How gravity works beyond the Event Horizon

[PeterDonis]

Is that why you call the EH an outgoing null surface?

Yes.

Are you talking about Kruskal coordinates?

No. In Kruskal coordinates on Schwarzschild spacetime, the sign of the $dt^2$ term does not change inside the horizon. But it does in standard Schwarzschild coordinates, Painleve coordinates, and Eddington-Finkelstein coordinates.

 

[PWiz]

Yes.
No. In Kruskal coordinates on Schwarzschild spacetime, the sign of the $dt^2$ term does not change inside the horizon. But it does in standard Schwarzschild coordinates, Painleve coordinates, and Eddington-Finkelstein coordinates.

All right, thanks.

 

[rrogers]

That's my question, as I should have worded it.

What's the difference between tidal force and gravity in this situation?

Think slope and height: You can have a steep slope (cliff) or a gentle one; height governs the potential energy involved. The slope is the local characteristic; the height determines the global result. The actual equations are considerably harder; but you might consider the slope/potential fall to be a simple case. A long gentle walk to pikes peak; or steep climb up the side of your house. In a certain sense they are independent. Or you can take the curvature representation; the tidal force amounts to separate parts or your body trying to go on different separating paths while "gravity" generally refers to the potential energy (total separation) difference between two locations (space-time points usually with the possibility of being connected).

 

[Justice Hunter]

Maybe a picture will clear things up :)

Edit: Just to be clear, gravity does get stronger as you approach the center,

It's just the escape velocity at the event horizon IS the speed of light. So as you approach the singularity, the escape velocity just gets more and more then the speed of light.

Tidal force is what I'm depicting in the picture here, and is the reason why you don't notice much until you get close to the singularity.

http://[ATTACH=full]199794[/ATTACH] [ATTACH=full]199795[/ATTACH]

 

[PeterDonis]

Think slope and height: You can have a steep slope (cliff) or a gentle one; height governs the potential energy involved.

This viewpoint only works outside the horizon. At and inside the horizon, the concept of "gravitational potential energy" is not well-defined; there is no "cliff" to have a slope or a height in your analogy.

gravity does get stronger as you approach the center

Again, this viewpoint only works outside the horizon. At and inside the horizon, there are no static observers, and "gravity" in the sense you're using it here is only well-defined relative to static observers.

It's just the escape velocity at the event horizon IS the speed of light. So as you approach the singularity, the escape velocity just gets more and more then the speed of light.

Same comment here: "escape velocity" is only well-defined relative to static observers, and there aren't any at or inside the horizon.

Tidal force is what I'm depicting in the picture here

Again, this depiction is only valid outside the horizon. At and inside the horizon, there is no way to draw a picture of "space", because that would require spacetime to be static, and spacetime is not static at or inside the horizon.

 

[Justice Hunter]

Same comment here: "escape velocity" is only well-defined relative to static observers, and there aren't any at or inside the horizon.

I see where your coming from, because everything is rushing towards an infinitely far away point in the singularity. That's cool.
Then again, it's a little abrasive to say that escape velocity is irrelevant just because we are talking about black holes.

The statement being made here is that the reason a black hole is black is because nothing escapes from it, because light cannot travel faster then the collapsing space-time. In essence is the same thing as escape velocity. you can for the sake of practicality, say that a black hole is just a really massive planet, with special properties.

Again, this depiction is only valid outside the horizon. At and inside the horizon, there is no way to draw a picture of "space", because that would require spacetime to be static, and space time is not static at or inside the horizon.

I just don't see how this relates to anything? How else am i supposed to draw a black hole? It's inferred that space-time is collapsing infinitely at the singularity, I'm not going to bend my mind to try and depict something as abstract as that...especially in MS Paint

 

[Nugatory]

I'm not going to bend my mind to try and depict something as abstract as that...especially in MS Paint

If you use MS Office or other software with a built in calendar, you're already doing it. The calendar display depicts events laid out along a timelike worldline just fine.

 

[PeterDonis]

everything is rushing towards an infinitely far away point in the singularity.

The singularity is not "infinitely far away". I don't know where you're getting that from, but it's not correct.

it's a little abrasive to say that escape velocity is irrelevant just because we are talking about black holes.

I didn't say it was irrelevant period; I said it wasn't well-defined at or inside the horizon.

The statement being made here is that the reason a black hole is black is because nothing escapes from it

Yes.

because light cannot travel faster then the collapsing space-time.

No, spacetime is not collapsing. Space is--at least, it is if you are using a popular interpretation of black holes called the "river model". Andrew Hamilton introduced the model in this paper, which I highly recommend reading:

http://arxiv.org/abs/gr-qc/0411060

In essence is the same thing as escape velocity.

Outside the horizon, yes. At or inside the horizon, not really, because, once again, there are no static observers relative to which this escape velocity can be measured.

