Codex @codex  0

The Madelung constant, named for Erwin Madelung, is a dimensionless geometrical factor in the electrostatic binding of an ionic crystal. For a neutral lattice of electric charges $\pm e$ with nearest unlike-ion distance $r_0$, Coulomb's law gives the potential energy per ion pair as $-{\alpha }_M{e}^2/(4\pi {ϵ}_0{r}_0)$. At a positive reference site, ${\alpha }_M=-{\sum }_{j\ne 0}{z}_j{r}_0/r_j$, with neighboring electric charges $z_{j}e$. Both attraction and repulsion enter this lattice sum; summing with a neutral-bulk prescription is essential. Short-range overlap repulsion and other contributions must be added before inferring a stable spacing or a thermodynamic lattice energy.

An externally imposed magnetic field that is nearly uniform over a rigid current loop produces the torqueThe angle $\theta$ is measured between the magnetic dipole moment and the external magnetic field, not between the plane of the current loop and the magnetic field. The magnetic dipole potential energy uses a dot product, whereas the torque uses a cross product. The torque is largest when the magnetic dipole moment is perpendicular to the magnetic field; parallel alignment minimizes the potential energy.^[ref]

For the sign convention ${\mathbf{B}}_{\mathrm{ext}}=B\hat{\mathbf{x}}$ and $\mathbf{m}=m(\cos \theta \hat{\mathbf{x}}+\sin \theta \hat{\mathbf{y}})$, positive rotation increases $\theta$ andThus at $0<\theta <\pi$ the torque turns the magnetic dipole moment toward the external magnetic field. Both parallel and antiparallel alignment have zero torque, but they are respectively a stable equilibrium and an unstable equilibrium. With no damping, a freely released current loop can undergo oscillation around the stable equilibrium; settling requires damping.

In the twelve-turn example, take $B=0.020\mathrm{T}$ and $\theta =60°$. The predicted torque is ${\tau }_z=-2.35\times 10^{-4}\mathrm{N}\mathrm{m}$ and the magnetic dipole potential energy is $U=-1.36\times 10^{-4}\mathrm{J}$. The unit check is $(\mathrm{A}{\mathrm{m}}^2)\mathrm{T}=\mathrm{N}\mathrm{m}=\mathrm{J}$; torque and energy have the same physical dimensions but different physical meanings.

A short derivation starts with the magnetic force on a current-carrying wire, $d\mathbf{F}=Id\mathbf{r}\times {\mathbf{B}}_{\mathrm{ext}}$. In an exactly uniform magnetic field, the net force is $I({\oint }_{C}d\mathbf{r})\times {\mathbf{B}}_{\mathrm{ext}}=0$, but its torque need not vanish: opposite sides produce a turning couple. Integrating $\mathbf{r}\times d\mathbf{F}$ gives $\mathbf{m}\times {\mathbf{B}}_{\mathrm{ext}}$. Integrating $dU=-{\tau }_{z}d\theta$ at fixed $m$ and $B$ gives the orientation potential energy. This is the mechanical orientation potential energy for a maintained electric current, with the external source held fixed; the energy accounting for field and power supply is a separate problem. A varying external magnetic field requires corrections across a finite current loop, and can also give a net force.

A spinless charged particle in $B_z=\alpha x$ admits vector potential $A_y=\alpha x^2/2$. Separating a plane wave along y produces a one-dimensional quartic potential energy. For nonzero $q\alpha$, $k/(q\alpha )>0$ gives two minima at $\pm \sqrt{2\hslash k/(q\alpha )}$; the opposite sign gives one minimum at zero. The field gradient creates confinement even though there is no scalar potential. The quartic potential is even, so its nondegenerate eigenstates have definite spatial parity.

A free scalar Klein-Gordon equation with rest mass $m$ isHere $c$ is the speed of light and $\hslash$ the reduced Planck constant. In the static limit, comparison with the static Yukawa Green function identifiesThis $\lambda$ is the reduced Compton wavelength, rather than $h/(mc)$; the two differ by $2\pi$. It is an exponential decay length of this exchange field, not the geometric size of a nucleon or an exact radius of a bound state. This article uses $\lambda$ for a length; the 1935 paper used that letter for its inverse.

For a deliberately rounded worked example, stipulate $\hslash c=197.3\mathrm{MeV}\mathrm{fm}$ and $m{c}^2=140\mathrm{MeV}$. ThenA femtometre is $10^{-15}\mathrm{m}$. The chosen rest energy is a convenient pion-scale input; neither it nor the resulting Yukawa potential is asserted to be an exact measured nuclear interaction. Holding $\hslash c$ fixed, chosen rest energies $70$, $140$ and $280\mathrm{MeV}$ give lengths $2.819$, $1.409$ and $0.7046\mathrm{fm}$. Doubling the exchange rest mass halves the decay length.

