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A generalized coordinate absent from the Lagrangian is cyclic. Its conjugate generalized momentum is conserved by the Euler-Lagrange equations.

A Lagrange top is a rigid body that is symmetric about a principal axis, has a point on that axis fixed in space, and has its center of mass on the same axis while gravity acts uniformly. Here $I_1=I_2$ is the transverse principal moment of inertia about the fixed point, $I_3$ is the moment about the symmetry axis, $M$ is the total mass, and $l$ is the distance from the fixed point to the center of mass.

The Euler angles for a symmetric top use $\theta$ for inclination, $\varphi$ for precession, and $\psi$ for spin about the body axis. Since $\psi$ is a cyclic coordinate, its generalized momentumis conserved. The coordinate $\varphi$ is also cyclic, sois a second integral. Finally, the Lagrangian has no explicit time dependence, and conservation of energy from time-translation invariance gives the independent integral

For steady precession set $\theta$ constant and $\dot{\varphi }=\Omega$ constant. The $\theta$ Euler-Lagrange equation, with $0<\theta <\pi /2$, reduces after division by $\sin \theta$ toThis quadratic has a real precession rate precisely when its quadratic discriminant is nonnegative. Hence Steady precession of a Lagrange top is possible if and only if

The generalized momenta conjugate to the two ignorable coordinates areandSince the Lagrangian contains neither $\varphi$ nor $\psi$, both momenta are constant in time.

Eliminating the two angular velocities givesConservation of energy then yields the heavy symmetric top reductionwhere

As $\theta \to 0$, the first term diverges unlessWriting their common value as $p$, the small-angle expansion isBy the effective potential stability criterion, the vertical state is stable exactly whenAt the vertical state $p=I_3(\dot{\psi }+\dot{\varphi })$, so this is the gyroscopic stabilization of an inverted symmetric top by sufficiently rapid spin.

Let $\hat{b}=B/|B|$. For a constant uniform magnetic field, choose the symmetric gaugewhich has $\nabla \times A=B$ and $A\cdot \hat{b}=0$. With $E=0$, one may take $\varphi =0$. The coordinate $r_{\parallel }=r\cdot \hat{b}$ is then absent from the Lagrangian, so it is an ignorable coordinate. Its generalized momentumis conserved. Hencewhich is the canonical momentum parallel to a uniform magnetic field in this gauge.

The generalized momentum conjugate to $q$ is

Stationarity of the action under variations with fixed endpoints gives the Euler-Lagrange equation. The generalized momentum is $p=\partial L/\partial \dot{q}$. The Legendre transform in mechanics isafter solving $p(q,\dot{q},t)$ for $\dot{q}$. This requires at least local invertibility, ${\partial }^{2}L/\partial {\dot{q}}^2\ne 0$; global single-valuedness requires global invertibility.