Black hole gravitational pull questions

[hyunxu]

I hear people say that even light cannot escape from black hole.So how big is this black hole?Why gravitational pull is very high? Does it have any escape velocity like Earth have?

 

[stefan r]

I hear people say that even light cannot escape from black hole.So how big is this black hole?Why gravitational pull is very high? Does it have any escape velocity like Earth have?

Yes, escape velocity is greater than the speed of light inside the event horizon. You can calculate the escape velocity in exactly the same way you calculate it for any other body:

For a spherically symmetric, massive body such as a star, or planet, the escape velocity for that body, at a given distance, is calculated by the formula[3]

v e = 2 G M r , {\displaystyle v_{e}={\sqrt {\frac {2GM}{r}}},}

where G is the universal gravitational constant (G ≈ 6.67×10−11 m3·kg−1·s−2), M the mass of the body to be escaped from, and r the distance from the center of mass of the body to the object

The radius of the event horizon any point where the escape velocity is equal to the speed of light.

 

[Ibix]

So how big is this black hole?

In principle, black holes can be any size. The ones that are formed from stars, though, don't form unless the star is above a certain mass because only big enough stars can compress their core enough to overcome internal pressure and collapse into a black hole. So there's a minimum size limit for black holes formed this way.

Why gravitational pull is very high?

Because all of its mass is concentrated in a small volume (there are several criticisms one could make of that answer, but it'll do). At the moment, you are feeling a 1g pull from the Earth. Some of its mass is very close to you, but some of it is over 12,000km away - so the pull from those bits is quite weak. It turns out to average out to the same pull as if all of the mass of the Earth were at a point in the centre of the Earth. But if you (somehow) compressed the Earth into a black hole, then all of the mass would actually be in a very small volume around the centre of the Earth and you could get very close to all of the mass, and it would all pull on you very strongly.

Does it have any escape velocity like Earth have?

No (contrary to the previous answer). At the "surface" of a black hole, which is called the event horizon (it's not a surface in any physical sense, but is the point of no return, where you can no longer escape no matter how powerful your rocket is) it is simply impossible to move away from the black hole without moving faster than light, which is impossible. This is a different concept from the escape velocity, which is the speed you need to be going at to get to infinite distance without using your engines any more. That is meaningless for a black hole because you cannot move away even slightly, let alone to infinite distance.

Escape velocity does make sense as a concept from points above the event horizon, but the formula provided by @stefan r is incorrect, because it uses Newtonian physics, and that isn't accurate close to a black hole. It's more or less a coincidence that the Newtonian formula for escape velocity being c matches the Schwarzschild radius.

 

[PeterDonis]

Escape velocity does make sense as a concept from points above the event horizon, but the formula provided by @stefan r is incorrect

Actually, the formula $v_e = \sqrt{2M / r}$ is exactly correct for the Schwarzschild metric in GR, for any point above the horizon. However, even though this approaches $1$ (the speed of light) as a limit as the horizon is approached, the interpretation as "escape velocity" breaks down at the horizon, which is why light can't escape at the horizon, even though $v_e$ equals the speed of light there, which would seem to indicate that light could escape.

 

[Ibix]

Actually, the formula $v_e = \sqrt{2M / r}$ is exactly correct for the Schwarzschild metric in GR, for any point above the horizon.

Is that correct? The only reference I can find online that actually derives it (https://link.springer.com/article/10.1134/S102833581506004X) arrives at that result for the weak-field solution (equation 14). Frustratingly, the last two pages where they appear to be deriving the full solution are paywalled.

I tried to derive it from Carroll chapter 7. His equations 7.47 & 7.48, using $\lambda=\tau$, setting L=0 for radial motion, $\epsilon=-1$ for a massive particle, and E=1 for it to come to rest at infinity, say that$$
\begin{eqnarray*}\frac 12\left(\frac{dr}{d\tau}\right)^2-\frac 12+\frac{GM}r&=&\frac 12\\
\left(\frac{dr}{d\tau}\right)^2&=&2-\frac{2GM}r\end{eqnarray*}$$Even noting that I think I actually want to know $\sqrt{g_{rr}(dr/d\tau)^2}$, I don't come up with your result.

 

[nikkkom]

I hear people say that even light cannot escape from black hole.So how big is this black hole?Why gravitational pull is very high? Does it have any escape velocity like Earth have?

I recommend going to wikipedia for basic information like this.

 

[PeterDonis]

His equations 7.47 & 7.48, using $\lambda=\tau$, setting L=0 for radial motion, $\epsilon=-1$ for a massive particle, and E=1 for it to come to rest at infinity

It should be $\epsilon = 1$ for a massive particle; see equation 7.39 and the text just following. Try it that way and you should get the result I gave.

