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Time-dependent coordinate maps transform the potential-part equation into the substituted Euler-Lagrange equation

Physical scope

Covariance of the model's equation of motion under a time-dependent coordinate map of the configuration domain: along a curve described by a C^3 map X with everywhere invertible configuration Jacobian, the Euler-Lagrange expression of the substituted Lagrangian is the transpose-Jacobian multiple of the inertial-chart expression, so the model's potential-part equation holds along the Cartesian description if and only if the substituted Euler-Lagrange equation holds along the coordinate description; when the chart condition holds, this is equivalent to stationarity of the substituted action. The statement is conditional on the adopted force branch, the potential presentation and the smooth invertibility of the coordinate map; it asserts nothing about constraints with fewer coordinates, which are treated separately, and no empirical claim.

Statement

Adopt the force model with potential presentation $(U, d)$ and mechanical Lagrangian $L = K - U$ of The action functional of a force model with a potential presentation: the mechanical Lagrangian, its chart instance and its dimensions, with masses $m_{1}, \dots, m_{N} > 0$ , open configuration domain $Q \subseteq R^{3 N}$ , time window $I_{0}$ , mass operator $M$ and $C^{2}$ potential $U : I_{0} \times Q \to R$ (Potential presentations of a force model, potential and mechanical energy, closed versus exact force fields, local potentials and equipotential level sets). Let $Ω \subseteq R \times R^{3 N}$ be open and nonempty and let $X : Ω \to Q$ be of class $C^{3}$ with $X_{q} (t, q)$ invertible for every $(t, q) \in Ω$ ( $C^{k}$ Euclidean maps and diffeomorphisms, Invertible Euclidean linear maps). Put

$\tilde{O} : = {(t, q, v) \in Ω \times R^{3 N} : (t, X (t, q), X_{t} (t, q) + X_{q} (t, q) v) \in I_{0} \times Q \times R^{3 N}},$

$\tilde{L} (t, q, v) : = L (t, X (t, q), X_{t} (t, q) + X_{q} (t, q) v) .$

Let $a < b$ and let $q : [a, b] \to R^{3 N}$ be a $C^{2}$ curve with $(t, q (t), q^{'} (t)) \in \tilde{O}$ for all $t \in [a, b]$ ; put $x (t) : = X (t, q (t))$ . Then:

$\tilde{O}$ is open, $\tilde{L}$ is of class $C^{2}$ on it, and

$E_{\tilde{L}} (q) (t) = X_{q} (t, q (t))^{T} E_{L} (x) (t) (t \in [a, b]);$

consequently $E_{\tilde{L}} (q) = 0$ on $[a, b]$ if and only if $E_{L} (x) = 0$ on $[a, b]$ , and (by The stationary-action criterion and the potential-part Euler-Lagrange equation) $q$ satisfies the substituted Euler-Lagrange equation if and only if the Cartesian description $x$ satisfies the model's potential-part equation $M x^{''} (t) = - \nabla_{x} U (t, x (t))$ on $[a, b]$ .

If in addition there is an open set $W \subseteq R^{3 N}$ with $q ([a, b]) \subseteq W$ and $[a, b] \times W \times R^{3 N} \subseteq \tilde{O}$ , then $(\tilde{O}, \tilde{L})$ is a time-dependent Lagrangian chart over $[a, b] \times W$ and: $q$ is stationary for the substituted action $\tilde{S} (q) = \int_{a}^{b} \tilde{L} (t, q, q^{'}) d t$ if and only if $x$ is stationary for $S$ .

The statement is conditional on the adopted force branch, on the declared potential presentation and on the smooth invertibility of $X$ ; the residual forces $d$ do not enter, so clause 1 concerns the potential part of the force map.

Facts & Assumptions

Given: The model data, presentation, the map $X$ and the curve $q$ of the statement, with $x = X \circ q$ .

[F1]

The covariance supplier (Coordinate covariance of the Euler-Lagrange expression under time-dependent point transformations): for $Ω \subseteq R \times R^{n}$ open, $X : Ω \to R^{n}$ of class $C^{3}$ with $X_{q}$ everywhere invertible, and $L$ of class $C^{2}$ on an open $O_{x} \subseteq R \times R^{n} \times R^{n}$ , the set $\tilde{O}$ defined as above is open, $\tilde{L}$ is $C^{2}$ on it, and for every $C^{2}$ curve $q$ with $(t, q (t), q^{'} (t)) \in \tilde{O}$ the transport identity holds along the curve; equivalently $E_{\tilde{L}} (q) = X_{q}^{T} E_{L} (x)$ in the notation of A time-dependent Lagrangian chart, its action functional and its fixed-endpoint first variations; and since $X_{q} (t, q (t))$ is invertible, the vanishing of one expression is equivalent to the vanishing of the other.

