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Form of Kruskal-Szekeres Completion

  1. Oct 17, 2012 #1
    In my general relativity class my professor derived the form of the Kruskal-Szekeres form of the Schwarzschild metric using several formulas without actually going through the algebra. I am trying to prove the answer using these formulas but I am having some trouble, especially using the Lambert W. function. In the file I have attached, you can see that equation (9) is the final result from the changes made in the previous equations. In my attempt I tried taking the derivative of the equation for 'r' and arrived at what I have shown below. However, when subbing this in and working with this I don't feel I am making progress. Any assistance is greatly appreciated, thanks.

    [itex]dr=2m\frac{dL}{dz}dz[/itex]

    where [itex]z=\frac{-uv}{e}[/itex]

    The link for my professor's paper is at this link, http://arxiv.org/pdf/1202.0860v2.pdf , but I will also include the .pdf
     

    Attached Files:

  2. jcsd
  3. Oct 19, 2012 #2
    That seems like a proper mess :)

    But basically what you have is
    [tex] r = 2m(\mathcal{L}(1+\Psi(u,v))) [/tex]
    so
    [tex] dr = \frac{\partial r}{\partial u} du + \frac{\partial r}{\partial v} dv = ... = \frac{2m \mathcal{L}}{1 + \mathcal{L}} (du/u + dv/v) [/tex]
    or something like that; I'm too lazy to do the work :) Then repeat for dt. Then you just plug them into the metric and pray everything works out.
     
  4. Oct 19, 2012 #3
    got it, thanks!
     
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