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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.

Use coordinates with the downward vertical as reference. The first rod's center is at $(a/2)(\sin \theta ,-\cos \theta )$, so its translational plus rotational kinetic energy is $m{a}^2{\dot{\theta }}^2/6$. The second center isIts squared speed is $a^2{\dot{\theta }}^2+a^2{\dot{\varphi }}^2/4+a^2\dot{\theta }\dot{\varphi }\cos (\theta -\varphi )$. Adding its center-of-mass rotation, using moment of inertia $m{a}^2/12$, gives the totalThe two center heights give $V=-mga(3\cos \theta +\cos \varphi )/2$. Hence $L=T-V$ is exactly the stated Lagrangian.

Small stable oscillations are about the downward equilibrium $\theta =\varphi =0$. Retaining quadratic terms gives $T_2=\frac{1}{2}{\dot{q}}^{T}M\dot{q}$ and $V_2=\frac{1}{2}{q}^{T}Kq$, whereA normal mode $q=v{e}^{i\omega t}$ satisfies $(K-{\omega }^{2}M)v=0$. With $\kappa =a{\omega }^2/g$, the determinant equation isThus the angular frequencies areBoth squared frequencies are positive. The amplitude ratio is $\varphi /\theta =(9-8\kappa )/(3\kappa )$: it is positive in the lower-frequency mode and negative in the higher-frequency mode. The rods therefore swing in phase in the first mode and in opposite phase in the second.

Write $q=(x,X{)}^T$, $M=\left(\begin{array}{cc}a& b\\ b& c\end{array}\right)$ and $K=\left(\begin{array}{cc}\alpha & \beta \\ \beta & \gamma \end{array}\right)$. The linear equations are $M\ddot{q}+Kq=0$. A normal mode has all coordinates oscillating with one frequency and fixed relative amplitudes, $q=v\cos (\omega t+\phi )$. Its squared frequency solvesand its amplitude vector is a nonzero null vector of $K-{\omega }^{2}M$. The assumed distinct positive roots give two normal frequencies, their positive square roots.

For the rods let ${\theta }_1,{\theta }_2$ be their small inclinations from the downward vertical. To first order, ${\theta }_1=x/a$ and ${\theta }_2=(X-x)/(2a)$. The first rod has kinetic energy $m{\dot{x}}^2/6$. The second has centre-of-mass horizontal velocity $(\dot{x}+\dot{X})/2$ and angular velocity $(\dot{X}-\dot{x})/(2a)$, givingTogether these give $T=m(3{\dot{x}}^2+2\dot{x}\dot{X}+2{\dot{X}}^2)/6$. The centre-of-mass height changes giveproving the displayed quadratic Lagrangian.

Here $M=(m/3)\left(\begin{array}{cc}3& 1\\ 1& 2\end{array}\right)$ and $K=(mg/(2a))\left(\begin{array}{cc}6& -1\\ -1& 1\end{array}\right)$. Set $\lambda =2a{\omega }^2/(3g)$. The determinant equation reduces to $5{\lambda }^2-17\lambda +5=0$, soThe first row of the mode equation gives $X/x=(6-3\lambda )/(1+\lambda )$. The actual rod-angle ratio is thereforeIt is positive for the smaller root and negative for the larger root. Thus the low-frequency mode has the rods in phase, and the high-frequency mode has them out of phase. Using the rod-angle ratio is necessary because $X$ is the displacement of the end of the second rod, not that rod's inclination.

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

For the elastic pendulum, set $r=l+x>0$ and measure $\theta$ from the downward vertical. In the vertical plane, Cartesian coordinates are $X=r\sin \theta$ and $Z=-r\cos \theta$, with $Z$ upward. Differentiating gives ${\dot{X}}^2+{\dot{Z}}^2={\dot{x}}^2+r^2{\dot{\theta }}^2$. Thus the Lagrangian isFor $x$, the Euler-Lagrange equation givesFor $\theta$, the canonical momentum is $p_{\theta }=m(l+x{)}^2\dot{\theta }$. Its Euler-Lagrange equation isor, for positive length,At an equilibrium, all time derivatives vanish, so $kx=mg\cos \theta$ and $\sin \theta =0$. Modulo full rotations, the two formal possibilities areThe lower configuration always has positive length. The upper one is a physical straight-spring configuration only if $l>mg/k$. For example, $l=mg/(2k)$ would give the purported upper state a negative length $-mg/(2k)$; at $l=mg/k$ its zero length makes the angular coordinate singular. Also, the straight compressed spring is an idealization; a real unsupported compressed spring can buckle.