[F2]

The model-terms identity (The stationary-action criterion and the potential-part Euler-Lagrange equation, The first variation of the mechanical action and the Euler-Lagrange expression in model terms): for the mechanical Lagrangian of the presentation, a $C^{2}$ admissible curve $z$ of the Cartesian chart is stationary for $S$ if and only if $E_{L} (z) (t) = 0$ for all $t$ of its interval, and the latter is equivalent to $M z^{''} (t) = - \nabla_{x} U (t, z (t))$ there.

[F3]

The chart dictionary (A time-dependent Lagrangian chart, its action functional and its fixed-endpoint first variations, The action functional of a force model with a potential presentation: the mechanical Lagrangian, its chart instance and its dimensions, Stationarity of the time-dependent action is equivalent to the vanishing of the Euler-Lagrange expression): a pair $(O^{'}, L)$ with $O^{'} \subseteq R \times R^{3 N} \times R^{3 N}$ open and $[a, b] \times W \times R^{3 N} \subseteq O^{'}$ is a time-dependent Lagrangian chart over $[a, b] \times W$ ; stationarity means the vanishing of the first variation $δ S (q) [η]$ for all variations $η$ with $η (a) = η (b) = 0$ , and it is equivalent to the vanishing of the Euler-Lagrange expression on $[a, b]$ .

[F4]

The matrix fact (Invertible Euclidean linear maps, Linear map between vector spaces over the same field, Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes): if $A \in M_{n} (R)$ is invertible and $u \in R^{n}$ , then $A^{T} u = 0$ implies $u = 0$ , because $A^{T}$ is surjective.

Proof

technique · direct; instantiate the mathematical covariance proposition at the mechanical Lagrangian, then read the result through the model-terms identity and the chart dictionary

1.1F1given

Clause 1, structure: the Cartesian Lagrangian $L$ is $C^{2}$ on the open set $O_{x} : = I_{0} \times Q \times R^{3 N} \subseteq R \times R^{3 N} \times R^{3 N}$ (The action functional of a force model with a potential presentation: the mechanical Lagrangian, its chart instance and its dimensions), the map $X$ and the set $Ω$ satisfy the hypotheses of [F1] with $n : = 3 N$ , and the displayed $\tilde{O}$ , $\tilde{L}$ are exactly the supplier's constructions; hence $\tilde{O}$ is open, $\tilde{L}$ is $C^{2}$ , and the transport identity holds along $q$ .

2.1F1F4step 1.1algebra

Clause 1, identity and equivalence: substituting the transport identity of [F1] gives $E_{\tilde{L}} (q) (t) = X_{q} (t, q (t))^{T} E_{L} (x) (t)$ ; since $X_{q} (t, q (t))$ is invertible, [F4] gives $E_{\tilde{L}} (q) (t) = 0 \Leftrightarrow E_{L} (x) (t) = 0$ at each $t \in [a, b]$ , hence the vanishing conditions on the whole interval agree.

3.1F2step 2.1

Clause 1, model reading: by [F2] applied to the Cartesian curve $x$ , the vanishing of $E_{L} (x)$ on $[a, b]$ is equivalent to $M x^{''} (t) = - \nabla_{x} U (t, x (t))$ on $[a, b]$ ; combining with step 2.1, the substituted equation $E_{\tilde{L}} (q) = 0$ holds exactly when the Cartesian description satisfies the model's potential-part equation.

4.1F3step 2.1given∎

Clause 2: under the additional chart hypothesis, $(\tilde{O}, \tilde{L})$ is a time-dependent Lagrangian chart over $[a, b] \times W$ in the sense of [F3], and $q$ has values in $W$ and is $C^{2}$ , so $q$ is an admissible curve of that chart; by [F3], $q$ is stationary for $\tilde{S}$ if and only if $E_{\tilde{L}} (q) = 0$ on $[a, b]$ . Similarly the Cartesian curve $x$ is stationary for $S$ if and only if $E_{L} (x) = 0$ on $[a, b]$ , and step 2.1 gives the equivalence of the two vanishing conditions.

Remarks

Time-dependent maps are included. The map $X$ may depend on $t$ , so the result covers moving charts of the same type as the moving-frame construction of the preceding page: the substituted Lagrangian reads the Cartesian velocity as $X_{t} + X_{q} v$ , and the equation of motion in the moving coordinates is the Euler-Lagrange equation of $\tilde{L}$ by clause 1.
Fewer coordinates are a different construction. Here $X$ is a local diffeomorphism of the configuration domain ( $X_{q}$ invertible), so the number of coordinates equals the number of Cartesian degrees of freedom. The reduction to a parametrisation of a holonomic constraint locus, where $X$ has fewer coordinates than the ambient configuration space, is treated separately on this page.

Time-dependent coordinate maps transform the potential-part equation into the substituted Euler-Lagrange equation

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

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