Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Help with showing the existance of the Schwarzschild singularity at r=0.

  1. Feb 28, 2008 #1

    blp

    User Avatar

    According to Wikipedia, if the metric was vacuum, spherically symmetric and static the Schwarzschild metric may be written in the form:
    ds^2=((1-2GM/(c^2r))^-1)*dr^2+r^2*(dtheta^2+sin(theta)^2*dphi^2)-c^2*(1-2GM/(c^2r))*dt^2
    I need someone to help me to derive an expression from the Schwarzschild metric that is a function of r that can be graphed, that clearly shows the singularity at r=0. Actually, their is apparently a pair of pos and neg singularities there. Is that true? Is that because there is a square root taken during it's derivation? Thanks.
     
  2. jcsd
  3. Feb 28, 2008 #2

    Mentz114

    User Avatar
    Gold Member

    Surely

    [tex]1- \frac{2GM}{c^2r}[/tex] clearly diverges when r = 0. And its inverse diverges when

    [tex]1 = \frac{2GM}{c^2r}[/tex].
     
  4. Feb 29, 2008 #3

    blp

    User Avatar

    Sorry, I should have been clearer. Obviously ds^2 diverges at r=0, but I want some way of graphing a 2D spacetime clearly curving towards infinity in the z axis direction as r approaches 0.

    Also concerning the pair of pos and neg singularities at r=0. I remember reading somewhere that shortly after Schwarzschild came up with the black hole solution to GR that Einstein and Rosen noticed the above pair of singularities and interpreted them as being the two ends of a wormhole and called it a Einstein-Rosen bridge. If that is true, I'm curious how the singularities were mathematically derived. Does any one know?
     
  5. Feb 29, 2008 #4

    Mentz114

    User Avatar
    Gold Member

    I think space-times are analysed with Kruskal or Penrose diagrams. If you want to take an equatorial slice, then rewrite the metric in cartesian coordinates. I can't say any more than that, not having studied the singularites because I don't think they can be physical. But that's just an opinion.
     
  6. Mar 3, 2008 #5

    blp

    User Avatar

    Thanks! I'm going to re post this with a different title to see if someone else might know about this.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Help with showing the existance of the Schwarzschild singularity at r=0.
  1. R=0 singularity (Replies: 1)

Loading...