Why Do Smaller Black Holes Have Higher Temperatures?

[meteor]

I really would like to know which of the two event horizons of a Kerr black hole is the fatal, that is the point where nothing can't scape
Also, the formula for the Bekenstein-Hawking entropy is calculated in the inner or the outer event horizon?

 

[marcus]

Originally posted by Tail
You're confusing me with all the maths.

Anyway, first I need to understand about the gravity at the event horizon. The event horizon is where light can no longer get out. If you're telling me that gravity at a big black hole's horizon is 4 times as small as that of a smaller hole's horizon, it seems impossible; if the gravity is even a bit smaller than that of the smaller hole's horizon, LIGHT CAN GET OUT. It's not the event horizon anymore.

Ok, WHERE am I wrong?

Hello Tail, According to my textbook (at this point it is only talking about the standard kind, not rotating "Kerr" ones)

the acceleration of gravity is INFINITE at the EH, whatever size of hole.

I think your reasoning would be correct, assuming it were finite. But it isn't finite.

There is this awkward fact of technical jargon however----there is a certain finite parameter which for decades has been called the
"surface gravity". Maybe Hawking introduced the term---I don't know. You have to remember that what they call the "surface gravity" in technical papers is not the real actual acceleration of gravity at the event horizon.

the real actual acceleration is infinity meters per second per second. Or infinity feet per second per second. Ah, but you said you don't like math

infinity is almost worthless for calculating with, anyway of very limited usefulness, so the have this OTHER (finite) quantity which is related to the gravity at the event horizon and which is useful in calculating stuff and which they call "surface gravity"

Thank god however you are not interested in that bit of jargon!
You do not like math so probably you don't like jargon either.

So what is bothering you?

It's infinite, right at the horizon
Light can't get out
What is left to worry about?
except, you know, war, corruption in government,
and the usual things like that

 

[jcsd]

I think you may be confused marcus, even the region inside the event horzion must obey the normal laws of physics, it's only at the singularity they break down, the accelartion should be infinite at the singularity not the evnt horizon.

 

[Tail]

Hmm... it seems to me too that it cannot be infinite, only perhaps in the sense than nothing can move faster than light anyway...

Ok, what's your version, jcsd? You posted that gravitation is the same at the event horizon, no matter how big or small the hole is, if I understand it correctly. So it IS so?


Wait... hmm, I've got an idea...

Let's see, if the gravitation becomes weaker more rapidly when we are dealing with small black holes, perhaps THAT's the reason? The hole radiated more because the differences in gravitation are bigger? I've no idea why that should matter, though...

Just an idea... well?

 

[jcsd]

Originally posted by Tail
Hmm... it seems to me too that it cannot be infinite, only perhaps in the sense than nothing can move faster than light anyway...

Ok, what's your version, jcsd? You posted that gravitation is the same at the event horizon, no matter how big or small the hole is, if I understand it correctly. So it IS so?


Wait... hmm, I've got an idea...

Let's see, if the gravitation becomes weaker more rapidly when we are dealing with small black holes, perhaps THAT's the reason? The hole radiated more because the differences in gravitation are bigger? I've no idea why that should matter, though...

Just an idea... well?

No, I posted that gravitation is stronger at the event horizon the smaller the black hole (see last post)

 

[Tail]

I'll try to explain how I understand your post and you tell me what I've got wrong, ok?

Originally posted by jcsd
the escape velocity is the same for all size black holes at the event horizon

Well, if the escape velocity is the same, then the gravitation at the event horizon must also be the same, as the escape velocity is directly proportional to gravitation, no?

So as you can see from the bottom equation the smaller the event horizon (R_BH) the larger the accleration due to graviation (a) at the event horizon as c² is constant.

Ok, here's how I see it: there can be no acceleration exactly at the event horizon (um... nowhere to accelerate?), it can be from the event horizon onward or something.

The equation (the smaller the hole, the larger the acceleration) would still be right, because (it seems to me so) the gravitation becomes stronger more rapidly for a small black hole than for a big one, so, naturally, the acceleration would be stronger...

Well?

 

[jcsd]

No, accelration is proportional to escape velocity squared over distance, so as the smaller black hole's event horizon is closer to it's centre of gravity the accelration due to gravity (which is obviously a linear function of the graviational force and the mass of the attracted object) is more.

 

[Tail]

I don't understand the mathematical side of this so well, but that was what I was saying too, actually.

What do you think about my first comment on your post (on escape velocity and gravitation)?

 

[jcsd]

By gravitaion I assume you meant graviational force, which is stronger at the event horizon of a smaller black hole than it is at a larger black hole.

 

[meteor]

By gravitaion I assume you meant graviational force, which is stronger at the event horizon of a smaller black hole than it is at a larger black hole

The gravitational force is the same at the event horizon of any black hole, the size of the black hole doesn't matter

 

[Tail]

But the escape velocity is the same! Doesn't it depend on gravity?

