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I need a non-mathemathical explanation...

Anyone? Pretty please?

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- Thread starter Tail
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I need a non-mathemathical explanation...

Anyone? Pretty please?

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According to quantum mechanics (and the Heisenburg uncertainty principle) the vacuum of space cannot be a true vacuum; that is, there has to be some energy there to satisfy the HUP. This comes from virtual particles pairs that spontaneously appear, then annihilate with each other almost immediately after (producing energy).

Now, if one of these virtual pairs forms near the event horizon, it is possible for one of the particles to fall into the black hole, while the other can escape, thus preventing their annihilation. The particles that escape radiate outwards -- this is Hawking radiation. To satisfy energy conservation, the "hole" left behind by this particle must be filled -- part of the mass of the black hole is converted to energy. This is why radiating black holes eventually evaporate and disappear.

As far as size, I'm not 100% sure why smaller ones are hotter, but perhaps one way of thinking of it is that the smaller the black hole, the less distance these particles have to travel, and the more likely they are to radiate away (and of course, the more particles that get radiated, the higher the temperature).

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marcus

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Originally posted by Tail

I need a non-mathemathical explanation...

Anyone? Pretty please?

I'll start the ball rolling on this one

there is a possibly confusing sense in which small holes have more extreme or intense surface gravity than large ones

for any BH the actual gees at the event horizon is infinite

but theorists have a finite quantity they CALL the surface gravity which is bigger for smaller holes

and the temp is proportional to that surface gravity parameter

I'll get the formula for it in a moment

anyway maybe this could be intuitive

if the surface grav is more extreme then more of that

Hawking radiation happens per square centimeter

(with gravity pulling one partner in and promoting the other

from virtual to real)

so the more intense radiation means higher temp

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marcus

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my textbook (Frank Shu---The Physical Universe)

defines the surface gravity parameter and proves that the Hawking temp is proportional to it

The intuitive content of his proof is (put very roughly) that

the more extreme the gravity is the more of this Hawking radiation-producing process goes on

notice that with the little BH the spatial gradient of the gravity is steeper-----it would ramp up faster as you approached

for those who happen to like formulas, in natural units the H. temp

is given by the formula

kT = 1/(8pi M) where M is the mass

also kT = (surface grav parameter)/(2pi)

the actual gravitational acceleration as you get within distance R

is given by

(M/R^{2}) divided by sqrt(1 - 2M/R)

The Schw. radius R_{Schw} = 2M

so the sqrt thing you divide by goes to zero as you approach

the event horizon making the fraction go to infinity.

Its the numerator (M/R^{2})

which they call the "surface gravity" and which the temp is proportional to.

Let's calculate the surface gravity at the event horizon (that is, where R = R_{Schw} = 2M

M/R^{2} = M/R_{Schw} ^{2} =

M/(2M)^{2} = 1/(4M)

So surface gravity divided by 2pi is 1/(8piM) which is

the Hawking formula for the temperature (kT) in terms of mass.

The smaller the mass the bigger that number 1/(8piM) gets.

As the hole evaporates it gets progressively hotter etc.

until ("pop") we've all heard the story

defines the surface gravity parameter and proves that the Hawking temp is proportional to it

The intuitive content of his proof is (put very roughly) that

the more extreme the gravity is the more of this Hawking radiation-producing process goes on

notice that with the little BH the spatial gradient of the gravity is steeper-----it would ramp up faster as you approached

for those who happen to like formulas, in natural units the H. temp

is given by the formula

kT = 1/(8pi M) where M is the mass

also kT = (surface grav parameter)/(2pi)

the actual gravitational acceleration as you get within distance R

is given by

(M/R

The Schw. radius R

so the sqrt thing you divide by goes to zero as you approach

the event horizon making the fraction go to infinity.

Its the numerator (M/R

which they call the "surface gravity" and which the temp is proportional to.

Let's calculate the surface gravity at the event horizon (that is, where R = R

M/R

M/(2M)

So surface gravity divided by 2pi is 1/(8piM) which is

the Hawking formula for the temperature (kT) in terms of mass.

The smaller the mass the bigger that number 1/(8piM) gets.

