Miscellaneous notes#

Curve length integral#

\[ L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx \]

Speed of waves in fluids#

Continuity equation:

\[\begin{split} \begin{align} (\rho u)_x & = -\rho_t \\ \rho_x u + \rho u_x & = -\rho_t \\ \rho u_x & = -\rho_t \\ \rho u_{xx} & = -\rho_{tx} = -\rho_{xt} \end{align} \end{split}\]

Pressure:

\[\begin{split} \begin{align} p_x & = -(\rho u)_t \\ \frac{\partial p}{\partial \rho} \rho_x & = -\rho_t u - \rho u_t = -\rho u_t \\ \frac{\partial p}{\partial \rho} \rho_{xt} & = -\rho_t u_t - \rho u_{tt} = -\rho u_{tt} \\ \end{align} \end{split}\]

Combining $(4)$ and $(7)$,

\[\begin{split} \begin{align} u_{tt} & = \frac{\partial p}{\partial \rho} u_{xx} \\ c & = \sqrt{\frac{\partial p}{\partial \rho}} \\ \end{align} \end{split}\]

Oscillations#

Simple harmonic motion:

\[ A = \sqrt{x_0^2 + \left(\frac{v_0}{\omega}\right)^2} \]

where $x_0$ and $v_0$ are the initial displacement and velocity, respectively.

Caution

Be careful with the signs of $x_0$ and $v_0$!!!

For the phase, draw a circle of trigonometry and be careful with the domain of the $\arcsin x$ and $\arccos x$ functions.

Collisions#

\[ v_1' = \frac{(m_1 - m_2)v_1 + 2m_2v_2}{m_1 + m_2} \]

For $v_2'$, simply replace $1$ with $2$ and $2$ with $1$.

Solid Angles#

\[ d\Omega = \sin\theta\,d\theta\,d\phi \]

A solid angle with apex angle $2\theta$

\[ \Omega = \int\sin\theta\,d\theta\,d\phi = \int_0^{2\pi}d\phi \int_0^\theta\sin\theta\,d\theta = 2\pi(1-\cos\theta) = 4\pi\sin^2\frac{\theta}{2} \]

Envelope of a family of curves#

Assume every curve of the family is an implicitly defined function $f_t(x, y) = 0$ where $t$ is the parameter. Let

\[ F(t, x, y) = f_t(x, y) \]

Then the envelope is the collection of points $(x, y)$ satisfying

\[ F(t, x, y) = 0\;\;\text{and}\;\;\frac{\partial F(t, x, y)}{\partial t} = 0 \]

Circles in polar coordinates#

\[\begin{split} \begin{align*} (x - a)^2 + (y - b)^2 & = R^2 \\ (r\cos{\theta} - a)^2 + (r\sin{\theta} - b)^2 & = R^2 \\ r^2 - (2a\cos{\theta} + 2b\sin{\theta})r + a^2 + b^2 - R^2 & = 0 \\ \end{align*} \end{split}\]

Satellite Orbits and Binet Equation#

Part I: integrating orbital motion#

\[\begin{split} \begin{align*} E & = \frac{L^2}{2mr^2} + \frac{1}{2}m\dot{r}^2 - \frac{GMm}{r} \\ \frac{dr}{d\theta} & = \frac{\dot{r}}{\dot{\theta}} \\ L & = mr^2\dot{\theta} \\ \end{align*} \end{split}\]

If we eliminate $\dot{\theta}$ and $\dot{r}$, and let $u = \dfrac{1}{r}$, then

\[\begin{split} \begin{align*} du & = -\sqrt{\frac{2mE}{L^2} + \frac{2GMm^2}{L^2}u - u^2}\;d\theta \\ & = -\sqrt{A^2 - (u - B)^2}\;d\theta \end{align*} \end{split}\]

where $A = \sqrt{\dfrac{2mE}{L^2} + B^2}$ and $B = \dfrac{GMm^2}{L^2}$

Substitute $u = Au' + B$, then

\[\begin{split} \begin{align*} \int - \frac{du'}{\sqrt{1-u'^2}} & = \int d\theta \\ \arccos \frac{u - B}{A} & = \theta - \theta_0 \\ r & = \frac{B^{-1}}{1 + \frac{A}{B}\cos(\theta - \theta_0)} \\ \end{align*} \end{split}\]

Thus semi-latus rectum

\[ l = B^{-1} = \frac{L^2}{GMm^2} = \frac{b^2}{a} \]

and eccentricity

\[ e = \frac{A}{B} = \sqrt{1 + \frac{2EL^2}{G^2M^2m^3}} \]

