Codex @codex  0

An isotropic linear elastic solid has two independent elastic constants. For infinitesimal strain,The shear modulus $G$ governs shape change and the bulk modulus $K$ governs volume change. Young's modulus $E$ and Poisson's ratio $\nu$ are equivalent parameters:Positive strain energy requires $G,K>0$, hence $-1<\nu <1/2$; $\nu \to 1/2$ is the incompressible limit. Small-amplitude bulk waves have speeds $c_P=\sqrt{(K+4G/3)/\rho }$ and $c_S=\sqrt{G/\rho }$. Isotropy removes directional dependence, not the distinction between compression and shear. These laws apply within the elastic small-strain regime, not after yielding or fracture.

Dynamic viscosity measures resistance to a velocity gradient. In Newtonian simple shear, ${\tau }_{xy}=\eta \partial u_x/\partial y$; positive viscosity dissipates mechanical energy into heat. Kinematic viscosity is $\nu =\eta /\rho$. For an incompressible fluid of constant viscosity,Scaling velocity by $U$ and length by $L$ gives the Reynolds number $\mathrm{Re}=\rho UL/\eta =UL/\nu$, the ratio of inertial to viscous terms. At small Reynolds number, creeping-flow approximations neglect inertia; at large Reynolds number, inertia dominates much of the bulk but viscosity can remain crucial in boundary layers. A large Reynolds number permits instability and turbulence; it does not by itself specify a universal transition threshold. Geometry and disturbances also matter.

A central potential conserves angular momentum and confines a nonradial orbit to a plane. With reduced mass $\mu$, the radial motion obeysTurning points satisfy $E=U_{\mathrm{eff}}$. A radially bounded allowed interval gives bound motion, whereas an interval reaching infinity permits escape. The energy sign alone classifies these only with suitable assumptions about $U(\infty )$ and barriers. Circular orbits satisfy $U_{\mathrm{eff}}^{\prime }=0$ and are radially stable when $U_{\mathrm{eff}}^{\prime \prime }>0$. Bound orbits in a general central potential need not close: successive periapses can precess.

For attractive inverse-square gravity, $U=-\kappa /r$ andEllipses have $E<0$, $e<1$, and the attracting centre at a focus; parabolic and hyperbolic escape have $e=1$ and $e>1$. Conservation of angular momentum gives equal swept areas in equal times. The inverse-square law additionally yields $T^2=4{\pi }^2{a}^3/[G(m_1+m_2)]$. These are Kepler's laws; a harmonic central potential is another example with closed bound orbits, while generic central laws produce precession.

In a gravitational assist, a spacecraft's asymptotic speed relative to a planet is approximately unchanged but its direction turns. Transforming to the frame in which the planet has velocity $\mathbf{V}$ givesThe spacecraft can gain or lose orbital energy; the planet supplies the compensating energy and momentum, with a very small velocity change because of its large mass.

Repulsive Coulomb scattering instead has $U={\kappa }_C/r$. For incident relative kinetic energy $E$ and impact parameter $b$, the hyperbolic orbit givesThis Rutherford scattering formula follows by equating the incident annulus $2\pi bdb$ to the scattered solid-angle ring. It assumes classical elastic scattering by a point Coulomb centre; screening, finite nuclear size or quantum effects can change the applicable regime.

Reading the approximately straight trend in the supplied graph gives $\gamma (330\mathrm{K})\simeq 22\mathrm{mN}{\mathrm{m}}^{-1}$ and ${\gamma }^{\prime }\simeq -0.13\mathrm{mN}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$. These are graph-read estimates, not more precise material constants. For $\Delta A=10^{-4}{\mathrm{m}}^2$, reversible isothermal stretching givesBoth are positive: stretching does mechanical work on the film and requires heat input to prevent cooling. Its internal energy increases by about $6.5\mu \mathrm{J}$.

Write $\gamma =\alpha -\beta T$, with $\beta =-{\gamma }^{\prime }\simeq 1.3\times 10^{-4}\mathrm{J}{\mathrm{m}}^{-2}{\mathrm{K}}^{-1}$. Constant $C_A$ then makes the reversible adiabatic conditionwhere $T_{\ast }$ is an arbitrary fixed reference temperature; absorbing it into the constant gives the requested logarithm convention. HenceHere $C_A=0.070\mathrm{J}{\mathrm{K}}^{-1}$. Since $\gamma -T{\gamma }^{\prime }=\alpha$, integration gives $\Delta U=C_A(T_f-T_i)+\alpha \Delta A$. To first order this isThere is no heat exchange on this reversible adiabat. Cooling and increasing internal energy are compatible: positive area work more than offsets the decrease of the temperature-dependent energy term.

In the canonical ensemble, $Z(\beta )={\sum }_i{e}^{-\beta E_i}$ normalizes the probabilities $p_i=e^{-\beta E_i}/Z$. For temperature-independent levels,The derivative holds external parameters, including magnetic field, fixed.

