# Gravitational Acceleration

1. Oct 13, 2011

### Slinkey

I had a quick look through the topics and I can't seem to find a thread that already addressed this question, but if it has been addressed then I apologise in advance for repetition.

If I drop an object from a specific height above an event horizon, how do I calculate the speed it will attain when it reaches the horizon? I know how to calculate it for a constant acceleration but as you get nearer to an event horizon it will experience an increasing acceleration.

This is not a course or homework question I'm just curious!

2. Oct 13, 2011

### pervect

Staff Emeritus
The answer is that if you measure your speed relative to a stationary observer, you'll approach the speed of light as you cross the event horizon. You can't have a stationary observer exactly at the event horizion, so its a limit.

Perhaps a better way of thinking about it is that in your spaceship, the event horizon is a trapped null surface, and will always approach you at 'c', being esseintally lightlike.

As far as how you go about computing it, you basically solve the geodesic equations. This is aided by certain conserved quantities which are essentially energy and momentum.

See for instance http://www.fourmilab.ch/gravitation/orbits/
http://en.wikipedia.org/w/index.php?title=Geodesic_(general_relativity)&oldid=452675802

There are several ways of getting the geodesic equatios, one technique is to look for the curve which maximzes (or more precisely, extremizes) proper time.

3. Oct 13, 2011

### Slinkey

Thanks for the info, Pervect, however that looks quite technical and a little too advanced for me at this time as I'm having trouble getting my head around it a bit. Do you know of a site where I can find info that is aimed more at beginners?

4. Oct 13, 2011

### Ben Niehoff

I disagree with Pervect. The answer is that the question makes no sense!

"Speed" measured relative to what? The speed of an object in its own reference frame is zero by definition. So we need some observer relative to whom we can measure the speed.

But in curved spacetime, you cannot compare velocities between distant points. So the observer in this case will have to be at the same location as the falling object.

Pervect mentions "stationary observers", which are well-defined only in certain kinds of spacetimes. It so happens that the Schwarzschild geometry allows one to define "stationary observers", so in this particular case, we can find an answer: compare the objects velocity to that of a stationary observer at the same location.

In this case, I guess I do agree with Pervect: the velocity defined in this way will approach c as the event horizon is approached.

5. Oct 13, 2011

### Slinkey

Ben, I don't see how the question makes no sense. If I drop an object towards a black hole under the influence of gravity from the black hole then obviously I am asking for it's speed relative to me. No?

6. Oct 13, 2011

### PAllen

Another take on this is that at every point above the event horizon, you can drop an object which initially has zero speed relative to the hovering observer. However, there is no such thing a hovering observer at the event horizon. Further, for an object dropped from any point above the horizon (even 10^-1000 meters), the horizon passes at c (simply because it is lightlike).

So, if one takes the limit of the velocity of test object dropped from a hovering observer at the moment of drop, as the distance to the horizon approaches zero, you get zero as the limit.

Thus, I agree with Ben Niehoff's initial intuition - the question has no well defined answer. You can get any value between zero and c as the limit depending on how you formulate the problem.

7. Oct 13, 2011

### pervect

Staff Emeritus
Ben's answer is a little more complete than mine, and I mostly agree with the technical points and cautions about measuring speed - i.e. to get a physically meaningful number you want to compare to objects, or worldlines, at the same point in spacetimes.

The reason for this is that to compare speeds at distant locations you need to transport the velocity vector from the distant location to your location. The standard method of doing this, called parallel transport, is well defined and independent of path in flat space-time, but is dependent on the path in curved space-time. Thus your answer (for relative velocity) will depend on unspecified details of how you do the comparison in a general curved space-time, making the question not have a single answer.

However, it does seem to me that it's valid to say that if you consider the frame of the space-ship (a frame in which you are stationary), it's reasonable to ask at what speed the event horizon appears to approach you. Assuming that you are sufficiently close to it , that is, so that you don't have to worry about the effects of curved space-time, the event horizon will approach you at 'c'.

We can consider a trapped photon that's hanging at the event horizon, and then realize we're essentially asking "what is the velocity of a light beam relative to you", and realize that the answer is 'c'. It's another special case, but it's one of interest to the OP.

I'm not quite sure if Ben finds anything objectionable in this analysis or not -at the moment I don't think we really have a serios disgareement, though.

8. Oct 13, 2011

### Passionflower

Since nobody is sticking out his neck to actually provide a formula, I will do it.

The local velocity wrt to a stationary observer for a radially falling test observer falling from a stationary state at r0 is:

$$\left( \sqrt {{\frac {{\it rs}}{r}}}-\sqrt {{\frac {{\it rs}}{{\it r0 }}}} \right) \left( 1-\sqrt {{\frac {{\it rs}}{r}}}\sqrt {{\frac {{ \it rs}}{{\it r0}}}} \right) ^{-1}$$

Where rs is the Schwarzschild radius, r the position for which one wants to get the local velocity wrt a stationary observer and r0 is the location from where the object is dropped.

If we graph this we get:
[PLAIN]http://img839.imageshack.us/img839/3871/velocityx.png [Broken]

Comparing the velocity of an object dropped from r=6 and r=infinity (free fall)
[PLAIN]http://img9.imageshack.us/img9/9528/velocity2.png [Broken]

Last edited by a moderator: May 5, 2017
9. Oct 13, 2011

### Ben Niehoff

By the way, there is more confusion here that needs addressed, and that is "acceleration".

