Black Hole Formation: Does it Start at the Center or Simultaneously?

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Saw somewhere that the force of gravity "cancels out" in that if in the approx. centre of the Earth you would feel no gravity (net zero).

Does a black hole "start" forming from a particular point? (ie where gravity is strongest) Or "start" to form from the centre out? Or does it all just happen simultaneously?

Last would the centre of a black hole have the same net zero gravity area as Earth?
 
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I believe there is not "in" or "out" in a black hole. You see, the reason that you feel less gravity as you move toward the center of the Earth is because gravity is determined by how much mass is "inside a sphere" between you and the Earth's center. As the 'sphere' inside you shrinks, so does the gravitational pull.

Now, with a black hole all the mass is concentrated as a very, very tiny center. So there is no way for you to have some of the mass "outside" of you, because of the way it exists. But someone else will surely clarify this, I hope!
 
schaefera said:
I believe there is not "in" or "out" in a black hole. You see, the reason that you feel less gravity as you move toward the center of the Earth is because gravity is determined by how much mass is "inside a sphere" between you and the Earth's center. As the 'sphere' inside you shrinks, so does the gravitational pull.

Now, with a black hole all the mass is concentrated as a very, very tiny center. So there is no way for you to have some of the mass "outside" of you, because of the way it exists. But someone else will surely clarify this, I hope!

I can help clarify my question.
I get there is no going into a black hole. But it does have a measurable size so Id guess math can figure out the "centre" and math models for the distrabution of gravity.

I guess to ask it differently does the direction of acceleration of Earth gravity all focus to some point in the centre of the earth?
 
Force and Mass are proportional, so, when you are at the center of the Earth, all the mass is (for the sake of ease we'll assume is distributed evenly) evenly around you, which is to say, you are being pulled from every direction equally, which causes the situation where you are experiencing all the acceleration canceling itself out.

Edit: ...you are experiencing... acceleration in every direction equally.
 
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I asked a question here, probably over 15 years ago on entanglement and I appreciated the thoughtful answers I received back then. The intervening years haven't made me any more knowledgeable in physics, so forgive my naïveté ! If a have a piece of paper in an area of high gravity, lets say near a black hole, and I draw a triangle on this paper and 'measure' the angles of the triangle, will they add to 180 degrees? How about if I'm looking at this paper outside of the (reasonable)...
From $$0 = \delta(g^{\alpha\mu}g_{\mu\nu}) = g^{\alpha\mu} \delta g_{\mu\nu} + g_{\mu\nu} \delta g^{\alpha\mu}$$ we have $$g^{\alpha\mu} \delta g_{\mu\nu} = -g_{\mu\nu} \delta g^{\alpha\mu} \,\, . $$ Multiply both sides by ##g_{\alpha\beta}## to get $$\delta g_{\beta\nu} = -g_{\alpha\beta} g_{\mu\nu} \delta g^{\alpha\mu} \qquad(*)$$ (This is Dirac's eq. (26.9) in "GTR".) On the other hand, the variation ##\delta g^{\alpha\mu} = \bar{g}^{\alpha\mu} - g^{\alpha\mu}## should be a tensor...
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