If you read the paper on the river model that I linked to just above, it is careful to state that the static "river bed" relative to which space is modeled as "flowing inward" is not real; it's just a mathematical construct used to help with visualization. Using that mathematical construct, you can indeed define "escape velocity" as the velocity at which the river is flowing inward relative to the river bed. This velocity is $c$ at the horizon and greater than $c$ inside the horizon. But no physical measurement will ever give that result, so you have to be very careful in using the concept.

How else am i supposed to draw a black hole?

Using a spacetime diagram. The Kruskal diagram is a good one:

https://en.wikipedia.org/wiki/Kruskal–Szekeres_coordinates

It's inferred that space-time is collapsing infinitely at the singularity

Using the "river model" referred to above, yes, but this view has limitations. See above.

 

[Justice Hunter]

Look, my picture was only trying to visually explain what tidal force is. Everything else thereafter mentioned is nothing more then technicality based on my crude representation of the concept.

The OP said he's not a physicists, why bother dragging in a relativity diagrams which need a firm understanding in relativity to actually explain something like tidal force, when it can easily be expressed in a "euclidean" manner.

But hey i won't argue with you! I'm all for proper physics, i just want to make sure the OP understands what's going on in the thread he started :)

 

[PWiz]

The OP said he's not a physicists, why bother dragging in a relativity diagrams which need a firm understanding in relativity to actually explain something like tidal force, when it can easily be expressed in a "euclidean" manner.

I don't want to sound pedantic, but I disagree. Spacetime diagrams can explain things much better than a Euclidean intuition can, although both of these strategies are "shortcuts" which avoid a lot of messy tensor algebra (which is not exactly the best visual description for most people).

 

[PeterDonis]

my picture was only trying to visually explain what tidal force is.

The OP's question was about how gravity works beyond the horizon. So showing him a diagram that doesn't apply beyond the horizon doesn't seem helpful.

Also, your diagram only illustrates tidal force in the radial direction. Tidal force also exists in the tangential direction.

 

[Justice Hunter]

I don't want to sound pedantic, but I disagree. Spacetime diagrams can explain things much better than a Euclidean intuition can, although both of these strategies are "shortcuts" which avoid a lot of messy tensor algebra (which is not exactly the best visual description for most people).

Any relativistic space-time diagram is of course going to be more accurate/precise. But none are visually intuitive and require a background in the subject. It just so happens that not everyone is a physicist. If you want to explain tidal forces with space-time geometry, then feel free to do so,

but consider this analogy; you can't teach a fishermen how to fish without first teaching him how to build a fishing pole, but you can at least teach him how to grab one from the water.

Also, your diagram only illustrates tidal force in the radial direction. Tidal force also exists in the tangential direction.

it was hard enough drawing the stick figures

 

[Tanelorn]

I thought the OP was asking about this kind of effect earlier:
http://www.quora.com/How-much-does-the-orbit-of-the-Moon-affect-my-weight

 

[PeterDonis]

Any relativistic space-time diagram is of course going to be more accurate/precise. But none are visually intuitive and require a background in the subject.

Unfortunately, there isn't a simple "visually intuitive" way to describe the region inside the horizon of a black hole. Our visual intuitions have certain expectations about space and time that are not true in that region (one of them is that space is static and doesn't change with time). That's why we have to resort to things like the Kruskal diagram to describe it.

 

[Stephanus]

Forgive my ignorance here, I'm not a physicist, but the exact nature of gravity beyond the Event Horizon has me a bit stumped.

Say I cross the Horizon of a Supermassive Black Hole, meaning the pull on me isn't really that much stronger than Earth's pull initially - why can't I then escape?

What exactly changes the escape velocity when I'm still only subjected to Earth-level gravitational pull?

I'll try. I'm no scientist either. The pull on you is very "BIG". But you'll not notice, because you go along with the pull. Imagine this. Try pulling your self on a tree branch with only three fingers. Very heavy isn't it. But then someone cut the branch. BUT DON'T TRY THIS, just imagine this. How much pull that you feel? Nothing, right. But the BH pull is actually very big.

 

[Stephanus]

Yes, that's because you are a member of a species that evolved in a place (the surface of the Earth) where tidal forces are trivial. There is no survival value in having any "intuition" about what happens near a small BH.

Just do the math.

EDIT: when I say "do the math", I mean

- look up the mass of a small BH
- figure out the "radius" to the EH
- calculate the force of gravity at the EH
- calculate the force of gravity 6' outside the EH
- compare the two

But to calculate the gravity on EH is not just $F = \frac{GM}{r^2}$
Someone in this forum, I forgot who and how, tell me this. Is it something like this?
$F=\frac{GM}{r^2*\sqrt{1-\frac{\frac{2GM}{r}}{c^2}}}$

 

[phinds]

But to calculate the gravity on EH is not just $F = \frac{GM}{r^2}$
Someone in this forum, I forgot who and how, tell me this. Is it something like this?
$F=\frac{GM}{r^2*\sqrt{1-\frac{\frac{2GM}{r}}{c^2}}}$

I don't know about that, but you can take the EH as being a limiting case of "just outside the EH" and just use the same math as you do out side the EH. Or, for that matter, use 1 foot outside the EH and 7 feet outside the EH. Do this for a one solar mass BH and then for a billion solar mass BH, and see what the difference is in tidal forces. Not having previously seen this, you will likely be surprised by the results.

