# Similarity between the Coulomb force and Newton's gravitational force

Coulomb force and gravitational force has the same governing equation. So they should be same in nature. A moving electric charge creates magnetic field, so a moving mass should create some force which will be analogous to magnetic force.

• See the notion of gravitoelectromagnetism, cf. e.g. Wikipedia. – Qmechanic Feb 24 '13 at 9:53
• More on gravitoelectromagnetism. Related: physics.stackexchange.com/q/944/2451 , physics.stackexchange.com/q/15990/2451 and links therein. – Qmechanic Feb 24 '13 at 10:07
• It's always possible in situations like this that the reason for same equations is same underlying nature, and physicists are always on the lookout for scenarios like that. But it's very specious reasoning to jump to that conclusion automatically, as is the case here. The two equations are only approximately valid, and it's entirely coincidental that the first order approximation of one equals that of the other. The same happens with ideal springs & pendulums, which obey the same eq.'s IF you use the small-angle approximation for the pendulum. – David H Apr 27 '13 at 8:02

It's a good observation that the electric and gravitational fields both satisfy Poisson's equation $$\nabla^2\Phi_G = 4\pi\rho_G, \qquad \nabla^2\Phi_E = -\frac{\rho_E}{\epsilon_0}$$ where $\Phi_G, \Phi_E$ are the gravitational and electric potentials and $\rho_G,\rho_E$ are the mass and charge densities. It would seem from the perspective of Newtonian gravitation that this is where the analogy stops. The magnetic field in electrodynamics arises due to moving charges because of the following Maxwell equation $$\nabla\times\mathbf B - \frac{1}{c^2}\frac{\partial \mathbf E}{\partial t} = \mu_0 \mathbf J$$ There are no such analogous equations for moving masses in Newtonian gravitation.

However using general relativity, and specifically Einstein's field equations, one can show in certain situations (where gravity is weak and the spacetime is reasonably flat), that gravity does in fact behave like electrodynamics. In fact, there are fields called gravitoelectric and gravitomagnetic fields that obey analogous equations to the Maxwell equations. The result is what is aptly called gravitoelectromagnetism (GEM).

This http://arxiv.org/pdf/gr-qc/0311030.pdf seems like a pretty nice overview.

• Of course, it should be said that <b>any</b> Lorentz-invariant theory of gravity will have gravetomagnetic effects, by the standard argument used to show why the magnetic field is necessary. – Jerry Schirmer May 12 '13 at 2:18

To add to the other responses, you can't generally represent gravity with Maxwell-esque equations for two very fundamental reasons: (1) gravity's field equations are non-linear, and (2) the gravitational field couples to a rank-2 tensor.

Maxwell's equations are linear, meaning the superposition principle applies and the field doesn't interact with itself. The Einstein field equation, however, are nonlinear, so the superposition principle does not apply and the field will self-interact.

The electromagnetic field couples to charge density and current density, which can be written as a Lorentz-invariant four-vector (a rank-1 tensor). The metric (gravitational) field couples to the stress-energy tensor, which is a Lorentz-invariant rank-2 tensor. This is why photons have spin-1, while theoretical gravitons have spin-2. Since the Maxwell-esque gravitomagnetic field equations involve only four components of the stress-energy tensor (mass-density and mass-current), they cannot possibly be Lorentz-invariant. Thus, a simple change in reference frame will lead to predictions that contradict experiment.

• Great comments +1. – joshphysics Feb 24 '13 at 19:32

In an inertial reference frame at rest to a charge, one only sees a Coulomb field. When transformed into another inertial reference frame in motion relative to the charge, one gets a magnetic field. The detailed transformation formula can be seen at this wiki page.

Now the question is, can we do the same for gravity (assume we are only in the regime of special relativity and pretend general relativity does not exist)?

The answer is "no" because of an essential difference between charge and mass: the charge is an invariant of Lorentz transformation, which means different inertial reference frames see the same amount of charge, while the mass is not. We don't know (do we?) what the "Coulomb" part of the gravitational force should be like when the mass is moving. Should we use the rest mass, or the "inflated mass" ($m_0/\sqrt{1-v^2/c^2}$)? And which distance should we put into the Coulomb's law? The delayed one or the instantaneous one? Even if we try to mimic in every aspect of the Coulomb force, namely, we use $m_0$ and the delayed distance, and arrive at a set of transformation rules for the "electromagnetic-like" gravitational force, the theory is not self-consistent, simply because the mass density and mass flux together do not form a relativistic four-vector, while the charge density and charge flux do. Intuitively speaking, the difference is, energy can create rest mass, while nothing can create a net charge.

Update: I just noted that my argument that (mass density, mass flux) does not form a four-vector while the (charge density, charge flux) does is kind of equivalent to the answer of joshphysics that the Maxwell equation involving $\mathrm{J}$ does not exist for gravity.

One may get something similar to electromagnetism in the framework of general relativity (as pointed out in the comments), but to me that's another story...

I kind of guess (but not sure because I don't know what really happened to Einstein during 1905-$\sim$1915; correct me if I am wrong) what I talked about above might be one reason for Einstein to abandon such an "apparent" approach to accomodate gravity into the framework of special relativity, and to choose the much more indirect, nontrivial, and revolutionary "geometrization" approach, namely, the theory of general relativity.