Conservation of Angular Momentum

Theorem

Newton's laws imply the conservation of angular momentum in systems of masses in which no net force is acting.

Proof

We start by stating Newton's Third Law in all its detail.

We consider a collection of massive bodies denoted by the subscripts $1$ to $N$.

These bodies interact with each other and exert forces on each other and these forces occur in conjugate pairs.

The force of object $i$ on $j$ ($F_{ij}$) is related to the force of object $j$ on $i$ by this formula:

$\vec{F_{ij}} = -\vec{F_{ji}}$

The final part of the third law is that these conjugate forces act on through the line that connects the two bodies in question. This can be stated thus:

$\vec{F_{ij}} = a_{ij}(\vec{r_{j}} - \vec{r_{i}})$

where $\vec{r_{i}}$ is position of position of body $i$ and $a_{ij}$ is just a force parameter that describes how hard two objects push on each other.

As a consequence of the first formula:

$a_{ji} = a_{ij}$

Now let's look at the total torque on the system assuming that there is no net force ($\vec{r_{0}}$ is the origin about which we measure the torque).

 $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle \vec{\tau_{\operatorname{total} } }$$ $$=$$ $$\displaystyle$$  $$\displaystyle$$ $$\displaystyle \sum_i \vec{\tau_i}$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$  $$\displaystyle$$ $$\displaystyle \sum_i((\vec{r_i} - \vec{r_0}) \times \sum_j \vec{F_{ji} })$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$  $$\displaystyle$$ $$\displaystyle \sum_{i,j}((\vec{r_j} - \vec{r_0}) - (\vec{r_j}-\vec{r_i})) \times \vec{F_{ji} })$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$  $$\displaystyle$$ $$\displaystyle \sum_{i,j}((\vec{r_j} - \vec{r_0}) - (\vec{r_j}-\vec{r_i})) \times a_{ij}(\vec{r_j} - \vec{r_i}))$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$  $$\displaystyle$$ $$\displaystyle \sum_{i,j}(\vec{r_j} - \vec{r_0}) \times a_{ij}(\vec{r_j} - \vec{r_i}))$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$  $$\displaystyle$$ $$\displaystyle \sum_{i,j} \vec{r_j}\times a_{ij}(\vec{r_j} - \vec{r_i}) - (\vec{r_0}\times \sum_{i,j} \vec{F_{ij} })$$ $$\displaystyle$$ $$\displaystyle$$

The second term disappears because of the assumption of no net force.

Thus:

 $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle \vec{\tau_{\operatorname{total} } }$$ $$=$$ $$\displaystyle$$  $$\displaystyle$$ $$\displaystyle \sum_{i,j} \vec{r_j}\times a_{ij}(\vec{r_j} - \vec{r_i})$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$  $$\displaystyle$$ $$\displaystyle \sum_{i>j} \vec{r_j}\times a_{ij}(\vec{r_j} - \vec{r_i}) + \vec{r_i}\times a_{ji}(\vec{r_j} - \vec{r_i})$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$  $$\displaystyle$$ $$\displaystyle \sum_{i>j} \vec{r_j}\times\vec{r_i}(a_{ji} - a_{ij})$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$  $$\displaystyle$$ $$\displaystyle \vec 0$$ $$\displaystyle$$ $$\displaystyle$$

In summary, in a system of masses in which there is no external force, the total torque on the system is equal to 0.

This is because the pair of torques between two objects must cancel out.

Since the rate of change of angular momentum is equal to the torque, the angular momentum of a system is conserved when no external force is applied.

$\blacksquare$