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Is there any case in classical (non relativistic) mechanics where the strong form of Newton's third law does not hold (that is, reaction forces are not collinear)? For example, if we consider a system of two point particles in equilibrium with each other upon which a constraint acts so that the reaction forces are directed in a direction that is not collinear. Is such a situation possible?

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If the forces are not precisely opposite one another, this would imply that conservation of (linear) momentum no longer holds; this has never been found in any experiment or observation.

Interestingly, the force must also point along the vector joining the two bodies: in other words, the cross product of the force with the position vector joining the two bodies at the point where the force acts, must be zero. If not, this would imply that conservation of angular momentum no longer holds. This, too, has never been found in any experiment or observation.

That's not to say it's impossible, or that designing experiments to look for it, in ways that are more sensitive than any yet done, are not worthwhile. We actually need people checking on this, to see if conservation of angular or linear momentum is all right, or just "mostly" right.

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  • $\begingroup$ Fair enough. Another question though; The (Weak) Form of Newton's Third Law says that, if we consider two particles, 1 and 2, the force of 1 acting on 2 is equal in magnitude and opposite in direction to the force exerted by 2 on 1. So that would mean that the reaction forces are in vectorial equilibrium. But for this to be the case, don't the reaction forces have to act in the collinear direction? If so, what is even the point of explicitly stating the strong form of this law? $\endgroup$ Commented Jan 21, 2013 at 4:21
  • $\begingroup$ Okay, I had never seen "weak" form applied to Newton's 3rd Law, and a quick googling doesn't give me the answer in the top 2 hits, but I assume you mean the case where the net force is zero, but the net torque is not. For example, suppose we have a 2-dimensional system, with particles at positions $\textbf{p}_1=1m \textbf{i}$ and $\textbf{p}_2=-1m \textbf{i}$, but forces $\textbf{F}_1=1N \textbf{j}$ and $\textbf{F}_2=-1N \textbf{j}$. This isn't made explicit, but it's not seen in nature. If seen, it would violate conservation of angular momentum. $\endgroup$
    – Will Cross
    Commented Jan 21, 2013 at 13:50
  • $\begingroup$ oops, used p rather than r, sorry. I was learning the LaTex to make it look good, and went over the 5 minute deadline for re-editing. For $\textbf{p}$, please pretend it's $\textbf{r}$ $\endgroup$
    – Will Cross
    Commented Jan 21, 2013 at 13:57
  • $\begingroup$ Look at this. Look at the first two lines of the answer where the strong form is mentioned (I didn't know about the 'strong' or 'weak' form until a couple of days ago either, hence the confusion.) $\endgroup$ Commented Jan 21, 2013 at 14:38
  • $\begingroup$ @SumukhAtreya Ah, yes, I was right in my guess. The weak form simply implies conservation of linear momentum, the strong form conservation of angular momentum. And for conservative, spherically symmetric fields centered on the point particles, you're right. You can't have a torque between two point particles if they can only push & pull each other along the line. Let them "shove" each other "sideways" (by which I mean perpendicular to the line joining them), and you still have an action/reaction pair, but it violates conservation of angular momentum. $\endgroup$
    – Will Cross
    Commented Jan 22, 2013 at 16:22
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Well I think it does if 1 charged particle is fixed and 1 another similarity charged particle is moving away with constant velocity, then the force on the rest particle at an instant will be inversely proportional to r ( where r is the distance bw them) but force measured on the moving particle will be proportional to r-dr at that same instant since information needs some time to travel and in that time moving particle would have moved dr distance, so the actual force is delayed.It is based on the fact that everything has a speed limit which is c and hence here Newton’s third law doesn’t hold valid.Please correct me if I am wrong.

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    $\begingroup$ Two charges don't form an action-reaction pair. You're forgetting about the electromagnetic field between them. Each charge forms an action-reaction pair with the electromagnetic field, where the force on the field is defined as the time derivative of the momentum contained in the field. Newton's Third Law still holds here, you've just misidentified the action-reaction pair. $\endgroup$ Commented Jun 1, 2020 at 22:36
  • $\begingroup$ How can we find the momentum in the field.? $\endgroup$
    – Ashpect
    Commented Jul 7, 2020 at 15:20
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    $\begingroup$ For a general field: find the stress-energy tensor $T^{\mu\nu}$ for the field, and integrate $T^{0i}$ over all space for $i=1,2,3$ to get $\vec{P}$ for the field. For the electromagnetic field in particular, the momentum density is $\vec{g}=\frac{1}{\mu_0 c^2}\vec{E}\times\vec{B}$. Integrating this over all space gives you the total momentum. $\endgroup$ Commented Jul 7, 2020 at 15:33

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