 

[Stephanus]

@phinds

You are doing the wrong calculation. You are evidently using the Newtonian formula, $a = G M / R^2$, to compute this. That formula is not correct; the correct GR formula is

$$
a = \frac{GM}{R^2 \sqrt{1 - \frac{2GM}{c^2 R}}}
$$

According to this formula, $a$ increases without bound as $R \rightarrow 2GM/c^2$, i.e., as the horizon is approached.
It is impossible for a rocket, or any object with nonzero rest mass, to remain static at the horizon. Only massless objects, which move at the speed of light, can stay static at the horizon.

 

[phinds]

@phinds

OK, as I said, I wasn't sure about that and PeterDonis is correct of course. I was not fully awake when I typed my previous post and wasn't thinking. So just use the correct equation. They use exactly the same variables. The point is to do the calculations.

 

[PWiz]

@Stephanus It is important to remember that this expression only gives the coordinate acceleration. The proper acceleration will always be zero at every point along a geodesic path in curved spacetime.

 

[phinds]

@Stephanus It is important to remember that this expression only gives the coordinate acceleration. The proper acceleration will always be zero at every point along a geodesic path in curved spacetime.

Yeah, but I think bringing that up just confuses the issue as far as the tidal force is concerned and the tidal force is the issue at hand.

EDIT: I see I should amend that statement. Your point IS a good one based on the original question, it's just that the discussion has veered of into more of a discussion of tidal forces.

 

[DrGreg]

@Stephanus It is important to remember that this expression only gives the coordinate acceleration.

Actually it's the proper acceleration of a particle hovering at constant R.

The proper acceleration will always be zero at every point along a geodesic path in curved spacetime.

True.

 

[PWiz]

Actually it's the proper acceleration of a particle hovering at constant R.

I was tempted to replace "coordinate acceleration" with "the acceleration of an object released from a point infinitely far away from the BH (where $g_{μν} \rightarrow η_{μν}$ as the spatial distance from the BH $R \rightarrow ∞$ ) as measured by a static (hovering) observer" while making that post, but I decided not to, because I wasn't 100% sure. Would this have been a legitimate statement?

 

[Stephanus]

@Stephanus It is important to remember that this expression only gives the coordinate acceleration. The proper acceleration will always be zero at every point along a geodesic path in curved spacetime.

Wow, that's new! At least for me

 

[PeterDonis]

I was tempted to replace "coordinate acceleration" with "the acceleration of an object released from a point infinitely far away from the BH (where $g_{μν} \rightarrow η_{μν}$ as the spatial distance from the BH $R \rightarrow \infty$ ) as measured by a static (hovering) observer" while making that post, but I decided not to, because I wasn't 100% sure. Would this have been a legitimate statement?

Technically, yes, but it's way more complicated than it needs to be. You're thinking of escape velocity, which is the velocity of an object that freely falls from rest at infinity, relative to a static observer. For acceleration, it doesn't matter where the object was released as long as it's freely falling; an object released from rest at infinity and an object released from rest by the static observer will both have the same coordinate acceleration relative to the static observer if they are both freely falling, even though they will have very different velocities. (Note also that, as I have just implied, this acceleration is still coordinate acceleration.)

 

[PWiz]

Technically, yes, but it's way more complicated than it needs to be. You're thinking of escape velocity, which is the velocity of an object that freely falls from rest at infinity, relative to a static observer. For acceleration, it doesn't matter where the object was released as long as it's freely falling; an object released from rest at infinity and an object released from rest by the static observer will both have the same coordinate acceleration relative to the static observer if they are both freely falling, even though they will have very different velocities. (Note also that, as I have just implied, this acceleration is still coordinate acceleration.)

I see. I guess if there's one thing I've learned hanging around these relativity threads is that you can never be too careful

 

[rrogers]

This viewpoint only works outside the horizon. At and inside the horizon, the concept of "gravitational potential energy" is not well-defined; there is no "cliff" to have a slope or a height in your analogy.
Again, this viewpoint only works outside the horizon. At and inside the horizon, there are no static observers, and "gravity" in the sense you're using it here is only well-defined relative to static observers.
Same comment here: "escape velocity" is only well-defined relative to static observers, and there aren't any at or inside the horizon.
Again, this depiction is only valid outside the horizon. At and inside the horizon, there is no way to draw a picture of "space", because that would require spacetime to be static, and spacetime is not static at or inside the horizon.

Thanks for the correction/elucidation; yes the analogy was only intended to illustrate the difference (and relationship) between local effects and global accumulation of them. There is indeed a definite "edge" beyond which it shouldn't be pushed.