 

[Ibix]

It should be $\epsilon = 1$ for a massive particle; see equation 7.39 and the text just following. Try it that way and you should get the result I gave.

D'oh! Careless reading of the text following 7.39. Correcting that, I now get $dr/d\tau=\sqrt{2M/r}$ since the halves now cancel and the Newtonian potential term has the opposite sign. I do indeed need to multiply by $\sqrt{-g_{rr}}$ to convert dr into something physically meaningful, but I also need to multiply by $\sqrt{g_{tt}}$ to convert from the escaping object's proper time to a hovering observer's proper time, and it happens that $-g_{rr}g_{tt}=1$.

Sorry for the confusion - @stefan r's formula is correct for escape velocity from above the horizon. My point about escape velocity from the horizon itself not being meaningful stands.

 

[PeterDonis]

Even noting that I think I actually want to know $\sqrt{g_{rr}(dr/d\tau)^2}$

It depends on how you want to define $v_e$ physically. The way I was defining it was the locally measured velocity relative to a static observer at the same $r$ coordinate. That is given by the ratio of two inner products: the inner product of $v^\mu$ with a unit vector $e^\nu$ pointing in the outward radial direction, and the inner product of the 4-velocity $v^\mu$ of the escaping object with the 4-velocity $u^\nu$ of the static observer. Physically, this amounts to taking the ratio $p / E$ of radial momentum over energy for the object, relative to a static observer (strictly speaking, both of these quantities are per unit mass as we are calculating them, but the object's mass cancels in the ratio anyway so we can just leave it out).

It is straightforward to obtain the components of these vectors (I'll only give the $t$ and $r$ components since we are restricting to radial motion):

$$
v^\mu = \left( \frac{1}{1 - 2M / r}, \sqrt{\frac{2M}{r}} \right)
$$
$$
u^\nu = \left( \frac{1}{\sqrt{1 - 2M / r}}, 0 \right)
$$
$$
e^\nu = \left( 0, \sqrt{1 - 2M / r} \right)
$$

We can then write (note the minus sign in the denominator, which arises because we are using the $- + + +$ metric signature convention and we want the energy to be positive):

$$
v_e = \frac{g_{\mu \nu} v^\mu e^\nu}{- g_{\mu \nu} v^\mu u^\nu}
$$

Substitution yields

$$
v_e = \frac{ \left( 1 - 2M / r \right)^{-1} \sqrt{2M / r} \left( 1 - 2M / r \right)^{1/2}}{\left(1 - 2M / r \right) \left( 1 - 2M / r \right)^{-1} \left( 1 - 2M / r \right)^{-1/2}}
$$

All of the weird factors cancel out and we are left with the result I gave. This might appear counterintuitive, but heuristically, the time dilation factor of the static observer relative to infinity cancels out the radial length factor relative to infinity, so $dr / d\tau$ ends up being the physically measured $v_e$ relative to the static observer.

 

[stefan r]

I hear people say that even light cannot escape from black hole.So how big is this black hole?Why gravitational pull is very high? Does it have any escape velocity like Earth have?

Some statements I believe can be said:

An Earth mass black hole is unlikely. If one existed an object like the international space station could orbit around 6800 km from the center at the same velocity that the ISS orbits earth. The gravity feel would be the same. A stationary object 3657 km from the singularity would start to fall. Would feel identical to falling at the north pole. The event horizon would be about the around the size of a ping pong ball. You could not land on the micro-black hole. Part of you would be in retrograde orbit and collide with other parts. Your atoms would become high energy cosmic radiation. If you fell from outside the tropics and had Earth's rotation you would miss the ping pong ball sized event horizon and you would have no problems (except vacuum etc). You would follow an elliptical orbit and get back to Earth's radius.

A object like Earth which has Earth's velocity would orbit a solar mass black hole the same way that Earth orbits the sun. Years would still be 365 days and you would need leap years. The gravity would be the same. If you could somehow get liquid surface water then the tides would be the same.

You also get 9.8 m/s² gravity around the sun, black holes, and the super massive black holes. Our sun has 9.8 m/s² at 5.3 solar radii.

If a black hole's event horizon was 4.6 x 10¹⁵ m radius then gravity at the event horizon would be 9.8 m/s². So around one light year diameter black holes have gravity just like the surface of Earth at the event horizon. The entire milky way compressed into a black hole would have radius 0.2 light years.

