Physical interpretation of the canonical scalar product in linear dynamics Consider a unforced, undamped, linear mechanical system with a finite number of degrees of freedom. Its (second order) dynamical equations can be gathered in a matrix equation
$$M\ddot X + K X=0$$
Three natural scalar products appear:


*

*The canonical scalar product $X,Y\in\mathbb{R}^n\mapsto X^\top Y$

*The scalar product which stems from the kinetic energy: $X,Y\in\mathbb{R}^n\mapsto X^\top M Y$

*The scalar product which stems from the potential energy: $X,Y\in\mathbb{R}^n\mapsto X^\top K Y$


If written in the first order form the ODE leads, in the same spirit, to:


*

*Canonical scalar product $X,Y\in\mathbb{R}^{2n}\mapsto X^\top Y$

*Scalar product from the quadratic form of the energy $X,Y\in\mathbb{R}^{2n}\mapsto X^\top\begin{bmatrix} K & 0 \\ 0 & M \end{bmatrix} Y$


Question While the physical meaning of the scalar product for the energy is obvious, how can the canonical scalar product be interpreted?
Edit The answer could be that there is none (what I tend to believe), but it should be supported.
 A: From the physics perspective, there is a dimensional (or unit analysis) problem with the inner product on ${\mathbb R}^{2n}$ of the form $(\dot X, X)^T.(\dot X, X)$.  Naively, I'm adding a velocity squared to a position squared, and the units are not compatible.  Or maybe, if I'm allowed to trade $\dot X$ for momentum $P$, then I'm adding momentum squared to position squared which also doesn't make any sense.  To make sense of this inner product, I need to introduce some kind of scale which will make the units compatible -- much the way inserting the matrices $M$ and $K$ will perform the same job and give me an interpretation in terms of the total energy.
A more natural inner product on ${\mathbb R}^{2n}$ in this case involves a symplectic bilinear form.  In the two dimensional case, such an inner product would involve the matrix
$$
\omega = 
\left(
\begin{matrix}
0 & 1 \\
-1 & 0 
\end{matrix}
\right)
$$
This inner product, from a quantum mechanics perspective anyway, has a physical interpretation in terms of the commutator of position and momentum.  
Back to classical mechanics, writing the ODE using a Hamiltonian, e.g. $H = (p^2 / m + m x^2)/2$, naturally involves this symplectic bilinear form. The ODE in first order form becomes
$$
\frac{\partial H}{\partial x} = -\dot p \ , \\
\frac{\partial H}{\partial p} = \dot x \ .
$$
which I can then recast as
$$
\frac{\partial H}{\partial Z} = - \omega \cdot \dot Z
$$
where I have defined $Z = (x,p)$.  There is tons of literature on symplectic forms and classical mechanics, for example Arnold's book.
