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Subtracting the two linearized equations eliminates their acceleration terms:For $g,k>0$, the assumed nontrivial simple harmonic motion therefore hasThe support displacement and pendulum angle oscillate in phase. Inserting this relation into either Euler-Lagrange equation yieldsso the one positive finite angular frequency isThe negative frequency describes the same real oscillation after a phase change. For a general phase, replace the sine in both coordinates by $A\cos (\omega t)+B\sin (\omega t)$.

There is no missing second finite branch. The quadratic Lagrangian iswhose kinetic matrix is $\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)$ and has rank one. Its characteristic determinant is ${\omega }_0^2{\omega }_1^2-({\omega }_0^2+{\omega }_1^2){\omega }^2$, not a quadratic polynomial in ${\omega }^2$. This singular kinetic matrix in small oscillations results from taking the support mass to be exactly zero. Adding a small support mass restores a second mode that runs to infinite frequency as that mass vanishes.

For a stiff spring,The hinge becomes fixed, recovering the simple pendulum frequency, as expected for the pendulum with a massless spring-mounted support. At finite $k$, horizontal support motion makes the frequency smaller.

If $L^{\prime }=L+dF(q,t)/dt$, the action changes by an endpoint term:In the principle of stationary action with fixed endpoints, $\delta q(t_i)=\delta q(t_f)=0$ and the endpoint times are fixed. Therefore $\delta (S^{\prime }-S)=0$, and the two actions give the same bulk Euler-Lagrange equations.

For a Lagrangian density, adding ${\partial }_{\mu }{K}^{\mu }$ changes the action by ${\int }_{\partial \Omega }{K}^{\mu }{n}_{\mu }d\Sigma$, by the divergence theorem. Its variation vanishes for compactly supported field variations, or when the appropriate boundary data are held fixed. These total divergences in a variational principle can therefore be discarded when deriving bulk equations under those conditions.

The qualification about boundaries matters: if $K^{\mu }$ depends on field derivatives, their variations on the boundary must also be controlled. A total derivative can change the canonical momenta or the chosen representative of a Noether current, even while leaving the bulk dynamics unchanged.

Constraint and example. A holonomic constraint can be written $f_{\alpha }(q_1,\dots ,q_n,t)=0$, with no irreducible dependence on velocities. A velocity relation is holonomic if it integrates to such an equation. For example, a particle constrained to a stationary circle satisfies $x^2+y^2={\ell }^2$, reducing its two Cartesian degrees of freedom to one angle. A nonholonomic constraint cannot be integrated in this way.

Kinematics and kinetic energy. Set $d=R-r>0$. In the plane of the original diagram, choose the orbital angle $\theta$ positive counterclockwise from vertically downward. The marked radius of the inner cylinder then has absolute orientation $\alpha =\theta -\varphi$ measured from the same downward direction. Its signed spin angular velocity is $\Omega =\dot{\theta }-\dot{\varphi }$, whileThe moment of inertia of a uniform solid cylinder is $I=m{r}^2/2$. Applying the kinetic energy decomposition about the center of mass gives

No slip. With tangential unit vector ${\mathbf{e}}_{\theta }=(\cos \theta ,\sin \theta )$, the inner contact point has velocity $(d\dot{\theta }+r\Omega ){\mathbf{e}}_{\theta }$. The fixed wall requires this to vanish. HenceBecause $r,R$ are constant, integration giveswhere the integration constant is zero with the marked-dot reference in the diagram and continuously unwrapped angles. This is a holonomic constraint, even though it was first obtained from a condition on velocities.

Oscillations. The gravitational potential energy above the lowest position is $V=mgd(1-\cos \theta )$. Eliminating $\varphi$ gives the reduced LagrangianThe Euler-Lagrange equation becomesThus the bottom is a stable equilibrium, and for $|\theta |\ll 1$ there is simple harmonic motion, withThe extra rotational kinetic energy reduces the frequency compared with a particle sliding without friction along the same circular path. The small oscillations inside a circular cylinder formula applies with inertia ratio $\gamma =1/2$.

Friction and energy. Let $F_t$ be the signed tangential contact force in the ${\mathbf{e}}_{\theta }$ direction. Applying Newton's second law to translation and the torque equation to rotation givesUsing the rolling without slipping condition yieldsExcept at special instants, static friction is needed to produce the spin acceleration. There is no sliding at the contact point, whose instantaneous velocity is zero; therefore this force does no total work. Its translational and rotational powers cancel: $F_{t}d\dot{\theta }+F_{t}r\Omega =0$. The stationary normal reaction likewise does no work. With a rigid wall, adequate static friction, and no rolling resistance, conservation of energy gives

Angular momentum. The cylinder's spin angular momentum $I\Omega$ is generally not conserved because the contact friction exerts a torque about its center of mass. About the fixed outer-cylinder center $O$, the angular momentum decomposition about the centre of mass givesGravity and friction exert the external torqueThus the rolling cylinder does not generally conserve angular momentum, although it conserves mechanical energy. There is a special geometric cancellation at $R=3r/2$: then $L_O$ and its torque vanish identically for this planar motion. Angular momentum of a larger isolated system including the wall and Earth is a separate conservation statement.

