PHYS104: Advanced Mechanics

Calculus of Variation

Concept

• function: mapping number to number. Ex. $y=f(x)$
• operator: mapping function to function. Ex. $\frac{d}{dx}$
• functional: mapping function to number. Ex. $max(f(x) = x^2)$

Newton’s Second Law

$\bm{F} = m\bm{a}$, is essentially stating a ordinary differential equation
$\ddot\bm{r} = f(t,\bm{r},\bm{\dot r})$
which has two degrees of freedom $\bm{r}(0)$ and $\bm{\dot r}(0)$

Stationary Point

For $f(x)$: $df = 0$ for any $dx$. Since $df = f'dx$, we have $f' = 0$.
For multi-variable function $F(x_1, x_2, \cdots, x_n)$,
$dF = \nabla F \cdot (dx_1, dx_2, \cdots, dx_n)^T = 0$
Change variable $dx_i = \eta_id\alpha$
$dF = d\alpha \sum_{i=1}^{n}\frac{\partial F}{\partial x_i}\eta_i = 0$
Since $dx_i$ is arbitrary, $\eta_i$ is also arbitrary. For the $j^{th}$ term, select $\eta_i = 0, i\neq j$ and $\eta_j = 0$, we can obtain
$\frac{\partial F}{\partial x_i} = 0, \forall i$

Variation

• Virtual Displacement: An imagined displacement with no time change, denoted as $\delta x$. Ex. the $\delta y$ in the graph.
• Virtual Path: an imagined path which is different from the right path. Ex. the pink path on the graph.
  
  Change variable: $\delta y(x) = \eta(x)d\alpha$, so that
  $Y(x) = y(x) + \delta y(x) = y(x) + \alpha\eta(x)$

Properties of Variation

$\delta F = F(x_1+\delta x_1, x_2+\delta x_2, \cdots, x_n+\delta x_n) - F(x_1, x_2, \cdots, x_n)$
$\delta F = \nabla F\cdot (\delta x_1, \delta x_2, \cdots, \delta x_n)^T$
$\delta(\frac{d}{dx}y) = \frac{d}{dx}(\delta y)$
$\delta\int_a^bfdx = \int_a^b\delta fdx$

Euler-Lagrange Equation

Find the right path making the following integral stationary, with start point and end point fixed.
$I = \int_a^b f(y,y',x)dx$

Book’s Method

$I$ is only dependent on $\alpha$ and it is stationary at $\alpha = 0, Y(x) = y(x)$, therefore
$\frac{dI}{d\alpha}\Big|_{\alpha=0} = \int_a^b\frac{\partial f(y,y',x)}{\partial\alpha}dx = 0$
Apply the chain rule
$\int_a^b(\frac{\partial f}{\partial y}\frac{\partial y}{\partial\alpha} + \frac{\partial f}{\partial y'}\frac{\partial y'}{\partial\alpha})dx = \int_a^b(\frac{\partial f}{\partial y}\eta + \frac{\partial f}{\partial y'}\eta')dx = 0$
Treat the second term in the integral with the Law of Integral by Parts
$\int_a^b \frac{\partial f}{\partial y'}\eta'dx = \frac{\partial f}{\partial y'}\eta'(x)\Big|_a^b - \int_a^b\eta\frac{d}{dx}\frac{\partial f}{\partial y'}dx$
Since the start point and end point are fixed, the variation $\eta(x_1) = \eta(x_2) = 0$, so the first term will vanish. Take the second term back into $\frac{dI}{d\alpha}$
$\frac{dI}{d\alpha}\Big|_{\alpha=0} = \int_a^b\eta(x)(\frac{\partial f}{\partial y}-\frac{d}{dx}\frac{\partial f}{\partial y'})dx = 0, \quad \forall \eta(x)$
Therefore, we get the Lagrange Equation
$\frac{\partial f}{\partial y}-\frac{d}{dx}\frac{\partial f}{\partial y'} = 0$

Lagrange’s Method (1755)

$\delta I = I(\alpha) - I(0) = \int_a^b \delta f(y,y',x)dx$
$\because \delta x = 0 \quad \therefore \delta f = \frac{\partial f}{\partial y}\delta y + \frac{\partial f}{\partial y'}\delta y'$
Use the change of variable for $\delta y$
$\frac{d\delta f}{d\alpha} = \frac{\partial f}{\partial y}\eta + \frac{\partial f}{\partial y'}\eta'$
Take back to the integral. And for the stationary right path, $\delta I$ should be zero for all infinitesimal $d\alpha$.
$\frac{d\delta I}{d\alpha} = \int_a^b(\frac{\partial f}{\partial y}\eta + \frac{\partial f}{\partial y'}\eta')dx = 0$
Use the same technique as in the Book’s Method, we can get the Euler-Lagrange Equation
$\frac{\partial f}{\partial y}-\frac{d}{dx}\frac{\partial f}{\partial y'} = 0$

Euler’s Method (1756)

Make partition between $x_0$ and $x_n$, with $h = x_j - x_{j-1} = (b-a)/n$. $y_1, y_2, \cdots, y_{n-1}$ are unknown variables.

By definition of Rieman Sum,
$I = \lim_{n\to \infin}S_n =\lim_{n\to \infin}\sum_{j=1}^nf(y_j,y'_j,x_j)h$
By definition of derivatives $y'_j = (y_j - y_{j-1})/h$. Therefore the Rieman Sum is only function of $y_j, \quad j\in\{1,\cdots,n\}$

For stationary $I$,
$\frac{\partial S}{\partial y_k} = 0, \quad \forall k\in \{1,\cdots,n\}$
only two terms ( $k^{th}$ and $(k+1)^{th}$) are involved. Apply chain rule
$\begin{aligned} \frac{\partial S}{\partial y_k} &= h(\frac{\partial f(y_k,y'_k,x_k)}{\partial y_k} + \frac{\partial f(y_k,y'_k,x_k)}{\partial y'_k}\frac{y'_k}{y_k} + \frac{\partial f(y_{k+1},y'_{k+1},x_{k+1})}{\partial y'_{k+1}}\frac{y'_{k+1}}{y_k}) \\ &= h(\frac{\partial f}{\partial y_k}\Big|_k + \frac{1}{h}\frac{\partial f}{\partial y'}\Big|_k - \frac{1}{h}\frac{\partial f}{\partial y'}\Big|_{k+1}) \\ &= h(\frac{\partial f}{\partial y_k}\Big|_k - \frac{\frac{\partial f}{\partial y'}\Big|_{k+1} - \frac{\partial f}{\partial y'}\Big|_k}{h}) \end{aligned}$
As it is for all $k$, we can conclude that
$\frac{\partial f}{\partial y}-\frac{d}{dx}\frac{\partial f}{\partial y'} = 0$

