BH question

[hoola]

I saw a discussion of black holes of different sizes having differing gravity forces at the event horizons, but if the horizon is established by C, and C is constant, shouldn't the gravity at the surfaces be equal?

Edited May 27, 2017 by hoola

[Strange]

The event horizon is defined by the Schwarzschild radius: [latex]r_s = \frac{2 G M}{c^2}[/latex]. It is not directly related to c. It is the point at which there are no paths from the inside to the outside.

 

The escape velocity at the horizon is [latex]v_e = \sqrt{\frac{2GM}{r}}[/latex]

 

If you substitute r with the Schwarzschild radius, you do get an escape velocity of c. But that is not the reason why nothing can escape from a black hole. After all, you can leave the surface of the Earth at less than the escape velocity.

[imatfaal]

I saw a discussion of black holes of different sizes having differing gravity forces at the event horizons, but if the horizon is established by C, and C is constant, shouldn't the gravity at the surfaces be equal?

 

Good question.

 

By the way - a word to the wise; most people will realise you mean the speed of light when you write C (upper case), especially in this context - but the speed of light should be designated by c (lower case). It seems pernickity to mention it - but misuse of terms, units, and symbols can snowball into a real misunderstanding very quickly (even at the highest levels); it is best to practice precision as much and, as soon as possible.

[MigL]

Not to be picky, Strange, but...

 

The Schwartzschild radius ( or size of the EH ) clearly has a 1/c^2 dependence.

( although you could argue that is an inverse, square relation and not a direct relation )

[imatfaal]

Not to be picky, Strange, but...

 

The Schwartzschild radius ( or size of the EH ) clearly has a 1/c^2 dependence.

( although you could argue that is an inverse, square relation and not a direct relation )

 

I was gonna post that too but decided that it was not quite right. Dependency is a term (normally) reserved for the relationship between a variable and another (set of) variables - but c is not a variable. I wuld say that that G/c^2 is the constant of proportionality of the relationship between r_S and M; I would not say that r_S depends on c in a mathematical sense - only in a plain language usage of the word.

[MigL]

I stand corrected.

[imatfaal]

I stand corrected.

 

Damn - I hope I am right.

[swansont]

Newtonian gravity shows us that F = GMm/r^2, so the acceleration depends on M as well as r. But as Strange has posted, the event horizon varies linearly with mass. If you double the mass you double r_s, but that means the strength of gravity at that distance has dropped in half.

 

edit: but this is inherently relativistic, so my analysis is moot

 

IOW, it's complicated

https://en.wikipedia.org/wiki/Surface_gravity#Surface_gravity_of_a_black_hole

 

but surface gravity still appears to vary inversely with mass

[Lord Antares]

As far as I know, the escape velocity for all event horizons is c and therefore, the force of gravity is the same. What isn't different is the gravity on a set distance between a singularity (or the centerpoint of the event horizon, if you wish) and the event horizon.

 

So, in other words, the event horizon is there BECAUSE the gravitational force at that point is the same, having an escape velocity of c, so the force at the event horizon must always be the same, by definition. Someone correct me if I'm wrong.

 

It would appear that the discussion was inccorect. Can you link to it?

Edited May 27, 2017 by Lord Antares

[swansont]

As far as I know, the escape velocity for all event horizons is c and therefore, the force of gravity is the same. What isn't different is the gravity on a set distance between a singularity (or the centerpoint of the event horizon, if you wish) and the event horizon.

 

So, in other words, the event horizon is there BECAUSE the gravitational force at that point is the same, having an escape velocity of c, so the force at the event horizon must always be the same, by definition. Someone correct me if I'm wrong.

 

It would appear that the discussion was inccorect. Can you link to it?

 

 

I suspect a problem here is that photons do not behave the same as massive objects.

[Lord Antares]

Good point, but doesn't the fact that photons always need v > c to escape from within the event horizon ultimately mean that there is always the same amount of force present at the horizon? To assume differently would be to assume that different photons in different event horizons need different escape velocities, which would imply that the forces at event horizons differ. Therefore, the gravitational force at the event horizon must be the same even for massive particles, no?

[Mordred]

As far as escape velocity is concerned. There isn't any difference. Photons use null geodesics while massive particles spacelike paths. However the escape velocity itself at a particular radius is the same value.

 

Probably the easiest way to understand it is when the escape velocity equals c. The escape path is a null geodesic. There is no spacelike path that a massive particle can take to escape a BH. Only massless particles can use null paths. As only massless particles can reach c.

 

For any value less than c the path is a spacelike path. The required escape velocity doesn't change just the type of geodesic.

Edited May 27, 2017 by Mordred

[hoola]

my thinking is that larger BHs have a smaller effective gravity felt at the approach of the EH due to the averaging effect on space shear, by it being closer to a flat surface than smaller ones....but that the gravity felt at the EH should be the same for all BHs or else the horizon wouldn't be where it was...strange, I don't understand your last remark about escape velocity of c not being why things can't get out...I thought that was a pretty settled issue...

Edited May 27, 2017 by hoola

[Mordred]

edit

Edited May 28, 2017 by Mordred

[Strange]

Srange, I don't understand your last remark about escape velocity of c not being why things can't get out...I thought that was a pretty settled issue...

