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Stationary action and Lagrangian dynamics — Examples

1 · Prerequisites

2 · Summary

Worked checks for cm07-stationary-action-and-lagrangian-dynamics. The free particle: the unique stationary curve between fixed endpoints is the straight uniform motion, and it minimises the action treats the free one-label model: the stationary curves are the affine uniform motions, the curve between two fixed endpoints is unique, and the exact identity $S (x_{*} + η) - S (x_{*}) = \frac{m}{2} \int ∣ η^{'} ∣^{2} d t$ shows that the stationary curve is a genuine minimiser in the fixed-endpoint class. The harmonic oscillator on a long interval: the stationary curve is neither a local minimum nor a local maximum computes the action of the sine-mode variations of the oscillator in closed form, $S (ε η_{n}) = \frac{m T ℓ_{0}^{2}}{4} ε^{2} ((n π / T)^{2} - ω^{2})$ , so on intervals longer than $π / ω$ arbitrarily small variations make the action both smaller and larger: stationarity is not minimisation. Cylindrical coordinates for one label: the kinetic metric m diag(1, r^2, 1) and the coordinate form of the model's equation verifies the coordinate-covariance theorem on cylindrical local coordinates off the $z$ -axis, computing the kinetic metric $m d i a g (1, r^{2}, 1)$ and identifying the Euler-Lagrange equations with the Cartesian equation in the orthonormal frame. A total-derivative Lagrangian on explicit data: the endpoint term and the unchanged equations uses a vector configuration and a dimensioned affine curve to exhibit the endpoint term $2 α t_{0}^{2} (e_{1} \cdot v_{0})$ while the Euler-Lagrange expression and the model equation stay fixed. The unrestricted action principle fails for the normalized nonholonomic relation $z_{s} = x y_{s}$ : the d'Alembert motion is not vakonomic uses normalized dimensionless coordinates and time for the relation $z_{s} = x y_{s}$ ; its stated d'Alembert motion has $y_{s} = (1 + s^{2})^{- 1 / 2} > 0$ and $y_{s s} \neq 0$ for $s > 0$ , so it cannot satisfy the vakonomic equations for those data. This shows that the prescriptions are not generally equivalent for velocity constraints. The planar pendulum by constraint elimination: kinetic metric, Lagrange equation and the ideal reaction recovered recovers the planar pendulum from the reduction theorem, with kinetic metric $G = m l^{2}$ , substituted Lagrangian $\frac{1}{2} m l^{2} {\dot{θ}}^{2} + m g l \cos θ$ , the reduced equation $θ^{''} = - (g / l) \sin θ$ , and the ideal reaction as the transpose constraint-Jacobian multiple $D_{x} g_{c} (X (θ))^{T} (λ_{1}, λ_{2})^{T}$ , where $λ_{2} = 0$ and the multiplier components have their declared, different dimensions.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

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The free particle: the unique stationary curve between fixed endpoints is the straight uniform motion, and it minimises the action

Physical scope

A worked check of the stationary-action equivalence on the free one-label model: zero force map, zero potential, Cartesian Lagrangian of a free particle; on every closed time interval the stationary curves are exactly the affine uniform motions, for each pair of endpoint configurations there is exactly one, and the action difference to the stationary curve is a nonnegative quadratic functional of the variation, so the stationary curve is a minimiser within the fixed-endpoint class. The mass, position, time and velocity carry dimensions mass, length, time and length per time, respectively. All quantities are model values under the declared idealization of a particle subject to no force.

Example

Then:

$x$ is stationary for the action $S$ if and only if $x$ is affine, $x (t) = x_{a} + v_{0} (t - a)$ with constant $v_{0} \in R^{3}$ ; consequently there is exactly one stationary curve with the prescribed endpoints, $x_{*} (t) : = x_{a} + \frac{t - a}{b - a} (x_{b} - x_{a}), x_{*}^{'} = \frac{x_{b} - x_{a}}{b - a} .$
For every variation $η$ of $x_{*}$ one has the exact identity $S (x_{*} + η) - S (x_{*}) = \frac{m}{2} \int_{a}^{b} ∣ η^{'} (t) ∣_{2}^{2} d t \geq 0,$ with equality if and only if $η = 0$ ; hence $x_{*}$ is the unique minimiser of $S$ in the class of $C^{2}$ curves with these endpoints.
The global affine curve $x_{\max} : R \to R^{3}$ defined by $x_{\max} (t) = x_{a} + v_{0} (t - a)$ , with $v_{0} : = (x_{b} - x_{a}) / (b - a)$ , is the unique maximal motion through the datum $(a, x_{a}, v_{0})$ , and $x_{*} = x_{\max} ∣_{[a, b]}$ .

All quantities are model values; no measurement is reported.

Facts & Assumptions

Given: The model data of the Example, $a < b$ , the endpoints $x_{a}, x_{b}$ , the curve $x_{*}$ and a variation $η$ with $η (a) = η (b) = 0$ of $x_{*}$ .

[F1]

The chart and the equivalence (The action functional of a force model with a potential presentation: the mechanical Lagrangian, its chart instance and its dimensions, The stationary-action criterion and the potential-part Euler-Lagrange equation): $(O, L)$ with $O = R \times R^{3} \times R^{3}$ is a time-dependent Lagrangian chart, and a $C^{2}$ curve $x$ is stationary for $S$ if and only if $m x^{''} (t) = - \nabla U (t, x (t)) = 0$ for every $t \in [a, b]$ .

[F2]

Motion and maximal motion (Stationary curves of the mechanical action are exactly the restrictions of the model's motions (global potential branch), The associated first-order system, motions, maximal motions and the Newtonian flow of a force model): the stationary curves of $S$ on $[a, b]$ are the restrictions of the motions of the model with interval containing $[a, b]$ ; every motion with $M x^{''} = 0$ is affine, and through each datum there is exactly one maximal motion.

[F3]

Integral and derivative calculus (The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ , Integrable functions on $[a, b]$ form a set closed under sums and scalar multiples, and $\int_{a}^{b} (λ f + μ g) = λ \int_{a}^{b} f + μ \int_{a}^{b} g$ , If $f \leq g$ on $[a, b]$ and both are integrable then $\int_{a}^{b} f \leq \int_{a}^{b} g$ ; and $m (b - a) \leq \int_{a}^{b} f \leq M (b - a)$ , A continuous $f \geq 0$ on $[a, b]$ with $\int_{a}^{b} f = 0$ is identically $0$ , A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion, The lower and upper Darboux integrals of a bounded $f$ on $[a, b]$ as $\sup_{P} L (f, P)$ and $\inf_{P} U (f, P)$ , Darboux integrability as their equality, and the notation $\int_{a}^{b} f$ , The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral, The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ ): the integral is linear and monotone on continuous integrands; a continuous nonnegative function with zero integral vanishes identically; and componentwise the fundamental theorem gives $\int_{u}^{v} f^{'} = f (v) - f (u)$ for the $C^{1}$ functions used here. The derivatives of a $C^{2}$ curve on $[a, b]$ are continuous and the usual derivative algebra applies.

Verification

technique · direct; solve the second-order equation of the free model, then expand the action of a varied curve and integrate by parts

1.1F1F3givenalgebra

Equation and uniqueness: by [F1], stationary means $x^{''} \equiv 0$ on $[a, b]$ because $m > 0$ . Componentwise, [F3] applied to $x^{'}$ gives $x^{'} (t) - x^{'} (a) = \int_{a}^{t} x^{''} (u) d u = 0$ for every $t \in [a, b]$ ; applying it to $x$ then gives $x (t) = x (a) + x^{'} (a) (t - a)$ , so $x$ is affine. Conversely an affine curve has $x^{''} \equiv 0$ and is stationary by [F1]. Its endpoint values force $x^{'} (a) = (x_{b} - x_{a}) / (b - a)$ , well defined because $b > a$ ; hence $x_{*}$ is the unique stationary curve with the prescribed endpoints.

2.1F3step 1.1algebra

Expansion of the action: let $η$ be a variation. Since $L$ is quadratic with no linear term, $S (x_{*} + η) - S (x_{*}) = \frac{m}{2} \int_{a}^{b} (∣ x_{*}^{'} + η^{'} ∣_{2}^{2} - ∣ x_{*}^{'} ∣_{2}^{2}) d t = \frac{m}{2} \int_{a}^{b} (2 x_{*}^{'} \cdot η^{'} + ∣ η^{'} ∣_{2}^{2}) d t .$ The first term vanishes: $x_{*}^{'}$ is the constant vector $v_{0}$ , so by [F3] $\int_{a}^{b} x_{*}^{'} \cdot η^{'} d t = v_{0} \cdot \int_{a}^{b} η^{'} d t = v_{0} \cdot (η (b) - η (a)) = 0$ because $η$ vanishes at both endpoints. Hence $S (x_{*} + η) - S (x_{*}) = \frac{m}{2} \int_{a}^{b} ∣ η^{'} ∣_{2}^{2} d t \geq 0$ by [F3].

