Testing Latex

This post tests the usage of $\LaTeX$ on this website. For example, the contravariant formulation of classical electromagnetism states that

\[\begin{equation} \label{eq:GaussAmpere} \partial_{\alpha} F^{\alpha \beta} = \mu_0 J^{\beta} \end{equation}\]

\[\begin{equation} \label{eq:GaussFaraday} \partial_{\alpha} \left( \frac{1}{2} \epsilon^{\alpha \beta \gamma \delta} F_{\gamma \delta} \right) = 0 \end{equation}\]

Where $F^{\alpha \beta}$ is the electromagnetic tensor, $J^{\alpha}$ the four-current and $\epsilon^{\alpha \beta \gamma \delta}$ the Levi-Civita symbol. Equations can even be referenced, like \eqref{eq:GaussAmpere} or \eqref{eq:GaussFaraday}.

Classical Mechanics Example

For example, one can write the Lagrangian of a point particle moving in one dimension space under a certain potential as $\mathcal{L}=T-V$, meaning

\[\mathcal{L}=\frac{m}{2} \dot{q}^2 - V(q) \]

where $m$ is the mass of the particle, $q$ is its position, and $\dot{q}$ is its velocity. In Hamiltonian mechanics, the momentum $p$ is defined as $p=\partial_q \mathcal{L}=m \dot{q}$. The Hamiltonian is defined as $H=p \dot{q} - \mathcal{L}$, so replacing $\dot{q}$, it yields

\[ H = \frac{p^2}{2 m} + V(q) \]

Now, one could use the Hamilton equations of motion to solve for the position and momentum, however, we’re taking the long way. Due to the simmetries of the Hamiltonian, $H’=H + \partial_t S$ gives the same equations of motion, where $S=S(q,p,t)$ is any function of the position, momentum and time. The idea is to look for an action $S$ such that the new Hamiltonian is zero, thus forming the Hamilton-Jacobi equation. This is a very general problem, so in this case it turns tedious and much longer than using the Euler-Lagrange equations. If one defines $\partial_q S=p$, then one gets a differential equation for the action $S$, of the form

\[ \frac{\left( \partial_q S \right)^2}{2 m} + V(q) + \partial_t S = 0 \]

This equation is generally hard to solve. However, identifying cyclic coordinates, or continuous symmetries of the Hamiltonian, one can simplify the problem. In this case, $\partial_t H=0$. This lets one choose an action $S=W(q) - E t$, where in this case $E$ corresponds to the total energy of the particle. Replacing this new $S$,

\[ \frac{W’^2}{2 m} + V(q) = E \Rightarrow W(q) = \int \sqrt{2 m (E - V(q))} dq \]

Then, having solved for $S$, the chosen canonical transformations result in $\partial_E S = \beta$, where $\beta$ is another constant. This implies

\[ \beta = -t + \int \frac{dq}{\sqrt{\frac{2}{m} \left( E - V(q) \right)}} \]

Replacing $V(q)$ for a specific potential does not assure that one can write $q$ as $q(t)$, however, for most simple cases, the integral can be solved, and the solution $q(t)$ is found.

For example, a harmonic oscillator has a potential that can be written as $V(q)=\frac{k}{2}q^2$, where $k$ is a constant. Replacing on the above equation and solving the integral, one gets

\[ \sqrt{k m}(t+\beta) = m \arcsin{\left( \sqrt{\frac{k}{2 E}} q \right)} \]

Finally, one can simplify for $q(t)$ to get

\[ q(t) = \sqrt{\frac{2 E}{k}} \sin{\left( \sqrt{\frac{k}{m}}(t + \beta) \right)} \]

Where $E$ can be interpreted as the initial energy of the system, and $\beta$ as the initial phase. This returns a sine wave, as expected.

Aren’t headings cool?

def function(x):
  return 'vale crack gracias'