The potential energy is $V=k{x}^2/2-mg(l+x)\cos \theta$. At either formal stationary point its mixed second derivative vanishes, whileBoth curvatures are positive at the lower state, making it a stable equilibrium; the upper state, when it has positive length, has negative angular curvature and is unstable.

To linearize about the stable state, define the small displacement $\xi =x-mg/k$, not $x$ itself, and $r_0=l+mg/k$. Products such as ${\dot{\theta }}^2$, $\dot{\xi }\dot{\theta }$ and $\xi \ddot{\theta }$ are second order, while $\cos \theta =1+O({\theta }^2)$ and $\sin \theta =\theta +O({\theta }^3)$. The first-order equations of motion therefore decouple:There are two independent normal modes. The radial one is vertical simple harmonic motion with angular frequency ${\omega }_r=\sqrt{k/m}$. The transverse one is a simple pendulum oscillation at fixed, statically stretched length $r_0$, with angular frequency ${\omega }_{\theta }=\sqrt{g/r_0}$. Their decoupling is a first-order small-oscillation result, not an assertion about the nonlinear motion of the elastic pendulum.

Each electromagnetic normal mode is a quantum harmonic oscillator with $E_n=(n+1/2)h\nu$. For $x=h\nu /(k_{B}T)$ its canonical partition function and occupation probabilities areThe geometric distribution is normalized and gives thermal mode energy $h\nu /(e^x-1)$. The zero-point term $h\nu /2$ is not part of the thermally emitted black-body intensity. The equilibrium photon chemical potential is zero because photon number is not conserved.

Counting two transverse polarizations gives $8\pi {\nu }^2/c^3$ modes per volume per frequency interval. Thus the spectral energy density is $u_{\nu }=8\pi h{\nu }^3/[c^3(e^x-1)]$. In an isotropic radiation field, spectral radiance is $B_{\nu }=c{u}_{\nu }/(4\pi )$. Transforming the spectral measure, $B_{\lambda }=B_{\nu }|d\nu /d\lambda |$, givesHere intensity means radiance per wavelength and solid angle; the hemispheric spectral exitance of a black surface is $M_{\lambda }=\pi B_{\lambda }$. The extra Jacobian is essential: the wavelength peak is not found by simply converting the frequency peak using $\lambda =c/\nu$.

At short wavelength the exponential suppresses radiation. At long wavelength the Rayleigh-Jeans limit is $B_{\lambda }\simeq 2c{k}_{B}T/{\lambda }^4$; using classical equipartition at every frequency instead would produce the ultraviolet catastrophe. The wavelength maximum solves $5(1-e^{-x})=x$, with nonzero root $x=4.9651$, yielding Wien's displacement law. For the specified temperature,Integrating the hemispheric flux gives the Stefan-Boltzmann law, $M=\sigma T^4\simeq 3.01\times 10^{-6}\mathrm{W}{\mathrm{m}}^{-2}$. The plot uses the specified ideal temperature, rather than substituting a later measured value. The physical black-body description of the cosmic microwave background is tested by comparing the sky spectrum with an absolute reference blackbody and separating instrumental and Galactic contributions.^[ref]

Write the real displacement as $U_n(t)=\mathrm{Re}[u{e}^{i\omega t-iqna}]$, choosing real positive peak amplitude $u$. Nearest-neighbour harmonic springs store energy in their extension $U_n-U_{n+1}$, each spring counted once. Therefore the instantaneous physical energy isThe printed complex displacement is a calculation device: squaring the complex phasor itself would not produce physical energy. The chain sketch in the accompanying figure shows masses at equilibrium spacing $a$, with longitudinal displacements along the chain.

