Schwarzschild radius of the inner part

Discussion in 'Astronomy, Exobiology, & Cosmology' started by RajeshTrivedi, Aug 16, 2017.

  1. RajeshTrivedi Valued Senior Member

    This is simple maths, but is quite nagging and indicates certain contradictions with GR.

    The Schwarzschild radius of a core Rs = 2GM/c2.

    1. assume that a star core is just at its schwarzschild radius r = Rs, that is at this moment Event Horizon is a surface.
    2. We can make one more safe assumption, that is the core is of uniform density sphere.

    Now consider the inner concentric spheres of this core, that is spheres of 0.9r, 0.8r, 0.7r, 0.6 r (pr where p is any number between 0 to 1 that is 0 < p < 1). The schwarzschild radius of any inner part sphere, say of radius pr, at this moment (when r = Rs) is p^3*r. That is Schwarzschild radius of a sphere of 0.9r radius is actually 0.9 * 0.9 * 0.9 r = 0.729r.

    So for a core of size 0.9 r the schwarzschild radius is 0.729 r, that means none of the inner part spheres (even 0.99999999r) is inside its schwarzschild radius as p^3*r < p*r.

    Now the crux, the GR says one inside the EH, all the paths lead to r = 0, that means no particle of photon can face away or move away towards the surface. May be right as per maths of GR, but physical reality could be different. Let us say a photon is emitted at 0.5 r, the schwarzschild radius of 0.5r sphere (mass inside) at this point is 0.125r, so escape velocity at this point is not c, it is less, so theoretically a photon or even a particle can move towards the circumference even inside the EH of the original star, this contradicting GR.

    Any comments? Any fallacy in this argument?
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  3. mathman Valued Senior Member

    Assumption of uniform density is questionable. However in your scenario, the photon can move around inside the sphere, but not escape. The assumption all paths lead to r=0 assumes all mass in concentrated at the center, not uniformly distributed.
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