Proof: Conservation of Momentum

I love this simple profound proof, and it’s worth understanding. For those unfamiliar with calculus, the notation $\dot{\bf{p}}$ or $\frac{d \bf{p}}{dt}$ denotes the rate of change of $\bf{p}$ with respect to time ($t$).

We need to show that the net force ($\mathbf{F}$) applied on an object is proportional to the rate of change of that object’s momentum with respect to time ($\dot{\mathbf{p}}$).

We know Newton’s second law: $\mathbf{F} = m\mathbf{a}$, and $\mathbf{p} = m \mathbf{v}$.

Differentiating $\mathbf{p} = m \mathbf{v}$ with respect to $t$, presuming constant mass ($m$) yields,

$$ \begin{aligned} \frac{d\mathbf{p}}{dt} = \dot{\mathbf{p}} &= \frac{d}{dt}(m\mathbf{v}) \\ \\[0.1pt] &= m \ \frac{d\mathbf{v}}{dt} \\ \\[0.1pt] &= m\mathbf{a} \\ \\[0.1pt] \therefore \quad \mathbf{F} &= \dot{\mathbf{p}} \quad \blacksquare \end{aligned} $$

With this lemma, we can proceed with the main proof.

Suppose we have a system with point particles $p_1$ and $p_2$, exerting forces $\mathbf{F}_{21}$ and $\mathbf{F}_{12}$, on each other.

We also approximate any external forces on these two particles as $\mathbf{F}^{\text{ext}}_1$ and $\mathbf{F}^{\text{ext}}_2$,

$$ \begin{aligned} \mathbf{F}_{p_1} &= \mathbf{F}_{12} + \mathbf{F}^{\text{ext}}_1 \\ \\[0.1pt] \mathbf{F}_{p_2} &= \mathbf{F}_{21} + \mathbf{F}^{\text{ext}}_2 \end{aligned} $$

By the lemma $\mathbf{F} = \dot{\mathbf{p}}$,

$$ \begin{aligned} \mathbf{F}_{p_1} &= \dot{\mathbf{p}}_1 \\\\ \mathbf{F}_{p_2} &= \dot{\mathbf{p}}_2 \end{aligned} $$

The total momentum of our two-particle system is:

$$ \mathbf{P}_{\text{total}} = \mathbf{p}_1 + \mathbf{p}_2 $$

Differentiating with respect to time yields,

$$ \begin{aligned} \dot{\mathbf{P}}_{\text{total}} &= \dot{\mathbf{p}}_1 + \dot{\mathbf{p}}_2 \\ \\[0.1pt] &= \mathbf{F}_{12} + \mathbf{F}_{21} + \mathbf{F}^{\text{ext}}_1 + \mathbf{F}^{\text{ext}}_2 \end{aligned} $$

Since $\mathbf{F}_{12} = -\mathbf{F}_{21}$, their sum cancels out. Letting the total external force be $\mathbf{F}_{\text{ext}} = \mathbf{F}^{\text{ext}}_1 + \mathbf{F}^{\text{ext}}_2$, we have:

$$ \dot{\mathbf{P}}_{\text{total}} = \mathbf{F}_{\text{ext}} $$

This important result shows that the rate of change of total momentum ($\dot{\mathbf{P}}_{\text{total}}$) in a two-particle system is determined solely by the net external force.

In the special case where $\mathbf{F}_{\text{ext}} = 0$, we conclude:

$$ \dot{\mathbf{P}}_{\text{total}} = 0 \quad \Rightarrow \quad \mathbf{P}_{\text{total}} = \text{constant} \quad \blacksquare $$