild r coordinate.

The units bothered me, so I worked it out with GRtensor, getting

a = $$\frac{m}{r^{\frac{3}{2}}\sqrt{r-2m}}$$

this gives the correct units for acceleration (in geometric units!). This also has the property that a-> m/r^2 when r is very large, which is correct in geometric units (where G=1).

In non-geometric units this is

a = $$\frac{Gm}{r^{\frac{3}{2}}\sqrt{r-r_{s}}}$$

where rs= Schwarzschild radius = 2Gm/c^2, and G and c are the gravitatioanl constant and the speed of light, respectively, and m is the mass of the black hole.

Last edited: Apr 23, 2005
5. May 3, 2005

Mozart

You used the word singularity. I am having trouble understanding what it really means. I always see that word popping up around these boards. I've seen it as singularity, mathematical singularity, quantum singularity. Theres gotta be more...anyways what does/do it/they mean.

6. May 6, 2005

z4955

3 entries found for singularity.
sin·gu·lar·i·ty ( P ) Pronunciation Key (snggy-lr-t)
n. pl. sin·gu·lar·i·ties
The quality or condition of being singular.
A trait marking one as distinct from others; a peculiarity.
Something uncommon or unusual.
Astrophysics. A point in space-time at which gravitational forces cause matter to have infinite density and infinitesimal volume, and space and time to become infinitely distorted.
Mathematics. A point at which the derivative does not exist for a given function but every neighborhood of which contains points for which the derivative exists. Also called singular point.

7. May 6, 2005

Mozart

Thanks for answering, but shortly after I posted it I ended up at dictionary.com finding out what it meant, and then regreted posting WHAT IS SINGULARITY.

8. May 18, 2005

Mariko

cosmic singularity is the big bang

9. May 19, 2005

Chronos

This is actually a fairly simple question. The velocity of an object attracted toward a gravitating body from infinity is the same as the escape velocity at any arbitrary distance from the center of the gravitating body - so the answer is slighly less than the speed of light at the event horizon of a black hole.

10. May 19, 2005

pervect

Staff Emeritus
This velocity (of the infalling observer) is measured relative to what?

In Schwarzschild coordinates, the velocity works out to be zero. In the Schwarzschild basis, the velocity at the horizon does turn out to be 'c' for an radially infalling object when the object is at rest at infinity, but not when the object starts with an inital radial velocity .

(Unlesss I've made some error in my calculations, I suppose).

The "schwarzschild basis" can be thought of as the coordinate system of an observer who is "holding station" at some particular Schwarzschild r,theta, phi value.

Because such an observer is necessarily accelerating, one might add that it's actually the local coordinate system of an instantaneously co-moving observer. Another way to describe it is that the local metric is Lorentzian (diagonal with all coefficients +/- 1).

Maybe there's some other way to measure velocity, but I can't think of what it could be. The Schwarzschild basis seems like the most logical candidate for what one means by the "velocity" of an infalling observer.

See for instance the thread

where I convince myself that the velocity of an obsrever is not always 'c' in the Schwarzschild basis after originally thinking it should be.

11. May 19, 2005

Zanket

It must be exactly c regardless how the object fell to arrive at the horizon. If < c at the horizon then the object (or its image or its transmission etc.) can escape, in which case it's not a horizon. If > c at the horizon then > c would be directly measureable by someone hovering just above the horizon, violating special relativity. Both a rocket accelerating in at any rate of acceleration, and a stone hurled from any altitude at any velocity, cross the horizon at a proper velocity of exactly c.

This leads to a problem where the velocity of an object relative to the horizon cannot change while crossing. Nothing special happens when you cross the horizon (e.g. tidal force smoothly increases). So whenever nothing special is happening, you could be crossing a horizon. Then you could cross head-on when you're passing another car on a freeway, in which case for a brief moment your car must be in lockstep with the other car (not passing). So implies general relativity.

12. May 19, 2005

Zanket

Infinite acceleration is needed by an object hovering at the horizon, as pervect noted. Because of that, an object cannot hover at or rise from the horizon and instead must fall in once it reaches it.

