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Vaidya/Lemaitre-Tolman metrics

  1. Aug 26, 2011 #1
    Is anyone here familiar with the Vaidya/Lemaitre-Tolman metrics? Wiki claims that they're both used to model radiating spherically symmetric dust, but I'm not so sure after grinding out the curvature tensors, etc.

    tl;dr a (non rotating) spherical star is ejecting a spherical shell of dust - what metric do I use (short of brute forcing a new solution)?
     
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  3. Aug 26, 2011 #2

    Mentz114

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    The Vaidya metric is

    [tex]
    ds^2=r^2d\Omega^2-2dudr-(1-2m(u)/r)du^2
    [/tex]
    where u is a null coordinate.

    I get the Einstein tensor for dust, only G00 being non-zero.
    [tex]
    G_{00}=\frac{2\,\left( \frac{d}{d\,u}\,m\right) }{{r}^{2}}
    [/tex]

    This paper may be of interest arXiv:gr-qc/0609036v2 .

    I'll see if I have the results for the Tolman metrics somewhere a bit later.

    [edit]
    After some calculation I found that the Einstein tensor calculated in a local frame basis is that of pure radiation. Confirmed in a textbook also. It seems strange that a change of basis can turn the dust tensor into radiation. But it certainly is the case.
     
    Last edited: Aug 26, 2011
  4. Aug 26, 2011 #3

    PeterDonis

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    Isn't the Vaidya dust null dust? (In other words, aren't the worldlines of the "dust particles" null worldlines?)
     
  5. Aug 27, 2011 #4

    Mentz114

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    I think the null coordinates ensure that. What I find strange is that that in the (holonomic) coordinate basis the Einstein tensor is the one I've quoted above, but in a local frame basis we get pure radiation with Gab being proportional to kakb where k is a null vector. I suppose if the 'dust' is null to begin with this is possible.
     
  6. Aug 27, 2011 #5

    Bill_K

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    I think you need to recheck G0r and Grr.
     
  7. Aug 27, 2011 #6

    K^2

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    For general m(u), I'm picking up non-zero terms in T as well. Is there some constraint on m(u) that forces these to zero? Or are they supposed to be non-zero?
     
  8. Aug 27, 2011 #7

    George Jones

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    I haven't done the calculations, but I have a couple of books that state that [itex]G_{uu}[/itex] is the only non-zero component of the Einstein tensor.
    Even though references say that [itex]u[/itex] is a null coordinate, I don't think that [itex]u[/itex] is a null coordinate. [itex]\partial_u[/itex] is the tangent vector to u-coordinate curves, and the above metric gives
    [tex]g \left( \partial_u , \partial_u \right) = -\left(1-2m\left(u\right)/r\right).[/tex]
    Consequently, [itex]u[/itex] is a timelike coordinate. The above metric also gives
    [tex]g \left( \partial_r , \partial_r \right) = 0,[/tex]
    and thus [itex]r[/itex] is a null coordinate.
    When I use the (inverse) metric to raise indices, I calculate that
    [tex]G^{rr}=\frac{2\,\left( \frac{d}{d\,u}\,m\right) }{{r}^{2}}[/tex]
    is the only non-zero (contravariant) component of the Einstein tensor. Therefore, as a tensor ( not just components)
    [tex]G = G^{rr} \partial_r \otimes \partial_r[/tex]

    with [itex]\partial_r[/itex] a lightlike vector.
     
  9. Aug 27, 2011 #8

    Mentz114

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    George and Bill_K,

    there is some ambiguity about what is null in this metric. Joshi ( see below) just refers to 'null coordinates'.

    According to Joshi in 'Global Aspects in Gravitation and cosmology' (Cambridge, 1993), the relevant tensors are those calculated in the frame basis, in which the Einstein tensor is that of pure radiation ( as I wrote above ).


    I used the inverse of the co-tetrad
    [tex]
    \Lambda=\pmatrix{-\frac{2\,m-2\,r}{\sqrt{3\,{r}^{2}-2\,r\,m}} & -\sqrt{\frac{r}{3\,r-2\,m}} & 0 & 0\cr -\sqrt{\frac{r}{3\,r-2\,m}} & -\frac{\sqrt{r}}{\sqrt{3\,r-2\,m}} & 0 & 0\cr 0 & 0 & r & 0\cr 0 & 0 & 0 & r\,sin\left( \theta\right) }
    [/tex]
    to transform the holonomic G into the frame basis and got this - which is pure radiation in the Vaidya zone.

    [tex]
    G_{ab}=\pmatrix{-\frac{2\,\left( \frac{d}{d\,u}\,m\right) }{2\,r\,m-3\,{r}^{2}} & \frac{2\,\left( \frac{d}{d\,u}\,m\right) }{2\,r\,m-3\,{r}^{2}} & 0 & 0\cr \frac{2\,\left( \frac{d}{d\,u}\,m\right) }{2\,r\,m-3\,{r}^{2}} & -\frac{2\,\left( \frac{d}{d\,u}\,m\right) }{2\,r\,m-3\,{r}^{2}} & 0 & 0\cr 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0}
    [/tex]
     
    Last edited: Aug 27, 2011
  10. Aug 28, 2011 #9

    Bill_K

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    Of course u is a null coordinate.
    grr = 0, gur = gru = 1, guu = 1-2m/r.
    grr = -1 + 2m/r, gur = gru = 1, guu = 0.
    u u = guu = 0, so u = const is a null surface.
    In fact for m = 0, you get Minkowski space in which the coordinate u = t-r is retarded null time. The Vaidya metric is an example of Kerr-Schild metric, defined to be flat space plus the square of a null vector: gμν = ημν + H kμkν.
     
  11. Aug 28, 2011 #10

    George Jones

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    Yes, and (I think) that the lightlike vector [itex]\partial_r[/itex] is normal to this null surface. Maybe my definition of null coordinate is non-standard. The character of coordinate is determined by letting the coordinate in question vary while holding all other coordinates constant. The character of the coordinate is given by the tangent vector to the resulting curve.
    If the coordinate transformation u = t - r and v = t + r is made, then u is a null coordinate. If the coordinate transformation u = t - r , r = r is made, then u is not a null coordinate. For more on this (possibly non-standard view), see

    https://www.physicsforums.com/showthread.php?p=2714132#post2714132.

    Unfortunately, I didn't follow up in this thread.
     
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