The two spring forces on site $n$ giveSubstituting the normal-mode phasor and cancelling its nonzero amplitude givesThe long-wavelength limit is acoustic, $\omega \simeq a\sqrt{\alpha /M}|q|$, and the first Brillouin zone ends at $|q|=\pi /a$. The maximum angular frequency is ${\omega }_m=2\sqrt{\alpha /M}$.

Time averaging uses $⟨[\mathrm{Re}(z{e}^{i\omega t}){]}^2⟩=|z{|}^2/2$. ConsequentlyThe dispersion relation makes these two contributions equal, givingThis is equality of time-averaged kinetic and potential energies in a harmonic normal mode; the instantaneous energy partition need not be equal.

Use periodic boundary conditions on length $L=Na$ to count modes: $q_j=2\pi j/L$. The density on the signed $q$ axis is $L/(2\pi )$. Using folded wave-number counting for a monatomic chain combines $q$ and $-q$ at positive magnitude and counts both directions, so the convention used by the requested expression isThis is a large-$N$ continuum density, not a second independent branch. It gives $N$ modes after integrating over that half-zone; endpoint counting changes only finite-size corrections.

For positive $q$, $\omega ={\omega }_m\sin (qa/2)$ andThe density of states of a monatomic harmonic chain vanishes outside the band, is constant to leading order near zero, and has an integrable Van Hove singularity at the upper band edge where group velocity vanishes. Normalization checks the directional counting:

Put type-A atoms at $2ja$ and type-B atoms at $(2j+1)a$. Their displacements $u_j,v_j$ obey Newton's second law,Choose normal modes $u_j=U{e}^{i(2jaq-\omega t)}$ and $v_j=V{e}^{i((2j+1)aq-\omega t)}$. Substitution givesA nonzero displacement vector requires zero determinant:Solving this quadratic in ${\omega }^2$ gives the diatomic-chain dispersion relation with equal springs,The primitive unit cell has length $2a$, so the first Brillouin zone is $|q|\le \pi /(2a)$.

A phonon is one quantum of a crystal's vibrational normal modes. In the harmonic approximation, each mode is a quantum harmonic oscillator with energy levels $(n+1/2)\hslash \omega$. A phonon carries excitation energy $\hslash \omega$ and crystal momentum $\hslash q$ modulo a reciprocal lattice vector. It is a collective lattice excitation, not one atom travelling along the chain. Phonons obey Bose-Einstein statistics; their number need not be conserved in thermal equilibrium.

Take nearest-neighbour separation $d$, a common spring constant $K$, and one scalar displacement per atom. For a monatomic chain, the harmonic approximation near a potential minimum, Newton's second law and a normal mode giveThe monatomic-chain dispersion relation has one acoustic phonon branch, with $\omega \simeq d\sqrt{K/m}|k|$ near the origin. The first Brillouin zone ends at $\boxed{k=\pm \pi /d}$, where $\omega =2\sqrt{K/m}$ and the group velocity vanishes.

For alternating masses $m,M$, choose a two-atom cell of length $2d$. The coupled normal modes give the diatomic-chain dispersion relation with equal springs,The first Brillouin zone now ends at $\boxed{k=\pm \pi /(2d)}$. The minus branch is the acoustic phonon branch; its two atoms move together at long wavelength. The plus branch is the optical phonon branch; at $k=0$ the atoms move oppositely with a stationary centre of mass and ${\omega }_{+}^2=2K(1/m+1/M)$. For $M>m$ the edge frequencies are $\sqrt{2K/M}$ and $\sqrt{2K/m}$, with a gap between them.

If $M=m$, the two branches reduce to $2\sqrt{K/m}|\sin (kd/2)|$ and $2\sqrt{K/m}|\cos (kd/2)|$, touching at $\sqrt{2K/m}$. This is zone folding in an equal-mass diatomic chain: identical sites actually have a primitive unit cell of length $d$. A doubled cell folds the monatomic branch but opens no gap. Equal masses alone do not specify all possible spring models; the sketches assume equal nearest-neighbour spring constants.

