Talk:PlanetPhysics/Conservation of Angular Momentum

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%%% This file is part of PlanetPhysics snapshot of 2011-09-01 %%% Primary Title: conservation of angular momentum %%% Primary Category Code: 45.50.-j %%% Filename: ConservationOfAngularMomentum.tex %%% Version: 1 %%% Owner: mdo %%% Author(s): mdo %%% PlanetPhysics is released under the GNU Free Documentation License. %%% You should have received a file called fdl.txt along with this file. %%% If not, please write to gnu@gnu.org. \documentclass[12pt]{article} \pagestyle{empty} \setlength{\paperwidth}{8.5in} \setlength{\paperheight}{11in}

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

The angular momentum, $\mathbf{L}$ of a particle with \htmladdnormallink{position vector}{http://planetphysics.us/encyclopedia/PositionVector.html}, $\mathbf{r}$, and total linear \htmladdnormallink{momentum}{http://planetphysics.us/encyclopedia/Momentum.html}, $\mathbf{p}$ is given by $\mathbf{L} = \mathbf{r}\times\mathbf{p}$. If some \htmladdnormallink{force}{http://planetphysics.us/encyclopedia/Thrust.html}, $\mathbf{F}$, 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{eqnarray*} \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{eqnarray*} 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 \htmladdnormallink{vector}{http://planetphysics.us/encyclopedia/Vectors.html}, $\mathbf{A}$, $\mathbf{A}\times\mathbf{A}=\mathbf{0}$ (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.

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