Scope of the holonomic reduction. The integrated constraint is exact at every angle for planar rolling without slipping with sustained contact; it is not restricted to small oscillations. Completing a full turn requires tracking angle winding rather than reducing each angle separately modulo $2\pi$. This global bookkeeping does not make the local velocity relation nonintegrable.

However, arbitrary initial conditions need not sustain this ideal rolling motion. The inward normal reaction and the required friction areContact requires $N\ge 0$, and a finite coefficient ${\mu }_s$ of static friction requires $|F_t|\le {\mu }_{s}N$. If either fails, the cylinder detaches or slips and the one-coordinate rolling Lagrangian no longer applies. Slipping contact allows $\theta$ and $\varphi$ to be independent; detachment removes the fixed center-distance constraint. These changes alone do not establish a nonholonomic constraint. Allowing general three-dimensional reorientation can introduce nonintegrable rolling constraints, but that motion is outside the diagram's two-angle planar model. Thus the stated planar rolling constraint is holonomic throughout its regime of validity; it cannot be imposed on every physically possible motion.

Use the angular coordinate $\theta$ for the position on the ring. The kinetic energy is $T=m{a}^2{\dot{\theta }}^2/2$. The gravitational potential energy is constant because the ring is horizontal, and the ideal constraint force has no tangential component. Thus the Lagrangian has no dependence on $\theta$ itself: it is a cyclic coordinate.

The Euler-Lagrange equation givesEquivalently, the conserved conjugate momentum $p_{\theta }=m{a}^2\dot{\theta }$ is the vertical angular momentum, as required by rotational symmetry.

The angular Euler-Lagrange equations areA static equilibrium point has both angular velocities zero and $\sin (\theta -\phi )=0$. Hence the complete equilibrium families, with angles taken modulo $2\pi$, areThe first family is aligned in azimuth and minimizes the spring length; the second is antipodal and maximizes it. Their potentials are $k[(a-A{)}^2+h^2]/2$ and $k[(a+A{)}^2+h^2]/2$, respectively. Common rotations generate a whole family rather than isolated equilibria.

Vary each field component independently and integrate the gradient term by parts, taking fixed boundary values or vanishing surface variations. With $\rho =m_1^2+m_2^2+m_3^2$, the Euler-Lagrange equations for the free-energy saddle areThe anisotropy shifts the first two inverse susceptibilities in opposite directions. The derivative of $u{\rho }^2$ is $4u\rho m_j$, which fixes the factor needed for both the magnetization and the free energies in c. Under natural free-boundary conditions the integration by parts instead supplies ${\partial }_n{m}_j=0$ at the surface.

The Euler-Lagrange equation for the real scalar field givesThus the conjugate momentum and Hamiltonian density areThe spatial Fourier transform reduces the Klein-Gordon equation to ${\ddot{\varphi }}_{\mathbf{p}}+{\omega }_{\mathbf{p}}^2{\varphi }_{\mathbf{p}}=0$, with positive frequency ${\omega }_{\mathbf{p}}=\sqrt{{\mathbf{p}}^2+m^2}$. Hence a general real solution is the mode expansion of a free fieldThe two coefficients are complex conjugates because the field is real. Each pair of initial data $\varphi (0,\mathbf{x}),\pi (0,\mathbf{x})$ fixes these coefficients. The normalization anticipates canonical quantization and the relativistic normalization of one-particle states.

Goldstone's theorem states that a spontaneously broken continuous internal global symmetry in field theory of a local relativistic quantum field theory produces massless Goldstone bosons. For a Lorentz invariance vacuum breaking $G$ to its stabilizer $H$, there are $\dim G-\dim H$ independent symmetry-required massless directions. Additional accidentally massless modes are possible. A gauge redundancy does not obey this physical-state counting, and nonrelativistic systems need not have one independent mode per broken generator.

For the classical proof, consider real scalar fields ${\varphi }_i$ with a positive canonical kinetic term and an invariant smooth scalar potential $V$. An infinitesimal generator $T^a$ satisfiesLet $v$ be a stationary vacuum and differentiate this identity with respect to ${\varphi }_j$. At $v$, the term proportional to ${\partial }_{i}V$ vanishes, leavingThe Hessian matrix ${\mathsf{M}}^2$ is the scalar mass matrix. An unbroken generator has $T^{a}v=0$; broken generators span the tangent space of the vacuum manifold $G/H$. Each independent nonzero tangent vector is a zero eigenvector of ${\mathsf{M}}^2$. The quadratic Euler-Lagrange equations in these directions are $\square {\pi }^a=0$, giving the massless dispersion relation ${\omega }^2={\mathbf{p}}^2$. This proves the classical result and explains why the Goldstone bosons describe motion along the degenerate minima rather than against a restoring curvature.

