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In Cartesian coordinates the Lagrangian isIntroduce a Lagrange multiplier $\lambda$ for $x^2+y^2+z^2=c^{-2}$. The Euler-Lagrange equations becomeorDifferentiating the constraint twice givesTaking the scalar product of the vector equation with $\mathbf{x}$ therefore yieldsSubstitution gives

For $c=0$, the equations have the first integralsWriting $X=x+B$ and $Y=y-A$ gives $\dot{X}=-Y$ and $\dot{Y}=X$. Hence every trajectory isThese are helices of unit angular frequency about a line parallel to the $z$ axis, including circles when $C=0$ and a straight line when $R=0$.

Define the canonical momentaWhen the velocity Hessian is nonsingular, these equations can be inverted for ${\dot{q}}_i(q,p,t)$, and the Hamiltonian is the Legendre transformIts differential isbecause the $p_{i}d{\dot{q}}_i$ terms cancel. The Euler-Lagrange equation identifies $\partial \mathcal{L}/\partial q_i={\dot{p}}_i$, so coefficient comparison gives Hamilton's equations

For the symmetric top, the momenta areThusThe Legendre transform gives

Both $\varphi$ and $\psi$ are cyclic coordinates, so $p_{\varphi }$ and $p_{\psi }$ are conserved. The Hamiltonian has no explicit time dependence and is also conserved. Since $H$ is independent of $\varphi ,\psi$,under the Poisson bracket. Hence three independent first integrals in involution are

Let a one-parameter infinitesimal transformation be $\delta q_i=ϵX_i(q,t)$ and suppose its change in the Lagrangian is a total derivative,With canonical momentum $p_i=\partial \mathcal{L}/\partial {\dot{q}}_i$, the chain rule and the Euler-Lagrange equations give, along a motion,Comparison with the assumed total derivative proves Noether theorem:

The given Lagrangian has no explicit time dependence, so its energy function is conserved:It is also invariant under the dilation $(x,y)\mapsto (e^{ϵ}x,e^{ϵ}y)$: the velocities and $y$ all scale by the same factor, while $x/y$ is unchanged. The generator is $(X_x,X_y)=(x,y)$ and $F=0$. SinceNoether's theorem supplies the second first integralThe integrals $E$ and $D$ are independent on an open dense subset of phase space.

The two Euler-Lagrange equations areDifferentiating $\kappa =f-u^{\prime }{f}_{u^{\prime }}-w^{\prime }{f}_{w^{\prime }}$ along an extremal and using these equations gives $d\kappa /dx=f_x$. Hence the Beltrami identity makes $\kappa$ constant when $f$ has no explicit dependence on $x$. If $f$ omits $u$ or $w$, that variable is a cyclic coordinate and the corresponding momentum $f_{u^{\prime }}$ or $f_{w^{\prime }}$ is an additional first integral; continuous symmetries give the analogous conserved quantities through Noether's theorem.

Put $A=1-m/u$. Since $w$ is cyclic,for a constant $\lambda$. The Beltrami integral isTherefore

If $\kappa =-{\lambda }^2$, then $u^{\prime 2}={\lambda }^{2}m/u$ andFor $z>1$, definethe hinted identity gives $F^{\prime }(z)=z^{3/2}/(z-1)$. Hence $w=C\pm mF(u/m)$. Because $F(z)\to +\infty$ as $z\to \infty$, the prescribed limit selects the minus sign:As $u\downarrow m$, the logarithm tends to $-\infty$, so every such solution satisfies $\boxed{w\to +\infty }$.