Beltrami’s Identity

$\frac{df}{dx} = \frac{\partial f}{\partial y}y' + \frac{\partial f}{\partial y'}\frac{dy'}{dx} + \frac{\partial f}{\partial x} = \frac{d}{dx}(\frac{\partial f}{\partial y'})y' + \frac{\partial f}{\partial y'}\frac{dy'}{dx} + \frac{\partial f}{\partial x}$
$\frac{\partial f}{\partial x} = \frac{df}{dx} - \frac{d}{dx}(\frac{\partial f}{\partial y'}y') = \frac{d}{dx}(f-\frac{\partial f}{\partial y'}y')$
when $f$ is independent of $x$, we have the Beltrami’s Identity:
$f-\frac{\partial f}{\partial y'}y' = constant$

Examples

Length of the curve on a plane
$I = \int_l dl = \int_l\sqrt{1+y'^2}dx$
Take into the Euler-Lagrange Equation
$0=\frac{d}{dx}\frac{y'}{\sqrt{1+y'^2}} = \frac{y''\sqrt{1+y'^2}-\frac{y'^2y''}{\sqrt{1+y'^2}}}{1+y'^2} = \frac{y''(1+y'^2)-y'^2y''}{(1+y'^2)^{3/2}}$
Which implies $y'' = 0$. Thus, the path that makes the length between two given point shortest is a segment of a line.

Brachistochrone
Under gravity $g$, figure out the path $x=x(y)$ between given points a and b, such that an object falls with the shortest time.
$T = \int_a^b\frac{dl}{v} = \int_a^b\frac{\sqrt{x'^2+1}}{\sqrt{2gy}}dy = \frac{1}{\sqrt{2g}}\int_a^b\sqrt{\frac{x'^2+1}{y}}dy$
Take into the Euler-Lagrange Equation (independent variable is y)
$0 = \frac{d}{dy}\frac{x'}{\sqrt{x'^2+1}\sqrt{y}}$
So that we can find the part within the derivative should be a constant. For convenience, square it and denote it as
$\frac{x'^2}{y(x'^2+1)} = \frac{1}{2a}$
Make a transformation
$x' = \sqrt{\frac{y}{2a-y}}$
Make substitution $y = \alpha(1-\cos\theta)$ and integrate. Without loss of generallity, fix initial point at $(0,0)$, we get
$x=\alpha(\theta-\sin\theta)$
Which is exactly the parametric funtion for cycloid.

Lagrangian Mechanics

Lagrangian

Sometimes for system with conservation of energy and all potential force, we can obtain the equation of motion by $\frac{dE}{dt} = 0$. But with multiple degrees of freedom, this cannot work, which requires further technique.

D’Alembert’s Principle

In dynamics, $\bm{F} - m\bm{a} = \bm{0}$. For virtual work $\delta W = (\bm{F} - m\bm{a})\cdot\delta\bm{r} = 0$
$\int_{t_1}^{t_2}\delta Wdt = \int_{t_1}^{t_2}[\bm{F}-\frac{d}{dt}(m\bm{v})]\delta\bm{r}dt$
For potential forces,
$\int_{t_1}^{t_2}\bm{F}\cdot\delta\bm{r}dt = -\delta\int_{t_1}^{t_2}Udt$
For fixed initial and final problems, we have $\delta\bm{r}(t_1) = \delta\bm{r}(t_2) = 0$. Thus
$-\int_{t_1}^{t_2}\frac{d}{dt}(m\bm{v})\delta\bm{r}dt = -m\bm{v}\delta\bm{r}\Big|_{t_1}^{t_2} + \int_{t_1}^{t_2}m\bm{v}\frac{d}{dt}(\delta\bm{r})dt = \int_{t_1}^{t_2}m\bm{v}\delta\bm{v}dt$

Note: $\delta(x^2) = 2x\delta x$

$-\int_{t_1}^{t_2}\frac{d}{dt}(m\bm{v})\delta\bm{r}dt = \delta\int_{t_1}^{t_2}\frac{m\bm{v}^2}{2}dt = \delta\int_{t_1}^{t_2}Tdt$
Take them back into $\int_{t_1}^{t_2}\delta Wdt$, we have
$\int_{t_1}^{t_2}\delta Wdt = \delta\int_{t_1}^{t_2}(T-U)dt = 0$
Define Lagrangian as
$\mathcal{L} = T-U$
Define Action as
$\mathcal{A} = \mathcal{S} = \int_{t_1}^{t_2}\mathcal{L}dt$

Generalized Coordinate

In Multi-objects system with multiple degree of freedom, Lagrangian may take the form of
$\mathcal{L} = \mathcal{L}(q_1, \dot q_1, q_2, \dot q_2, \cdots, q_n, \dot q_n, t)$
in which $q_1, q_2, \cdots, q_n$ are generalized coordinates.

Each position vector is a function of generalized coordinates and maybe time
$\bm{r}_a = \bm{r}_a(q_1, q_2, \cdots, q_n, t)$

• Scleronomic: $\bm{r}$ is not time-dependent (nature).
• Rheonomic: $\bm{r}$ is time-dependent (not nature).
• Holonomic: The degree of freedom is the same as the number of coordinates needed to describe the system.
• Nonholonomic: Ex. a ball rolls without sliding on a plane has 2 degree of freedom, but its orientation changes are not same when rolling along different paths.

For $j\in\{1,\cdots,n\}$, define
$\textbf{Generalized Momentum: } \mathcal{P_j} = \frac{\partial\mathcal{L}}{\partial\dot q_j}$
$\textbf{Generalized Force: } \mathcal{Q_j} = \frac{\partial\mathcal{L}}{\partial q_j}$

Lagrange Equation and Hamilton’s Principle

D’Alembert’s Principle implies that the action
$\mathcal{A} = \int_{t_1}^{t_2}\mathcal{L}dt$
should be stationary, which is called Hamilton’s Principle.

