# Is Newton's third law merely a consequence of the laws of universal gravitation and Coulomb?

Can it be said that Newton's third law is simply the fact that gravity and electromagnetism do obey an action/reaction principle (as per $$\vec{f}_{grav,12}=G\frac{mm'}{r^2}\vec{e}_{12}$$ and $$\vec{f}_{elec,12}=\frac{qq'}{4\pi\varepsilon_0}\vec{e}_{12}$$ which are experimental facts, at least in classical physics), and that at the all the forces that we consider in Newtonian mechanics (if one rules out magnetic and time-dependant electric phenomena) originate from these two? This would certainly explain why the third law fails for magnetic forces (see Newton's Third Law Exceptions?). In this picture the 'correct third law' would be the conservation of momentum, field momentum included, from which one could retrieve action/reaction under certain hypotheses (in the static regim for instance).

• It's the other way around. Newton explicitly formulated his theory of gravitation so that it would be consistent with conservation of momentum (Newton's third law). Coulomb did the same 100 years later when he formulated what is now called Coulomb's law. – David Hammen Jun 30 '19 at 19:39
• @DavidHammen That is interesting to know, but if I'm not mistaken the heuristic that they used to find a possible law has no bearing on the logic of the theory, and thus this doesn't answer my question (the formulas are valid because they are verified by experiment, not because they were derived using, among other things, Newton's third law). – user235800 Jul 1 '19 at 13:19

• @my2cts I'm aware that the statement 'in a closed system momentum is conserved' is sometimes taken to be that of Newton's third law. However I happen to have come across the following reasoning to prove conservation of angular momentum for a system $\Sigma$ of $N$ ponctual objects : take two points $1$ and $2$. By Newton's third law, we have $\vec{f}_{12}=f\vec{e}_{12}=-f\vec{e}_{21}=-\vec{f}_{21}$ (none of the precedent equalities are obvious), thus, if we call $\Gamma_{\{1,2\}}$ the torque exerted on $\Sigma$ due to the interaction of $1$ and $2$, we have $\Gamma_{\{1,2\}}=0$. – user235800 Jul 1 '19 at 13:28