A waveguide confines the transverse structure of a wave while permitting propagation along a chosen route. It can deliver energy to a distant receiver without the same geometric spreading as an unguided beam, protect a signal from unwanted coupling, and select transverse normal modes. It also makes propagation paths and coupling to devices more controllable. Confinement does not eliminate wave absorption or scattering loss, and it can introduce wave dispersion and cutoff frequencies of a membrane waveguide or corresponding cutoffs for other wave types.

Two examples are an optical fiber, which guides light by its core-and-cladding refractive index structure, and a hollow electromagnetic waveguide used to carry microwaves between a source and an antenna. A duct for acoustic waves is a further example. Each uses different material and boundary conditions, but the common idea is controlled transverse confinement.

The fixed edge at $y=0$ is automatically satisfied by $\sin (qy)$. At $y=a$, the displacement must vanish at every $x,t$, so $\sin (qa)=0$. Thus the allowed transverse normal modes haveThe case $n=0$ gives zero displacement for this sine solution and is not a physical propagating mode. Negative $n$ only reverses the amplitude sign and does not create a distinct transverse mode. This discrete transverse spectrum is a consequence of the Dirichlet boundary conditions, rather than quantization of the longitudinal wavenumber in an indefinitely long guide.

The Debye model approximates a three-dimensional crystal as an elastic continuum supporting quantized normal modes, or phonons. At long wavelengths, the acoustic phonon branches have approximately linear dispersion relations ${\omega }_s(k)=v_{s}k$: one longitudinal branch and two transverse branches. Replace their complicated actual spectra by these linear forms, then introduce a cutoff chosen to preserve the total $3N$ vibrational degrees of freedom for $N$ atoms. This is an approximation to the spectrum, not a claim that a real crystal has no non-acoustic branches.

For periodic boundary conditions in volume $V$, each allowed wavevector occupies $(2\pi {)}^3/V$ in the space of wavevectors. Counting a spherical shell for each branch gives the density of states per unit angular frequencyAn effective speed of sound $v_D$ can be defined by $3/v_D^3=1/v_l^3+2/v_t^3$. In the single-speed Debye model, choose ${\omega }_D=v_D{k}_D$ so thatThus the Debye wavevector, Debye frequency and Debye temperature satisfyHere $g$ has units of seconds when angular frequency is measured in ${\mathrm{s}}^{-1}$; its integral is the dimensionless number of normal modes. The spherical cutoff replaces the real Brillouin zone by an equal-count continuum construction.

Each phonon mode is a quantum harmonic oscillator. Since phonon number is not conserved at thermal equilibrium, the chemical potential is zero and its Bose-Einstein distribution gives mean thermal energy $\hslash \omega /[e^{\hslash \omega /(k_{B}T)}-1]$. The zero-point energy is independent of temperature in the harmonic, fixed-volume approximation and does not contribute to the heat capacity at constant volume. Excluding that constant term,Differentiating the first integral at fixed volume avoids losing the temperature-dependent upper limit in the second one. The resulting Debye heat capacity isFor $T\ll {\Theta }_D$, extend the upper limit to infinity. Expanding $1/(e^x-1)={\sum }_{r\ge 1}{e}^{-rx}$ and integrating term by term gives ${\int }_0^{\infty }{x}^3/(e^x-1)dx=6\zeta (4)={\pi }^4/15$. ThereforeThis $T^3$ law follows from the three-dimensional ${\omega }^2$ density of states and the availability only of low-frequency modes at low temperature. It is a central success of the Debye model.

For $T\gg {\Theta }_D$, the Bose-Einstein distribution reduces to the classical equipartition theorem: every mode contributes thermal energy $k_{B}T$. Consequentlythe Dulong-Petit law, or $3R$ per mole of atoms. The simple model omits detailed phonon dispersion relation, distinct optical phonon branches, and anharmonic effects such as thermal expansion. In a metal, the low-temperature total heat capacity additionally contains a term approximately proportional to $T$ from electrons, so the lattice $T^3$ term should not be mistaken for the whole measurement.

