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A mechanical chart Lagrangian with positive definite kinetic matrix is hyperregular, with the explicit kinetic-plus-potential Hamiltonian

Statement

Let $n \geq 1$ , let $J \subseteq R$ and $U \subseteq R^{n}$ be open and nonempty, and let $G : J \times U \to M_{n} (R)$ , $A : J \times U \to R^{n}$ and $V : J \times U \to R$ be of class $C^{2}$ ( $C^{k}$ Euclidean maps and diffeomorphisms). Assume that $G (t, q)$ is symmetric and positive definite for every $(t, q)$ :

$G (t, q)^{T} = G (t, q), w^{T} G (t, q) w > 0 for every 0 \neq w \in R^{n} .$

Define the mechanical chart Lagrangian

$L (t, q, v) : = \frac{1}{2} v^{T} G (t, q) v + A (t, q) \cdot v - V (t, q),$

with the Euclidean inner product of The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ . Then:

$L$ is of class $C^{2}$ on $J \times U \times R^{n}$ , with $L_{v} (t, q, v) = G (t, q) v + A (t, q)$ and $L_{v v} (t, q, v) = G (t, q)$ ; in particular $L$ is regular on the chart;
the Legendre map $F_{L} (t, q, v) = (t, q, G (t, q) v + A (t, q))$ is a $C^{2}$ bijection of $J \times U \times R^{n}$ onto itself with $C^{2}$ inverse

$v (t, q, p) = G (t, q)^{- 1} (p - A (t, q));$

hence $L$ is hyperregular in the sense of The Legendre map of a time-dependent Lagrangian: regularity and hyperregularity; 3. the Hamiltonian of the inverse branch is

$H (t, q, p) = \frac{1}{2} (p - A (t, q))^{T} G (t, q)^{- 1} (p - A (t, q)) + V (t, q),$

a $C^{2}$ function on $J \times U \times R^{n}$ .

Facts & Assumptions

Given: The $C^{2}$ fields $G, A, V$ with $G (t, q)$ symmetric positive definite, the Lagrangian $L$ displayed above, and the Legendre map $F_{L} (t, q, v) = (t, q, L_{v} (t, q, v))$ .

[F1]

Partial derivatives of a quadratic form: if $G$ is symmetric, then the gradient of $v \mapsto \frac{1}{2} v^{T} G v$ is $v \mapsto G v$ , and the second derivative in $v$ is the constant matrix $G$ ; this is the sum and product rule applied to $\frac{1}{2} \sum_{i, j} G_{i j} v_{i} v_{j}$ together with $G_{i j} = G_{j i}$ (The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case, Directional derivatives and partial derivatives of a map $U \subseteq R^{m} \to R^{n}$ , Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes).

[F2]

Finite sums and products of $C^{r}$ Euclidean maps are $C^{r}$ , and composites of composable $C^{r}$ maps are $C^{r}$ ( $C^{k}$ Euclidean maps are closed under componentwise algebra and composition, $C^{k}$ Euclidean maps and diffeomorphisms). In particular $L$ is $C^{2}$ because $G, A, V$ are.

[F3]

A symmetric positive definite matrix is invertible: if $G w = 0$ then $w^{T} G w = 0$ , so $w = 0$ , and a linear map of $R^{n}$ with trivial kernel is bijective by Rank-nullity: $\dim_{F} V = nullity T + rank T$ (zero kernel gives image dimension $n$ ); its linear inverse is two-sided (Invertible matrices and the general linear group ${GL}_{n} (F)$ , Linear map between vector spaces over the same field).

