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A Why is all the mass of a BH in the singularity?

  1. Dec 22, 2017 #1
    Hi everybody, and thanks in advance. My doubt is: why should the mass go in the singularity? I'm thinking about this situation: imagine a sphere with radius R<2M; then that sphere generates a BH. Schwarzschild tensor metric is valid only in the exterior region of the mass and for this reason i can't use it to "achieve" the singularity and to explain how the matter behaves there... So how can I say that the matter of the sphere falls inside the BH to the singularity? And why? Thanks.
     
  2. jcsd
  3. Dec 22, 2017 #2

    Dale

    Staff: Mentor

    Basically, it is because we don’t know of anything strong enough to stop it. The highest limit we know, the neutron degeneracy pressure, is still insufficient.
     
  4. Dec 22, 2017 #3

    Grinkle

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    @Dale At a "B" level, is the OP's question equivalent to asking why we model such a thing as a closed region of space-time? Once one contemplates such a model, the reason all particles have the singularity in their future is because that is that nature of the model. If there is some force that can change this, would that imply the region isn't closed after all?
     
  5. Dec 22, 2017 #4
    Ok, but "mathematically" speaking, how can i say it? Which equation tells me "the mass has to fall down to the singularity?" if I can't use the Schwarzschild metric inside the mass itself? Is it a supposition?
     
  6. Dec 22, 2017 #5
    Ok, but the model is true only in an exterior region of the mass. How can i give prediction inside the mass? The singularity is inside the sphere...
     
  7. Dec 22, 2017 #6

    Grinkle

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    Is the below what you are asking?

    From:

    http://www.jimhaldenwang.com/black_hole.htm

    I put a snip -


    Inside the Black Hole


    Now let's consider the Schwarzschild solution for 0 < r < 2M (inside a black hole). A small but very important change must be made to the metric for this case. When r > 2M, the coefficient (1 − 2M/r) is positive. However, for 0 < r < 2M, this coefficient is negative. In order to work with positive coefficients for this case, we use

    eqn16g.gif

    The metric then becomes

    eqn17.gif

    Notice how the minus sign has moved from the t coordinate to the r coordinate. This means that inside the event horizon, r is the timelike coordinate, not t. In relativity, the paths of material particles are restricted to timelike world lines. Recall the discussion of timelike separation earlier in this paper (2). It is the coordinate with the minus sign that determines the meaning of "timelike." According to relativity, inside a black hole time is defined by the r coordinate, not the t coordinate. It follows that the inevitability of moving forward in time becomes, inside a black hole, the inevitability of moving toward r = 0. This swapping of space and time occurs at r = 2M. Thus, r = 2M marks a boundary, the point where space and time change roles. For the observer inside this boundary, the inevitability of moving forward in time means that he must always move inward toward the center of the black hole at r = 0.
     
  8. Dec 22, 2017 #7

    Dale

    Staff: Mentor

    The Schwarzschild metric is the unique metric for a spherically symmetric vacuum spacetime. However, you can also calculate non vacuum spherically symmetric spacetimes. When you do so, you also include an equation of state which describes the stress energy tensor of the matter. At a certain point, it can no longer be static for any material.
     
    Last edited: Dec 22, 2017
  9. Dec 22, 2017 #8

    PAllen

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    The key is actually Buchdahl’s theorem, which, without assuming any theory of matter, establishes that once the radius of a body falls to 9/8 the Schwarzschild radius, the central pressure becomes infinite, unless some matter inside is moving on spacelike trajectories (i.e. locally FTL).
     
  10. Dec 22, 2017 #9

    PAllen

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    I should add that Buchdahl’s result assumes spherical symmetry. A more general type of result are the Penrose Hawking singularity theorems, which assume nothing about symmetry, and establish that once matter is inside the horizon, some type of singularity must form. They do not require that all the matter ends up in the singularity, and for realistic collapses, it is an open question what happens, even classically. The singularity theorems also make use of energy conditions which are true of all fundamental classical laws considered to apply to matter, but are not strictly true for quantum theories.

    So, in the real world, we have no idea what happens in the interior of a BH.
     
  11. Dec 22, 2017 #10
    Ok perfect, thank you everybody!
     
  12. Dec 23, 2017 #11
    Sorry if I came back here, but I still have a little doubt: what we can say about Kerr black-hole? The space-time it's only axisymmetric and so is there any method to demonstrate that matter would collapse in the singularity (i'm always considering the destiny of the mass that generates the black hole)? Thank you again.
     
  13. Dec 23, 2017 #12

    PAllen

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    The Kerr interior is not stable against small perturbations, and even classically, it is unknown what a plausible interior would be like for a collapse with significant angular momentum. However, for the fun of it, a Kerr interior allows stable orbits between the inner and outer horizons, suggesting that not all matter need reach the singularity.

    The general singularity theorems only state that some type of singularity forms, not that all matter reaches it, nor that all interior geodesics end on it.
     
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