PlanetPhysics/Conservation of Angular Momentum

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

The angular momentum, $𝐋$ of a particle with [[../PositionVector/|position vector]], $𝐫$, and total linear [[../Momentum/|momentum]], $𝐩$ is given by $𝐋=𝐫\times 𝐩$. If some [[../Thrust/|force]], $𝐅$, acts on that particles, then the torque is defined similarily as $𝐍=𝐫\times 𝐅=𝐫\times d𝐩/dt$.

Taking the time derivative of the angular momentum equation,

$\begin{array}{ccc}\frac{d𝐋}{dt}& =& \frac{d}{dt}(𝐫\times 𝐩)\\ & =& (\frac{d𝐫}{dt}\times 𝐩)+(𝐫\times \frac{d𝐩}{dt}).\end{array}$

Consider the term, $d𝐫/dt\times 𝐩$. Since $𝐩=md𝐫/dt$, it follows that $\frac{d𝐫}{dt}\times 𝐩=m(\frac{d𝐫}{dt}\times \frac{d𝐫}{dt}).$ But, given an arbitrary [[../Vectors/|vector]], $𝐀$, $𝐀\times 𝐀=\mathrm{𝟎}$ (the zero vector), so the expression for the time derivative of the angular momentum becomes, $\frac{d𝐋}{dt}=(𝐫\times \frac{d𝐩}{dt})=𝐍.$ Writing the above simplistically as $d𝐋/dt=𝐍$ is is clear that when the torque is zero, then the angular momentum is constant in time; it is conserved.