2.4.6 The Gravitational Constant
We have shown above that the two processes, Hubble expansion and Newtonian gravitational attraction, are really two aspects of the same basic process. Since Hubble expansion is governed by the Hubble constant, \(H\), and Newtonian gravity is governed by the gravitational constant, \(G\), it is to be expected that these two constants are related to each other in some way.
We consider a region of the energy continuum small enough so that, initially, the inherent curvature of the hyper-sphere may be ignored and this region of interest is essentially flat. We also assume that the density of the energy continuum, \(\rho\) is constant throughout this region, and can be regarded as having an equivalent mass density, \(\mu\), such that \(\rho = \mu c^2\). We imagine a hypothetical sphere in this region and consider the gradient of the continuum with reference to its center point.
If the gradient on the surface of the sphere is negative, then Hubble expansion takes place and the radius of the sphere will tend to expand with a velocity of \(Hr\) and an acceleration of \(H^2 r\). On the other hand, if the gradient on the spherical surface is positive, then Newtonian gravitation applies with a centripetal acceleration of \(MG/r^2\).
We imagine a very thin, spherical shell, of radius, \(r\), and thickness, \(\Delta r\), such that the energy content of this shell is given by \(4\pi r^2 \Delta r \rho\). The mass equivalent of this shell, \(\Delta m\), is given by \(\Delta m =4 \pi \mu r^2 \Delta r\). The question we now ask is how does the gravitational field 'emitted' by this shell affect the shell itself.
Newton showed that for a uniform spherical shell of radius \(r_o\), the gravitational field within the shell, \(r<r_o\), is zero. He also showed that the field outside the shell, \(r>r_o\), is equivalent to that of a point mass of the same value at the center of the sphere. Hence, the gravitational acceleration acting on the spherical shell, due to its own mass, is given by
\begin{align}
acceleration\ at\ radius\ r\ =&\frac{G\Delta m}{r^2} \nonumber \\
=& \frac{G4\pi\mu r^2 \Delta r}{r^2} \nonumber \\
=&4\pi \mu G \Delta r \nonumber
\end{align}
This acceleration is towards the center of the spherical shell, so that this hypothetical sphere will tend to shrink. Opposing this is the Hubble expansion of the universe, which we have shown is exponential and applies to all lengths. Hence, the radius of this sphere will tend to expand, with velocity $Hr$ and acceleration \(H^2r\). In order that the continuum remains stable, these two accelerating processes must cancel out, so that we have the relationship,
\begin{align}
4\pi \mu G \Delta r =& H^2r \nonumber \\
G \Delta r =& \frac{H^2r}{4\pi \mu} \nonumber
\end{align}
For an infinitesimally small radius, \(r = \Delta r\), so that for each point in the continuum, we have the relationship
\begin{align}
G =& \frac{H^2}{4\pi \mu} \nonumber
\end{align}
or in terms of energy density, \(\rho_o\)
\begin{align}
G =& \frac{H^2c^2}{4\pi \rho_o} \nonumber
\end{align}
The same result can be obtained more directly by using Guass's representation of a gravitational acceleration field, \(g\),
\begin{equation}
Divergence (g) = \nabla . g = - 4\pi \mu G \nonumber
\end{equation}
Since gravitational attraction and Hubble expansion are essentially the same process, we can apply Gauss's representation to Hubble acceleration, \(a=H^2 r\). That is
\begin{equation}
Divergence (a) = \nabla . a = + 4\pi \mu G \nonumber
\end{equation}
In this case the sign is positive since Hubble acceleration and gravity act in opposite directions.
The divergence of \(a\) is given by \(\nabla. a = \frac{\partial}{\partial r} H^2 r = H^2 \). Hence we find that
\begin{align}
H^2 =& 4\pi \mu G \nonumber \\
so\ that\ G =& \frac{H^2}{4\pi \mu} \nonumber
\end{align}
and in terms of energy density, \(\rho_o\)
\begin{align}
G =& \frac{H^2c^2}{4\pi \rho_o} \nonumber
\end{align}
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