How gravity works beyond the Event Horizon

[Gaz1982]

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?

 

[phinds]

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?

That is incorrect. The TIDAL FORCES on you are very small as you cross the EH of a supermassive BH, but the gravity is so strong that not even light can escape to back outside the EH.

 

[Gaz1982]

That is incorrect. The TIDAL FORCES on you are very small as you cross the EH of a supermassive BH, but the gravity is so strong that not even light can escape to back outside the EH.

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

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

 

[phinds]

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

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

The tidal force is the difference in gravitational attraction between two points not far from each other ... say from your head to your feet. As you enter the EH of a supermassive BH, the tidal force on you is trivial. Long before you enter the EH of a very small BH, the tidal forces will have turned you into a long thin strand of bloody mess. This is called sphagetification.

 

[Gaz1982]

Sure.

But tidal force weak but gravity inescapable. I'm struggling to compute

 

[phinds]

Sure.

But tidal force weak but gravity inescapable. I'm struggling to compute

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

 

[Gaz1982]

Right, sorry to sound thick. Bear with me.

To me the tidal forces on Earth aren't trivial. We're being pulled towards the centre of the Earth quite strongly. A 15 feet fall can kill someone.

The escape velocity of Earth is about 25,000 miles an hour

As a non-physicist I'm struggling to reconcile tidal force with gravity.

 

[Gaz1982]

Yes I know the moon exercises a tidal force on the Earth for example, but doesn't drag people upwards. I'm struggling to define tidal force.

I'll check Wiki

 

[Quds Akbar]

the pull on me isn't really that much stronger than Earth's pull initially

That is not true, why would a supermassive black hole subject you to the same amount of gravitational pull as Earth? A black hole definitely has stronger gravitational pull than Earth's gravity, that is why it is a black hole, because light cannot escape at a certain point, the gravity is so strong that you reach a point where the escape velocity and the speed required to escape the pull is far greater than the speed of light, which violates relativity.
So I think what you missed here is that a black hole has much, much stronger gravitational force than our Earth that is nothing compared to a supermassive black hole's gravity.

 

[Noctisdark]

The force of gravity changes when varying the distance, closer objects will feel more attracted than further objects, In classical mechanics it varies with the inverse square of distance, [The math of general relativity is different, some math I totally don't understand], anyway when you calculate the rate of change of that force when you change the distance, you'll find that it changes very quicky when you are very close to the center of mass, and the same "should" happen near a black hole, when you are close the force on your feet is very different and larger than the force on your head, these are tides waves that tear you apart, so keep away from a black hole and just one thing, crossing an event horison is the END, the pull is way stronger that the pull of earth, it so strong that it made me say this sentence, even light cannot escape from it, and don't worry, no one will shout on you, good luck

 

[phinds]

To me the tidal forces on Earth aren't trivial. We're being pulled towards the centre of the Earth quite strongly...

Which has absolutely NOTHING to do with tidal forces. That's just gravity.

I'll say it again. DO THE MATH.

EDIT: and if you can't be bothered to do the math at least look up sphagetification where it will all be explained.

 

[cosmik debris]

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 think you are confusing the pull of gravity with the tidal force which is the difference in gravity between two points. If you imagine Gravity to be a field with values at each point then each point can have a high value which will cause an acceleration, but it is only the difference between two points that you will "feel". If this difference is large and your head is at one point and your feet another you will feel unwell.

 

[phinds]

I think you are confusing the pull of gravity with the tidal force which is the difference in gravity between two points. If you imagine Gravity to be a field with values at each point then each point can have a high value which will cause an acceleration, but it is only the difference between two points that you will "feel". If this difference is large and your head is at one point and your feet another you will feel unwell.

Yes, that's what I told him in post #4. It doesn't seem to be clear to him yet. It SEEMS like a trivially clear distinction, but I can recall not "getting" stuff that later became crystal clear, so I sympathize.

 

[DaveC426913]

To me the tidal forces on Earth aren't trivial. We're being pulled towards the centre of the Earth quite strongly. A 15 feet fall can kill someone.

That is not tidal force; that is gravitational force.
Tidal force on Earth is the difference between the force at your head and the very slightly stronger force at your feet (which are closer to Earth's centre of mass). If you were in free-fall, you would find your feet being pulled toward Earth just a little stronger than your head. You'd get taller!

Satellites in orbit around Earth are sometimes large enough to feel this very gentle tug along their lengths. Unless corrected, they will have a tendency to orient themselves with their long axis pointed toward the Earth, as their near end is pulled down from the centre of mass and their far end is pulled up from the centre of mass. (The sat's centre of mass is in a stable orbit, but the near end of the satellite is actually going too low for its slightly lower orbit, so it tends to drop. The far end of the satellite is actually going too fast for its slightly higher orbit and this pulls it upward).

This is also why the Moon is tidally locked. It is not a perfect sphere, and a tiny tidal force over billions of years has oriented the Moon with its larger diameter pointed along the Earth-Moon axis.

 

[phinds]

Well, we've explained it pretty much exactly the same way at least 3 times now. @Gaz1982 did you get it?