Part II: Binet equation#

\[ E = \frac{L^2}{2m}u^2 + \frac{1}{2}m\left(\frac{d}{dt}\frac{1}{u}\right)^2 + V(r) \]

where

\[ \frac{d}{dt}\frac{1}{u} = \frac{d}{du}\left(\frac{1}{u}\right)\frac{du}{d\theta}\frac{d\theta}{dt} = -\frac{1}{u^2}u'\frac{L}{m}u^2 = -\frac{L}{m}u' \]

According to the conservation of energy

\[\begin{split} \begin{align*} \frac{dE}{du} & = \frac{L^2}{2m}\left(2u + \frac{d(u'^2)}{d\theta}\frac{d\theta}{du}\right) + \frac{d}{dr}V(r)\frac{dr}{du} \\ & = \frac{L^2}{m}(u + u'') + \frac{F(u^{-1})}{u^2} \\ & = 0 \end{align*} \end{split}\]

Thus we have Binet equation

\[ \frac{L^2}{m}(u + u'') + \frac{F(u^{-1})}{u^2} = 0 \]

Part III: the Laplace-Runge-Lenz vector#

This vector is somehow a bit exotic. However, it sometimes appears in problems.

\[ \vec{B} = \vec{p}\times\vec{L} - \alpha m\hat{r} \]

(where $\vec{p}$ is the momentum, $\vec{L}$ is the angular momentum and $\alpha = GMm$)

\[\begin{split} \begin{align*} \frac{d\vec{B}}{dt} & = \frac{d\vec{p}}{dt}\times\vec{L} - \alpha m \frac{d\hat{r}}{dt} \\ & = -\frac{\alpha}{r^2}\hat{r}\times(mr^2\dot{\theta}\hat{z}) - \alpha m\dot{\theta}\hat{\theta} \\ & = \vec{0} \end{align*} \end{split}\]

\[ \vec{B}\cdot\vec{L} = 0 \]

, $\vec{B}$ is in the plane of motion.

\[\begin{split} \begin{align*} \vec{r}\cdot\vec{B} & = rB\cos\theta \\ & = \vec{L}\cdot(\vec{r}\times\vec{p}) - \alpha mr \\ & = L^2 - \alpha mr \\ r & = \frac{L^2}{\alpha m + B\cos\theta} \end{align*} \end{split}\]

which is a conic section.

Multi-variable linear recurrence relations#

Suppose matrix $A$ has eigenvalue matrix $\Lambda$ and eigenvector matrix $X$, then

\[\begin{split} \begin{align*} \vec{v}_{n+1} & = A\vec{v}_n + \vec{b} \\ X^{-1}\vec{v}_{n+1} & = X^{-1}A\vec{v}_n + X^{-1}\vec{b} \\ & = \Lambda X^{-1}\vec{v}_n + X^{-1}\vec{b} \\ \end{align*} \end{split}\]

Let $X^{-1}\vec{v}_n = \vec{v}'_n$ and $X^{-1}\vec{b} = \vec{b}'$, then

\[ \vec{v}'_{n+1} = \Lambda\vec{v}'_n + \vec{b}' \]

is a recurrence relation that can be solved one by one.

Finally,

\[ \vec{v}_n = X\vec{v}'_n \]

Second derivative test of multi-variable functions#

\[ f_x(a, b) = f_y(a, b) = 0 \]

then

local maximum at $(a, b)$ if $f_{xx} < 0$ and $f_{xx}f_{yy} - f_{xy}^2 > 0$
local minimum at $(a, b)$ if $f_{xx} > 0$ and $f_{xx}f_{yy} - f_{xy}^2 > 0$
local saddle at $(a, b)$ if $f_{xx}f_{yy} - f_{xy}^2 < 0$
inconclusive if $f_{xx}f_{yy} - f_{xy}^2 = 0$

Vector calculus#

\[\begin{split} \begin{align*} \nabla t & = \sum_{cyc}\frac{1}{f}\frac{\partial t}{\partial u}\hat{u} \\ \nabla\cdot\vec{A} & = \frac{1}{fgh}\sum_{cyc}\frac{\partial}{\partial u}(ghA_u) \\ \nabla\times\vec{A} & = \frac{1}{fgh}\begin{vmatrix} f\hat{u} & g\hat{v} & h\hat{w} \\ \frac{\partial}{\partial u} & \frac{\partial}{\partial v} & \frac{\partial}{\partial w} \\ fA_u & gA_v & hA_w \\ \end{vmatrix} \end{align*} \end{split}\]

Cross product in spherical coordinates#

There seems to be some dispute over this topic on the Internet. For the time being, I will stick to Cartesian coordinates.