Apply the Euler-Lagrange Equation, we can obtain the Lagrange Equation
$\frac{\partial\mathcal{L}}{\partial q_j} - \frac{d}{dt}\frac{\partial\mathcal{L}}{\partial\dot q_j}$

Constraint System

Sometimes the system is constraint, like on some surface. For a virtual path from $t_1$ to $t_2$.
$\bm{R}(t) = \bm{r}(t) + \bm{\epsilon}(t), \quad \bm{\epsilon}(t_1) = \bm{\epsilon}(t_2) = 0$
$\bm{\epsilon}(t)$ is infinitesimal variation in the constraint surface. Therefore,
$\begin{aligned} \delta\mathcal{L} &= \frac{1}{2}m[(\bm{\dot r}+\bm{\dot \epsilon})^2-\bm{\dot r}^2] - [U(\bm{r}+\bm{\epsilon},t)-U(\bm{r},t)] \\ &= \frac{1}{2}m(2\bm{\dot \epsilon}\bm{\dot r}+\bm{\dot \epsilon}^2) - (\nabla U\cdot\bm{\epsilon}+O(\bm{\epsilon}^2)) \end{aligned}$
Omit all the higher order terms for infinitesimal variation $\bm{\epsilon}(t)$, and note that $-\nabla U$ is the non-conservative potential force $F$ of the system. Thus, we have $\delta\mathcal{L} = m\bm{\dot \epsilon}\bm{\dot r} - \bm{F}\cdot\bm{\epsilon}$

Hence, for the Action of the system
$\begin{aligned} \delta\mathcal{A} &= \int_{t_1}^{t_2}\delta\mathcal{L}dt = \int_{t_1}^{t_2}(m\bm{\dot \epsilon}\bm{\dot r} - \bm{F}\cdot\bm{\epsilon})dt \\ &= m\bm{\dot r}\bm{\epsilon}(t)\Big|_{t_1}^{t_2} - \bm{\epsilon}\int_{t_1}^{t_2}(m\ddot\bm{r} - F)dt \\ \end{aligned}$
The first term is zero due to the boundary condition. The second term is equivalent to $\bm{F}_{net} - \bm{F}$ which is the constraint force of the system. As the constraint force is perpendicular to the surface. So $\bm{\epsilon}\cdot(\bm{F}_{net} - \bm{F})$ is zero as well.

Therefore, we have proved that for the constraint system the action is stationary.

Lagrange Multiplier

In the proof above, we assume that the generalized coordinates follows the contraint conditions. Ex. in simple pendulum, fix the length of the string and use $\theta$ represents the only degree of freedom. However, when constraint condition takes the form of constraint equation, we have to modify the Lagrange Equation with Lagrange Multiplier.

If there is serveral constraint $\phi_k(q_1,q_2,\cdots,q_n) = 0$, then the Lagrange Equation is modified as
$\frac{\partial\mathcal{L}}{\partial q_j}-\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial\dot q_j} + \sum_k\lambda_k\frac{\partial\phi_k}{\partial q_j} = 0$

Interestingly, in this modified Lagrange Equation, $\lambda_k\frac{\partial\phi_k}{\partial q_j}$ is the contraint force in the particular coordinate $q_j$.

Symmetry

Following two aspects of symmetry is known as Emmy Neother theorem (1915), which indicates that when action doesn’t change symmetry leads to conservation law.

Symmetry in Space

If we move every particle in a system by $\delta x$ and the Lagrangian doesn’t change, the system is symmetric in space.

$\delta \mathcal{L} = \sum\frac{\partial\mathcal{L}}{\partial\dot x_a}\dot{\delta x} + \sum\frac{\partial\mathcal{L}}{\partial x_a}\delta x = \sum\frac{\partial\mathcal{L}}{\partial x_a}\delta x$

Use the Lagrange Equation, we have
$\frac{\delta\mathcal{L}}{\delta x} = \sum(\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial\dot x_a}) = \frac{d}{dt}(\sum\mathcal{P}_a) = 0$

Therefore, the total generalized momentum is constant.

Symmetry in Time

If the Lagrangian is not time-dependent

$d\mathcal{L} = \sum\frac{\partial\mathcal{L}}{\partial\dot q_j}d\dot q_j + \sum\frac{\partial\mathcal{L}}{\partial q_j}dq_j + \frac{\partial\mathcal{L}}{\partial t}dt$

Use the Lagrange Equation in the second term, we have
$-\frac{\partial\mathcal{L}}{\partial t} = \sum(\frac{\partial\mathcal{L}}{\partial\dot q_j}\frac{d\dot q_j}{dt} + \frac{d}{dt}\frac{\partial\mathcal{L}}{\partial\dot q_j}\dot q_j) - \frac{d\mathcal{L}}{dt} = \frac{d}{dt}(\sum\frac{\partial\mathcal{L}}{\partial\dot q_j}\dot q_j - \mathcal{L}) = 0$

Define Hamiltonian as
$\mathcal{H} = \sum\frac{\partial\mathcal{L}}{\partial\dot q_j}\dot q_j - \mathcal{L}$

If every object has a position vector independent from time, $T$ is a second order homogeneous function for $\dot q_1, \dot q_2, \cdots, \dot q_n$

If function $f(x_1, x_2, \cdots, x_n)$ has the property
$f(kx_1, kx_2, \cdots, kx_n) = k^nf(x_1, x_2, \cdots, x_n)$
Then it is called homogeneous function with the order of $n$. And it’s proved that for this function
$\sum\frac{\partial f}{\partial x_j}x_j = nf(x_1, x_2, \cdots, x_n)$

$\mathcal{H} = \sum\frac{\partial T}{\partial\dot q_j}\dot q_j - \mathcal{L} = 2T - T + U = E$

Which implies the conservation of energy.

Electro-Magnetic Modification

$m\bm{a} = q(\bm{E}+\bm{v}\times\bm{B})$
$\mathcal{L} = \frac{1}{2}m\bm{v}^2 - q(\phi - \bm{v}\cdot\bm{A})$

Non-Inertia Frame of Reference

Translational Motion

The unit vectors don’t change. Thus, $\bm{r} = \bm{r}_{o'} + \bm{r}'$
Get time derivative of both sides, $\bm{a} = \bm{a}_{o'} + \bm{a}'$ Therefore, we have the Newton’s Second Law with the Inertial Force or Frictional Force
$\sum{\bm{F}} - m\bm{a}_{o'} = m\bm{a}'$

Tidal Motion

Ignore the rotation of the Earth.

