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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 $E,F,G$.

If a straight line segment lies in the surface, parameterize it byThen $\ddot{\gamma }=0$, so its acceleration has zero tangential component. The two displayed equations hold directly, and the segment is a geodesic.

For the one-sheeted hyperboloidputTwo distinct ruling lines through $P$ areIndeed, the direction $d_{\pm }$ satisfiesso substitution shows that every point of the line lies in $H$. These are geodesics by the straight-line argument.

A third geodesic is the meridian through $P$. Choose $s_0$ with $\sinh s_0=z_0$; thenIts acceleration is ${\gamma }_m$, which is normal to $H$, so it is a geodesic. If $z_0\ne 0$, these give the required three distinct subsets.

If $z_0=0$, there is also the equatorial circle. Writing $P=(\cos {\theta }_0,\sin {\theta }_0,0)$,Its acceleration is normal to $H$ along $z=0$, so it is a fourth geodesic distinct from the meridian and the two rulings.

Finally, write $\rho =\sqrt{x^2+y^2}=\sqrt{1+z^2}$. Clairaut's relation gives the conserved quantityChoose initial data in $z>0$ withSince $|\sin \psi |\le 1$, every point of the resulting geodesic satisfies $\rho \ge c$, and thereforeContinuity keeps the geodesic in the component $z>0$, and the stated completeness assumption defines it for every real time. The continuum of choices of $c$ supplies infinitely many such geodesics.

Forthe Euler-Lagrange equation isorThe general solution of this inhomogeneous linear differential equation isThe condition $y(0)=-1$ gives $C+D=0$, and the condition at $x=1$ gives $C=1$. Therefore

Let $y_{\epsilon }=y+\epsilon \eta$, where the variation $\eta$ satisfies $\eta (a)=\eta (b)=0$. Differentiating the functional at $\epsilon =0$ gives its first variationUsing integration by parts and the fixed endpoint conditions,If this vanishes for every admissible $\eta$, the fundamental lemma of the calculus of variations gives the Euler-Lagrange equation

Forone has $L_y=0$ andA circle of radius $R$ centred at $(0,c)$ on the $y$-axis satisfiesOn any arc that does not cross $y=c$,which is constant. Hence $d{L}_{y^{\prime }}/dx=0=L_y$, so the Euler-Lagrange equation holds and $\delta I=0$. These circles are the coordinate-swapped form of a Geodesic in the Poincare half-plane model.

Replacing a minimizer by its absolute value does not increase its gradient norm, so choose a nonnegative minimizer $P$. The Euler-Lagrange equation isfor a Lagrange multiplier $\mu$. Multiplication by $P$ and integration show thatso $\mu >0$. Set $P_{\eta }=\sqrt{\mu }P$. Then $P_{\eta }$ is nonzero, belongs to $H^1({\mathbb{R}}^2)$, and satisfies the trapped focusing cubic ground-state equationStandard elliptic regularity and the strong minimum principle for elliptic operators make the nonnegative solution positive.

For an infinitesimal field variation $\delta \varphi$ such that $\delta \mathcal{L}={\partial }_{\mu }{K}^{\mu }$, integration by parts givesOn the Euler-Lagrange equations, Noether's theorem therefore givesFor transformations of spacetime, the variation at fixed coordinates and the change of the integration measure give the corresponding energy-momentum tensor terms.

Write the action density asUnder a compactly supported variation $\varphi \mapsto \varphi +ϵ\eta$,Integrating the first term by parts in time and the second in space, and discarding the boundary terms, givesThe principle of stationary action requires this to vanish for every $\eta$. Sincedivision by $-a^3$ yields the field Euler-Lagrange equationThis is the equation generated by the inflaton action in an expanding universe.

The Euler--Lagrange equations for the quadratic Lagrangian areFor a normal mode $x(t)=a{e}^{i\omega t}$, they become the generalized eigenvalue problem for small oscillationsPut $\lambda =m{\omega }^2/k$. With the stated spring constants,The final matrix has characteristic polynomialso its eigenvalue are ${\nu }_{\pm }=\pm \sqrt{1+{\delta }^2}$. HenceThe two angular frequencies therefore satisfyThis is the spectrum recorded by normal modes of two equal masses between three springs.

Forthe Euler-Lagrange equation isHere this becomesBecause the forcing is resonant with the complementary solution, a particular integral is $(x/2)\cos x$. HenceandThe Neumann boundary conditions giveThus

For fixed endpoint values, the first variation of the functionalvanishes exactly when the Euler-Lagrange equationholds. If $f$ has no explicit $x$ dependence, differentiation along an extremal gives the Beltrami identityFor a further constraint $J[z]=J_0$, introduce a constant Lagrange multiplier $\lambda$ and apply the Euler-Lagrange equation to $I+\lambda J$; $\lambda$ is then chosen so that the constraint holds.

For variations $y+\epsilon \eta$ with $\eta ={\eta }^{\prime }=0$ at both endpoints, integrating the ${\eta }^{\prime }$ term once and the ${\eta }^{\prime \prime }$ term twice givesThe fundamental lemma yields the higher-order Euler-Lagrange equation

WriteDifferentiating $\log J(u+th)$ at $t=0$ for a real test function $h$ givesAfter integration by parts,Rescaling the dependent and independent variables reduces this Euler-Lagrange equation to the ground-state equation for $Q$. The uniqueness of its positive radial solution and the equality cases in rearrangement show that all minimizers areThis is the classification of Weinstein-functional minimizers.

Suppose an infinitesimal continuous transformation has variations $\delta \psi ,\delta \bar{\psi }$ and changes the Lagrangian density by $\delta \mathcal{L}={\partial }_{\mu }{K}^{\mu }$. Expanding by the chain rule and integrating derivatives by parts givesOn solutions of the Euler-Lagrange equations, the first two brackets vanish. Hence Noether's theorem gives the conserved currentFor spacetime transformations, the coordinate variation supplies the corresponding energy-momentum term.