PX285 - G2 - stiffness matrix

note: from typed notes

equilibrium conditions

considering the equilibrium position:

{\dot{q}}_{i} = 0 \forall q_{i}

this requires $\vec{p} = 0$ and the system remains at equilibrium only if $\dot{\vec{p}} = 0$
from hamilton's equation, this is only possible if the generalized forces also vanish, ie: $- \frac{\partial H}{\partial q_{i}} = 0$
since all generalized momenta vanish at equilibrium, $T = 0 :$

\begin{matrix} (1) & \frac{\partial V}{\partial q_{i}} = 0 \end{matrix}

this corresponds to the newtonian condition of forces vanishing at equilibrium

examining the motion

the equilibrium position:

{\vec{q}}^{(0)} = {q_{1}^{(0)}, q_{2}^{(o)} \dots, q_{i}^{(0)} \dots}

where, $\frac{\partial V}{\partial q_{i}} |_{{\vec{q}}^{(0)}} = 0$

changing the coordinate variables by the translation:

x = \vec{q} - {\vec{q}}^{(0)} ⟺ x_{i} = q_{i} - q_{i}^{(0)} \forall i

here, $\dot{\vec{x}} = \dot{\vec{q}}$ , so the definitions of momentum are unchanged
the taylor expansion of the potential:

V (x) = V (0) + \sum_{i} x_{i} \frac{\partial V}{\partial x_{i}} |_{0} + \frac{1}{2} \sum_{i j} x_{i} x_{j} \frac{\partial^{2} V}{\partial x_{i} \partial x_{j}} |_{0} + \dots

where, $V (0)$ is the constant value at equilibrium

from equation $(1)$ , the first derivative is zero
the generalized forces are given by:

\begin{aligned} F_{i} & = - \frac{\partial V (\vec{x})}{\partial x_{i}} \\ = - \sum_{j} K_{i j} x_{j} \end{aligned}

the stiffness/force matrix is defined as:

K_{i j} = \frac{\partial^{2} V}{\partial x_{i} \partial x_{j}} |_{0}

considering a spring with $V = \frac{1}{2} k x^{2}$ , with a restoring force, $F = - k x$ , $K$ is a $1 \times 1$ matric with $K_{11} = k$ , the 'stiffness' of the spring