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).
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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
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@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). – Jul 01 '19 at 13:19
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Newton's third law expresses conservation of momentum and is of general validity. It only requires that the system is translation invariant. Force is defined as rate of change of momentum, thus the sum of all forces is zero. An exception is the magnetic force, which is not the derivative of momentum.
my2cts
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An exception is the magnetic force, which is not the derivative of momentum. Can you expand on this point? Does this mean Newton's second law doesn't apply for magnetic forces? – BioPhysicist Jun 30 '19 at 22:44
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The second law must expanded anyway even for electrostatic forces. What it means is that the magnetic force, hence the Lorentz force, is not the time derivative of momentum. You have to include field momentum to recover momentum conservation or you have to abandon gauge invariance, which is the choice I worked out in a published paper. – my2cts Jun 30 '19 at 23:50
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Right... I just figured this point should be mentioned in your answer. – BioPhysicist Jun 30 '19 at 23:52
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@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 $\Gamma_{{1,2}}=0$. – Jul 01 '19 at 13:28
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This requires the full strength of the 'action-reaction principle', not merely conservation of momentum. – Jul 01 '19 at 13:30
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@RaphaelPicovschi Momentum conservatiin us all there is to the third law. – my2cts Jul 02 '19 at 08:44
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@AaronStevens min this way momentum conservation is recovered but the fact remains that the Lorentz force itself does not conserve it. Therefore it is not the rate of change of momentum. – my2cts Jul 02 '19 at 08:45
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@my2cts I was asking my questions as a way to show how your answer could be improved to be better understood for people who are not aware of this. I was not looking for an explanation. I thought it would be useful to go into more detail in your answer. – BioPhysicist Jul 02 '19 at 09:44
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In other words, I would initially be confused by magnetic force is not the derivative of momentum because if I have a charged particle moving in a magnetic field would it not be valid to say $$\mathbf F_B=m\mathbf a=\mathbf{\dot p}$$ – BioPhysicist Jul 02 '19 at 09:50
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@RaphaelPicovschi Why do you ask and what do you mean by "the full strength of the action-reaction principle" other than momentum conservation? – my2cts Jul 02 '19 at 16:08
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@AaronStevens In general for two moving charges the total Lorentz force is not zero and mv1+mv2 is not constant. – my2cts Jul 02 '19 at 16:11
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@my2cts Yes I know this. Once again, I am not asking for clarification for myself. I am saying your answer is unclear. If the Lorentz force is not a rate of change of momentum then you are invalidating newton's second law, not the third law. The third law comes into play when you notice that the changes in momentum do not add up to zero. That is what I am getting at. I am not asking for you to teach me about Lorentz forces. – BioPhysicist Jul 02 '19 at 18:47
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@AaronStevens The third law states in effect the conservation of momentum. That is the same thing as saying that the changes in momentum add up to zero. Apologies for teaching you. – my2cts Jul 02 '19 at 18:54
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Yes... I know this. I am talking about your claim that they are not derivatives of momentum. If you are looking at the Lorentz force acting on a single particle, then it will be the derivative of its momentum. That is the second law. Violation of the third law (which I agree is violated) does not say the Lorentz force is not a derivative of momentum. The violation says those derivatives do not add up to zero when taking into account all of he particles in the system. – BioPhysicist Jul 02 '19 at 19:00
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@AaronStevens The pint is that mv is not the momentum of the charge, so ma is not the force. Thus is clear from the fact that, as stated, for a pair of moving charges in general mv1+mv2 is not conserved, hence not the total momentum. – my2cts Jul 02 '19 at 19:08
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@my2cts What I meant is that the equations $\vec{f}{12}=f\vec{e}{12}=-f\vec{e}{21}=-$\vec{f}{21}$, that I saw in the textbook I'm referring to, are not implied by mere conservation of momentum. – Jul 02 '19 at 22:17