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Label the equilibrium vertices in the order $+x,-x,+y,-y,+z,-z$, and write their displacements as ${\mathbf{u}}_i(t)\in {\mathbb{R}}^3$. Their eighteen Cartesian components are generalized coordinates. Linearize the axial Hookean springs about the unstressed equilibrium. For an edge with unit direction ${\mathbf{n}}_{ij}$, its first-order extension is ${\mathbf{n}}_{ij}\cdot ({\mathbf{u}}_i-{\mathbf{u}}_j)$, so the quadratic potential energy isThree zero modes are translations: for each Cartesian unit vector ${\mathbf{e}}_{\alpha }$, take ${\mathbf{u}}_i={\mathbf{e}}_{\alpha }$ at every vertex. Three more are infinitesimal rotations ${\mathbf{u}}_i={\mathbf{e}}_{\alpha }\times {\mathbf{r}}_i$. Explicitly, in the stated vertex order, their displacement vectors areTranslations leave each displacement difference zero. For rotations,because the equilibrium edge vector is parallel to ${\mathbf{n}}_{ij}$. Hence every spring has zero first-order extension in all six patterns, and the stiffness matrix annihilates them.

The translations have nonzero summed displacement, while all the rotations have zero sum. The three rotational patterns are themselves independent because the vertices do not lie on a single line. ThusThese are the rigid-motion zero modes of a spring framework. The rotational displacements describe infinitesimal rigid rotations; an arbitrary finite rotation would require its nonlinear displacement field.