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Is this line element known to anyone ?

  1. Jul 11, 2007 #1

    Mentz114

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    This is based on the metric of the surface of a 3D sphere. A and B are constants with dimension (length)^2. Coordinates are

    [tex] x^0 = t, x^1 = \theta, x^2 = \phi[/tex]

    [tex]ds^2 = -c^2AB^{-1}dt^2 + Bd\theta^2 + Bsin^2(\theta)d\phi^2[/tex]

    It satisfies the Einstein field equations with only one component of the energy-momentum tensor non-zero -

    [tex]T^{00} = \frac{c^2A}{B^2}[/tex]

    The curvature scalar is

    [tex]R^\mu_{ \mu} = \frac{2}{B}[/tex]

    This looks like a two-dimensional static cosmos with constant curvature ( radius sqrt(B)) and total energy A, and no energy currents.

    Is this a viable interpretation ?

    [ Assuming I haven't made any gross errors in the calculation...]
     
    Last edited: Jul 11, 2007
  2. jcsd
  3. Jul 12, 2007 #2
    I checked your two results with "Great.m", it is correct.
    (see http://library.wolfram.com/infocenter/MathSource/4781/ )

    It is funny (and obvious) that if you add a third spacial coordinate r = sqrt(B),
    the energy momentum and the curvature drop to zero.

    Why is it that constraining motion to a sphere leads to Too =/= 0 ?
     
    Last edited: Jul 12, 2007
  4. Jul 12, 2007 #3

    Mentz114

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    Thanks, lalbatross. I tried adding a radius spatial coord with g_rr = 1 and
    get the same values as the 3D except for T11 which is now 1/B. Looks like a constant pressure term. I now think this is an interior solution of a fluid of constant pressure.

    Maybe because the constraint induces constant curvature which requires a source ?
     
    Last edited: Jul 12, 2007
  5. Jul 12, 2007 #4
    If g_rr=1 and r²=B, you get no curvature and Too=0.
    For the interpretation, maybe you could get it by writing the equation of motion of a particle.
    This Too must be what keeps a particle on the surface of a sphere.
    What is the physical interpretation of Too and how does it relate to keeping a particle constrained to a sphere?
     
  6. Jul 12, 2007 #5

    Mentz114

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    Using

    [tex]ds^2 = -c^2AB^{-1}dt^2 + dr^2 + Bd\theta^2 + Bsin^2(\theta)d\phi^2[/tex]

    I get the same curvature, the same T_00 and T_11 = 1/B.

    I have written out the equations of motion for the 3D and 4D case but I can't solve them. Obviously the null geodesics will be like great circles in the 3D case.

    The rest mass/energy that causes the cuvature ?
     
    Last edited: Jul 12, 2007
  7. Jul 13, 2007 #6
    Sorry for being a bit lazy with latex.
    I was also a bit too fast in my answer.
    What I meant is:

    [tex]ds^2 = -c^2dt^2 + dr^2 + r^2d\theta^2 + r^2sin^2(\theta)d\phi^2[/tex]

    which meant B=r² and A=B, which i only a special case.

    Anyway, in this special case, it is clear that the 3D nergy momentum should be related to the centripedal force needed to keep a particle moving on a sphere. When A is different from B, an additional interpretation would be needed.
     
    Last edited: Jul 13, 2007
  8. Jul 13, 2007 #7

    Mentz114

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    Yes, that metric has no curvature. If you're keeping the motion on the surface then dr = 0 and you get back to two spatial dimensions, where I started from.

    I don't think that the 4D metric in my post#5 is physical.
     
  9. Jul 13, 2007 #8
    It could be physical.
    You just need the T to be so.
    You could check if such a metric could not occur in a central potential.
    Is it not possible, for some A and B, to get the Swchartchild metric restricted on the dr=0 surface?
     
  10. Jul 13, 2007 #9

    Mentz114

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    Hi lalbatros.

    I've worked out T for the 4D metric and I get

    [tex]T^{00}=\frac{1}{Ac^2}, T^{11}= -\frac{1}{B}[/tex] all other components zero. Curvature scalar is still 2/B.

    T_11, which we can interpret as radial momentum, depends only on B which seems very weird.

    So it could be the interior r < sqrt(B) of a sphere of matter. Interestingly, light rays can be trapped to surfaces of constant r, that is great circles. I'm not sure if light could escape.

    If this is so, then the exterior solution must be Schwarzschild and the joining will probably give the relation between A and B.

    I'm sure I've seen this in a textbook. I'm going to look for it.
     
    Last edited: Jul 13, 2007
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