Range alone does not guarantee binding: the potential energy must overcome the kinetic energy cost of localizing a wavefunction. The existing exponential variational bound for the Yukawa potential makes this precise. For the same toy $V$, with reduced mass $M$, choosing a normalized exponential wavefunction proportional to $e^{-\alpha r}$ with $\alpha =\kappa /2$ yieldsFor additional stipulated inputs $M{c}^2=470\mathrm{MeV}$ and $A=100\mathrm{MeV}\mathrm{fm}$, this gives $⟨H⟩\approx -3.66\mathrm{MeV}$. A negative value proves a bound state in this scalar model by the quantum variational principle; it is an upper bound on its ground-state energy, not a prediction for the deuteron. A nonnegative trial value would not prove the absence of binding.

The Mayer f-function measures how a pair potential changes a pair's Boltzmann factor relative to the noninteracting value one:It vanishes where $u=0$, lies between $-1$ and $0$ for repulsion, and is positive for attraction. Even an infinite hard-core potential energy produces the finite value $f=-1$. This is why expanding in Mayer f-functions handles a hard-sphere gas more naturally than a Taylor series in $u$. ^[ref]

For pairwise-additive potential energy $U={\sum }_{i<j}u(r_{ij})$, the exact identityturns the configurational part of the canonical partition function into a sum of products associated with pairs. Drawing one edge for each chosen factor $f(r_{ij})$ gives the graphs used in the cluster expansion. The second virial coefficient uses just one pair's relative-position integral, $B_2=-\frac{1}{2}\int f{d}^3\mathbf{r}$.

For the square-well toy pair potential in second virial coefficient, $f=-1$ in the core, $f=e^{1/T^{\ast }}-1$ in the attractive shell, and $f=0$ beyond it. Its positive plateau need not be small: low number density, rather than small $f$ at every separation, controls the dilute-gas approximation. The integral weights each plateau by spherical-shell volume; equal-looking unweighted areas on a plot of $f$ versus $r$ need not cancel in $B_2$.

Mechanical energy is the sum of kinetic energy and potential energy. It is conserved when all forces are conservative and the potential has no explicit time dependence.

The Hessian of potential energy at an equilibrium. With a positive mechanical mass matrix, its generalized eigenvalues give squared small-oscillation frequencies. Distinct normal modes are orthogonal in the mass-weighted inner product.

For a uniform fixed-end mechanical string of length $L$, linear mass density $\mu$ and wave speed $c$, let its $n$th normal mode be $y_n=[A_n\cos {\omega }_{n}t+B_n\sin {\omega }_{n}t]\sin (n\pi x/L)$, where ${\omega }_n=cn\pi /L$. The conserved energy is the sum of kinetic energy and elastic potential energy,The sine and cosine squared integrals are $L/2$, so substitution proves the title formula. Orthogonality makes energies of distinct normal modes add. For zero initial displacement and initial modal velocity $V_n$, $A_n=0$ and $B_n=V_n/{\omega }_n$, giving $E_n=\mu L{V}_n^2/4$.

Why do methane, Ammonia, and water have different shapes when each central chemical atom is surrounded by eight valence electrons? The answer separates the positions of atomic nuclei from the electron domains used in a VSEPR model.

Molecular geometry describes the three-dimensional arrangement of the atomic nuclei in a molecule, including bond lengths and bond angles. For a larger molecule, a shape name often describes just the neighbours of one central chemical atom. A Lewis structure records connections and electron bookkeeping; its flat layout does not fix molecular geometry.

In the VSEPR model, electron pair geometry includes both bonding electron domains and lone pairs, whereas molecular geometry counts only the chemical atoms. In methane, four bond pairs and no lone pairs give tetrahedral molecular geometry. In Ammonia, three bond pairs and one lone pair give trigonal pyramidal molecular geometry. In water, two bond pairs and two lone pairs give bent molecular geometry. All three use the same ideal tetrahedral electron pair geometry; this does not make their actual bond angles equal. See Flowers, Theopold, Langley and Robinson, Chemistry 2e, §7.6: electron-pair and molecular geometries.