Write the admissible variation as $y_{\epsilon }=y+\epsilon \eta$, where $\eta (x)\to 0$ at both ends. Differentiating the functional under the integral sign gives the first variationAfter integration by parts, with the boundary term vanishing,Thus the Euler-Lagrange equation isA second differentiation gives the second variation

For unit mass, the Lagrangian in cylindrical coordinates isIntroduce shearing sheet coordinates by $r=r_0+x$ and $\phi ={\Omega }_{0}t+y/r_0$. Evaluate all derivatives of $\Phi$ at $(r_0,0)$. The reference circular orbit obeys ${\Phi }_r=r_0{\Omega }_0^2$, while midplane symmetry gives ${\Phi }_z={\Phi }_{rz}=0$. Its Taylor expansion isExpanding the kinetic energy to the same order givesThe terms linear in $x$ cancel by circular-orbit balance. Discard the constant and the term $r_0{\Omega }_0\dot{y}$ by total-time-derivative invariance of a Lagrangian; these do not change the Euler-Lagrange equations. The particle Lagrangian in a shearing sheet is thereforewith shearing-sheet tidal potentialThe second form follows from ${\Phi }_r=r{\Omega }^2$, the orbital shear parameter $q_0=-d\log \Omega /d\log r$ and the vertical epicyclic frequency ${\Omega }_z^2={\Phi }_{zz}$.

The Euler-Lagrange equations of the particle Lagrangian in a shearing sheet areThe terms coupling $x$ and $y$ are the Coriolis acceleration. The cyclic coordinate $y$ has conserved canonical momentumFor a Newtonian potential of a point mass, $q_0=3/2$ and ${\Omega }_z={\kappa }_r={\Omega }_0$. Set $p_y={\Omega }_0{x}_0/2$ and substitute $\dot{y}=p_y-2{\Omega }_{0}x$ in the radial equation. It becomes $\ddot{x}+{\Omega }_0^2(x-x_0)=0$, a harmonic oscillator equation about the epicyclic guiding center $x_0$. Integration givesThe four real constants in $A,y_0,x_0$ and the two in $B$ account for the six initial position and velocity data.

Expanding the inertial specific angular momentum gives $h=h_0+r_0{p}_y+O(2)$. Thus $p_y$ measures the angular-momentum offset from the reference circular orbit, and $x_0$ specifies the radius of its associated epicyclic guiding center.

The conserved horizontal energy in the rotating frame isIt is the horizontal part of the local Jacobi energy in a shearing sheet, rather than the inertial specific orbital energy. Its positive term is the epicyclic energy; its negative term is the energy of the background shear at guiding-center position $x_0$. Independently, the vertical harmonic oscillator has conserved energy

Letbe the infinitesimal generator of the continuous symmetry, and letbe the canonical momentum. Noether theorem states that along every trajectory satisfying the Euler-Lagrange equations,is conserved.

Indeed, differentiate the invariance of $L$ with respect to $s$ at zero:On an Euler-Lagrange trajectory, $\partial L/\partial q_i=d{p}_i/dt$, soThis proves the Noether conserved quantity for a mechanical point symmetry.

The canonical momentum isIn components, the Euler-Lagrange equation isThusThe last term is the $i$th component of $\mathbf{v}\times (\nabla \times \mathbf{A})$, sothe Lorentz force law.

Under the gauge transformation,because the curl of a gradient vanishes, andThe transformed Lagrangian isA total time derivative changes the action only by endpoint terms, whose variation vanishes for fixed endpoints, so the Euler-Lagrange equations are unchanged.

The new momentum isUse the type-two generating function for a canonical transformationIt gives $\mathbf{p}=\partial F_2/\partial \mathbf{r}=\tilde{\mathbf{p}}-q\nabla f$, $\tilde{\mathbf{r}}=\partial F_2/\partial \tilde{\mathbf{p}}=\mathbf{r}$, andThis equals $|\tilde{\mathbf{p}}-q\tilde{\mathbf{A}}{|}^2/(2m)+q\tilde{\varphi }$, so the gauge change is a time-dependent canonical transformation.

The first fundamental form of a surface of revolution isThe geodesic Lagrangian is consequentlyThe rotation angle $u$ is a cyclic coordinate, so its conjugate momentum is conserved by the Euler-Lagrange equation:Thus $f(v(s){)}^2{u}^{\prime }(s)$ is constant.

The positively oriented unit tangent to the parallel of a surface of revolution through $\gamma (s)$ is $e_u={\varphi }_u/f(v)$. Since $\gamma$ has unit speed and $\theta$ is its angle with that parallel,Multiplication by $f(v(s))$ yieldswhich is Clairaut's relation.

