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A common rotation changes neither the separation nor the spring energy. Its infinitesimal direction is $(1,1)$, exactly the zero-frequency normal mode. Thus the continuum of rotated equilibrium points requires that zero mode; a nonzero restoring frequency in that direction would contradict rotational symmetry.

The second mode has $I_1{\eta }_1+I_2{\eta }_2=0$, so it changes the relative angle while leaving the inertia-weighted common angle unchanged. It therefore carries zero perturbation in the conserved angular momentum associated with common motion. Indeed the coordinatesseparate the Lagrangian exactly:The cyclic coordinate $q$ is free, while $\delta$ has restoring or inverted curvature. This also explains why all members of either equilibrium family have the same frequencies. Reflection symmetry makes the potential even in $\delta$, and the stable oscillatory mode is consistent with time reversal in classical mechanics through its two opposite frequency signs.

The dispersion relation shows that an infinite rod has unstable long-wave modes. A finite rod excludes sufficiently long wavelengths, so stability depends on its length and the precise end conditions. The fourth spatial derivative requires two conditions at each end; fixing transverse positions alone does not specify whether slopes are constrained.

First take displacement-fixed ends with free rotation. Setting $\varphi (0)=\varphi (l)=0$ leaves slope variations free; the bending boundary term in the first variation then gives the corresponding natural boundary conditions on the slope, ${\varphi }_{xx}(0)={\varphi }_{xx}(l)=0$. The spatial normal modes are $\sin (n\pi x/l)$, andThe first mode destabilizes first. ConsequentlyAt equality, the first mode has zero restoring force and is marginal, with a static component and a possible linear-in-time component in the linear theory. Every integer $n<l\sqrt{\alpha /\beta }/\pi$ gives an exponentially growing mode.

If "fixed" instead means clamped, the clamped boundary condition is $\varphi ={\varphi }_x=0$ at both ends. The critical compression is thenFor example, the static critical shape $1-\cos (2\pi x/l)$ obeys the four clamp conditions and the zero-frequency equation at this load. To establish that it is the first threshold, write $v={\varphi }_x$: it has zero endpoints and zero mean, and periodic extension gives ${\int }_0^l{v}_x^{2}dx\ge (2\pi /l{)}^2{\int }_0^l{v}^{2}dx$ by the zero-mean periodic Poincaré inequality, with equality at $v\propto \sin (2\pi x/l)$. This proves the critical Rayleigh quotient. Both cases are instances of the compressed Euler-Bernoulli beam and its endpoint-dependent buckling threshold. The source does not specify the endpoint rotation, so both interpretations are stated explicitly.

Quanta of collective motion. An elementary excitation is a low-energy departure from the many-body ground state that can often be treated as a quasiparticle. Near an ordered equilibrium, small collective motions form normal modes. Quantization turns a harmonic mode of frequency ${\omega }_{\mathbf{k}}$ into bosonic excitations with energy $\hslash {\omega }_{\mathbf{k}}$. Their number is not an independently conserved thermodynamic variable in equilibrium, so their chemical potential is zero andHere $\nu$ labels branches, and $U_0$ contains the ground-state contribution, including phonon zero-point energy where appropriate. The approximation is useful while interactions and damping of the excitations are weak enough that the modes remain identifiable.

Phonons. A phonon is a quantum of an elastic lattice-vibration mode. Long-wavelength acoustic phonon branches have $\omega \simeq c_{\nu }k$, with one longitudinal and two transverse polarizations in a three-dimensional monatomic solid. A multi-atom primitive cell can also support optical phonon branches, generally with nonzero frequency at $k=0$. Linear acoustic dispersion gives $D(ϵ)\propto {ϵ}^2$, so the energy integral has $U-U_0\propto T^4$ and the heat capacity obeys the Debye $T^3$ law. In the Debye model,At temperatures high compared with all phonon energies but below structural changes, the classical harmonic limit gives approximately $C_V=3N{k}_B$. Optical modes are frozen out at temperatures well below their gaps.

Spin waves. A spin wave is a spatially varying collective precession of ordered spins; its quantum is a magnon. In a simple isotropic ferromagnet, a long-wavelength gapless branch has energy ${ϵ}_{\mathbf{k}}\simeq D_s{k}^2$, where the spin stiffness $D_s$ has units $\mathrm{J}{\mathrm{m}}^2$. State counting gives $D(ϵ)\propto {ϵ}^{1/2}$. For one isotropic branch in three dimensions,This exponent is specific to a gapless quadratic ferromagnetic branch. Long-wavelength antiferromagnetic spin waves can be linear and instead give a $T^3$ heat-capacity law in three dimensions. Magnetic anisotropy or an applied field can open a gap, causing activated behaviour at still lower temperatures. Near a magnetic ordering transition, a dilute noninteracting spin-wave description is no longer adequate.

These examples illustrate the low-temperature energy of gapless bosonic excitations: for $ϵ\propto k^s$ in $d$ dimensions, $D(ϵ)\propto {ϵ}^{d/s-1}$, hence $U-U_0\propto T^{1+d/s}$ and $C_V\propto T^{d/s}$. The exponent records both the spatial dimension and the dispersion, making calorimetry a probe of elementary excitations.

