Question of event horizons

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Wondering if anyone can help. Is there likely to be any difference in the gravitational properties of an event horizon around a quantum black hole and that around a supermassive bearing in mind the differences in mass inside the event horizon.
 

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  • #2
bcrowell
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When you say a quantum black hole, do you mean one whose mass is on the order of the Planck mass? If so, then I don't think anyone knows the answer to your question, since we don't have a theory of quantum gravity.
 
  • #4
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Smallest either thought to or known to exist.
 
  • #5
if there is a difference in mass, then surely there is a difference in gravitational strength
 
  • #6
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The reason I asked is because as I understand it an event horizon is where gravity overtakes the speed of light, as the speed of light is constant then the gravity at any event horizon should be the same but according to the maths the greater the mass the greater the gravity and even the earth has a theoretical event horizon (shwartzchild radius) but going by its mass how can the earths gravity ever overtake the speed of light.
 
  • #7
Nabeshin
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The mathematical solution for a black hole simply scales quite simply with the mass of the hole. In GR, there is no preferred scale, so a hole with mass 10^-20kg behaves exactly the same as a hole with mass 10^50kg. If you want, you can think of it simply as a rescaling of your units, so changing the mass really changes nothing.

(Of course, once we deviate from classical GR this is no longer true.)
 
  • #8
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That is interesting because I was talking to someone on another site who tried to tell me mass was the cause of gravity, the greater the mass the higher the gravitational pull
but it seems when taken to extreme i.e black hole mass has no further effect on gravity, at least at the EH other than to increase r. So do Newton and GR just decribe effect and not cause, if just effect then where does gravity origionate. Can gravity have a maximum value, in reality, rather than the extremes to which maths can take it to
 
  • #9
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That is interesting because I was talking to someone on another site who tried to tell me mass was the cause of gravity, the greater the mass the higher the gravitational pull
but it seems when taken to extreme i.e black hole mass has no further effect on gravity, at least at the EH other than to increase r. So do Newton and GR just decribe effect and not cause, if just effect then where does gravity origionate. Can gravity have a maximum value, in reality, rather than the extremes to which maths can take it to
That someone is right. For example, if earth collapses into the black hole (Rs~9 mm), moon would still orbit it, like nothing happened. Black holes have three properties that can be seen from 'outside'. Mass, charge and spin. More massive BH, greater the pull.
As far as math is concerned, it was clear since the earliest days of relativity, that if you force enough mass into small enough volume (half Planck mass into the sphere with radius of one Planck length, or any multiplicative of that) singularity forms. Now, singularities are really, really, nasty and counter intuitive states, so people generally don't believe that they exist. They are regarded as breakdown of theory. All we know for sure is that black holes have event horizons.
 
  • #10
jambaugh
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A better way of thinking of event horizons is to realize you are surrounded by them. In flat space there is an event horizon around the future light-cone of a point event. No object inside can get out.

Gravity bends the light cones so that a black hole has a "static event horizon" at the Schwarzchild radius. The light cone becomes a "light-tube" if you will. But an observer falling across this event horizon doesn't see anything locally different than you do as you pass across the light-cones of various events around you.

I don't think that's appreciably different for the case of a hypothesized "quantum black hole" excepting that "locally" will not have much meaning at that scale.
 
  • #11
Janus
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The reason I asked is because as I understand it an event horizon is where gravity overtakes the speed of light, as the speed of light is constant then the gravity at any event horizon should be the same but according to the maths the greater the mass the greater the gravity and even the earth has a theoretical event horizon (shwartzchild radius) but going by its mass how can the earths gravity ever overtake the speed of light.
The event horizon is where the "escape velocity" equals the speed of light, and this is not the same as the "strength of gravity" at the event horizon. The pull of gravity in increases by the mass ands falls off by the square of the distance, while escape velocity increases by the squareroot of the mass and falls off by the squareroot of the distance,

The radius of the event horizon increases proportionally to the mass.

So, if you double the mass of the BH, you have to double the radius of the BH to keep the escape velocity equal.

But doubling the mass and doubling the radius means that the pull of gravity at the event horizon decreases by 1/2.
 
  • #12
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Still a bit confused. I assume you mean distance from center. If the gravitational pull decreases by 1/2 then light would escape so it woulde'nt be an EH. As a matter of interest why cant escape velocity be used to describe gravity at surface of mass what is the difference between escape velocity and gravitational pull apart from units used they just pull in opposite directions but by the same ammount.
Also have come accross r=0 as singularity, it does not make sense to me to use any number less than 1 to describe r because above 1 r^2 is greater than r, at r=1 then r & r^2 are the same but less than 1 r^2 is less than r which does not make sense.
I just have a feeling that something about gravity as we know it does not add up, I think its in interpretation rather than maths which obviously works.
 
  • #13
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John it is really simple.

Escape velocity: [tex]v_{0}=\sqrt{\frac{2GM}{r_{0}}}[/tex]

[tex]r_{0}[/tex] - is the radius of the body from which surface you want to escape.

Take Earth, for example. Earth radius is 6 350 000 meters. Earth mass is 5.97 * 10^24 kg. If you calculate, you will get ~11 200 m/s for escape velocity.

Now, lets calculate what would be the escape velocity from the surface of the body that has exactly same mass as Earth, but radius of only 9 mm. We get v0=c.

Edit: Thanks Janus. Edited for accuracy.
 