For a fixed transverse normal mode, define its cutoff angular frequencyThe longitudinal phase velocity and group velocity, for propagation in $+x$, areThus $v_p{v}_g=v^2$, $v_p>v$ and $v_g<v$. The large longitudinal phase velocity reflects the geometry of the transverse standing wave and does not imply faster transport of energy than the membrane wave speed.

The cutoff frequency of a membrane waveguide for mode $n$ isAt cutoff $k_x=0$ and $v_g=0$; the longitudinal phase velocity has no finite value. Below cutoff,and a bounded field excited at $x=0$ decays as $e^{-{\gamma }_{n}x}$. It is an evanescent wave, rather than a freely propagating mode carrying net time-averaged energy flux down an indefinitely long uniform strip. A finite section can still transmit through coupling of evanescent fields.

Twice differentiating the proposed normal mode givesIt therefore solves the two-dimensional wave equation precisely whenThe fixed-edge boundary conditions require $z(x,0,t)=z(x,b,t)=0$. The first is automatic, and the second requires $\sin (k_{y}b)=0$. The nonzero normal modes are thereforeThe apparent $n=0$ choice gives identically zero displacement for this sine solution. There are no imposed end conditions in $x$, so $k_x$ is continuous rather than quantized by a second length. For a propagating mode it is real; the plus sign in the stated $\cos (\omega t-k_{x}x)$ convention describes propagation in $+x$. Each positive integer $n$ labels a transverse standing wave across the strip.

The natural interpretation is the fundamental transverse normal mode, $n=1$. Indeedso only $n=1$ propagates over the main part of the given Gaussian distribution of angular frequencies. Its centre travels atDifferentiate the group velocity with respect to angular frequency:At the centre this is $0.010149\mathrm{m}$. The one-standard-deviation group velocity spread is consequentlyOver the central travel time the associated spatial shear isEquivalently, use representative components at ${\omega }_0\pm {\sigma }_{\omega }$, for which $v_g(95)=1.5001\mathrm{m}{\mathrm{s}}^{-1}$ and $v_g(105)=1.6024\mathrm{m}{\mathrm{s}}^{-1}$. Their separation after $t_0$ isAt a fixed detector $10\mathrm{m}$ away, the corresponding arrival-time difference is $10/v_g(95)-10/v_g(105)=0.426\mathrm{s}$.

The paper asks for an estimate of spreading, without specifying a spatial width convention or the initial spectral phase. A Gaussian distribution has no finite full range. Thus $0.326\mathrm{m}$ is a one-standard-deviation dispersive contribution, whereas about $0.65\mathrm{m}$ describes the separation of the $\pm {\sigma }_{\omega }$ representatives. Neither is automatically the increase in the packet's original width. For an initially unchirped Gaussian wave packet in the narrow-band approximation, independent initial and dispersive contributions combine as ${\sigma }_{x,f}^2\simeq {\sigma }_{x,0}^2+{\sigma }_{x,\mathrm{disp}}^2$; a packet with an initial chirp can behave differently. The modal assumption and width convention must accompany the numerical answer.

Different transverse normal modes have different cutoff frequencies and group velocities. A pulse exciting several of them reaches the output in delayed modal components, causing intermodal dispersion, pulse distortion and overlap of successive signals. Their interference also makes the transverse wave intensity pattern depend on propagation distance; irregularities can transfer energy between modes.

Mitigations are to operate between the first and second cutoff frequencies of a membrane waveguide, excite only the desired mode shape with a suitable source, and use smooth uniform boundaries or gradual transitions to reduce mode conversion. Reducing the signal bandwidth keeps its spectrum inside the single-mode interval and reduces group velocity dispersion within the selected mode. Single-mode operation removes intermodal dispersion but does not remove the dispersion derived in part (b).

The trihydrogen cation is a nonlinear three-nucleus molecule. Removing three translations and three rigid rotations leaves $3(3)-6=3$ vibrational degrees of freedom. One normal mode is the symmetric in-plane breathing motion: every nucleus moves radially toward or away from the centre. The remaining two normal modes form a degenerate in-plane distortion pair, stretching one direction while compressing another.

