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Translate the generalized coordinates so that the equilibrium is at zero. Its potential gradient vanishes, and its quadratic potential energy is $q^T\mathsf{V}q/2$, where $\mathsf{V}$ is the symmetric Hessian matrix. The Euler-Lagrange equations for small motion areA normal mode has fixed component ratios, $q(t)=v\cos (\omega t+\varphi )$, withThe angular frequency $\omega$ is its eigenfrequency. A zero mode instead admits a constant or linear time factor. For modes $v_{\alpha },v_{\beta }$, symmetry givesHence mass-matrix orthogonality of normal modes isWithin a degenerate eigenspace, a basis can be chosen orthogonal in this positive weighted inner product.

For three masses connected pairwise, the exact energies areA fixed common translation $x_i\mapsto x_i+a$ preserves all differences and velocities. A fixed rigid rotation $x_i\mapsto Q{x}_i$ preserves their Euclidean norms. Thus both energies have these symmetries. Here the transformation is fixed in time; a moving reference frame would change the kinetic energy. Every equilateral configuration of side $\sqrt{3}$ has zero spring extensions, hence zero force and minimum potential, proving that it is an equilibrium.

For the specified displacement coordinates, writeThe exact potential is $V=(k/2){\sum }_{ij=12,13,23}(d_{ij}-\sqrt{3}{)}^2$. For an equilibrium edge vector $d$ and small perturbation $\delta d$,Thus only each first-order extension is needed for the quadratic potential. These extensions, with an immaterial sign choice for the horizontal edge, areConsequently $V_2=(k/2)(s_1^2+s_2^2+s_3^2)$, with remainder $O(|q{|}^3)$. If $s=Bq$, then $\mathsf{V}=k{B}^{T}B$. Expanding the squares givesThe small-oscillation mass and stiffness matrices are $\mathsf{T}=m{I}_6$ and the displayed $\mathsf{V}$. Hence

The two common translations give zero-mode vectorsAn infinitesimal counterclockwise rotation sends each equilibrium vertex $r_i$ to $J{r}_i$, where $J(x,y)=(-y,x)$. Therefore a third zero-mode vector isEach vector makes all three $s_j$ vanish, so lies in the kernel of a linear map $B$ and therefore in the kernel of $\mathsf{V}$. The rows of $B$ are independent: the $q_4$ and $q_6$ entries first force the coefficients of its last two rows to vanish, then the first coefficient vanishes. Thus $\mathrm{rank}B=3$, and the rank-nullity theorem proves that these are exactly the three zero modes.

For radial motion takeso that $x_i(t)=(1+a(t))r_i$ to first order. The force on mass $i$ is $-ka{\sum }_{j\ne i}(r_i-r_j)=-3ka{r}_i$, since ${\sum }_i{r}_i=0$. Newton's equation therefore gives $m\ddot{a}=-3ka$. Equivalently $\mathsf{V}{v}_b=3k{v}_b$. Hence the breathing normal mode hasAll three masses move radially outward together, then inward together, preserving the triangular shape. The remaining two normal modes of an equilateral spring triangle are shear deformations with squared frequency $3k/(2m)$; together with the breathing mode and three rigid zero modes they account for all six degrees of freedom.

The Lagrangian is $\mathcal{L}=T-\mathcal{V}$, and the Euler-Lagrange equations areAt a rest equilibrium $q=0$, ${\partial }_i\mathcal{V}(0)=0$ for all three coordinates. The Taylor expansion therefore starts asThe symmetric Hessian matrix is the matrix $\mathsf{V}$ of the quadratic potential energy. If a quadratic coefficient vanishes along a direction, higher-order terms can be the first nonconstant terms there; equilibrium by itself does not imply strict stability.