 

[Ibix]

If a black hole's event horizon was 4.6 x 10¹⁵ m radius then gravity at the event horizon would be 9.8 m/s².

This part is not correct, or, at least, is potentially misleading. The proper acceleration needed to hover at the event horizon is infinite, whatever the mass of the black hole.

 

[stefan r]

This part is not correct. The proper acceleration needed to hover at the event horizon is infinite, whatever the mass of the black hole.

Would not be "hovering".

You can stand on Earth because the forces from the floor balance the force of gravity. There is nothing to stand on at the event horizon. Falling would briefly have the same acceleration rate.

 

[PeterDonis]

You can stand on Earth because the forces from the floor balance the force of gravity. There is nothing to stand on at the event horizon.

By "hovering" I think @Ibix meant "using rocket engines". You don't need to stand on anything to hover.

However, it is impossible to "hover" even using rocket engines at the event horizon, because you would have to move at the speed of light, and that's impossible.

Falling would briefly have the same acceleration rate.

No, free-falling means zero proper acceleration.

 

[stefan r]

No, free-falling means zero proper acceleration.

Please explain. When I jump out of an airplane or jump off a diving board I get the impression that I am free falling and also appear to be accelerating. [disregarding air drag]

 

[rootone]

I hear people say that even light cannot escape from black hole.So how big is this black hole?Why gravitational pull is very high? Does it have any escape velocity like Earth have?

There is no size of black hole, that's why it is a singularity in mathematical terms.
However a black hole does have an event horizon which has a size.
Nothing inside that horizon can ever get outside of it.

 

[PeterDonis]

When I jump out of an airplane or jump off a diving board I get the impression that I am free falling and also appear to be accelerating.

The acceleration you appear to be undergoing is coordinate acceleration, not proper acceleration. Coordinate acceleration means you've chosen coordinates (in this case, coordinates fixed to the surface of the Earth) in which you appear to be accelerating. Proper acceleration means you feel weight.

In relativity we focus on proper acceleration because it's invariant; it doesn't depend on your choice of coordinates, whereas coordinate acceleration does (you can just as easily choose coordinates in which you, free-falling from the airplane, are at rest, but you can't change whether you feel weight or not by doing that). And a primary principle in relativity is that physics is contained in invariants; things that depend on your choice of coordinates aren't telling you about the actual physics, only about your choice of coordinates.

 

[Ibix]

You can stand on Earth because the forces from the floor balance the force of gravity. There is nothing to stand on at the event horizon. Falling would briefly have the same acceleration rate.

As Peter says, this formulation of gravitational acceleration doesn't have an unambiguous definition in general relativity. If you are in a closed box, no experiment will tell you the acceleration due to gravity that you are undergoing - if you are in free fall you cannot tell if you are seconds from impact on the ground or are far out in deep space. This leads us (well, Einstein) to the conclusion that gravitational acceleration is a fictitious acceleration, similar to the centrifugal acceleration that appears in rotating reference frames. What its value is depends on the frame you pick. So "the acceleration rate of an observer crossing the event horizon" - in this sense - can take any value you like.

The closest concept to the Newtonian acceleration due to gravity is the proper acceleration experienced by an observer at constant altitude. Proper acceleration is basically the reading on a weighing scale, so cannot be made to change by a choice of frame. And the reading when you are at a constant altitude is the same whether you are hovering on rockets, stood on a solid surface, or any other means of staying at the same altitude. I used "hovering" as a catch-all term for this. And this value, which is $g=GM/r^2\sqrt {1-r_S/r}$ where $r_S=2GM/c^2$ is the Schwarzschild radius, goes to infinity at the event horizon. Which is another way to say that once you are at the event horizon you are going down.

 

[russ_watters]

An Earth mass black hole is unlikely.

This isn't correct. Black holes of a wide variety of sizes, including very small, are likely...though tthere are more common sizes.

If a black hole's event horizon was 4.6 x 10¹⁵ m radius then gravity at the event horizon would be 9.8 m/s².

Some people tried to address this, but not succinctly/directly enough:

A black hole (and its size) is defined by the event horizon, which has an escape velocity of C and infinite "surface" g. So saying g=9.8 isn't correct/is a contradiction.

The error is you are mixing together an earth-mass black hole and an earth-radius black hole. A black hole of Earth's mass is only 2 cm in diameter. An Earth radius black hole has a mass much larger than any single star.

 

[Ibix]

A black hole is defined by the event horizon, which has an escape velocity of "C"

As you get closer and closer to the event horizon the escape velocity approaches c, yes. But I think it's worth repeating that there isn't really an escape velocity from the event horizon because you can't escape. An object may hover at the event horizon if it was created there and moves at c. But nothing can reach infinity.

