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Let the particle have mass $m$, and assume its force is derived from a time-independent potential energy $V(x)$ during the stated interval. Differentiation givesFor $n\ne 0$, choose a branch on which $t=(x/A{)}^{1/n}$ is real. Conservation of energy is the quickest integration of Newton's second law:The additive constant $E$ is arbitrary. For instance, the displayed expression applies directly for $t>0$, $A\ne 0$ and $x/A>0$, without raising a possibly negative $A$ to a fractional power. Its derivative gives $-V^{\prime }(x)=mn(n-1)A{t}^{n-2}$ along the trajectory, as required. The second term has units of kinetic energy because $A$ has units $\mathrm{m}{\mathrm{s}}^{-\mathrm{n}}$.

For $n=1$, the velocity is constant, so the potential energy is constant on the visited interval; the expression above reduces to such a constant. For $n=0$, or $A=0$, the particle is stationary: the trajectory only fixes $V^{\prime }$ to vanish at its position and does not determine the potential energy elsewhere. These exceptional cases cannot determine a unique nonconstant function.

Magnetic field and force. A steady electric current produces a magnetic field. In vacuum the Biot-Savart law givesThe Lorentz force on the charge carriers transfers force to the material carrying them. The resulting force on a current element of another circuit is $d{\mathbf{F}}_2=I_{2}d{\mathbf{\ell }}_2\times {\mathbf{B}}_1$, or, for a continuous current distribution, the magnetic force per unit volume is $\mathbf{f}=\mathbf{J}\times \mathbf{B}$.

For two long parallel straight wires a distance $r$ apart, Ampère's law gives $B_1={\mu }_0{I}_1/(2\pi r)$. ThereforeThe force is attractive for parallel currents and repulsive for opposite currents, by the cross product rule. Its units are $\mathrm{N}{\mathrm{m}}^{-1}$. This approximation neglects wire thickness and end effects; a complete circuit includes its return current. Element-by-element forces need not form an isolated Newton's-third-law pair, but complete stationary circuits have balanced mutual forces when the entire system is included.

Energy description. For two linear coupled circuits the magnetic energy isHere $q$ is a mechanical coordinate and $M$ the signed mutual inductance. At prescribed currents, the generalized force is $I_1{I}_2\partial M/\partial q$ if the self-inductances do not vary. This positive derivative is obtained from magnetic coenergy, or by including the electrical sources' work; differentiating $W_B$ with a minus sign at fixed currents would omit that source work. At prescribed flux linkages the force instead equals the negative derivative of stored magnetic energy with those linkages fixed. The geometry thus determines both inductance and mechanical force.

An electrical transformer transfers alternating electrical power between windings through a shared changing magnetic flux. A core with large magnetic permeability confines most of the flux, so both windings link approximately the same flux $\Phi (t)$. Faraday's law gives induced electromotive forcesFor an ideal electrical transformer, negligible winding resistance and leakage flux imply the terminal-voltage magnitude ratioThe relative phase depends on the winding sense and terminal convention. At a loaded secondary, Lenz's law makes the secondary current oppose the flux change. The primary draws an additional current which approximately cancels the secondary magnetomotive force: $N_p{I}_p\simeq N_s{I}_s$ in magnitude when the magnetizing current is negligible. Consequentlyfor a resistive load, using root mean square values. More generally the ideal device conserves instantaneous power, with the appropriate current signs. A load impedance $Z_L$ appears at the primary as $Z_{\mathrm{in}}=(N_p/N_s{)}^2{Z}_L$, enabling impedance matching and electrical isolation.

For sinusoidal flux, $\Phi ={\Phi }_{\max }\sin \omega t$ gives $V_{\mathrm{rms}}=\omega N{\Phi }_{\max }/\sqrt{2}$. At a given supply voltage, reducing frequency increases the required peak flux and can drive the core into magnetic saturation. Real losses include winding Joule heating, magnetic hysteresis, eddy currents and leakage flux; a laminated core suppresses eddy-current loops. A steady direct current produces no continuing secondary electromotive force, while its small winding resistance can allow a damaging primary current after switching transients.

From molecular dipoles to macroscopic fields. A dielectric contains charges which remain bound to its atoms or molecules. An applied electric field displaces positive and negative charge centres and can orient permanent electric dipoles. The electric polarization $\mathbf{P}$ is electric dipole moment per unit volume, in $\mathrm{C}{\mathrm{m}}^{-2}$. Its associated bound charge isat the boundary of the polarized material, where $\hat{\mathbf{n}}$ points outward. A uniform electric polarization has no volume bound charge, but generally has surface bound charge. The electric field remains the force per unit test charge; it includes the fields of both free and bound charges.

