In the Wikipedia article Classical Field Theory (Gravitation), it says

After Newtonian gravitation was found to be inconsistent with special relativity, . . .

I don't see how Newtonian gravitation itself is inconsistent with special relativity. After all, Newton's Universal Law of Gravitation is completely analogous to Coulomb's Law, so it would seem that, if there were an analogous "gravitational magnetic field", one could formulate a theory of gravitation in exact analogy with Maxwell's Theory of Electromagnetism, and of course, this would automatically be consistent with Special Relativity.

What about this approach to gravitation does not work? The only problem I could see with this is the lack of evidence for a "gravitational magnetic field". That being said, gravity is "weak" as it is, and my guess is that it would be extremely difficult to set up an experiment in which a massive body were moving fast enough for this "magnetic field effect" to be observable.

EDIT: As has been pointed out to me in the answers, Newton itself is inconsistent with SR.
Similarly, however, so is Coulomb's Law, yet electromagnetism is still consistent with SR. Thus, how do we know that it is not the case that Newton's Law is the special "static" case of a more general gravitomagnetic theory exactly analogous to Maxwell's Theory?

Let $\mathbf{G}$ be the gravitation field, let $\rho$ be the mass density, and $\mathbf{J}$ be the mass current density, let $\gamma _0$ be defined so that $\frac{1}{4\pi \gamma _0}=G$, the Universal Gravitation Constant, let $\nu _0$ be defined so that $\frac{1}{\sqrt{\gamma _0\nu _0}}=c$, and suppose there exists a field $\mathbf{M}$, the gravitomagnetic field, so that the following equations hold: $$ \vec{\nabla}\cdot \mathbf{G}=-\frac{\rho}{\gamma _0} $$ $$ \vec{\nabla}\cdot \mathbf{M}=0 $$ $$ \vec{\nabla}\times \mathbf{G}=-\frac{\partial \mathbf{M}}{\partial t} $$ $$ \vec{\nabla}\times \mathbf{M}=\nu _0\mathbf{J}+\gamma _0\nu _0\frac{\partial \mathbf{G}}{\partial t} $$ where these fields would produce on a mass $m$ moving with velocity $\mathbf{v}$ in our inertial frame the Lorentz force $\mathbf{F}=m\left( \mathbf{G}+\mathbf{v}\times \mathbf{M}\right)$. This theory would automatically be consistent with SR and would reduce to Newton's Law of Gravitation for the case of gravitostatics: $\frac{\partial \mathbf{G}}{\partial t}=\frac{\partial \mathbf{M}}{\partial t}=\mathbf{J}=0$. (To be clear, you can't just set the time derivatives equal to $\mathbf{0}$ as it seems I have done. Upon doing so, you obtain the corresponding static theory, which is technically incorrect (as you can easily see because it won't be relativistically invariant), but is nevertheless often a useful approximation.)

My question can thus be phrased as: without appealing directly to GR, what is wrong with this theory?

  • 4
    $\begingroup$ For gravitoelectromagnetism, see also Wikipedia. $\endgroup$
    – Qmechanic
    Oct 21, 2011 at 10:22
  • 5
    $\begingroup$ If you ever get the chance to read Misner, Thorne and Wheeler please do. They devote a fair amount of space to building up successively better relativistic theories of gravity, arriving ultimately at GR. But they diagnose the pathologies of the other theories at every step along the way. I'm not sure from memory if your proposal is in there, but there's a good chance you'll find some useful discussion. $\endgroup$
    – Michael
    Mar 29, 2013 at 14:43
  • 3
    $\begingroup$ Your equations are inconsistent with matter conservation. Take the div of the last equation. You'll get a continuity equation with the wrong sign, which implies either statics or the explosive generation of matter. If you change the sign of the current then you get a compatible system, but it's equivalent to Maxwell EM where like charges repel instead of attract. $\endgroup$
    – Michael
    Mar 29, 2013 at 16:32
  • 1
    $\begingroup$ See physicsforums.com/showthread.php?p=4024260 $\endgroup$
    – user4552
    Mar 29, 2013 at 20:19
  • $\begingroup$ Its gravitoelectromagnetism you' are talking a bout, but it isn't Lorentz Invariants. $\endgroup$ Jul 19, 2013 at 6:28

10 Answers 10


Newtonian gravitation is just the statement that the gravitational force between two objects obeys an inverse-square distance law, is proportional to the masses and is directed along the line that joins them. As such, it implies that the interaction between the objects is transmitted instantaneously and it must be inconsistent with special relativity (SR).

