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What is geometric mechanics? At first blush, it is an approach to studying mechanics:
Let us look briefly at each in turn.
Classical mechanics \[ F = \ddt{t}{(mv)} = \ddt{t}{p} \]
Classical mechanics - clarifications \[ \underbrace{F}_{{\color{red}{\textrm{what is a force?}}}} = \underbrace{\ddt{t}{\overbrace{(mv)}^{{\color{red}{\textrm{what is mass? velocity?}}}}}}_{{\color{red}{\textrm{what is the rate of change of velocity?}}}} = \overbrace{\ddt{t}{p}}^{{\color{red}{\textrm{what is momentum?}}}} \]
\begin{align*}
A \ddot{x} &= -S(x-y) - R(x-z) \\
B \ddot{y} &= -S(y-x) - T(y-z) \\
C \ddot{z} &= -R(z-x) - T(z-y)
\end{align*}
\begin{align*}
\dfrac{d\,}{dt}
\begin{bmatrix}
A & 0 & 0\\0 & B & 0\\0 & 0 & C
\end{bmatrix}
\begin{bmatrix}
\dot{x}\\\dot{y}\\\dot{z}
\end{bmatrix}
&=
-
\begin{bmatrix}
+S+R & -S & -R\\
-S & +S+T & -T\\
-R & -T & +R+T
\end{bmatrix}
\begin{bmatrix}
x\\y\\x
\end{bmatrix}
\\
\dfrac{d\,}{dt} \mathbf{M} \dot{\mathbf{q}} &= -\mathbf{Q} \mathbf{q}
\end{align*}
A Force is something we integrate over a path to compute Work done.
\begin{align*}
W &= \int\limits_{\sigma}^{} \ip{F}{d \sigma}
\end{align*}
A Force is a differential 1-form.
Let's compute the exterior derivative of the force 1-form $F$ for the spring-mass system:
\begin{align*} F &= \sum_i (\mathbf{Qq})_i \D{\mathbf{q}}_i = \sum_{i, j} \mathbf{Q}_{ij}\mathbf{q}_j \D{\mathbf{q}}_i \\ \D{F} &= \sum_{i, j} \mathbf{Q}_{ij} \D{\mathbf{q}}_j \wedge \D{\mathbf{q}}_i = 0 \end{align*}
\begin{align*} f(\mathbf{q}) &= \wint{\mathbf{0}}{\mathbf{q}} \ip{F(\sigma(t))}{\dot{\sigma}(t)} \D t = \wint{0}{1} \ip{F(t \mathbf{q})}{\mathbf{q}} \D t\\ &= - \wint{0}{1} \ip{\mathbf{Qq}}{\mathbf{q}} t \D t = - \wint{0}{1} t \D t \times \ip{\mathbf{Qq}}{\mathbf{q}} = - \frac{1}{2} \ip{\mathbf{Qq}}{\mathbf{q}}. \end{align*}
By definition $U = -f$ is the potential energy of the system.
How to measure velocity. |
We need to do calculus. To do calculus, we need differentiable manifolds.
We need to do calculus. To do calculus, we need differentiable manifolds.
Coordinate charts |
Let $\sphere{2} \subset \R^3$ be the unit sphere.
Let $A$ be the set of all lines $\R^3$.
Let $M$ be a manifold with a coordinate system $x : U \subset M \to \R^n$.
We compute that \begin{align*} \ip{\D{x_i}}{\didi{}{x_j}} &= \didi{x_i}{x_j} = \delta_{ij} \\ \ip{\D{x_i}}{V} &= \lieder{V}{x_i} = \sum_j V_j \times \ip{\D{x_i}}{\didi{}{x_j}} = V_i \\ \ip{F}{V} &= \sum_i F_i(x) V_i(x). \end{align*}