Take the Earth as the Frame of Reference

$m\ddot\bm{r} = \sum\bm{F}-m\bm{a}_{o'} = m\bm{g} - G\frac{M_mm}{s^2}\bm{\hat{s}} + G\frac{M_mm}{L^2}\bm{\hat{L}} + \bm{F}_{other}$

In which the third term is the inertial force. And we define
$F_{tidal} = - G\frac{M_mm}{s^2}\bm{\hat{s}} + G\frac{M_mm}{L^2}\bm{\hat{L}}$
The inertial force has fixed direction and fixed magnitude, so it is potential. Thus, all the potential energy:
$\begin{aligned} &U_L = -G\frac{M_mm}{L^2}x = -G\frac{M_mm}{L^2}R\cos\phi \\ &U_g = mgh(\phi) \\ &U_s = -G\frac{M_mm}{s} \\ \therefore\; &\frac{U}{m} = -G\frac{M_m}{s}-G\frac{M_m}{L^2}R\cos\phi+gh(\phi) \end{aligned}$
With Law of Cosine and Taylor Expansion
$\begin{aligned} \frac{1}{s} &= \frac{1}{L}(1+(\frac{R}{L})^2 + 2(\frac{R}{L})\cos\phi)^{-\frac{1}{2}} \\ &= \frac{1}{L}\Big\{ 1+(-\frac{1}{2})\big[(\frac{R}{L})^2 + 2(\frac{R}{L})\cos\phi\big] + \frac{3}{8}\big[(\frac{R}{L})^2 + 2(\frac{R}{L})\cos\phi\big]^2 \Big\} \\ &= \frac{1}{L}\Big\{ 1-\frac{1}{2}(\frac{R}{L})^2 - (\frac{R}{L})\cos\phi + \frac{3}{2}(\frac{R}{L})^2\cos^2\phi + O(\frac{R}{L})^3 \Big\} \end{aligned}$
Take this back into $\frac{U}{m}$, which should be a constant for the surface of ocean.
$\frac{U}{m} = -G\frac{M_m}{L} + \frac{GM_mR^2}{2L^3} -\frac{3GM_mR^2}{2L^3}\cos^2\phi + gh(\phi)$
Select $h = 0$ at $\phi = 90\degree$, and substitute $g = GM_e/R^2$
$h(\phi) = \frac{3M_mR^4}{2M_eL^3}\cos^2\phi$

Rotational Motion

For rotational motion:
$\bm{v} = \bm{\omega}\times\bm{r}$

For a changing unit vector (for example $\bm{i}$), we have
$\bm{\dot i} = (\bm{\dot i}\cdot\bm{i})\bm{i} + (\bm{\dot i}\cdot\bm{j})\bm{j} + (\bm{\dot i}\cdot\bm{k})\bm{k}$

Note: $\bm{i}\cdot\bm{i} = 1$, derivative of both sides: $2\bm{\dot i}\cdot\bm{i} = 0$. Similarly, $\bm{\dot i}\cdot\bm{j} = -\bm{\dot j}\cdot\bm{i}$

Any position vector can express in two frames of reference (inertial and rotational)
$\bm{r} = x\bm{i} + y\bm{j} + z\bm{k} = x_0\bm{i_0} + y_0\bm{j_0} + z_0\bm{k_0}$

If $\bm{r}$ is constant ( $\dot x=\dot y=\dot z=0$), we have
$\begin{aligned} \bm{\dot r} &= x\bm{\dot i} + y\bm{\dot j} + z\bm{\dot k} \\ &= x[(\bm{\dot i}\cdot\bm{j})\bm{j}+(\bm{\dot i}\cdot\bm{k})\bm{k}] + y[(\bm{\dot j}\cdot\bm{i})\bm{i}+(\bm{\dot j}\cdot\bm{k})\bm{k}] + z[(\bm{\dot k}\cdot\bm{i})\bm{i}+(\bm{\dot k}\cdot\bm{j})\bm{j}] \\ &= [(\bm{\dot j}\cdot\bm{i})y + (\bm{\dot k}\cdot\bm{i})z]\bm{i} + [(\bm{\dot i}\cdot\bm{j})x + (\bm{\dot k}\cdot\bm{j})z]\bm{j} + [(\bm{\dot i}\cdot\bm{k})x + (\bm{\dot j}\cdot\bm{k})y]\bm{k} \\ &= [(\bm{\dot k}\cdot\bm{i})z - (\bm{\dot i}\cdot\bm{j})y]\bm{i} - [(\bm{\dot j}\cdot\bm{k})z - (\bm{\dot i}\cdot\bm{j})x]\bm{j} + [(\bm{\dot j}\cdot\bm{k})y - (\bm{\dot k}\cdot\bm{i})x]\bm{k} \end{aligned}$
Therefore, define $\bm{\Omega} = (\bm{\dot j}\cdot\bm{k}, \bm{\dot k}\cdot\bm{i}, \bm{\dot i}\cdot\bm{j})$
$\bm{\dot r} = \left|\begin{matrix} \bm{i} & \bm{j} & \bm{k} \\ \bm{\dot j}\cdot\bm{k} & \bm{\dot k}\cdot\bm{i} & \bm{\dot i}\cdot\bm{j} \\ x & y & z \end{matrix}\right| = \bm{\Omega}\times\bm{r}$

Index Notation $x_l\bm{e}_l$ is defined as
$\sum_{l = 1}^{3}x_l\bm{e}_l$

If $\bm{r}$ is not constant, $\bm{\dot r} = \dot x_l\bm{e}_l + x_l\bm{\dot e}_l$. Therefore, for any vector $r$, we have:

$\bm{\dot r}\big|_{in} = \bm{\dot r}\big|_{rot} + \bm{\Omega}\times\bm{r}$

Lab frame = inertial frame, Moving frame = rotational frame

$\bm{\dot r_L}\big|_{in} = \bm{\dot r_{OM}}\big|_{in} + \bm{\dot r_M}\big|_{in} = \bm{\dot r_{OM}}\big|_{in} + \bm{\Omega}\times\bm{r_M} + \bm{\dot r_M}\big|_{rot}$

Also, for unit vectors
$\bm{\dot e_l}\big|_{in} = \bm{\Omega}\times\bm{\dot e_l}\big|_{rot}$

