LaTeX Fixes

docs/physics/mechanics/13-hamiltonian-mechanics.md

@@ -0,0 +1,29 @@
+# Chapter 13 - Hamiltonian Mechanics
+
+## Section 13.1 - The Basic Variables
+
+**Definition**. Consider a Laplacian defined as $\mathcal{L} = \mathcal{L}(q_1, \ldots, q_n, \dot{q}_1, \ldots, \dot{q}_n, t)$. Then, the set of coordinates $q_1, \ldots, q_n$ are the *configuration space* while the set of coordinates $q_1, \ldots, q_n, \dot{q}_1, \ldots, \dot{q}_n$ are known as the *state space*.
+
+Recall that the generalized momenta $p_i$ is also defined such that
+
+$$p_i = \frac{\partial \mathcal{L}}{\partial \dot{q}_i}$$
+
+**Definition**. The generalized momenta is also called the *canonical momentum* or the *momentum conjugate to $q_i$*.
+
+**Definition**.  The *Hamiltonian* is defined as
+
+$$\mathcal{H} = \sum_{i = 1}^n p_i \dot{q_i} - \mathcal{L}$$
+
+
+
+## Section 13.2 - Hamilton's Equations for One-Dimensional Systems
+
+## Section 13.3 - Hamilton's Equations in Several Dimensions
+
+## Section 13.4 - Ignorable Coordinates
+
+## Section 13.5 - Lagrange's Equations vs. Hamilton's Equations
+
+## Section 13.6 - Phase-Space Orbits
+
+## Section 13.7 - Lioville's Theorem

docs/physics/mechanics/4-energy.md

@@ -60,7 +60,7 @@ Here, we can define a potential $U(s)$ such that $F_{tang} = -dU/ds$ and the tot
 
 **Definition**. In an Atwood machine, there are two masses of mass $m_1$ and m_2$, suspended with an inextensible massless string over a pulley. The system can be constrained by a single parameter $x$, where $x$ is the vertical distance from the center of the pulley and the center of mass of $m_1$.
 
-Then, we can see that $\Delta T_1  + \Delta U_1 = W_1^{tension}$, and respectively $\Delta T_2 + \Delta U_2 = W_2^{tension}$. Then, we can see that $\W_1^{tension} = -W_2^{tension}$, so $\Delta(T_1 + U_1 + T_2 + U_2) = 0$. That is, $E = T_1 + U_1 + T_2 + U_2$, which is conserved.
+Then, we can see that $\Delta T_1  + \Delta U_1 = W_1^{tension}$, and respectively $\Delta T_2 + \Delta U_2 = W_2^{tension}$. Then, we can see that $W_1^{tension} = -W_2^{tension}$, so $\Delta(T_1 + U_1 + T_2 + U_2) = 0$. That is, $E = T_1 + U_1 + T_2 + U_2$, which is conserved.
 
 Notably, if all forces are conservative, we can define a potential $U_\alpha$ for each particle $\alpha$ such that
 

docs/physics/mechanics/5-oscillations.md

@@ -102,7 +102,7 @@ Notably, if we try and force the oscillator to move at a frequency $\omega$, whe
 
 Then, we can see that $\omega_2 = \sqrt{\omega_0^2 - 2 \beta^2}$ is the frequency at which the response is maximum. This then lets us see that $A_{max} \approx \frac{f_0}{2\beta \omega_0}$.
 
-We can then calculate the FWHM, or full-width at half maximum, and the HWHM, or the half width at half maximum. These are the distance between the two points in which $A^2$ is at half its maximum value. Note that $\omega \approx \omega_0 \pm \beta$, so $\text {FWHM} \approx 2\beta$ and $\txt{HWHM} \approx \beta$.
+We can then calculate the FWHM, or full-width at half maximum, and the HWHM, or the half width at half maximum. These are the distance between the two points in which $A^2$ is at half its maximum value. Note that $\omega \approx \omega_0 \pm \beta$, so $\text{FWHM} \approx 2\beta$ and $\text{HWHM} \approx \beta$.
 
 We can then calculate the sharpness of the peak as the natural frequency over the FWHM, or $Q = \frac{\omega_0}{2\beta} = \pi \frac{1 / \beta}{2\pi \omega_0} = \pi \frac{\text{decay time}}{\text{period}}$.
 
@@ -122,7 +122,7 @@ We then find that $\omega = 2\pi / \tau$.
 
 We can then set $x_n(t) = A_n \cos(n  \omega t - \delta_n)$, where
 
-$$A_n = \frac{f_n}{\sqrt{(\omega_0^2 - n^2 \omegaa^2)^2 + 4\beta^2n^2\omega^2}}$$
+$$A_n = \frac{f_n}{\sqrt{(\omega_0^2 - n^2 \omega^2)^2 + 4\beta^2n^2\omega^2}}$$
 
 We also see that