Introducing the electric displacement field $\mathbf{D}={ϵ}_0\mathbf{E}+\mathbf{P}$ packages the free-charge source separately:For a homogeneous linear isotropic dielectric,The electric susceptibility ${\chi }_e$ is dimensionless; the permittivity $ϵ$ has units $\mathrm{F}{\mathrm{m}}^{-1}$. This is a constitutive approximation. An anisotropic dielectric requires a permittivity tensor, while a nonlinear dielectric need not have electric polarization proportional to electric field. The field in these equations is the macroscopic field, not necessarily the local field at an individual molecule.

Interfaces. Let $\hat{\mathbf{n}}$ point from medium 1 into medium 2. Integrating Gauss's law over a thin pillbox, and the electrostatic curl equation around a thin loop, gives the electrostatic interface boundary conditionsThus normal electric displacement is continuous when there is no free surface charge, whereas tangential electric field is continuous even if the permittivity changes. Normal electric field generally jumps because the interface carries bound charge. The electric potential is continuous for an ordinary interface without an ideal dipole sheet. These conditions, together with Laplace's equation in each homogeneous charge-free region, determine the field.

Energy and electrical sources. Reversibly assembling free charge requires incremental work $\delta W=\int \mathbf{E}\cdot \delta \mathbf{D}dV$. For a linear dielectric whose response can be reversibly integrated from zero field, the electrostatic energy isThis includes the material polarization response, rather than simply using ${ϵ}_0{E}^2/2$ everywhere. In a capacitor, $W=Q^2/(2C)=C{V}^2/2$. Inserting a dielectric increases capacitance. At fixed charge the stored energy decreases. At fixed voltage the stored field energy increases, but the voltage source supplies $VdQ$; the appropriate mechanical energy including that source changes by $-V^{2}dC/2$. Both cases predict attraction of dielectric into the capacitor. A force calculation must state which electrical quantity is fixed. A lossy or hysteretic dielectric also dissipates energy, so the reversible quadratic expression does not describe its full history.

Shape and depolarization. Suppose the dielectric is surrounded by vacuum and the distant imposed electric field is ${\mathbf{E}}_0$. Bound surface charge produces a depolarization field. For a uniformly polarized ellipsoid, along a principal axis,Here $N_d$ is the depolarization factor of a dielectric ellipsoid. This self-consistent field explains why the same material responds differently in different shapes.

An infinite slab with electric field normal to its faces has $N_d=1$, giving $E_{\mathrm{in}}=E_0/{ϵ}_r$; its opposite surface bound charges reduce the internal field. For electric field parallel to the faces, $N_d=0$, giving $E_{\mathrm{in}}=E_0$. End effects are neglected.

A very long circular rod in an axial electric field has negligible depolarization away from its ends, so $E_{\mathrm{in}}\simeq E_0$. In a transverse electric field the infinite-cylinder solution has $N_d=1/2$, giving $E_{\mathrm{in}}=2E_0/({ϵ}_r+1)$. To see the latter directly, take the applied field along $x$. The interior and exterior electric potentials can be written ${\varphi }_{\mathrm{in}}=-Br\cos \theta$ and ${\varphi }_{\mathrm{out}}=-E_{0}r\cos \theta +A\cos \theta /r$. Continuity of potential and normal electric displacement at $r=a$ gives $B=2E_0/({ϵ}_r+1)$.

For a sphere of radius $a$, $N_d=1/3$ andFor example, write ${\varphi }_{\mathrm{in}}=-Br\cos \theta$ and ${\varphi }_{\mathrm{out}}=-E_{0}r\cos \theta +C\cos \theta /r^2$. Continuity of potential gives $B=E_0-C/a^3$, and continuity of normal electric displacement gives $ϵB={ϵ}_0(E_0+2C/a^3)$. Solving yields $B=3E_0/({ϵ}_r+2)$ and $C=a^3{E}_0({ϵ}_r-1)/({ϵ}_r+2)$. The $1/r^2$ potential is that of the electric dipole moment above. For ${ϵ}_r=1$ all these induced effects vanish; as ${ϵ}_r$ becomes very large the normal-field sphere and slab approach the corresponding conductor field exclusion. The ideal infinitely long rod in an exactly axial imposed field has a different, singular end-free limit.

Let a coordinate frame rotate with angular velocity $\mathbf{\Omega }$ relative to an inertial frame. For any vector $\mathbf{u}$,Applying this twice to position, with coincident nonaccelerating origins, givesNewton's second law written using rotating coordinates therefore contains the Coriolis forceThe other inertial corrections are the centrifugal force and, for changing rotation rate, the Euler force. An accelerating origin introduces an additional translational fictitious force.

The Coriolis force depends on velocity and is perpendicular to it, so ${\mathbf{F}}_C\cdot {\mathbf{v}}_{\mathrm{rotating}}=0$: the Coriolis force does no work in the rotating frame. It bends trajectories without directly changing their rotating-frame speed. It is absent for motion parallel to the rotation axis.

