Why Do Smaller Black Holes Have Higher Temperatures?

[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!

 

[jcsd]

Originally posted by 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!

'course the decelration matters 'cos if you get decelrated enough you'll start coming back to your orginal point and you wouldn't of acheived your escape velocity.

Light can't get away at the event horizon it sits there static and infinitely red-shifted.

I've demonstrated it mathematically using only basic physics, what more do you want? blood?

 

[Tail]

Yes, blood sounds good... ok, ok, just kidding!

Ok, let's say light can get out of x gravity. If the gravity is weaker than that afterwards, can it still get pulled back (yes, I know light cannot be "pulled", but you understand...)?

 

[jcsd]

No, because light travels at a constant velocity*


We're starting to get more into GR now than the basic Newtonian physics above.

 

[Tail]

Ok. That's good. Then the deceleration DOESN'T matter after all? (We're talking just about light, because that's how you determine where the event horizon is!)

 

[jcsd]

Originally posted by Tail
Ok. That's good. Then the deceleration DOESN'T matter after all? (We're talking just about light, because that's how you determine where the event horizon is!)

No, because as soona s we start to bring light into the equation with have to start to talk about how gravity warps space-time, and at the event horizon of a smaller hole space-time is more is warped to a greater degree.

 

[Tail]

Well, as it's the same you're once again saying that gravitation is stronger at the event horizon of a small black hole than at the event horizon of a big black hole.

But look at it, if it is so, then light cannot escape the smaller hole at it's event horizon, which is a contraction of terms basically!

Event horizon - the exact place where light CAN escape because the gravitation is weak enough (doesn't hold light in) and not too weak (well, it isn't the event horizon if the gravitation is weaker).
The speed of light - always the same.
If the speed is always the same, shouldn't the gravity it can escape ALSO be always the same?

 

[jcsd]

Okay, bringing light into the matter only complicates things tho' as we have to start looking at GR.


The remote observer co-ordinate speed of light for light moving radially away from a a Scwarzchild black hole is given by:

v_r = ±(1 - 2GM_BH/rc²)c


From the formula from escape velocity we can establish that:

R_BH = 2GM_BH/c²

Now at the event horizon r = R_BH, therefore substituting this in:

V_r = (1 - 2GM_BHc²/2GM_BHc²)c = 0

So the remote observer co-ordiante speed of light at the event horizon is zero.

So at the event horizon to a remote observer the light is hovering on the edge.

 

[Tail]

I agree with your conclusions, but I think we both know what the event horizon is and what light does when it's there. The point is not to define it, but agree about the gravitation at the event horizon.

 

[meteor]

how do YOU explain the different temperatures of black holes?

I've been digging in "A brief history of time" of S. Hawking, and he explains the story this way:
At the event horizon of a black hole a pair particle-antiparticle is created (one with positive energy and the other with negative energy). There are 3 possibilities: the two scape, the two are absorbed or one is absorbed an the other scapes. In this last case, the particle absorbed is always the particle that have negative energy, but this particle will not become a real particle until traveling certain distance inside the event horizon. The needed distance is minor in smaller black holes, so in these the virtual negative particles become real particles sooner than in black holes with greater mass. The virtual positive particle that was hanging about outside the event horizon is not radiated until the negative particle becomes real. So in smaller BH in each second there's a greater number of negative particles that are becoming real (thus major the number of positive particles that are radiated)

 

[Tail]

Wow, I'd read the book, but hadn't noticed that! Well, sounds cool!

Ok, I'll have a lokk at the book again.

I have to go away a bit, I'll be back in 3 days or something, let's carry on then!