Classical Mechanics

Euler-Lagrange Equation

$\displaystyle \delta I = \int_a^b f(z, \dot{z}, x) dx = 0$    
perturb slightly; $\displaystyle z \rightarrow z + \epsilon n$    
$\displaystyle \frac{dI}{d\epsilon} = \int_a^b \bigg( \frac{\partial f}{\partial...
...ot{z}}{d\epsilon} + \frac{\partial f}{\partial x}\frac{dx}{d\epsilon} \bigg) dx$    
$\displaystyle \frac{dI}{d\epsilon} \vert _{\epsilon = 0} = \int_a^b \bigg( \fra...
...artial f}{\partial z} \eta + \frac{\partial f}{\partial\dot{z}} \eta' \bigg) dx$    


Integrate by parts;

$\displaystyle \int_a^b \bigg( \frac{\partial f}{\partial z} - \frac{d}{dx} \fra...
...\eta dx + \bigg[\frac{\partial f}{\partial\dot{z}}\eta \bigg]^b_a \rightarrow 0$    
if stationary.    
$\displaystyle 0 = \int_a^b \bigg(\frac{\partial f}{\partial z} = \frac{d}{dx} \frac{\partial f}{\partial\dot{z}}\eta \bigg) dx$    
$\displaystyle \boxed{\frac{\partial f}{\partial z} = \frac{d}{dx}\frac{\partial f}{\partial\dot{z}} = 0}$    


A differentiable functional is stationary at a local maximum or minimum.

Lagrange Multipliers
Lagrange Multipliers allow optimization without explicit parameterization. Look for the stationary points of a function $f(x)$ subject to the constraint $g(x) = c$

$\displaystyle \mathcal{L}(x,\lambda) = f(x) + \lambda g(x)$ (1)

This can also be formed by modifying Lagrange's equations, rather than the lagrangian itself.

$\displaystyle \frac{d}{dt} \frac{\partial\mathcal{L}}{\partial\dot{x}} - \frac{\partial\mathcal{L}}{\partial x} = \lambda \frac{\partial f}{\partial x}$ (2)


Example: Atwood's Machine

$\displaystyle \mathcal{L} = \frac{1}{2}m_1\dot{x}_1^2 + \frac{1}{2}m_2\dot{x}_2^2 + m_1 g x_1 + m_2 g x_2$ (3)
$\displaystyle x_1 + x_2 = l =$   const. (4)
Turn the E-L crank.    
$\displaystyle m_1 \ddot{x_1} - m_1 g = \lambda$    
$\displaystyle m_2 \ddot{x_2} - m_2 g = \lambda$    
$\displaystyle \ddot{x}_1 = - \ddot{x}_2;$ (5)
$\displaystyle m_1 \ddot{x}_1 - m_1 g = -m_2 \ddot{x}_1 - m_2 g$    
$\displaystyle \rightarrow \boxed{\ddot{x}_1 = \frac{(m_1 - m_2) g}{m_1 + m_2}}$ (6)


This is also of extreme use in statistical mechanics, where lagrange multipliers often appear in the form of the chemical potential $\mu$.

Hamiltonian Mechanics
One can obtain the Hamiltonian via a Legendre transform of the Lagrangian:

$\displaystyle \mathcal{H}(q,p) = p\dot{q} - \mathcal{L}$ (7)


The equations of motion are
The Hamiltonian is the cotangent bundle over t.

Poisson Bracket

$\displaystyle \{A,B\}_{q,p} = \frac{\partial A}{\partial q}\frac{\partial B}{\partial p} - \frac{\partial A}{\partial p}\frac{\partial B}{\partial q}$ (8)
$\displaystyle \frac{df}{dt} = \{f,\mathcal{H} \} + \frac{\partial f}{\partial t}$ (9)


There is a strong relation between the poisson bracket of classical mechanics and the commutator of quantum mechanics.

Moment of Inertia

$\displaystyle I = \int r^2 dm = \int r^2 \rho(x) dV$ (10)
$\displaystyle I_{ij} = \int \rho(x) [\delta^i_j \left\vert\mathbf{x}\right\vert^2 - x_ix_j] d^3x$ (11)
Parallel Axis Theorem: $\displaystyle I_{parallel} = I_{cm} + MR^2$ (12)

Euler Equations
For rigid body rotations:

