I often heard that conservation of momentum is nothing else than Newton's third law.

Ok, If you have only two interacting particles in the universe, this seems to be quite obvious.

However if you have an isolated system of $n$ ($n > 2$) interacting particles (no external forces). Then clearly Newton's third law implies conservation of total momentum of the system. However presuppose conservation of total momentum you only get:

$$ \sum_{i\neq j}^n \mathbf F_{ij} = \frac{d}{d t} \mathbf P = 0 $$

Where $\mathbf F_{ij}$ is the forced acted by the $i$th particle upon the $j$th particle and $\mathbf P$ is the total linear momentum.

But this doesn't imply that $\mathbf F_{ij} = -\mathbf F_{ji}$ for $j \neq i$.

So does conservation of momentum implies Newton's third law in general or doesn't it? Why?


4 Answers 4


Right, you could satisfy the momentum conservation by forces that don't satisfy "action vs reaction" law $F_{ij}=-F_{ji}$ but the relevant formulae would have to depend on coordinates and momenta of all the particles. If you assume that the particles are controlled by two-body forces only, the momentum conservation does imply that $F_{ij}=-F_{ji}$.

  • $\begingroup$ Are there any "non two body forces" in classical physics? $\endgroup$
    – martin
    Oct 25, 2011 at 13:09
  • 2
    $\begingroup$ Yes, there are. $\endgroup$
    – Ron Maimon
    Oct 25, 2011 at 20:28

It doesn't, and there are a list of examples in this nearly identical question: Deriving Newton's Third Law from homogeneity of Space

There are no examples of fundamental classical three body forces where the forces are contact forces and linear gravity/EM, because linear fields are two-body interactions. The most obvious non two-body force is in the strong nonlinear gravitational regime.

Other physical examples are things like nucleon-nucleon 3-body forces, which are unfortuately completely quantum.


The force used in Newton's laws is not a "real" thing, it's just a mathematical tool. If you combine Newton's laws, it is not to hard to see how the conservation of momentum is inherent in the picture. Thus,

$$\frac{d}{dt} p = 0$$

For a many body system this becomes

$$\frac{d}{dt} \left( \sum_{i=1} ^N p_i \right) = 0$$

Moving all but one of the momentum terms to the other side

$$\frac{d}{dt} p_1 = - \frac{d}{dt} \left( \sum_{i=2} ^N p_i \right)$$

and voila. That's the physics of the situation so in reality they are inextricably linked and it doesn't even make sense to talk about the force caused by one object or another.

But... if you just care about the theoretical side of things, whether or not the Law of Conservation of Momentum implies Newton's Third Law depends on how you define the Law of Conservation of Momentum. It is always defined as applying to an isolated system. If you stipulate that for a system that is not isolated, the portion of the total change in momentum due to any particular part is that change that would take place within that portion in the if it were isolated, you then get the Third Law.

For example, in the situation described above, the Law of Conservation of momentum is applied to the whole system. If I draw a boundary around objects $1$ and $2$, and say that the contribution to the total change in momentum of the system from these objects is zero then I get

$$\frac{d}{dt} p_{12} = - \frac{d}{dt} p_{21} $$

which is what we expect. But again, when the system is not truly isolated it becomes just an intellectual exercise.


No, but Newton's third law of motion implies the conservation of momentum. In other words, Newton's third law is a special case of the more general law, which is the conservation of momentum.


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