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Derivation of Schwarzschild radius from escape velocity

  1. Dec 3, 2013 #1
    Hi,

    Is it pure coincidence that if you put ##c=v_e=\sqrt{2GM/R}## in the escape velocity, you end up with the Schwarzschild radius ##R=2GM/c^2##?

    The derivation of the escape velocity is purely classical mechanics. It involves ##E_{kin}=mv^2/2## which is incorrect in special relativity even for massive particles and is entirely useless for massless photons.
    One could argue that with the constants ##G,M,c## at hand the form of the Schwarzschild radius follows from dimensional analysis, but that doesn't explain why the factor ##2## is present in both derivations.
     
  2. jcsd
  3. Dec 3, 2013 #2

    PAllen

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    Newtonian escape velocity can be derived without use energy.

    That Newtonian escape velocity = c matches SC radius is generally considered a coincidence. Note that in Newtonian mechanics, there is nothing special about c, and nothing preventing a body having escape velocity > c, nor preventing projectiles with v>c that can escape.
     
  4. Dec 4, 2013 #3

    dvf

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