Broken third law

Problem

What would happen to conservation of momentum if Newton’s third law did not hold? Build a concrete example with two interacting particles and show where the balance breaks down.

Solution

Start from the link between the third law and conservation. Take an isolated system of two particles $A$ and $B$ that interact only with each other. The rate of change of total momentum is determined by the sum of the internal forces: $\frac{d\vv{p}_\text{tot}}{dt} = \vv{F}_{AB} + \vv{F}_{BA}$ where $\vv{F}_{AB}$ is the force $A$ exerts on $B$ and $\vv{F}_{BA}$ the one $B$ exerts on $A$. The third law states $\vv{F}_{AB} = -\vv{F}_{BA}$: the two forces cancel, so $\dfrac{d\vv{p}_\text{tot}}{dt} = 0$ and total momentum is conserved. Example with the third law broken. Imagine that $A$ pushes $B$ with $\vv{F}_{AB} = (5;\, 0)\ \mathrm{N}$ but that $B$ reacts on $A$ with a force that is not equal and opposite, for example $\vv{F}_{BA} = (-3;\, 0)\ \mathrm{N}$ Now the sum of the internal forces is no longer zero: $\frac{d\vv{p}_\text{tot}}{dt} = (5;\, 0) + (-3;\, 0) = (2;\, 0)\ \mathrm{N} \neq 0$ The system, despite being isolated (no external forces), would gain momentum out of nowhere: it would “self-propel” without expelling anything, violating all physical experience. So conservation of $\vv{p}$ for an isolated system is exactly equivalent to the third law: without equal and opposite action and reaction, the momentum balance collapses. $\ev{\text{Without the 3rd law the internal forces don't cancel: an isolated system gains } \vv{p} \text{ out of nowhere}}$

Links

Topics: Momentum and collisions Concepts: Newton’s third law · Conservation of momentum · Momentum Skills: Limiting-case and thought-experiment analysis