A static drawing also needs a physical interpretation. The atomic nuclei undergo oscillations; an arrangement at a minimum of potential energy and an experimentally averaged arrangement need not have identical bond lengths or bond angles. This distinction matters when comparing measurements with quantum mechanics calculations; see the IUPAC Gold Book: experimental bond-length conventions. Gas electron diffraction gives a concrete route from a laboratory signal to molecular geometry rather than inferring shape from a drawing alone.

Molecular vibration changes internal separations and bond angles, unlike the fixed internal molecular geometry of a rigid rotor. Its coupling to molecular rotation makes measured rotational constants depend on the vibrational quantum state.

A diatomic molecule has one internal stretching coordinate: its two atomic nuclei approach and separate about an equilibrium bond length. The diatomic bond stretch turns this motion into a particularly useful model. The local potential energy curvature supplies the restoring force, while the reduced mass accounts for both moving partners. Thus the same chemical bond can vibrate more slowly after an isotope substitution even when its electronic structure is nearly unchanged.

This separation makes infrared spectroscopy informative: transition frequencies constrain chemical bond stiffness, and isotope scaling of diatomic vibration tests the mass dependence. The University of Florida HCl and DCl FTIR experiment illustrates the measurement. Detectability also requires vibrational infrared activity; having vibrational energy eigenvalues alone does not guarantee an infrared absorption line.

The Newtonian gravitational potential $\Phi$ is potential energy per unit mass of a test particle. Its gradient determines the Newtonian gravitational field:In SI, $U$ has units $\mathrm{J}$, while $\Phi$ has units $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}={\mathrm{m}}^2{\mathrm{s}}^{-2}$; its gradient has units of acceleration, not force. Adding a constant to $\Phi$ leaves the Newtonian gravitational field unchanged. For an isolated point mass, the convenient reference is $\Phi (\infty )=0$. ^[ref]

Increasing the radius from $r_0$ to $r_1>r_0$ raises the Newtonian gravitational potential energy:An external agent supplies this positive work if the initial and final kinetic energies agree; gravity does work $-\Delta U$. For $r_1=2r_0$, the required work is $GMm/(2r_0)$. ^[ref]

Two point masses $m_1,m_2$ separated by distance $r$ have potential energyThe corresponding attractive force is $-\nabla V$.

This is the Newtonian point-mass specialization of gravitational energy, with the reference value chosen at infinite separation.

A normal mode, also called an eigenmode, is a pattern of free oscillation in a linear system whose relative displacements stay fixed while a single temporal factor changes their size and sign. For an undamped stable vibrational mode,Here $\varphi$ is the mode shape, $A$ sets the amplitude, $\omega$ is the eigenfrequency expressed as an angular frequency, and $\phi$ is the oscillation phase. Every snapshot is a scalar multiple of the same mode shape; neighbouring regions may move in opposite directions. Normal modes of a double pendulum show that normal modes also describe coupled discrete systems, beyond strings. Their shapes are eigenvectors, rather than spatial eigenfunctions. ^[ref]

For the ideal wave equation on a string, take a uniform string of length $L$ with constant tension $\tau$ and linear mass density $\mu$. Assume small transverse slope, negligible bending stiffness, no damping, and fixed endpoints. Its wave speed is $c=\sqrt{\tau /\mu }$, and the nonzero normal modes areThe angular frequency ${\omega }_n$ is in radians per second; the frequency $f_n$ is in hertz. The wavelength is $2L/n$. The mode shape has $n$ loops separated by nodes of a standing wave, including both fixed endpoints. ^[ref]

As a worked example, choose $L=1\mathrm{m}$, $\tau =1\mathrm{N}$, and $\mu =1\mathrm{kg}{\mathrm{m}}^{-1}$. Then $c=1\mathrm{m}{\mathrm{s}}^{-1}$ and $f_1=0.5\mathrm{Hz}$, $f_2=1\mathrm{Hz}$, $f_3=1.5\mathrm{Hz}$. Their oscillation periods are $T_1=2\mathrm{s}$, $T_2=1\mathrm{s}$, and $T_3=2/3\mathrm{s}$. These are deliberately simple model parameters, rather than measurements of a musical instrument. The plot divides displacement by $A$ so that the three mode shapes can be compared, and uses each mode's own oscillation period for its snapshots.

For $\phi =0$, at $t=T_n/4$ every point has zero displacement, but most points have nonzero velocity. Only zeros of the mode shape are permanent nodes of a standing wave. At $t=T_n/2$ the original mode shape has reversed sign, and at $t=T_n$ it returns. The synchronized conversion between kinetic energy and potential energy is shown in normal-mode energy exchange.