 

[stefan r]

...
The error is you are mixing together an earth-mass black hole and an earth-radius black hole. A black hole of Earth's mass is only 2 cm in diameter. An Earth radius black hole has a mass much larger than any single star.

I certainly did not make that error.

But I would like to check the facts. I would not consider myself an expert on gravity.

...Some people tried to address this, but not succinctly/directly enough:

A black hole (and its size) is defined by the event horizon, which has an escape velocity of C and infinite "surface" g. So saying g=9.8 isn't correct/is a contradiction.

Saturn's surface gravity is 10.55 m/s2 but escape velocity is 35.5 km/s. "Escape velocity c" is not the same as "infinite surface gravity". The singularity probably does have infinite "surface" gravity.

If you are a few meters above the event horizon (4.6 x 10¹⁵ +10 m ) and you turn on a light what happens to the photons?

On Earth the 9.8 m/s² gravity has a small but greater than 0 effect on the wavelength of light. The photon would be longer if measured on the moon for example. Gravity's effect at 4 Earth radii is twice the effect at 2 Earth radii. From a long distance the red shift is not much.
Near this event horizon gravity has the same effect on the photon. However, the gravity would still be 4.9 m/s2 at 1.4 times the radius from the hole. The gravity would be stretching the photon's wavelength for a very long time-distance (1.9 x 10¹⁵m) . So from a long distance the red shift is very large.

Time also stretches. The astronaut turning on a light would not get a message home on a reasonable time scale. However, two astronauts shining flashlights at each other would see something much like two skydivers shining flash lights at each other. If you were holding the light in front of you it would not look red shifted even if your hand crossed the event horizon first.

 

[jbriggs444]

"Escape velocity c" is not the same as "infinite surface gravity".

You need to define what you mean by infinite surface gravity before you can tell us what values it takes on. Gravity is not a force.

The singularity probably does have infinite "surface" gravity.

This claim is even more problematic.

 

[Ibix]

"Escape velocity c" is not the same as "infinite surface gravity".

Well, as we keep saying, nothing has an escape velocity of c. Escape velocity doesn't make sense at or below the event horizon because you cannot escape. I don't think it's clear what "surface gravity" would mean in a general relativistic context. The closest analogy to the Newtonian concept is the proper acceleration needed to remain at constant altitude. So, with the caveats that the escape velocity doesn't really mean "escape" and that "surface gravity" is to be interpreted as the proper acceleration needed to stay at a height then both "escape velocity c" and "infinite surface gravity" are synonyms for "at the event horizon". All three are the same.

The singularity probably does have infinite "surface" gravity.

I'm pretty certain that isn't a coherent concept. The singularity isn't a place - it's a time in the future of anyone crossing the event horizon. Does next Friday have a surface gravity? (Edit: although Mondays do suck...)

If you are a few meters above the event horizon (4.6 x 1015 +10 m ) and you turn on a light what happens to the photons?

Assuming you're pointing the lamp straight up, they go straight up. Otherwise they follow curving paths that may continue to infinity or may curve back down into the black hole - it depends on the angle. They'll be either red- or blue-shifted when detected, depending on whether the detector is at a higher or lower altitude. Other Doppler effects may occur depending on the states of motion of the emitter and receiver.

On Earth the 9.8 m/s2 gravity has a small but greater than 0 effect on the wavelength of light. The photon would be longer if measured on the moon for example.

You don't really specify what experiment you are doing here. I think you're shining a light upwards and detecting it at a point higher in the gravitational field. In that case, the receiver will measure a longer wavelength, yes. As you say, the effect is not extreme in the Earth's gravity. Please note, however, that the effect does not depend on the gravitational acceleration - it cannot, since that isn't well defined. Gravitational redshift depends on the difference in gravitational potential between the receiver and emitter. So I think this bit:

Near this event horizon gravity has the same effect on the photon. However, the gravity would still be 4.9 m/s2 at 1.4 times the radius from the hole. The gravity would be stretching the photon's wavelength for a very long time-distance (1.9 x 1015m) . So from a long distance the red shift is very large

...is wrong.

Time also stretches.

That's not a good way to look at it. Better to say that a clock hovering near a black hole runs slower than a clock hovering far away. It's more precise, and avoids complications about "whose time do you mean?".

The astronaut turning on a light would not get a message home on a reasonable time scale.

Depends what you mean by a reasonable timescale, who's doing the measuring, and where home is. But yes, since clocks near a black hole run slowly, a distant observer will regard the astronaut as taking a long time to send a message - part and parcel of the redshift.