If say the Sun suddenly started moving away from the Earth at a speed very close to the speed of light, SR tells you that the Earth must still move as if the Sun were in its old position until about 8 minutes after it started moving. In contrast, Newtonian gravitation would predict an instantaneous deviation of Earth from its old orbit.

What you have discovered in your reasoning is that indeed, Coulomb's Law is NOT relativistically invariant either. But Maxwell electromagnetism is not Coulomb's Law.

As a matter of fact, Coulomb's Law is deduced from Maxwell equations as a particular case. The assumptions are those of electrostatics, namely that the magnetic field is zero and that the electric field is constant in time. These assumptions lead to the Coulomb field but they are NOT consistent with SR in the sense that they can not be valid in every reference frame since if the electric field is constant in a reference frame, then there exists another frame in which it will be varying and the magnetic field will be differnent from zero. For more you can start reading this. Maxwell's electromagnetism IS consistent with SR since the full Maxwell's equations apply in all reference frames, no matter whether the particle is moving or not.

General Relativity is the analogous for gravity of Maxwell's electromagnetism and, as it has already been said, it leads to equations for the gravitational field (the metric) analogous to those of Maxwell. Thus, it is not strange that something that resembles gravitational magnetism should appear.

  • 2
    $\begingroup$ Great answer. I would only add that Newtonian Gravity also assumes all of Newtonian Mechanics which is definitely not consistent with Einstein's special relativity mechanics. In particular, Newtonian Mechanics and Gravity both assume absolute space and time which is not consistent with Einstein's relative space and time. $\endgroup$
    – FrankH
    Oct 21, 2011 at 15:44
  • $\begingroup$ THe OP's proposal avoids this problem, because those field equations are only valid for retarded potentials. $\endgroup$ Mar 29, 2013 at 14:52
  • 2
    $\begingroup$ This is not a complete answer. The original author already realizes that Newtonian gravity with instant action is as inconsistent as Coulomb's law with instant action. The question is why should a direct analogy of Maxwell's equations for gravitation not be the exactly correct extension of gravitation to the relativistic case as the Maxwell's equations are for EM? $\endgroup$ Mar 18, 2018 at 4:32
  • 2
    $\begingroup$ Other ways to phrase this may be: why is the correct generalization of gravity necessarily nonlinear, why is gravitation necessarily a tensor field instead of a vector field, and why is a Maxwell-like analogy valid in weak field limit only. The given post provides no answers to these questions as is. $\endgroup$ Mar 18, 2018 at 4:32

The reason your theory fails is because like charges repel, while like masses attract. You are using a spin-1 theory, and to get attraction, you need spin 0 or spin 2. The spin 0 theory of gravity leads to no light scattering, and it is called Nordstrom gravity. The spin-2 theory is GR, and it has more "magnetic" fields than just EM.

  • 3
    $\begingroup$ Can't you easily remedy this by setting $\frac{1}{4\pi \gamma _0}=-G$ rather than $\frac{1}{4\pi \gamma _0}=G$ as Jonathan Gleason originally wrote? Making this constant negative would ensure that in the "electrostatic" case, like charges would attract rather than repel. $\endgroup$
    – Hypnosifl
    Mar 25, 2016 at 7:08

There is gravitational magnetic field, if you move through a static field very fast. Google for gravitomagnetism.

The main reason that general relativity is not the same is because the equivalent Gauss' law does not hold. General relativity is non-linear -- gravitational fields have energy and so act as sources for more field.

