Value of g near a black hole (re-visited)

pawprint

I've engaged with several threads concerning simple (i.e. non-rotating, uncharged) black holes. My general line of argument has been that, as apparent time becomes infinitely stretched at the event horizon nothing can be observed to enter the BH in finite time. jambaugh wrote two lengthy dissertations on this at the end of the thread "Can a blackhole suck in another blackhole?"

I even began a thread- "Value of g near a black hole" directly asking for confirmation. It read- "On approach to a simple (non-rotating, uncharged) singularity, does g increase asymptotically near...
a) the singularity,
b) its event horizon,
or c) no?"
...to which I understood the answers to point towards b).

Nobody has disputed these assertions, unless it was in mathematics beyond my understanding. Now please lets keep this thread related only to 'simple' BHs.

I now believe my prior understanding (which is intuitive rather than mathematical) was flawed. I propose that gravity approaches infinity asymptotically at the singularity, not the EH. The EH of a simple BH can surely be defined as the distance from the singularity at which escape velocity = c.

Can a purely Newtonian approach be used for calculations at discernable distances from the singularity?

Comments please :{)

PeterDonis

I assume the previous thread you are referring to is this one?

http://physicsforums.com/showthread.php?t=576973

If so, yes, the answer you were given is b), and it is the correct answer.

Why do you think the answer you were given in the previous thread is wrong? If it's because of this...

...then it's not a valid reason, because the answer to the question just quoted is "no". A "purely Newtonian" approach will give incorrect answers unless you are at a radial coordinate r that is much, much larger than the Schwarzschild radius (2M), so that the error in the Newtonian formulas becomes too small to measure.

pawprint

Yes, that is the thread referred to. Sorry I did not know how to find it as showthread.php?t=576973.
I would prefer to wait before addressing questions raised, as they are likely to come in bunches. However I certainly wouldn't apply Newton to orbits. Only F=(m1*m2)/D².

PeterDonis

If you're trying to find a thread you have participated in, the easiest way I know of is to click on "My PF", and then click on the "List Subscriptions" link on the left of the screen. That will show you a list of all the threads you have posted in, in reverse chronological order (most recent at the top).

pervect

If you take the time to look this up in some GR textbooks, or some of the GR Faq's, you should be easily able to find out that objects can fall past the event horizon in finite proper time.

Depending on your background (I don't know what it is) you may or may not be able to follow the calculations yourself, but you should be able to find the answer written down in a number of places.

FOr instance the sci.physics.faq: http://math.ucr.edu/home/baez/physic...s/fall_in.html

bcrowell

I disputed it. As I explained in the previous thread, g has any value you like, depending on your coordinates.

No, it's not correct without some further qualification, such as specifying coordinates or a frame of reference.

What the OP needs to understand at this point, and doesn't seem to, is that this becomes a crucial point inside the horizon, where there are no stationary observers.

Nabeshin

Again, I think my post in your previous thread perfectly describes the situation. But it does have qualifications and you have to read it carefully! I am not knitpicking!

Furthermore, when you say 'gravity' goes to infinity, you need to be specific in what you mean. Most (relativists) would take this to mean the invariant curvature goes to infinity, which is indeed true at the singularity (so this is not a coordinate effect). However, nothing haywire takes place with the curvature at the event horizon.

Just a general note about applying Newtonian reasoning to objects like black holes... although it gives some correct answers (i.e. schwarzschild radius), it doesn't make much sense. For example, in the Newtonian picture, a light ray shot away from the event horizon will continue along OUT TO INFINITY. This is of course absurd, you can't communicate from beyond the event horizon! A truer picture would be, you shoot a light ray 'radially outward' but in fact the light cones are so tipped that this direction corresponds to the interior of the black hole. My point is, unfortunately, you cannot apply Newtonian reasoning and trying to use Newtonian intuition is likely to lead you to wrong conclusions. (Edit: This is actually the apparent horizon I am describing, which corresponds to the event horizon in the case of a stationary spacetime)

pawprint

I have read all responses and suggested references. I don't want to change the original question but merely re-frame it.

1) The EH is at a radius from the singularity where escape velocity on the inner side is greater than c, and on the outer side is less than c. Correct?

2) How can a 'non-infinite' escape velocity be reconciled with a practically infinite value of gravity at the same radius? This is what you appear to be asserting.

If the answers require great mathematical insight then I withdraw. Surely they can be expressed more simply.

