angular momentum

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Conservation of angular momentum

Theorem: The angular momentum of a closed system (no external forces and conserved mass) of particles is conserved.

Proof: The total angular momentum of the system is

$\mathbf{L} = \sum_{i=1}^N \mathbf{r}_i \times \mathbf{p}_i\,$ ,

where $\mathbf{r}_i\,$ and $\mathbf{p}_i\,$ are the position and momentum of the $i^{th}\,$ particle, respectively. The center of mass is defined by $M \mathbf{R}_{com} = \sum_i m_i \mathbf{r}_i \,$ , where $M = \sum_i m_i\,$ is the total mass of the system. Define the relative position and momentum of the $i^{th}\,$ particle through $\mathbf{r}_i = \mathbf{R}_{com} +\mathbf{r}'_i\,$ and $\mathbf{p}_i = m_i \dot{\mathbf{R}}_{com} + \mathbf{p}'_i\,$

Then

$\mathbf{L}\,$	$= \sum_i \left( \mathbf{R}_{com} +\mathbf{r}'_i \right) \times \left( m_i \dot{\mathbf{R}}_{com} + \mathbf{p}'_i\right)\,$ ,
	$= \mathbf{R}_{com} \times \left( \sum_i m_i \right) \dot{\mathbf{R}} + \sum_i\mathbf{r}'_i \times \mathbf{p}'_i\,$ since $\sum_i m_i\mathbf{r}'_i = 0\,$ ,
	$= \mathbf{L}_{com} + \sum_i \mathbf{L}_{i,around\mbox{ }com}\,$ .

Now

$\frac{d}{dt} \mathbf{L} = \frac{d}{dt} \mathbf{L}_{com} + \sum_i \frac{d}{dt}\mathbf{L}_{i,around\mbox{ }com}\,$ ,

and for a closed system, the first term vanishes, since there is no net external force acting on the system. Thus

$\frac{d}{dt} \mathbf{L}\,$	$= \sum_i m_i \mathbf{r}'_i \times \ddot{\mathbf{r}}'_i\,$ ,
	$= \sum_i m_i \mathbf{r}'_i \times \mathbf{F}_i\,$ , where $\mathbf{F}_i\,$ is the internal force acting on the $i^{th}\,$ particle,
	$= \sum_{ij} m_i \mathbf{r}'_i \times \mathbf{F}_{ij}\,$ , where $\mathbf{F}_{ij}\,$ is the force acting on the $i^{th}\,$ particle due to the $j^{th}\,$ ,
	$= \frac{1}{2}\sum_{ij} m_i \left( \mathbf{r}'_i + \mathbf{r}'_i\right)\times \mathbf{F}_{ij}\,$
	$= \frac{1}{2}\sum_{ij} m_i \left( \mathbf{r}'_i - \mathbf{r}'_j\right)\times \mathbf{F}_{ij}\,$ , by relabling of indices and because $\mathbf{F}_{ij} = -\mathbf{F}_{ji}\,$ by Newton's third law,
	$= 0\,$ , since $\mathbf{F}_{ij}\,$ is colinear with $\mathbf{r}_i - \mathbf{r}_j\,$ , also by Newton's third law.

Thus the angular momentum of the system is conserved. As can be seen, the key idea is that the forces occur in pairs and these forces are along the lines joining pairs of particles. Taking each pair of particles into consideration, the torque they produce relative to the pair's center of mass is zero (since $\mathbf{F}_{ij}\,$ is colinear with $\mathbf{r}_i - \mathbf{r}_j\,$ ) as well as the change in their center of mass momentum.

Relation to momentum

Suppose a particle is in motion about a fixed point, and that

$\mathbf{L} = \mathbf{r}\times\mathbf{p}\,$

is the angular momentum of the particle. Note that

$\frac{L^2}{r^2}\,$	$= \left( \hat\mathbf{r}\times\mathbf{p} \right)^2\,$ ,
	$=( \epsilon_{ijk} \hat{r}_j p_k) (\epsilon_{imn} \hat{r}_m p_n ) \,$ (Einstein summation notation implied),
	$= ( \delta_{jm} \delta_{kn} - \delta_{jn} \delta_{km} ) \hat{r}_j p_k \hat{r}_m p_n \,$ ("contracted epsilon identity"),
	$= p^2 - (\hat\mathbf{r} \cdot \mathbf{p})^2\,$ ,
	$= p^2 - p_r^2\,$ ,