# Conservation of Total Angular Momentum Proof

For this post, I want to prove that in the absence of external forces, the total angular momentum of an N-particle system is conserved.

I start with  $\vec{P} = \displaystyle \sum_{\alpha} \vec{p}_{\alpha}$ which is the total momentum of an N-particle system. Now I can vectorially multiple the total momentum by $\vec{r}$ which is the position vector measured from the same origin $O$ for each particle. This will give me the total angular momentum of the system which can be written as $\vec{L} = \displaystyle \sum_{\alpha = 1}^{N} \vec{\ell}_{\alpha} = \displaystyle\sum_{\alpha = 1}^{N} \vec{r}_{\alpha} \times \vec{p}_{\alpha}$. After differentiating with respect to $t$, I obtain $\dot{\vec{L}} = \displaystyle \sum_{\alpha} \dot{\vec{\ell}}_{\alpha} = \displaystyle \sum_{\alpha} \frac{d}{dt} (\vec{r} \times \vec{p}) = \displaystyle \sum_{\alpha} ( \dot{\vec{r}} \times \vec{p}) + (\vec{r} \times \dot{\vec{p}})$. In the first cross product, I can substitute $\vec{p}$ with $m\dot{\vec{r}}$, and since the cross product of any two parallel vectors is zero, the first term becomes zero. This leaves implies that my sum becomes $\displaystyle\sum_{\alpha}\vec{r}_{\alpha} \times \vec{F}_{\alpha}$. Now, I can rewrite the net force on particle $\alpha$ as $\vec{F}_{\alpha} = \displaystyle \sum_{\beta \neq \alpha} \vec{F}_{\alpha \beta}$, where $\vec{F}_{\alpha \beta}$ represents the force exerted on particle $\alpha$ by particle $\beta$. Now I can make a substitution for $\vec{F}_{\alpha}$ to give me $\dot{\vec{L}} = \displaystyle \sum_{\alpha} \displaystyle \sum_{\beta \neq \alpha} \vec{r}_{\alpha} \times \vec{F}_{\alpha \beta}$

…I will finish the rest of this at some point. I seem to have misplaced the book.

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