As the hole evaporates it gets progressively hotter etc.

until ("pop") we've all heard the story

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LURCH

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marcus

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Originally posted by LURCH

I say right, to that. It is a different perspective----I was saying the gravity is more extreme near the event horizon of a small BH and you put it more geometrically by saying the space is more radically curved---and it makes sense to me that would breed more virtual particles. Maybe we will get yet another perspective on this. I find Hawking radiation hard to understand so could use whatever other viewpoints on it.

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How comes the surface gravity of smaller black holes is bigger than that of bigger ones?

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marcus

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Originally posted by Tail

How comes the surface gravity of smaller black holes is bigger than that of bigger ones?

Glad you asked

well actually I was wondering if and when you might respond

(disclaimer as usual: cant give an authoritative answer etc etc)

The gravity from ANYTHING has this term M/R

in it----proportional, that is, to the mass of the thing

and falling off as the square of the distance

and the point about the little hole-lets is that you can get in very very close and so even tho M is less

M/R

EXPOSITORY BRAINSTORM: maybe instead of saying that gravity "falls off as the square of the distance"

physicists should say that it

"increases with the square of the closeness"

the event horizon radius which gives an idea of the size of the BH

(also called "Schwarzschild radius") is proportional to the mass

So if you cut the mass in half you can get TWICE AS CLOSE

so that means gravity per unit mass is four times stronger

which more than compensates for having only half as much mass

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I didn'e get it. I mean, aren't the mass and the radius of a black hole directly proportional?

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marcus

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-----------

the event horizon radius which gives an idea of the size of the BH

(also called "Schwarzschild radius") is proportional to the mass

So if you cut the mass in half you can get TWICE AS CLOSE

so that means gravity per unit mass is four times stronger

which more than compensates for having only half as much mass

---------------

In response (I think) to this, you said:

I didn'e get it. I mean, aren't the mass and the radius of a black hole directly proportional?

Indeed they are

so if you reduce the mass by half

you cut the radius down by half

so you are "twice as close" when you are at the boundary

(if you like thinking of one-over-the-distance as the closeness)

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The gravitation just outside a big black hole should be bigger than the gravitation just outside a small black hole, no?

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marcus

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Originally posted by Tail

The gravitation just outside a big black hole should be bigger than the gravitation just outside a small black hole, no?

No. or maybe I should say Why?

gravity depends on nearness

the pull between two massive planets can be less than the pull between two small less-massive planets if the small ones are much closer together

two big things dont necessarily have the strongest attractive force between them, becausse being big they cannot get so close together----their bulk gets in the way

if you had infinitely strong strain-gauges and lab apparatus so you could measure really strong forces, you could make the strongest gravitational force in the universe only using a couple of small black holes-----they could attract each other more than two neutron stars or anything else, more than two galaxies-----because they could get closer together

the force is proportional to the two masses multiplied together and DIVIDED by the square of the separation

(so you can make the force big by making the denominator of the fraction small----making the separation, and thus its square, small)

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The distance to the event horizon is directly proportional to mass of the black hole. Which means that the gravitation right on the horizon is the same for every black hole. Am I wrong?

And it takes longer for the gravitation of a big black hole to lessen over distance than that of a smaller black hole, no?

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LURCH

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Originally posted by Tail

The distance to the event horizon is directly proportional to mass of the black hole. Which means that the gravitation right on the horizon is the same for every black hole. Am I wrong?

And it takes longer for the gravitation of a big black hole to lessen over distance than that of a smaller black hole, no?

This one has always had me spun too, Tail. But the big idea here is that if you cut the mass in half, you can get twice as close (half the mass = half the Schwartzchild radius) and getting twice as close means FOUR TIMES the gravitational pull. That's becuase as you approach, the pull of gravity grows exponentially. So one-half the distance means four times the pull, one-third the distance means nine times the pull, etc.

This brings up the very counterintuitive situation once mentioned in the Astronomy Q&A game; appearently, a black hole of about 1 billion SM wuold have a gravitational pull at its event horizon of one G. I'm still having trouble reconciling this with my own preconceptions about what an event horizon is.

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Hurkyl

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The earth has a mass of 5.97*10^24 kg and a radius of 6.38*10^6 m. If we compute the surface gravity as Gm/r^2, we get 9.78 m/s^2 (roundoff error. boo!)