  • $\begingroup$ +1 Very interesting, I didn't know anything about this. Funny to see there is a "gravitoelectric" charge but no "gravitomagnetic" one in the equations. $\endgroup$
    – Heidar
    Oct 21, 2011 at 11:57
  • 2
    $\begingroup$ @Heidar: there are no "magnetic" charges in electrodynamics either. A small magnet, like the moment on fundamental particles, can be modelled as an orbiting charge. Similarly, rotating masses will give off gravitomagnetic fields. $\endgroup$
    – genneth
    Oct 21, 2011 at 12:09
  • $\begingroup$ Let $\mathbf{G}$ be the gravitational field, let $\gamma _0$ be defined so that $\frac{1}{4\pi \gamma _0}=G$, the Universal Gravitational Constant, and let $\rho$ be the mass density. Then, the equivalent of Gauss's Law would read $\nabla \cdot \mathbf{G}=\frac{\rho}{\gamma _0}$. You say that this does not hold. Why? Is there experimental evidence that contradicts this? $\endgroup$ Oct 21, 2011 at 14:11
  • $\begingroup$ @JonathanGleason: consider two point masses inside your (space-like) surface. These two masses have some mass, but the binding energy due to gravitation modifies this mass. Now try and make this self-consistent, and you find yourself contorting to define "local energy", which is known to be difficult in GR (Google for energy-momentum pseudo-tensors). $\endgroup$
    – genneth
    Oct 21, 2011 at 16:48
  • $\begingroup$ @genneth But isn't this based on the assumption of the equivalence of gravitation mass and inertial mass? It should modify their inertial mass, but if these two concepts are in fact distinct, then there is no contradiction. How much evidence is there supporting their equivalence? Are there any compelling theoretical reasons for their equivalence? In my view, their equivalence, if it is true, is the single most compelling reason for gravitation and electromagnetism to not be formulated on the same footing. $\endgroup$ Oct 21, 2011 at 16:51

It is a very interesting question that you pose and indeed that is the spirit of being a physicist. As a matter of fact there are many things wrong with that new theory you wrote and they can all be summarized by saying, 'your theory is in flat contradiction with experiment' which, of course, is what is wrong with every wrong theory.

For instance, without leaving electrostatics, your theory predicts that a static point mass gives rise to a $1/r^2$ gravitational field, that is to say, a $1/r$ gravitational potential. Therefore, your theory predicts Keplerian orbits and this we know to be not true. In the correct theory of gravitation, General Relativity, the gravitational potential of a point mass turns out to be (loosely speaking) $1/r$ plus some correction terms that go like $1/r^2$ and $1/r^3$. These terms are admittedly proportional to the probe particle's angular momentum but a particle not moving radially won't describe a conic.

The important point is that this has been beautifully confirmed by measuring to a high precission the orbits of Mercury! So, your theory must be wrong.

More importantly, your theory couples the "gravitomagnetic field" with the mass current. Therefore, classically, your theory has no effect on anything massless so your theory can't affect photons! This is again in flat contradiction with photon deflection by large masses. You might try to remedy this by coupling the theory to an energy current, for instance, instead of $\rho$ being the mass density you could take the energy density. I would need to check it but I think you would still get the wrong deflection.

Up to what extent your theory resembles gravity can be answered accurately but it would take some time. A bit more technical, your theory is actually a $U(1)$ gauge theory of gravity. Probably someone has thought of this before. One should start from the Lagrangian for both your theory and GR and find what relations may exist between your four-potential and the metric.

  • 3
    $\begingroup$ This theory does not predict Keplerian orbits, because the gravitomagnetic field and radiation reaction would create spin-orbit effects. I doubt it will agree with post-newtonian gravity, but this reasoning isn't why. $\endgroup$ Mar 29, 2013 at 14:54

I want to address in more detail why we can't take any obvious modified form of Maxwell's equations and assume it applies to the gravitational field $\vec{g}$ along with a hypothetical field $\vec{m}$ that accompanies $\vec{g}$ the same way the magnetic field accompanies the electric field, with gravitational mass as charge. There are valid reasons involving the equivalence principle, the precession of Mercury's orbit, or deflection of light by gravity, but I want to try to spell out something different to all that. I haven't seen anyone spell this out explicitly as well, so I hope anyone reading this can give a critical look and tell me if there are any mistakes.

To put it shortly, the most straightforward repackaging of Maxwell's theory with attracting like charges is incompatible with a stable universe. Such a theory would fail to have positive-definite energy and it would have catastrophic run-away solutions. My post is based on this other post, although here I will be slightly more general to accommodate OP's possibility. I apologize if some parts in the argument look a bit handwavy.

Suppose we have fields $\vec{E}$, $\vec{B}$ such that \begin{align*} \nabla\cdot\vec{E} &= s_{1}\frac{\rho}{\epsilon_{0}},\qquad\qquad\qquad && \nabla\times\vec{E} = s_{2}\partial_{t}\vec{B}, \\ \nabla\cdot\vec{B} &= 0, && \nabla\times\vec{B} = s_{3}\mu_{0}\vec{J} + s_{4}\mu_{0}\epsilon_{0}\partial_{t}\vec{E} \end{align*} where $\epsilon_{0}, \mu_{0}$ are positive constants and $s_{1}, s_{2}, s_{3}, s_{4}\in\{-1, +1\}$, and suppose charges respond to the fields via the Lorentz force law $\vec{F} = q\vec{E} + q(\vec{v}\times\vec{B})$.

