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The notion that newtonian gravity is incompatible with special relativity is often suggested by declaring the familiar equation $$F_g=\frac{Gm_1m_2}{|\vec{r}_1(t)-\vec{r}_2(t)|^2}$$ and stating that this requires the notion of both particles instantaneously being at certain positions, which doesn't hold up when you allow moving observers to measure time differently.

It is often said that electromagnetism (in the form described by Maxwell's equations) is compatible with special relativity since the equations are symmetric under a Lorentz boost; there exists a reformulation of Maxwell's equations in terms of the field tensor $F_{\mu\nu}$ and the 4-current $(\rho,\vec{J})$. However, we can use Maxwell's equations and the Lorentz force law to find an expression very familiar to the one mentioned above: $$F_e=\frac{k_eq_1q_2}{|\vec{r}_1(t)-\vec{r}_2(t)|^2}.$$ Why doesn't the same special relativity-based objection apply to this equation?


A common answer seems to be that the form of Coulomb's law written above is only valid for static cases. I don't believe that's correct: Coulomb's law accurately describes the electric attraction/repulsion even in a non-relativistic approximation if the charges are moving (or at the minimum, it is mathematically well-defined because a universal time is defined); to find the actual net force, one would simply need to add a magnetic correction, in accordance with the complete Lorentz force equation. I understand that once you do that, you're involving relativity under the hood, even though it looks newtonian, but if that's the case, why can't the Newtonian gravitational force expression be similarly modified using a similar velocity-dependent "correction"?

Qmechanic
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Dispeah
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  • Essentially, Coulomb's force law doesn't apply in non-electrostatic cases. I think the full equations are the monstrous Jefimenko's equations. – Maximal Ideal May 02 '22 at 03:37
  • This is beginning to look a lot like https://physics.stackexchange.com/q/15990/333327; I'm struggling to see how the top answer there addresses the uniform time issue though. – Dispeah May 02 '22 at 03:54
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    "Coulomb's law accurately describes the electric attraction/repulsion even in a non-relativistic approximation if the charges are moving" <-- this is completely false. – Prahar May 02 '22 at 07:24
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    "why can't the Newtonian gravitational force expression be similarly modified using a similar velocity-dependent "correction"?" -- In a sense it can, this is called the post-Newtonian expansion. it's just that the modification is more complicated and leads you to general relativity. – Andrew May 02 '22 at 11:33
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    Coulombs law ≠ maxwells equations. It is a specific solution to maxwells equations. The lorentz transformation is formulated, assuming maxwells equations hold in all inertial frames – jensen paull May 02 '22 at 12:26
  • To add onto prahar, see https://en.m.wikipedia.org/wiki/Li%C3%A9nard%E2%80%93Wiechert_potential even when there is zero acceleration, it is different to coulombs law. – jensen paull May 02 '22 at 12:30
  • @Dispeah Your edit poses a question too different from your original one imo. In any case, take a look at the answer I wrote here just now. – Maximal Ideal May 02 '22 at 17:45

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Probably because the Maxwell equation you furnish holds only if the charges are not moving relative to one another. If they are, then magnetic forces will come into play and the situation gets more complicated.

My way of looking at this is as follows: Maxwell's equations can be used to derive the speed of electromagnetic interactions, and we find this answer does not contain any dependence on the relative motion of the source and the detector. In this sense, Maxwell's equations contain special relativity.

However, the speed with which gravitational interactions are propagated through space (i.e., c) is not extractable from Newton's law of gravity. This means there are truths contained in GR which Newton's gravity cannot furnish us, and hence that Newtonian gravity is incomplete.

niels nielsen
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Why doesn't the same special relativity-based objection apply to this equation?

The same objection doesn’t apply because this formula (Coulomb’s law) is only valid in static situations. The simultaneity issues you mention do not arise in a static scenario.

we can use Maxwell's equations and the Lorentz force law to find an expression very familiar to the one mentioned above

To obtain Coulomb’s law from Maxwell’s equations you must set $\frac{\partial}{\partial t} \vec B =0$ and $\frac{\partial}{\partial t} \vec E =0$ and $\vec J =0$. So it is a special case which is only valid for static scenarios.

You can also derive Coulomb’s law from Gauss’ law and spherical symmetry, but a charge’s field is only spherically symmetric if it is at rest. So again, the static assumption is built in to Coulomb’s law.

Dale
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  • I actually thought that it does work in non-static situations: within a non-relativistic context, one frequently implicitly uses that form of coulomb's law as the electric force part of the Lorentz force equation. For instance, if you considered the problem of two charges moving in empty space, you could consider the electric field instantaneously created by each charge (a radial distribution centered at the time-dependent position) and the magnetic field created by the instantaneous velocity, and then find the action each charge's two fields on the other chage using the lorentz force. – Dispeah May 02 '22 at 03:40
  • @Dispeah Coulomb's law may hold approximately in non-static situations, much like Newtonian gravity approximates general relativity in weak gravity. For an exact solution you need to use Jefimenko's equations, or the Heaviside-Feynman formula for point charges. The use the retarded time in these equations solves the causality issues. – Puk May 02 '22 at 07:00
  • @Dispeah I updated my answer with details – Dale May 02 '22 at 11:32
  • @Dispeah the field due to a uniformly moving charge is not uniform but varies with the angle from the direction of motion. The field due to an accelerating charge does not even point to the instantaneous position of the charge in general - that it does for a uniformly moving charge is an accident. – jacob1729 May 02 '22 at 11:42