In Einsteins Field equations(adsbygoogle = window.adsbygoogle || []).push({});

[itex] R_{\mu \nu} -\frac{1}{2}R\,g_{\mu \nu} + \Lambda\,g_{\mu \nu} = {8 \pi G \over c^4} T_{\mu \nu} [/itex]

if you make an ansatz that the gravitational constant G is not constant, but a function of either the Ricci scalar R or the Ricci tensor [itex] R_{\mu \nu} [/itex], would you still require the cosmological constant λ to balance out the unsteady nature of the universe? The procedure would be either to solve for G's function or make a guess as to what it is in order to get a steady-state universe without λ.

[itex] R_{\mu \nu} -\frac{1}{2}R\,g_{\mu \nu} = {8 \pi G(R) \over c^4} T_{\mu \nu} [/itex]

Actually the more I think of this, the more dumb I think I am, because (and correct me if I'm wrong because I probably am!) we can express the field equations with the cosmological constant this way

[itex] R_{\mu \nu} -\frac{1}{2}R\,g_{\mu \nu} = {8 \pi \over c^4} \left(G-\frac{c^4}{8 \pi}\Lambda\,g_{\mu \nu}T^{\mu \nu}\right)T_{\mu \nu} [/itex]

Which seems to imply that the cosmological constant may be looked at as modifying G, such that G varies as

[itex]G=G_{0}-\frac{c^4}{8 \pi}\Lambda\,g_{\mu \nu}T^{\mu \nu}[/itex]

in Einstein's original derivation of

[itex] R_{\mu \nu} -\frac{1}{2}R\,g_{\mu \nu} = {8 \pi G \over c^4} T_{\mu \nu} [/itex]

Where perhaps [itex]G_{0}[/itex] is the gravitational constant in flat space.

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# What if G is a function of R?

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