For locally horizontal motion on Earth at latitude $\lambda$, take $x$ east, $y$ north and $z$ up. Keeping the vertical component $\Omega \sin \lambda$ givesHorizontal motion is deflected right in the northern hemisphere and left in the southern hemisphere. In the ideal absence of other horizontal forces these equations give inertial circles with angular frequency $|f|$. The same velocity-dependent correction explains the precession of a Foucault pendulum, whose precession-rate magnitude is $\Omega |\sin \lambda |$. Terrestrial vertical motions also involve Earth's horizontal rotation component, so the reduced horizontal equations are not the full force law.

In Lagrangian mechanics, generalized coordinates $q_i$ describe the independent configurations allowed by the constraints. For particles with ordinary conservative forces in an inertial frame, use $L=T-V$, with kinetic energy $T$ and potential energy $V$. Holonomic constraints can often be built into the coordinates, avoiding explicit constraint forces.

The principle of stationary action makes the action $\mathcal{S}={\int }_{t_1}^{t_2}L(q,\dot{q},t)dt$ stationary for variations vanishing at the endpoints. Integrating the velocity variation by parts givesSince each independent variation is arbitrary, the Euler-Lagrange equations areIf nonconservative generalized forces are present, their components appear on the right. The formalism is invariant under suitable coordinate changes and adding a total time derivative to $L$ leaves the fixed-endpoint equations unchanged.

The generalized momentum is $p_i=\partial L/\partial {\dot{q}}_i$. If $q_i$ is cyclic, $\partial L/\partial q_i=0$ and $p_i$ is conserved. If $L$ has no explicit time dependence, the conserved energy function isFor a kinetic energy quadratic in velocities and a velocity-independent potential energy, this equals $T+V$. This connects symmetry to conservation laws.

For example, a simple pendulum of mass $m$ and length $\ell$ hasIts tension does not appear because the fixed-length constraint is already imposed. Expanding for small $\theta$ gives simple harmonic motion with $\omega =\sqrt{g/\ell }$. More generally, expansion of $T$ and $V$ to quadratic order about a stable equilibrium leads directly to the normal-mode eigenvalue problem.

For a slender elastic beam undergoing small bending, assume plane cross-sections remain plane and normal to the neutral axis. Fibres a signed distance $z$ from that axis have longitudinal strain ${\epsilon }_{xx}=-z/R$, where $R$ is the local radius of curvature. Hooke's law gives stress ${\sigma }_{xx}=-Ez/R$, where $E$ is Young's modulus. Zero resultant axial force places the neutral axis through the section's centroid for a homogeneous beam.

The bending moment is the moment of this stress distribution. Its magnitude isThe second moment of area $I_A$ has units ${\mathrm{m}}^4$, and the bending stiffness $E{I}_A$ has units $\mathrm{N}{\mathrm{m}}^2$. A rectangular section of width $b$ and depth $h$ has $I_A=b{h}^3/12$, so placing more material far from the neutral axis strongly resists bending. This is the purpose of an I-section.

For small deflection $y(x)$, the signed curvature is approximately $y^{\prime \prime }$. A consistent variational sign convention follows from the elastic energy and a transverse load $q(x)$:For constant bending stiffness this is the Euler-Bernoulli beam equation $E{I}_A{y}^{\prime \prime \prime \prime }=q$. A clamped end fixes $y=y^{\prime }=0$; a simply supported end has $y=0$ and zero bending moment; an unloaded free end has zero moment and shear, $y^{\prime \prime }=y^{\prime \prime \prime }=0$ for uniform $E{I}_A$. Endpoint loads replace the corresponding zero-force boundary conditions.

For a cantilever of length $L$ carrying a tip force $F$, choose positive deflection along $F$. The internal bending gives $E{I}_A{y}^{\prime \prime }=F(L-x)$, with $y(0)=y^{\prime }(0)=0$. Two integrations giveThis approximation requires small slopes, elastic strains and a length large compared with section dimensions. Shear deformation, rotary inertia, plastic yielding and large rotations require extensions. Adding inertia gives flexural waves with ${\omega }^2=(E{I}_A/{\rho }_m{A}_s)k^4$ for a uniform beam, where ${\rho }_m{A}_s$ is mass per unit length.

Definition and equations. A normal mode is a motion in which every component of a linear system has the same time dependence with fixed amplitude ratios and phases. For a conservative system near equilibrium, the most useful form isThe vector $\mathbf{u}$ describes the mode shape, while $\omega$ is its angular frequency. Linear combinations of modes describe arbitrary sufficiently small motions when the modes form a complete basis.

