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?
[Physics] ny case in classical mechanics where Newton’s (strong) third law doesn’t hold
classical-mechanicsforcesnewtonian-mechanics
Related Solutions
The full mathematical statement is as follows:
Theorem
If two particles exert a mutual conservative force $\mathbf{F}_{12}$ and $\mathbf{F}_{21}$ which is independent of any other degree of freedom of any bigger system they're part of, and obeys Newton's third law as $\mathbf{F}_{12}+\mathbf{F}_{21}=\mathbf{0}$, with the forces collinear to the particles' relative orientation, then this mutual force can be written in the form $$\mathbf{F}_{ij}=-\nabla_j V(|\mathbf{r}_1-\mathbf{r}_2|)$$ for some appropriate potential $V$.
It is important to assume that the force is independent of any degrees of freedom other than the particles' DOFs. This is quite natural, as otherwise it would be very difficult to interpret the force as one caused by one particle on another.
Under this assumption, the requirement that the force be conservative requires the existence of some potential $V$, which depends only on $\mathbf{r}_1$ and $\mathbf{r}_2$, such that $$\mathbf{F}_{ij}=-\nabla_j V(\mathbf{r}_1,\mathbf{r}_2).$$
This function can be transformed, without losing any information, into a function of the relative position $\mathbf{r}=\mathbf{r}_1-\mathbf{r}_2$ and some sort of "mean" position, say the average $\mathbf{R}=\tfrac12\mathbf{r}_1+\tfrac12\mathbf{r}_2$, though any linear combination (e.g. COM position) independent to $\mathbf{r}$ is acceptable. Under this change of variables, the gradients transform as $$ \left\{\begin{array}{} \nabla_{\mathbf{r}_1}=\phantom- \nabla_{\mathbf{r}}+\frac12\nabla_{\mathbf{R}},\\ \nabla_{\mathbf{r}_2}=-\nabla_{\mathbf{r}}+\frac12\nabla_{\mathbf{R}}, \end{array}\right. $$ so that the statement of Newton's third law in this setting becomes $$ \mathbf{0} =\nabla_{\mathbf{r}_1}V(\mathbf{r},\mathbf{R})+\nabla_{\mathbf{r}_2}V(\mathbf{r},\mathbf{R}) =\nabla_{\mathbf{R}}V(\mathbf{r},\mathbf{R}). $$
There is thus no dependence on the absolute position $\mathbf{R}$, and the force can be written as $$\mathbf{F}_{ij}=-\nabla_j V(\mathbf{r}).$$
For the final step, as QMechanic pointed out, you use the fact that the force is collinear to the relative orientations. If $V$ depends on the angular variables of $\mathbf{r}$, then its gradient will not be purely radial. If you require a formal proof, note that forcing the non-radial components of the gradient in spherical coordinates, $$ \nabla f =\frac{\partial f}{\partial r}\hat{{r}} +\frac1r\frac{\partial f}{\partial \theta}\hat{{\theta}} +\frac{1}{r\sin\theta}\frac{\partial f}{\partial r}\hat{{\phi}}, $$ to vanish, requires the partial derivatives with respect to $\theta$ and $\phi$ to vanish; there is thus only a dependence in $r$.
With this, then, you have your final result: the force can be written as $$\mathbf{F}_{ij}=-\nabla_j V(|\mathbf{r}_1-\mathbf{r}_2|).$$
Let charge A be at the origin, moving to the right (along the positive x axis). Let charge B be at coordinates (1,0), moving in the positive y direction.
A's magnetic force on B vanishes, since by symmetry the magnetic field due to A is zero at B's position.
B's magnetic force on A doesn't vanish.
Does the law already fails in magnetostatics? I guess not, but how to prove it?
In magnetostatics there can't be any radiation. If there's no radiation, then mechanical momentum is the only form of momentum we have. If Newton's third law fails, then mechanical momentum isn't conserved. This would lead to a violation of conservation of momentum, which is impossible. So no, the third law can't fail in magnetostatics.
This is my main question: Is there any experiment, ideally something which can be accomplished with high school lab equipment, which shows in a convincing way that Newton's third law doesn't hold for magnetic forces in general?
It would have to be an experiment in which a large amount of momentum was carried away by radiation. Seems tough to me. Even if you build a very powerful and directional radio transmitter, the amount of momentum carried away is tiny in mechanical terms.
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Best Answer
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.