PAllen

As several people have already said, everything depends on what you mean by gravity, especially as you are thinking by analogy to Newtonian gravity.

One concept of gravity is called 'tidal gravity': even if a moon is in free fall or orbiting a massive body, different parts of it are pulled in different directions so it is under stress: tidal stress. This is the sense of gravity that, in GR, corresponds to curvature and is not a coordinate dependent feature. It is the feature that causes a star approaching a small black hole to be torn apart. Here, I mean small in size compared to a star, but with mass of several stars. And by size, I mean the event horizon. This sense of gravity only becomes infinite on approach to the singularity itself. For a super-massive black hole (a billion suns, for example), tidal gravity at the horizon is quite small.

A different concept of gravity is what you think of as how hard you are pulled to the ground, which is better viewed as how hard the ground is pushing to keep you from moving on a free fall path. For a simple, static black hole, the notion of static observer is well defined outside the event horizon, while being impossible on or inside the event horizon. For a static observer very slowly approaching an event horizon, the thrust needed to hold a static position goes to infinity as you approach the horizon. Note that the velocity needed relative to a static observer required to escape to infinity approaches c as a static observer approaches the event horizon.

pawprint

Thank you PAllen. I understand both concepts and have no difficulty with their differences. I would not presume to post on PhysicsForums otherwise.

The nub of my question, however, may point to a lack of understanding. A clock on a floor on Earth runs a little slower (about 1 part in 10¹⁷ I think) than one on a metre-high table.

I would expect a clock near a singularity to run near infinitely slowly regardless of observation. Yet 'common knowledge', (including many posts in PF), says that this occurs at the event horizon of a black hole, where, as you point out, the gravitational gradient may be quite low.

I don't see why, in this instance, 'tidal gravity' should behave differently, because it is the same (Einsteinian) sort of gravity which is responsible for the Earth-bound clocks' disagreements. What am I missing?

Nabeshin

You have to compare two clocks to do what you are talking about. I can compare my clock to a clock a meter higher, or a far away observer can compare his clock to that of his buddy, who is hovering just outside the event horizon. Once you put the clock past the event horizon, it is no longer possible to compare it to the outside clock! The two are causally disconnected! Keep in mind, that if YOU were the one falling into a black hole (and you've got really stiff bones, so you don't get ripped apart easily) watching your own clock, you would see nothing strange at all as you approached the singularity. Time will tick by always at the rate of 1 second per second.

pawprint

*This has crossed over Nabeshin's last post.*

Eureka!

I think I have it. The clock does not slow except as expected. It is only the outside observer who sees it do so, and that is not a phenomenon of gravity, but of light.

Do I win my own prize? If so I wouldn't have realised this without your help.

pawprint

Clarification: "The clock does not slow except as expected." means as expected by Einsteinian gravity.

If this is correct then my question (previous post noted in first post of this thread) should be answered a) not b). It only LOOKS like b).

PeterDonis

You're right, I should have qualified my answer by saying that b) is the answer if "g" means "the proper acceleration of an observer who is at a constant radial coordinate r for all time". From reading the previous thread it appeared that that was the definition of "g" that had been settled on.

Good point; that's often a stumbling block.

questionpost

How come I don't get the feeling of infinitely stretch space time from a finitely deep gravitational well who's escape velocity just coincidentally happens to exceed light at some random value? We don't even know if there actually is a fabric of space, it's just a mathematical representation of how gravitational fields change over distance.

pervect

A stationary clock near a singularity would run slow when compared with another stationary clock that's far away from any singularity, and it'd approach stopping as the stationary clock got closer and closer to the event horizon.

There's no such thing as a stationary clock at the event horizon, however. In fact, any clock crossing the event horizon must be moving at the speed of light - or rather, since the event horizon can be thought of as trapped light, any physical infalling clock, which is stationary in its own frame, will see the event horizon approaching it at the speed of light.

This motion causes signficant SR effects. If you neglect the velocity effects, it would be correct to say that from the point of view of an infalling observer, a clock at infinity would run faster and faster, without bound, as one approached the event horizon,.

When you include the velocity effects, though, the clock at infinity doesn't run infinitely fast.

You will run into the usual special relativity (SR) issues associated with the twin paradox when you include the velocity effects - I'm not sure what yoru background is in SR.

questionpost

Why are people making all this hype about the event horizon? That's just an escape velocity boundary, what we should really worry about a singularity. Also, I thought it was impossible for matter to travel at the speed of light, I even had a separate topic just for that and someone posted a Lawrence transformation.