Let $q_i$ be displacements of independent generalized coordinates from equilibrium. Expand the Lagrangian to quadratic order:The mass matrix $M$ defines a positive-definite quadratic form on the independent coordinates, while the stiffness matrix $K$ is the Hessian of the potential energy. Substituting a mode gives the generalized eigenvalue problem for small oscillationsA practical calculation identifies the equilibrium, chooses independent coordinates, computes the quadratic kinetic and potential energies, solves for the eigenvalues and then obtains their eigenvectors. One can equivalently diagonalize the real symmetric matrix $M^{-1/2}K{M}^{-1/2}$. This guarantees real ${\omega }^2$ and a complete orthogonal basis in mass-weighted coordinates.

For distinct eigenvalues, the fact that $K$ and $M$ are real symmetric matrices impliesThus distinct modes are orthogonal in the inner product weighted by the mass matrix. Choosing ${\mathbf{u}}_i^{\mathsf{T}}M{\mathbf{u}}_j={\delta }_{ij}$ decouples the motion into ${\ddot{Q}}_i+{\omega }_i^2{Q}_i=0$, with total energy ${\sum }_i({\dot{Q}}_i^2+{\omega }_i^2{Q}_i^2)/2$. A positive-definite matrix $K$ gives positive eigenvalues and oscillatory stability. Zero modes can describe rigid translations or rotations without restoring force; a negative eigenvalue gives exponential instability rather than an oscillatory mode.

Symmetry. A symmetry preserving both quadratic energies maps modes into modes of the same frequency. Coordinates adapted to symmetry divide the eigenvalue problem into smaller independent blocks. For two identical subsystems, exchanging them gives symmetric and antisymmetric modes. The degeneracy of normal modes means that independent modes share one frequency. Higher-dimensional group representations of a symmetry can enforce such degeneracies; symmetry does not require every pair of modes to be degenerate, and accidental degeneracies can also occur. Within a degenerate eigenspace, orthogonal linear combinations are equally valid mode shapes.

For two equal masses $m$, each attached to a fixed support by a spring of stiffness $k$ and coupled to the other by a spring of stiffness $k_c$,The symmetric mode ${\mathbf{u}}_{+}\propto (1,1)$ leaves the coupling spring unstretched and has ${\omega }_{+}^2=k/m$. The antisymmetric mode ${\mathbf{u}}_{-}\propto (1,-1)$ stretches it and has ${\omega }_{-}^2=(k+2k_c)/m$. Superposing the two gives beats when their frequencies are close.

Continuous systems. The same idea replaces matrices by differential operators and mode vectors by functions satisfying boundary conditions. For a taut string of length $L$, tension $\mathcal{T}$ and mass per unit length ${\mu }_{\ell }$, the wave equation and fixed endpoints giveOrthogonality of the sine functions allows an arbitrary initial displacement to be decomposed into modes. Changing the endpoints changes the spectrum: boundary conditions are part of the physical mode problem.

A uniform chain of equal masses $m$ joined by springs $k_s$, with spacing $a$, has travelling normal modes $q_j\propto e^{i(kja-\omega t)}$ withTheir quantization gives phonons; the long-wavelength limit is a sound wave. Rectangular electromagnetic cavity modes satisfy Maxwell's equations and conductor boundary conditions, with $\omega =c|\mathbf{k}|$ in vacuum. An elastic beam instead has a fourth-order spatial equation and flexural dispersion $\omega \propto k^2$. These examples show how restoring forces and geometry determine the frequency spectrum.

Real systems. Small-amplitude motion is often approximately linear even when the full equations are nonlinear. Large amplitudes produce frequency shifts, mode coupling and transfer of energy between modes. Friction broadens resonances and makes free oscillations decay. A damping matrix proportional to $M$ or $K$ can retain the undamped mode shapes; general damping need not, and the appropriate eigenfrequencies and modes can be complex. External periodic forcing predominantly excites modes near resonance, with amplitude set by coupling to the forcing and dissipation. Imperfections split ideal degeneracies, while radiation or coupling to a surrounding fluid can remove energy from the chosen subsystem. Normal modes remain useful when linearization, the boundary conditions and the treatment of damping accurately describe the motion being measured.

The radial force is $F_r=-dU/dr=-3A{r}^2$. A nontrivial circular orbit therefore requires $A>0$, so that the force is inward. Balancing its magnitude against the required centripetal acceleration givesThusThis is the circular-orbit energy ratio in a power-law potential, $2K=r{U}^{\prime }(r)=sU$ for $U=A{r}^s$, with $s=3$. It uses the zero of potential energy specified by $U=A{r}^3$; adding a constant would change the ratio without changing the motion. For $A<0$, the force is outward and no such free circular orbit exists.