Yes, assuming you mean gravitational acceleration.

You can use it at the horizon.

Yes, if beamed precisely at the horizon.

13. May 20, 2005

pervect

Staff Emeritus
I probably need to think about this some more, but I do have a few more thoughts. This post got a little long, so let me summarize by saying that Zanket has made some excellent points.

I do think it's important to figure out what we mean by "the velocity" of an observer falling into a black hole, though - velocities are relative, we need to define what the velocity is relative to. We can't , for instance, measure the velocity by taking the rate of change of the distance to the singularity - the later isn't well-defined at all.

I think it's reasonable to define the infalling velocity as the velocity measured by a "station-keeping" observer at a constant Schwarzschild coordinates (R, theta, phi). In this problem I think we're mainly interested in the case in which the infalling observer has no angular momentum, and the only component of the velocity is the radial component.

This may be a coordinate dependent definition, but it's the best one I can think of. I'm open to other suggestions as to what the term "infalling velocity" really means.

The station-keeping observer is necessarily accelerating, but the relative velocity should be well-defined when the station-keeping observer is at the same point in space as the infalling observer.

The relative velocity of an observer and an accelerating observer is a function of time - thus it's not well defined when the observers are not at the same point in space-time, because of issues of the relativity of simultaneity

Given that we can agree on this as a defintion of just what the "velocity of an observer falling into a black hole" actually is, I have to agree with Zanket's point, more or less. At any distance R greater than the Schwarzschild radius, the infalling velocity by the defintion above will be less than 'c' because it's an actual "physical" velocity between two objects. The observer exactly at the event horizon isn't "physical", it requires an infinite proper acceleration to hold station there. So we need to take the limit of the velocity as R->R_s to make any sense of the velocity "at the horizon".

Also, the velocity should increase as R decreases.

Given this, it seems almost certain that the limit as R-> schwarzschild radius of the "infalling velocity" should be equal to "c" if it exists.

So I need to go back over my calculations when I have more time to figure out what's going wrong.

Last edited: May 20, 2005
14. May 21, 2005

Zanket

Yes. The same is noted in the book Exploring Black Holes. On pg. 3-15 it says:

15. May 21, 2005

Chronos

I think you should use the derivative with respect to 'c', not infinity.

16. Jun 8, 2005

Flatland

Ok I have a curious question. What if half my body was above the event horizon and half my body was beneath it. Would I be able to lift my arm?

17. Jun 8, 2005

Zanket

You could have crossed an event horizon while you wrote your post, so yes. You can't hover draped across the horizon, according to the theory. You must move across it at the speed of light.

18. Jun 8, 2005

Labguy

Agreed that you can't hover, but the rest (moving at c) is just not correct. Massless photons move at c whether near a BH or not, but any particle or object cannot be accellerated to c no matter how strong the gravity. A BH is just a big source of a strong gravitational field, and one of the basics of GR is that nothing with mass can be accellerated to c as that would require an infinite amount of energy.

Infinite energy is not available from any source. Even neutrinos, with their very small mass, do not travel at c. It is close, but not quite c. A lot of outdated websites are still out there stating that neutrinos are massles and travel at c, but ignore them.

Last edited: Jun 8, 2005
19. Jun 8, 2005

Zanket

In relativity theory, the speed of light limit applies only to direct measurements of velocity.

Objects (assume material; i.e. not photons) at the horizon have been accelerated to c in the limit. All objects must move inward at exactly c at the horizon, precluding any velocity <> 0 relative to another object to be directly-measurable. But an infalling object can directly measure the velocity of a object hovering arbitrarily close to and above the horizon to be arbitrarily close to c.

To say that one cannot move at c at the horizon puts the cart before the horse. The reason that c is not directly measurable there is because all objects must move inward at c there, because gravity has there accelerated to c all objects.

Last edited: Jun 8, 2005
20. Jun 8, 2005

Labguy

Sorry, but this is simply not the case at all. Maybe some other "expert" will chime in with an opinion.