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?


1 Answer 1


$\frac{\phi}{c^2}$ 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 $1/G$.

  • 1
    $\begingroup$ Relativists normally use units in which $G=1$. $\endgroup$
    – user4552
    Mar 15, 2018 at 21:24
  • $\begingroup$ Yeah Ben, they do normally use those units, but Sciama is not clear as he jumps from not using it to using it, then there was also the question of how the inverse G arises from the construction of phi/c^2 $\endgroup$ Mar 15, 2018 at 21:45
  • $\begingroup$ @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 $-M/r$. $\endgroup$
    – Mike
    Mar 16, 2018 at 0:30
  • $\begingroup$ Still true, except Sciama is using a linearized form of gravity, which is why he wrote it this way. $\endgroup$ Jun 11, 2019 at 1:56

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