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.

Proof of Summation Identity

Let g_{\mu\nu} = g^{\mu\nu} be defined by the following relations g_{00} = 1; g_{kk} = -1, k = 1,2,3; g_{\mu\nu} = 0, \mu \neq \nu; and \gamma_{\nu} = \sum g_{\nu\mu}\gamma^{\mu} . I want to show that \sum \gamma_{\mu} \gamma^{\alpha} \gamma^{\mu} = -2 \gamma^{\alpha} , where \alpha = 2 . It must be noted also that the summation is over \mu = 0,1,2,3 .

I can start by writing the summation out as \gamma_{0} \gamma^{2} \gamma^{0} + \gamma_{1} \gamma^{2} \gamma^{1} + \gamma_{2} \gamma^{2} \gamma ^{2} + \gamma_{3} \gamma^{2} \gamma^{3} . This can be rewritten as \gamma^{0} \gamma^{2} \gamma^{0} - \gamma^{1} \gamma^{2} \gamma^{1} - \gamma^{2} \gamma^{2} \gamma^{2} - \gamma^{3} \gamma^{2} \gamma^{3} due to the fact that \gamma_{0} = \sum_{\mu} g_{00} \gamma^{0} which implies \gamma_{0} = \gamma^{0} , \gamma_{1} = \sum_{\mu} g_{11} \gamma^{1} which implies \gamma_{1} = -\gamma^{1} , \gamma_{2} = \sum_{\mu} g_{22} \gamma^{2} which implies \gamma_{2} = -\gamma^{2} , and \gamma_{3} = \sum_{\mu} g_{33} \gamma^{3} which implies that \gamma_{3} = -\gamma^{3} . Now, I can use another identity, namely that \gamma^{\mu} \gamma^{i} = -\gamma^{i}\gamma^{\mu} to permute some of the terms so that the expression looks like - \gamma^{0} \gamma^{0} \gamma^{2} + \gamma^{1} \gamma^{1} \gamma^{2} - \gamma^{2} \gamma^{2} \gamma^{2} + \gamma^{3} \gamma^{3} \gamma^{2} . Now, I can use two final identities, that (\gamma^{0})^{2} = 1, (\gamma^{i})^{2} = -1 to write the expression as -(\gamma^{0})^{2} \gamma^{2} + (\gamma^{1})^{2}\gamma^{2} - (\gamma^{2})^{2} \gamma^{2} + (\gamma^{3})^{2} \gamma^{2} , which is equivalent to -\gamma^{2} - \gamma^{2} + \gamma^{2} - \gamma^{2} . After canceling terms, I obtain -2 \gamma^{2} which is what I wanted to show.