For an explicit sketch, let ${\mathbf{r}}_j$ be the equilibrium vertices relative to their centre. Displacements proportional togive the breathing mode and a basis of the degenerate pair. Linear combinations within the latter pair are equally valid. Their actual frequencies require molecular force constants, which are not supplied. An apparent out-of-plane motion of three point nuclei is a translation or infinitesimal rigid rotation, rather than an additional independent vibration.

A normal mode has all displacement coordinates proportional to a common time dependence, with fixed relative amplitudes. For a nonzero frequency the time dependence is sinusoidal. A zero-frequency mode is a free displacement or uniform drift rather than an oscillation.

Use the printed mass ordering $M,m,M$. Displacements are measured from the unstretched equilibrium positions, givingEquivalently,This is the generalized eigenvalue problem for normal modes of three masses connected by two equal springs. Substitute $\mathbf{x}=\mathbf{u}{e}^{i\omega t}$ to obtain $\mathsf{K}\mathbf{u}={\omega }^2\mathsf{M}\mathbf{u}$. Reflection symmetry gives an antisymmetric vector $(1,0,-1)$ and symmetric vectors $(1,b,1)$. The former has ${\omega }^2=k/M$. For a symmetric vector, the end and centre equations give $M{\omega }^2=k(1-b)$ and $m{\omega }^{2}b=2k(b-1)$. Their solutions are $b=1$, $\omega =0$, or $b=-2M/m$, ${\omega }^2=k/M+2k/m$. HenceThese are all three eigenmodes. The last two keep the centre of mass fixed and are orthogonal in the mass-weighted inner product. The general zero-mode contribution is $(A+Bt)(1,1,1)$, accounting for both a centre-of-mass shift and a conserved total momentum. There are no fixed end springs in this system.

The original amplitude profiles illustrate $M=2m$; the symmetric internal shape is rescaled for display. The formula above gives every positive mass ratio.

Two further physical examples are standing-wave modes of a string fixed at both ends, with spatial shapes $\sin (n\pi x/L)$ and frequencies $n\pi c/L$, and electrical normal modes of two coupled LC circuits, in which the capacitor charges oscillate at fixed relative amplitudes. Independent normal mode coordinates simplify both systems in the same way as the spring-chain eigenvectors.

Orthogonality means zero inner product, not merely that two objects look different. For complex vectors, $⟨u,v⟩={\sum }_i{u}_i^{\ast }{v}_i$; for functions a common choice is $⟨u,v⟩=\int w(x)u(x{)}^{\ast }v(x)dx$ with positive weight $w$. An orthonormal set has $⟨u_m,u_n⟩={\delta }_{mn}$.

For an orthogonal set, the component of $f$ along $u_n$ has coefficientA complete orthonormal set gives $f={\sum }_n{c}_n{u}_n$ in the appropriate norm and Parseval's identity $\Vert f{\Vert }^2={\sum }_n|c_n{|}^2$. Without completeness there can be a nonzero residual orthogonal to the entire chosen set. Orthogonal nonzero vectors are linearly independent, but independence alone does not imply orthogonality.

Fourier coefficients are projections becauseLikewise, for a self-adjoint operator, $A{u}_n={\lambda }_n{u}_n$ gives $({\lambda }_n-{\lambda }_m)⟨u_m,u_n⟩=0$, so distinct eigenvalues have orthogonal eigenvectors. Degeneracy does not itself force a chosen pair to be orthogonal, but an orthonormal basis can be selected within each eigenspace. The differential version uses the correct Sturm-Liouville weight and boundary conditions.

For mechanical normal modes with unequal masses the relevant inner product is $⟨u,v{⟩}_{\mathsf{M}}=u^{†}\mathsf{M}v$. This makes the modes in D14 orthogonal even when an ordinary unweighted dot product does not. Expanding in such coordinates diagonalizes the kinetic and potential quadratic forms, explaining why modal motions evolve independently and their energies add.