Define $A=2\cot \alpha /3+8\mathrm{Re}/15$. The linearization about $h=1$ givesThe Galilean transformation $\xi =x-2t$, $\tau =t$ has ${\partial }_t={\partial }_{\tau }-2{\partial }_{\xi }$ and ${\partial }_x={\partial }_{\xi }$. It removes the leading advection, leavingFor a normal mode proportional to $e^{\sigma \tau +ik\xi }$, the second and fourth derivatives contribute $-k^2$ and $k^4$. The dispersion relation is thereforeFor $A>0$ and $S>0$, the unstable wavenumbers satisfy $0<|k|<\sqrt{A/S}$. In the laboratory coordinates, the temporal exponent is $\sigma -2ik$, so this linear stability calculation predicts growth together with downslope transport at dimensionless speed $2$.

For the $n$th normal mode, the denominator just obtained isIts forcing coefficient is nonzero for $c\ne 0$ and $\omega >0$. Therefore the periodic response grows like $({\omega }_n^2-{\omega }^2{)}^{-1}$ as $\omega$ approaches ${\omega }_n$: this is resonance. At exact resonance there is no bounded periodic particular solution in the linear inviscid model; the resonant amplitude instead contains a secular term proportional to $t\sin ({\omega }_{n}t)$ or $t\cos ({\omega }_{n}t)$. Physical damping and nonlinear effects eventually matter. Away from resonance, undamped free oscillations can persist alongside the periodic particular solution.

Substitute the proposed Fourier sine series into the heat equation:Projecting onto each sine using orthogonality gives $T_n^{\prime }=-D{k}_n^2{T}_n$ for every mode with $b_n\ne 0$. The initial condition fixes $T_n(0)=1$, since $b_n$ already describes the initial amplitude. HenceFor a mode with $b_n=0$, the value of $T_n$ is immaterial and the same convention is convenient. The positive thermal diffusivity $D$ damps every mode, with higher frequencies decaying faster. For square-integrable initial data the damping justifies termwise differentiation for $t>0$ and recovery of the initial data in L2 norm convergence as $t\downarrow 0$.

The left panel illustrates the Fourier series in part (b). The right panel is a self-generated example with $D=L=1$ and ${\Theta }_0(x)=\sin (\pi x)+\frac{1}{2}\sin (3\pi x)$, showing the two normal modes decaying at rates ${\pi }^2$ and $9{\pi }^2$.

For a nonzero, time-independent vector $\hat{\mathbf{u}}$, substitute $\mathbf{u}(t)=\hat{\mathbf{u}}{e}^{i\omega t}$. Taking two time derivatives givesCancel the nonzero exponential to obtain the eigenvalue problemThus an eigenvector specifies a normal mode shape and its negative eigenvalue specifies the square of its angular frequency, $\omega =\sqrt{-\lambda }$. Real displacements are obtained by taking the real part, or by combining the two temporal solutions as $\hat{\mathbf{u}}(a\cos \omega t+b\sin \omega t)$.

Each supplied eigenvector gives relative displacements at $x=1,2,3$ in a standing wave. Insert zero at $x=0,4$ to describe the whole sampled string. A real normal mode is $u_j(t)=C{v}_j\cos (\omega t+\varphi )$: its shape stays fixed while its amplitude oscillates.

These semidiscrete fixed-string normal modes are samples of $\sin (k\pi x/4)$, with $k=3,2,1$ in the order listed. More precisely, their interior vectors are $v_j^{(k)}=2^{-1/2}\sin (k\pi j/4)$, which have unit Euclidean norm and form an orthonormal basis of ${\mathbb{R}}^3$. Their eigenvalues and angular frequencies obeyThus larger absolute value of a negative eigenvalue means faster oscillation, rather than faster decay. The numerical angular frequencies approximate the continuous fixed-string values $k\pi c/L$; the finite difference spatial approximation changes them. Only three independent interior normal modes exist on this grid, and any interior displacement or initial velocity vector can be expanded in this orthonormal basis.

The wave equation is second order in time. Even one normal mode obeys a second-order ordinary differential equation, so its initial displacement and initial velocity are independent data. Physically they specify stored elastic deformation and the initial motion.

The heat equation is first order in time; its initial temperature already fixes the instantaneous time derivative through the spatial Laplacian. It needs one initial field. In either case spatial boundary conditions remain necessary and are not substitutes for initial data. Classical fixed-end string data must satisfy the appropriate endpoint compatibility, including $f(0)=f(L)=0$ and, when endpoint velocities exist, $g(0)=g(L)=0$.

For $k>0$, the harmonic oscillator equation gives $X=C\sin (kx)+D\cos (kx)$. The condition at zero requires $D=0$, and the condition at $L$ requires $C\sin (kL)=0$. A nonzero normal mode therefore hasA multiplicative normalization has been absorbed into the temporal amplitude. The $k=0$ candidate is affine and the two zero endpoint values force it to vanish. A positive separation constant gives hyperbolic functions, whose two zero endpoint values likewise give only the trivial solution. Negative $n$ adds no new eigenfunctions because it only reverses their signs.