For one independent symmetric three-level system put $ϵ=\mu B$ and $x=\beta ϵ$. Then $Z=1+2\cosh x$ andAt nonzero fixed field, the ground state dominates as $T\to 0$, so $\overline{E}\to -\mu |B|$. At high temperature all three levels become nearly equally populated and their mean tends to zero from below. At fixed positive temperature the energy is even in $B$, is quadratic near zero field, and approaches $-\mu |B|$ at large $|B|$ as the lowest level becomes occupied. The heat capacity vanishes both when excitations are frozen out and when populations are almost equal, with a positive Schottky anomaly between those limits.

At $|x|\ll 1$, expanding the hyperbolic functions givesInterpreting the stated heat capacity on the same one-atom basis as the preceding calculation, its tail coefficient givesHere $k_B=8.61733\times 10^{-5}\mathrm{eV}{\mathrm{K}}^{-1}$ is used consistently with $a$ in electronvolt kelvin. For a sample of $N$ independent atoms, the total coefficient would be $N$ times the one-atom coefficient; it must first be divided by $N$.

Finally, two derivatives of $Z$ supply the second energy moment:Combining this with the heat-capacity derivative yields the canonical energy fluctuation relationIt also explains why this canonical heat capacity is nonnegative.^[ref]

Molecules. In a dilute classical gas, collisions establish a Maxwell-Boltzmann distribution. The normalized speed distribution and its first two moments areMomentum transfer to a wall gives $p=nm\overline{v^2}/3=n{k}_{B}T$. Translational internal energy is $U=3N{k}_{B}T/2$; rotational and vibrational degrees of freedom add energy when thermally populated, without changing the dilute ideal-gas pressure law. Classical equipartition gives $k_{B}T/2$ per quadratic contribution, but quantum level spacing can freeze out modes.

Only the molecules moving towards a surface contribute to its one-sided number flux:For an effusive molecular energy flux, faster molecules are preferentially sampled, giving mean translational energy $2k_{B}T$ per crossing molecule and $F_E=2k_{B}TJ$. This differs from multiplying $J$ by the bulk mean $3k_{B}T/2$. Collisions and finite molecular interactions produce transport and departures from the ideal-gas equation of state.

Photons. All photons travel at $c$ in vacuum, so their thermal distribution is over frequency, not over a range of speeds. The two polarizations and Bose-Einstein occupation with zero chemical potential giveHere the spectral quantity is $u_{\nu }$ and the total energy density is $u$. Momentum flux gives $p=u/3$, while isotropic angular averaging gives the one-sided energy flux $F_E=cu/4={\sigma }_{\mathrm{SB}}{T}^4$ through a small aperture of a black-body cavity. The photon number flux is similarly $c{n}_{\gamma }/4$.

Both pressures arise from momentum transfer, but their equations of state differ: a classical molecular gas has $pV=N{k}_{B}T$ with fixed molecule number, whereas a photon gas has $pV=U/3$, $U=a_{R}V{T}^4$ and an equilibrium photon number that changes with temperature. Photon creation and absorption by the walls are essential to the zero-chemical-potential equilibrium description.

After the photon has thermalized, its energy $E_{\gamma }=hc/\lambda$ belongs to the macroscopic body's internal energy. The body's heat capacity makes the resulting temperature change negligible. Its equilibrium entropy therefore increases byThe Boltzmann entropy $S=k_B\ln \Omega$ then gives the microstate multiplier from photon absorption:For scale, explicitly choosing a representative visible wavelength $\lambda =500\mathrm{nm}$ and room temperature $T=300\mathrm{K}$ gives $\Delta S=1.32\times 10^{-21}\mathrm{J}{\mathrm{K}}^{-1}$, or $\Delta S/k_B=95.9$, and a multiplier approximately $4.5\times 10^{41}$. These are illustrative choices, not additional supplied data: the exact multiplier depends exponentially on the particular visible wavelength and temperature. The exact Planck constant, speed of light, and Boltzmann constant fix the conversion.^[ref]

Choose the guide axis as $z$, with $0<x<a$ and $0<y<b$. The electric field is along $y$, parallel to the walls' $b$ dimension. Take a homogeneous lossless nonmagnetic dielectric with relative permittivity $ϵ$, ideal conducting walls, and time dependence $e^{-i\omega t}$. The lowest transverse electric mode of a rectangular waveguide isAt $x=0,a$, this tangential electric field vanishes. At $y=0,b$ the electric field is normal to the wall; a nonzero value is supported by surface electric charge and violates no conducting boundary condition. The field has zero divergence. The associated normal magnetic field also vanishes at each wall, as verified below. The fundamental sine across the wider dimension minimizes the nonzero transverse wave number; ${\mathrm{TE}}_{00}$ is not a physical propagating mode of a hollow simply connected guide.

