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Singularities and Kasner solution

  1. Feb 1, 2007 #1
    1. The problem statement, all variables and given/known data
    Investigate the possible behaviour of the singularity as [tex]t \rightarrow 0[/tex] in the Kasner solution.


    2. Relevant equations
    The metric for the Kasner solution is given by

    [tex]ds^2 = c^2dt^2 - X_1^2(t)dx_1^2 - X_2^2(t)dx_2^2 - X_3^2(t)dx_3^2[/tex]


    3. The attempt at a solution
    I have no clue...
     
  2. jcsd
  3. Feb 1, 2007 #2

    Dick

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    Try writing down a more specific form of the Kasner soln. The spatial metric functions can be written as powers of t. What are the conditions on the exponents coming from Einsteins eqns?
     
  4. Feb 1, 2007 #3
    I don't get it.
     
  5. Feb 1, 2007 #4

    Dick

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    You can't do anything with the metric as written - it's too general. What ARE X_1,X_2 and X_3 in the Kasner soln. You may have to look it up...
     
  6. Feb 1, 2007 #5
    All I can find about this is that substituting the metric into the Einstein equations gives

    [tex]\frac{\ddot{X}_i}{X_i} - \left(\frac{\dot{X}_i}{X_i}\right)^2 +3\left(\frac{\dot{X}_i}{X_i}\right)\left(\frac{\dot{a}}{a}\right) = \frac{4\pi G}{c^4}\left(\rho - \frac{p}{c^2}\right)[/tex]

    in which [tex]a^3 = X_1X_2X_3[/tex].
     
  7. Feb 1, 2007 #6

    Dick

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    That's a start. You shouldn't have a loose index i floating around though. Assuming you can get the correct Einstein equations (and there should be two), Kasner is a vacuum solution, so put rho=p=0. Put X_i=t^p_i. Turn this into equations in the constants p_i. Are you supposed to actually derive Kasner or just describe it's properties? It might be a good idea to just look up the solution to see what you are aiming for.
     
  8. Feb 1, 2007 #7
    I think I'm just supposed to describe the behaviour of the singularity. I don't like this book, it has too many errors. ;) Why should I use [tex]X_i = t^{p_i}[/tex]? According to my book this is a perfect fluid model.
     
  9. Feb 1, 2007 #8
    Ok, I found that this is a particulary simple behaviour. Also

    [tex]\frac{\dot{X}_1 \dot{X}_2}{X_1X_2} + \frac{\dot{X}_2 \dot{X}_3}{X_2X_3} + \frac{\dot{X}_3 \dot{X}_1}{X_3X_1} = \frac{8 \pi G}{c^4}\rho[/tex]

    Putting in [tex]X_i = t^{p_i}[/tex] and using [tex]\rho=0[/tex] gives me

    [tex]0 \propto \frac{1}{t^2}[/tex]

    Is this true?
     
  10. Feb 1, 2007 #9

    Dick

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    It gives you an algebraic condition on the p's that must vanish. What is it? Again there is another Einstein equation. It will give you another algebraic condition.
     
  11. Feb 1, 2007 #10
    So [tex]p_1p_2 + p_2p_3 + p_3p_1 = 0[/tex]?
    I also got that

    [tex]\frac{\dot{a}}{a} = \frac{1}{3} \left( \frac{\dot{X}_1}{X_1} + \frac{\dot{X}_2}{X_2} + \frac{\dot{X}_3}{X_3} \right)[/tex]

    but then I get some a aswell..?
     
  12. Feb 1, 2007 #11

    Dick

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    That is one alright. Since we are not actually trying to derive this you should find

    [tex]p_1+p_2+p_3=1[/tex] and
    [tex]{p_1}^2+{p_2}^2+{p_3}^2=1[/tex].
     
  13. Feb 1, 2007 #12
    Yeah that I got..
     
  14. Feb 1, 2007 #13

    Dick

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    Good. So there are actually lots of Kasner solutions corresponding to different values of the p's. Find a specific example and describe it's behavior.
     
  15. Feb 1, 2007 #14
    Well, what I don't understand is HOW to describe it's behaviour.
     
  16. Feb 1, 2007 #15

    Dick

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    Ok, here's a set of p's.

    [tex]p_1=1/3, p_2=(1+\sqrt{3})/3, p_3=(1-\sqrt{3})/3[/tex]

    How would you describe this behavior as t->0?
     
  17. Feb 1, 2007 #16
    Well, I will always get

    [tex]\frac{\dot{a}}{a} = \frac{1}{3t}[/tex]

    so for [tex]t \rightarrow 0[/tex]

    [tex]\dot{a} \rightarrow \infty[/tex]
     
  18. Feb 1, 2007 #17

    Dick

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    Sure. But describe qualitatively the behavior of the scale factors themselves (the X_i's).
     
  19. Feb 1, 2007 #18
    Sorry, I don't know what to say about them.
     
  20. Feb 1, 2007 #19

    Dick

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    That's ok. In a matter dominated universe (a=t^(2/3), I think) a->0 as t->0. Is that true here?
     
  21. Feb 1, 2007 #20
    Humm, if I solve

    [tex]\frac{\dot{a}}{a} = \frac{1}{3t}[/tex]

    I get

    [tex]a = t^{1/3} \rightarrow 0[/tex]

    for [tex]t \rightarrow 0[/tex].

    But if I use [tex]a_0[/tex] instead, I get

    [tex]a = a_0 \frac{1}{3} ln(t) \rightarrow - \infty[/tex].
     
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