3.1F3step 2.1algebra

Equality case: if the integral vanishes, the continuous nonnegative integrand $∣ η^{'} ∣_{2}^{2}$ vanishes identically on $[a, b]$ by [F3], so $η^{'} \equiv 0$ . The fundamental theorem in [F3] gives $η (t) - η (a) = \int_{a}^{t} η^{'} (u) d u = 0$ componentwise; since $η (a) = 0$ , $η \equiv 0$ . Thus the stationary curve is the unique minimiser in the fixed-endpoint class.

4.1F2step 1.1∎

Consistency with the flow: by [F2] the curve $x_{*}$ is the restriction of the maximal motion through $(a, x_{a}, v_{0})$ ; the affine curve $t \mapsto x_{a} + v_{0} (t - a)$ is a motion on all of $R$ , hence the maximal motion is defined for all times and equals $x_{*}$ on $[a, b]$ .

Remarks

The exact identity of step 2.1 shows that for the free particle the stationary curve is a genuine minimiser, not a saddle; the oscillator example of this page exhibits the opposite phenomenon for large intervals. Both computations are model checks of the stationarity theorem; neither is a claim that nature extremises an action.

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The harmonic oscillator on a long interval: the stationary curve is neither a local minimum nor a local maximum

Physical scope

A worked check that stationarity is not minimisation, on a harmonic oscillator embedded in three-dimensional configuration space and moving along one fixed unit vector, with positive stiffness: on intervals longer than pi divided by the angular frequency, arbitrarily small fixed-endpoint variations in the declared dimensionless C^2 metric around the zero stationary curve produce both strictly larger and strictly smaller actions, so the stationary curve is neither a local minimiser nor a local maximiser; on shorter intervals every sine-mode variation in the checked family increases the action. The computation uses exact model actions and exact trigonometric integrals; no experimental values are reported.

Example

Fix a duration $T > 0$ and declare a one-label model of the force branch of Adopted force model: force maps, the equation of motion, additive composition of force contributions and frame covariance with $N = 1$ , mass $m > 0$ , time window $I_{0} = R$ , configuration domain $Q = R^{3}$ , a fixed unit vector $e \in R^{3}$ with dimensionless components, and force map $F (t, x, v) = - k (e \cdot x) e$ with constant $k > 0$ . This is the global potential model with $U (x) = \frac{1}{2} k (e \cdot x)^{2}$ and no residual forces (Potential presentations of a force model, potential and mechanical energy, closed versus exact force fields, local potentials and equipotential level sets), with mechanical Lagrangian $L (t, x, v) = \frac{1}{2} m ∣ v ∣_{2}^{2} - \frac{1}{2} k (e \cdot x)^{2}$ (The action functional of a force model with a potential presentation: the mechanical Lagrangian, its chart instance and its dimensions). Under Dimensions and SI unit conventions of the non-relativistic classical mechanics framework, $[m] = μ$ , $[e] = 1$ , $[x] = ℓ$ , $[t] = [T] = τ$ , $[k] = μ τ^{- 2}$ , $[ω] = τ^{- 1}$ and $[ℓ_{0}] = ℓ$ , where $ω : = \sqrt{k / m} > 0$ . Fix a length scale $ℓ_{0} > 0$ and let $η_{n} (t) : = ℓ_{0} \sin (n π t / T) e$ ; the scalar $ε$ below is dimensionless. Consider curves on $[0, T]$ .

Then:

the zero curve $x_{*} \equiv 0$ is stationary for the action $S$ ;
for every integer $n \geq 1$ the map $η_{n}$ is a variation of $x_{*}$ , and for every dimensionless real $ε$ one has the exact identity $S (ε η_{n}) = \frac{m T ℓ_{0}^{2}}{4} ε^{2} ((\frac{n π}{T})^{2} - ω^{2});$
if $T > π / ω$ , then $S (ε η_{1}) < 0$ and $S (ε η_{n}) > 0$ for every integer $n > ω T / π$ and every $ε \neq 0$ ; since $ε η_{n} \to 0$ in the dimensionless $C^{2}$ metric defined below as $ε \to 0$ , the stationary curve $x_{*}$ is neither a local minimiser nor a local maximiser in that topology;
if $0 < T < π / ω$ , then $S (ε η_{n}) > 0$ for every $n \geq 1$ and every $ε \neq 0$ : every variation in this family increases the action, and no claim of global minimality is made.

Facts & Assumptions

Given: The model data of the Example, $T > 0$ , the zero curve $x_{*}$ , the maps $η_{n} (t) = ℓ_{0} \sin (n π t / T) e$ , and a dimensionless real $ε$ .

[F1]

The model and its stationarity criterion (The action functional of a force model with a potential presentation: the mechanical Lagrangian, its chart instance and its dimensions, The stationary-action criterion and the potential-part Euler-Lagrange equation): $x_{*}$ is stationary if and only if $m x^{''} (t) = - k (e \cdot x (t)) e$ for all $t \in [0, T]$ ; the zero curve satisfies this.

[F2]

Trigonometric facts (Sine and cosine defined by their real power series, The derivatives of sine and cosine are cosine and minus sine, A function differentiable at $c$ is continuous at $c$ , The addition formulas for sine and cosine, The zero sets of sine and cosine and the least positive common period 2 pi, Parity and the Pythagorean identity for sine and cosine): $\sin^{'} = \cos$ , $\cos^{'} = - \sin$ , the product-to-sum identity $2 \sin^{2} u = 1 - \cos (2 u)$ (the addition formula for the cosine with $u = u$ ), the basic values $\sin 0 = 0$ , $\cos 0 = 1$ and $\sin (m π) = 0$ for every integer $m$ , and $\sin^{2} + \cos^{2} = 1$ ; iterating the derivative identities twice gives $η_{n} \in C^{2}$ , since the derivatives through order two are scalar multiples of $\sin$ or $\cos$ , which are continuous by differentiability and A function differentiable at $c$ is continuous at $c$ . Consequently $e \cdot η_{n} (t) = ℓ_{0} \sin (n π t / T)$ and $η_{n}^{'} (t) = ℓ_{0} (n π / T) \cos (n π t / T) e$ , while $η_{n}$ vanishes at $t = 0$ and $t = T$ for every integer $n \geq 1$ .

[F3]

Integral and order calculus (The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ , The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , Integrable functions on $[a, b]$ form a set closed under sums and scalar multiples, and $\int_{a}^{b} (λ f + μ g) = λ \int_{a}^{b} f + μ \int_{a}^{b} g$ , A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion, The lower and upper Darboux integrals of a bounded $f$ on $[a, b]$ as $\sup_{P} L (f, P)$ and $\inf_{P} U (f, P)$ , Darboux integrability as their equality, and the notation $\int_{a}^{b} f$ , Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral, Absolute value in an ordered field): the integral is linear, continuous integrands are integrable, and for $α \neq 0$ the function $t \mapsto \sin (α t) / α$ is a primitive of $t \mapsto \cos (α t)$ on $[0, T]$ by the chain rule and the fundamental theorem; hence $\int_{0}^{T} \cos (α t) d t = (\sin (α T) - \sin 0) / α$ . The Archimedean property gives, for $T > π / ω$ , an integer $n > ω T / π$ (Every complete ordered field is Archimedean); and $ω = \sqrt{k / m}$ exists uniquely by Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ .

[F4]

The topology used for local extrema: on $C^{2}$ curves $h : [0, T] \to R^{3}$ define $d_{T, ℓ_{0}} (f, g) : = \sup_{t \in [0, T]} (\frac{∣ f (t) - g (t) ∣_{2}}{ℓ_{0}} + \frac{T ∣ f^{'} (t) - g^{'} (t) ∣_{2}}{ℓ_{0}} + \frac{T^{2} ∣ f^{''} (t) - g^{''} (t) ∣_{2}}{ℓ_{0}}) .$ This is a dimensionless metric: finiteness follows from continuity on the compact interval $[0, T]$ (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line, A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value), and the metric axioms follow componentwise from the Euclidean norm and its triangle inequality (Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation). A local minimiser or maximiser means one relative to the fixed-endpoint $C^{2}$ curve class with this metric. For fixed $n$ , $d_{T, ℓ_{0}} (ε η_{n}, 0) \leq ∣ ε ∣ (1 + n π + n^{2} π^{2})$ by the trigonometric bounds, so it tends to zero as $ε \to 0$ .

Verification

technique · direct; evaluate the action of the scaled sine variations in closed form and read off the two signs

1.1F1F2given

Stationarity of the zero curve: $M x^{''} = m \cdot 0 = - k (e \cdot 0) e$ on $[0, T]$ , so by [F1] $x_{*}$ is stationary; and each $η_{n}$ is a variation because it is $C^{\infty}$ on $[0, T]$ with $η_{n} (0) = η_{n} (T) = 0$ by [F2].