 

[PeterDonis]

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?

Because spacetime is curved at the horizon to the point where radially outgoing light no longer moves outward. In other words, outside the horizon, an object that "stays in the same place" is moving slower than light. But at the horizon, an object that "stays in the same place" must be moving at the speed of light, so only light can do it. And inside the horizon, nothing, not even light, can "stay in the same place"; everything falls inward, including radially "outgoing" light.

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

Escape velocity isn't the same as "gravitational pull". This is true even in the Newtonian regime, i.e., for objects much less compact than black holes. Just look at the Newtonian formulas:

Escape velocity: $v_e = \sqrt{2GM/r}$

Acceleration due to gravity: $a = GM / r^2$.

So it's perfectly possible for two different planets, say, to have the same $v_e$ but different $a$ at their surface, or the same $a$ but different $v_e$.

 

[Gaz1982]

Well, we've explained it pretty much exactly the same way at least 3 times now. @Gaz1982 did you get it?

Yes,

No need to be rude

 

[PWiz]

@OP I would just like to add that gravity is not a force (in GR). You would feel no force at all in free fall, because your body would be following a geodesic path (objects always follow straight path in spacetime, and when spacetime becomes curved, objects continue to follow straight paths along the curved spacetime). When you are near a black hole, what really tears you up is not this gravitational "pull," (curvature) but rather the tidal effects - you're head is trying to follow a slightly different geodesic compared to your feet (if you're falling feet/head first), and these differences become really noticeable the closer you get to a black hole where curvature (measured by the Riemann tensor) varies much more over shorter spatial distances.

The escape velocity (which becomes $c$ at the EH) at a point is the minimum velocity at which an object must be moving to move "against" the geodesic path at that point. These are different concepts.

 

[Gaz1982]

@OP I would just like to add that gravity is not a force (in GR). You would feel no force at all in free fall, because your body would be following a geodesic path (objects always follow straight path in spacetime, and when spacetime becomes curved, objects continue to follow straight paths along the curved spacetime). When you are near a black hole, what really tears you up is not this gravitational "pull," (curvature) but rather the tidal effects - you're head is trying to follow a slightly different geodesic compared to your feet (if you're falling feet/head first), and these differences become really noticeable the closer you get to a black hole where curvature (measured by the Riemann tensor) varies much more over shorter spatial distances.

The escape velocity (which becomes $c$ at the EH) at a point is the minimum velocity at which an object must be moving to move "against" the geodesic path at that point. These are different concepts.

That makes sense

Thank you

 

[phinds]

Yes,

No need to be rude

I was kidding US more than you (and I WAS kidding, not being rude). We have a knack for saying the same thing over and over with each person thinking that his particular way of saying it is better than the previous n versions.

 

[PeterDonis]

The escape velocity (which becomes $c$ at the EH) at a point is the minimum velocity at which an object must be moving to move "against" the geodesic path at that point.

No, it isn't. An object moving at escape velocity is moving on a geodesic path. The definition of escape velocity is the velocity, relative to a static observer (one who is "hovering" at a constant altitude), at which an object traveling on a geodesic path would have to be moving in order to escape to infinity. Strictly speaking, escape velocity is undefined at (and inside) the horizon, because there are no static observers there. The statement that "escape velocity is $c$ at the horizon" is really a (somewhat sloppy) shorthand for "the limit of escape velocity as $r \rightarrow r_s$ is $c$", where $r_s$ is the radial coordinate of the horizon.

 

[PWiz]

No, it isn't. An object moving at escape velocity is moving on a geodesic path. The definition of escape velocity is the velocity, relative to a static observer (one who is "hovering" at a constant altitude), at which an object traveling on a geodesic path would have to be moving in order to escape to infinity. Strictly speaking, escape velocity is undefined at (and inside) the horizon, because there are no static observers there. The statement that "escape velocity is $c$ at the horizon" is really a (somewhat sloppy) shorthand for "the limit of escape velocity as $r \rightarrow r_s$ is $c$", where $r_s$ is the radial coordinate of the horizon.

Well yes, an object moving at escape velocity is moving along a geodesic path, but in the opposite direction in which it would otherwise move (away from the BH). Isn't that right? (Just to be clear, I mean A to B and B to A are the same paths but in opposite direction.)

 

[PeterDonis]

an object moving at escape velocity is moving along a geodesic path, but in the opposite direction in which it would otherwise move (away from the BH).

It's moving in the opposite direction from an object moving on a geodesic path that falls into the hole. Neither geodesic path is "preferred"; they're both just geodesic paths.

What you may be trying to say here is that an object falling into the hole is moving in the same direction as the "gravitational pull" of the hole, whereas the object on a geodesic escape trajectory is moving in the opposite direction. Or, if we want to avoid the problematic phrase "gravitational pull" (since gravity in GR is not a force), we could say that the object on an infalling geodesic trajectory increases its velocity relative to static observers as it falls, while an object on an escape geodesic trajectory decreases its velocity relative to static observers as it rises. That's true.

 

[A.T.]

Well yes, an object moving at escape velocity is moving along a geodesic path, but in the opposite direction in which it would otherwise move (away from the BH).