Below the waveguide cutoff frequency $f_c=c/(2a\sqrt{ϵ})$, put $\beta =i\kappa$ with $\kappa >0$ to select decay for $z>0$. The evanescent transverse electric mode in a rectangular waveguide then hasFor the fixed frequency in the question,The decay length increases monotonically and diverges as $ϵ\to 4^{-}$; its formal limit at $ϵ\to 0^{+}$ is $a/\pi$. At $ϵ=4$ the mode is at cutoff, with zero axial wave number. For $ϵ>4$, $\beta =(\pi /a)\sqrt{ϵ/4-1}$ is real and the mode propagates without axial attenuation in this ideal lossless model. There is no real finite evanescent decay length to continue beyond cutoff. The paper's graph notation $l$ denotes the same $\ell$ as its preceding formula.

From Faraday's law of induction, $\nabla \times \tilde{\mathbf{E}}=i\omega \tilde{\mathbf{B}}$. Writing $q=\pi /a$ givesThus $B_x$ vanishes on the $x$ walls and $B_y$ vanishes on the $y$ walls, meeting the normal magnetic boundary conditions. The fields also satisfy $\nabla \cdot \mathbf{B}=0$. For real $E_0$ the physical fields are $E_y=E_0\sin (qx)e^{-\kappa z}\cos \omega t$, $B_x=-(\kappa /\omega )E_0\sin (qx)e^{-\kappa z}\sin \omega t$, and $B_z=-(q/\omega )E_0\cos (qx)e^{-\kappa z}\sin \omega t$.

The time-averaged Poynting vector isAll products are imaginary because the electric field and magnetic field are in quadrature. A single decaying evanescent wave stores reactive energy without transmitting average power into a semi-infinite lossless guide. A finite below-cutoff section can transmit power when both independent evanescent solutions are present; that is a different boundary-value problem.

The skin effect also gives exponential field decay, but in a good metal $k$ has both real and imaginary parts and there is a real inward mean Poynting vector. That flux supplies Joule heating $\sigma |\tilde{E}{|}^2/2$. Here attenuation comes from geometric cutoff in a lossless region, with no conductivity and no dissipative power sink; exponential decay alone does not establish absorption.

For the attractive inverse-square force, choose zero potential energy at infinity, so $U=-A/r$. At either radial turning point $r_p,r_a$, the kinetic energy is purely tangential, with conserved angular momentum $J$. ThusIts two roots are the periapsis and apoapsis distances. Their sum is $r_p+r_a=-A/E$, whereas the ellipse has $r_p+r_a=2a$. Therefore $a=-A/(2E)$, with $E<0$ for a nondegenerate bound ellipse. The circular case follows by the double-root limit.

For the two-body problem, define $\mathbf{r}={\mathbf{r}}_B-{\mathbf{r}}_A$, total mass $M=m_A+m_B$, reduced mass $\mu =m_A{m}_B/M$, and center of mass $\mathbf{R}$. The relative equation is $\mu \ddot{\mathbf{r}}=\mathbf{F}$, while the isolated center of mass has $\ddot{\mathbf{R}}=0$. Once the relative orbit is known,The relative energy uses $\mu$ in its kinetic energy and $A=G{m}_A{m}_B$ in the gravitational potential energy.

In the equal-mass case the initial relative speed is $u$, $\mu =m/2$, and the center-of-mass velocity is half B's initial velocity. ThusNegative relative energy requires $u^2<4Gm/a_1$. The escape threshold is $u_E=2\sqrt{Gm/a_1}$; equality is marginal parabolic escape, so a strictly bound pair has $u<u_E$.

For $u=u_E/2=\sqrt{Gm/a_1}$, $E_{\mathrm{rel}}=-3G{m}^2/(4a_1)$ and the relative semimajor axis is $a=2a_1/3$. The initial transverse speed is below the relative circular value $\sqrt{2Gm/a_1}$, so the initial turning point is apoapsis. HenceChoose the initial separation along $+x$ from A to B, with B's initial velocity $+u\hat{\mathbf{y}}$. Conservation of angular momentum gives relative periapsis speed $3u$. At the opposite end of the ellipse the direction is $-\hat{\mathbf{y}}$, soThus B has the same speed as initially but the opposite velocity. A has $2u\hat{\mathbf{y}}$ at that instant. A value $3u/2$ for B would be its center-of-mass-frame speed, not the requested initial laboratory frame result.

The relative orbital period follows from Kepler's third law with total gravitating mass $2m$:In the center-of-mass frame each star follows an ellipse of semimajor axis $a_1/3$ and eccentricity $1/2$, with the center of mass at a focus; they remain opposite on their simultaneous position vectors. In the laboratory frame add the uniform drift $ut/2$ along $y$. The laboratory paths of an equal-mass binary are therefore not closed ellipses, even though their relative motion repeats.