Write ${\mathsf{T}}_0=\mathsf{T}(0)$. Terms involving the variation of $\mathsf{T}(q)$ are cubic in small displacements and velocities, so the quadratic small oscillation Lagrangian isA normal mode has fixed displacement ratios, $q(t)=v\cos (\omega t+\varphi )$. Its angular frequency $\omega$ is an eigenfrequency, or normal frequency, and satisfies the generalized eigenvalue problem for small oscillationsHere ${\mathsf{T}}_0$ is positive definite for independent physical coordinates and positive masses. A stable quadratic minimum makes ${\omega }^2\ge 0$; a negative value would describe exponential instability rather than oscillation.

For the suspended bar, let ${\kappa }_0$ and ${\kappa }_1$ denote the actual incremental spring constants of the upper and each lower spring. The physical displacement of the bar is $q_1$, while the lower spring extensions change by $q_2-q_1$ and $q_3-q_1$. ConsequentlyThe linear gravitational and elastic terms cancel at the loaded equilibrium. Expanding the remaining squares gives the small-oscillation mass and stiffness matricesThusSource convention: the printed matrix is obtained by setting ${\kappa }_0=\lambda /L$ and ${\kappa }_1=\mu /l$. Although the PDF prose calls $\lambda ,\mu$ spring constants, that matrix treats them as elastic parameters whose ratios to the stated lengths are the incremental stiffnesses. If they are read literally as Hooke-law spring constants, the divisions by $L,l$ should instead be omitted. The numerical case has $L=l=1$, so this discrepancy does not affect its results.

With the numerical parameters, ${\mathsf{T}}_0=\mathrm{diag}(2,1,1)$ andUse the exchange symmetry of the lower masses to find the vertical modes of a horizontally constrained suspended bar. The antisymmetric vector $v_a=(0,1,-1{)}^T$ obeys $\mathsf{V}{v}_a={\mathsf{T}}_0{v}_a$, giving ${\omega }_a=1$. In the symmetric sector put $v=(A,B,B{)}^T$ and $s={\omega }^2$. The two independent equations areTheir determinant gives $s^2-4s+2=0$, so $s=2\pm \sqrt{2}$. Taking $A=1$, the full set isIn the lowest normal mode, all three masses move in phase, with the lower masses moving about $2.414$ times as far as the bar. In the antisymmetric normal mode, the bar stays still and the lower masses move equally in opposite directions; their changing forces on the bar cancel. In the highest normal mode, the lower masses move together against the bar, with magnitude about $0.414$ of its displacement. Any small motion is a superposition of these three modes.

The angular frequency of an underdamped mode isFor ordinary damping, $\alpha \ge 0$, mode $n$ oscillates precisely when $\alpha <cn\pi /L$. In particular, all modes, including the fundamental, oscillate when $\alpha <c\pi /L$. There are some oscillatory modes for every finite $\alpha$, since the allowed $n$ are unbounded; no finite damping coefficient overdamps the entire ideal string spectrum.

If $\alpha <0$, the same mode has negative damping: oscillatory amplitudes grow as $e^{|\alpha |t/2}$. For modes with real characteristic roots, both roots are positive, since $|\alpha |/2>\sqrt{{\alpha }^2/4-c^2{k}_n^2}$. Hence those modes also grow. Negative damping makes the equilibrium unstable.

With the stated parameters, the mass and restoring matrices areThe generalized eigenvalue problem givesTherefore the two normal angular frequencies areThe second row of the modal equation is $-{\omega }^{2}x+(1-{\omega }^2)\varphi =0$. It gives the mode shapesEach normal mode has arbitrary small amplitude $a$ and phase $\delta$. The bob's horizontal displacement is $X=x+\varphi$ to first order. Thus in the lower mode, bob and cart move in phase, with bob amplitude $1+\sqrt{2}$ times the cart amplitude. In the higher mode they move oppositely, because $X=(1-\sqrt{2})x$; the bob amplitude is $\sqrt{2}-1$ times the cart amplitude. Its vertical displacement is quadratic in the angle and is absent at this order.