If you mean the geodesic paths in space-time (free fall condition) then no, they aren't opposite, because that would require one of them moving back in time.

If you mean the paths in space, which happen to be geodesics in this special case (but aren’t in general for free fall), then yes, "up" is the opposite of "down". Obviously.

 

[pervect]

If you try to hover over a large black hole, and you measure tidal forces by measuring the difference of (proper) acceleration between your head and your feet, the "tidal force" measured in this way becomes infinite. But that's not what's people mean when they talk about the tidal forces of a black hole in the context of GR.

If you free-fall through the event horizon of a large black hole, rather than try to hover over it, there is no such artifact, there is no infinite force trying to pull you apart. The difference in proper acceleration between your head and your feet, which is the force you feel trying to pull your body apart can be negligibly small for a large enough black hole. The later case corresponds to what people mean when they say the "tidal forces" vanish near a large black hole. More technically, it also demonstrates that the interpretation of the Riemann tensor as a tidal force can be just a tiny bit misleading if taken too seriously, it's really not quite right. The Riemann curvature tensor is a tensor and independent of the observer's acceleration. The head-toe force definition is NOT independent of the observer's acceleration, the Riemann tensor notion of tidal force IS independent of the observer's acceleration. THis issue comes up a lot in the context of Bell's spaceship paradox, the force on a spaceship of constant proper length that is accelerating.

Hovering close to the event horizon of a black hole isn't really possible/practical, the required accelerations are way too high. We had a longish thread on this a while back. Using a particular notion of distance (the ruler notion of distance), it takes about 10^16 g's to hover 1 meter (ruler distance) above the event horizon of a larae black hole, independently of the black holes mass.

Note that when you look at GR textbooks that calculate the tidal force, they're really computing tensor quantity via geodesic deviation,rather than the non-tensor version correspoinding to the head-toe acceleration difference. The two are normally not very different under normal circumstances, but at accelerations of 10^16 g, the tiny differences become significant.

 

[PeterDonis]

The two are normally not very different under normal circumstances, but at accelerations of 10^16 g, the tiny differences become significant.

It's not the magnitude of the acceleration that makes the difference, it's the amount of spacetime curvature. For example, consider a Rindler congruence of accelerated observers in flat spacetime. Neighboring observers in such a congruence will have slightly different proper accelerations, even though spacetime curvature is zero (i.e., zero geodesic deviation); and this will be true regardless of the magnitude of the proper acceleration.

 

[cosmik debris]

I was kidding US more than you (and I WAS kidding, not being rude). We have a knack for saying the same thing over and over with each person thinking that his particular way of saying it is better than the previous n versions.

Not better but different; a variety of answers often aids understanding. Also some people don't seem to be sensitive to the level of understanding of a poster and give answers that although correct are far too advanced for the OP.

e.g.

Young boy: Dad where did I come from?
Father: (Long winded birds and bees explanation.)
Young boy: Yes I know that, but Jimmy comes from London, where do I come from?

 

[PWiz]

@A.T. Ah yes, I really should've made the manifold in which I was describing the geodesic clear. Thanks.
@PeterDonis That is what I meant (although my translation from math to words wasn't so great!). Does this mean that the escape velocity beyond the EH is undefined, as a static observer cannot exist beyond the EH (I haven't studied the Kerr metric very well, but I'm pretty sure that for a non-rotating black hole, the $dt$ terms becomes positive beyond the EH, and I guess this means that moving towards the singularity is the same as moving ahead in time) ? I'm asking this because I've frequently seen users here at PF say that escape from a BH once beyond the EH is impossible because the escape velocity exceeds $c$ , but in accordance with the definition you've posted, this would be an incorrect statement.

 

[PeterDonis]

Does this mean that the escape velocity beyond the EH is undefined, as a static observer cannot exist beyond the EH

Yes. (It is also undefined, strictly speaking, at the horizon, since only light can remain static there.)

for a non-rotating black hole, the $dt$ terms becomes positive beyond the EH,

Only in certain coordinates. But the non-staticity of spacetime at and inside the EH is not coordinate-dependent; it's an invariant geometric feature of the spacetime.

I guess this means that moving towards the singularity is the same as moving ahead in time

This is true, but again, it's not dependent on coordinates; it's an invariant geometric feature of the spacetime. (A better way of describing it would be to say that every future-directed timelike and null curve inside the horizon intersects the singularity.)

I've frequently seen users here at PF say that escape from a BH once beyond the EH is impossible because the escape velocity exceeds cc , but in accordance with the definition you've posted, this would be an incorrect statement.

Yes, although it's a fairly common incorrect statement. The formula for escape velocity is well-defined mathematically at any radius greater than zero, so one can calculate a number that one is tempted to call "escape velocity" at or inside the horizon even if there isn't a physical interpretation of that number in that region that justifies the name the way there is outside the horizon.

 

[PWiz]

Yes. (It is also undefined, strictly speaking, at the horizon, since only light can remain static there.)

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

Only in certain coordinates.

Are you talking about Kruskal coordinates?

 

[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 seperation) 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.