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Mass vs Density of Black Holes

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#1 Ira

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Posted 18 April 2013 - 06:02 AM

I have a simple black hole confusion, but it's driving me crazy. When a black hole forms, matter is compressed into a singularity with an event horizon. But the AMOUNT of matter in the original object doesn't change,just its density. So why does it become a black hole at all? Its gravitational pull should remain exactly the same since gravitational force is determined by mass alone, not density:
HELP!

/Ira

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#2 llanitedave

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Posted 18 April 2013 - 09:48 AM

The gravitational pull DOES remain the same per unit distance. The tremendous gravitational effects come to play at small values of r, near the event horizon.

#3 EJN

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Posted 18 April 2013 - 10:21 AM

First off, that is the Newtonian equation in your post which says nothing
about spacetime curvature, for a black hole the simplest solution is the
Schwarzschild metric in general relativity. And even in the Newtonian
equation, as r^2 approaches 0, F approaches infinity. In Newtonian
gravity, you could have a massive object with a sufficiently small radius
so that the escape velocity at the surface is greater than c. These were
called "dark stars," but are *not* black holes because they have no event
horizon. Newtonian gravitation assumes absolute space and absolute time,
so the concept of a black hole requires general relativity.

Second, in general relativity the stress-energy tensor contains components
which correspond to the density & pressure of an object - and eventually
pressure will become a major factor in gravitational collapse because it contributes
to spacetime curvature. For an idealized frictionless fluid, the stress-energy
tensor would have the components:
| p 0 0 0 |
| 0 p 0 0 |
| 0 0 p 0 |
| 0 0 0 p |

where p = mass energy density,
and p are the x, y, & z components of the pressure vector.

#4 Ira

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Posted 20 April 2013 - 06:46 PM

So, how far away from the center of a black hole is the event horizon, anyway? Is there a formula for locating the event horizon?

/Ira

#5 EJN

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Posted 20 April 2013 - 07:25 PM

Yes,

r = 2Gm/c^2

where
r = radius of event horizon
G = gravitational constant (the same G in Newton's equation)
m = mass
c = speed of light

#6 Ira

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Posted 20 April 2013 - 07:30 PM

Thanks. Can you help with some intuitions about this?

How far from the center is a 10x solar mass black hole's event horizon?
1Mx solar mass black hole?

/Ira

#7 llanitedave

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Posted 20 April 2013 - 09:57 PM

Here's a hint:

c is in meters/second
m is in kilograms
G is 6.67384 × 10^-11 m^3 kg^-1 sec^-2

All those units should cancel each other out leaving you with meters for r

The mass of the Sun is 1.989 x 10^30 kg

#8 Ira

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Posted 21 April 2013 - 10:14 AM

Well, I still don't know the answer, but apparently I can buy an event horizon on eBay:

"Results for Radius of black hole event horizon?
www.ebay.com
Event Horizon On EBay
Great Deals on Event Horizon. eBay! Buy it new. Buy it now.™ "

/Ira

#9 Ira

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Posted 21 April 2013 - 05:56 PM

I found this nice rule of thumb for calculating a black hole's event horizon:

"A quick way to calculate the event horizon for a non-spinning black hole would be to increase the radius of the event horizon by 2.7 kilometers for every solar mass (the mass of our sun) that is contained within the black hole. A black hole the mass of the Earth would have an event horizon about .81 centimeters across."

/Ira

#10 Pess

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Posted 22 April 2013 - 04:09 PM

In a simple, non-mathmatical, explanation you need a certain amount of 'pull' against an escaping photon so that the gravitational attraction is as great as the escape velocity of the photon.

The way our sun is now, as we approach its center of mass we get within its circumference before the gravitational pull exceeds the escape velocity of light. As we enter the sun there is mass 'behind' us also pulling so we never approach a spot where gravity has enough pull to keep light in.

Compressing the mass of the sun down now puts ore and more mass in a concentrated area so we can ride our photon much closer and still have the entire mass gravitationally pulling on us.

At some point the mass is all in front of us, and we are close enough, so that the pull exceeds the escape velocity of our photon. So a trapped photon is bent into a circular orbit around the BH mass center. Since no photon can escape this grip of mass it would appear 'black' at the event horizon. Hence the term 'Black Hole'

The original mass is all there and unchanged, just compacted. In fact you can grow a Black Hole by dumping mass into it. So the mass doesn't 'go' anywhere..no pouring out some White Hole on the other side of the galaxy etc.

Pesse (Probably just one giant sub-quarkian pile in the center.) Mist

#11 Qwickdraw

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Posted 23 April 2013 - 04:57 AM

Reading something here made me ask the question that seems applicable to at least large massive bodies like the Earth, sun, etc. I believe the gravitational effects in the center of a spherical mass cancel each other out due to the fact that equal gravity surrounds the center. Would this also apply to a black hole?

#12 Carl Coker

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Posted 23 April 2013 - 10:45 AM

Reading something here made me ask the question that seems applicable to at least large massive bodies like the Earth, sun, etc. I believe the gravitational effects in the center of a spherical mass cancel each other out due to the fact that equal gravity surrounds the center. Would this also apply to a black hole?


No, because all the mass in a black hole is concentrated in the singularity at its center, we think. No known force can provide enough pressure to withstand gravity once you compress something within its Schwarschild radius, and the collapse should continue down to a point. Whatever turns out to be the correct theory of quantum gravity will presumably halt the collapse before it reaches an actual mathematical point, but the resulting object will still be very small compared to the size of the event horizon. Nothing at all special happens to you once you cross the event horizon of a black hole. See here for videos simulating what happens when you fall into one.

Also, if you are anywhere (not just at the center) within a uniform spherical shell of matter, the gravitational force on you is zero. In a solid body, there is still mass below you, and you feel a pull towards the center.

#13 Pess

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Posted 23 April 2013 - 01:18 PM

I remember a thought experiement of what would happen if a minature Black Hole collided with the Earth.

The Black Hole would essentially 'reverberate' back and forth until it settled down at the exact center and starting knawing away at the Earth material around it.

Of course, Steven hawking told us that such minature BH's would not exist very long because they would evaporate very quickly.

We know (or surmise) that virtual particle pairs pop in and then almost instantly annilate each other and pop back out of existence. But if these pairs pop next to an event horizon it is possible one of the pair would be captured before they annilate each other.

The remaining one of the pair then shoots off into the Universe as a 'real' particle and carries a bit of the BH's energy with it and thus decreases the mass of the BH by a tiny amount.

Pesse (So always keep your BH covered least it dissappear) Mist






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