1.2F2F3algebra

The two integrals: with $α : = 2 n π / T \neq 0$ and the product-to-sum identity $2 \sin^{2} u = 1 - \cos (2 u)$ of [F2] applied to $u = n π t / T$ , the linearity of the integral and the primitive of [F3] give $\int_{0}^{T} (e \cdot η_{n} (t))^{2} d t = \frac{ℓ_{0}^{2}}{2} \int_{0}^{T} (1 - \cos (α t)) d t = ℓ_{0}^{2} T / 2,$ because $α T = 2 n π$ is an integer multiple of $2 π$ and $\sin (2 n π) = 0 = \sin 0$ by [F2]; and $\int_{0}^{T} ∣ η_{n}^{'} (t) ∣_{2}^{2} d t = ℓ_{0}^{2} (n π / T)^{2} \int_{0}^{T} \cos (n π t / T)^{2} d t = ℓ_{0}^{2} (n π / T)^{2} T / 2,$ using $\cos^{2} = 1 - \sin^{2}$ of [F2] and the value just computed.

2.1F3step 1.2algebra

Exact action of the scaled variation: since $x_{*} = 0$ and the integrand is quadratic, for every dimensionless real $ε$ one has $S (ε η_{n}) = \frac{ε^{2}}{2} \int_{0}^{T} (m ∣ η_{n}^{'} (t) ∣_{2}^{2} - k (e \cdot η_{n} (t))^{2}) d t = \frac{ε^{2}}{2} (m \int_{0}^{T} ∣ η_{n}^{'} ∣_{2}^{2} d t - k \int_{0}^{T} (e \cdot η_{n})^{2} d t) .$ Inserting the two integrals of step 1.2 gives $S (ε η_{n}) = \frac{ε^{2}}{2} \cdot \frac{ℓ_{0}^{2} T}{2} (m (\frac{n π}{T})^{2} - k) = \frac{m T ℓ_{0}^{2}}{4} ε^{2} ((\frac{n π}{T})^{2} - ω^{2}),$ using $k = m ω^{2}$ with $ω = \sqrt{k / m}$ of [F3]. This proves clause 2.

3.1F3step 2.1algebra

Signs for $T > π / ω$ : the coefficient $((n π / T)^{2} - ω^{2})$ is negative for $n = 1$ because $π / T < ω$ ; it is positive for every integer $n > ω T / π$ because then $n π / T > ω$ ; both statements are read off the identity of step 2.1, and [F3] provides such an integer. Hence for $ε \neq 0$ the actions $S (ε η_{1})$ and $S (ε η_{n})$ differ in sign from $S (x_{*}) = 0$ .

3.2F3step 2.1algebra

Signs for $0 < T < π / ω$ : then $π / T > ω$ , hence $(n π / T)^{2} \geq (π / T)^{2} > ω^{2}$ for every $n \geq 1$ , and step 2.1 gives $S (ε η_{n}) > 0$ for $ε \neq 0$ . This is a statement about this family of variations only; no global minimality is asserted.

4.1F4step 3.1

Smallness: by [F4] the curves $ε η_{n}$ converge to $x_{*}$ in the declared dimensionless $C^{2}$ metric as $ε \to 0$ , so every neighbourhood of $x_{*}$ in the fixed-endpoint curve class contains curves of strictly smaller and of strictly larger action by step 3.1; the zero curve is neither a local minimiser nor a local maximiser.

5.1step 1.1step 2.1step 3.1step 4.1step 3.2∎

Clauses 1-4 are proved; all quantities are exact model values under the declared oscillator idealization.

Remarks

The computation shows the explicit threshold: for $T < π / ω$ the oscillator's zero curve increases the action along every sine-mode variation, while for $T > π / ω$ both signs occur. The mathematical analogue with the dimensionless Lagrangian $\frac{1}{2} (v^{2} - ω^{2} q^{2})$ and its second variation is recorded in the verification page of the mathematics track.

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Cylindrical coordinates for one label: the kinetic metric m diag(1, r^2, 1) and the coordinate form of the model's equation

Physical scope

A worked check of the coordinate-covariance theorem on cylindrical coordinates for one labelled particle of mass m in a declared potential presentation: the map covers the open region off the z-axis periodically and supplies local coordinates there, but is not a global chart of all configuration space; the substituted Lagrangian has kinetic metric m diag(1, r^2, 1), the coordinate Euler-Lagrange equations are exactly the Cartesian equation written in the cylindrical frame, and the Cartesian stationary curves correspond to coordinate lifts. The radial and axial coordinates have length dimension, the angle is dimensionless, and no measurement is reported.

Example

Declare a one-label model of the force branch of Adopted force model: force maps, the equation of motion, additive composition of force contributions and frame covariance with $N = 1$ , mass $m > 0$ , time window $I_{0} = R$ , configuration domain $Q = R^{3}$ and a global potential presentation $F (t, x, v) = - \nabla U (t, x)$ with $C^{2}$ potential $U : I_{0} \times Q \to R$ and no residual forces (Potential presentations of a force model, potential and mechanical energy, closed versus exact force fields, local potentials and equipotential level sets); the mechanical Lagrangian is $L (t, x, v) = \frac{1}{2} m ∣ v ∣_{2}^{2} - U (t, x)$ (The action functional of a force model with a potential presentation: the mechanical Lagrangian, its chart instance and its dimensions). Let $Ω : = {(r, θ, z) \in R^{3} : r > 0}$ and define the cylindrical coordinate map $X (r, θ, z) : = (r \cos θ, r \sin θ, z), X : Ω ⟶ Q,$ with substituted Lagrangian and kinetic metric $\tilde{L} (t, r, θ, z, \dot{r}, \dot{θ}, \dot{z}) = \frac{1}{2} m ({\dot{r}}^{2} + r^{2} {\dot{θ}}^{2} + {\dot{z}}^{2}) - \tilde{U} (t, r, θ, z), \tilde{U} (t, r, θ, z) : = U (t, X (r, θ, z)),$ $G (r, θ, z) = D X (r, θ, z)^{T} (m I_{3}) D X (r, θ, z) = m (\begin{array}{ccc} 1 & 0 & 0 \\ 0 & r^{2} & 0 \\ 0 & 0 & 1 \end{array}) .$

Under Dimensions and SI unit conventions of the non-relativistic classical mechanics framework, $[m] = μ$ , $[r] = [z] = ℓ$ , $[t] = τ$ and $[θ] = 1$ (dimensionless angle); $[\dot{r}] = [\dot{z}] = ℓ τ^{- 1}$ , $[\dot{θ}] = τ^{- 1}$ , and $[U] = μ ℓ^{2} τ^{- 2}$ . The configuration image is $Q_{c y l} : = {(x, y, z) \in R^{3} : x^{2} + y^{2} > 0}$ , which omits the $z$ -axis.

Then:

$X$ is a $C^{\infty}$ map with everywhere invertible Jacobian, hence a local diffeomorphism at every point of $Ω$ , so the covariance theorem applies to it; its image is $Q_{c y l}$ , it is periodic in $θ$ , and it is not a global chart of $Q$ . The coordinate equations are interpreted on these local charts or along a chosen coordinate lift.
the Euler-Lagrange equations of $\tilde{L}$ are exactly $m (\ddot{r} - r {\dot{θ}}^{2}) = - \partial_{r} \tilde{U}, m (r^{2} \ddot{θ} + 2 r \dot{r} \dot{θ}) = - \partial_{θ} \tilde{U}, m \ddot{z} = - \partial_{z} \tilde{U},$ which are the coordinate components of $m x^{''} = - \nabla U (t, x)$ along $x = X (r, θ, z)$ ;
consequently, for $a < b$ and a $C^{2}$ curve $(r, θ, z) : [a, b] \to Ω$ with $x : = X \circ (r, θ, z)$ , the coordinate curve is stationary for the substituted action if and only if the Cartesian curve $x$ satisfies the model's equation of motion on $[a, b]$ .

Facts & Assumptions

Given: The model data of the Example, the map $X$ and a $C^{2}$ curve $q (t) = (r (t), θ (t), z (t))$ with values in $Ω$ and $x : = X \circ q$ .

[F1]

Regularity and invertibility ( $C^{k}$ Euclidean maps and diffeomorphisms, The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case, Invertible Euclidean linear maps, The derivatives of sine and cosine are cosine and minus sine, The principle of mathematical induction, A function differentiable at $c$ is continuous at $c$ , Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes, The transpose $A^{T}$ of a matrix, The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ): repeated differentiation cycles through $\pm \sin$ and $\pm \cos$ ; induction and their continuity show these functions, and therefore the coordinate functions of $X$ , are $C^{\infty}$ . The Jacobian is $D X (r, θ, z) = (\begin{array}{ccc} \cos θ & - r \sin θ & 0 \\ \sin θ & r \cos θ & 0 \\ 0 & 0 & 1 \end{array}), \det D X (r, θ, z) = r > 0,$ so $D X$ is invertible on $Ω$ ; and $D X^{T} (m I_{3}) D X$ is the displayed diagonal matrix because its columns are orthogonal with squared norms $1, r^{2}, 1$ by $\sin^{2} + \cos^{2} = 1$ (Parity and the Pythagorean identity for sine and cosine).

