PHYS104: Advanced Mechanics

Index

PHYS104: Advanced Mechanics

Calculus of Variation
Euler-Lagrange Equation
Lagrangian Mechanics

Lagrangian

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
$\ddot\bm{r} = f(t,\bm{r},\dot\bm{r})$
which has two degrees of freedom $\bm{r}(0)$ and $\dot\bm{r}(0)$

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

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
$\begin{aligned} \frac{\partial S}{\partial y_k} &= h(\frac{\partial f(y_k,y'_k,x_k)}{\partial y_k} + \frac{\partial f(y_k,y'_k,x_k)}{\partial y'_k}\frac{y'_k}{y_k} + \frac{\partial f(y_{k+1},y'_{k+1},x_{k+1})}{\partial y'_{k+1}}\frac{y'_{k+1}}{y_k}) \\ &= h(\frac{\partial f}{\partial y_k}\Big|_k + \frac{1}{h}\frac{\partial f}{\partial y'}\Big|_k - \frac{1}{h}\frac{\partial f}{\partial y'}\Big|_{k+1}) \\ &= h(\frac{\partial f}{\partial y_k}\Big|_k - \frac{\frac{\partial f}{\partial y'}\Big|_{k+1} - \frac{\partial f}{\partial y'}\Big|_k}{h}) \end{aligned}$ ∂yk∂S=h(∂yk∂f(yk,yk′,xk)+∂yk′∂f(yk,yk′,xk)ykyk′+∂yk+1′∂f(yk+1,yk+1′,xk+1)ykyk+1′)=h(∂yk∂f∣∣∣k+h1∂y′∂f∣∣∣k−h1∂y′∂f∣∣∣k+1)=h(∂yk∂f∣∣∣k−h∂y′∂f∣∣∣k+1−∂y′∂f∣∣∣k)
As it is for all $k$ , we can conclude that
$\frac{\partial f}{\partial y}-\frac{d}{dx}\frac{\partial f}{\partial y'} = 0$

Ex. 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.

Ex. 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 integral. 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.

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:
$f-\frac{\partial f}{\partial y'}y' = constant$

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
$A = \int_{t_1}^{t_2}\mathcal{L}dt$