Consider either the angular momentum of the earth around the sun or equivalently swinging a ball horizontally on a string.
I know that with respect to the point of rotation of the swinging ball, angular momentum is constant. Consider now an origin outside of the 'orbit'. It seems to me that the angular momentum is different as the orbit progresses but there is no external torque (the physical scenario is unaltered!). The easiest way to see this is to consider when the ball is at opposite ends of the circle – the radius changes but the velocity only flips sign.
In addition to this, how about we consider an event where we cut the string. Now place our reference point on the line of motion when the ball flies away in a straight line. By definition the ball will have zero angular momentum with reference to any point on this line ($m|r||v|sin\theta=0$) but certainly before the event it had some angular momentum.
The main question is in the title, but essentially what am I missing conceptually about this problem?
Best Answer
Yes. For any system of particles, the following statement is true:
If the net torque on a system of particles is zero, and if the interactions between particles of the system point along the lines joining them, then the total angular momentum of the system is conserved.
The proof in the context of classical mechanics is below.
For the ball on the string example, if you are only considering the ball, then there is an external torque on the ball: that of the string. One subtlety is that if you pick the origin of your coordinates to be the center of the circle about which it rotates, then in that case there is no torque and the angular momentum of the ball is, in fact, conserved. However, if you pick a different point as your origin, then it's not the case that the position vector is always along the line of the tension vector, and therefore there will be a nonzero torque. Remember that when you calculate the angular momentum and the torque, you need to use the same origin for both to be consistent.
For the orbiting example, you need to consider the system consisting of both planets, then there is no external torque on this system and the total angular momentum is conserved.
proof.
Let $m_i$ denote the mass of particle $i$ and let $\mathbf x_i$ denote the position of particle $i$, then the total angular momentum of the system is defined as $$ \mathbf L = \sum_i \mathbf x_i\times\mathbf (m_i \dot{\mathbf x}_i) $$ Taking a time-derivative gives $$ \dot{\mathbf L} = \sum_i\Big(m_i\dot{\mathbf x}_i\times\dot{\mathbf x_i} + \mathbf x_i\times(m_i\ddot{\mathbf{x}}_i)\Big) = \sum_i\mathbf x_i\times\mathbf F_i $$ where $\mathbf F_i$ is the net force on each particle. Now split the force on each particle into the net external force $\mathbf F_i^e$ and the net force due to all of the other particles in the system $$ \mathbf F_i = \mathbf F_i^e + \sum_j \mathbf f_{ij} $$ where $\mathbf f_{ij}$ denotes the force on particle $i$ due to particle $j$. Then we have $$ \dot{\mathbf L} = \sum_i\mathbf x_i\times\mathbf F_i^e + \sum_{ij}\mathbf x_i\times\mathbf f_{ij} $$ by Newton's third law, we have $\mathbf f_{ij} = -\mathbf f_{ji}$ which causes the last sum to vanish provided that the interactions between particles point along the lines connecting them, we are left with $$ \dot{\mathbf L} = \sum_i\mathbf x_i\times\mathbf F_i^e $$ where the expression on the right is precisely the net external torque on the system.
Note. The necessary assumption that the interactions between particles need to point along the lines joining them is often reasonable because in classical mechanical systems in the real world, these forces are often the coulomb or gravitational interactions which have this property.