Unlike velocity, it is possible to give an unambiguous definition of the acceleration experienced by any given object. This is called the "proper acceleration", which just means the acceleration in the object's own frame of reference.

However, in GR there is a bit of a trick. A free-falling object experiences no acceleration! This is because of the equivalence principle. No acceleration is felt at any time during free fall, and in particular the object feels no acceleration as it crosses the horizon.

It is the stationary observer that experiences acceleration. This may sound counterintuitive, but in fact you can feel this very acceleration right now. Your chair exerts an upward force on your body, which prevents you from falling through the floor. It is this force which gives you a constant acceleration away from the center of the Earth, preventing you from following a free-fall path to its center.

10. Oct 13, 2011

### Slinkey

Passionflower,

thanks for the equation. Would I be right in assuming that this is the same for any black hole regardless of its mass?

Ben,

thanks for pointing that out. This is something, as a layman, that doesn't become apparent because we are so used to thinking of ourselves as not moving but, as you point out, the reason I am sitting here on my chair is because I am being accelerated and the chair/floor/earth is in the way of my motion.

However, something just came to mind. When I accelerate in a car at say, 3m/s^2, this is a linear acceleration. But whilst in free fall the acceleration is not linear. So wouldn't I feel that change?

11. Oct 13, 2011

### Ben Niehoff

When you're sitting in your chair, the acceleration is constant. If you ride a rollercoaster, you'll feel it change.

12. Oct 13, 2011

### Slinkey

Ok. Understood. Thanks. :)

13. Oct 13, 2011

### Passionflower

The variable rs is the Schwarzschild radius which is equal to double the mass of the BH.

14. Oct 13, 2011

### Passionflower

You can ask that question but unlike the local velocity, e.g. the velocity of something zooming by at the same location, the velocity at a distance is at least subject to interpretation due to the curvature of spacetime.

15. Oct 14, 2011

### Slinkey

Yep, I got that. Thanks. I've run some numbers through it now and found the curve remains the same regardless of the mass of the black hole.

The thing I find intriguing is that the escape velocity at the EH of a black hole remains constant (ie. = c) regardless of the mass of the black hole. But when you work out the acceleration at the EH that as the mass of a black hole increases the acceleration you would feel reduces.

For example, for a 10 solar mass black hole the acceleration at the horizon works out at 1.233*10^6 m/s^2. But for a 20 solar mass black hole the acceleration at the EH works out at 8.722*10^5 m/s^2.

I was under the impression that the acceleration you would feel is what actually increases your velocity but from your equation it seems the escape velocity is the important factor.

Am I missing something here?

EDIT: Scratch that question. I've just realised what I am missing I think. If I am falling from 6*sr in each case, 6*sr for a 10 mass BH is nearer to the EH than 6*sr for a 20 mass BH. So although I will experience a less extreme acceleration for the larger BH I will be falling for longer and they are equivalent once we reach the EH.

Last edited: Oct 14, 2011
16. Oct 14, 2011

### Slinkey

Hmm. When you say subject to interpretation you mean that if I don't take into account the curvature then it's apparent velocity would be different from what I calculate?

17. Oct 15, 2011

### Slinkey

Cough... correction. 10 solar mass black hole has 1.521*10^12 m/s^2 and 20 solar mass has 7.607*10^11 m/s^2 at their respective event horizons.

18. Oct 16, 2011

### pervect

Staff Emeritus
In general, there isn't any way to define the notion of velocity or "apparent velocity" that's covariant and doesn't require additional specification of how you do the measurement.

I mentioned before that parallel transporting the velocity winds up being path dependent, I would guess that that statement is unfamiliar and that you don't have a background in differential geometry needed to appreciate it.

But in practice, since you have a static geometry, if you stick to using the descriptions used by static observers, you probably won't wind up with any problems. Therfore, I'd simply reccomend that you just say that you're computing the "velocity according to a static observer" rather than just "velocity" and I think everyone will get on track with what you're doing and be able to say whether the answers are numerically right or not without getting off on any interesting, but rather advanced, side-tracks.

It's also mildly important that you understand what you are doing to the extent that you realize the underlying definitions of clock synchronziation are taken from said static observers, along with the associated notions of distance and time.

19. Oct 16, 2011

### Slinkey

Correct. I will be doing differential and integral calculus later on my current course so will be taking my first steps into understanding it, but I'm thinking that when we are talking about the curved space around a black hole the path you take is very important? As opposed to say calculating a drop from 1000m to the ground in Earth gravity where the curvature is negligible and can be mostly ignored.

OK, I understand. I have to be more specific about the question I am asking in order to leave out room for ambiguities in the respose. Although, that being said it's sometimes hard to be unambiguous in a scenario where you don't fully appreciate the complexities of your question! All part of the learning curve I guess.

Is this alluding to the fact that if someone is static at another point around the same black hole that if they watched the same event they would come out with different answers because our respective clocks and distances are affected by the curvature and any answers we give are dependent on our observation frame (if that makes sense in my non-technical language)?

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