# Is Newton's universal gravitational constant the inverse of permittivity of mass in vacuum?

Is it possible to consider Newton's universal gravitational constant, $G$, as inverse of vacuum permittivity of mass?

$$\epsilon_m=\frac {1}{4\pi G}$$

if so, then vacuum permeability of mass will be:

$$\mu_m=\frac {1}{\epsilon_m c^2}$$

then Gravitational force will be:

$$F_G=\frac {1}{4\pi \epsilon_m}\frac {m.M}{r^2}$$

Yes, and this formulation is the accepted one in gravitoelectromagnetism, an approximation to general relativity. Just to give you a quick introduction to the theory, gravitoelectromagnetism separates "gravity due to mass" ($T^{00}$, accounting for Newtonian gravity) and the "gravity due to momentum" ($T^{0i}=T^{i0}$) into two separate forces "gravitoelectricty" and "gravitomagnetism" and unify the two much like Maxwell's electromagnetism, so that Newton's gravity only accounts for the "gravitoelectricity".

The results of the analogy are pretty splendid -- in Newtonian gravity, you have Poisson's equation for the "gravitoelectric field", $\nabla\cdot\vec G_E=4\pi G\rho_m$. Compare this to Gauss's law for electricity, $\nabla\cdot\vec E=\frac{\rho_q}{\epsilon_q}$ -- it is thus natural to set:

$$\epsilon_m=\frac{1}{4\pi G}$$

So Poisson's law for Newton's gravity becomes Gauss's law for gravitoelectricity:

$$\nabla\cdot\vec G_E=\frac{\rho_m}{\epsilon_m}$$

And there's the Ampere-Maxwell equation -- $$\nabla \times {\vec B_q} = \mu_q {\vec{J_q}} + \frac{1}{{{c^2}}}\frac{{\partial {\vec{E_q}}}}{{\partial t}}$$

In gravitoelectromagnetism,

$$\nabla \times {{\vec G_B}} = { - \mu_m {\vec{J}_m} + \frac{1}{{{c^2}}}\frac{{\partial {{\vec{G}}_{E}}}}{{\partial t}}}$$

(Sometimes people to add a factor of 4 to multiply the right-hand-side of the gravitoelectromagnetic Ampere-Maxwell equation -- an alternative, however, is to use the form given above and define the gravitoelectromagnetic Lorentz force as $\vec F_g = m(\vec G_E+4\vec v\times\vec G_B)$.)

• The Wiki page you link says $\nabla\cdot\vec{E}_G=-4\pi G\rho_g$, so wouldn't that take care of the negative sign in the ${\rm curl}\vec{B}_G$ equation? – Kyle Kanos Jun 29 '13 at 16:57
• @dimension10 what I am saying is that $\varepsilon_m=-1/4\pi G$, then the minus sign is taken care of and $\mu_g$ remains positive. – Kyle Kanos Jun 29 '13 at 17:21
• I see now, but I would think that (a) forcing g to be negative can be confusing to most who do not assume it is negative and (b) it would be more analogous to E&M if you opted for the more standard notation. – Kyle Kanos Jun 29 '13 at 19:01
• Fair enough. You can do it your way, I can do it mine :D – Kyle Kanos Jun 29 '13 at 19:09
• @Dimension10 Wow! What an amazing link - thanks so much! Quick question: the "eigenmodes" of the gravitoelectromagnetic (GEM) equations will be circularly polarized plane waves - I can't quite see how this fits in with the "quadrupole" polarization of gravitational waves that arises from the conventional treatment of the weak field Einstein equations (WFEE): is there some way to make this aspect of the two theories agree, or are GEM and WFEE essentially different approximations and, if so, what are the differences between the situations that make each good? – Selene Routley Aug 21 '13 at 3:33

Dimension10 answer shows that your idea is applicable to GravitoElectroMagnetism.Ehmm.., go even further, In another viewpoint we can replace vacuum permittivity and permeability of mass with electromagnetism one then Einstein Gravitational Field Equation

$$G_{\mu\nu}=2\mu_m^2\epsilon_mT_{\mu\nu}$$

will be a kind of Curvature due to Electromagnetic Stress-Energy tensor.

$$G_{\mu\nu}=2\mu_0^2\epsilon_0T_{\mu\nu}$$

where $T_{\mu\nu}$ denotes EM stress-energy tensor.

• Huh? Curvature due to EM? Do you mean curvature on the $U(1)$ bundle? – Abhimanyu Pallavi Sudhir Jun 29 '13 at 13:15