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Denominator in cosmological constant term in Newtonian gravity

  1. Apr 2, 2012 #1
    In General relativity: an introduction for physicists, the authors derive Newtonian gravity from the EFE, but then they also give a short statement that inserting in the cosmological constant derives down to:
    [tex]
    \vec{g}=-\nabla\Phi=-\frac{GM}{r^{2}}\hat{\vec{r}}+\frac{\Lambda c^{2}r}{3}\hat{\vec{r}}.
    [/tex]

    I don't see how they end up with a 3 in the denominator. Probably am missing something simple. Any help?
     
  2. jcsd
  3. Apr 3, 2012 #2

    Bill_K

    User Avatar
    Science Advisor

    Sure, the factor of 1/3 comes from integrating Λr2. :smile:

    Seriously, I don't think you can expect an intuitive motivation for each factor of two or three. Intuition is great for order of magnitude estimates, or determining minus signs, but to get right down to an actual numerical factor you usually just need to plow through the math.

    One of Einstein's equations for spherical symmetry happens to be d(r g00)/dr = 1 - Λr2, and the solution for g00 has a 1/3 in it.
     
  4. Apr 3, 2012 #3
    Heh, I wish it was just the integration (and maybe it just is :redface:). Anyway, I should probably post how I think it is done here and maybe someone can pinpoint where my confusion lies (or more likely I will just end up embarrassing myself :tongue2:).
     
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