Friedmann Equations with varying G?

Question

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 $G$?

This post imported from StackExchange Physics at 2014-05-04 11:16 (UCT), posted by SE-user John Eastmond

Comments

Why don't you try to investigate? Do you know how the Friedmann equations are usually derived?

This post imported from StackExchange Physics at 2014-05-04 11:16 (UCT), posted by SE-user Danu

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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.

This post imported from StackExchange Physics at 2014-05-04 11:16 (UCT), posted by SE-user John Eastmond

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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.

This post imported from StackExchange Physics at 2014-05-04 11:16 (UCT), posted by SE-user Danu

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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.

This post imported from StackExchange Physics at 2014-05-04 11:16 (UCT), posted by SE-user Jerry Schirmer

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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.

This post imported from StackExchange Physics at 2014-05-04 11:16 (UCT), posted by SE-user John Eastmond

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@JohnEastmond: yes it does. that gradient includes a time derivative. if the time derivative of $G$ is nonzero, then you either have to abandon the gradient of $T_{ab}$ being zero, which is unphysical, or you have to modify Einsteins' equation. Brans and Dicke did the latter, and that theory is observationally ruled out.

This post imported from StackExchange Physics at 2014-05-04 11:16 (UCT), posted by SE-user Jerry Schirmer

Answer 1

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 $G=G(t)$.

This post imported from StackExchange Physics at 2014-05-04 11:16 (UCT), posted by SE-user Photon

Comments

Great! Thanks very much

This post imported from StackExchange Physics at 2014-05-04 11:16 (UCT), posted by SE-user John Eastmond

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Note that if you take this approach, it just amounts to rescaling the values of $\rho(t)$ and $P(t)$ by a factor of $G(t)$

This post imported from StackExchange Physics at 2014-05-04 11:17 (UCT), posted by SE-user Jerry Schirmer