For an ordinary conservative mechanical system, choose generalized coordinates $q_i$ measured from a static equilibrium point. Assume the kinetic energy is quadratic in velocities and there are no velocity-dependent forces. Expanding about equilibrium, the linear term in the potential energy vanishes. Retaining only terms of total degree two gives the small oscillation LagrangianAn irrelevant constant has been dropped. Coordinate dependence of $M$ contributes only terms of third or higher total degree. The matrices $M,K$ are symmetric, $M$ is positive definite for independent coordinates, and $K$ is positive semidefinite at a stable equilibrium. The Euler-Lagrange equations areA normal mode has a fixed mode shape $\mathbf{v}$ and one time factor: $\mathbf{q}=\mathbf{v}Q(t)$. Its angular frequency $\omega$ solves the generalized eigenvalue problem $K\mathbf{v}={\omega }^{2}M\mathbf{v}$, with $\ddot{Q}+{\omega }^{2}Q=0$. Thus $\omega$ is the normal angular frequency, and the frequency in cycles per unit time is $\omega /(2\pi )$. A zero mode satisfies $K\mathbf{v}=0$; its time dependence is $Q=A+Bt$, describing an unopposed displacement or uniform drift rather than a restoring oscillation.

For a breathing displacement take ${\mathbf{u}}_i=q(t){\mathbf{r}}_i$. Every spring has equilibrium length $\sqrt{2}$ in the given length units, so its linear extension is $\sqrt{2}q$. At the vertex ${\mathbf{r}}_i$, the four adjacent equilibrium vertices sum to zero. Therefore the spring force there isIt remains radial and proportional to the original displacement at every vertex, proving that this symmetric displacement is a normal mode. Newton's second law givesThe first is angular frequency and the second is cycles per unit time. For $k$ in $\mathrm{N}{\mathrm{m}}^{-1}$ and $m$ in $\mathrm{kg}$, these have units $\mathrm{rad}{\mathrm{s}}^{-1}$ and $\mathrm{Hz}$ respectively. An energy check gives $T=3m{\dot{q}}^2$ and $V=12k{q}^2$, leading to the same ratio ${\omega }_b^2=24k/(6m)$.

For two bars, after subtracting the gravitational equilibrium,A normal mode $\mathbf{\eta }=\mathbf{v}{e}^{i\omega t}$ requires the generalized eigenvalue problem $(K-{\omega }^{2}M)\mathbf{v}=0$. With $z={\omega }^2$,Thus, for $m_2>0$,Both squared angular frequencies are positive. A corresponding eigenvector is $(1,2-{\omega }_j^2{)}^T$, from the first row of the matrix equation.

For three unit masses, $M=I$ andThe uniform vector has $11^{T}1=31$, whereas every vector with coordinate sum zero is annihilated by $11^T$. One orthonormal choice of normal modes is thereforeIf $k=1$ is retained from (b), the angular frequencies are $1,2,2$. The equal upper frequencies follow from degenerate normal modes from permutation symmetry: the stiffness is scalar on the two-dimensional sum-zero subspace. In the uniform mode the inter-bar strings do not change length, but all three beam connections stretch, giving the nonzero lower frequency. The motion is an arbitrary sine-cosine superposition of these modes about the gravitational equilibrium.

The signal is continuous at zero with $f(0)=1$. Its envelope is $e^{-\lambda |t|}$, but its angular frequency falls from $2$ on the incoming side to $1$ on the outgoing side. Its one-sided slopes are $f^{\prime }(0^{-})=\lambda$ and $f^{\prime }(0^{+})=-\lambda$, producing a cusp at the peak. Both tails decay to zero.

Use $\tilde{f}(\omega )={\int }_{\mathbb{R}}{e}^{-i\omega t}f(t)dt$. Expanding each cosine into two exponentials and integrating its half-line givesEquivalently,The positive $\lambda$ guarantees convergence of the Fourier transform integrals.

For the stated geometry and numerical masses, the inertia and stiffness matrices areA normal mode $\mathbf{\theta }=\mathbf{v}{e}^{i\omega t}$ requires $(K-{\omega }^{2}M)\mathbf{v}=0$. Writing $z={\omega }^2$, its characteristic equation isAt the claimed $z=6$, this determinant is $-27/50\ne 0$. Hence the printed claim that $\sqrt{6}$ is a normal frequency is false for the stated data, and there is no corresponding nonzero normal mode.

