05.00.E1 · symplectic / lagrangian-mechanics

Lagrangian and variational mechanics exercise pack (Arnold Part II supplement)

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Formal definition of the pack Intermediate

Arnold's Part II builds classical mechanics on the configuration manifold $M$ : a Lagrangian is a smooth function $L : T M \to R$ , a motion is a critical point of the action $S [γ] = \int_{a}^{b} L (γ, \overset{γ}{˙}) d t$ , and the Euler-Lagrange equation $\frac{d}{d t} \partial_{v} L - \partial_{q} L = 0$ is the condition for criticality. The Legendre transform (fibre derivative) $F L : (q, v) \mapsto (q, \partial_{v} L)$ carries this to the Hamiltonian side, and Noether's theorem ties each one-parameter symmetry group of $L$ to a conserved quantity. Several of Arnold's Part II problems cut across these results at once — a single computation will set up the action, derive the equations of motion, and read off a conservation law from an invariance.

This pack collects nine such exercises — three easy, four medium, two hard — each with a hint and a full solution. It is meant to be read alongside its prerequisite units 05.00.01, 05.00.02, 05.00.03, 05.00.04, and the worked-examples unit 05.00.09, not as a standalone development. The problems are grouped by Arnold section: the Euler-Lagrange machinery and elementary variational problems (easy), the Legendre transform and worked mechanical systems (medium), and Noether-theorem applications (hard).

The conventions throughout are Arnold's: $q$ denotes generalised coordinates on $M$ , $v = \overset{q}{˙}$ the fibre coordinate on $T M$ , $p = \partial_{v} L$ the conjugate momentum, $ω = d p \land d q = - d θ$ with $θ = p d q$ , and the energy is $E = p \cdot v - L$ .

Key theorem with full solution Intermediate

Before the pack proper, we work one exercise in full as an exemplar of the format. The remaining eight follow the same structure (problem, hint, full answer in <details> blocks).

Lead exercise. Derive the Euler-Lagrange equation from Hamilton's principle, and verify that $L = \frac{1}{2} m ∣ \overset{q}{˙} ∣^{2} - V (q)$ recovers Newton's equation.

Solution. Let $γ : [a, b] \to M$ be a motion and consider a variation $γ_{s}$ with $γ_{0} = γ$ , fixed endpoints $γ_{s} (a) = γ (a)$ , $γ_{s} (b) = γ (b)$ , and variation field $δ q = \partial_{s} γ_{s}_{s = 0}$ , which vanishes at $a$ and $b$ . The action is $S [γ_{s}] = \int_{a}^{b} L (γ_{s}, \overset{γ}{˙}_{s}) d t$ . Differentiate under the integral sign at $s = 0$ in a chart:

\frac{d}{d s} S [γ_{s}]_{s = 0} = \int_{a}^{b} (\frac{\partial L}{\partial q} δ q + \frac{\partial L}{\partial v} δ \overset{q}{˙}) d t .

Since $δ \overset{q}{˙} = \frac{d}{d t} δ q$ , integrate the second term by parts:

\int_{a}^{b} \frac{\partial L}{\partial v} \frac{d}{d t} δ q d t = [\frac{\partial L}{\partial v} δ q]_{a}^{b} - \int_{a}^{b} \frac{d}{d t} (\frac{\partial L}{\partial v}) δ q d t .

The boundary term vanishes because $δ q (a) = δ q (b) = 0$ . Collecting,

δ S = \int_{a}^{b} (\frac{\partial L}{\partial q} - \frac{d}{d t} \frac{\partial L}{\partial v}) δ q d t .

A motion is a critical point, so $δ S = 0$ for every admissible $δ q$ . By the fundamental lemma of the calculus of variations, the integrand's coefficient vanishes identically:

\frac{d}{d t} \frac{\partial L}{\partial v} - \frac{\partial L}{\partial q} = 0.

For $L = \frac{1}{2} m ∣ \overset{q}{˙} ∣^{2} - V (q)$ in $R^{n}$ : $\partial_{v} L = m \overset{q}{˙}$ and $\partial_{q} L = - \nabla V$ . The Euler-Lagrange equation becomes $\frac{d}{d t} (m \overset{q}{˙}) + \nabla V = 0$ , that is $m \overset{q}{¨} = - \nabla V (q)$ — Newton's second law with force $- \nabla V$ . $□$

This is the canonical bridge from the variational principle to the equations of motion. Every problem downstream that "derives the equations of motion" runs this same first-variation argument, then specialises the Lagrangian.

Exercises Intermediate

Exercise 1 (easy, symbolic). Euler-Lagrange for the shortest path.

