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Why vacuum Einstein equations are hyperbolic equations?

  1. Jan 16, 2010 #1
    Hello friends,

    I am now interested in Einstein scalar field equations with a very little knowledge about Physics. I would like to ask why vacuum Einstein equations are hyperbolic equations? As far as I know that for [tex]3+1[/tex]-dimensional manifold [tex]V[/tex], we can convert the vacuum Einstein equations as a system of time derivative w.r.t. the induced metric and the second fundamental form of a [tex]3[/tex]-dimensional manifold, called hypersurface [tex]\Sigma[/tex]. What happen to non-vacuum case since we have an extra term, called stress-energy momentum tensor. Can you explain in details?

    Thank you.
     
  2. jcsd
  3. Jan 18, 2010 #2
    I learn from a book saying that the vacuum Einstein equation can be rewritten as

    [tex]{\partial _t}{g_{\alpha \beta }} = - 2N{k_{\alpha \beta }} + {\mathbb{L}_X}{g_{\alpha \beta }}[/tex]

    and

    [tex]{\partial _t}{k_{\alpha \beta }} = - {\nabla _\alpha }{\nabla _\beta }N + N\left( {{R_{\alpha \beta }} + trk{k_{\alpha \beta }} - 2{k_{\alpha \gamma }}k_\beta ^\gamma } \right) + {\mathbb{L}_X}{k_{\alpha \beta }}[/tex]

    however I am not sure how to make sure that the above system is hyperbolic. Thanks.
     
  4. Jan 18, 2010 #3
    Actually a hyperbolic partial differential equation is a partial differential equation (PDE) that, roughly speaking, has a well-posed initial value problem, so the Cauchy problem for GR can simply answer your question. But historically, it is ineresting to know that the first scalar theory of gravitation, known as the 1th theory of Nordstrom's gravitation, had the field equation [tex]\square \phi = 0[/tex] with [tex]\square[/tex] being the D'Alembert's operator. As you can see, this is exactly the equation of a wave in vacuum that is hyperbolic because if on the line t=0, you have the first derivative of \phi w.r.t t and itself specified arbitrarily as the initial values, then there exists a solution for all time.

    AB
     
  5. Jan 18, 2010 #4
    Thanks, what I am expecting is to show that studying Cauchy problems for Einstein equations make sense. My advisor suggests me to show that the vacuum Einstein equation is indeed a hyperbolic equation but I am still not able. Having this fact, it is reasonable to study Cauchy problem for the Einstein equations.
     
  6. Jan 18, 2010 #5
    Yeah it is! In the scalar theory of Gravitation, the scalar curvature R vanishes in vacuum, i.e. R=0. So introducing this into the Einstein and Fokker [*] definition of the Ricci scalar,

    [tex]-6(\square \phi)/\phi^3 = R[/tex] results in the wave equation again which is hyperbolic as discussed earlier. Note that here the vacu solution can be obtained from their field equation,

    [tex]R=24\pi GT[/tex]

    where you put T=0. (T, here, is the trace of the energy-momentum tensor and G is Newton's gravitational constant).

    I think now it is clear why the vacu scalar field equation would be hyperbolic.

    But in the tensor theory of Gravitation or GR, this is a little bit hard to explain as you are required to take advanced lessons for the Cauchy problem of GR which elegantly shows what is behind the name 'hyperbolic' for field equations not only in vacuum, but anywhere else.

    For a full study of this problem, take a look at:

    Witten L. (ed.) Gravitation: an introduction to current research (Wiley, 1962), chapter 4.

    [*] A. Einstein and A.D. Fokker, Ann. d. Phys. 44, 321 (1914).
    AB
     
  7. Jan 19, 2010 #6
    Oh thanks, I am reading. BTW, I have another approach: Consider the Lorentzian metric
    [tex]d{s^2} = - dt \otimes dt + {g_{ij}}(x,t)d{x^i} \otimes d{x^j}[/tex].​
    Einstein equations in vacuum, i.e., [tex]G_{ij} = 0[/tex] become
    [tex]\frac{{{\partial ^2}{g_{ij}}}}{{\partial {t^2}}} + 2{R_{ij}} + \frac{1}{2}{g^{pq}}\frac{{\partial {g_{ij}}}}{{\partial t}}\frac{{{\partial _{pq}}}}{{\partial t}}{ - g^{pq}}\frac{{\partial {g_{ip}}}}{{\partial t}}\frac{{{\partial _{jq}}}}{{\partial t}} = 0.[/tex]​
    It seems that the above eqn. is hyperbolic but I am not sure how to check. Thanks.
     
    Last edited: Jan 19, 2010
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