Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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 completley 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 obsevable.

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 theses 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?

share|improve this question
1  
what's a gravitational magnetic field?? –  Vineet Menon Oct 21 '11 at 7:06
4  
For gravitomagnetism, see also Wikipedia en.wikipedia.org/wiki/Gravitomagnetism and physics.stackexchange.com/q/11096/2451 –  Qmechanic Oct 21 '11 at 10:22
    
Is there any time dependence in Newtons or Coulombs laws? –  Georg Oct 21 '11 at 11:06
2  
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. –  Michael Brown Mar 29 '13 at 14:43
    
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. –  Michael Brown Mar 29 '13 at 16:32

7 Answers 7

up vote 13 down vote accepted

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.

share|improve this answer
1  
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. –  FrankH Oct 21 '11 at 15:44
    
THe OP's proposal avoids this problem, because those field equations are only valid for retarded potentials. –  Jerry Schirmer Mar 29 '13 at 14:52

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.

share|improve this answer

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.

share|improve this answer

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.

share|improve this answer

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...

share|improve this answer

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.

share|improve this answer
1  
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. –  Jerry Schirmer Mar 29 '13 at 14:54

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.

share|improve this answer
    
+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. –  Heidar Oct 21 '11 at 11:57
1  
@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. –  genneth Oct 21 '11 at 12:09
    
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? –  Jonathan Gleason Oct 21 '11 at 14:11
    
@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). –  genneth Oct 21 '11 at 16:48
    
@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. –  Jonathan Gleason Oct 21 '11 at 16:51

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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