The temperature of black holes

In summary, black holes have a temperature because of emitted radiation, which follows a blackbody spectrum. This radiation is quantum mechanical in nature and is caused by virtual particles pairs that spontaneously appear near the event horizon and get separated, with one falling into the black hole and the other escaping. This process, known as Hawking radiation, causes the black hole to radiate energy and eventually evaporate. Smaller black holes have a higher temperature because their event horizons have a sharper curvature, leading to a higher production of virtual particles and thus more intense radiation. The surface gravity of a black hole is also larger for smaller black holes, contributing to their higher temperature.
  • #71
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?
 
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  • #72
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
 
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  • #73
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!
 
  • #74
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.
 
  • #75
1) yes, you could

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

3) yes.
 
  • #76
I think LURCH's point is that the Earth has the gravity of one G.

Hmm?
 
  • #77
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?
 
  • #78
Wouldn't light be able to get away from a point the gravity of which is just 1G...?
 
  • #79
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
 
  • #80
Tail, we're talking about an escape velocity that's the speed of light, so no you couldn't.
 
  • #81
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.
 
  • #82
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.
 
  • #83
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.
 
<h2>1. What is the temperature of a black hole?</h2><p>The temperature of a black hole is determined by its size and the amount of matter it has consumed. Smaller black holes have a higher temperature than larger ones. The temperature of a black hole is inversely proportional to its mass, meaning that as a black hole grows in size, its temperature decreases.</p><h2>2. How is the temperature of a black hole measured?</h2><p>The temperature of a black hole is measured using a formula known as the Hawking temperature. This formula takes into account the mass and surface area of the black hole. It is a theoretical temperature, as black holes do not emit thermal radiation that can be directly measured.</p><h2>3. Can the temperature of a black hole change?</h2><p>Yes, the temperature of a black hole can change over time. As a black hole consumes more matter, its temperature decreases. However, if a black hole stops consuming matter, its temperature will remain constant. Additionally, as a black hole evaporates, its temperature will increase.</p><h2>4. How does the temperature of a black hole relate to its event horizon?</h2><p>The temperature of a black hole is directly related to its event horizon, which is the point of no return for anything that gets too close to the black hole. The event horizon is the boundary where the escape velocity is equal to the speed of light. As the temperature of a black hole increases, its event horizon also expands.</p><h2>5. What is the significance of the temperature of a black hole?</h2><p>The temperature of a black hole is significant because it is a key factor in understanding the behavior and evolution of black holes. It is also an important concept in the study of quantum mechanics and the Hawking radiation that is emitted by black holes. Additionally, the temperature of a black hole can provide insight into the mass and size of the black hole, as well as its surroundings.</p>

1. What is the temperature of a black hole?

The temperature of a black hole is determined by its size and the amount of matter it has consumed. Smaller black holes have a higher temperature than larger ones. The temperature of a black hole is inversely proportional to its mass, meaning that as a black hole grows in size, its temperature decreases.

2. How is the temperature of a black hole measured?

The temperature of a black hole is measured using a formula known as the Hawking temperature. This formula takes into account the mass and surface area of the black hole. It is a theoretical temperature, as black holes do not emit thermal radiation that can be directly measured.

3. Can the temperature of a black hole change?

Yes, the temperature of a black hole can change over time. As a black hole consumes more matter, its temperature decreases. However, if a black hole stops consuming matter, its temperature will remain constant. Additionally, as a black hole evaporates, its temperature will increase.

4. How does the temperature of a black hole relate to its event horizon?

The temperature of a black hole is directly related to its event horizon, which is the point of no return for anything that gets too close to the black hole. The event horizon is the boundary where the escape velocity is equal to the speed of light. As the temperature of a black hole increases, its event horizon also expands.

5. What is the significance of the temperature of a black hole?

The temperature of a black hole is significant because it is a key factor in understanding the behavior and evolution of black holes. It is also an important concept in the study of quantum mechanics and the Hawking radiation that is emitted by black holes. Additionally, the temperature of a black hole can provide insight into the mass and size of the black hole, as well as its surroundings.

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