A dielectric responds to an electric field mainly by displacing or orienting bound charges, rather than by sustaining a steady conduction current. The polarization density $\mathbf{P}$ is electric dipole moment per unit volume. It creates bound chargeSeparating these charges from free charge motivatesFor an isotropic linear dielectric, $\mathbf{P}={ϵ}_0{\chi }_e\mathbf{E}$, soThe electric susceptibility ${\chi }_e$ describes material polarization and is not the magnetic susceptibility. More generally it is a tensor and can depend on frequency, field strength and temperature.

Electronic polarization displaces an electron cloud relative to its nucleus; ionic polarization displaces positive and negative ions; orientational polarization partially aligns permanent molecular dipoles against thermal disorder. For independent permanent moments of magnitude $p$, weak-field orientational response gives ${\chi }_e\simeq n{p}^2/(3{ϵ}_0{k}_{B}T)$ when local-field corrections and interactions are negligible. This explains a common inverse-temperature contribution without asserting that every dielectric follows that law.

At a static interface, $\mathbf{n}\cdot ({\mathbf{D}}_2-{\mathbf{D}}_1)={\sigma }_f$ and the tangential electric field is continuous. Bound surface charge is already included through $\mathbf{P}$. For a homogeneous filled capacitor, $C=ϵA/d$, and a reversible linear dielectric stores energy density $u=\mathbf{E}\cdot \mathbf{D}/2$. At fixed voltage, insertion raises the stored field energy but the voltage source also supplies energy; force arguments must include that source. At fixed charge the capacitor energy instead decreases on insertion.

At frequencies too high for a polarization mechanism to follow the field, its contribution weakens. A complex permittivity represents dissipative dielectric loss as well as phase lag; the sign of its imaginary part depends on the time convention. Strong fields can cause nonlinear response or dielectric breakdown, beyond the reversible linear model.

Magnetization and field response. A magnetic material contains microscopic magnetic moments associated with electron spin and orbital motion. The magnetization $\mathbf{M}$ is magnetic dipole moment per unit volume. In SI the macroscopic fields satisfyHere $\mathbf{H}$ separates free-current excitation from the material response. A uniform isotropic linear material has $\mathbf{M}={\chi }_m\mathbf{H}$, givingThis constant scalar magnetic permeability model is useful only within its linear range. Crystals can have a susceptibility tensor; strongly magnetic materials depend on field and magnetic history. For a finite specimen the internal $\mathbf{H}$ also includes its demagnetizing field. In a uniformly magnetized ellipsoid, $H_{\mathrm{int}}=H_{\mathrm{applied}}-N_{d}M$ along a principal axis, so sample shape affects the apparent susceptibility.

Diamagnetic and paramagnetic response. Diamagnetism is a weak negative susceptibility caused by induced orbital response opposing the applied field. It occurs in all matter but can be masked by larger positive contributions. Closed-shell materials such as bismuth or many organic compounds show it clearly. Ordinary diamagnetism is often only weakly temperature dependent; superconducting flux exclusion is a separate collective phenomenon and is not described by an arbitrarily small negative susceptibility.

Paramagnetism arises when available magnetic moments partially align with a field. Thermal disorder prevents complete alignment at weak fields. For independent localized moments of magnitude $m$, the classical weak-field result iswhere the second expression uses $B\simeq {\mu }_{0}H$ in a weakly magnetic sample. Thus localized moments obey a Curie law in this regime. Quantum angular momentum changes the coefficient to the appropriate $J(J+1)$ expression and gives saturation at strong field. Conduction electrons instead exhibit Pauli paramagnetism: only states near the Fermi surface can change spin populations, so a degenerate gas has a nearly temperature-independent response. Its orbital diamagnetism must also be included for a full metallic susceptibility.

Collective magnetic order. Exchange interactions can favor parallel alignment, producing ferromagnetism below the Curie temperature. Iron, cobalt and nickel are familiar examples. A material may contain regions of spontaneous magnetization, or magnetic domains, while having little net external moment because the domains point in different directions. Applied fields move domain walls and rotate moments; at large fields the magnetization approaches saturation. Pinning and energy barriers make this process hysteretic. Remanence is the magnetization left after removing the driving field, and coercivity measures the reverse field needed to bring the chosen response back to zero.

The area of a closed $B$–$H$ loop is the energy dissipated per unit volume per quasistatic cycle, $\oint HdB$, with positive dissipated magnitude. Consequently the magnetic permeability is neither one universal constant nor enough information to describe a ferromagnet: differential magnetic permeability $dB/dH$, saturation and hysteresis all matter. Above the ordering temperature the spontaneous moment disappears, and a Curie–Weiss law is often an appropriate approximate paramagnetic description away from the critical region.

In antiferromagnetism, ordered opposing sublattices have equal moments and cancel macroscopically in the ideal compensated case. The order disappears above the Néel temperature. In ferrimagnetism, opposing sublattices have unequal moments and leave a net magnetization. Ferrites combine ferrimagnetic response with relatively high electrical resistivity, making them valuable when eddy-current losses would be substantial in a metallic core.