Velocity: (assume the two frames of reference share the same origin)
$\bm{v} = \bm{\dot r}\big|_{in} = \bm{\dot r}\big|_{rot} + \bm{\Omega}\times\bm{r} = \bm{v}_{rot} + \bm{\Omega}\times\bm{r}$

Acceleration: (assume $\bm{\Omega}$ is constant)
$\begin{aligned} \bm{a} = \bm{\dot v}\big|_{in} &= \bm{\dot v}\big|_{rot} + \bm{\Omega}\times\bm{v} \\ &= \frac{d}{dt}(\bm{v}_{rot} + \bm{\Omega}\times\bm{r})\Big|_{rot} + \bm{\Omega}\times(\bm{v}_{rot} + \bm{\Omega}\times\bm{r}) \\ &= \bm{a}_{rot} + 2\bm{\Omega}\times\bm{v}_{rot} + \bm{\Omega}\times(\bm{\Omega}\times\bm{r}) \end{aligned}$

Note:
$\bm{\Omega}\times(\bm{\Omega}\times\bm{r}) = \bm{\Omega}(\bm{\Omega}\cdot\bm{r}) - \bm{r}(\bm{\Omega}\cdot\bm{\Omega}) = \Omega^2\bm{r}_{\parallel} - \Omega^2\bm{r} = -\Omega^2\bm{r}_{\perp}$
Which is known as the centripetal acceleration.
$2\bm{\Omega}\times\bm{v}_{rot}$
is known as the Coriolis acceleration.

Newton’s Second Law in non-inertial frame

$\sum\bm{F} + 2m\bm{v}_{rot}\times\bm{\Omega} + m\Omega^2\bm{r}_{\perp} = m\bm{a}_{rot}$

Ficticious forces/forces of inertial: $\bm{F}_{fict} = 2m\bm{v}_{rot}\times\bm{\Omega} + m\Omega^2\bm{r}_{\perp}$

• the first term is called the Coriolis force
• the second term is called the centrifugal force.

E.x. of Coriolis force:

• Cyclon
• Foucalt Pendulum

Foucalt Pendulum

With the Newton’s Second Law:
$m\ddot x = -T\frac{x}{L} + 2m\dot y\Omega\cos\theta \quad\text{and}\quad m\ddot y = -T\frac{y}{L} - 2m\dot x\Omega\cos\theta$
Take $T \approx mg$, and multiply the second expression by $i$ and make it imaginary, we have
$\ddot\eta = -\frac{g}{L}\dot\eta - 2i\Omega\cos\theta\eta, \quad\text{with } \eta = x + iy$

Denote $\omega_1^2 = g/L$ and $\omega_2^2 = \Omega\cos\theta$ while $\omega_2 << \omega_1$

Therefore, we have the solution
$\eta = e^{i\omega t}, \quad\text{with } \omega = -\omega_2\plusmn\omega_1[1+(\frac{\omega_2}{\omega_1})^2]^{\frac{1}{2}} \approx -\omega_2\plusmn\omega_1(1+\frac{1}{2}\frac{\omega_2^2}{\omega_1^2}) = -\omega_2\plusmn\omega\plusmn\frac{1}{2}\frac{\omega_2^2}{\omega_1}$

For approximation, we have $\omega = -\omega_2\plusmn\omega_1$. Thus
$\eta(t) = e^{-i\omega_2t}(C_1e^{i\omega_1t} + C_2e^{-i\omega_1t})$

If $\omega_2 = 0$ (ignore earth rotation), we have
$\tilde\eta = C_1e^{i\omega_1t} + C_2e^{-i\omega_1t} = \tilde x + i\tilde y$

Therefore,
$\eta(t) = e^{-i\omega_2t}(\tilde x + i\tilde y) = (\cos\omega_2t + i\sin\omega_2t)(\tilde x + i\tilde y)$

Hence,
$\left[\begin{matrix} x \\ y \end{matrix}\right] = \left[\begin{matrix} \cos\omega_2t & \sin\omega_2t \\ -\sin\omega_2t & \cos\omega_2t \end{matrix}\right]\left[\begin{matrix} \tilde x \\ \tilde y \end{matrix}\right]$
which is clearly a rotational matrix with angle $\omega_2 = \Omega\cos\theta$
$T' = \frac{2\pi}{\omega_2} = \frac{2\pi}{\Omega\cos\theta} = \frac{T_0}{\cos\theta}, \quad\text{with } T_0 = 24h$

Rigid Body Motion

Johan König (1712-1751)
$\bm{L} = \bm{r}_{cm}\times\bm{p}_{cm} + \bm{L}' = \bm{L}_{orb} + \bm{L}_{spin}$
$T = \frac{1}{2}M\bm{\dot r}_{cm}^2 + T' = T_{cm} + T_{rel}$
In which
$T' = \frac{1}{2}\sum mv_a^2 = \frac{1}{2}\sum m(\bm{\omega}\times\bm{r}_a)^2$

Note $(\bm{a}\times\bm{b})^2 = a^2b^2-(\bm{a}\cdot\bm{b})^2$
$T' = \frac{1}{2}\sum m[\omega^2r_a^2-(\bm{\omega}\cdot\bm{r}_a)^2]$
Use the index notation,
$T' = \frac{1}{2}\sum m[\omega_i\omega_ix_{aj}x_{aj} - \omega_ix_{ai}\omega_{j}x_{aj}] = \frac{1}{2}\sum m\omega_i\omega_j[\delta_{ij}x_{ak}x_{ak} - x_{ai}x_{aj}]$
Define
$I_{ij} = \sum m(\delta_{ij}x_{ak}x_{ak} - x_{ai}x_{aj})$
we have
$T' = \frac{1}{2}I_{ij}\omega_i\omega_j$

Tensor of Inertia

In the last part, we defined the Tensor of Inertia
$\bm{I} = \left[ \begin{matrix} \sum m(y^2+z^2) & -\sum mxy & -\sum mxz \\ -\sum myx & \sum m(x^2+z^2) & -\sum myz \\ -\sum mzx & -\sum mzy & \sum m(x^2+y^2) \end{matrix} \right]$
On the diagonal line, $I_{xx}$ is called Moment of Inertia, while the others are called Product of Inertia.

