Topics in Geometric Mechanics: Week 8

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Week 8

This week we will look at

• the Euler-Lagrange equations
• systems with symmetry
• conserved quantities

The Euler-Lagrange Equations

Recall

$L : TM \to \R$ is a smooth function of the position $x$ and velocity $\dot{x}$. The Euler-Lagrange equations are \begin{align*} \eulerlagrange{t}{x}{\dot{x}}{L} &= 0 \end{align*}

The Central Force Problem

Conservative, central force \begin{align*} F(x) &= -\lambda(r)x &&\implies&& F(x) = -\nabla V(r), &&&& r=\norm{x} \end{align*}

The central force problem.

Conservative, central force \begin{align*} L &= \frac{1}{2} \norm{\dot{x}}^2 - V(r)\\ \eulerlagrange{t}{x}{\dot{x}}{L} &= 0 &&\implies&& \ddot{x} = - \lambda(r)x. \end{align*}

The central force problem.

Quadratic Potential

Suppose that

\begin{align*} V(x) &= \frac{1}{2} \norm{x}^2 = \frac{1}{2} r^2, && \ddot{x} = -x. \end{align*}

Newtonian Potential

Suppose that

\begin{align*} V(x) &= -\dfrac{1}{r}, && \ddot{x} = - \dfrac{x}{r^3}. \end{align*} Then \begin{align*} L &= \frac{1}{2} (\dot{r}^2 + r^2 \dot{\theta}^2) + \dfrac{1}{r}. \end{align*} So \begin{align*} \ddt{t}{} r^2 \dot{\theta} &= 0 && \ddot{r} = r \dot{\theta}^2 - \dfrac{1}{r^2}. \end{align*}

Solutions

Substitute \begin{align*} \ddt{t}{\theta} &= \dfrac{c}{r^2}, && r = u^{-1} && \ddt{t}{r} = c u^2 \ddt{\theta}{r} = -c \ddt{\theta}{u}. \end{align*} This gives \begin{align*} %%-c^2 u^2 \dndtn{\theta}{u}{2} &= c^2 u^3 - u^2,\\ \dndtn{\theta}{u}{2} &= c^{-2} - u. \end{align*}

The solutions are conic sections

\[ r = \dfrac{1}{c^{-2} + e \cos(\theta-\theta_0)} \]

An elliptical, parabolic and hyperbolic orbit.

Tensoriality

The Euler-Lagrange equations \[ \eulerlagrange{t}{x}{\dot{x}}{L} = 0 \qquad (*) \] are tensorial. That is, if $y = \phi(x)$ is a change of coordinates, then (*) takes the same form in the $y$ system of coordinates.

Proof 1

Boring.

Let $v=\dot{x}$ and $w=\dot{y}$, so $w=\D{\phi}\cdot v$. \begin{align*} \didi{L}{v_i} &= \didi{L}{w_{\alpha}} \didi{\phi_{\alpha}}{x_i} && \didi{L}{x_i} = \didi{L}{y_{\alpha}} \didi{\phi_{\alpha}}{x_i} + \didi{L}{w_{\alpha}} \didi{w_{\alpha}}{x_i} \end{align*} Therefore \begin{align*} \underbrace{% \eulerlagrange{t}{x_i}{v_i}{L}% }_{\text{LHS}} &= \underbrace{% \left[ \eulerlagrange{t}{y_{\alpha}}{w_{\alpha}}{L} \right]% }_{\text{RHS}}% \didi{\phi_{\alpha}}{x_i} + \didi{L}{w_{\alpha}}% \underbrace{% \left[ \ddt{t}{}\didi{\phi_{\alpha}}{x_i} - \didi{w_{\alpha}}{x_i} \right]% }_{\text{NIL}} \end{align*} But $y_{\alpha} = \phi_{\alpha}$ and $w_{\alpha} = \displaystyle\ddt{t}{y_{\alpha}}$, so NIL vanishes. Since $\phi$ is a diffeomorphism, the LHS vanishes for all $i$ iff RHS vanishes for all $\alpha$.

Proof 2

Let $a \lt b$ and let $p, q \in M$. Define \[ \Omega_{p, q}^{a, b} = \set{ c : [a, b] \to M \st c(a)=p, c(b)=q, c \textrm{ is piecewise AC} }. \] Define the action functional $A : \Omega_{p, q}^{a, b} \to \R$ by \[ A(c) := \int_a^b L(c(t), \dot{c}(t)) \D{t}. \]

If $c$ is a critical point of $A$, then $c$ solves the Euler-Lagrange equations.

This implies that the Euler-Lagrange equations are tensorial.

Variations of a curve.