Some Notes.

  • If we want to consider a modified version of the Lorentz force law with minus signs (e.g. suppose $\vec{F} = -q\vec{E} + q(\vec{v}\times\vec{B})$), we can redefine the E- and B-fields so that the minus signs in the force law are transferred to $s_{1}, s_{2}, s_{3}, s_{4}$ accordingly. Thus it's safe to take the original Lorentz force law for granted.
  • As you can guess, in a hypothetical theory of gravity $\vec{E}$ would serve as the gravitational field $\vec{g}$, $s_{1}$ would be $-1$, and $\epsilon_{0}$ would be $1/4\pi G$ where $G$ is Newton's gravitational constant. In this theory, charges stand for gravitational masses.

We proceed as follows:

  1. By applying divergence to the fourth equation (lower right equation), then substituting the first equation (upper left equation) in, and then dividing out by $\mu_{0}$, we obtain $$ 0 = s_{3}\nabla\cdot\vec{J} + s_{4}s_{1}\partial_{t}\rho. $$ Now this looks a lot like the charge conservation equation (or the continuity equation). It is possible that gravitational mass is not conserved (after all we know it is not conserved by the equivalence principle and mass-energy equivalence). However, if the two terms in the continuity equation had opposite signs, it would render the equation as nonsensical, because if positive charge were accumulating in one location ($\nabla\cdot\vec{J} < 0$), that would cause charge to decrease in that location ($\partial_{t}\rho < 0$). In order for the equation to be sensible, we must have $s_{3} = s_{4}s_{1}$.

  2. Consider the vacuum solutions with $\rho=0$ and $\vec{J} = 0$. By applying curl to the third equation (upper right equation) and substituting the fourth equation (lower right equation), we get $$ \nabla\times(\nabla\times\vec{E}) = s_{2}s_{4}\mu_{0}\epsilon_{0}\partial_{t}^{2}\vec{E}. $$ The LHS is $\nabla(\nabla\cdot\vec{E}) - \nabla^{2}\vec{E}$, and by assuming no charges are present, it reduces to $-\nabla^{2}\vec{E}$. Then we get $$ - \nabla^{2}\vec{E} = s_{2}s_{4}\mu_{0}\epsilon_{0}\partial_{t}^{2}\vec{E}. $$ Now if $s_{2}s_{4} = 1$, then if any component of $\vec{E}$ is concave down (or concave up) at any point in space, the component will be driven up (respectively down), and loosely speaking this leads to a further exaggeration of the concavity. This leads to a run-away solution where components of $\vec{E}$ can send themselves to $\pm\infty$ at any point in space. This is one of the instabilities that renders the field theory not sensible. Thus we are forced to conclude $$ s_{2}s_{4} = -1. $$

  3. Consider some charge going in a circle. In the case of electrodynamics, this is natural to consider when you have a loop of current. In the case of gravity, this is natural to consider when you have a rotating planet. Suppose we have charge going counterclockwise "from the top view." If $s_{3} = -1$, then this generates a B-field such that it goes into the loop from the top. If furthermore $s_{2} = -1$, then this in turn creates an E-field going counterclockwise from the top view. Since $\vec{F} = q\vec{E}$, this drives charge to go counterclockwise faster, which in turns creates more of the B-field, and this creates a run-away feedback that renders the loop of current or, in the case of gravity, the rotating planet as unstable. This is again an instability that renders the field theory as not sensible. Similar reasoning shows that the same thing happens if $s_{3} = s_{2} = +1$. Thus we are forced to conclude that $s_{2}$ and $s_{3}$ have opposite signs and so $$ s_{2}s_{3} = -1. $$

So far we've shown that if gravity is to follow Maxwell's equations with modified signs (and modified constants), then the conclusions of #1, #2, and #3 must hold. But now consider the following. By #2 and #3, we have $s_{3} = -s_{2} = s_{4}$. By #1, we have $s_{1}s_{3}s_{4} = +1$. Since $s_{3}$ and $s_{4}$ are equal (as $\pm 1$), it follows that $s_{3}s_{4} = +1$. Hence $s_{1} = +1$.