Similar to the rotational kinetic energy, we have
$\bm{L}' = \bm{I}\bm{\omega} \quad\text{and}\quad T' = \frac{1}{2}\bm{L}'\cdot\bm{\omega}$

For integral,
$I_{ij} = \iiint\rho(\bm{r})(\delta_{ij}x_{ak}x_{ak} - x_{ai}x_{aj})dxdydz$

Principle Axis

For the Matrix of Tensor of Inertia, if we find the eigenvector $\bm{\tilde\omega}$ of it
$\bm{I}\bm{\tilde\omega} = I\bm{\tilde\omega}$
then the direction of $\bm{\tilde\omega}$ is the Principle Axis of the object, with the eigenvalue $I$ as the Principle Moment of Inertia. The principle axis must exist, as the matrix of inertia tensor is symmetric and therefore orthogonally diagonalizable.

The Euler’s Equations

Let the rotational frame take the coordinates of the principle axes $\bm{e}_1,\bm{e}_2,\bm{e}_3$, with the moment of inertia $I_1, I_2, I_3$ respectively.
$\bm{L} = (I_1\omega_1, I_2\omega_2, I_3\omega_3)$
$\bm{\Gamma} = \frac{d\bm{L}}{dt}\Big|_{in} = \frac{d\bm{L}}{dt}\Big|_{rot} + \bm{\omega}\times\bm{L}$
Then we get the Euler’s Equations
$\left\{ \begin{aligned} &\Gamma_1 = I_1\dot\omega_1 - (I_2 - I_3)\omega_2\omega_3 \\ &\Gamma_2 = I_2\dot\omega_2 - (I_3 - I_1)\omega_3\omega_1 \\ &\Gamma_3 = I_3\dot\omega_3 - (I_1 - I_2)\omega_1\omega_2 \end{aligned} \right.$
Use the tensor operator
$\sum_{k=1}^{3}\epsilon_{ijk}I_k\dot\omega_k - (I_i-I_j)\omega_i\omega_j = \sum_{k=1}^{3}\epsilon_{ijk}\Gamma_k$

Types of Rigid Body

1. $I_1 = I_2 = I_3$: Spherical
2. $I_1 = I_2 \ne I_3$: Symmetrical
   • $I_1 = I_2 < I_3$: Oblate symmetrical (disk shaped)
   • $I_3 < I_1 = I_2$: Prolate symmetrical (cigar shaped)
   • $I_3 << I_1 = I_2$: Linear rotor
3. $I_1 \ne I_2 \ne I_3$: Asymmetrical

Examples of motion

Type	Torque	Angular Velocity	Conclusion
	$\Gamma_1 \ne 0,\Gamma_2=\Gamma_3=0$	$\omega_2=\omega_3=0$	$\dot\omega_2=\dot\omega_3=0$
Spherical	$\bm{\Gamma}=\bm{0}$		$\bm{\omega}=\bm{0}$
Symmetrical	$\bm{\Gamma}=\bm{0}$		$\dot\omega_3=0$ and below
Asymmetrical	$\bm{\Gamma}=\bm{0}$	$(\epsilon_1,\epsilon_2,\omega)$	below

Symmetrical Free: ( $\bm{\Gamma}=\bm{0}$)
By the Euler’s Equations
$\dot\omega_1 = \Omega\omega_2 \quad\text{and}\quad \dot\omega_2 = -\Omega\omega_1 \quad\text{where}\quad \Omega = \frac{I-I_3}{I}\omega_3$
Solve it by $\eta=\omega_1+i\omega_2$ and ignore phares
$\bm{\omega} = (|\eta_0|\cos\Omega t, -|\eta_0|\sin\Omega t, \omega_3)$
Particularly, $(\bm{e}_3\times\bm{\omega})\cdot\bm{L} = 0$, which means these three vectors are coplanar.

Asymmetrical Free: ( $\bm{\Gamma}=\bm{0}$)
For a rotational motion w.r.t. the third principle axis with a little variation $\bm{\omega} = (\epsilon_1, \epsilon_2, \omega)$. With the Euler’s Equation, we have
$\omega \approx \text{constant} \quad\text{with}\quad \dot\epsilon_1 = \frac{I_2-I_3}{I_1}\omega\epsilon_2 \quad\text{and}\quad \dot\epsilon_2 = \frac{I_3-I_1}{I_2}\omega\epsilon_1$
Take the time derivative to the third equation and substitute by the second equation, we get
$\ddot\epsilon_2 = \frac{(I_3-I_1)(I_2-I_3)}{I_1I_2}\omega^2\epsilon_2$

• $I_3 < I_1,I_2$ or $I_3 > I_1,I_2$: oscillation, stable
• $I_1 < I_3 < I_2$: exponentially growth, instable

Euler’s Angle

A way to describe the orientation of object by rotation from the lab frame

1. Rotate around z-axis by $\phi$
2. Rotate around new x-axis by $\theta$
3. Rotate around new z-axis by $\psi$

Mathematically, we have the transformation from the lab frame coordinates
$\left[\begin{matrix}x \\ y \\ z\end{matrix}\right] = \left[\begin{matrix} \cos\psi & \sin\psi & 0 \\ -\sin\psi & \cos\psi & 0 \\ 0 & 0 & 1 \end{matrix}\right] \left[\begin{matrix} 1 & 0 & 0 \\ 0 & \cos\theta & \sin\theta \\ 0 & -\sin\theta & \cos\theta \end{matrix}\right] \left[\begin{matrix} \cos\phi & \sin\phi & 0 \\ -\sin\phi & \cos\phi & 0 \\ 0 & 0 & 1 \end{matrix}\right] \left[\begin{matrix}x_L \\ y_L \\ z_L\end{matrix}\right]$

Therefore, $\bm{\omega} = \bm{\dot\phi} + \bm{\dot\theta} + \bm{\dot\psi}$

In the there step, we have three rotation. Transform there angular velocity individually into the final rotational frame and add the results together, we get
$\bm{\omega} = (\dot\phi\sin\theta\sin\psi+\dot\theta\cos\psi, \dot\phi\sin\theta\cos\psi-\dot\theta\sin\psi,\dot\phi\cos\theta+\dot\psi)$