Let $L (y, y^{'}) = 1 + (y^{'})^{2}$ on $R^{2}$ , so $S [y] = \int_{a}^{b} 1 + (y^{'})^{2} d x$ is arc length. Show that the Euler-Lagrange equation forces $y$ to be a straight line.

Hint

$L$ has no explicit $y$ -dependence, so $\partial_{y} L = 0$ . The Euler-Lagrange equation then says $\partial_{y^{'}} L$ is constant in $x$ .

Answer

Since $\partial_{y} L = 0$ , the Euler-Lagrange equation $\frac{d}{d x} \partial_{y^{'}} L - \partial_{y} L = 0$ reduces to $\frac{d}{d x} \partial_{y^{'}} L = 0$ , so $\partial_{y^{'}} L$ is a constant $c$ :

\frac{y ^{'}}{1 + ( y ^{'} ) ^{2}} = c .

Solving for $y^{'}$ : $(y^{'})^{2} = c^{2} (1 + (y^{'})^{2})$ , so $(y^{'})^{2} (1 - c^{2}) = c^{2}$ and $y^{'} = c / 1 - c^{2}$ , a constant. Therefore $y (x) = m x + b$ is a straight line, with slope $m = c / 1 - c^{2}$ .

This is the variational proof that straight lines minimise length in the plane — the prototype Euler-Lagrange problem (Arnold §12).

Exercise 2 (easy, symbolic). Conserved energy when $L$ is time-independent.

Suppose $L = L (q, v)$ has no explicit time dependence. Show that the energy $E = v \cdot \partial_{v} L - L$ is conserved along motions.

Hint

Differentiate $E$ along a solution and use the Euler-Lagrange equation $\frac{d}{d t} \partial_{v} L = \partial_{q} L$ to cancel terms.

Answer

Compute $\frac{d E}{d t}$ along a motion, with $v = \overset{q}{˙}$ :

\frac{d E}{d t} = \overset{v}{˙} \cdot \partial_{v} L + v \cdot \frac{d}{d t} \partial_{v} L - \frac{d L}{d t} .

By the chain rule (no explicit $t$ ), $\frac{d L}{d t} = \partial_{q} L \cdot \overset{q}{˙} + \partial_{v} L \cdot \overset{v}{˙} = \partial_{q} L \cdot v + \partial_{v} L \cdot \overset{v}{˙}$ . Substituting and cancelling $\partial_{v} L \cdot \overset{v}{˙}$ :

\frac{d E}{d t} = v \cdot \frac{d}{d t} \partial_{v} L - \partial_{q} L \cdot v = v \cdot (\frac{d}{d t} \partial_{v} L - \partial_{q} L) = 0

by the Euler-Lagrange equation. So $E$ is constant along motions. This is the energy integral; it is the Noether conserved quantity for time-translation symmetry (Arnold §16).

Exercise 3 (easy, numeric). Momentum conjugate to a cyclic coordinate.

A particle moves in the plane under a central potential, with Lagrangian $L = \frac{1}{2} m (\overset{r}{˙}^{2} + r^{2} \overset{φ}{˙}^{2}) - V (r)$ in polar coordinates. Compute the momentum $p_{φ}$ conjugate to $φ$ and show it is conserved.

Hint

$φ$ is cyclic — it does not appear in $L$ . The conjugate momentum $p_{φ} = \partial L / \partial \overset{φ}{˙}$ is then constant.

Answer

The conjugate momentum is

p_{φ} = \frac{\partial L}{\partial φ ˙} = m r^{2} \overset{φ}{˙} .

Because $L$ does not depend on $φ$ explicitly ( $\partial_{φ} L = 0$ ), the Euler-Lagrange equation for $φ$ gives $\frac{d}{d t} p_{φ} = \partial_{φ} L = 0$ , so $p_{φ} = m r^{2} \overset{φ}{˙}$ is conserved.

This is the angular momentum, and its conservation is Kepler's second law: $r^{2} \overset{φ}{˙} =$ const means the radius vector sweeps equal areas in equal times (the areal velocity is $\frac{1}{2} r^{2} \overset{φ}{˙}$ ). Arnold §16, central-force example.

Exercise 4 (medium, symbolic). Legendre transform of a hyper-regular Lagrangian.

Let $L = \frac{1}{2} v^{⊤} A v - V (q)$ on $T R^{n}$ , with $A$ a constant symmetric positive-definite matrix. Compute the Hamiltonian $H (q, p)$ via the Legendre transform $H = p \cdot v - L$ with $p = \partial_{v} L$ .