However, two astronauts shining flashlights at each other would see something much like two skydivers shining flash lights at each other. If you were holding the light in front of you it would not look red shifted even if your hand crossed the event horizon first.

Correct, assuming the astronauts aren't so far apart that tidal effects become apparent. Ditto your hand - near a small black hole tidal effects will spaghettify you.

 

[PeterDonis]

I would not consider myself an expert on gravity.

Then you should not be posting with so much confidence. You should be asking questions, not answering them incorrectly.

The singularity probably does have infinite "surface" gravity.

No, it doesn't. The concept of "surface gravity" is not even well-defined below a black hole's horizon.

If you are a few meters above the event horizon (4.6 x 1015 +10 m ) and you turn on a light what happens to the photons?

Um, they move away from you at the speed of light?

If you're asking about the redshift of radially outgoing photons, yes, they will be very strongly redshifted by the time they reach a static observer far away from the hole.

Time also stretches.

This is not a good way of describing what's going on in ordinary language. Please take the time to learn the actual math.

 

[russ_watters]

I certainly did not make that error.

Then I don't know what error caused you to say what I quoted. Regardless, it was wrong.

Saturn's surface gravity is 10.55 m/s2 but escape velocity is 35.5 km/s. "Escape velocity c" is not the same as "infinite surface gravity".

This is true; surface gravit[ational acceleration, or g] and escape velocity are not the same thing. But at the event horizon of a black hole, infinite g and escape velocity of c coincide. For other objects, the relationship depends on the mass and radius of the object, which can vary with respect to each other. For a black hole, each mass has only one radius (you can't have black holes of the same mass with different radii).

The singularity probably does have infinite "surface" gravity.

Others have pointed out this doesn't make sense, but I'll assume you meant "event horizon", not "singularity". In that case, unlike your previously quoted statement, it is true.

 

[russ_watters]

As you get closer and closer to the event horizon the escape velocity approaches c, yes. But I think it's worth repeating that there isn't really an escape velocity from the event horizon because you can't escape. An object may hover at the event horizon if it was created there and moves at c. But nothing can reach infinity.

Since nothing can reach infinity, I don't see "you can't escape" and "escape velocity of c" as contradicting each other because you can't reach c. The main difference I see between that and other examples of escape velocity is there isn't even any ballistic trajectory that takes you away from the event horizon at all because no matter how fast you are going, you are stationary with respect to c.

 

[Monsterboy]

I have a question, a single black hole finds itself between three other black holes who are more massive and are equidistant from it and from each other, can the three black holes rip the central black hole apart ?

 

[Martin0001]

I have a question, a single black hole finds itself between three other black holes who are more massive and are equidistant from it and from each other, can the three black holes rip the central black hole apart ?

No.
However such a system would be very unstable. How would you keep 3 larger BH apart?

 

[Monsterboy]

No.
However such a system would be very unstable. How would you keep 3 larger BH apart?

They won't be apart for long, I wanted to know the effect on the central black hole before all of them combine.

 

[stefan r]

They won't be apart for long, I wanted to know the effect on the central black hole before all of them combine.

Before the combination the black hole's event horizon has a radius. After the combination the event horizon has a larger radius. To say that "ripped the event horizon outward" would be a strange way to look at it.

Rotating black holes have a ring singularity. If 3 black holes merge they should become another rotating black hole. I have know idea which part of the new ring came from each black hole. Could that be your question?

There is a no hair theorem. The effect is that it does not matter how the black hole got its mass, momentum, and spin. They are all identical to someone on the outside. That would not need to mean that there was no ripping going on. It just cannot be observed.

 

[Martin0001]

Before the combination the black hole's event horizon has a radius. After the combination the event horizon has a larger radius. To say that "ripped the event horizon outward" would be a strange way to look at it.

Rotating black holes have a ring singularity. If 3 black holes merge they should become another rotating black hole. I have know idea which part of the new ring came from each black hole. Could that be your question?

There is a no hair theorem. The effect is that it does not matter how the black hole got its mass, momentum, and spin. They are all identical to someone on the outside. That would not need to mean that there was no ripping going on. It just cannot be observed.

No hair theorem is increasingly questioned. It implies information loss and undermines one of sacrest cows of QM, unitarity principle.
Read also about BH fireball paradox.

 

[PeterDonis]

a single black hole finds itself between three other black holes who are more massive and are equidistant from it and from each other, can the three black holes rip the central black hole apart ?

No. What would probably happen is that the central hole would merge with one of the others (and over time the whole system would probably merge into a single black hole).