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Timelike and null coordinate

  1. May 11, 2010 #1
    Now I'm study the Schwarzschild geometry from "General Relativity (M.P. Hobson)".

    Since the Schwarzschild metric has coordinate singularity at [tex]r=2M[/tex] so to remove this singularity they use the Eddington-Finkelstein coordinate,

    first they begin with introduces new time parameter "p"

    [tex]p=ct+r+2M ln\left |\frac{r}{2M}-1 \right |[/tex]
    which is

    [tex]dp=c dt+\frac{r}{r-2M}dr[/tex]
    and they said that it's a null coordinate

    after that , they said "since p is a null coordinate, which might be intuitively unfamiliar, it is common practice to work instead with the related timelike coordinate [itex]t^\prime[/itex]defined by"

    [tex]ct^{\prime}=p-r=ct+2M ln\left |\frac{r}{2M}-1 \right |[/tex]

    and it is a timelike coordinate which called "advanced Eddingtion-Finkelstein coordinate"

    My question is how can I check that which coordinate are timelike nulllike or spacelike? Is there any explicit calculation to check this?

    What wrong with the former coordinate which defined as p? Why should we use the new one instead?
     
    Last edited by a moderator: May 12, 2010
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  3. May 11, 2010 #2

    George Jones

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    Are you familiar with vectors represented by partial derivative operators? If you are, then use the chain rule.
     
  4. May 12, 2010 #3
    "vectors represented by partial derivative operators"

    like this? [tex]V=V^{\mu}\partial_{\mu}[/tex]

    sorry, but I have no idea about what you are said. Could you please tell me more about it?
     
  5. May 12, 2010 #4

    George Jones

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    Yes!

    Unfortunately, there is some subtlety here, and this subtlety seems to have confused Hobson, Efstathiou, and Lasenby (HEL). Most of the subtlety has to do with Woodhouse's "second fundamental confusion of calculus."

    By HEL's own definition on page 248,
    [itex]p[/itex] in Eddington-FinkelStein coordinates [itex]\left(p,r,\theta,\phi \right)[/itex] is a timelike coordinate, not a null coordinate. To see this, apply HEL's prescription on page 248 to equation (11.6). Varing [itex]p[/itex] while holding [itex]r[/itex], [itex]\theta[/itex], and [itex]\phi[/itex] constant gives [itex]dr = d\theta = d\phi = 0[/itex] and

    [tex]ds^2 = \left( 1 - \frac{2M}{r} \right) dp^2.[/tex]

    Hence, (when [itex]r > 2M[/itex]) [itex]ds^2[/itex] is positive, and [itex]p[/itex] is a timelike coordinate.

    HEL are thinking of [itex]p[/itex] in Kruskal coordinates [itex]\left(p,q,\theta,\phi \right). [/itex]. In this case, applying the page 248 prescription to equation (11.16) gives that [itex]p[/itex] is a null coordinate. Do you see why?

    What type of coordinate is [itex]r[/itex] in Eddington-FinkelStein coordinates [itex]\left(p,r,\theta,\phi \right)[/itex]?

    By now, you should be thoroughly confused! How can the "same" [itex]p[/itex] be timelike in one set of coordinates and null in another set of coordinates? If you want, I am willing to spend some time explaining in detail what is going on here, and what Woodhouse's "second fundamental confusion of calculus" is.
     
  6. May 12, 2010 #5
    So , according to your comment when I apply the prescription in P.248 to the metric in Kruskal Coordinates [tex](p,r,\theta,\phi)[/tex].


    [tex]ds^{2}=\left(1-\frac{2M}{r}\right)dp dq-r^{2}d\Omega^{2}_{s^{2}}[/tex]

    then as [tex]dq=d\theta=d\phi=0[/tex] so we conclude that [tex]p[/tex] is null coordinate.

    Am I right?

    If that were the case, then in P.255 they made some mistake because they told that [tex]p[/tex] is null coordinate. In stead, as your suggestion then coordinate [tex]p[/tex] as defined in (11.5) is already timelike coordinate. so What is HEL's propose to introduce new coordinate called "advanced Eddington-Finkelstein"(11.8) which claim to be a timelike coordinate.

    I found in Black Hole Physics (Frolov & Novikov) they also said that [tex]v[/tex] is null coordinate,which defined slightly different from HEL but still get the same metric as HEL (11.6). So right now I'm so confused about that

    Yes, sure I want to. Thank you so much I appreciated that
     
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