Why Do Smaller Black Holes Have Higher Temperatures?

[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!

 

[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.