The complete geometry-based answer is obtained from the three positive roots $z_j$:An exact closed form for the squared angular frequencies isThey are approximately $19.555611657$, $6.209018836$ and $1.235369507$, respectively. For any of these roots, a corresponding eigenvector can be chosen asThe last two rows of $(K-zM)\mathbf{v}=0$ give this expression directly; the first is the characteristic equation. These three modes span all motion, with arbitrary sine and cosine amplitudes. Positivity of $M$ and of the spring potential energy ensures that all squared angular frequencies are positive.

If instead one takes the PDF's printed $I_1=m_1{a}^2/6$ as a stipulated inertia, then $M=\mathrm{diag}(1/2,1/4,2/25)$ andThe corresponding angular frequencies are approximatelyThe same formula for $\mathbf{v}(z)$ applies, because its derivation used only the second and third rows. Here the determinant at $z=6$ is $-63/25\ne 0$, so following the printed inertia does not repair the frequency claim. The PDF therefore needs a further correction to its numerical data or asserted frequency; the intended correction cannot be inferred uniquely.

Put ${\theta }_i={\theta }_i^0+{\eta }_i$. Differentiating the spring length givesThe springs are unextended at equilibrium, so the quadratic potential energy contains only the squares of these first-order extensions:The linearized Euler-Lagrange equations are thereforeThe stiffness matrix is $(k/4)(3I-11^T)$. Its eigenvector $1=(1,1,1{)}^T$ has zero eigenvalue, and every vector whose components sum to zero has eigenvalue $3k/4$. One convenient orthonormal set of normal modes isAny orthonormal basis of the sum-zero plane is equally valid for the two degenerate normal modes. The general small motion isThe linear drift in the zero mode describes a slow uniform rotation. These are angular frequencies; the ordinary frequencies of the two vibrational modes are ${\omega }_1/(2\pi )$.

A common shift of all three angles leaves every spring length unchanged. This continuous rotational symmetry gives the zero mode: there is no restoring force against rigid rotation.

The equilateral equilibrium also has the dihedral group symmetry $D_3$. Its action on the three angular displacements preserves the uniform one-dimensional subspace and the two-dimensional sum-zero subspace. The latter is an irreducible representation over the real numbers; a reflection may reverse all angular signs as well as exchange bead labels, which does not change these invariant subspaces. The symmetric stiffness matrix must act as a scalar on the sum-zero subspace, explaining the equal vibrational angular frequencies. Directly, each sum-zero vector satisfies $(3I-11^T)\mathbf{v}=3\mathbf{v}$. This is an example of degenerate normal modes from permutation symmetry, and of normal modes of three equal masses with a symmetric gap potential.

The quadratic Lagrangian is diagonal in $x,y,\theta$, so the Euler-Lagrange equations give two pure translations and one pure rotation. Their angular frequencies areA horizontal translation has $(x,y,\theta )=(A\cos {\omega }_{x}t,0,0)$; a vertical translation has $(0,A\cos {\omega }_{y}t,0)$; and the rotational mode has $(0,0,A\cos {\omega }_{\theta }t)$. Any fixed-direction translation is another choice in the same two-dimensional degenerate normal mode space. The rotation keeps the centre fixed.

Since energy and angular frequency are related by $E=\hslash \omega$, the given eigenvalue relation yieldsWith $E_P=4.3\mathrm{eV}$,The ${\lambda }_i$ are dimensionless modal eigenvalues, not optical wavelengths despite the symbol. The same $\hslash$ multiplies both mode and bulk angular frequencies and cancels in the energy ratio. These are bare-core predictions of the stipulated lossless Drude model; adding a dielectric/metallic shell, interband response or damping changes the actual resonances.