[F2]

The covariance theorem (Time-dependent coordinate maps transform the potential-part equation into the substituted Euler-Lagrange equation): for this coordinate map the Euler-Lagrange expression of $\tilde{L}$ along $q$ satisfies $E_{\tilde{L}} (q) = D X^{T} E_{L} (x)$ , and the vanishing of the substituted Euler-Lagrange equation is equivalent to the Cartesian curve solving $m x^{''} (t) = - \nabla U (t, x (t))$ on $[a, b]$ (The stationary-action criterion and the potential-part Euler-Lagrange equation for the Cartesian equivalence).

[F3]

Chain rule for the coordinate description (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , The derivatives of sine and cosine are cosine and minus sine, The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length): differentiating $x (t) = X (q (t))$ twice gives $x^{'} = \dot{r} e_{r} + r \dot{θ} e_{θ} + \dot{z} e_{3}, x^{''} = (\ddot{r} - r {\dot{θ}}^{2}) e_{r} + (r \ddot{θ} + 2 \dot{r} \dot{θ}) e_{θ} + \ddot{z} e_{3},$ where $e_{r} : = (\cos θ, \sin θ, 0)$ , $e_{θ} : = (- \sin θ, \cos θ, 0)$ , $e_{3} : = (0, 0, 1)$ are the displayed orthonormal basis vectors; and $\nabla U (t, x) = \partial_{r} \tilde{U} e_{r} + \frac{1}{r} \partial_{θ} \tilde{U} e_{θ} + \partial_{z} \tilde{U} e_{3}$ along $x$ .

[F4]

The trigonometric parameterization needed for the global image and periodicity (The derivatives of sine and cosine are cosine and minus sine, A function differentiable at $c$ is continuous at $c$ , Intermediate value theorem, by bisection with a canonical left-half rule: a continuous function on $[a, b]$ takes every value between $f (a)$ and $f (b)$ , Pi is the first positive zero of sine, Quarter-turn values and shifts by pi/2 and pi, Parity and the Pythagorean identity for sine and cosine, The zero sets of sine and cosine and the least positive common period 2 pi, Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ , Squares of nonzero elements are positive, Absolute value in an ordered field): cosine is continuous on $[0, π]$ , with $\cos 0 = 1$ and $\cos π = - 1$ , and $\sin θ \geq 0$ on this interval. Thus, for $u, v \in R$ with $u^{2} + v^{2} = 1$ , $v^{2} \geq 0$ gives $u^{2} \leq 1$ , hence $u \in [- 1, 1]$ ; the intermediate value theorem gives $θ_{0} \in [0, π]$ with $\cos θ_{0} = u$ . The Pythagorean identity and sine's sign then give $\sin θ_{0} = ∣ v ∣$ . Taking $θ = θ_{0}$ when $v \geq 0$ and $θ = - θ_{0}$ when $v < 0$ gives $(\cos θ, \sin θ) = (u, v)$ . The period theorem gives period $2 π$ for sine and cosine.

Verification

technique · direct; compute the Jacobian, the kinetic metric and the Euler-Lagrange equations of the substituted Lagrangian, and identify them with the Cartesian equation in the orthonormal cylindrical frame

1.1F1F4algebra

Clause 1: the displayed Jacobian and its determinant $r > 0$ are computed from the derivatives of sine and cosine, so $D X (r, θ, z)$ is invertible for every $(r, θ, z) \in Ω$ and $X$ has the invertibility hypothesis of the covariance theorem. By The Euclidean inverse function theorem and A local inverse of a $C^{k}$ regular map is $C^{k}$ , its restriction near any point is a smooth diffeomorphism onto an open part of its image; the inverse is a local cylindrical coordinate chart, and these local images show $X (Ω)$ is open. The image is contained in $Q_{c y l}$ because $(r \cos θ)^{2} + (r \sin θ)^{2} = r^{2} > 0$ . Conversely, for $(a, b, c) \in Q_{c y l}$ put $ρ : = \sqrt{a^{2} + b^{2}} > 0$ , $u : = a / ρ$ and $v : = b / ρ$ . Then $u^{2} + v^{2} = 1$ , so [F4] supplies $θ$ with $\cos θ = u$ and $\sin θ = v$ , giving $X (ρ, θ, c) = (a, b, c)$ . Thus $X (Ω) = Q_{c y l}$ . By [F4], $X (r, θ + 2 π, z) = X (r, θ, z)$ with distinct inputs, so $X$ is not injective and is not a global chart of $Q$ . The kinetic metric is computed by matrix multiplication: the columns of $D X$ are $e_{r}$ , $r e_{θ}$ , $e_{3}$ , an orthogonal triple of norms $1$ , $r$ , $1$ , so $D X^{T} (m I_{3}) D X = m d i a g (1, r^{2}, 1)$ as displayed.

2.1F1step 1.1algebra

Substituted Lagrangian: from clause (b) of the generalized-coordinate construction, $\tilde{L} = L \circ (t, X, D X v)$ equals $\frac{1}{2} m ({\dot{r}}^{2} + r^{2} {\dot{θ}}^{2} + {\dot{z}}^{2}) - \tilde{U}$ , matching the computed kinetic metric.

3.1F2step 2.1algebra

Euler-Lagrange equations: with $\partial \tilde{L} / \partial \dot{r} = m \dot{r}$ , $\partial \tilde{L} / \partial r = m r {\dot{θ}}^{2} - \partial_{r} \tilde{U}$ , $\partial \tilde{L} / \partial \dot{θ} = m r^{2} \dot{θ}$ , $\partial \tilde{L} / \partial θ = - \partial_{θ} \tilde{U}$ , $\partial \tilde{L} / \partial \dot{z} = m \dot{z}$ and $\partial \tilde{L} / \partial z = - \partial_{z} \tilde{U}$ , the equations $\frac{d}{d t} \partial_{\dot{q}} \tilde{L} - \partial_{q} \tilde{L} = 0$ read $m \ddot{r} - m r {\dot{θ}}^{2} + \partial_{r} \tilde{U} = 0, m r^{2} \ddot{θ} + 2 m r \dot{r} \dot{θ} + \partial_{θ} \tilde{U} = 0, m \ddot{z} + \partial_{z} \tilde{U} = 0,$ which are the displayed coordinate equations.

4.1F3step 3.1algebra

Cartesian identification: by [F3], $m x^{''} = - \nabla U (t, x)$ is equivalent, in the orthonormal frame $(e_{r}, e_{θ}, e_{3})$ , to the three scalar equations $m (\ddot{r} - r {\dot{θ}}^{2}) = - \partial_{r} \tilde{U}$ , $m (r \ddot{θ} + 2 \dot{r} \dot{θ}) = - \frac{1}{r} \partial_{θ} \tilde{U}$ , and $m \ddot{z} = - \partial_{z} \tilde{U}$ ; the second equation of step 3.1 is this one multiplied by $r$ . Hence the coordinate Euler-Lagrange equations are exactly the Cartesian equation written in the cylindrical frame.

5.1F2step 3.1step 4.1∎

Clause 3: regard the time-independent map as $\hat{X} : R \times Ω \to Q$ , $\hat{X} (t, q) : = X (q)$ . Then its substituted domain is $\tilde{O} = R \times Ω \times R^{3}$ , which is a time-dependent Lagrangian chart over $[a, b] \times Ω$ because $Ω$ is open and nonempty. By clause 2 of [F2], stationarity of $q$ for the substituted action is equivalent to stationarity of $x$ for $S$ ; by clause 1 of [F2] and step 4.1, the latter is equivalent to the Cartesian equation along $x = X \circ q$ .

Remarks

The example displays the two ingredients of the generalized-coordinate construction: the kinetic metric is the Jacobian-transpose multiple of the mass operator, and the coordinate equations carry the centripetal and Coriolis terms of the Cartesian acceleration exactly as computed by the chain rule. No new dynamical content is added by the change of coordinates.
For a $C^{2}$ curve with $z \equiv z_{0}$ satisfying the first two equations, the third equation holds exactly when $\partial_{z} \tilde{U} (t, r (t), θ (t), z_{0}) = 0$ throughout its interval, because $\ddot{z} = 0$ . Under this condition the plane-polar equations describe a motion of the declared three-dimensional model in that plane. The displayed angular Euler-Lagrange equation divided by $r > 0$ is $m (2 \dot{r} \dot{θ} + r \ddot{θ}) = - \frac{1}{r} \partial_{θ} \tilde{U}$ .

