# May the gravitational constant $G$ be a *functional* of all fields on spacetime?

Is it theoretically possible that the $G$ constant in Einstein's equation be a functional of all fields present in a given spacetime ? $$\tag{1} G_{\mu \nu} + \Lambda \, g_{\mu \nu} = -\, \frac{8 \pi}{c^4} \, G[g, \phi] \, T_{\mu \nu}.$$ Since it's by hypothesis a global functional of some sort defined over all of spacetime, it's still a constant independant of position (so the Bianchi identity and local conservation of energy-momentum still apply), but $G$ may change for different spacetimes and fields configurations. In a sense, it's "scale" dependant.

In other words : Given an asymptotically flat spacetime with matter and total energy $E$ (ADM or Tolman mass), could $G$ be dependant on energy : $G(E)$ ?

And as a generalisation, what about the other "constants" of nature ?

Could the cosmological constant $\Lambda$ and the fine structure constant $\alpha \equiv k \, e^2 / \hbar c$ also be some functionals of fields over spacetime ?

What would be the arguments against this idea ?

Note : I'm not asking about position dependance, which isn't the same thing at all : $G = G[g, \phi] \ne G(x)$ (notice the square brackets!). I'm talking about a functional, like the fields action : $S \equiv S[g, \phi]$. So maybe $G$ is proportional to $S$, or any other functional.

EDIT : May the dark matter be explained by such an hypothesis ? If $G$ depends on the scale (i.e the energy involved), then gravity isn't responding the same at our solar system's scale, and at a galactic scale.

This idea has some machian feel, in a way, since properties of curved spacetime may depend on the amount of matter/energy in it, from its $G(E)$ !

Since some people appear to have difficulties with the idea of a functional (not a function), I'm giving a naive example of what $G$ may look like, according to the idea above. For some real scalar field $\phi(x) \sim \mathrm{L}^{-1}$ : $$\tag{2} G[g, \phi] = G_0 \, \sqrt{ 1 + G_0^2 \int_{\mathcal{M}} \big( g^{\mu \nu} (\partial_{\mu} \, \phi)(\partial_{\nu} \, \phi) \big)^{2} \, \sqrt{- g} \: d^4 x + \ldots },$$ where $G_0 \sim \mathrm{L}^2$ is a "naked" gravitationnal constant. The other terms are "scale dependant corrections". So the real gravitational coupling constant $G$ depends globaly on the matter content in the whole of spacetime, or on the scale we consider to do the calculations.

EDIT 2 : Here's a small argument in favor of the previous idea. Ever noticed that both $G$ and $\alpha$ have physical dimensions (i.e units) that depend on the spacetime dimensions $D$ ? (this is well known. Just examine the Poisson equation : $\nabla^2 \phi = 4 \pi G \rho$, where the density $\rho$ depends on the $D - 1$ space dimensions) : \begin{align}\tag{3} G &\sim \mathrm{L}^{D - 2}, &\alpha &\sim \mathrm{L}^{D - 4}. \end{align} Then if their value necessarily changes with the dimensionality $D$ of spacetime, why should they stay the same for all spacetimes of a given dimensionality ?

• And what about the possibility that our problems with dark matter - which happens at some large scale - may be explained by a different $G$ at that scale ?
– Cham
Jul 17 '17 at 17:58
• Doesn't condensed matter physics do stuff like this, introduce a new matter field $\phi$, then pretend $\langle\phi\rangle$ is a coupling constant of some sort? Jul 17 '17 at 18:35
• As per arguments against this, it's horribly nonlocal... Jul 17 '17 at 18:48
• I agree it's highly non-local, but since $G$ is still just a constant, we could in principle solve for any $G$, then compute it after we know the dynamics of the fields. This is viable only for analytical solutions, I guess (like the Schwarzschild solution), which is pretty limited. This non-locality would be essentially the same for any machian theory. Or maybe it's viable for successive approximations, if we already know the functional dependance of $G$.
– Cham
Jul 17 '17 at 19:17
• Is the notation $g$ supposed to refer to the metric? If so, then you have a problem with the equivalence principle. Is $\phi$ referring to matter fields? If so, then why can't this just be absorbed into $T$?
– user4552
Jul 17 '17 at 20:05

Unfortunately, however, it seems there are theoretical difficulties with a running $G$, namely there's no consistent "universal" theoretic way to do it below the Planck scale. For more on this problem, see:
• What do you mean by "running" constant ? A scalar field slowly changing with position or time ? $G(t)$ ?? Geez, again, this is not what I'm talking about. I'm talking about a constant that depends on the scale we choose to make the calculations : $G(E) \ne G(t)$.
• @Cham in QFT, certain fields have their coupling constant "run" (or change values) with the energy scale. That is to say, have $G=G(E)$...which is what you're talking about... Jul 18 '17 at 14:25