An accelerated charge radiates. For ultrarelativistic circular motion, the emitted radiation is concentrated into a cone of angular width of order ${\gamma }^{-1}$ about the instantaneous velocity. As the cone sweeps past a distant observer, propagation-time compression produces a short pulse of width of order $\rho /({\gamma }^{3}c)$, where $\rho$ is orbit curvature radius. Its Fourier spectrum extends to high harmonics, with conventional critical angular frequencyThis is a spectral scale, not a sharp upper cutoff. For transverse acceleration and $v\simeq c$, the classical radiated power isThe radiation is linearly polarized for an observer in the orbit plane, with elliptical polarization away from it. The two sides of the orbit plane have opposite handedness. The strong forward preference and high-frequency spectrum follow from the relativistic radiation calculation.^[ref]

Magnetic bending fields steer the charge but do no work, so maintaining a storage-ring beam requires energy replacement by accelerating fields. Bright electron beams provide useful radiation from infrared to X-rays; stronger curvature or higher electron energy shifts the spectral scale. The classical formulas assume negligible quantum recoil.

wave phase labels the stage of a physical oscillation at a spacetime event. Two observers comparing the same event must identify the same crest, trough or intermediate stage: a change of coordinates cannot change the number of completed cycles between physically specified wavefronts. The continuous wave phase can therefore be treated as a Lorentz scalar, up to a common choice of wave phase origin. This does not assert invariance of the measured angular frequency, which is the rate of change of that wave phase along the observer's trajectory.

For Lorentz invariance of plane-wave phase, writeusing the metric with signature $(+,-,-,-)$. The gradient $K_{\mu }={\partial }_{\mu }\Phi$ transforms as a covector because $\Phi$ is a scalar. Raising its index therefore gives the four-wavevector. Explicitly, substituting the inverse boost $t=\gamma (t^{\prime }+v{x}^{\prime }/c^2)$, $x=\gamma (x^{\prime }+v{t}^{\prime })$ givesThese are exactly the Lorentz transformation laws for $(\omega /c,\mathbf{k})$. In vacuum, $K_{\mu }{K}^{\mu }=0$; the Doppler effect and aberration of light are two consequences of transforming the same four-vector.

Take $u>0$ to mean that the mirror recedes along the incident plane electromagnetic wave. Put $\beta =u/c$, with $|\beta |<1$. The moving-mirror Doppler shift can be obtained directly by matching wave phase on the moving surface. Write the incident wave phase as $k_{i}z-{\omega }_{i}t$ and the reflected wave phase as $-k_{r}z-{\omega }_{r}t$, where $k_i={\omega }_i/c$ and $k_r={\omega }_r/c$ are positive wavenumbers. At the mirror, $z=z_0+ut$, the two phases must have the same time dependence; their constant offset incorporates the reflection phase shift. ThusConsequently the observed reflected angular frequency and ordinary frequency areThis is a redshift for a receding mirror and a blueshift for an approaching mirror. At small velocity, ${\omega }_r\simeq {\omega }_i(1-2u/c)$.

Equivalently, the relativistic Doppler effect gives ${\omega }_i^{\prime }=\gamma (1-\beta ){\omega }_i$ in the mirror rest inertial frame. Reflection there reverses propagation and preserves frequency. Transforming the backward-going reflected wave to the laboratory gives another factor $\gamma (1-\beta )$, so ${\omega }_r={\gamma }^2(1-\beta {)}^2{\omega }_i={\omega }_i(1-\beta )/(1+\beta )$. A single Doppler factor would miss the effect of the return path.

Let $\mathcal{I}$ be the sphere's mechanical moment of inertia about its rotation axis, and let ${\omega }_i=\omega (0)$. For an undriven rigid sphere the rotational energy is $E_{\mathrm{rot}}=\mathcal{I}{\omega }^2/2$. In the weak-radiation-loss approximation, conservation of energy givesIntegrating $d({\omega }^{-2})/dt=2K/\mathcal{I}$ givesEquivalently, the characteristic spin-down time isThe radiative spin-down of a rotating electric dipole therefore has angular frequency that decays as $t^{-1/2}$ at late times within this model, rather than exponentially. The original paper gives no mass or mass-density profile, so $\mathcal{I}$ cannot be determined from $R$ and the relative permittivity. If a homogeneous solid sphere of mass $M$ is additionally assumed, insert $\mathcal{I}=2M{R}^2/5$; uniform dielectric response alone does not imply this mechanical mass distribution.