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A total-derivative Lagrangian on explicit data: the endpoint term and the unchanged equations

Physical scope

A worked gauge check on the free one-label model in three-dimensional configuration space: adding the total time derivative of F(t,x)=alphatx_1 to the free Lagrangian changes the action by endpoint values and leaves the Euler-Lagrange expression equal to the original one, hence leaves stationary curves and the model equation unchanged. The position is a vector of length dimension, time has time dimension, alpha has force dimension, and F has action dimension. The potential normalization U + gammat is the special case F = -gammat^2/2, with gamma of energy-per-time dimension. The values are exact model values of a declared computation.

Example

Declare a one-label model of the force branch of Adopted force model: force maps, the equation of motion, additive composition of force contributions and frame covariance with $N = 1$ , mass $m > 0$ , time window $I_{0} = R$ , configuration domain $Q = R^{3}$ and zero force map; the mechanical Lagrangian is $L (t, x, v) = \frac{1}{2} m ∣ v ∣_{2}^{2}$ (The action functional of a force model with a potential presentation: the mechanical Lagrangian, its chart instance and its dimensions). Let $t_{0} > 0$ be a time scale, fix the interval $[t_{0}, 2 t_{0}]$ , a dimensioned parameter $α$ with $[α] = μ ℓ τ^{- 2}$ , and the $C^{2}$ function

$F (t, x) : = α t x_{1}, L^{'} (t, x, v) : = L (t, x, v) + F_{t} (t, x) + F_{x} (t, x) \cdot v = \frac{1}{2} m ∣ v ∣_{2}^{2} + α x_{1} + α t v_{1},$

so that $L^{'}$ is the Lagrangian $L$ plus the total time derivative of $F$ (Total-derivative Lagrangians, the potential representative and scaling: invariant stationary curves with endpoint terms). Under Dimensions and SI unit conventions of the non-relativistic classical mechanics framework, $[m] = μ$ , $[x] = ℓ$ , $[t] = [t_{0}] = τ$ , and $[F] = [S] = μ ℓ^{2} τ^{- 1}$ . Fix a unit vector $e_{1} = (1, 0, 0)$ and a velocity $v_{0}$ with $[v_{0}] = ℓ τ^{- 1}$ , and consider the dimensionally consistent affine curve $x_{*} (t) : = v_{0} (t - t_{0})$ on $[t_{0}, 2 t_{0}]$ .

Then:

for every $C^{2}$ curve $x : [t_{0}, 2 t_{0}] \to R^{3}$ one has $E_{L^{'}} (x) (t) = E_{L} (x) (t) = - m x^{''} (t)$ at every $t \in [t_{0}, 2 t_{0}]$ , and the stationary curves of $S$ and $S^{'}$ coincide;
for the curve $x_{*}$ the actions satisfy $S^{'} (x_{*}) - S (x_{*}) = F (2 t_{0}, x_{*} (2 t_{0})) - F (t_{0}, x_{*} (t_{0})) = 2 α t_{0}^{2} (e_{1} \cdot v_{0}),$ with $S (x_{*}) = \frac{1}{2} m ∣ v_{0} ∣_{2}^{2} t_{0}$ and $S^{'} (x_{*}) = S (x_{*}) + 2 α t_{0}^{2} (e_{1} \cdot v_{0})$ ;
for the reference-unit choice $m = 1 k g$ , $t_{0} = 1 s$ , $v_{0} = (1 m s^{- 1}) e_{1}$ and $α = 1 k g m s^{- 2}$ , these values are $S = 0.5 k g m^{2} s^{- 1}$ , $S^{'} = 2.5 k g m^{2} s^{- 1}$ and endpoint term $2 k g m^{2} s^{- 1}$ , all of dimension $μ ℓ^{2} τ^{- 1}$ , while the equation of motion and stationary curves are unchanged.

Facts & Assumptions

Given: The model and functions of the Example, the curve $x_{*}$ , and a $C^{2}$ curve $x : [t_{0}, 2 t_{0}] \to R^{3}$ .

[F1]

The gauge theorem (Total-derivative Lagrangians, the potential representative and scaling: invariant stationary curves with endpoint terms): for the mechanical Lagrangian $L$ and the $C^{3}$ function $F$ defined throughout the chart time-configuration domain with action dimension, the Lagrangian $L^{'} = L + F_{t} + \nabla_{x} F \cdot v$ is $C^{2}$ , satisfies $E_{L^{'}} = E_{L}$ and $S^{'} (x) = S (x) + F (b, x (b)) - F (a, x (a))$ for every $C^{2}$ admissible curve, and has the same stationary curves; a potential normalization $\tilde{U} = U + γ t$ uses $F (t, x) = - γ t^{2} / 2$ , where $[γ] = μ ℓ^{2} τ^{- 3}$ .

[F2]

Stationarity criterion (The stationary-action criterion and the potential-part Euler-Lagrange equation, The action functional of a force model with a potential presentation: the mechanical Lagrangian, its chart instance and its dimensions): a $C^{2}$ curve is stationary for the action of a mechanical Lagrangian if and only if it satisfies $m x^{''} = - \nabla_{x} U$ ; for the free model this is $x^{''} = 0$ , whose solutions on $[t_{0}, 2 t_{0}]$ are the affine vector curves.

[F3]

Calculus facts (The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ , The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , Integrable functions on $[a, b]$ form a set closed under sums and scalar multiples, and $\int_{a}^{b} (λ f + μ g) = λ \int_{a}^{b} f + μ \int_{a}^{b} g$ , A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion, The lower and upper Darboux integrals of a bounded $f$ on $[a, b]$ as $\sup_{P} L (f, P)$ and $\inf_{P} U (f, P)$ , Darboux integrability as their equality, and the notation $\int_{a}^{b} f$ , The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case, Directional derivatives and partial derivatives of a map $U \subseteq R^{m} \to R^{n}$ , The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral, The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length): for this $F$ one has $F_{t} = α x_{1}$ and $\nabla_{x} F = α t e_{1}$ , so $\frac{d}{d t} F (t, x (t)) = F_{t} + \nabla_{x} F \cdot x^{'}$ ; its integral over $[t_{0}, 2 t_{0}]$ is the endpoint difference. Also $S (x_{*}) = \int_{t_{0}}^{2 t_{0}} \frac{1}{2} m ∣ v_{0} ∣_{2}^{2} d t = \frac{1}{2} m ∣ v_{0} ∣_{2}^{2} t_{0}$ .

Verification

technique · direct; compute the two Euler-Lagrange expressions on a general curve and evaluate the endpoint term and both actions on explicit data

1.1F1F2F3givenalgebra

The displayed $F (t, x) = α t x_{1}$ is a polynomial, hence $C^{\infty}$ (in particular $C^{3}$ ), throughout the chart projection $R \times R^{3}$ . For a $C^{2}$ curve $x$ on $[t_{0}, 2 t_{0}]$ , $\nabla_{x} L^{'} = α e_{1}$ and $L_{v}^{'} = m x^{'} + α t e_{1}$ , so $E_{L^{'}} (x) (t) = α e_{1} - \frac{d}{d t} (m x^{'} (t) + α t e_{1}) = - m x^{''} (t) = E_{L} (x) (t)$ pointwise; hence the vanishing sets on this interval coincide and, by [F2], so do the stationary curves.

1.2F1F3givenalgebra

Endpoint term on the explicit curve: $x_{*} (t_{0}) = 0$ and $x_{*} (2 t_{0}) = v_{0} t_{0}$ , so $F (t_{0}, x_{*} (t_{0})) = 0$ and $F (2 t_{0}, x_{*} (2 t_{0})) = 2 α t_{0}^{2} (e_{1} \cdot v_{0})$ . By [F1] this is $S^{'} (x_{*}) - S (x_{*})$ .

2.1F1F3step 1.2algebra

Actions: $S (x_{*}) = \frac{1}{2} m ∣ v_{0} ∣_{2}^{2} t_{0}$ by [F3], and step 1.2 gives $S^{'} (x_{*}) = S (x_{*}) + 2 α t_{0}^{2} (e_{1} \cdot v_{0})$ .

3.1F2step 1.1step 2.1algebra

Reference-unit data: substituting the specified $m, t_{0}, v_{0}, α$ into step 2.1 gives $S = (1 / 2) k g m^{2} s^{- 1}$ and $S^{'} = (5 / 2) k g m^{2} s^{- 1}$ , differing by endpoint term $2 k g m^{2} s^{- 1}$ . Their numerical values in that SI action unit are $0.5$ , $2.5$ and $2$ , respectively; the dimension is $μ ℓ^{2} τ^{- 1}$ . The curve $x_{*}$ is affine, hence stationary for both actions by [F2], and the model equation $x^{''} = 0$ is the same for both.