Hint

Invert $p = A v$ to get $v = A^{- 1} p$ , then substitute into $H = p \cdot v - L$ .

Answer

The fibre derivative is $p = \partial_{v} L = A v$ . Since $A$ is positive-definite it is invertible, so $L$ is hyper-regular and $v = A^{- 1} p$ . Substitute into $H = p \cdot v - L$ :

H = p \cdot A^{- 1} p - (\frac{1}{2} (A^{- 1} p)^{⊤} A (A^{- 1} p) - V (q)) .

Using $A^{⊤} = A$ , the kinetic term is $\frac{1}{2} (A^{- 1} p)^{⊤} A (A^{- 1} p) = \frac{1}{2} p^{⊤} A^{- 1} p$ . Therefore

H (q, p) = p^{⊤} A^{- 1} p - \frac{1}{2} p^{⊤} A^{- 1} p + V (q) = \frac{1}{2} p^{⊤} A^{- 1} p + V (q) .

The Legendre transform inverts the mass matrix: $\frac{1}{2} v^{⊤} A v$ on the Lagrangian side becomes $\frac{1}{2} p^{⊤} A^{- 1} p$ on the Hamiltonian side, and $- V$ flips sign to $+ V$ . For $A = m I$ this is $H = ∣ p ∣^{2} / (2 m) + V$ . Arnold §15 (Legendre transform), unit 05.00.03.

Exercise 5 (medium, symbolic). Legendre transform is an involution.

Let $f : R \to R$ be smooth and strictly convex ( $f^{''} > 0$ ). Its Legendre transform is $g (p) = sup_{v} (p v - f (v))$ . Show that the Legendre transform of $g$ is $f$ again.

Hint

The supremum is attained where $p = f^{'} (v)$ . Differentiate $g (p) = p v (p) - f (v (p))$ to find $g^{'} (p) = v (p)$ , then transform again.

Answer

For fixed $p$ , the function $v \mapsto p v - f (v)$ has derivative $p - f^{'} (v)$ , vanishing where $p = f^{'} (v)$ . Strict convexity makes $f^{'}$ strictly increasing, hence invertible; write $v = v (p)$ for the solution. The supremum is attained there (the second derivative is $- f^{''} < 0$ ), so $g (p) = p v (p) - f (v (p))$ .

Differentiate, using $p = f^{'} (v (p))$ :

g^{'} (p) = v (p) + p v^{'} (p) - f^{'} (v (p)) v^{'} (p) = v (p) + (p - f^{'} (v (p))) v^{'} (p) = v (p) .

Now form the Legendre transform of $g$ : $h (w) = sup_{p} (w p - g (p))$ , attained where $w = g^{'} (p) = v (p)$ , i.e. $p = f^{'} (w)$ . Then

h (w) = w p - g (p) = w f^{'} (w) - (f^{'} (w) w - f (w)) = f (w) .

So $h = f$ : the Legendre transform is an involution on strictly convex functions. This is why the Lagrangian-to-Hamiltonian and Hamiltonian-to-Lagrangian passages are mutually inverse (Arnold §14).

Exercise 6 (medium, proof). Noether: rotational symmetry gives angular momentum.

Let $L = \frac{1}{2} m ∣ \dot{r} ∣^{2} - V (∣ r ∣)$ for $r \in R^{3}$ , invariant under the rotation group $S O (3)$ . Using Noether's theorem, show that angular momentum $L = m r \times \dot{r}$ is conserved.

Hint

Take the one-parameter rotation $g_{s}$ about a fixed axis $n$ . The Noether quantity is $I = \partial_{v} L \cdot \frac{d}{d s} g_{s} (r)_{s = 0}$ .

Answer

Let $g_{s}$ be rotation by angle $s$ about the unit axis $n$ , so $\frac{d}{d s} g_{s} (r)_{s = 0} = n \times r$ . Since $V$ depends only on $∣ r ∣$ and rotations preserve length and speed, $L (g_{s} r, g_{s} \dot{r}) = L (r, \dot{r})$ : the action is invariant.

Noether's theorem states that the quantity

I = \frac{\partial L}{\partial r ˙} \cdot \frac{d}{d s} g_{s} (r)_{s = 0}

is conserved. Here $\partial_{\dot{r}} L = m \dot{r}$ , so

I = m \dot{r} \cdot (n \times r) = n \cdot (r \times m \dot{r}) = n \cdot L,

using the cyclic property of the scalar triple product. Since $n$ is an arbitrary axis, every component $n \cdot L$ is conserved, hence the full vector $L = m r \times \dot{r}$ is conserved. Rotational symmetry yields angular-momentum conservation — Arnold §20, unit 05.00.04.

