If Newton's constant $G$ actually varies with cosmological time $t$ would a suitably modified form of the Einstein field equations:

$$G_{\mu \nu} + \Lambda g_{\mu \nu} = \frac{8 \pi G(t)}{c^4} T_{\mu \nu},$$

together with the standard cosmological assumptions, lead to equations that look like the standard Friedmann equations but with the varying function $G(t)$ in place of the constant $G$?

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    $\begingroup$ Why don't you try to investigate? Do you know how the Friedmann equations are usually derived? $\endgroup$ – Danu May 2 '14 at 14:21
  • $\begingroup$ Not really no. You're right I should investigate myself but I'm lazy. Maybe I'll try to find some cosmology lecture notes online that might give me some clues. $\endgroup$ – John Eastmond May 2 '14 at 15:07
  • $\begingroup$ Energy conservation + Einstein's equations is the way to go. I found that Carroll's book on General Relativity did a good job at explaining it. $\endgroup$ – Danu May 2 '14 at 15:21
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    $\begingroup$ Note that this ansatz breaks energy conservation. The reason Einstein picked the equation he did was because $\nabla_{a}\left(R^{ab} - \frac{1}{2}Rg^{ab}\right) = 0$, which satisfies the flux requirement on the stress-energy tensor, $\nabla_{a}T^{ab} = 0$. If you make $G$ a function of spacetime coordinates, you break this. Also note that this is similar to the approach taken by Brans-Dicke theory, where $G$ is promoted to a scalar field, and is given its own dynamics. Brans-Dicke has been heavily constrained by solar system observations. $\endgroup$ – Jerry Schirmer May 2 '14 at 16:35
  • $\begingroup$ But is it acceptable for the global energy scale to change with cosmological time so that $G$ is a function of cosmological time alone? This might not imply a breakdown of local energy conservation. $\endgroup$ – John Eastmond May 2 '14 at 18:40

The Friedmann equations are differential equations for the scale factor $a(t)$. You can derive them by plugging in the Friedmann metric $$ds^2=dt^2-a^2\left(\frac{dr^2}{1-kr^2}-r^2d\Omega^2\right)$$ where $d\Omega^2=d\vartheta^2+\sin^2\vartheta d\varphi^2$ into the Einstein equations you posted above.

Since the gravitational constant only enters in front of the energy momentum tensor, no derivatives of $G$ occur throughout the derivation. Therefore you can just use the standard Friedmann equations and put $G=G(t)$.

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  • $\begingroup$ Note that if you take this approach, it just amounts to rescaling the values of $\rho(t)$ and $P(t)$ by a factor of $G(t)$ $\endgroup$ – Jerry Schirmer May 2 '14 at 22:23

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