# Conservation of Angular Momentum

## Theorem

Newton's Laws of Motion imply the conservation of angular momentum in systems of masses in which no external force is acting.

## Proof

We start by stating Newton's Third Law of Motion 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 equal and opposite pairs.

The force $F_{i j}$ exerted by body $i$ on body $j$ is related to the force exerted by body $j$ on body $i$ by:

$(1): \quad \vec {F_{i j}} = -\vec {F_{j i}}$

The final part of Newton's Third Law of Motion is that these equal and opposite forces act through the line that connects the two bodies in question.

This can be stated thus:

$(2): \quad \vec {F_{i j}} = a_{i j} \paren {\vec {r_j} - \vec {r_i} }$

where:

$\vec{r_i}$ is the position of body $i$
$a_{i j}$ is the magnitude of the force.

As a consequence of $(1)$:

$a_{j i} = a_{i j}$

Let the total torque $\vec {\tau_{\operatorname {total} } }$ on the system be measured about an origin located at $\vec {r_0}$.

Thus:

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

By hypothesis there is no external force.

Thus the second term disappears, and:

 $\displaystyle \vec {\tau_{\operatorname {total} } }$ $=$ $\displaystyle \sum_{i, j} \vec {r_j} \times a_{i j} \paren {\vec {r_j} - \vec {r_i} }$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \sum_{i \mathop > j} \vec {r_j} \times a_{i j} \paren {\vec {r_j} - \vec {r_i} } + \vec {r_i} \times a_{j i} \paren {\vec {r_j} - \vec {r_i} }$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \sum_{i \mathop > j} \vec {r_j} \times \vec {r_i} \paren {a_{j i} - a_{i j} }$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \vec 0$ $\quad$ $\quad$

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 torque between two bodies must cancel out.

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

$\blacksquare$