Exercise 7 (medium, numeric). Brachistochrone reduces to a cycloid.

A bead slides without friction under gravity from $(0, 0)$ to $(x_{1}, y_{1})$ along a curve $y (x)$ (with $y$ measured downward). The descent time is $T [y] = \int_{0}^{x_{1}} \frac{1 + ( y ^{'} ) ^{2}}{2 g y} d x$ . Show the minimising curve satisfies $y (1 + (y^{'})^{2}) =$ const, the cycloid equation.

Hint

The integrand $F = (1 + (y^{'})^{2}) / (2 g y)$ has no explicit $x$ -dependence. Use the Beltrami identity $F - y^{'} \partial_{y^{'}} F =$ const.

Answer

Because $F$ does not depend explicitly on $x$ , the Euler-Lagrange equation has a first integral, the Beltrami identity: $F - y^{'} \partial_{y^{'}} F = C$ . Compute

\partial_{y^{'}} F = \frac{1}{2 g y} \cdot \frac{y ^{'}}{1 + ( y ^{'} ) ^{2}} .

Then

F - y^{'} \partial_{y^{'}} F = \frac{1}{2 g y} (1 + (y^{'})^{2} - \frac{( y ^{'} ) ^{2}}{1 + ( y ^{'} ) ^{2}}) = \frac{1}{2 g y} \cdot \frac{1}{1 + ( y ^{'} ) ^{2}} = C .

Squaring and rearranging: $\frac{1}{2 g y ( 1 + ( y ^{'} ) ^{2} )} = C^{2}$ , that is

y (1 + (y^{'})^{2}) = \frac{1}{2 g C ^{2}} =: k,

a constant. This is the differential equation of a cycloid; parametrising $x = \frac{k}{2} (θ - sin θ)$ , $y = \frac{k}{2} (1 - cos θ)$ solves it. The brachistochrone is a cycloid — the classical Bernoulli problem, Arnold §12.

Exercise 8 (hard, proof). Noether's theorem in full, with a divergence term.

Suppose a one-parameter family of maps $q \mapsto Q (q, s)$ with $Q (q, 0) = q$ changes the Lagrangian only by a total time derivative: $L (Q, \dot{Q}) = L (q, \overset{q}{˙}) + s \frac{d F}{d t} + O (s^{2})$ for some function $F (q, t)$ . Show that $I = \partial_{v} L \cdot \frac{\partial Q}{\partial s}_{s = 0} - F$ is conserved along motions.

Hint

Differentiate the invariance relation in $s$ at $s = 0$ , write the result as $\partial_{q} L \cdot ξ + \partial_{v} L \cdot \dot{ξ} = \frac{d F}{d t}$ with $ξ = \partial_{s} Q ∣_{s = 0}$ , and use the Euler-Lagrange equation to turn $\partial_{q} L$ into $\frac{d}{d t} \partial_{v} L$ .

Answer

Let $ξ = \frac{\partial Q}{\partial s}_{s = 0}$ , the infinitesimal generator. Differentiating the quasi-invariance relation $L (Q, \dot{Q}) = L + s \frac{d F}{d t} + O (s^{2})$ at $s = 0$ :

\frac{\partial L}{\partial q} \cdot ξ + \frac{\partial L}{\partial v} \cdot \dot{ξ} = \frac{d F}{d t},

where $\dot{ξ} = \frac{d}{d t} ξ$ . Along a motion, the Euler-Lagrange equation gives $\partial_{q} L = \frac{d}{d t} \partial_{v} L$ . Substitute:

\frac{d}{d t} (\frac{\partial L}{\partial v}) \cdot ξ + \frac{\partial L}{\partial v} \cdot \dot{ξ} = \frac{d}{d t} (\frac{\partial L}{\partial v} \cdot ξ) = \frac{d F}{d t} .

Therefore $\frac{d}{d t} (\partial_{v} L \cdot ξ - F) = 0$ , so

I = \frac{\partial L}{\partial v} \cdot ξ - F

is conserved. The divergence term $F$ matters precisely for quasi-symmetries that change $L$ by a total derivative — the most important case being Galilean boosts, where $F = m v_{boost} \cdot r$ and the conserved quantity $I$ is the initial position of the centre of mass. When $F = 0$ this reduces to the strict-symmetry Noether theorem. Arnold §20; the boost case anchors 05.00.07.

Exercise 9 (hard, symbolic). Geodesic flow as a Lagrangian system.