Electromagnets and engineering choice. A coil supplies free-current magnetomotive force $NI$. For a simple uniform magnetic path, Ampère's law gives $H\ell \simeq NI$ and flux $\Phi \simeq \mu ANI/\ell$. A high-magnetic permeability soft magnetic core concentrates flux and reduces the current required for a given field. With an air gap $g$,Even a short gap can dominate the magnetic reluctance. The approximate attractive force across one uniform gap is $F\simeq B^{2}A/(2{\mu }_0)$, obtained from the field energy, with fringing neglected.

Soft magnetic materials have small coercivity and narrow hysteresis loops; they suit relays, electrical transformer cores and controllable electromagnets. Hard magnetic materials retain their magnetization and suit permanent magnets. Laminating metallic cores breaks large eddy-current loops; ferrites are useful at higher frequencies because of their resistivity. Increasing current cannot increase $B$ indefinitely through the linear formula: core saturation, coil heating and leakage limit performance. A good design selects material for the field amplitude, switching frequency and acceptable loss, rather than simply choosing the largest quoted magnetic permeability.

Take a fixed sphere of radius $R$ in an unbounded incompressible inviscid fluid, with uniform velocity $U{\mathbf{e}}_z$ at infinity. Assume irrotational flow, so $\mathbf{v}=\nabla \varphi$ and incompressibility gives ${\nabla }^2\varphi =0$. Axisymmetry and the dipolar angular dependence of the far field suggestMatching the far field gives $A=U$. The no-penetration condition $v_r(R)=0$ then gives $B=U{R}^3/2$. ThusThe surface has stagnation points at its front and rear poles, while the equatorial speed is $3U/2$. Fluid slips tangentially past the sphere; imposing a no-slip condition would contradict this inviscid potential-flow model.

With gravity omitted or the hydrostatic pressure subtracted, steady Bernoulli givesAt the surface,The front–rear symmetry makes the integral of pressure force along the flow vanish: terms proportional to $\cos \theta$ integrate to zero over the sphere. This is D'Alembert's paradox, not a prediction that real spheres have no drag. Viscosity creates boundary layers; separation and a wake break the fore–aft symmetry. Potential flow remains valuable for the outer velocity field when viscous regions are thin, but it cannot supply their drag by itself.

A spatially uniform external gravitational acceleration moves a body's centre of mass without stretching it. Tidal forces arise from the variation of that acceleration across the body. Let a distant mass $M$ lie at distance $D$ along unit vector $\mathbf{n}$, and let $|\mathbf{r}|\ll D$ denote position relative to the body's centre. Subtract the centre's acceleration. To first order,The tidal tensor is $T_{ij}=GM(3n_i{n}_j-{\delta }_{ij})/D^3$. It stretches along the line to the source with eigenvalue $2GM/D^3$ and compresses in the two perpendicular directions with eigenvalues $-GM/D^3$. Its trace vanishes outside the source, consistently with Laplace's equation for the gravitational potential.

After removal of the constant and uniform-acceleration terms, the tidal potential isAn ideal equilibrium ocean on a sphere of radius $R$ follows an equipotential of the body's gravity plus this perturbation. For surface gravity $g$, the leading displacement isup to a constant fixing total water volume. The quadrupolar form produces bulges on both the near and far sides; the far-side bulge follows from differential acceleration, without postulating a repulsive gravitational force.

The Moon and Sun both raise terrestrial tides, with relative strength set mainly by $M/D^3$, rather than by the total gravitational pull $M/D^2$. Their contributions reinforce near new and full Moon to produce spring tides, and partly cancel near quadrature to produce neap tides. Earth's rotation relative to the tidal pattern gives a strong semidiurnal component in the simplest geometry. Real tide heights and timing depend on ocean depth, coastlines, basin resonances, Coriolis acceleration, friction and the declination of the forcing bodies. Tidal dissipation produces phase lags and torques, exchanging rotational and orbital angular momentum and gradually altering the spin and orbit. The equilibrium bulge is therefore a useful starting model, not an exact local tide forecast.

Configuration count. Choose the end-to-end direction as positive and let $N_R,N_L$ count rightward and leftward links. ThenA configuration is determined by choosing the positions of its leftward links, soThe counts must be nonnegative integers; in particular, allowed signed extensions occur in steps of $2a$. If distance means $|L|$ with neither direction selected, there are two corresponding sets for $L>0$, adding a factor of two. That constant does not affect the tension. There is no excluded-volume penalty in this one-dimensional idealized count.

Entropy from the multiplicity. Put $x=L/(Na)$, so $|x|<1$. Applying the leading Stirling's approximation to all three large factorials givesTerms subleading in $N$, including logarithmic factorial prefactors, have been omitted consistently. Expanding near zero extension givesStretching reduces the number of available configurations and hence the entropy.