For example, you can throw a ball or fire a missile up at less than escape velocity and it will go up and then fall back. So escape velocity doesn't stop something leaving the surface, it just means it won't keep going. On the other hand, nothing can leave the event horizon even temporarily

[swansont]

For example, you can throw a ball or fire a missile up at less than escape velocity and it will go up and then fall back. So escape velocity doesn't stop something leaving the surface, it just means it won't keep going. On the other hand, nothing can leave the event horizon even temporarily

 

 

There's also the situation that a missile can escape because it's continually generating thrust, and has far less than the escape velocity. But would still not escape a black hole.

[Janus]

As far as I know, the escape velocity for all event horizons is c and therefore, the force of gravity is the same. What isn't different is the gravity on a set distance between a singularity (or the centerpoint of the event horizon, if you wish) and the event horizon.

 

So, in other words, the event horizon is there BECAUSE the gravitational force at that point is the same, having an escape velocity of c, so the force at the event horizon must always be the same, by definition. Someone correct me if I'm wrong.

 

It would appear that the discussion was inccorect. Can you link to it?

There is no direct relationship between the force of gravity at a given point and the escape velocity at that same point. In Newtonian physics an example is Earth and Uranus. The surface gravity of Uranus is 0.889 g (compared to the 1 g for the Earth), yet the escape velocity for Uranus is 21.3 km/sec compared to the 11.19 km/sec fro the Earth. It has a lower surface gravity, but a nearly twice the escape velocity.

[MigL]

But the event horizon is not a surface.

It is a mathematical construct.

By definition, the radial distance at which light/information can no longer escape to the observable universe as a result of gravity or space-time curvature. And this happens at equivalent gravity or space-time curvature.

As a result, the event horizon is dark, and we lose all information regarding the former matter other than mass/energy, charge and ang. momentum.

( at least until Quantum Gravity Theory tells us differently )

Edited June 15, 2017 by MigL

[imatfaal]

!

Moderator Note

 

Discussion on photon hovering etc moved to thread already discussing this matter - this thread to remain for OP

 

http://www.scienceforums.net/topic/107085-light-near-a-black-holes-event-horizon/

 

[imatfaal]

But the event horizon is not a surface.

It is a mathematical construct.

By definition, the radial distance at which light/information can no longer escape to the observable universe as a result of gravity or space-time curvature. And this happens at equivalent gravity or space-time curvature.

As a result, the event horizon is dark, and we lose all information regarding the former matter other than mass/energy, charge and ang. momentum.

( at least until Quantum Gravity Theory tells us differently )

 

Sorry about moving this.

 

Not sure I agree with the content - should have read it properly the first time then a) I would not have mistakenly moved it and b) would have responded.

 

Surface Gravity - ie gravity of a black hole when at Event Horizon is not constant and varies with the size of the black hole. If we dente surface gravity as g then

 

[latex]g=\frac{1}{m} \cdot \frac{c^4}{4G}[/latex]

 

It is inversely proportional to mass and thus to radius - which in my layman's mind makes little sense. I can almost rationalise it thus; the curvature needed for there to be no exit can be mild (low gravity) within a big black hole but within a smaller black hole that curvature (and thus gravity) needs to be greater. In utterly simplistic analogistic terms that I cannot believe to be true - you can draw gentler curves wholly within a big circle than you can within a smaller one

[hoola]

If two astronauts were near a large black hole, holding onto a mirror which was affixed so as to remain still, and one astronaut was to let go of the mirror and descend into the hole waving goodbye at a steady rate, would the falling astronaut see the identical wave slowing of his reflection in the mirror that his friend would see of him directly?

Edited June 19, 2017 by hoola

[beecee]

If two astronauts were near a large black hole, holding onto a mirror which was affixed so as to remain still, and one astronaut was to let go of the mirror and descend into the hole waving goodbye at a steady rate, would the falling astronaut see the identical wave slowing of his reflection in the mirror that his friend would see of him directly?

Hmmmm...That's a sticky one......I would say he would see his image blueshifted and that of his mate speeded up, as it was reflected back towards a stronger gravitational potential, but he would also cross the EH in pretty quick time I imagine.

Looking outside from inside the EH, he would see the whole universe focused in front of him due to incredible gravitational lensing.

The meeting with the singularity would be in even quicker time, and when he is spaghetiffied and torn apart, depends on the size of the BH....a stellar mass BH, would have him possibly torn apart even before he crossed the EH...A SMBH like Sagitarius A, would see the tidal effects not evident until he was well on the way to the singularity.

Edited June 19, 2017 by beecee

[MigL]

Light is both , a clock and a signal. It is climbing out of the gravity well, and its wavelength stretched/ frequency reduced, and then falling back into the gravity well, with the reverse effects. You would think there would be no difference to what the infalling observer sees reflected from the mirror, as opposed to the faraway observer who only sees the signal climbing out.

Of course this only applies before the infalling observer crosses the event horizon, after he crosses no signal/ information can travel to the mirror anymore.

 

Strange's post #2 indicates that the Event Horizon is defined by the radius Rs=2GM/c^2, while the escape velocity is Ve=root[2GM/R].

Simple substitution when R=Rs gives Ve=root[c^2]=c.

I.E. When radius is equal to the event horizon, the escape velocity is c.

( no matter what g, the gravitational field strength, is )

Edited June 19, 2017 by MigL

[hoola]

I would think that the rate of waving would be seen to be slower to the infalling observer's reflection, as the time slowing in a high gravity field upon the light beam as it travels twice, just as a fast traveller leaving the earth and coming back would be aged more slowly. The fact that the traveler comes back to the earth does not reverse the time slowing effect as he is leaving it....

Edited June 19, 2017 by hoola

[MigL]

No, gravitational time dilation doesn't work that way.