On a Riemannian manifold $(M, g)$ take $L (q, v) = \frac{1}{2} g_{ij} (q) v^{i} v^{j}$ . Show the Euler-Lagrange equations are the geodesic equations $\overset{q}{¨}^{k} + Γ_{ij}^{k} \overset{q}{˙}^{i} \overset{q}{˙}^{j} = 0$ , with $Γ_{ij}^{k}$ the Christoffel symbols of $g$ .

Hint

Compute $\partial_{v} L$ , $\partial_{q} L$ , and $\frac{d}{d t} \partial_{v} L$ in coordinates. Use the symmetry $g_{ij} = g_{j i}$ and the definition $Γ_{ij}^{k} = \frac{1}{2} g^{k l} (\partial_{i} g_{l j} + \partial_{j} g_{l i} - \partial_{l} g_{ij})$ .

Answer

With $L = \frac{1}{2} g_{ij} (q) v^{i} v^{j}$ :

\frac{\partial L}{\partial v ^{k}} = g_{k j} v^{j}, \frac{\partial L}{\partial q ^{k}} = \frac{1}{2} \partial_{k} g_{ij} v^{i} v^{j} .

Differentiate the momentum in time along a curve ( $v = \overset{q}{˙}$ ):

\frac{d}{d t} \frac{\partial L}{\partial v ^{k}} = g_{k j} \overset{q}{¨}^{j} + \partial_{i} g_{k j} \overset{q}{˙}^{i} \overset{q}{˙}^{j} .

The Euler-Lagrange equation $\frac{d}{d t} \partial_{v} L - \partial_{q} L = 0$ reads

g_{k j} \overset{q}{¨}^{j} + \partial_{i} g_{k j} \overset{q}{˙}^{i} \overset{q}{˙}^{j} - \frac{1}{2} \partial_{k} g_{ij} \overset{q}{˙}^{i} \overset{q}{˙}^{j} = 0.

Symmetrise the middle term in $i \leftrightarrow j$ (legal since $\overset{q}{˙}^{i} \overset{q}{˙}^{j}$ is symmetric): $\partial_{i} g_{k j} \overset{q}{˙}^{i} \overset{q}{˙}^{j} = \frac{1}{2} (\partial_{i} g_{k j} + \partial_{j} g_{k i}) \overset{q}{˙}^{i} \overset{q}{˙}^{j}$ . Then

g_{k j} \overset{q}{¨}^{j} + \frac{1}{2} (\partial_{i} g_{k j} + \partial_{j} g_{k i} - \partial_{k} g_{ij}) \overset{q}{˙}^{i} \overset{q}{˙}^{j} = 0.

Contract with the inverse metric $g^{l k}$ and use $g^{l k} g_{k j} = δ_{j}^{l}$ :

\overset{q}{¨}^{l} + \frac{1}{2} g^{l k} (\partial_{i} g_{k j} + \partial_{j} g_{k i} - \partial_{k} g_{ij}) \overset{q}{˙}^{i} \overset{q}{˙}^{j} = \overset{q}{¨}^{l} + Γ_{ij}^{l} \overset{q}{˙}^{i} \overset{q}{˙}^{j} = 0.

These are the geodesic equations. So free motion on $(M, g)$ — the Lagrangian with kinetic energy alone — traces geodesics; this is the Lagrangian shadow of the geodesic Hamiltonian flow 05.02.06. Arnold §17.

Exercise pack EP. Arnold Mathematical Methods of Classical Mechanics Part II supplement: Euler-Lagrange equations, the Legendre transform, Noether's theorem, and worked mechanical systems across §12-§20.

Prerequisites

05.00.01 pending
05.00.02 pending
05.00.03 pending
05.00.04 pending
05.00.09 pending

Tier anchors

beginner
intermediate: Arnold *Mathematical Methods of Classical Mechanics* Part II exercises (§12-§20: Euler-Lagrange, least action, Legendre transform, Noether's theorem); Goldstein *Classical Mechanics* Ch. 1-2 problems
master

References

Arnold — Mathematical Methods of Classical Mechanics (2nd ed., GTM 60) · Part II (Ch. 3-5) exercises: §12 (variational calculus), §13 (Hamilton's principle), §15 (Legendre transform), §16 (Euler-Lagrange on a manifold), §20 (Noether's theorem); worked systems in §22, §25
Goldstein, Poole, Safko — Classical Mechanics (3rd ed.) · Ch. 1-2 derivations and problems — Euler-Lagrange equations, constraints, conservation laws
Gelfand-Fomin — Calculus of Variations · §1-§4 — first variation, Euler-Lagrange equation, conservation laws from invariance

Estimated time

intermediate: 5h