$\displaystyle \boxed{\frac{d}{dt} \bigg\vert _I = \frac{d}{dt} \bigg\vert _B + \mathbf{\omega} \times ( \ \ \ ) }$ (13)
$\displaystyle \frac{d}{dt} \bigg\vert _I \mathbf{L} = \frac{d}{dt}\bigg\vert _B \mathbf{L} + \mathbf{\omega}\times \mathbf{L}$    
$\displaystyle \rightarrow \mathbf{\Gamma}^{(e)} = \mathbf{I}\cdot \frac{\mathbf{\omega}}{dt} + \mathbf{\omega} \times \mathbf{I}' \mathbf{\omega}'$    
$\displaystyle \Gamma_1^{(e)} = I_1 \frac{d\omega_1}{dt} + \omega_2 I_3 \omega_3 - \omega_3 I_2 \omega_2$    
$\displaystyle \Gamma_1^{(e)} = I_1 \dot{\omega_1} + \omega_2 \omega_3 (I_3 - I_2)$    
cyclic; $\displaystyle \boxed{I_i \dot{\omega_i} = \omega_j \omega_k (I_j - I_k) + \Gamma_i^{(e)} }$ (14)


This is especially useful when motion is constrained to be torque free; the $\Gamma$ dependence drops out and the equations are much simpler.

Euler Angles TODO: PICTURE
The dashed line is the line of nodes.

$\displaystyle \boxed{\omega_1 = -\dot{\alpha} \sin{\beta} \cos{\gamma} + \dot{\beta} \cos{\gamma} }$ (15)
$\displaystyle \boxed{\omega_2 = \dot{\alpha} \sin{\beta} \sin{\gamma} + \dot{\beta} \cos{\gamma} }$ (16)
$\displaystyle \boxed{\omega_3 = \dot{\alpha} \cos{\beta} + \dot{\gamma}}$ (17)


Note: The angle projections always depend on the previous angle's rate of change, never a later one; i.e., $\dot{\alpha} \cos{\beta}$ will occur, but NOT $\dot{\beta}\cos{\alpha}$

Differential Scattering Cross Section

$\displaystyle \frac{d\sigma}{d\Omega} = \frac{b}{\sin{\theta}} \left\vert\frac{db}{d\theta}\right\vert$ (18)
$\displaystyle d\sigma = 2\pi b db$    
$\displaystyle d\Omega = 2 \pi \sin{\theta} d\theta$    
$\displaystyle \textbf{TODO: DIAGRAM}$    


Gauss' Law for Newtonian Gravity

$\displaystyle \mathcal{L}_{g} = -\rho(\mathbf{x},t)\phi(\mathbf{x},t) - \frac{1}{8\pi G}(\nabla\mathbf{\phi}(\mathbf{x},t))^2$ (19)
$\displaystyle \rightarrow -\rho = \frac{2}{8\pi G} \nabla^2{\phi}$    
$\displaystyle \boxed{\nabla^2\phi = 4\pi G \rho}$ (20)
$\displaystyle \boxed{\mathbf{g} = -\nabla\mathbf{\phi}}$ (21)
Gauss' Law differential form: $\displaystyle \nabla\cdot\mathbf{\mathbf{g}} = -4\pi G \rho$ (22)
Gauss' Law integral form: $\displaystyle \oint \nabla\cdot\mathbf{\mathbf{g}} dV$ (23)
$\displaystyle = \oint \mathbf{g} \cdot d\mathbf{\sigma} = -4\pi G M$    


Relativistic Doppler Shift

$\displaystyle p^{\mu} = \{ E, p^x, p^y, 0 \}$ (24)
$\displaystyle v_{obs}^{\mu} = \{ \frac{dt}{d\tau}, \frac{dx}{d\tau} \} = \{ \frac{dt}{d\tau}, \gamma v_x \}$ (25)
$\displaystyle E = p^{\mu}V_{\mu} = \omega \frac{dt}{d\tau} - \omega \cos{\theta}v_x\gamma$ (26)
$\displaystyle \omega' = \omega(\gamma - \cos{\theta}\gamma v_x)$ (27)
transverse: $\displaystyle \boxed{ \frac{\omega'}{\omega} = \frac{1 - v\cos{\theta}}{\sqrt{1-v^2}} }$ (28)
parallel: $\displaystyle \boxed{ \frac{\omega'}{\omega} = \sqrt{ \frac{1 - v}{1 + v}}}$ (29)


Schwarzschild Metric

$\displaystyle ds^2 = -\bigg( 1 - \frac{2GM}{r} \bigg) dt^2 + \bigg(1 + \frac{2GM}{r} \bigg)^{-1} dr^2 + r^2 d\Omega^2$ (30)


Gravitational field outside a spherical mass; no charge/ angular momentum. Universal cosmological constant = 0.

FLRW Metric
Friedmann-LemaƮtre-Robertson-Walker metric for the curvature of the universe:

$\displaystyle ds^2 = -dt^2 + \frac{dr^2}{1-kr^2} + r^2 d\Omega^2$ (31)