This link shows the extra requirements for strong law of action and reaction. Why is Newton's third law known as weak law of action and reaction? Is the strong law of action and reaction not Newton's third law? But while working with central forces we apply Newton's third law (which is known as weak law) and not specifically strong law of action and reaction. At least I never heard people saying like "from the strong law of action and reaction the two forces are equal and opposite", they say "from Newton's third law…" . Why is it not something like "there are two categories under Newton's third law- weak law and strong law"? What is the difference?
[Physics] Why is Newton’s third law known as *weak law of action and reaction*
definitionforcesnewtonian-mechanicsterminology
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 $$\mathbf{F}_{ij}=-\nabla_j V(|\mathbf{r}_1-\mathbf{r}_2|).$$
I have Marion-Thornton 4th ed. around here somewhere. It is an older book and presents some material differently than we are used to in more modern books (for instance they even use the old imaginary time method when discussing some things in special relativity, which I personally dislike). However I agree with DanielSank, different pedagogy does not equal "nonsense".
Newton's laws are presented slightly differently by different books. For instance, it can be argued Newton meant his second law to be $F=dp/dt$ (although he didn't write it in this modern notation), although many books present it as $F=ma$. Some people go even further and try to extract a modern meaning, as I've seen some people say Newton's third law is the conservation of momentum. This may be pedagogically useful, but not historically accurate. It is worth reminding that some debate over the exact statements translated to modern language is understandable. Even though Newton invented calculus, some concepts in mechanics still took long after Newton to come into their modern understanding, such as the concept of kinetic energy was put in its modern form much later.
Thus to answer this question requires agreeing on a statement for Newton's third law. I don't have Marion-Thornton handy, so using wikipedia
When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body.
The force between two particles in electromagnetism can violate this. For a concrete example consider a positive charged particle A pulled along the x axis at a constant velocity in the positive direction, and another positive charged particle B pulled along the y axis at a constant velocity in the positive direction. If it is arranged such that when A is at (0,0), B is at (0,1), then we can calculate the fields and find:
- the electric forces on the particles will be in opposing directions
- the magnetic force on A is zero
- the magnetic force on B is in the -x direction
Does this mean momentum is not conserved here? No.
If we include the person or device pulling these charges along as part of the system (so there are no external forces), then we should expect the momentum of the system to be conserved.
Where is the missing momentum then? It is in the fields!
I constructed this scenario specially to also help break a bad habit of some descriptions of this phenomena. Because the charges are moving at a constant velocity, there is no radiation. We don't need radiation to provide a force back on the partices or something to solve this. Momentum can be stored in the fields themselves. (While not shown in this example, even static fields can have non-zero momentum.)
Best Answer
Not all forces are central. For example the Lorentz force is not a central force.
However I suspect most of us would regard the distinction between the weak and strong versions of the third law as rather pointless. The third law is a statement that momentum is conserved, which is itself the result of a fundamental symmetry. Whether the force is central or not makes no difference to this fundamental principle. I would file this one in the some physicists have too much free time category.