4.1step 1.1step 1.2step 2.1step 3.1F1∎

Clauses 1-3 are verified: the total-derivative modification changes the action by the endpoint value of $F$ and leaves the Euler-Lagrange expression, the stationary curves and the model equation unchanged; the potential normalization is the case $F = - γ t^{2} / 2$ by [F1].

Remarks

The endpoint value $2 α t_{0}^{2} (e_{1} \cdot v_{0})$ is not the difference of a physical energy: it depends on the chosen $F$ and the endpoints. The illustration complements the gauge theorem of this page, which licenses reading the potential's arbitrary $c (t)$ as such a total derivative.

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The unrestricted action principle fails for the normalized nonholonomic relation $z_{s} = x y_{s}$ : the d'Alembert motion is not vakonomic

Physical scope

A worked check in dimensionless normalized coordinates and time, with physical length and time scales declared, on the nonholonomic relation z_s = x y_s: the d'Alembert motion from the stated normalized initial data has strictly positive y_s and nonzero y_{ss} for s>0, so it cannot satisfy the vakonomic (unrestricted-action) equations of the same constraint, which force y_{ss} to vanish wherever x_{ss} does. Thus these two prescriptions differ for this model and data, so no general equivalence holds. The example reports exact model computations, not measurements.

Example

Adopt the constraint model of The velocity constraint $z^{'} = x y^{'}$ : virtual displacements, the d'Alembert law and the vakonomic law in dimensionless normalized variables: let $s = t / T_{0}$ be dimensionless time and $(x, y, z) = r / L_{0} \in R^{3}$ be dimensionless coordinates, where $T_{0} > 0$ , $L_{0} > 0$ and $m_{0} > 0$ are reference quantities with $[T_{0}] = τ$ , $[L_{0}] = ℓ$ and $[m_{0}] = μ$ . Primes below mean differentiation with respect to $s$ . The normalized unit-mass equations have zero applied force and velocity constraint $z_{s} = x y_{s}$ ; in physical variables this is ${\dot{r}}_{z} = (r_{x} / L_{0}) {\dot{r}}_{y}$ , which is dimensionally consistent. The physical kinetic action is $m_{0} L_{0}^{2} / T_{0}$ times the normalized action; scaling the augmented multiplier by $m_{0} L_{0} / T_{0}$ makes its augmented action the same positive multiple of the normalized one, preserving stationarity. Physical reactions scale as $m_{0} L_{0} / T_{0}^{2}$ times normalized reactions, and physical virtual displacements scale as $L_{0}$ , so the zero-virtual-work condition is preserved. In these normalized variables the d'Alembert prescription is $x_{s s} = 0$ , $y_{s s} = - λ x$ , $z_{s s} = λ$ with $λ = x_{s} y_{s} / (1 + x^{2})$ along motions, and the vakonomic prescription is $x_{s s} = - η y_{s}$ , $y_{s s} = η_{s} x + η x_{s}$ , $z_{s s} = - η_{s}$ for the normalized multiplier $η$ . Consider the normalized initial data $(x, y, z) (0) = (0, 0, 0), (x_{s}, y_{s}, z_{s}) (0) = (1, 1, 0) .$

Then:

the curve $x (s) : = s, y (s) : = \int_{0}^{s} \frac{d u}{\sqrt{1 + u^{2}}}, z (s) : = \int_{0}^{s} u y_{s} (u) d u = \sqrt{1 + s^{2}} - 1$ is a d'Alembert motion with these data on $R$ , and it satisfies $y_{s} (s) = (1 + s^{2})^{- 1 / 2} > 0$ and $y_{s s} (s) = - s (1 + s^{2})^{- 3 / 2}$ for every $s \in R$ , so $y_{s s} (s) \neq 0$ whenever $s \neq 0$ ;
$y_{s s} (1 / 2) = - 4 / (5 \sqrt{5}) \neq 0$ ;
consequently this d'Alembert motion is not a vakonomic motion of the same constraint with the same data: on any nondegenerate interval where a vakonomic pair for this curve is stationary, $x_{s s} = 0$ and $y_{s} > 0$ force $η \equiv 0$ , and then the vakonomic equation forces $y_{s s} \equiv 0$ . Every nondegenerate interval contains some $s \neq 0$ , which contradicts clause 1;
hence, for this normalized model and initial data, the vakonomic stationary-action prescription does not reproduce the d'Alembert motion. This conclusion concerns the augmented-action prescription defined by the vakonomic law above; it makes no claim about a separately specified minimization problem or other variational prescriptions.

Facts & Assumptions

Given: The normalized constraint model and initial data of the Example, the scales $L_{0}, T_{0}, m_{0}$ and the displayed curve.

[F1]

The constraint model (The velocity constraint $z^{'} = x y^{'}$ : virtual displacements, the d'Alembert law and the vakonomic law): in normalized variables the allowed velocities and virtual displacements are $D (q) = {(w_{x}, w_{y}, w_{z}) : w_{z} = x w_{y}}$ ; the normalized sample Lagrangian is $\frac{1}{2} ∣ q_{s} ∣^{2}$ . The displayed d'Alembert equations follow from acceleration orthogonal to $D (q)$ and differentiation of the constraint; the displayed vakonomic equations are necessary for augmented-action stationarity, as derived in the Counterexample paragraphs headed "the d'Alembert system" and "the vakonomic equations" of The d'Alembert and vakonomic motions of $z^{'} = x y^{'}$ differ from the same initial data.

[F2]

The comparison result (The d'Alembert and vakonomic motions of $z^{'} = x y^{'}$ differ from the same initial data): every d'Alembert motion with the stated data is claimed to be a vakonomic motion of the same constraint with the same data; the statement is refuted there by exhibiting the difference. In particular the two prescriptions are not equivalent for this constraint and data. The present example re-derives the discrepancy on explicit quantities.

[F3]

Real-power and integral calculus (Continuity and derivatives of positive-base real powers, Real powers for positive bases, with the zero-base positive-exponent convention, Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ , The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , The first fundamental theorem: if $f$ is integrable on $[a, b]$ and continuous at $c$ , then $F^{'} (c) = f (c)$ ; in particular a continuous $f$ has $F$ as a primitive, The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ , For $a < c < b$ : $f$ is integrable on $[a, b]$ if and only if it is integrable on $[a, c]$ and on $[c, b]$ , and then $\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$ ; with the oriented form for arbitrary $a, b, c$ , A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion, The integral with oriented limits: $\int_{a}^{a} f : = 0$ and $\int_{b}^{a} f : = - \int_{a}^{b} f$ , Squares of nonzero elements are positive, Absolute value in an ordered field, The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral, The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length): $s \mapsto (1 + s^{2})^{- 1 / 2}$ is continuously differentiable on $R$ with derivative $- s (1 + s^{2})^{- 3 / 2}$ , the expression $1 + s^{2}$ is positive, and $\sqrt{5} > 0$ . Continuous integrands are integrable on every nondegenerate compact interval. Oriented additivity gives $\int_{u}^{v} f + \int_{v}^{w} f = \int_{u}^{w} f$ for arbitrary endpoints in such an interval. The first part of the fundamental theorem differentiates $v \mapsto \int_{A}^{v} f$ on a fixed $[A, B]$ , and its second part evaluates the integral of an integrable derivative.

[F4]

The model setting (Configuration domain of $N$ labeled particles: open admissible sets, regular holonomic constraints and the mass operator, clause (e), Dimensions and SI unit conventions of the non-relativistic classical mechanics framework, The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ): realize the separately adopted d'Alembert prescription of [F1] by one particle of mass $m_{0}$ with $C^{2}$ position curve $r (t) = L_{0} q (t / T_{0})$ , where $q = (x, y, z) : R \to R^{3}$ . The applied force input $F_{a p p} : R \times R^{3} \times R^{3} \to R^{3}$ is identically zero. Along a d'Alembert motion the ideal reaction is $R (t) = (m_{0} L_{0} / T_{0}^{2}) λ (t / T_{0}) (0, - x (t / T_{0}), 1)$ , a continuous vector of force dimension, so $m_{0} \ddot{r} = F_{a p p} + R$ . Its pairing with every physical virtual displacement $L_{0} δ$ , $δ \in D (q)$ , vanishes because $δ_{z} = x δ_{y}$ . Thus zero applied force does not mean zero total force. This constrained prescription is outside Adopted force model: force maps, the equation of motion, additive composition of force contributions and frame covariance, clause (i), which explicitly excludes constrained dynamics; that postulate supplies no equation with zero total force for this curve.