Hence when we go back to the modified Maxwell equations, we see that the first equation must have a positive sign $s_{1} = +1$. But in the case of statics, this implies that the corresponding force law $$ \vec{F} = s_{1}\frac{q_{1}q_{2}}{4\pi\epsilon_{0}r^{2}} \hat{r} $$ must have $s_{1}/(4\pi\epsilon_{0}) > 0$, meaning like charges repel. Since in the case of gravity we must have attracting like charges, we see that the modified Maxwell equations cannot model gravity.

The only way out is to drop the assumptions going into #1, #2, or #3, which would render the theory as not sensible. Thus, the example theory that OP proposed would not be physically sensible (to be more specific it would lead to a universe with unstable run-away solutions).


In addition to Ron Maimon's correct assertion that this theory will give rise to attractive and repulsive forces (the latter of which is not observed), this theory also does not predict gravitational lensing, and of the different geodesics travelled by null and timelike particles. Thus, it cannot be a correct theory of gravity.

  • $\begingroup$ See my comment to Ron Maimon, if you make Coulomb's constant be a negative number and also assume all gravitational charges are positive, I think you should only have attractive forces in the "electrostatic" case. The point about lensing would rule the theory out, as would observations about the precession of Mercury's orbit, but Einstein was motivated to come up with GR before lensing observations and possibly before the precession was measured precisely enough to rule out other explanations like dust in the neighborhood of the Sun causing drag, so I wonder why he ruled out this theory. $\endgroup$
    – Hypnosifl
    Mar 25, 2016 at 7:14

Somewhat cautiously, since it seems that some of the posters here know more about « gravito-magentronic-whatevers » than I do, I want to answer with a few comments. Even while Einstein was developing GR, a rival non-tensorial, scalar general relativistic theory of gravity was developed, but predicted different answers for the typical GR effects and so is wrong.

The equality of inertial mass and gravitational mass is also a feature of Newtonian theory, as an assumption. In GR it is a consequence, not an assumption. This is a theoretical superiority. But even as to the empirics, this equality is very well established by experiment.

Minkowski and others did make the minimal modifications to Newton's theory needed to make gravity compatible with Special Relativity. But, after all, Special Relativity is wrong: the speed of light in vacuum is not constant. General Relativity corrects Special Relativity in this regard.

Finally, having a principle of relativity only for linear transformations is philosophically unsatisfying and so making gravitation compatible with Special Relativity is a waste of time... anyway the answer to the OP is yes, but we don't want that kind of gravity anyway...


Your basic idea is sound, even if the details need work. Newton's theory needs to be made Lorentz-covariant, but that can be done. The result is a theory that treats gravity as a potential in flat space-time and agrees with the weak-field measurements (precession, light bending, spectral shift) that have provided confirmation of the general theory. See:

Biswas, Special Relativistic Newtonian Gravity. Foundations of Physics 24(4), 1994, pp. 513 - 524.

Surely a Galilean-covariant theory could be made, analogous to Galilean electromagnetism (see: Bellac and Levy-Leblond, Galilean Electromagnetism. Il Nuovo Cimento 14B (2), 217-233), but that doesn't seem to support radiation.


It's the Doppler equations

Einstein had an argument (paraphrased from memory), that if you drop a flashlight from a height, and have it switch on just before it hits, the light emitted immediately before the impact is blueshifted by the flashlight's downward velocity. However, if the flashlight is instead switched on briefly at the top of its fall, while the torch is still effectively stationary, and the same lightenergy is instead blueshifted by the gravitational blueshift effect.

If the energy crossing the same gravitational differential undergoes the same change, regardless of whether it travels as free light or as chemical energy in the flashlight battery, the gravitational shift on light crossing a gravitational differential with terminal velocity v, can be calculated from the conventional motion shift of a body moving towards or away from the observer at v.

This is where the Newtonian and Einsteinian calculations start to diverge.

In a system based on the Newtonian relationships, the recession redshift (and therefore the gravitational redshift for light climbing out of a gravitywell), is E'/E = (c-v)/c

Under an SR-based system like Einstein's 1916 theory, the recession redshift is instead E'/E = sqrt[(c-v)/(c+v)]

Both systems generate a gravitational horizon at r=2Mg/c2 (commonly abbreviated "r=2M"), but some of the other properties of these horizons are different.