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  • #14
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Many thanks. How does that compare with Gravitation at surface using same 2 examples.
Can only find info on attraction between masses not for surface of single mass.
If it makes sense from what Janus said gravity changed by squared and EV by squareroot. At r=1 square and squareroot are same. If we take from above example r=9mm (theoretical Event horizon) as the r=1 and work from middle out rather than surface down how would that affect figures if at all
 
  • #15
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Well, you can calculate it with simple gravitational acceleration equation. Non vector form:
[tex]g=-\frac{GM}{r^{2}}[/tex].

For Earth, you get usual 9.82 m/s^2. For Earth-mass black hole you will get acceleration of 4.9*10^18 m/s^2 at the event horizon.

Changing the units want change things at all, of course if you do it properly.

EDIT: In both examples that I gave you here and in previous post, you can see that both escape velocity and g depends only of mass and distance. Even if Earth colapses into the black hole, but you are 6350 km away (Earth radius), you will still accelerate 9.82 m/s^2 at the begining, and you will still need to have velocity of 11.2 km/s to escape it from that point.
 
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  • #16
Janus
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Still a bit confused. I assume you mean distance from center. If the gravitational pull decreases by 1/2 then light would escape so it woulde'nt be an EH. As a matter of interest why cant escape velocity be used to describe gravity at surface of mass what is the difference between escape velocity and gravitational pull apart from units used they just pull in opposite directions but by the same ammount.
The gravity at the surface is found by
[tex]\frac{GM}{r^2}[/tex]
And is equal to the acceleration an object dropped will have. On the surface of the Earth this works out to be ~9.8 m/s^2. This value determines how much you would weigh on the Surface.

Escape velocity tells you how fast an object would have to leave the surface so that it would never fall back. It is found by the formula
[tex]V_e = \sqrt{\frac{2GM}{r}}[/tex]
(The formula given by Calimero is actually the orbital velocity)

On the surface of the Earth this works out to ~ 11 km/sec. It is where this value equals c that the BH event horizon exists. It should be pointed out that all of the mass M needs to be inside of the distance r to use this formula. Note that these are tow different types of measurement and that they do not scale the same if M or r changes.

If M doubles and r stays the same, the strength of gravity at the surface doubles, but the escape velocity only increases by 1.414
If r doubles and M stays the same, the strength of gravity decreases by a factor of 4, and the escape velocity decreases by a factor of 1.414.
If they both double, the strength of gravity decrease by a factor of 2 and the escape velocity stays the same.

Also have come accross r=0 as singularity, it does not make sense to me to use any number less than 1 to describe r because above 1 r^2 is greater than r, at r=1 then r & r^2 are the same but less than 1 r^2 is less than r which does not make sense.
I just have a feeling that something about gravity as we know it does not add up, I think its in interpretation rather than maths which obviously works.
Again, strength of surface gravity and Escape velocity do not scale together.

If you halve the radius, you quadruple the strength of gravity, but increase the escape velocity by 1.414. In other words, all this means is that decreasing the radius increases the surface gravity faster than it does the escape velocity.
 
  • #17
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Many thanks. How does that compare with Gravitation at surface using same 2 examples.
Can only find info on attraction between masses not for surface of single mass.
If it makes sense from what Janus said gravity changed by squared and EV by squareroot. At r=1 square and squareroot are same. If we take from above example r=9mm (theoretical Event horizon) as the r=1 and work from middle out rather than surface down how would that affect figures if at all
It's worth noting that the relativistic local equation for gravity for a static non-charged BH is-

[tex]a_g=\frac{Gm}{r^2} \cdot \frac{1}{\sqrt{1-\frac{2Gm}{rc^2}}}[/tex]

where the first part relates to the stress-energy tensor and the second part represents the metric tensor. The above shows that gravity becomes infinite at [itex]r=2Gm/c^2[/itex] (i.e. the event horizon) regardless of the mass involved.
 
  • #18
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Quick question regarding Relativistic Mass: If a proton travelled very, very close to the speed of light, does it become sufficently massive so as to make it possible to collapse into a black hole, OR is it that because in its own frame of reference it is no more massive than it was when it was stationary? Would other nearby objects detect the increased relativistic gravitiation? Would they need to be stationary to detect the increase in relativistic gravitation? Do Galaxies and stars moving at high relative velocities show the effects of relativisitic mass and relativistic gravitation, perhaps more readily observed during collision?
 
  • #19
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Not to sure about gravity becoming infinite. Problem with infinity is that if one thing is infinite then nothing else can exist with it, in other words if c was infinite then a photon travelling at infinite speed would effectively fill an infinite space. So in order for our universe to exist nothing in it, apart possibly from volume, can be infinite. I firmly believe that all in the universe has mathematical boundaries such as c, we just havent found all of them yet. I think its possible that c is where gravity becomes 0 somehow.
 
  • #20
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The whole notion of infinite gravity is only relative to the person crossing the event horizon, you could say that gravity is synonymous with infall velocity and if we consider proper velocity for an object that has fallen from rest at infinity, then the equation is-

[tex]v=\sqrt{\frac{2M}{r}}[/tex]

but for an observer at infinity, the velocity will be observed as-

[tex]v=\left(1-\frac{2M}{r}\right)\sqrt{\frac{2M}{r}}[/tex]

as the infalling object approaches the event horizon, to the observer at infinity, the object appears to come to a stand still. Once the object reaches the event horizon, the time dilation becomes zero [itex](d\tau=dt\sqrt(1-2M/r))[/itex] and at the moment it experiences infinite gravity & v=1, the exterior universes 'ends' (though the infalling object wouldn't witness this, only an object that hovers exactly at the event horizon would see this which isn't possible) so the infalling object doesn't experience infinite gravity relative to outside universe. On a side note, light falling in with the object would still be measured as travelling at c relative to the object.
 

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