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Understanding the Lemaitre metric

  1. Dec 17, 2009 #1
    With reference to this wikipedia article http://en.wikipedia.org/wiki/Lemaitre_metric
    it states

    [tex]r = \left(\frac{3}{2}(p-\tau)\right)^{2/3}\ r_g^{1/3} [/tex]

    and

    [tex] r_g = \frac{3}{2}(p-\tau) [/tex]

    Those two statements put together imply

    [tex]r = \left(\frac{3}{2}(p-\tau)\right)^{2/3} \left(\frac{3}{2}(p-\tau)\right)^{1/3} [/tex]

    [tex]r = r_g[/tex]

    This in turn implies the Schwarzschild radial variable r is equal to the constant Schwarzschild radius [itex]r_s = 2gm/c^2[/itex] and this leads to the Lemaitre metric being equivalent to

    [tex] ds^2 = d\tau^2 - dp^2 [/tex]

    and when [itex]r_g/r = 1[/itex] is inserted into the Lemaitre coordinate definitions:

    [tex]\begin{cases}
    d\tau = dt + \sqrt{\frac{r_{g}}{r}}\frac{1}{(1-\frac{r_{g}}{r})}dr~,\\
    d\rho = dt + \sqrt{\frac{r}{r_{g}}}\frac{1}{(1-\frac{r_{g}}{r})}dr~.
    \end{cases}[/tex]

    the result is:

    [tex]\begin{cases}
    d\tau = dt \pm\ \frac{dr}{0}~,\\
    d\rho = dt \pm\ \frac{dr}{0}~.
    \end{cases}[/tex]


    Obviously I am missing something important here. Can anyone clarify?
     
    Last edited: Dec 17, 2009
  2. jcsd
  3. Dec 17, 2009 #2

    Hurkyl

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    Those points are not covered by a Schwarzchild coordinate chart. Your observation is a manifestation of the coordinate singularity of Schwarzchild coordinates that prevents them from being extended any further.
     
  4. Dec 17, 2009 #3
    What exactly do you mean by "those points"?

    When the article states

    [tex]r = \left(\frac{3}{2}(p-\tau)\right)^{2/3}\ r_g^{1/3} [/tex]

    I assume it is referring to r everywhere (including outside the event horizon where the Schwarzschild metric is valid) and not just to the event horizon or points below the event horizon. Is that wrong?
     
  5. Dec 17, 2009 #4

    Hurkyl

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    The ones at the "gravitational radius" -- [itex]
    r_g = \frac{3}{2}(p-\tau)
    [/itex].
     
  6. Dec 17, 2009 #5
    OK. So if I want to plot a constant Schwarzschild radius in Lemaitre [itex](p,\tau)[/itex] coordinates I should use a constant value for [itex]r_g[/itex] in the equation:

    [tex]r = \left(\frac{3}{2}(p-\tau)\right)^{2/3}\ r_g^{1/3}\ [/tex]

    which results in a diagonal line parallel to the Lemaitre gravitational radius?
     
  7. Dec 17, 2009 #6

    Hurkyl

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    No, you should plot the points satisfying [itex]

    r_g = \frac{3}{2}(p-\tau)

    [/itex].
     
  8. Dec 17, 2009 #7
    So a line of constant Schwarzschild radius (other than the gravitational radius) can not be transformed into a Lemaitre chart??
     
  9. Dec 17, 2009 #8

    Hurkyl

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    Oh, you're talking about the Schwarzschild radial coordinate! I thought you were talking about the Schwarzschild radius.

    Yes, you had the right equation.
     
  10. Dec 17, 2009 #9
    But what is the right way to use it?

    If I substitute the definition given for [itex]r_g[/itex] into the equation for the definition of the Schwarzschild radial coordinate in terms of Lemaitre coordinates I end up with [itex] r = r_g[/itex] as mentioned in the OP, which is most unsatisfactory. :confused: I have never seen null worldlines and Schwarzschild radial coordinates plotted on a lemaitre chart so I am trying to do it for myself. Maybe everyone else has hit upon the same problems.
     
  11. Dec 17, 2009 #10

    Hurkyl

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    rg is the Schwarzschild radius. [itex]

    r_g = \frac{3}{2}(p-\tau)

    [/itex] is not a definition of rg. It is the equation of the event horizon -- it is valid only for those [itex](\tau, p)[/itex]-pairs lying on the event horizon, which is why, if you invoke that equation, you derive r=rg.
     
  12. Dec 17, 2009 #11
    OK, that seems reasonable. Using rg as a constant, lines of constant Schwarzschild radius are parallel but diagonal lines in Lemaitre coordinates.

    Now I want to obtain an equation for null worldlines that can be plotted on the Lemaitre chart. Starting with the Lemaitre metric:

    [tex]ds^{2} = d\tau^{2} - \frac{r_{g}}{r} dp^{2} [/tex]

    and taking ds = 0 for a null worldline, then

    [tex]dp/dt = \sqrt{ \frac{r}{r_g}} [/tex]

    By inspection it can be seen that for r = 0, r = rg and r = [itex] \infty[/itex] that dp/dt for a lightlike path is 0, [itex]\pm 1[/itex] and [itex]\pm \infty[/itex] respectively.

    Substituting

    [tex]r = \left(\frac{3}{2}(p-\tau)\right)^{2/3}\ r_g^{1/3} [/tex]

    into

    [tex]d\tau = \sqrt{ \frac{r_g}{r}} \ dp [/tex]

    gives

    [tex]d\tau = \left(\frac{2\ r_g}{3(p-\tau)}\right)^{1/3} \ dp[/tex]

    now it looks like multi variable integration is required at this point to obtain an expression that can be plotted. Is there anyone here who is handy with advanced calculus that would be kind enough to do that calculation?
     
    Last edited: Dec 17, 2009
  13. Dec 22, 2009 #12
    In Lemaitre coordinates,

    [tex]dS^2 = dt_{(proper)}^2 =\ \left( d\tau^2 - \frac{r_g}{r}dp^2 \right) [/tex]

    a "stationary observer" is free falling and when dp=0,

    [tex]\frac{dt_{(proper)}}{d\tau} = 1[/tex]

    as you would expect.

    The speed of light in these coordinates when dS=0 is

    [tex]\frac{dp}{dt} =\sqrt{{\frac{r}{r_g}[/tex]

    so at r = infinity, the speed of light is infinite also. At asymptotic infinity, the spacetime becomes almost flat or Minkowskian and the free falling velocity of an observer is almost zero, so it seems very strange that the observer should measure the speed of light to be infinite when he is almost stationary in flat space. I guess that is a quirk of how distance is defined in these coordinates?

    Lemaitre coordinates claim to "prove" that light can only travel inwards below the event horizon and yet no one on this forum is able to derive the equation for a null path in these coordinates or even to quote an equation for the null path from a text book?
     
    Last edited: Dec 22, 2009
  14. Dec 23, 2009 #13
    Perhaps it isn't clear what you are asking. The equation for a null path, in terms of any coordinate system, is ds = 0. Lemaitre coordinates don't claim to prove anything, they are simply one of infinitely many possible systems of coordinates that satisfy the field equations with spherical symmetry. Anyone who understands differential manifolds with semi-definite metrics can infer the light cone structure from any such system of coordinates. Of course, you have to be careful with statements like "light can only travel inwards..." because this ignores the "white hole" portion of the fully extended vacuum solution. I think it's best to learn the basics of calculus and physics before trying to evaluate the validity of general relativity.
     
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