Uranus has a mass of 8.66*10^25 kg and a radius of 2.56*10^7 m, making it over 10 times heaver and twice as big as earth. However, when we compute the surface gravity, we get 8.81 m/s^2

!

If you look at the equation for g:

g = Gm/r^2

you see that if you simultaneously double the mass and the radius, g gets cut in half. So larger black holes have less "surface" gravity. (Of course, Newton's formula is only an approximation, but it demonstrates the idea)

One thing that might help understand this phenomenon is if one considers the

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marcus

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Originally posted by LURCH

This brings up the very counterintuitive situation once mentioned in the Astronomy Q&A game; appearently, a black hole of about 1 billion SM wuold have a gravitational pull at its event horizon of one G. I'm still having trouble reconciling this with my own preconceptions about what an event horizon is.

Lurch, I was wrong when I said that back in the Q&A game! The actual acceleration of gravity at the event horizon of any BH is infinite.

But astrophysicists have a jargon term "surface gravity" for a useful parameter describing the black hole.

In fact you can have a black hole massive enough that its "surface gravity" is one gee.

But the actual acceleration due to gravity as you approach the event horizon is equal to "surface gravity" divided by a term that goes to zero! So the actual gravity goes to infinity as you approach the horizon.

The surface gravity parameter = GM/R

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marcus

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Originally posted by marcus

...the actual acceleration due to gravity as you approach the event horizon is equal to "surface gravity" divided by a term that goes to zero! So the actual gravity goes to infinity as you approach the horizon.

The surface gravity parameter = GM/R_{Schw}^{2}

One handy formula is for the temperature, which is

proportional to the "surface gravity" parameter. Infinity is not

very useful for calculation purposes, so it makes sense to have

a finite parameter as a handle on gravity at the event horizon

The actual grav. accel as you approach the surface is

(GM/r

As r --> R

the denominator goes to sqrt( 1 - 1) = 0

and the numerator goes to GM/R

so the fraction goes to "surface gravity"/0 = infinity

sorry about the confusion, at one point a while ago I was confused by the terminology in something I was reading about this

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jcsd

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marcus

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Originally posted by jcsd

This seems like a good kind of calculation to make. So I will imitate you and do this for a trillion solar mass galaxy

(I'm not sure about mass of our galaxy but trillion seems in ballpark)

I believe the Schw. radius for a solarmass black hole is 3 kilometer

So the radius for a trillion solarmass hole is 3 trillion km.

that seems to be one solar mass per E26 cubic km.

2E30 kilograms per E26 cubic km

20, 000 kg per cubic km.

that is 20 billionths of the density of water

we differ by a factor of 50, which could be accounted for

by different assumptions, esp about mass of galaxy.

I guess the density goes as the inverse square of the mass of the black hole. You may have assume a galaxy some 7 times more massive which is hardly any different for rough calculation purposes

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jcsd

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marcus

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Originally posted by Tail

And it takes longer for the gravitation of a big black hole to lessen over distance than that of a smaller black hole, no?

Tail you are saying that gravity ramps up more steeply as you approch a small BH, yes?

Or falls off more sharply as you go away from a small one.

I think that is so. Maybe Jcsd can confirm it, if someone hasnt already.

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Hurkyl

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But the actual acceleration due to gravity as you approach the event horizon is equal to "surface gravity" divided by a term that goes to zero! So the actual gravity goes to infinity as you approach the horizon.

In what coordinate chart?

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marcus

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Originally posted by marcus

The actual grav. accel as you approach the surface is

(GM/r^{2})/sqrt(1 - R_{Schw}/r)

As r --> R_{Schw}

the denominator goes to sqrt( 1 - 1) = 0

and the numerator goes to GM/R_{Schw}^{2}= surface gravity parameter

so the fraction goes to "surface gravity"/0 = infinity

Hi Hurkyl,

I copied this from page 138 of Frank Shu's textbook and, I should have explained, he was using a special position parameter r

which is just given by the Euclidean formula for the circumference---C = 2 pi r

In other words, to tell how far from the center of the BH you are in "r" terms, You measure the circumference of a circle around the BH at your distance from it, and then divide by 2pi = 6.28

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Hurkyl

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How does this

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