PlanetPhysics/Conservation of Angular Momentum

If a \htmladdnormallink{particle {http://planetphysics.us/encyclopedia/Particle.html} is subject to no torque then the angular momentum is conserved}

The angular momentum, $\mathbf {L}$ of a particle with position vector, $\mathbf {r}$, and total linear momentum, $\mathbf {p}$ is given by $\mathbf {L} =\mathbf {r} \times \mathbf {p}$. If some force, $\mathbf {F}$, acts on that particles, then the torque is defined similarily as $\mathbf {N} =\mathbf {r} \times \mathbf {F} =\mathbf {r} \times d\mathbf {p} /dt$.

Taking the time derivative of the angular momentum equation,

${\begin{matrix}{\frac {d\mathbf {L} }{dt}}&=&{\frac {d}{dt}}\left(\mathbf {r} \times \mathbf {p} \right)\\&=&\left({\frac {d\mathbf {r} }{dt}}\times \mathbf {p} \right)+\left(\mathbf {r} \times {\frac {d\mathbf {p} }{dt}}\right).\end{matrix}}$

Consider the term, $d\mathbf {r} /dt\times \mathbf {p}$. Since $\mathbf {p} =md\mathbf {r} /dt$, it follows that ${\frac {d\mathbf {r} }{dt}}\times \mathbf {p} =m\left({\frac {d\mathbf {r} }{dt}}\times {\frac {d\mathbf {r} }{dt}}\right).$ But, given an arbitrary vector, $\mathbf {A}$, $\mathbf {A} \times \mathbf {A} =\mathbf {0}$ (the zero vector), so the expression for the time derivative of the angular momentum becomes, ${\frac {d\mathbf {L} }{dt}}=\left(\mathbf {r} \times {\frac {d\mathbf {p} }{dt}}\right)=\mathbf {N} .$ Writing the above simplistically as $d\mathbf {L} /dt=\mathbf {N}$ is is clear that when the torque is zero, then the angular momentum is constant in time; it is conserved.