Proof 2 - cont'd

Let $x=(x_i)$ be a a coordinate system on $M$ and let $c_s \in \Omega_{p, q}^{a, b}$ be a smooth path of curves. Then \begin{align*} \ddt{s}{} A(c_s) &= \int_a^b \left[ \didi{L}{x_{\alpha}} \didi{}{s} c_s^{\alpha} + \overbrace{% \didi{L}{\dot{x}_{\alpha}}% }^{u} \underbrace{% \ddt{t}{} \didi{}{s} c^{\alpha}_s% }_{\D{v}}% \right] \D{t} \\ &= \int_a^b \underbrace{% \left[ \didi{L}{x_{\alpha}} - \ddt{t}{} \didi{L}{\dot{x}_{\alpha}} \right]% }_{E_{\alpha}} \underbrace{% \didi{c^{\alpha}_s}{s}% }_{v^{\alpha}} \D{t} + \underbrace{% \left[ \didi{L}{\dot{x}_{\alpha}} \didi{c^{\alpha}_s}{s} \right]_{t=a}^{t=b} }_{\text{NIL}} \end{align*}

Fundamental Theorem

Let $f : [a, b] \to \R$ be a continuous function. If, for all continuous functions $g$ such that $g(a)=g(b)=0$, \[ \int_a^b f(t)g(t) \D{t} = 0 \] then $f \equiv 0$.

If there is an $x \in [a, b]$ such that $f(x) \gt 0$, then by continuity, there is an $x_0 \in (a, b)$ such that $f(x_0) \gt 0$. By continuity, there is a $\delta \gt 0$ such that $f(x)^2 \gt \frac{1}{2} f(x_0)^2$ for all $x \in [x_0 - 2\delta, x_0 + 2\delta] \subset (a, b)$. Let $g : [a, b] \to \R$ be a non-negative continuous function such that $g = f$ on $[x_0 - \delta, x_0 + \delta]$ and $g(a)=g(b)=0$. Therefore \[ \int_a^b f(t)g(t) \D{t} \geq \delta f(x_0)^2 \gt 0. \]

Proof 2 - finished

The variation vector field $v = (v_{\alpha})$ can be an arbitrary continuous function that vanishes at $a$ and $b$. Now apply the fundamental theorem.

Legendre Transform

Let $L : TM \to \R$ be $C^2$. Define the Legendre transform $F : TM \to T^*M$ by \[ F(x, v) = (x, \didi{L}{v}). \]

If $L$ is $C^2$ and convex, then $F$ is a $C^1$ diffeomorphism. Moreover, the function \[ H(x, p) = \max_{v \in T_xM} \ip{p}{v} - L(x, v) \] is $C^2$ and $c$ solves the Euler-Lagrange equations for $L$ iff $\gamma = F \circ c$ satisfies Hamilton's equations for $H$.

Proof 3

If $L$ is convex, i.e. \[ \dindin{L}{v}{2} \gt 0 \] then there is a unique $v \in T_xM$ maximizing $\ip{p}{v} - L$. It solves \[ p = \didi{L}{v} \] for $v$. This implies $H$ is $C^2$. Also, $L$ satisfies \[ L(x, v) = \max_{p \in T^*_xM} \ip{p}{v} - H(x, p). \]

Proof 3 - cont'd

In sum \begin{align*} p &= \didi{L}{v} && v = \didi{H}{p} && \didi{H}{x} = -\didi{L}{x} \end{align*} Which implies \begin{align*} \dot{p} &= \underbrace{% \ddt{t}{} \didi{L}{v} = \didi{L}{x}% }_{\text{Euler-Lagrange}}% = -\didi{H}{x} && \dot{x} = v = \didi{H}{p}. \end{align*}

Let \begin{align*} \Omega &= \sum_i \D{x}_i \wedge \D{p}_i && X_H = \sum_i \didi{H}{p_i} \didi{}{x_i} - \didi{H}{x_i} \didi{}{p_i} \end{align*} We see that $\interiorproduc{X_H}{\Omega} = \D{H}$.

It suffices to prove that $\Omega$ is a tensor field.

The Symplectic Form

The $2$-form $\Omega = \sum_i \D{x}_i \wedge \D{p}_i$ is a tensor field on $T^*M$.

Let $y = \phi(x)$ be a change of coordinates, so that $p = (\D{\phi})^* q$ where $\D{\phi}^* : T^*_y M \to T^*_x M$. Then \begin{align*} \sum_{\alpha} \D{y_\alpha} \wedge \D{q_{\alpha}} &= \sum_{\alpha, \beta} \didi{\phi_{\alpha}}{x_{\beta}} \D{x_{\beta}} \wedge \D{q_{\alpha}} \\ &= \sum_{\beta} \D{x_{\beta}} \wedge \left[ \sum_{\alpha} \didi{\phi_{\alpha}}{x_{\beta}} \D{q_{\alpha}} \right] \\ &= \sum_{\beta} \D{x_{\beta}} \wedge \D{p_{\beta}}. \end{align*}