With the Newtonian horizon-bounded object, known as a dark star (described by John Michell in 1783), the surface is the surface at which the escape velocity equals the velocity of light. Although there are no outward unaccelerated ballistic trajectories that totally escape the surface, light and subluminal particles and bodies can briefly venture outside r=2M, for a limited distance and limited time before being inevitably turned back by gravity. While on the outside, they can be knocked free by collisions with passing matter (or each other), and escape along partially accelerated paths.

So the Newtonian object is dark, but not black. Its horizon is an "effective", projected, observer-centric horizon that fluctuates and radiates and leaks massenergy and information, in a manner statistically reminiscent of Hawking radiation under quantum mechanics (Thorne, 1994).

With the SR- and GR1916-based horizon-bounded object, championed by John Wheeler and known as a black hole, the horizon is totally black. The change to the SR Doppler equations means that for a body to even be momentarily stationary at the horizon forces it to confront infalling radiation that is infinitely gravitationally blueshifted. It faces an infinite inward radiation pressure and an infinite temperature, the blueshift gets even worse if it attempts to accelerate outwards, and even if it did somehow magically escape in any more than zero local time, an infinite amount of outsider-time would already have elapsed. (paper)

So with the "new" equations, nothing can even hover at the horizon, much less escape, the black hole is provably totally inescapable, and no event within r=2M can send a signal to the outside, by any means. The horizon is absolute, and is an event horizon.


The most basic interaction laws of Nature are the Coulomb and Newton Laws. They can be used safely in a universe where there is no motion.
The electrostatic and the gravitational fields are set in space by the particles. The fields propagate away of them at 'c' speed, and act at distance in a delayed way.

As they are forces the bodies are attracted/repeled and motion do happen.
Due to the finite speed of propagation of the field, different observers see a different reality dependent on their relative motion.

One must conclude that there is no such entity as a magnetic 'field' but only a magnetic 'force' and is perceived in the same sense as the Coriollis force is -Hand de Vries, 2008- and 'The Magnetic Force Between Two Currents Explained Using Only Coulomb's Law', Jan Olof Jonson, 1997, Chinese Journal of Physics ).

With the motion within a gravitational field happens the same as with the electrostatic field.

Compare your equations with those already present in WP (the online book 'motionmountain' has a chapter dedicated to the subject).

The sentence 'After Newtonian gravitation was found to be inconsistent with special relativity, . . .' is misleading.

Gerber treated the advancement of Mercury's perihelium (18 years before GR) using Newton gravity but paying attention to the delay of the propagation of the field (see WP Liénard–Wiechert potentials detailed in the 2nd chapter of the online book of Vries). His explanation of his derivation was incomplete (discussion at mathpages) and the solution rested in peace until Walter Orlov, 2011, explained why the Gerber solution is correct (and fit the observations). This issue probably will remain death, until the community ...

In short: Nothing is wrong with your line of reasoning. Gravity and Electromagnetism must be treated in the same way. The differences are minimal:
1 - scale - A pair of parallel electromagnetic radiating elements in opposition of phase and at a small distance emulates the small scale of gravity.
2 - always atractive - pairs of dipoles always atract. Because this force is proportional to $1/r^3$ I will assume that the propagating medium is of the type of 'Polarizable Vacuum' and the integration all along the path gives the usual $1/r^2$ force.

The Newton position was not favourable to Action at a distance. Because he did not know the why ("hypoteses non fingo"). Quoting from : Action-at-a-Distance and Local Action in Gravitation: Discussion and Possible Solution of the Dilemma

clearly expressed in a letter to Bentley in 1693 (Newton 1958, pp. 302 sq; Cohen 1980, p. 117): “that one body may act upon another at a distance through a vacuum without a mediation of any thing else by or through which their action of force may be conveyed from one to another is to me so great an absurdity that I believe no man who has in philosophical matters any competent faculty of thinking can ever fall into it”. Principia was a mathematical treatise of mechanics, and not, as Newton carefully pointed out, a physical treatise (Cohen 1980, Section 3). His physical ideas were published posthumously as A Treatise of the System of the World (1728) and were presented more openly in his letters. His physical viewpoint remained much less known in the scientific community than the mathematical treatise of Principia.

It can be shown that there is no anomaly in the Mercury's orbit and there is no need of Einstein's GR neither of Gerber. Newton alone is 100% right, provided that there is not an instant AAAD.


Not the answer you're looking for? Browse other questions tagged or ask your own question.