Oh Yeah

[Untitled] (14)


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.