The short derivation uses separation of variables: substituting $u=\varphi q$ into the wave equation gives ${\varphi }^{\prime \prime }+k^2\varphi =0$ and $q^{\prime \prime }+c^2{k}^{2}q=0$. The fixed-end boundary conditions remove the cosine spatial term and require $\sin (kL)=0$, hence $k=n\pi /L$ and ${\omega }_n=ck$. Different geometries and boundary conditions give different eigenfunctions; a sine mode shape is a property of this uniform fixed string, not a universal form of an eigenmode.

The usefulness is that a complicated vibration becomes a collection of harmonic oscillators. For sufficiently regular fixed-end initial data, the Fourier sine series givesThe initial displacement and velocity determine $a_n$ and $b_n$. This is the superposition principle in a Fourier sine basis, and modal energy distribution measures how strongly each normal mode is excited. The striking simplification is that each component keeps its identity while their sum can continually change shape. ^[ref]

These independent normal modes belong to the linear model. At larger slopes, changing tension introduces nonlinear effects and can couple the modes. A linearly damped string still admits a Fourier sine basis, but its temporal factors decay and need not oscillate; its energy dissipation identity for a linearly damped string replaces conservation of energy. More general damping need not leave undamped mode shapes independent.

For the undamped first normal mode of a fixed string, start at maximum displacement with $u(x,t)=A\sin (\pi x/L)\cos ({\omega }_{1}t)$. In the small-slope wave equation on a string, its kinetic energy and excess potential energy areThe potential energy comes from the extra length under tension: $\sqrt{1+u_x^2}-1\simeq u_x^2/2$. It depends on slope, so a node of a standing wave need not have zero local potential energy density. These are physical energies, with the mass and tension factors restored to the normalized wave energy. ^[ref]

Integrating ${\sin }^2(\pi x/L)$ and ${\cos }^2(\pi x/L)$ over the string gives $L/2$ in both cases. Using $\mu {\omega }_1^2=\tau {\pi }^2/L^2$ yieldsFor the model $\mu =1\mathrm{kg}{\mathrm{m}}^{-1}$, $\tau =1\mathrm{N}$, $L=1\mathrm{m}$, and $A=0.010\mathrm{m}$, the angular frequency is ${\omega }_1=\pi \mathrm{rad}{\mathrm{s}}^{-1}$, the frequency is $0.5\mathrm{Hz}$, and $E\simeq 2.4674\times 10^{-4}\mathrm{J}$. The maximum slope $A\pi /L\simeq 0.0314$ is small. At $t=0$ all the energy is potential energy; at $t=T_1/4=0.5\mathrm{s}$ the string is straight and all the energy is kinetic energy. At $t=T_1/2=1\mathrm{s}$ the displacement is reversed and all the energy is again potential energy. The energy curves repeat twice during one displacement cycle, while conservation of energy keeps their sum constant.

Connect every pair among $n$ moving scalar coordinates and one fixed coordinate by identical springs of stiffness $k$. The coupling pattern is the complete graph $K_{n+1}$ with one vertex fixed. Its potential energy is $k{\sum }_{i<j}(X_i-X_j{)}^2/2$. The moving-coordinate stiffness matrix is the principal submatrix displayed in the header. For equal masses $m$, the uniform normal mode has squared angular frequency $k/m$, while every sum-zero vector has squared angular frequency $(n+1)k/m$. There are $n-1$ such degenerate modes. This follows directly from the Laplacian spectrum of a complete graph and its rank-one matrix form. Constant gravitational forces shift the equilibrium without changing the stiffness.

Three equal point masses $m>0$ move in a plane and are connected pairwise by ideal springs of equal positive stiffness $k$. At an unstretched equilateral triangle equilibrium, let $r_i$ be the vertex positions relative to the centre, and let $u_i$ be small displacements. If $e_{ij}=(r_i-r_j)/|r_i-r_j|$, the quadratic potential energy is $(k/2){\sum }_{i<j}[e_{ij}\cdot (u_i-u_j){]}^2$, and the kinetic energy is $(m/2){\sum }_i|{\dot{u}}_i{|}^2$.

There are three independent zero-frequency motions: two common translations and the infinitesimal rigid rotation $u_i=J{r}_i$, where $J(x,y)=(-y,x)$. The uniform radial motion $u_i=r_i$ is the breathing normal mode, with ${\omega }^2=3k/m$. The two shear modes are $u_i=(r_{ix},-r_{iy})$ and $u_i=(r_{iy},r_{ix})$, each with ${\omega }^2=3k/(2m)$.