Verification

technique · direct; substitute the displayed curve into the d'Alembert equations, evaluate the curvature terms, and derive the contradiction with the vakonomic system

1.1F3givenalgebra

Clause 1, integral derivatives: for any $s_{0} \in R$ , choose $A < \min (0, s_{0})$ and $B > \max (0, s_{0})$ . For either continuous integrand $f (u) = (1 + u^{2})^{- 1 / 2}$ or $f (u) = u (1 + u^{2})^{- 1 / 2}$ , oriented additivity gives $\int_{0}^{s} f = \int_{A}^{s} f - \int_{A}^{0} f$ for $s \in [A, B]$ . The first part of the fundamental theorem differentiates the first term at the interior point $s_{0}$ and the second term is constant, so the integral has derivative $f (s_{0})$ , including when $s_{0} \leq 0$ . Thus $y_{s} = (1 + s^{2})^{- 1 / 2}$ and $z_{s} = s y_{s}$ on $R$ . The derivative of $\sqrt{1 + s^{2}}$ is $s (1 + s^{2})^{- 1 / 2}$ ; the second part of the fundamental theorem on the interval between $0$ and $s$ , with oriented reversal for $s < 0$ and the zero-integral convention for $s = 0$ , gives $z (s) = \sqrt{1 + s^{2}} - 1$ .

2.1F1F3F4step 1.1givenalgebra

Clause 1, d'Alembert equations: with $x (s) = s$ one has $x_{s} \equiv 1$ and $x_{s s} \equiv 0$ ; differentiating step 1.1 by [F3] gives $y_{s s} = - s (1 + s^{2})^{- 3 / 2}$ and $z_{s s} = y_{s} + s y_{s s} = (1 + s^{2})^{- 3 / 2}$ , both continuous, so the curve is $C^{2}$ . The d'Alembert multiplier is $λ = x_{s} y_{s} / (1 + x^{2}) = (1 + s^{2})^{- 3 / 2}$ , so $y_{s s} = - λ x$ and $z_{s s} = λ$ ; the constraint holds by step 1.1, and the initial data match. Thus the curve is a d'Alembert motion on $R$ with the stated data, $y_{s} > 0$ everywhere, and $y_{s s}$ vanishes only at $s = 0$ .

3.1F3step 2.1algebra

Clause 2: at $s = 1 / 2$ , $1 + s^{2} = 5 / 4$ , so by [F3] $y_{s s} (1 / 2) = - (1 / 2) (5 / 4)^{- 3 / 2} = - (1 / 2) \cdot 8 / (5 \sqrt{5}) = - 4 / (5 \sqrt{5}) \neq 0$ , because $4 \neq 0$ and $5 \sqrt{5} > 0$ .

3.2F1F2step 2.1algebra

Clause 3, contradiction: suppose the same curve were also vakonomic, with a $C^{2}$ multiplier $η$ making the augmented pair stationary on some nondegenerate interval $[a, b]$ . Its first vakonomic equation reads $x_{s s} = - η y_{s}$ ; since $x_{s s} = 0$ and $y_{s} > 0$ everywhere by step 2.1, this forces $η = 0$ on $[a, b]$ . The second equation then gives $y_{s s} = 0$ on $(a, b)$ . Choose $s_{0} \in (a, b)$ with $s_{0} \neq 0$ , which exists because $(a, b)$ is a nonempty open interval; step 2.1 gives $y_{s s} (s_{0}) \neq 0$ , a contradiction. Hence no interval in the vakonomic definition makes this curve stationary, so it is not a vakonomic motion with these normalized data.

4.1F1F2step 3.2∎

Clause 4: since the d'Alembert motion with these normalized data is not vakonomic by step 3.2, the vakonomic stationary-action prescription does not reproduce that motion for this model and data. The proof asserts no conclusion about minimization or other variational prescriptions, and no universal claim about all velocity constraints follows; the holonomic reduction of this page concerns the separate holonomic branch.

Remarks

The verification uses only the two displayed prescriptions and exact calculus; the underlying separation of the two dynamics is recorded in the cited counterexample, and the model-theoretic warning belongs to the scope remark of this pair's main page.

ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passOpen item page →

The planar pendulum by constraint elimination: kinetic metric, Lagrange equation and the ideal reaction recovered

Physical scope

A worked check of the constraint-elimination theorem on the planar pendulum model: one particle of mass m constrained without friction to a circle of radius l in the plane z=0 under a uniform downward field; the periodic angle map from R is a surjective immersion onto the circle, and multiple angle representatives are allowed. It gives kinetic metric m l^2, substituted potential -m g l cos(theta), the pendulum equation theta-double-dot = -(g/l) sin(theta), and the reaction as the transpose constraint-Jacobian multiple with its two component multipliers. The constraint components have dimensions length squared and length, so their multipliers have different dimensions. All quantities are model values of the declared idealization.

Example

Declare a one-label model of the force branch of Adopted force model: force maps, the equation of motion, additive composition of force contributions and frame covariance with $N = 1$ , mass $m > 0$ , time window $I_{0} = R$ , configuration domain $Q = R^{3}$ and force map $F (t, (x, y, z), v) : = (0, - m g, 0)$ with constant $g > 0$ : the global potential model of potential $U (x, y, z) = m g y$ and no residual forces (Potential presentations of a force model, potential and mechanical energy, closed versus exact force fields, local potentials and equipotential level sets). Under Dimensions and SI unit conventions of the non-relativistic classical mechanics framework, $[m] = μ$ , $[x] = [y] = [z] = [l] = ℓ$ , $[t] = τ$ , $[g] = ℓ τ^{- 2}$ , and $[U] = μ ℓ^{2} τ^{- 2}$ ; the angle $θ$ is dimensionless. Fix $l > 0$ and constrain the label to the circle of radius $l$ in the plane $z = 0$ by the regular holonomic-constraint datum of Holonomic-constraint models of the force branch: constraint data, the constrained equation, virtual displacements and the scleronomic/rheonomic distinction $O : = R \times R^{3}, k : = 2, g_{c} (x, y, z) : = (x^{2} + y^{2} - l^{2}, z),$ and adopt the ideal-reaction law of Adopted ideal-reaction law for holonomic constraints: the d'Alembert condition, its scope and its limits. The locus is the time-independent set $C = {(x, y, z) : x^{2} + y^{2} = l^{2}, z = 0}$ , so the datum is scleronomic, and the map $X : R ⟶ Q, X (θ) : = (l \sin θ, - l \cos θ, 0)$ is a surjective immersion of the constrained configurations in the sense of Generalized-coordinate maps of the model, the substituted Lagrangian, the kinetic metric and generalized forces. It is periodic and not one-to-one; the definition and reduction theorem allow these multiple angle representatives.

Then, with $r = 3 N - k = 1$ , kinetic metric $G (θ) = m l^{2}$ , substituted potential $\tilde{U} (θ) = - m g l \cos θ$ and substituted Lagrangian $\tilde{L} (θ, \dot{θ}) = \frac{1}{2} m l^{2} {\dot{θ}}^{2} + m g l \cos θ,$ the following hold:

for every interval $J \subseteq I_{0} = R$ with nonempty interior and every $C^{2}$ curve $θ : J \to R$ with $x : = X \circ θ$ , the curve $θ$ solves the Lagrange equation $Λ (\tilde{L}) (θ) = 0$ , i.e. $m l^{2} θ^{''} + m g l \sin θ = 0,$ if and only if $x$ is a constrained motion with ideal reaction;
when this holds, the reaction is $R (t) = D_{x} g_{c} (x (t))^{T} (\begin{array}{c} λ_{1} (t) \\ λ_{2} (t) \end{array}) = - m (l θ^{'} (t)^{2} + g \cos θ (t)) (\sin θ (t), - \cos θ (t), 0),$ where $λ_{1} (t) = - \frac{m}{2 l} (l θ^{'} (t)^{2} + g \cos θ (t))$ and $λ_{2} (t) = 0$ ; the multipliers have dimensions $[λ_{1}] = μ τ^{- 2}$ and $[λ_{2}] = μ ℓ τ^{- 2}$ , so the reaction has force dimension, annihilates the virtual displacements, and has a unique multiplier;
in particular the reduced pendulum equation $θ^{''} = - (g / l) \sin θ$ is exactly the Lagrange equation of $\tilde{L}$ , and every $C^{2}$ solution of it on such an interval $J$ , with $x = X \circ θ$ , is an ideal constrained motion of the model.

Facts & Assumptions

Given: The model data of the Example, the datum $(O, 2, g_{c})$ , the surjective immersion $X$ , a $C^{2}$ curve $θ : J \to R$ with $x = X \circ θ$ , and the constants $m, l, g > 0$ with the declared dimensions.

[F1]

The reduction theorem (Ideal scleronomic holonomic constraints: the substituted Lagrange equations are equivalent to the constrained model equation): for a scleronomic regular holonomic datum with a global-coverage parametrization $X (Ω) = C \cap Q$ and a potential presentation $(U, d)$ , a $C^{2}$ coordinate curve solves the Lagrange equations of $\tilde{L}$ with residual generalized force if and only if $x = X \circ θ$ is a constrained motion with ideal reaction; injectivity of $X$ is unnecessary, the virtual displacements at $x$ equal $D X (θ) (R)$ , and the reaction is unique with a unique multiplier vector.

