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For three ordered angular positions $q_1,q_2,q_3$, define consecutive gapsand the collective angle $\Phi =(q_1+q_2+q_3)/3$. Then $\alpha +\beta +\gamma =2\pi$. For three equal masses on a circle of radius $r$,Thus the ignorable coordinate $\Phi$ decouples from the relative gap dynamics whenever the potential depends only on the gaps.

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.

The gravitational three-body Lagrangian is

Put $M_{12}=m_1+m_2$ and $M=M_{12}+m_3$. Solving the definitions of the Jacobi coordinates for three particles givesSubstitution into the kinetic energy makes all cross terms cancel:where

The pair separations areHence the gravitational potential energy iswhich is independent of $\mathbf{c}$. Thus $\mathbf{c}$ is an ignorable coordinate andThe center of mass moves uniformly because the isolated system has no external force.

Finally, substituting the inverse coordinate transformation intoagain cancels every cross term and yieldsThis is the angular momentum decomposition in three-body Jacobi coordinates.

Write $q^1=u$, $q^2=v$, ${\sigma }_i={\partial }_i\sigma$, and let $g_{ij}={\sigma }_i\cdot {\sigma }_j$ be the first fundamental form. Sincedifferentiating ${\sigma }_u\cdot \dot{\gamma }$ and ${\sigma }_v\cdot \dot{\gamma }$, or equivalently taking the two inner products of the second formula with ${\sigma }_k$, givesThe matrix $(g_{k\ell })$ is invertible because $\sigma$ is a regular embedded surface parametrization. Consequently the two geodesic equations hold exactly when $\ddot{\gamma }$ is orthogonal to both tangent vectors ${\sigma }_u$ and ${\sigma }_v$, which says precisely that $\ddot{\gamma }$ is a normal vector. This is the ambient acceleration criterion for a surface geodesic.

Because $\dot{\gamma }$ is tangent and $\ddot{\gamma }$ is normal,Thus an affinely parametrized geodesic has constant speed, as recorded by constant speed of an affinely parametrized geodesic.

For a surface of revolution, use profile arc length $s$ and azimuth $\varphi$. Its Riemannian metric isThe azimuth is an ignorable coordinate, so the corresponding geodesic equation has the first integralIf $v=|\dot{\gamma }|$ and $\theta$ is the oriented angle with the parallel, then the component of velocity along the parallel isSince $v$ is constant,This is the Clairaut first integral for a surface of revolution.

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.

Choose ordered particle angles $q_1,q_2,q_3$ withand put $\Phi =(q_1+q_2+q_3)/3$. Solving for the individual angles givesThe kinetic energy is thereforeThe collective angle is an ignorable coordinate and decouples from the gaps, so its term may be omitted from the relative Lagrangian. Substituting $\gamma =2\pi -\alpha -\beta$ into the potential energy gives

A coordinate $q$ is an ignorable coordinate when $L$ has no explicit dependence on it, so $\partial L/\partial q=0$. Its Euler--Lagrange equation iswhich proves that $p_q$ is conserved.