[F4]

Inversion preserves regularity: if the entries of $B : J \times U \to M_{n} (R)$ are $C^{r}$ and $\det B$ never vanishes, then the entries of $B^{- 1}$ are $C^{r}$ (Matrix inversion preserves $C^{k}$ regularity where the determinant is nonzero). For $B = G$ , invertibility holds by [F3], and $\det G \neq 0$ because $\det (G) \det (G^{- 1}) = \det (G G^{- 1}) = \det (I_{n}) = 1$ for an invertible $G$ with inverse $G^{- 1}$ (For same-sized finite square matrices over a commutative ring, $\det (A B) = \det (A) \det (B)$ , A positive-sized square matrix over a commutative ring is invertible if and only if its determinant is a unit).

Proof

technique · direct; differentiate the quadratic Lagrangian, invert the velocity--momentum relation by positive definiteness, and complete the square

1.1F1F2F3algebra

By [F1] and [F2] the map $L$ is $C^{2}$ , its gradient in the velocity slot is $L_{v} (t, q, v) = G (t, q) v + A (t, q)$ and the velocity Hessian is $L_{v v} (t, q, v) = G (t, q)$ . Since $G (t, q)$ is positive definite, it is invertible by [F3], so $\det L_{v v} (t, q, v) = \det G (t, q) \neq 0$ ; this proves clause 1.

1.2F3algebra

Clause 2, bijectivity: fix $(t, q, p) \in J \times U \times R^{n}$ . By [F3] the matrix $G (t, q)$ is invertible, so the equation $G (t, q) v + A (t, q) = p$ has the unique solution $v = G (t, q)^{- 1} (p - A (t, q))$ ; hence $F_{L}$ is bijective with the displayed inverse.

2.1F4step 1.2algebra

Clause 2, regularity of the inverse: the entries of $G$ are $C^{2}$ and $\det G$ never vanishes, so by [F4] the entries of $(t, q) \mapsto G (t, q)^{- 1}$ are $C^{2}$ ; the formula for $v$ is then a finite combination of $C^{2}$ maps and is $C^{2}$ ( $C^{k}$ Euclidean maps are closed under componentwise algebra and composition). By The Legendre map of a time-dependent Lagrangian: regularity and hyperregularity this makes $L$ hyperregular.

3.1F1F2step 2.1algebra∎

Clause 3: with $v = v (t, q, p) = G^{- 1} (p - A)$ one has $p - A = G v$ , hence $p \cdot v - L (t, q, v) = (G v + A) \cdot v - \frac{1}{2} v^{T} G v - A \cdot v + V = \frac{1}{2} v^{T} G v + V = \frac{1}{2} (p - A)^{T} G^{- 1} (p - A) + V,$ where the last step uses $(G v) \cdot v = v^{T} G v = (p - A)^{T} G^{- 1} (p - A)$ and the symmetry of $G^{- 1}$ (the inverse of a symmetric matrix is symmetric). Thus clause 3 holds, and $H$ is $C^{2}$ by the same closure properties.

Remarks

The case of a constant diagonal matrix $G = diag (m_{1}, \dots, m_{n})$ with $m_{i} > 0$ and $A = 0$ gives $L = \frac{1}{2} \sum_{i} m_{i} {\dot{q}}_{i}^{2} - V (q)$ and $H = \frac{1}{2} \sum_{i} p_{i}^{2} / m_{i} + V (q)$ , the classical kinetic-plus-potential form. The proposition is stated for a configuration- and time-dependent positive definite $G$ , which is what the consuming classical-mechanics items need for moving constraints and rotating frames.
The covector field $A$ is a mathematical datum of the chart: it is allowed to depend on $(t, q)$ and enters the momentum as $p = G v + A$ . It is not an electromagnetic potential and no physical interpretation of $A$ is asserted here; in particular $p$ is not $mass \times velocity$ unless $G$ is constant and $A = 0$ . The scope of these identifications is recorded in Scope of the Legendre map: regular versus hyperregular, the meaning of the covector datum, and the limits of the branch description.

A mechanical chart Lagrangian with positive definite kinetic matrix is hyperregular, with the explicit kinetic-plus-potential Hamiltonian

Statement

Facts & Assumptions

Proof

Remarks

Depends on

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