Tension and the range of Hooke's law. The model assigns no extension-dependent internal energy. At fixed temperature, the externally applied tension is $\mathcal{F}=(\partial F/\partial L{)}_T=-T(\partial S/\partial L{)}_T$. Differentiating the full leading entropy givesIt is the external force needed to oppose the chain's restoring entropic force. Only when $|L|\ll Na$ can one put $\mathrm{artanh}x\simeq x$, yieldingThe paper's displayed linear law therefore requires this small-extension approximation, which is not explicitly stated in its premise $L<L_0$. It is not the exact force law up to full extension. The entropic spring constant grows with temperature because stronger thermal disorder makes a given reduction in configurational entropy cost more free energy. This is the one-dimensional entropic chain force-extension relation.

A sparse number of backward links. Write $q=N_L\ll N$, so $N_R=N-q$ and $L=a(N-2q)$. Exactly,For fixed $q$ and large $N$, each numerator factor approaches $N$, givingMore precisely, the logarithm of the product is $-q(q-1)/(2N)+O(q^3/N^2)$. A small relative error in the multiplicity requires $q^2/N\ll 1$; merely $q/N\ll 1$ need not make the omitted product close to one. The weaker sparse-link condition does suffice for the leading entropy density and leading high-extension tension. These are sparse backward links near full chain extension.

Tension in that limit. In the overlap regime $1\ll q\ll N$, use Stirling also for $q!$:ThusThe full leading entropy would give $(k_{B}T/(2a))\ln [(N-N_L)/N_L]$, which has the same dominant logarithm when $N_L/N\to 0$.

Approaching full extension. The continuum force law diverges logarithmically as $N_L\to 0$, or $L\to Na$. One must not literally apply Stirling to a zero factorial: at the discrete endpoint there is exactly one directed configuration, $\Omega =1$, and its entropy is zero. A continuous fixed-extension derivative is not defined at that endpoint.

An independent force-ensemble calculation makes the limiting physical statement precise. For applied tension $\mathcal{F}$, a link has partition function $z=2\cosh (\beta \mathcal{F}a)$, giving the exact mean extensionIt approaches $Na$ only as $\mathcal{F}\to \infty$ at positive temperature, and inversion reproduces the logarithmic force law for mean extension. In a real chain, bond extension or failure eventually replaces the ideal inextensible-link model.

The electric field is approximately radial in the long interior, but near an electrode end its lines spread into a fringing field of a capacitor. The electric field magnitude falls on moving outward through that end region. The dielectric's induced electric dipoles experience a net force toward the stronger field, pulling the material into the capacitor. For a weakly perturbing polarizable element this is expressed by $\mathbf{F}=\frac{1}{2}\alpha \nabla E^2$, with positive electric polarizability $\alpha$.

In the sketch the field lines run from the positive electrode to the negative one, curving outward near their ends. The material is attracted toward the interior where the field is stronger. A perfectly uniform interior field by itself would not give a longitudinal dipole-gradient force; the end region is essential to the local explanation. Neglecting its small contribution to the total capacitance does not justify omitting the finite insertion force that it transmits. The energy argument of part (d) gives that force without needing the detailed fringe solution.

This is the one-mole, or equivalently molar volume, form of the Van der Waals equation, an approximate equation of state for a real gas:Here $p$ is the gas pressure, $T$ its absolute temperature, $R$ the molar gas constant, and $V$ the volume occupied by one mole. If $V$ denotes molar volume explicitly, its units are ${\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$.

The positive parameter $b$ represents excluded volume arising from the finite size and short-range repulsion of the molecules. The volume available for translational motion is reduced from $V$ to $V-b$. This is an effective excluded volume, not simply the sum of the molecules' geometric volumes. Its molar units are ${\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$.

The positive parameter $a$ measures intermolecular attraction in a mean-field approximation. Attractions reduce the force exerted on the walls, lowering the measured pressure by $a/V^2$ relative to the repulsion-only expression. Thus $p+a/V^2$ is the pressure corrected for this attraction. The squared inverse-volume dependence reflects a contribution proportional to the square of the molecular number density. The molar units of $a$ are $\mathrm{Pa}{\mathrm{m}}^6\mathrm{mo}{\mathrm{l}}^{-2}$.

For an arbitrary amount $n$ measured in moles, with total container volume $V_{\mathrm{tot}}$, the corresponding formula isAt low density, when $b\ll V$ and $a/V^2\ll RT/V$, it reduces to the ideal gas law. The model captures qualitative attraction, repulsion and liquid-gas behavior, but does not supply an exact equation of state for every material.

