Sciama's paper, On the origin of inertia

In his paper "On the origin of inertia", Sciama identifies

$\frac{\Phi + \phi}{c^2} = -\frac{1}{G}$

This identity has confused me because I wonder how the right hand side arises since $\frac{\phi}{c^2}$ is a dimensionless quantity, often seen in red shift formula. I take it he really did mean this formula since he goes on to establish

$G\Phi = -c^2$

(or may be also seen as)

$\phi =- \frac{c^2}{G}$

Taking a look at an application, he gives

$\frac{m}{r^2} = -(\frac{\Phi + \phi}{c^2})\frac{dv}{dt}$

since velocity is $\frac{dx}{dt}$ then $\frac{dv}{dt} = \frac{d^2x}{dt^2}$ which is an acceleration term. The term on the left hand side, if that is all there is (no other additional constants set to 1) is calculated by stating

$F = ma = GM \frac{m}{r^2}$

Divide off $GM$ and it gives

$\frac{a}{G} = \frac{m}{r^2}$

In which we do indeed get an inverse Newtonian constant $G^{-1}$ and an acceleration term $a$ which means $-(\frac{\Phi + \phi}{c^2})$ has to also be defined in units of $-\frac{1}{G}$. This seems to be because of the difference of dimensions contained in the potential known as the Voltage and that usually attributed to the Newtonian gravitational $\phi$.

Leter the dimensions seem to make sense to me. For Sciama relationship:

$\frac{m}{r^2} = \omega^2r$

To be true, must be assuming $G=1$, with it in normal units

$\frac{m}{r^2} = \frac{\omega^2r}{G}$

Sciama uses the gravitational definition of the potential in the following way:

$-\frac{M}{r^2} - \frac{\phi}{c^2} \frac{\partial v}{\partial t}$

and $\phi = \frac{Gm}{r}$ the scalar potential. These dimensions make sense, where Sciama has set Newtons constant to 1. So in regards to

$\frac{\Phi + \phi}{c^2} = -\frac{1}{G}$

how does this arise dimensionally?

$\frac{\phi}{c^2}$ is a dimensionless quantity, often seen in red shift formula
This is usually true when you have something like $\phi = -GM/r$. But for some reason, Sciama chose to leave the factor of $G$ out of his definition of the potential. You can see this in his equation (1): \begin{equation} \Phi = - \int_V \frac{\rho}{r}\, dV. \tag{1} \end{equation} Normally, we would see a factor of $G$ out front. Similarly, when he introduces $\phi$ just below his equation (4), you see that he defines $\phi = -M/r$. Again, we're surprised to not see the $G$. It's not just that he's using geometric units; he uses the actual cgs value of $G$, but he really doesn't have $G$ in these expressions.
I suspect that the reason he takes $G$ out in this way is because — as he mentions in the abstract — his theory enables the amount of matter in the universe to be estimated from a knowledge of the gravitational constant. So the gravitational constant isn't so much a fundamental proportionality constant of the theory like it is in Newton's and Einstein's theories, but a quantity to be measured and related to another (not-so-absolute) property of the universe.
Just for completeness, here's a list of the units of various quantities as Sciama uses them. (He works explicitly in cgs starting in section 4.) \begin{gather} [G] = \frac{\mathrm{cm}^3}{\mathrm{g}\ \mathrm{s}} \\ [c] = \frac{\mathrm{cm}}{\mathrm{s}} \\ [M] = \mathrm{g} \\ [r] = \mathrm{cm} \\ [\phi] = \left[ -\frac{M}{r} \right] = \frac{\mathrm{g}}{\mathrm{cm}} \\ \left[\frac{\phi}{c^2}\right] = \frac{\mathrm{g}\ \mathrm{s}^2}{\mathrm{cm}^3} = \left[ \frac{1}{G} \right] \end{gather} You can see that it does work out so that $\phi/c^2$ is dimensionful and does indeed have the same units as $1/G$.
• Relativists normally use units in which $G=1$. – Ben Crowell Mar 15 '18 at 21:24
• @BenCrowell As a relativist myself, I know that quite well. But Sciama is specifically not using $G=1$. And yet he still poses the potential as $-M/r$. – Mike Mar 16 '18 at 0:30