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## Why the black hole density is called the infinite density.

 Quote by alexg Peter, a minor point, almost a diversion, but doesn't the Shell Theorem state that the gravity inside the earth remains the same until you reach the center? While there is more mass above you, you are closer to the rest of the mass on the other side, and it balances out.
The shell theorem (that you are thinking of) only applies when you are in the cavity of a hollow shell and in that case it states there is no net gravitational acceleration anywhere inside the cavity due to the cancelling effects. (There is however a lower gravitational potential inside the cavity than outside.)

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 Quote by alexg Peter, a minor point, almost a diversion, but doesn't the Shell Theorem state that the gravity inside the earth remains the same until you reach the center? While there is more mass above you, you are closer to the rest of the mass on the other side, and it balances out.
No. I assume you mean the Shell Theorem as stated, for example, on this Wiki page:

http://en.wikipedia.org/wiki/Shell_theorem

The theorem basically says two things:

(1) If you're on the outside of (i.e., above, further from the center than) a spherically symmetric mass distribution, you feel the gravity of the entire mass as if it were concentrated at the center of the sphere.

(2) If you're on the inside of (i.e., below, closer to the center than) a spherically symmetric mass distribution, you feel *no* net gravity from it, because the contributions from all different parts of it cancel out.

If you're in the interior of an idealized spherically symmetric body (like an ideal non-rotating spherical Earth), both (1) and (2) above apply. (2) applies to the part of the body that's above you, and says that that part contributes nothing to the gravity you feel. (1) applies to the part of the body that's below you, and says that you feel the gravity of that part normally, i.e., just as if the mass that's below you were concentrated at the center.

As you descend through the body, more and more of its mass is above you (and so doesn't contribute to the gravity you feel), and less and less is below you. So the gravity you feel gets less and less, reaching zero just as you reach the center.

 Quote by PeterDonis No. I assume you mean the Shell Theorem as stated, for example, on this Wiki page: http://en.wikipedia.org/wiki/Shell_theorem The theorem basically says two things: (1) If you're on the outside of (i.e., above, further from the center than) a spherically symmetric mass distribution, you feel the gravity of the entire mass as if it were concentrated at the center of the sphere. (2) If you're on the inside of (i.e., below, closer to the center than) a spherically symmetric mass distribution, you feel *no* net gravity from it, because the contributions from all different parts of it cancel out. If you're in the interior of an idealized spherically symmetric body (like an ideal non-rotating spherical Earth), both (1) and (2) above apply. (2) applies to the part of the body that's above you, and says that that part contributes nothing to the gravity you feel. (1) applies to the part of the body that's below you, and says that you feel the gravity of that part normally, i.e., just as if the mass that's below you were concentrated at the center. As you descend through the body, more and more of its mass is above you (and so doesn't contribute to the gravity you feel), and less and less is below you. So the gravity you feel gets less and less, reaching zero just as you reach the center.
Does density have a significant effect???
What is the difference between a solid spherical body and a uniformly distributed spherical particle cloud ,which of course can have a fairly dense distribution if we want???

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 Quote by Austin0 Does density have a significant effect??? What is the difference between a solid spherical body and a uniformly distributed spherical particle cloud ,which of course can have a fairly dense distribution if we want???
Density doesn't matter, nor does the exact state of matter in the body (solid, liquid, gas, dust, etc.), as long as the mass distribution is exactly spherically symmetric. That means any variable, such as density, can only be a function of radius (i.e., distance from the center). That's the key condition that leads to the conclusions of the theorem.

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 Quote by Austin0 Does density have a significant effect??? What is the difference between a solid spherical body and a uniformly distributed spherical particle cloud ,which of course can have a fairly dense distribution if we want???
If we have a low density body and a high density body with equal masses, then the effects of gravity on an external test particle are identical (in Newtonian terms). Both behave as if they are a point particle containing all the mass. The lower density body will have a larger radius. For two bodies with equal radius and differing densities, the lower density body will of course have a lower lower mass and therefore less of a gravitational effect.

In General Relativity, if we have a solid body in equilibrium such that there are stresses and pressures within the body, then its external gravitational effects will be different from a lower density particle cloud that is not in equilibrium even if it has the same total mass. The stresses and pressures add to the effective total mass of the solid body and the motion of the particles in the particle cloud will also have an effect.

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 Quote by Passionflower Well even that question is not simple. The class of observers that are stationary in a Schwarzschild spacetime can certainly chart the event horizon as a sphere with a certain distance removed from them but not a sphere with a distance removed from a center, because there is no center.
Why do we need a centre? We could have a spherical shell around a vacuum with no clearly defined centre and still define the sphere in terms of its diameter or its circumference divided by 2pi or in terms of its surface area (A) as r = sqrt(A/4pi) or in terms of its volume (V) as r = (3V/(4pi))^(1/3) etc.

Lets say we added mass to the Sun until it collapsed into a black hole and continued to add mass until it engulfed Mercury, Earth, Mars etc until only Pluto remained orbiting it. Would you agree that the black hole occupied a volume of space? Would you agree if occupies a volume then we can in some sense say it has a radius? Would you agree that to say it has no radius is to suggest that all points on the event horizon are at the same spatial point? Would you agree that the finite proper time taken by free falling observer to fall from the event horizon to the singularity contradicts the idea of their being volume or radius? Would you agree that no radius implies that the singularity and the event horizon are at the same spatial location?

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 Quote by yuiop In General Relativity, if we have a solid body in equilibrium such that there are stresses and pressures within the body, then its external gravitational effects will be different from a lower density particle cloud that is not in equilibrium even if it has the same total mass.
Careful! This is not correct as you state it. The "total mass" is one of the "external gravitational effects", so it *is* affected by the things you mention. I think you realize this, since you next say:

 Quote by yuiop The stresses and pressures add to the effective total mass of the solid body and the motion of the particles in the particle cloud will also have an effect.
"Effective total mass" and "total mass" are really the same thing. I think what you are getting at is that the same "amount of stuff" can have a different "effective total mass", but you have to be careful how you characterize "amount of stuff" if you want to make it distinct from "total mass". For example, you could say "the same total number of atoms can have a different total mass, depending on its configuration"--"total number of atoms" is a more precise way of defining "amount of stuff". (MTW uses "total number of baryons" in the same way.)

I should note that there is a lot more lurking in the details of how "total mass" is calculated, as shown, for example, in this thread: