I've been working with some dust solutions in General Relativity, practicing calculating the Riemann curvature tensor, and I came across an odd metric: the Tolman-Bondi-de Sitter metric. A quick internet search (to supplement the book I'm reading) can tell you that it describes spherical dust, while accounting for a cosmological constant. It's a pretty simple solution, with a line element of the form

$$ds^2=dt^2-e^{-2\Psi(t,r) } dr^2-R^2 (t,r)d\theta^2-R^2 (t,r) \sin^2⁡\theta d\phi^2$$

There's one term in there that has me a bit befuddled, and that I can't find an explanation for in a book or on the Internet: $\Psi(t,r)$

At first, I thought it had to be a simple wavefunction, but after looking at it more, I'm not quite sure. What is it, and what is its significance in the metric?

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    $\begingroup$ Could you please write your line element using LaTeX? As is, your solution looks to be a pretty generic spherically symmetric solution. $\endgroup$ – Jerry Schirmer Aug 7 '14 at 18:24
  • $\begingroup$ Actually, I'm not sure how to do that. How could I? $\endgroup$ – HDE 226868 Aug 7 '14 at 18:25
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    $\begingroup$ @HDE226868: See this link for more help on the Latex engine used on Physics.SE $\endgroup$ – Kyle Kanos Aug 7 '14 at 18:28
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    $\begingroup$ $e^{-2\Psi}$ is a function that allows you to express a spherically symmetric metric in its most general form. It takes specific values depending on extra assumptions you make, e.g. static Einstein, Friedmann etc. The reprinted article has more examples. $\endgroup$ – auxsvr Aug 7 '14 at 23:08
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    $\begingroup$ In the diagonal metrics of form $ds^2 = \alpha(r,t) dt^2 - \beta(r,t) dr^2+...$, you have to note that $\alpha$ and $\beta$ are positive, so you may always express then as exponentials of functions : $ds^2 = e^{\lambda(r,t)} dt^2 - e^{\mu(r,t)} dr^2+...$. This sometimes simplifies calculus of the Christoffel symbols, at least some of them could be expressed as first derivatives of $\lambda$ and $\mu$ (the expression of some components of the curvature or the Ricci tensor could be simpler, too). There is no general physical interpretation of $\lambda$ and $\mu$. $\endgroup$ – Trimok Aug 8 '14 at 11:02

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