 

[jcsd]

Originally posted by meteor
The gravitational force is the same at the event horizon of any black hole, the size of the black hole doesn't matter

No the force is different the escape velocity is the same, from one of my previous posts (a is the accleration due to gravity at the event horizon):

a = c²/2R_BH


F= ma =>

F = mc²/2R_BH

So you can see from the above an object of mass, m, will be subject to a greater force the smaller the radius of the event horizon (R_BH).

 

[Tail]

Um, that was for jcsd.


meteor, since you agree with me on the gravitation part, how do YOU explain the different temperatures of black holes?

 

[jcsd]

Originally posted by Tail
But the escape velocity is the same! Doesn't it depend on gravity?

the escape velocity is dpendant on the mass and the distance from the centre of gravity, like the graviational force, but the two are not directly proportional.

 

[Tail]

jcsd,
think about it, that formula is not for the event horizon, but for the distance from the event horizon to the singularity!

 

[Tail]

Originally posted by jcsd
the escape velocity is dpendant on the mass and the distance from the centre of gravity, like the graviational force, but the two are not directly proportional.

Well, how else can it be dependent on gravity if it's the only thing that the escape velocity depends on?

 

[jcsd]

Originally posted by Tail
jcsd,
think about it, that formula is not for the event horizon, but for the distance from the event horizon to the singularity!

I don't see what you're trying to say, the first formula gives you accelration due to graviation at the event horizon and the last one gives you the graviational force exerted on an object of mass, m, at the event horizon.

 

[jcsd]

Originally posted by Tail
Well, how else can it be dependent on gravity if it's the only thing that the escape velocity depends on?

No, the escape velocity is not dependant on the graviational force but the mass of the object and the distance to the centre of gravity:


F = Gm₁m₂/r²


escape velocity = (2GM/R)^1/2

 

[Tail]

Originally posted by jcsd
No, the escape velocity is not dependant on the graviational force but the mass of the object and the distance to the centre of gravity:

Um... yes, but mass is why gravity exists! So gravity is directly proportional to mass, and the distance to the center of the gravity ALSO depends on gravity in this case because it's where light can no longer escape...

 

[meteor]

No the force is different the escape velocity is the same, from one of my previous posts (a is the accleration due to gravity at the event horizon):

a = c2/2RBH


F= ma =>

F = mc2/2RBH

So you can see from the above an object of mass, m, will be subject to a greater force the smaller the radius of the event horizon (RBH).

Sorry, you are right

 

[jcsd]

Just look at the equations and you'll see that they are depedent on mass and distance in different ways. I've demonstrated mathematically that the graviational force at the event horizon of differen sized holes exertered on an object of mas m is different. Remember escape velocity is a function of how much energy is needed to escape a graviational field.

 

[jcsd]

sorry that last post was to Tail

 

[Tail]

How can it be as hard to escape stronger gravity as it is to escape weaker gravity?

If the speed is x, it is harder to escape the gravitational field of the Earth than that of the Moon!

 

[meteor]

Sorry, i,ve deleted this (whas a bad post)

 

[jcsd]

Originally posted by Tail
How can it be as hard to escape stronger gravity as it is to escape weaker gravity?

If the speed is x, it is harder to escape the gravitational field of the Earth than that of the Moon!

escape velocity depends also on how near you areto objects centre of gravity (assuming point masses) and as you get further away from the Earth the less it's escape velocity becomes, the same for a black hole.

 

[Tail]

Ok, let's say you only know that in one place gravity is x strong and in another gravity is 2x strong. Would you say it as easy to move aways from the second point as it is from the first?

 

[jcsd]

Originally posted by Tail
Ok, let's say you only know that in one place gravity is x strong and in another gravity is 2x strong. Would you say it as easy to move aways from the second point as it is from the first?

Intially it will be easier to move away from x, but overall it would depend on how the graviational potential dropped off over distance.

 

[Tail]

And now let's put it this way: if you can move fast enough to get away from x, but no faster (in both cases the gravitation decreases as you move away, doesn't matter how much exactly), will you be able to get away from 2x?

 

[jcsd]

Originally posted by Tail
And now let's put it this way: if you can move fast enough to get away from x, but no faster (in both cases the gravitation decreases as you move away, doesn't matter how much exactly), will you be able to get away from 2x?

But you see it does matter how graviataion decreases as it will decelrate you so if it drops off quicker you'll suffer from less overall decelration than if it dropped off slower.

 

[Tail]

The deceleration comes afterwards, I don't think it matters.

The point is that light either CAN get out (is fast enough) or CANNOT get out (is not fast enough). If a smaller black hole has more event horizon gravitation, light just won't be able to get away at the horizon!