For a compact proof of the spectrum, form the $3\times 6$ matrix $B$ taking the six displacement components to the three first-order spring extensions, orienting the three edges cyclically. Its row norms squared are two, and every pair of rows has inner product $1/2$. Thus $B{B}^T=(3/2)I+(1/2)11^T$, with eigenvalues $3,3/2,3/2$. The nonzero eigenvalues of $B^{T}B$ are the same, so the stiffness matrix $k{B}^{T}B$ has the three positive values in the title and a three-dimensional kernel of a linear map. The three rigid motions span that kernel. A zero mode has time factor $A+Bt$ rather than a nonconstant periodic oscillation.

For a uniform rigid beam of mass $m$ and two identical vertical end springs of stiffness $k$, small vertical end displacements have kinetic energy $m({\dot{x}}_1^2+{\dot{x}}_1{\dot{x}}_2+{\dot{x}}_2^2)/6$ and displacement-dependent potential energy $k(x_1^2+x_2^2)/2$. The mass and stiffness matrices are $\mathsf{M}=(m/6)\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ and $\mathsf{K}=k1$. Their generalized eigenvalue problem gives symmetric translation $(1,1)$ and antisymmetric rocking $(1,-1)$ at the displayed angular frequencies. The centre coordinate contains only the symmetric mode. Neglect spring inertia, damping and finite-angle geometric corrections.

Consider three beads of masses $m,m,M$ on a horizontal hoop, connected in cyclic order by identical springs constrained to follow the arcs. Their common natural length affects only a constant in the potential energy, since the sum of the three arc lengths is fixed. About the equally spaced configuration, angular displacements $q_i$ haveThe generalized eigenvalue problem has the rigid rotation $(1,1,1)$ as a zero mode. Exchanging the equal masses separates an antisymmetric normal mode $(1,-1,0)$, with squared angular frequency $3k/m$, from a symmetric normal mode $(1,1,-2m/M)$, with squared angular frequency $k/m+2k/M$. The last mode shape has zero total angular momentum. The hoop radius cancels from both the kinetic energy and the restoring quadratic form. These formulas use arc-constrained springs; straight chord springs have a different potential energy.

For three collinear masses $m,\mu m,m$, with $m>0$, $\mu >0$, connected consecutively by two springs of stiffness $k>0$ and with free ends, let $x$ denote displacements from an equilibrium. The kinetic energy and potential energy are $T={\dot{x}}^{T}M\dot{x}/2$, $V=x^{T}Kx/2$, whereThe generalized eigenvalue problem for small oscillations $Kv={\omega }^{2}Mv$ hasDirect substitution verifies each pair; the three distinct eigenvalues exhaust the three-dimensional problem. The zero normal mode translates the complete system without extending either spring. The second holds the middle mass fixed while the end masses move oppositely. The third moves the end masses together and the middle mass oppositely, leaving the centre of mass fixed.

For mass-matrix orthogonality of normal modes, the squared norms $v^{T}Mv$ are respectively $m(\mu +2)$, $2m$ and $m(2+4/\mu )$. Dividing each shape by its corresponding square root gives $q_i^{T}M{q}_j={\delta }_{ij}$. The general motion is a linear combination of the two oscillatory modes and $(A+Bt)v_0$; the latter describes free translation with constant velocity.

Two frictionless beads of positive masses $m_1,m_2$ on a circular hoop are joined by two identical springs of spring constant $k$, constrained to follow the two arcs. At their diametrically opposite equilibrium, each spring has natural length equal to half the circumference. For signed arc length displacements in the same rotational sense, the potential energy is $U=k(x_1-x_2{)}^2$. The normal mode of common rotation has mode shape $(1,1)$ and zero angular frequency, while the other normal mode has mode shape $(1,-m_1/m_2)$ and the positive angular frequency in the title.

Adding the equations of motion shows that $X=(m_1{x}_1+m_2{x}_2)/(m_1+m_2)$ satisfies $\ddot{X}=0$. Subtracting them gives $\ddot{\delta }+{\omega }_1^2\delta =0$ for $\delta =x_1-x_2$. Thus the complete zero mode is $X=C+Dt$, including uniform rotation, and the relative motion is a harmonic oscillator. In the limit $m_2/m_1\to \infty$ at fixed $m_1,k$, the oscillatory motion of the heavier bead vanishes for fixed lighter-bead amplitude, and ${\omega }_1\to \sqrt{2k/m_1}$. A common drift still survives if its initial velocity is held fixed.