Lagrangian mechanics expresses dynamics through generalized coordinates $q_i$ that respect the constraints. For a system with conservative, velocity-independent forces and standard kinetic energy, the Lagrangian is $L=T-V$. Its equations follow from the principle of stationary action:for variations that vanish at the endpoints. Varying the integral and integrating the $\delta {\dot{q}}_i$ term by parts givesIndependent variations therefore yield the Euler-Lagrange equations:Additional nonconservative generalized forces appear on the right-hand side as $Q_i$. The action is stationary, not necessarily a minimum. For ideal holonomic constraints, using suitable generalized coordinates removes the need to solve explicitly for constraint forces.

The conjugate momentum is $p_i=\partial L/\partial {\dot{q}}_i$. If a coordinate is cyclic, $\partial L/\partial q_i=0$ and its conjugate momentum is conserved. If $L$ has no explicit time dependence, the energy functionis conserved. It equals $T+V$ for the usual quadratic kinetic energy with time-independent geometry, but that identification is not universal for every Lagrangian or constraint.

For example, a simple pendulum with angle $\theta$, length $\ell$ and bob mass $m$ hasThe tension does no virtual work in the allowed angular displacement and never needs to be calculated to obtain the angular equation. The method scales naturally to coupled systems and coordinates with complicated geometry.

For a slender elastic beam, the Euler-Bernoulli beam equation assumes small strain, small deflection slopes and cross-sections that remain plane and normal to the deflected centreline. A fibre at signed distance $y$ from the neutral axis has longitudinal strain ${ϵ}_{xx}=-y\kappa$, where $\kappa$ is the centreline curvature. Hooke's law gives ${\sigma }_{xx}=-Ey\kappa$, with Young's modulus $E$.

Choose the internal bending-moment sign so that $M=-\int y{\sigma }_{xx}dA$. ThenHere $I_A$ is the second moment of area, with units ${\mathrm{m}}^4$, not the mass moment of inertia of a rotating body. For a uniform material under pure bending, zero axial resultant places the neutral axis through the section's centroid.

For small slopes, $\kappa \simeq w^{\prime \prime }(x)$. Combining the moment-curvature relation with transverse force and moment balance gives, with a consistent load sign convention,For constant stiffness this becomes $E{I}_A{w}^{(4)}=q$. The bending stiffness $E{I}_A$ has units $\mathrm{N}{\mathrm{m}}^2$. Boundary conditions specify displacement and slope at a clamped end, or bending moment and shear force at a free end.

For a cantilever of length $\ell$ carrying a transverse end load $F$, take deflection positive towards the load. The moment magnitude is $M(x)=F(\ell -x)$, and integration with $w(0)=w^{\prime }(0)=0$ givesThe free-end moment vanishes and the tip shear supplies the applied load. For a rectangular section of width $b$ and bending thickness $h$, $I_A=b{h}^3/12$, so increasing thickness strongly increases bending resistance. The model ceases to be reliable for large deflections, plastic strains or short thick beams where shear deformation matters.

The Magnus effect is the transverse force produced when a spinning body moves relative to a fluid. Rotation changes the flow and circulation around the body; the resulting asymmetric pressure and wake deflect the fluid, with the opposite transverse force on the body. Examples include the curved flight of spinning balls and the lift of rotating-cylinder sails.

A useful quantitative model is steady two-dimensional incompressible flow past a cylinder of radius $a$, with fluid velocity $\mathbf{U}=U\hat{\mathbf{x}}$ far away. Let circulation $\Gamma$ be positive anticlockwise as viewed from $+z$. The potential-flow velocity at the cylinder surface isThe Bernoulli equation outside the boundary layer givesIntegrating the pressure force around the cylinder, per unit span, yieldsThus the Kutta–Joukowski theorem givesThe sign here refers to the fluid velocity relative to the cylinder. For a body moving through otherwise stationary fluid, that relative fluid velocity points opposite to the body's translational velocity.

Real viscosity and the no-slip boundary condition transmit the surface rotation to the fluid and help set the circulation and separation pattern. Pure inviscid flow with only an impermeability condition does not determine $\Gamma$ from the cylinder's angular speed. Therefore the circulation must not be assigned a universal spin-rate formula without additional assumptions. The cylinder result illustrates the lift mechanism; quantitative forces on a finite spinning ball also depend on its geometry, Reynolds number, rotation rate, surface roughness and wake.

The bending moment $B(x)$ is the internal torque transmitted across a cut through an elastic beam, about the transverse bending axis. It balances the moments of the external forces on the isolated portion on one side of the cut. Its sign depends on the chosen convention, and its units are $\mathrm{N}\mathrm{m}$.

The relevant “moment of area” is the second moment of area of the cross-section about its neutral axis:where $z$ is perpendicular distance from that axis within the section. It has units ${\mathrm{m}}^4$ and is a geometrical quantity, unlike a mass moment of inertia. For a homogeneous beam under pure bending, the neutral axis passes through the section's centroid, and its longitudinal strain is zero.