# Is there a relation between the gravitational constant G and the Hubble constant H?

I know there is a whole collection of question about the relation between constant X and constant Y in physics. So I should explain why this is not just a gratuitous question on some of the remaining constant combinatorics.

My original motivation was to understand what could be the effect of universe expansion, or of inflation, on the various force fields. This is clearly too general a question, and I had to chose a force that still matters at the scales where expansion makes sense (at least to me). Gravity is the obvious candidate,

The problem is that any question about gravitation runs the risk of getting into specifics I am trying to avoid, such as remarks on the curvature of space (I read carefully answers on other questions about gravitation, questions that had more or less my level of naiveness).

And the best I could com up with, hoping it makes sense, is

Is there a relation between the gravitational constant G and the Hubble constant H ?

i.e. would G be affected, and how, if there was a variation of H for whatever reason ?

And of course: what about a very fast expansion like inflation, possibly considering other forces? But this may already be too much for one question.

• I don't think the answer is known. The gravitational constant is simply a measured value. It is assumed to be constant in time, but measuring its change over time is difficult. The Hubble parameter is not constant in time, but the way it changes depends on the cosmological model. – safesphere Nov 9 '17 at 21:34
• Like I said, the Hubble parameter depends on the cosmological model. The answer below is specific to the Friedmann model. While this model is currently accepted, it has more holes in it than a slice of Swiss cheese and is not supported by observation. Specifically, it requires huge amounts of unobserved "dark energy" and "dark matter" plus it produces non physical singular solutions for our flat universe (infinite total mass). In the Milne model, the Hubble parameter is constant. Other models can be constructed with this parameter increasing independently of the gravitational constant. – safesphere Nov 10 '17 at 17:53
• To be clear, this site is for the "mainstream" physics. The Friedmann universe is the "mainstream". So the answer below is correct within the scope of this site. – safesphere Nov 10 '17 at 17:56

## 1 Answer

The answer is given by the Friedmann's equation:

$$H^2=\dfrac{8\pi G}{3}\rho-\dfrac{kc^2}{a^2}=\dfrac{\dot{a}^2}{a^2}$$

and

$$\dot{H}+H^2=\dfrac{\ddot{a}}{a}=-\dfrac{4\pi G}{3}\left(\rho+\dfrac{3p}{c^2}\right)$$ Remarks:

1. The Hubble "constant" is not really "constant", it varies along the cosmic history since it is really the rate $H=\dot{a}/a$, where $a(t)$ is the so called scale factor of the Universe in a Robertson-Walker metric.
2. The Hubble parameter squared is related with $G=G_N$, the Newton gravitational constant, throught the relations above. It can also depend on the global "geometry" of the Universe. If the Universe is "flat", $k=0$ and you see that the Hubble parameter squared is proportional to the density of matter-content of the Universe and the proportionality constant is really "constant" if $G$ is constant, as we believe it is.
3. The time variation of $H$ plus $H$ squared gives also a relationship with $\rho$ and the pressure $p$. You need some equation of state for $p$, $p=p(\rho)$, in general.