Why Do Smaller Black Holes Have Higher Temperatures?

[Tail]

Ok, I'm back.

I'm sorry for being so very, um, sure I was right, jcsd. Looks like I wasn't. It seems that the gravity at event horizons of different black holes does indeed differ. I'll be damned if I understand why, though, so if somebody can help me...

So... the gravitation is stronger therefore there are more particle pairs forming at the event horizon? Or is it that they are just more easily turned to real particles?



meteor, where exactly in the book did you find that?

 

[meteor]

Hi Tail
I was using a online version of Hawking's book

www.thegeekgirl.net/Library/science/[/URL]
You can find it in chapter 7
[quote]
I'll be damned if I understand why, though, so if somebody can help me...So... the gravitation is stronger therefore there are more particle pairs forming at the event horizon? Or is it that they are just more easily turned to real particles?

[/quote]
Just use Newton's formula
F= MmG/(r^2)

The formula for the Schwarzschild radius (schwarzschild radius=event horizon) is=
r=2GM/(c^2)
Then substitute in the anterior equation and simplify:
F=((c^4)*m)/(4*G*M)
F is the force at the event horizon and as you can see, is inversely proportional to the mass of the BH

[quote]
So... the gravitation is stronger therefore there are more particle pairs forming at the event horizon? Or is it that they are just more easily turned to real particles?
[/quote]

The number of virtual particles forming in any volume of space is the same in all the universe.But the virtual particle that falls in the BH has to travel less distance in small black holes to become real

 

[Tail]

Originally posted by meteor
Hi Tail
I was using a online version of Hawking's book

Can you quote the paragraph or tell me after which "figure" that part is? I've got a blind spot, it seems...

Just use Newton's formula
F= MmG/(r^2)

The formula for the Schwarzschild radius (schwarzschild radius=event horizon) is=
r=2GM/(c^2)
Then substitute in the anterior equation and simplify:
F=((c^4)*m)/(4*G*M)
F is the force at the event horizon and as you can see, is inversely proportional to the mass of the BH

See? That's why I hate mathematics. One can have a formula, but not an explanation... no answer to the question "Why?"...

The number of virtual particles forming in any volume of space is the same in all the universe.But the virtual particle that falls in the BH has to travel less distance in small black holes to become real

Interesting... I'd love to see that quote!

 

[LURCH]

Originally posted by Tail
Ok, I'm back.

I'm sorry for being so very, um, sure I was right, jcsd. Looks like I wasn't. It seems that the gravity at event horizons of different black holes does indeed differ. I'll be damned if I understand why, though, so if somebody can help me...

I'm afraid that's where I get lost, Tail. I've seen the mathematical proofs that it's true, but this doea not resolve the problem. Perhaps if I present the paradox in the following manner, someone could resolve it. The paradox consists of the following three points;

1) A black hole (given enough mass) can have an event horizon where the gravitational pull towards the center of just 1G.

2) I can move outward from a gravitational field of 1G.

3) Nothing can move outward once inside the event hrizon of a black hole.

All three of these can be mathematically proven. All three cannot be true. Something is definitely wrong, here.

 

[jcsd]

1) yes, you could

2) No you couldn't because you'd find the energy needed would be infinite.

3) yes.

 

[Tail]

I think LURCH's point is that the Earth has the gravity of one G.

Hmm?

 

[Tail]

Also, given enough mass, a black should have to be able to have a surface gravity of 0.000000001 G, no? Wouldn't I be able to escape that, especially if I had a good spaceship?

 

[Tail]

Wouldn't light be able to get away from a point the gravity of which is just 1G...?

 

[meteor]

Interesting... I'd love to see that quote!

It's in the middle of the page. Just look below Fig 7.4

To an observer at a distance, it will appear to have been emitted from the black hole. The smaller the black hole, the shorter the distance the particle with negative energy will have to go before it becomes a real particle, and thus the greater the rate of emission, and the apparent temperature, of the black hole.

Tell me what part of the mathematical proof you don't understand

 

[jcsd]

Tail, we're talking about an escape velocity that's the speed of light, so no you couldn't.

 

[LURCH]

Originally posted by Tail
I think LURCH's point is that the Earth has the gravity of one G.

Hmm?

Pricessly. Every time I rise from a seated position, I move away from a center of gravity against a resistance of 1G. Same goes for climbing a ladder, walking up a staircase, etc. If the foot of the staircase were inside the event horizon, what force would prevent me from walking up it? Sorry to distract from the original Topic like this, but this is really bugging me.

Speaking of the original topic:

To an observer at a distance, it will appear to have been emitted from the black hole. The smaller the black hole, the shorter the distance the particle with negative energy will have to go before it becomes a real particle, and thus the greater the rate of emission, and the apparent temperature, of the black hole.

This I get. This might be a clearer explanation; pairs of virtual particles form at the EH of two black holes of different sizes. In each pair, one travels inward, the other out. We will focus our attention on the "inside" virtual particle from each pair. Each of these travels the same distance inward. Because the gravitational gradiant for a smaller black whole is much steeper, the particle traveling into the smaller black hole experiences a much greater increase in gravity's effect. Therefore it is more likely to get "trapped".

BTW, does this not also work in reverse? Can it be said that, in the same way, the "outside" member of the pair (the pair near the smaller black hole) experiences a greater drop in gravity's pull for the same distance travelled? And therefore it is more likely to escape. Both are saying the same thing, if I understand the concept as well as I think I do.

 

[Tail]

Thanks, meteor, I finally found the place!

Originally posted by LURCH
This I get. This might be a clearer explanation; pairs of virtual particles form at the EH of two black holes of different sizes. In each pair, one travels inward, the other out. We will focus our attention on the "inside" virtual particle from each pair. Each of these travels the same distance inward. Because the gravitational gradiant for a smaller black whole is much steeper, the particle traveling into the smaller black hole experiences a much greater increase in gravity's effect. Therefore it is more likely to get "trapped".

I think that when it's in, it's in. No way out. I agree with everything you said except for the very last sentence, which is the conclusion. I think it's already trapped, and that the stronger gravity is important because the gravity is what makes the virtual particle into a real one. In a smaller black hole, the particle has to travel a shorter distance to reach gravity strong enough to make it into a real one.

 

[Hurkyl]

If you're inside the event horizon, then no matter how fast you travel (even if you travel the speed of light), the event horizon is always moving away from you.