Choose a coarse-grained magnetization $m(\mathbf{x})$ as the order parameter: it vanishes in the paramagnetism phase and develops a spontaneous nonzero value in the ferromagnetism phase. In zero applied magnetic field, spin reversal sends $m\mapsto -m$, so the local Landau free energy contains only even powers. A minimal spatially inhomogeneous free-energy functional isThe positive gradient coefficient penalizes rapid spatial variations. Near an ordinary transition take $a(T)=a_0(T-T_0)$, $a_0>0$, and positive $b$; retain the sextic term when $b$ can approach or cross zero. The source $h$ is conjugate to the magnetization and explicitly breaks spin reversal.

Determine equilibrium by minimizing this free-energy functional, imposing the appropriate boundary conditions. Its Euler-Lagrange equation isFor a uniform system at $h=0$, compare the values of all stable stationary solutions, rather than accepting every solution of this equation. The selected global minimum gives the equilibrium magnetization; derivatives of its free energy give entropy, heat capacity and magnetic susceptibility. At an ordinary transition the two ordered minima emerge continuously from zero. Spatial dependence also permits domain walls and an estimate of the correlation length.

This is a mean-field approximation: it selects a minimizing profile and neglects statistical fluctuations around it. The symmetry constrains the expansion, but its numerical coefficients are phenomenological. Close enough to a transition, thermal fluctuations near a critical point can invalidate the resulting mean-field critical exponents.^[ref]

Treat $B_{\mu }$ and $B_{\mu }^{\ast }$ as independent variables in the Euler-Lagrange equation. Varying $B_{\nu }^{\ast }$ gives the Proca equationThe divergence of its first term vanishes by antisymmetry and commuting partial derivatives. Since $M\ne 0$, it follows that ${\partial }_{\nu }{B}^{\nu }=0$. Substitution into the original equation givesThe complex-conjugate variation gives ${\partial }_{\mu }{G}^{\ast \mu \nu }+M^2{B}^{\ast \nu }=0$, hence $(\square +M^2)B^{\ast \nu }=0$ and ${\partial }_{\nu }{B}^{\ast \nu }=0$.

For an on-shell plane wave $B^{\mu }={\epsilon }^{\mu }{e}^{-ipx}$, the Lorenz constraint in Proca theory is $p_{\mu }{\epsilon }^{\mu }=0$. In the rest frame it sets ${\epsilon }^0=0$, leaving three complex particle polarizations. Thus the complex Proca field has three independent complex dynamical components, or six real ones; its quantized particle and antiparticle each have three physical particle polarizations. The conjugate field is not another unrelated classical field.

For a first-derivative Lagrangian density with no explicit spacetime dependence, introduce the canonical energy-momentum tensorDirect differentiation giveson solutions of the Euler-Lagrange equations. The four Noether currents are the four columns of this tensor, whereas their spatial integrals are the conserved four-momentum components:For $\mu =0$ this is the Hamiltonian; for a spatial lower index it is $P_k=\int {\pi }_a{\partial }_k{\varphi }_a$. Under the paper's passive-translation convention, the corresponding generator is $Q=-{ϵ}^{\mu }{P}_{\mu }$. Conservation of the integrated quantities also uses the stated vanishing boundary flux.

Put $V(\varphi )=m^2{\varphi }^2/2+{\sum }_{n>2}{\lambda }_n{\varphi }^n/n!$. The Euler-Lagrange equation for the scalar field is $\square \varphi +V^{\prime }(\varphi )=0$, or explicitlyThe derivative of ${\varphi }^n/n!$ is ${\varphi }^{n-1}/(n-1)!$; all interaction terms therefore enter with the displayed plus sign for the chosen Minkowski metric.

For a bob of mass $m$ on a massless rod of length $\ell$, let a massless support slide horizontally against a spring of stiffness $k$. With support displacement $s=\ell \eta$ and angle $\theta$ from downward vertical, the quadratic Lagrangian, divided by $m{\ell }^2$, is $(\dot{\eta }+\dot{\theta }{)}^2/2-(k/m){\eta }^2/2-(g/\ell ){\theta }^2/2$. Subtracting its two Euler-Lagrange equations gives $\eta =mg\theta /(k\ell )$. There is one finite normal mode, with the displayed frequency. A stiff spring fixes the support and recovers the simple pendulum frequency. The missing second finite frequency reflects a singular kinetic matrix in small oscillations, not an omitted oscillation branch.

