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About LTB metric

  1. Sep 26, 2009 #1
    I was reading some papers about Lemaitre-Tolman-Bondi model these days, and was confused about the dimension of this metric.
    As we know, the parabolic LTB line element takes the form:[tex]$ ds^{2}=-c^{2}dt^{2}+(R')^{2}dr^{2}+R^{2}d\Omega^{2}$[/tex].
    In my GR lessons I was told that the metric is dimensionless. But here something seems to be paradoxical. If the coefficient of the second term is dimensionless, then we can deduce that R must has a dimension of length, which would conflict with the fact that the coefficient fo the thrid term, R, is required to be dimensionless. And vice versa.


    Forgive my poor English. lol.
     
    Last edited: Sep 26, 2009
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  3. Sep 26, 2009 #2

    tiny-tim

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    Hi micomaco86572! :smile:

    A metric isn't dimensionless … it has dimensions of length-squared. :wink:

    c R' and Ω are dimensionless, and everything else is length-squared.
     
  4. Sep 26, 2009 #3

    Thx for your reply! :smile:

    I may put it in a wrong way and didn't make it clear. I actually meant the component of the metric tensor is dimensionless. Of course the line element has a dimension of length. But why didi u say c is dimensionless? It should has the dimension of lenght/time, shouldn't it? And I am still not very sure about whether the [tex]\Omega[/tex] has a dimension or R. Could u show me some proof or evidence?

    Thx again for your reply!
     
  5. Sep 27, 2009 #4

    tiny-tim

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    Hi micomaco86572! :smile:

    (just got up … :zzz:)

    (oh, and have an omega: Ω and try using the X2 tag just above the Reply box :wink:)
    c is dimensionless because length and time are the same dimension (just think about dt2 - dx2 :wink:).

    Ω is dimensionless because it's area/radius2 (similarly, ordinary angle, = arc-length/radius, is dimensionless).

    And R' is dimensionless because it's ∂R/∂r … see http://en.wikipedia.org/wiki/Lemaitre–Tolman_metric" [Broken] :smile:
     
    Last edited by a moderator: May 4, 2017
  6. Sep 27, 2009 #5

    DrGreg

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    With respect, I think that's a slightly weird way of thinking of it, in the context of the equation being discussed in this thread. It is true that relativists often use the convention that c = 1, by measuring time and distance in appropriate units (e.g. years and light-years). Under that convention, you can regard "time" and "distance" as being dimensionally the same. But under that convention, the letter c doesn't appear in any equations, as it is 1.

    If you have an equation with an explicit c in it, then you have to regard time and distance as being dimensionally different, otherwise why would you bother writing the c? So I don't really buy "c is dimensionless" but I do accept "1 is dimensionless" (when c = 1).
     
  7. Sep 27, 2009 #6

    tiny-tim

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    Hi DrGreg! :smile:

    Yes, I always use c = 1, so I get confused when it isn't. :redface:

    hmm … let's think …

    although my reasoning was a bit iffy (in particular, I should have written "c2dt2 - dx2" :redface:),

    I'm still going to maintain that length and time have the same dimensions, and that c is a dimensionless constant, like the 12 for converting feet to inches, or like the 4π for converting from some cgs units to SI units.

    What do other people think? :smile:
     
  8. Sep 28, 2009 #7
    I don't think c is dimensionless. As DrGreg said, 1 is dimensionless, but c is not. In natural unit c is set to be 1, so it is dimensionless, but in the SI units, it has to have a dimension like the gravitational constant G or some other constants.
     
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