# How to prove the law of equal action and reaction?

Suppose $$F=ma$$ was known. For a system of two massive point objects, $$a$$ and $$b$$, suppose that $$a$$ act a force on $$b$$, $$F_{ab}$$, prove that $$b$$ also act a force on $$a$$, $$F_{ba}$$, and that $$F_{ab}=-F_{ba}$$.

I thought that, if the force was a conservative force, we might do it rather easily through conserved quantity. However, if the force was not conserved, how to prove the law of equal action and reaction mathematically?

(One may prove with the condition that the system was closed, however, a general prove was preferred.)

You can't prove it from $$F=ma$$. This is why Newton's second law and Newton's third law are specified as two separate things.

You either have to take it as a given (somewhat like an axiom), or you have to assume momentum is conserved between interactions and then show that Newton's third law must hold.

As you go deeper into physics, you find that most people take momentum conservation to be the more fundamental idea since there exist forces where Newton's third law does not hold but you can still show momentum is conserved (although not in the classical physics sense).

• Thank you. $F=ma$ was known, I want to prove that $F_{ba}$ exist and that $F_{ba}=-F_{ab}$. Jul 7, 2019 at 15:18
• @user9976437 I didn't say anything about the force being conservative or not. That distinction has nothing to do with Newton's third law. I think you need to carefully read my answer, since I said nothing about this point. Jul 7, 2019 at 15:23
• @user9976437 I see several missconceptions. 1) Conservative force doesn't mean that the force is conserved, 2) The fact that $F_{ab}$ is the opposite of $F_{ba}$ is independent of the forces involved been conservative or not, 3) The fact that $F_{ab}$ is the opposite of $F_{ba}$ does not mean that force is conserved (you can totally create or destroy a force in this universe), 4) You can indeed deduce $F_{ab}=-F_{ba}$ from the law of conservation of momentum (but for that you have to assume the conservation of momentum), or alternatively you can deduce conservation of momentum if you assume Jul 7, 2019 at 15:45
• that $F_{ab}=-F_{ba}$. But either way you have to assume one of them as a basic statement that can't be mathematically proven, you can only prove it experimentally as its consequences are observable in the real world. 5) Since $F_{ab}=-F_{ba}$ is logically equivalent to the conservation of momentum you cannot make the statements independent of each other, if one holds the other also holds, if we could deduce $F_{ab}=-F_{ba}$ from the absence of conservation of momentum then there would be no point at all to say they are logically equivalent. Jul 7, 2019 at 15:49

There is no proof of this phenomenon of a force always appearing paired with an equal but opposite reaction force.

Just like there is no proof of $$\sum F=ma$$. Roughly speaking, someone back in the days (Newton) just noticed this phenomenon. And then he performed many, many experiments to see if this really is a pattern.

If you after many, many experiments always see the same result, then this result might be how the world works generally.

Eventually, you can call it a law and say that this phenomenon is indeed how the world works. Your many, many experiments have proved it emperically or experimentally, you could say. But you cannot prove it any more accurately or analytically.

Well, maybe it is possible to prove it further. As the other answer suggest, momentum might be considered a more fundamental property, whose conservation law could prove Newton's third law. But then that momentum law is the one which is only emperically proven. In esssense, Newton's 3rd as well as his 2nd and 1st law, are emperical laws that are only experimentally, emperically, proven.

You cannot 'prove' this-it is specified as a separate law from the others for the same reason. What you can do is to look for other 'laws' with the same physical content, and see that they are the same as the third law. A simple example would be conservation of linear momentum in absence of any external force. This is plausible from the 2nd law itself, as $$dp/dt\equiv F=0\implies p(t)=const$$. If the system is, for example, made of balls $$A$$ and $$B$$, then the above implies $$mv_A(t)+mv_B(t)=mv_A(t')+mv_B(t')=const$$, and so $$\frac{mv_A(t')-mv_A(t)}{t'-t}=-\frac{mv_B(t')-mv_B(t)}{t'-t}$$, and choosing $$t\to t'$$, we have from the definition of force(rate of change of momentum)-$$F_{A,B}=-F_{B,A}$$(we know that the only possible forces are due to them exerting them on each other as the external for was zero).

It is interesting to note that we haven't said anything about the action-reaction forces being in the same line as the line joining the particle(they could, for example, both be perpendicular to it in opposite directions). This is called the WEAK form of the second law. The strong form-which says that the action reaction pair acts along the same line as the line joining the particle-is mathematically written as $$\vec{F_{AB}}=-\vec{F_{BA}}$$ $$\vec{r_{AB}}\times \vec{F_{AB}}=0$$. The sufficient condition for this is that the intearction potential depend only on the separation-$$U_{AB}=U_{AB}(|\vec{r}_{AB}|)$$

The point of this was that one could equally well postulate that 'the interaction potential between 2 particles is only a function of their distance', and derive the 3rd law from there. But both these statements have the same content-you are just finding alternate ways to express them.

And finally, decide for yourself which form of those statements would be simpler.