For each spatial normal mode, the temporal equation is $T_n^{\prime \prime }+{\omega }_n^2{T}_n=0$, with angular frequency ${\omega }_n=c{k}_n=cn\pi /L$. HenceThe ordinary frequency is ${\omega }_n/(2\pi )=cn/(2L)$; it should not be confused with the angular frequency used in the sine and cosine arguments.

The wave equation and its homogeneous Dirichlet boundary conditions are linear, so any finite linear combination of the separated normal modes is another solution. More strongly, the Fourier sine basis is complete in L2 space on $(0,L)$. Expanding the initial fields in this basis yieldsCompleteness of the Fourier sine basis is the reason the expansion covers general data; linearity alone would not prove that every solution has this form. The convergence and differentiation are understood in a function space appropriate to the data. For finite wave energy, one can take $f\in H_0^1(0,L)$ and $g\in L^2(0,L)$, obtaining an energy solution; stronger smoothness and endpoint compatibility give a classical solution.

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.

Define positive angular frequency magnitudes ${\omega }_c=|Q|B/m$ and ${\omega }_z>0$, with uniform $\mathbf{B}=B\hat{\mathbf{z}}$, $B>0$. Choose the static electrode polarity through the potential energyIts axial force gives $\ddot{z}=-{\omega }_z^{2}z$, while its transverse electric force points outwards. For $Q>0$, let $w=x+iy$. The Lorentz force givesTrying $w\propto e^{-i\omega t}$ gives the two radial normal mode magnitudes ${\omega }_{\pm }$ displayed above: the fast modified cyclotron motion and slow magnetron motion. For $Q<0$ the rotation directions reverse. Bounded confinement in this ideal model requires ${\omega }_c^2>2{\omega }_z^2>0$; the repeated-root boundary needs exclusion because it allows a linearly growing solution. At large ${\omega }_c/{\omega }_z$, ${\omega }_{-}\simeq {\omega }_z^2/(2{\omega }_c)$ and ${\omega }_{+}\simeq {\omega }_c-{\omega }_{-}$.

The ideal identitiesfollow directly from that quadratic. They show why the measured modified cyclotron frequency must not simply be substituted for the free cyclotron frequency. These are ideal-model deductions; the response to real electrode and magnetic field imperfections needs a separate correction analysis.

For a circle of circumference $2\pi$, the heat equation $u_t=D{u}_{xx}$ has periodic matching of both $u$ and $u_x$ at $x=\pm \pi$. For $D>0$, the displayed Fourier series solves the initial-value problem, with $\bar{f}=(2\pi {)}^{-1}{\int }_{-\pi }^{\pi }f(x)dx$ and the usual real Fourier coefficients. Every nonconstant normal mode decays, leaving the conserved mean. In particular $f(x)=(\pi -|x|{)}^2$ gives $\bar{f}={\pi }^2/3$, $a_n=4/n^2$ and $b_n=0$. The initial cusp is smoothed for every positive time.

A continuous family of normal modes labelled by the wavevector and a branch index $s$. For $r$ chemical atoms per three-dimensional primitive unit cell, the dynamical matrix has $3r$ eigenvalues, giving $3r$ branches: three acoustic phonon branches and $3r-3$ optical phonon branches for a mechanically stable ordinary crystal structure.

The displacement eigenvector of a lattice normal mode, including relative amplitudes and phases for the atoms in a primitive cell. Longitudinal and transverse describe its orientation relative to the phonon wavevector in simple isotropic cases.

Phonon scattering changes the occupation or propagation of lattice-vibration normal modes through interactions with boundaries, defects or other phonons. Anharmonic terms permit phonon-phonon scattering. Different processes need not relax heat flow equally: normal processes conserve total crystal momentum within the first Brillouin zone, whereas Umklapp scattering changes its representative by a nonzero reciprocal lattice vector.

Two identical vertical springs connect a driven pivot to two equal masses in sequence. After subtracting the static gravitational equilibrium, the laboratory displacement equation is $m\ddot{\mathbf{x}}+k\left(\begin{array}{cc}2& -1\\ -1& 1\end{array}\right)\mathbf{x}=k(z,0{)}^{\mathsf{T}}$. For $z=z_0\cos {\omega }_{0}t$ and ${\omega }_0^2=k/m$, a particular solution is $\mathbf{x}=(0,-z_0\cos {\omega }_{0}t{)}^{\mathsf{T}}$. Thus the upper mass has zero driven response even though this frequency is not an eigenfrequency. The general solution also contains both free normal modes.

The linear response relating an induced electric dipole moment to an applied electric field. In ordinary nonresonant Raman spectroscopy, changes in its symmetric components during a normal mode determine activity. These components transform like $x^2,y^2,z^2,xy,xz,yz$.