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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:

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

Abhimanyu Pallavi Sudhir
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    you could surely define a term such as permittivity of mass, but what physical significance would it hold? – udiboy1209 Jun 29 '13 at 12:01
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    @udiboy And what will be the meaning of the permeability of mass also ? Which interest does it has ? Why not making something more poetic like an Ahmad-bility ? – FraSchelle Jun 29 '13 at 12:18
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    @Oaoa,@udiboy, ;: The OP's question is more interesting than you think: http://en.wikipedia.org/wiki/Gravitoelectromagnetism – Abhimanyu Pallavi Sudhir Jun 29 '13 at 12:26
  • Related: http://physics.stackexchange.com/q/944/2451 , http://physics.stackexchange.com/q/15990/2451 , http://physics.stackexchange.com/q/54942/2451 and links therein. – Qmechanic Jun 26 '18 at 10:35

2 Answers2

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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 $\vec F_g = m(\vec G_E+4\vec v\times\vec G_B)$.)

Abhimanyu Pallavi Sudhir
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  • 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
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    @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 03:33
  • @WetSavannaAnimalakaRodVance That should be its own question! – Michael Aug 21 '13 at 03:50
  • @MichaelBrown Okay, shall ask it in the next day, after I've had a chance to do some reading. – Selene Routley Aug 21 '13 at 05:22
  • @MichaelBrown Actually, I'd like to hold off, unless what you mean is that you'd like to give an answer, in which case I will, but at the moment I've got my head into a few papers on GEM (which I have never heard of) e.g. this one and this one and so (1) I may well answer my own questions and (2) am so overwhelmed and excited by these articles right now that I don't think I can ask a cogent question! – Selene Routley Aug 22 '13 at 07:50
  • @Dimension10 Do you recommend any overview papers reviewing GEM? Aside from this one and this one (what do you think of these, BTM?) – Selene Routley Aug 22 '13 at 07:52
  • @WetSavannaAnimalakaRodVance I actually don't know a whole lot about GEM, so I'd love to see an answer, rather than give one. Sadly I don't have the time to study it in depth right now though. Good luck navigating the literature. :) – Michael Aug 22 '13 at 10:24
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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.

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