Saturday, January 21, 2012

Typesetting, again

Here are the equations from the last post, in LaTeX. It took about two hours, fairly painstaking work, but I think it looks pretty good..

UPDATE:

Here is a nice simple reference for LaTeX. I started with "LaTeX Bootcamp" (pdf).



$\mathbf{F} = m\mathbf{a}= m\ddot{\mathbf{r}}
= m\frac{d^2}{dt^2}\mathbf{r}$

$\mathbf{F} = -$$\nabla$$V(\mathbf{r})$

$E = \frac{1}{2}m\norm{\dot{\mathbf{r}}}^2+ V$

$\frac{d}{dt} E = \frac{d}{dt} (\frac{1}{2}m
\norm{\dot{\mathbf{r}}}^2 + V)$

\hspace*{2em}$ = ?$\\

$\norm{\dot{\mathbf{r}}}^2 = \norm{\dot{\mathbf{r}}}
\norm{\dot{\mathbf{r}}} = \dot{\mathbf{r}}
\cdot\dot{\mathbf{r}}$

$\frac{d}{dt}\frac{1}{2}m\norm{\dot{\mathbf{r}}}^2
= \frac{1}{2}m\frac{d}{dt}(\dot{\mathbf{r}}
\cdot\dot{\mathbf{r}})$

\hspace*{2em}$ = m\dot{\mathbf{r}}\cdot
\ddot{\mathbf{r}}$

\hspace*{2em}$ = \dot{\mathbf{r}}\cdot-($
$\nabla$$V$$)$\\

$\nabla$$V$$ = <\frac{\partial{V}} {\partial{x}}, \frac{\partial{V}} {\partial{y}},\frac{\partial{V}} {\partial{z}}>$

$\dot{\mathbf{r}}$ = $<$$\frac{dx}{dt}$, $\frac{dy}{dt}$,$\frac{dz}{dt}$$>$

$\nabla$$V$$\cdot$$\dot{\mathbf{r}}$$=< \frac{\partial{V}} {\partial{x}} \frac{dx}{dt}, \frac{\partial{V}} {\partial{y}} \frac{dy}{dt}, \frac{\partial{V}} {\partial{z}}\frac{dz}{dt}>$

\hspace*{2em}$=\frac{d}{dt}V$\\

$\frac{d}{dt}$E = $\dot{\mathbf{r}}$$\cdot$(-
$\nabla$V ) + $\nabla$V $\cdot$ $\dot{\mathbf{r}}$ = 0

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