[F2]

The datum and its locus (Holonomic-constraint models of the force branch: constraint data, the constrained equation, virtual displacements and the scleronomic/rheonomic distinction, Adopted ideal-reaction law for holonomic constraints: the d'Alembert condition, its scope and its limits, For an $m \times n$ matrix $A$ , $rank (A) + \dim N (A) = n$ , Submersions and immersions between Euclidean open sets, The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case): $g_{c}$ is smooth with $D_{x} g_{c} (x, y, z) = (\begin{array}{ccc} 2 x & 2 y & 0 \\ 0 & 0 & 1 \end{array})$ of rank $2$ on $C$ (the last row is independent; on $x^{2} + y^{2} = l^{2}$ not both $2 x, 2 y$ vanish), its component dimensions are $ℓ^{2}$ and $ℓ$ , the locus is nonempty and time-independent, and the virtual displacements at $X (θ)$ are $\ker D_{x} g_{c} = R (\cos θ, \sin θ, 0)$ .

[F3]

The parametrization and its derivatives (Generalized-coordinate maps of the model, the substituted Lagrangian, the kinetic metric and generalized forces, The derivatives of sine and cosine are cosine and minus sine, Sine and cosine defined by their real power series, A function differentiable at $c$ is continuous at $c$ , The principle of mathematical induction, Parity and the Pythagorean identity for sine and cosine, The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes, The transpose $A^{T}$ of a matrix, The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral): repeated differentiation cycles through $\pm \sin$ and $\pm \cos$ ; induction shows all iterated derivatives of $X$ exist and are continuous, so $X$ is $C^{\infty}$ . Also $D X (θ) = (l \cos θ, l \sin θ, 0)$ , $G (θ) = D X (θ)^{T} (m I_{3}) D X (θ) = m l^{2}$ , and $\tilde{U} (θ) = U (X (θ)) = - m g l \cos θ$ ; the displayed vectors $e_{θ} = (\cos θ, \sin θ, 0)$ and $e_{r} = (\sin θ, - \cos θ, 0) = X (θ) / l$ are orthonormal.

[F4]

The angle coverage and period (The derivatives of sine and cosine are cosine and minus sine, A function differentiable at $c$ is continuous at $c$ , Intermediate value theorem, by bisection with a canonical left-half rule: a continuous function on $[a, b]$ takes every value between $f (a)$ and $f (b)$ , Pi is the first positive zero of sine, Quarter-turn values and shifts by pi/2 and pi, Parity and the Pythagorean identity for sine and cosine, The zero sets of sine and cosine and the least positive common period 2 pi, Squares of nonzero elements are positive, Absolute value in an ordered field): cosine is continuous on $[0, π]$ , with $\cos 0 = 1$ and $\cos π = - 1$ , and sine is nonnegative there. For $u^{2} + v^{2} = 1$ , $v^{2} \geq 0$ gives $u^{2} \leq 1$ , hence $u \in [- 1, 1]$ ; the intermediate value theorem gives $θ_{0} \in [0, π]$ with $\cos θ_{0} = u$ . The Pythagorean identity and sine's sign give $\sin θ_{0} = ∣ v ∣$ ; use $θ = θ_{0}$ for $v \geq 0$ and $θ = - θ_{0}$ for $v < 0$ , with parity, to obtain $(\cos θ, \sin θ) = (u, v)$ . Sine and cosine have period $2 π$ .

Verification

technique · direct; verify the hypotheses of the reduction theorem on the pendulum datum, compute the substituted Lagrangian and its Lagrange equation, and evaluate the reaction and multiplier

1.1F2F3F4givenalgebra

Hypotheses of the reduction theorem: $g_{c}$ is smooth and $D_{x} g_{c}$ has rank $2$ on $C$ by [F2], so $(O, 2, g_{c})$ is a regular holonomic-constraint datum with nonempty locus, and it is scleronomic with $C = {(x, y, z) : x^{2} + y^{2} = l^{2}, z = 0}$ . The map $X$ is $C^{3}$ by [F3], with $D X (θ) = (l \cos θ, l \sin θ, 0)$ injective for all $θ$ because its norm is $l > 0$ . For any $(x, y, 0) \in C$ , put $u = - y / l$ and $v = x / l$ ; then $u^{2} + v^{2} = 1$ , so [F4] gives $θ$ with $\sin θ = v$ and $\cos θ = u$ , hence $X (θ) = (x, y, 0)$ . Conversely every $X (θ)$ lies in $C$ by the Pythagorean identity, so $X (R) = C \cap Q$ . The period statement in [F4] gives $X (θ + 2 π) = X (θ)$ with distinct inputs, so the map is periodic and not one-to-one. Moreover $(t, X (θ)) \in O$ for all $t, θ$ . Thus $X$ is a global-coverage parametrization with repeated angle representatives allowed, and $r = 3 N - k = 1$ ; the presentation $(U, d) = (m g y, 0)$ has no residual forces.

1.2F3algebra

Substituted Lagrangian and Lagrange equation: by [F3], $G (θ) = m l^{2}$ and $\tilde{U} (θ) = - m g l \cos θ$ , so $\tilde{L} = \frac{1}{2} m l^{2} {\dot{θ}}^{2} + m g l \cos θ$ and $Λ (\tilde{L}) (θ) = \frac{d}{d t} (m l^{2} θ^{'}) - \frac{\partial \tilde{L}}{\partial θ} = m l^{2} θ^{''} + m g l \sin θ .$ The Lagrange equation $Λ (\tilde{L}) (θ) = 0$ is therefore $m l^{2} θ^{''} + m g l \sin θ = 0$ , equivalent to $θ^{''} = - (g / l) \sin θ$ because $m l^{2} > 0$ .

2.1F1F2step 1.2algebra

Equivalence and reaction: since $d \equiv 0$ , the residual generalized force vanishes, and by [F1] the equation of step 1.2 holds along $θ$ if and only if $x = X \circ θ$ is a constrained motion with ideal reaction. In that case $R = m x^{''} - F$ and, differentiating $x$ twice, $x^{''} = l (- θ^{' 2} \sin θ + θ^{''} \cos θ, θ^{' 2} \cos θ + θ^{''} \sin θ, 0);$ substituting $θ^{''} = - (g / l) \sin θ$ and $F = (0, - m g, 0)$ gives $R = - m (l θ^{' 2} + g \cos θ) (\sin θ, - \cos θ, 0)$ . Since $D_{x} g_{c} (x)^{T} (λ_{1}, λ_{2}) = 2 λ_{1} (x, y, 0) + (0, 0, λ_{2})$ and $(x, y, z) = X (θ)$ , this equals $D_{x} g_{c} (x)^{T} (λ_{1}, 0)$ with $λ_{1} = - \frac{m}{2 l} (l θ^{' 2} + g \cos θ)$ , as stated. The reaction is normal to $\ker D_{x} g_{c}$ from [F2], so it annihilates the virtual displacements; [F1] supplies uniqueness of the multiplier vector.

3.1step 1.1step 1.2step 2.1∎

Clauses 1-3 are verified: the datum and parametrization satisfy the hypotheses of the reduction theorem; the Lagrange equation of $\tilde{L}$ is the reduced pendulum equation; and the reaction is the normal multiple with the displayed multiplier. All quantities are model values under the declared idealizations (frictionless circular track, rigid radius, uniform downward field), not measurements.

Remarks

The multiplier and reaction computed here agree with the multiplier-balance account of the preceding page on the same idealized apparatus; the present verification shows that the same motion is obtained by eliminating the constraint in the Lagrangian rather than by solving for the multiplier. Neither route adds an empirical claim.

Stationary action and Lagrangian dynamics — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

5 · Examples, experiments and counterexamples

The free particle: the unique stationary curve between fixed endpoints is the straight uniform motion, and it minimises the action

Example

Facts & Assumptions

Verification

Remarks

The harmonic oscillator on a long interval: the stationary curve is neither a local minimum nor a local maximum

Example

Facts & Assumptions

Verification

Remarks

Cylindrical coordinates for one label: the kinetic metric m diag(1, r^2, 1) and the coordinate form of the model's equation

Example

Facts & Assumptions

Verification

Remarks

A total-derivative Lagrangian on explicit data: the endpoint term and the unchanged equations

Example

Facts & Assumptions

Verification

Remarks

The unrestricted action principle fails for the normalized nonholonomic relation zs=xyszs​=xys​: the d'Alembert motion is not vakonomic

Example

Facts & Assumptions

Verification

Remarks

The planar pendulum by constraint elimination: kinetic metric, Lagrange equation and the ideal reaction recovered

Example

Facts & Assumptions

Verification

Remarks

Sources

The unrestricted action principle fails for the normalized nonholonomic relation $z_{s} = x y_{s}$ : the d'Alembert motion is not vakonomic