Hamiltonian Dynamics

Legendre Transformation

For a function $f(x,y)$, if we choose a new independent variable $u = \partial f/\partial x$, and write $x=x(u,y)$. Now introduce a new function
$g(u,y) = ux(u,y)-f(x(u,y),y)$
Therefore,
$\frac{\partial g}{\partial u} = x + u\frac{\partial x}{\partial u} - \frac{\partial f}{\partial x}\frac{\partial x}{\partial u} = x$
$\frac{\partial g}{\partial y} = u\frac{\partial x}{\partial y} - \frac{\partial f}{\partial x}\frac{\partial x}{\partial y} - \frac{\partial f}{\partial y} = - \frac{\partial f}{\partial y}$
Thus,
$dg = xdu-\frac{\partial f}{\partial y}dy$

Hamiltonian

From Lagrangian Mechanics
$d\mathcal{L} = \sum\frac{\partial\mathcal{L}}{\partial\dot q_j}d\dot q_j + \sum\frac{\partial\mathcal{L}}{\partial q_j}dq_j + \frac{\partial\mathcal{L}}{\partial t}dt, \qquad \frac{\partial\mathcal{L}}{\partial\dot q_j} = p_j$
Follow the Legendre Transformation ( $\dot q_j\to p_j$), define Hamiltonian as
$\mathcal{H}(p_j,q_j,t) = \sum p_jq_j - \mathcal{L}(\dot q_j, q_j, t)$
And therefore
$d\mathcal{H} = \sum\dot q_jdp_j + \sum\frac{\partial\mathcal{L}}{\partial q_j}dq_j-\frac{\partial\mathcal{L}}{\partial t}dt$
which implies
$\frac{\partial\mathcal{H}}{\partial p_j} = \dot q_j \qquad \frac{\partial\mathcal{H}}{\partial q_j} = -\frac{\partial\mathcal{L}}{\partial q_j} \qquad \frac{\partial\mathcal{H}}{\partial t} = -\frac{\partial\mathcal{L}}{\partial t}$

Dynamics

From the Lagrange Equations
$\frac{\partial\mathcal{L}}{\partial q_j} = \frac{d}{dt}\frac{\partial\mathcal{L}}{\partial\dot q_j}$
Bring this into the derivatives of Hamiltonian, we get the equation of motion for Hamiltonian: Canonical Equations
$\dot q_j = \frac{\partial\mathcal{H}}{\partial p_j} \qquad \dot p_j = -\frac{\partial\mathcal{H}}{\partial q_j}$

Hence,
$\frac{d\mathcal{H}}{dt} = \sum\frac{\partial\mathcal{H}}{\partial q_j}\frac{\partial\mathcal{H}}{\partial p_j} - \sum\frac{\partial\mathcal{H}}{\partial p_j}\frac{\partial\mathcal{H}}{\partial q_j} +\frac{\partial\mathcal{H}}{\partial t} = \frac{\partial\mathcal{H}}{\partial t}$
If $\mathcal{H}$ is independent of $t$, Hamiltonian is conserved. In scleronomic system, energy is constant.

For a function $f=f(q,p,t)$, we have
$\frac{df}{dt} = \sum\frac{\partial f}{\partial q_j}\frac{\partial\mathcal{H}}{\partial p_j} - \sum\frac{\partial f}{\partial p_j}\frac{\partial\mathcal{H}}{\partial q_j} +\frac{\partial f}{\partial t} = [f,\mathcal{H}] + \frac{\partial f}{\partial t}$

Note: Poissom’s Bracket
$[f,g] = \sum\Big(\frac{\partial f}{\partial q_j}\frac{\partial g}{\partial p_j} - \frac{\partial f}{\partial p_j}\frac{\partial g}{\partial q_j}\Big)$
Jacobi Identity:
$[[f,g],h] + [[h,f],g] + [[g,h],f] = 0$
Poissom’s Theorem: if $f,g$ are constant w.r.t. time and is not explicit dependent on time,
$[f,g] = \text{constant}$

If $f,g$ is constant w.r.t. time and is not explicit dependent on time, then
$[g,\mathcal{H}] = [f,\mathcal{H}] = 0 \qquad [[f,g],\mathcal{H}] + [[\mathcal{H},f],g] + [[g,\mathcal{H}],f] = 0$

Lagrangian from Hamiltonian

For $\mathcal{H} = p^2/2m + A(q)p + B(q)$, we can obtain $p = m[\dot q-A(q)]$ from the Hamiltonian Equations. Therefore,
$\mathcal{L} = p\dot q - \mathcal{H} = \frac{m[\dot q-A(q)]^2}{2} - B(q)$

Continuity Equation

From fluid dynamics, For a infinitesimal cube with volume $d\tau = dxdydz$,
$\frac{\partial dm}{\partial t} = \frac{\partial \rho dt}{\partial t} = dt\frac{\partial \rho}{\partial t}$

For fluid travels from face A to face B (opposite sides), with Taylor expansion, we have
$(\rho v_x)_B = (\rho v_x)_A + \frac{\partial (\rho v_x)}{\partial x}dx$
and therefore
$\Delta dm_A-\Delta dm_B = -\frac{\partial (\rho v_x)}{\partial x}d\tau dt$
Similarly for the other two pairs of faces, so
$\frac{dm}{dt} = - \nabla\cdot(\rho\bm{v})d\tau$
From another aspect, any geometric shape $G$
$\frac{dm}{dt} = - \iiint_{G}\frac{\partial\rho}{\partial t}d\tau$
when reduced to infinitesimal cube, we can get the continuity equation
$\frac{\partial\rho}{\partial t} + \nabla\cdot(\rho\bm{v}) = 0$
With divergence theorem
$\frac{\partial\rho}{\partial t} = -\oiint_{S} \rho\bm{v}\cdot d\bm{A}$

Liouville’s Theorem

Imagine N copies of system in p-q 2N dimensional space form a “gas” following the continuity equation ( $N \to \infin$)
$\frac{\partial\rho}{\partial t} + \sum_{j}\Big(\frac{\partial(\rho\dot q_j)}{\partial q_j} + \frac{\partial(\rho\dot p_j)}{\partial p_j}\Big) = 0, \qquad\text{where } \rho = \frac{\text{number of points}}{\text{volume}}$
By the product law
$\frac{\partial\rho}{\partial t} + \sum_{j}\Big(\frac{\partial\rho}{\partial q_j}\dot q_j + \frac{\partial\rho}{\partial p_j}\dot p_j + \rho(\frac{\partial\dot q_j}{\partial q_j} + \frac{\partial\dot p_j}{\partial p_j})\Big) = 0$
From the Hamiltonian Equation, the third term in the summation vanishes, and the first two terms change into a Poissom’s Bracket
$\frac{\partial\rho}{\partial t} + [\rho, \mathcal{H}] = 0 = \frac{d\rho}{dt}$
With the evolusion of the system, the points occupy the same spacial “volume” in the p-q space.

