4
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Suppose that you have a metric $g_{\mu\nu}=\phi^2\eta_{\mu\nu}$ for some function $\phi$. There is a standard formula for what the scalar curvature $R$ looks like in terms of $\phi$, which is given by (in dimension 4)

$$ R=-\frac{6}{\phi^3}\eta^{\mu\nu}\partial_\mu\partial_\nu\phi. $$

In trying to derive this result, I keep ending up with

$$ R=-\frac{1}{\phi^4}\eta^{\mu\nu}\big(\phi\partial_\mu\partial_\nu\phi-(\partial_\mu\phi)(\partial_\nu\phi)\big). $$

Is there something obvious I'm missing that allows one to rewrite the latter as the former? Or (what is far more likely, given the long, tedious nature of this derivation) did I simply manage to mix something up somewhere along the line? (In which case I'll just have to go back through the computation yet again, since it would be awful to type it all up here...)


EDIT: Since I'm calling it a "standard formula," here are some references it shows up in:

-- This paper by Einstein and Fokker

-- It is implied by an exercise from Misner, Wheeler, Thorne, which is all about getting the point of the above paper:

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For which the relevant equations are:

enter image description here

-- More generally, it is a standard result about what happens under a conformal change in metric, as can be found in most references on the subject, e.g. on the Wiki page for formulas from Riemannian geometry:

enter image description here

The latter equation, in dimension 4 and with the original metric being $\eta$, gives the my "standard" result (where $\phi$ is what they denote by $e^{\varphi}$).

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    $\begingroup$ if $\phi$ is strictly a function of time, then you have the flat robertson walker universe. I'll take a look and hopefully get back to you. $\endgroup$ – R. Rankin May 26 '16 at 20:09
  • $\begingroup$ Unfortuantely, I believe $\phi$ is just a general function of spacetime here. $\endgroup$ – user99292 May 27 '16 at 2:14
  • $\begingroup$ The problem I'm working on is find the Ricci scalar for: $\bar{g_{\mu\nu}}=a(t)^{2}g_{\mu\nu}$ $\endgroup$ – R. Rankin May 27 '16 at 3:48
  • $\begingroup$ for a general metric, just so you know, I treated it like a full function until the end, and got the the same answer as you (times 3) in taking all metric derivatives to be zero (your case). What source is your standard result from? $\endgroup$ – R. Rankin May 27 '16 at 3:54
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    $\begingroup$ Updated to include a few references. :) $\endgroup$ – user99292 May 27 '16 at 7:05

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