Spinning charged spherical shell and magnetization#

By means of integration of the magnetic vector potential, the magnetic field of a spinning charged spherical shell corresponds to a uniformly magnetized ball.

\[ \mathbf{K}_b = \mathbf{M}\times\mathbf{\hat{n}} = M\sin\theta\,\mathbf{\hat{\phi}} = \sigma\omega r\sin\theta\,\mathbf{\hat{\phi}} \]

Thus $\mathbf{M} = \sigma\mathbf{\omega} r$ and

\[ \mathbf{B} = \frac{2}{3}\mu_0\mathbf{M} \]

inside the sphere

and

\[ \mathbf{m} = \frac{4}{3}\pi R^3\mathbf{M} \]

outside the sphere.

\[ \mathbf{B}_{dip} = \frac{\mu_0}{4\pi}\frac{m}{r^3}(2\cos\theta\hat{r} + \sin\theta\hat{\theta}) \]

Compare:

\[ \mathbf{E}_{dip} = \frac{1}{4\pi\epsilon_0}\frac{p}{r^3}(2\cos\theta\hat{r} + \sin\theta\hat{\theta}) \]

Lagrangian and Hamiltonian Mechanics#

Lagrangian#

\[ L = T - V \]

Euler-Lagrange equation:

\[ \frac{d}{dt}\left(\frac{\partial L}{\partial\dot{q_i}}\right) = \frac{\partial L}{\partial q_i} \]

Generalized force and momentum:

\[\begin{split} \begin{align*} F_i &= \frac{\partial L}{\partial q_i} \\ p_i &= \frac{\partial L}{\partial\dot{q_i}} \\ \end{align*} \end{split}\]

Hamiltonian#

\[ H = \sum_i \dot{q}_i\frac{\partial L}{\partial \dot{q}_i} - L \]

(and rewrite it in terms of the generalized momentum $p_i$)

Hamiltonian’s equations:

\[\begin{split} \begin{align*} \dot{q}_i &= \frac{\partial H}{\partial p_i} \\ \dot{p}_i &= -\frac{\partial H}{\partial q_i} \\ \end{align*} \end{split}\]

Lagrangian and Hamiltonian Mechanics#

Lagrangian#

\[ L = T - V \]

Euler-Lagrange equation:

\[ \frac{d}{dt}\left(\frac{\partial L}{\partial\dot{q_i}}\right) = \frac{\partial L}{\partial q_i} \]

Generalized force and momentum:

\[\begin{split} \begin{align*} F_i &= \frac{\partial L}{\partial q_i} \\ p_i &= \frac{\partial L}{\partial\dot{q_i}} \\ \end{align*} \end{split}\]

Hamiltonian#

\[ H = \sum_i \dot{q}_i\frac{\partial L}{\partial \dot{q}_i} - L \]

(and rewrite it in terms of the generalized momentum $p_i$)

Hamiltonian’s equations:

\[\begin{split} \begin{align*} \dot{q}_i &= \frac{\partial H}{\partial p_i} \\ \dot{p}_i &= -\frac{\partial H}{\partial q_i} \\ \end{align*} \end{split}\]

Inverse matrices#

2x2 $$ \begin{align*} A & = \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} \\ A^{-1} & = \frac{1}{\det A}\begin{pmatrix} d & -b \\ -c & a \\ \end{pmatrix} \\ \end{align*} $$
other

Row-reduce

\[\begin{split} \left( \begin{array}{ccc|ccc} a & b & c & 1 & 0 & 0 \\ d & e & f & 0 & 1 & 0 \\ g & h & i & 0 & 0 & 1 \\ \end{array} \right) \end{split}\]

to get

\[\begin{split} \left( \begin{array}{ccc|ccc} 1 & 0 & 0 & j & k & l \\ 0 & 1 & 0 & m & n & o \\ 0 & 0 & 1 & p & q & r \\ \end{array} \right) \end{split}\]

Uncertainty in measurements#

Uncertainty is calulated with the corrected sample standard deviation:

\[ s = \sqrt{\frac{\sum_{i=1}^N (x_i - \bar{x})^2}{N - 1}} \]

To determine the uncertainty of a dependent variable $y = f(\mathbf{x})$, use the following approximation:

\[ s_y = \sqrt{\sum_{j=1}^m\left(\frac{\partial y}{\partial x_j}\right)^2 s_{x_j}^2} \]