For support displacement $\mathbf{a}(t)$ and angle measured from the downward vertical, the mass position is $(a_x+l\sin \theta ,a_y-l\cos \theta )$. Its Euler-Lagrange equation is $l\ddot{\theta }+{\ddot{a}}_x\cos \theta +(g+{\ddot{a}}_y)\sin \theta =0$. In the support frame the effective acceleration is $\mathbf{g}-\ddot{\mathbf{a}}$; taking its torque about the hinge gives $l^2\ddot{\theta }\hat{\mathbf{z}}=\mathbf{l}\times (\mathbf{g}-\ddot{\mathbf{a}})$. The prescribed support can do work, so pendulum energy need not be conserved.

A uniform rigid rod of length $l$ and mass $M_r$, pivoted at one end with a point mass $m$ at the other, has pivot moment of inertia $I=l^2(m+M_r/3)$. Its gravitational potential energy is $-gl(m+M_r/2)\cos \theta$. The planar Euler-Lagrange equation is $I\ddot{\theta }+gl(m+M_r/2)\sin \theta =0$, giving the displayed small-amplitude oscillation period. The limit $M_r=0$ gives a simple pendulum, whereas $m=0$ gives the uniform-rod physical pendulum. Thinness does not specify a rod's mass, so both contributions must be stated or explicitly neglected.

The p-Laplacian is the nonlinear divergence-form operatorIt is the Euler-Lagrange operator of the p-energy. For $p>2$ it is degenerately elliptic where $Du=0$.

For $V=m^2{\varphi }^2/2+\lambda ({\varphi }^2{)}^2/4$, $m^2<0$, the vacuum manifold is a sphere of radius $v=\sqrt{-m^2/\lambda }$. At a chosen vacuum the Hessian matrix is $2\lambda v^{2}n{n}^T$, where $n$ is its unit direction. The radial scalar has mass squared $2\lambda v^2$; the $n-1$ angular modes are physical Goldstone bosons when the symmetry is global.

For the two-component case, write $\mathbf{\varphi }=(v+\rho )(\cos \theta ,\sin \theta )$. The kinetic term becomes $\frac{1}{2}({\partial }_{\mu }\rho {)}^2+\frac{1}{2}(v+\rho {)}^2({\partial }_{\mu }\theta {)}^2$. At quadratic order the canonically normalized angular field is $\pi =v\theta$, with no mass term. Its Euler-Lagrange equation is the wave equation, so ${\omega }^2=|\mathbf{k}{|}^2$ in units where the propagation speed is one. The radial field instead obeys ${\omega }^2=|\mathbf{k}{|}^2+2\lambda v^2$. Thus a common rotation costs no energy, whereas a slowly varying rotation propagates as a collective wave. With the selecting source of continuous vacuum selection by an external source, use $\pi =w\theta$ and obtain ${\omega }^2=|\mathbf{k}{|}^2+h/w$. This elementary dispersion relation separates a Goldstone boson, a pseudo-Goldstone boson and a massive radial excitation within one example.

For axisymmetric heat conduction in a unit disk, separated modes satisfy $-(r\kappa R^{\prime }{)}^{\prime }=\nu rR$, with weight $r$, outer Dirichlet boundary condition $R(1)=0$, and zero radial flux at the origin. Integration by parts gives the displayed Rayleigh quotient. Its Euler-Lagrange equation under fixed $\int r{R}^2$ recovers the mode equation; the natural endpoint term is $r\kappa R^{\prime }\delta R$. The Rayleigh-Ritz variational principle says that every admissible nonzero trial profile gives an upper bound on the bottom of the spectrum, and on the lowest eigenvalue when this bottom is attained by a fundamental diffusion mode. Finite weighted energy, rather than pointwise existence of a second derivative at the origin, is the relevant trial admissibility requirement.

An integral constraint need not admit a normal Lagrange multiplier unless its first variation is nonzero. For the quadratic normalization in the title, the variation in direction $y$ has derivative $2\int w{y}^2=2$, so the constraint is always regular at an admissible function. With free endpoint values this justifies the Euler-Lagrange equation for $F-\lambda w{y}^2$ and its natural endpoint conditions. For $F=p{y}^{\prime 2}-q{y}^2$, multiplication of the differential equation by $y$ gives $\lambda =J[y]$.

The Euler-Lagrange equation is $y^{\prime \prime }+ky=0$. For positive $k$ a nonzero stationary function exists exactly when $\sqrt{k}$ is a positive integer $n$, and all such functions are $B\sin (nx)$. These values are the Dirichlet eigenvalues of $-d^2/d{x}^2$ on the interval. At $k=0$ the solutions are affine and the endpoint conditions leave only zero; negative $k$ likewise leaves only zero. Stationarity alone does not assert minimization of the quadratic functional.