Material Derivative

Stream Line is a continuous line that a particle travels, following the velocity field. Derivative w.r.t. time along the stream line is called Material Derivative.

For function $\phi(\bm{r}, t)$, the partial derivative w.r.t. time is
$\frac{\partial\phi}{\partial t} = \frac{\phi(\bm{r}, t+dt)-\phi(\bm{r}, t)}{dt}$
but the Material Derivative should be
$\frac{\mathcal{D}\phi}{\mathcal{D} t} = \frac{\phi(\bm{r}+\bm{v}dt, t+dt)-\phi(\bm{r}, t)}{dt} = (\bm{v}\cdot\nabla)\phi + \frac{\partial\phi}{\partial t}$
The expression is same when $\phi$ is a vector-value function.

For fluid dynamics, we can write the acceleration and the continuity equation as
$\bm{a} = \frac{\mathcal{D}\bm{v}}{\mathcal{D} t} \qquad \frac{\mathcal{D}\rho}{\mathcal{D} t} + \rho\nabla\cdot\bm{v} = 0$

Fluid Dynamic

With $d\tau = dxdydz$, the Newton’s Second Law
$dm\bm{a} = d\bm{F}_{net} \to \rho\bm{a} = \bm{f}_{net}, \qquad\text{where}\quad \bm{f}_{net} = \frac{d\bm{F}_{net}}{d\tau}$
For force due to pressure or gravity, we have
$\bm{f}_p = -\nabla p \qquad \bm{f}_g = \rho\bm{g}$
Therefore, we get the Euler’s Equation for fluid
$\rho(\bm{v}\cdot\nabla)\phi + \rho\frac{\partial\bm{v}}{\partial t} = -\nabla p + \rho\bm{g}$

Emmy Neother’s Theorem

For symmetry, i.e. spacial or time shifting with the action stays still

$d\tilde q_j = dq_j + \epsilon d\psi_j = dq_j + \epsilon \frac{d\psi_j}{dt}dt$
$d\tilde t = dt + \epsilon d\chi = (1 + \epsilon\frac{d\chi}{dt})dt$
Therefore, with the Taylor Expansion to the first order,
$\frac{d\tilde q_j}{d\tilde t} = (\dot q_j + \epsilon\frac{d\psi_j}{dt})(1 - \epsilon \frac{d\chi}{dt}) = \dot q_j(1 - \epsilon\frac{d\chi}{dt}) + \epsilon\frac{d\psi_j}{dt}$
Omit subscript $j$, write the action of the system
$\begin{aligned} \tilde S &= \int_{\tilde t_1}^{\tilde t_2}\mathcal{L}(\tilde q,\tilde{\dot q}, \tilde t)d\tilde t = \int_{t_1}^{t_2}\mathcal{L}(\tilde q, \tilde{\dot q}, \tilde t)(1 + \epsilon\frac{d\chi}{dt})dt \\ &= \int_{t_1}^{t_2}\mathcal{L}(q + \epsilon\psi, \dot q + \epsilon(\frac{d\chi}{dt} - \dot q \frac{d\chi}{dt}), t+\epsilon\chi)(1 + \epsilon\frac{d\chi}{dt})dt \end{aligned}$
where the second term is the substitution of $d\tilde q/dt$. Thus,
$\tilde S - S = \epsilon\int_{t_1}^{t_2}\Big[\frac{\partial\mathcal{L}}{\partial q}\psi + \frac{\partial\mathcal{L}}{\partial\dot q}\frac{d\psi}{dt} - \frac{\partial\mathcal{L}}{\partial\dot q}\dot q\frac{d\chi}{dt} + \frac{\partial\mathcal{L}}{\partial t}\chi + \mathcal{L}\frac{d\chi}{dt} \Big]dt$

Use the integral by part, we can deal with several terms,
$\int_{t_1}^{t_2}\frac{\partial\mathcal{L}}{\partial\dot q}\frac{d\psi}{dt}dt = \frac{\partial\mathcal{L}}{\partial\dot q}\psi\Big|_{t_1}^{t_2} - \int_{t_1}^{t_2}\frac{d}{dt}(\frac{\partial\mathcal{L}}{\partial\dot q})\psi dt$
$\int_{t_1}^{t_2}\frac{\partial\mathcal{L}}{\partial\dot q}\dot q\frac{d\chi}{dt}dt = \frac{\partial\mathcal{L}}{\partial\dot q}\dot q\chi\Big|_{t_1}^{t_2} - \int_{t_1}^{t_2}\frac{d\mathcal{L}}{dt}\chi dt$
$\int_{t_1}^{t_2}\mathcal{L}\frac{d\chi}{dt}dt = \mathcal{L}\chi\Big|_{t_1}^{t_2} - \int_{t_1}^{t_2}\frac{d\mathcal{L}}{dt}\chi dt$

Take these together, and for the contant action, we have
$\begin{aligned} 0 &= \frac{\tilde S - S}{\epsilon} &= \Big[\frac{\partial\mathcal{L}}{\partial\dot q}\psi - \frac{\partial\mathcal{L}}{\partial\dot q}\dot q\chi + \mathcal{L}\chi \Big]_{t_1}^{t_2} + \int_{t_1}^{t_2}\psi(\frac{\partial\mathcal{L}}{\partial q} - \frac{d}{dt}\frac{\partial\mathcal{L}}{\partial\dot q})dt \\ &&+ \int_{t_1}^{t_2}\chi(\frac{d}{dt}(\dot q\frac{\partial\mathcal{L}}{\partial\dot q}) + \frac{\partial\mathcal{L}}{\partial t} - \frac{d\mathcal{L}}{dt})dt \\ \end{aligned}$
From the Lagrangian Equation, we can prove that the integral term equals to zero. Hence,

$\Big[\frac{\partial\mathcal{L}}{\partial\dot q}\psi - \frac{\partial\mathcal{L}}{\partial\dot q}\dot q\chi + \mathcal{L}\chi \Big]_{t_1}^{t_2} = 0$
which can be rewrited as the Emmy Neother’s Theorem (1915)
$\mathcal{H}\chi + \sum p\psi = \text{constant}$