The approximation is important. If the dipole angle is $\varphi (t)$ with $\dot{\varphi }=\omega (t)$, direct differentiation gives $|\ddot{\mathbf{p}}{|}^2=p_0^2({\omega }^4+{\dot{\omega }}^2)$. Using the constant-frequency radiation law instantaneously neglects the second term and assumes $|\dot{\omega }|\ll {\omega }^2$. This gives the standard leading slow spin-down solution above. An exact nonadiabatic radiation reaction evolution would require the mechanical and near-field dynamics beyond the information supplied, rather than treating this leading law as exact at arbitrarily rapid damping.

In SI units the Maxwell's equations in vacuum, with sources, areTaking the divergence of the last equation and using Gauss's law gives ${\partial }_t\rho +\nabla \cdot j=0$, the local conservation of electric charge. Thus sources that violate charge conservation cannot satisfy all the Maxwell's equations. Equivalently, with this conservation law the curl evolution preserves the electric divergence constraint if it holds initially; Faraday's law similarly preserves $\nabla \cdot B=0$.

With $\rho =j=0$, use $\nabla \times (\nabla \times E)=\nabla (\nabla \cdot E)-\Delta E$ and substitute the two curl equations. This gives $-\Delta E=-{\mu }_0{ϵ}_0{\partial }_t^{2}E$. Applying the same calculation to $B$ yieldsA monochromatic plane wave has one angular frequency $\omega$ and constant-phase planes $\mathbf{K}\cdot r=\omega t-\text{constant}$, normal to the wavevector $\mathbf{K}$. Its wavelength is $2\pi /|\mathbf{K}|$.

For a nontrivial traveling wave with $\omega \ne 0$, substitution of the displayed electric field into the divergence equation gives $\mathbf{K}\cdot e_1=\mathbf{K}\cdot e_2=0$. Faraday's law is satisfied byIndeed ${\partial }_{t}B=\mathbf{K}\times E_{\varphi }$ and $\nabla \times E=-\mathbf{K}\times E_{\varphi }$. The magnetic divergence is zero, while Ampère's law reduces to $|\mathbf{K}{|}^2{E}_{\varphi }/\omega =(\omega /c^2)E_{\varphi }$, giving the dispersion relation. The amplitudes may be arbitrary transverse vectors; they need not be mutually orthogonal or have equal lengths. Their combination describes the polarization.

The lower particle has position $(x+l\sin \theta ,-l\cos \theta )$, so its squared speed is ${\dot{x}}^2+2l\cos \theta \dot{x}\dot{\theta }+l^2{\dot{\theta }}^2$. The bead contributes $m_1{\dot{x}}^2/2$ to the kinetic energy. The two equal springs have combined potential energy $k{x}^2$ up to a constant: their lengths are $a+x$ and $a-x$, and equal natural lengths cancel the linear terms in the sum of their squared extensions. The gravitational potential is $-m_{2}gl\cos \theta$. ThereforeThere is no explicit time dependence, so the total energyis conserved. The Euler-Lagrange equations areAt rest these require $x=0$ and $\sin \theta =0$, giving the downward and upward positions $\theta =0,\pi$ modulo $2\pi$. The downward position has positive-definite quadratic potential and is stable, whereas the upward position has a negative angular potential curvature.

Keeping terms of second order about the stable position and dropping a constant givesFor unit parameters the mass matrix and stiffness matrix are $M=\left(\begin{array}{cc}2& 1\\ 1& 1\end{array}\right)$ and $K=\left(\begin{array}{cc}2& 0\\ 0& 1\end{array}\right)$. A normal mode with squared frequency $s$ requiresThus the positive angular frequencies are