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

Use the geodesic Lagrangianwhere here the overdot on a coordinate means differentiation with respect to the affine parameter. Applying the Euler-Lagrange equation to each coordinate and comparing with the geodesic equationdetermines the Christoffel symbols. To display derivatives of the metric coefficients unambiguously, write ${\lambda }_t={\partial }_t\lambda$, ${\lambda }_r={\partial }_r\lambda$, and similarly for $\mu$. The nonzero symbols, together with those obtained by symmetry in the two lower indices, areThis is the Christoffel symbols of a diagonal spherical spacetime metric.

The momentum conjugate to $x^{\mu }$ iswhileThe Euler-Lagrange equation therefore givesSince $dot{x}^{\mu }=(d\tau /d\lambda )u^{\mu }$ and $\sqrt{-{\dot{x}}^2}=c,d\tau /d\lambda$, division by $d\tau /d\lambda$ gives the Lorentz-force equation from the worldline actionRaising the free index yields exactly

Let each mass be $m$ and each spring constant be $k$, with the $x_i$ denoting longitudinal displacements from equilibrium. The Lagrangian isThe inverse coordinate transformation isSubstitution givesThus the Euler-Lagrange equations separate into a symmetric pair,and an antisymmetric pair,

The symmetric normal modes are:

For antisymmetric modes, the squared dimensionless frequencies are the eigenvalues ofnamely $2\mp \sqrt{2}$. They giveandIn the original coordinates their mode shapes are respectivelyand

If $\mathbf{r}$ is measured in the rotating frame, its inertial velocity is $\dot{\mathbf{r}}+\mathbf{\omega }\times \mathbf{r}$. The masses producing the gravitational field are stationary in this frame, so their potential energy is a time-independent function $V(\mathbf{r})$. Thus kinetic minus potential energy gives the stated LagrangianPut $\mathbf{v}=\dot{\mathbf{r}}+\mathbf{\omega }\times \mathbf{r}$. ThenSubstitution in the Euler-Lagrange equation yieldsThe extra terms are the Coriolis acceleration, Euler acceleration, and centrifugal acceleration of a rotating reference frame.

The canonical momentum and Hamiltonian areAccordingly, Hamilton's equations are

Introduce a Lagrange multiplier $\lambda$ for the normalization and varyFor a smooth variation $u+\epsilon \eta$ with $\eta =0$ on $\partial D$, integration by parts givesThe fundamental lemma of the calculus of variations therefore yields the Euler-Lagrange equation

Multiplying by $u$ and integrating, while usingthe divergence theorem and $u=0$ on the boundary giveThe normalization is one, so the multiplier equals the stationary value:

Take a variation $x+\epsilon \xi$ with $\xi (a)=\xi (b)=0$. Expanding the action givesAn integration by parts and the fixed endpoint conditions give the first variationThe fundamental lemma of the calculus of variations therefore yields the Euler-Lagrange equationand the second variation is

Linearizing the equation of motion about $x$ givesso the Jacobi equation isWhen $u$ has no zero on $[a,b]$,The total derivative integrates to zero because $\xi (a)=\xi (b)=0$, hence

For the simple harmonic oscillator, the Jacobi equation is $\ddot{u}+{\omega }^{2}u=0$. Put $t_0=(a+b)/2$ and chooseIf $b-a<\pi /\omega$, then $|\omega (t-t_0)|<\pi /2$ throughout $[a,b]$, so $u$ is positive there. The preceding square identity proves that the classical path is a local minimum of the action whenever the elapsed time is less than half an oscillation period.

For a smooth curve $\gamma (t)$ on a Riemannian surface, its energy of a curve isChoose a local parameterization $\mathbf{X}(u,v)$ and write $\gamma (t)=\mathbf{X}(u(t),v(t))$. With coefficients of the first fundamental formthe Lagrangian isThe Euler-Lagrange equations are equivalentlyAfter multiplying by the inverse metric these become the geodesic equationwhere the Christoffel symbols are determined by $E,F,G$.