Oh Yeah Dat Boi Dis Boi   Given a vector function, I show that $\vec{F}(x,y,z,t)$, I show that $d\vec{F} = (d\vec{r} \cdot \nabla) \vec{F} + \frac{\partial \vec{F}}{\partial t}$  by writing $d\vec{F}$ in terms of independent variations in the $x, y, z$ and $t$ direction. I write $d\vec{F}$ as a sum of four increments, one purely in the $x$ direction, the $y$ direction, the $z$ direction and the $t$ direction as follows:

\$latex dF(x,y,z,t)

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.