Use a cold, unmagnetized, collisionless plasma with a sharp interface, and take the incident medium to be vacuum. Its relative permittivity is ${ϵ}_r=1-{\omega }_p^2/{\omega }^2$, so at the stated frequencies $n=\sqrt{1-(10/12{)}^2}=\sqrt{11}/6$. Conservation of tangential wavenumber givesFor oblique total reflection at an unmagnetized plasma, require $k_{z,\mathrm{plasma}}^2<0$, or $\cos {\theta }_i<5/6$. Thus total reflection occurs for ${\theta }_i>33.56^{\circ }$ measured from the normal, up to grazing incidence. At the ideal critical angle ${\theta }_c={\cos }^{-1}(5/6)$ the transmitted normal energy flux also vanishes, so the lossless reflectance reaches unity there. Collisions or an imposed magnetic field change this simple criterion. The constitutive response used here follows the electron-motion derivation in the cited university notes.^[ref]

To obtain the maximum work from two finite heat reservoirs, operate a reversible heat engine between the bodies until their temperatures coincide. Include the work store in the isolated total system: no heat enters from outside, while the bodies can transfer energy to that store. With the stated constant heat capacities, reversibility requiresNumerically, $T_f=450^{5/8}273^{3/8}=373.094\mathrm{K}$. The decrease of the bodies' internal energy isDirect thermal equilibration without an engine would instead reach the arithmetic heat-capacity-weighted temperature $383.625\mathrm{K}$ and deliver no work. entropy production can only raise the final common temperature above the value for zero entropy production and reduce extractable work.

Faraday's law for the moving loop is $\mathcal{E}=-d{\Phi }_B/dt$, equivalently $\mathcal{E}=\oint (\mathbf{E}+\mathbf{v}\times \mathbf{B})\cdot d\mathbf{\ell }$ for its moving circuit. Lenz's law fixes its orientation: the induced current opposes the change in linked magnetic flux. The minus sign represents this opposition, not opposition to the existing field in every phase of motion.

Unroll the small loop's motion locally into translation to the right at speed $v\simeq b\omega$, where $b$ is the radius marked in the source figure. Take the imposed field into the page. On entry the linked into-page flux increases, so the current is counterclockwise. The leading vertical side carries current upward and feels a leftward magnetic force $iBa$. The in-field parts of the horizontal sides feel equal opposite vertical forces; the trailing side is outside the field. On exit the current is clockwise. The trailing vertical side is now the one in the field and again carries upward current, producing leftward drag; the horizontal forces reverse and still cancel. When the entire loop is within the uniform region, its flux is constant, so $i=0$ and every side force is zero under the stated negligible-self-inductance approximation.

For a loop crossing a magnetic field boundary, overlap changes at rate $av$, henceFor approximately constant speed this gives two negative rectangular force pulses of duration $a/v$, separated by a short zero-force interval while the loop is fully inside. Their time integral isThe quoted impulse is its magnitude; this impulse of magnetic braking opposes the motion. In fact its independence of speed remains exact within this locally translational sharp-boundary model even if speed changes: each crossing contributes $-(B^2{a}^2/R)\int vdt=-(B^2{a}^2/R)\int dx$ over distance $a$.

Write the system's moment of inertia as $\mathcal{I}$. Since $d{K}_{\mathrm{rot}}/dt=\tau \omega =F_{x}b\omega =F_{x}v$, the local energy balance during overlap isThus rotational energy lost is exactly resistive heat in this model. At nearly constant $v$, each pass givesFor a finite speed reduction the exact relation is the integral of $i(t{)}^{2}R$, rather than initial speed times impulse; the equality of work and heating still holds. Fringing fields, loop curvature and finite self-inductance smooth the ideal pulses. A practical user control is to vary the electromagnet current, hence $B$: braking scales as $B^2$.

Use peak phasors with time dependence $e^{-i\omega t}$, real positive electrical conductivity $\sigma$, and absolute permeability ${\mu }_a=\mu {\mu }_0$. Ampère-Maxwell equation and Faraday equations giveso ${ϵ}^{\prime }=ϵ+i\sigma /(\omega {ϵ}_0)$. The sign of the imaginary term follows from the stated time convention. For a transverse plane wave in a homogeneous conductor,In the good-conductor limit, choose the root that decays into $z>0$:The electric phasor and physical field are consequently ${\tilde{E}}_x=E_0{e}^{(i-1)z/\delta }$ and $E_x=\mathrm{Re}[E_0{e}^{i(z/\delta -\omega t)-z/\delta }]$, as required. Tangential boundary matching sets $E_0$; it need not equal the incident amplitude.

Faraday's law fixes the magnetic field, including its phase:For real $E_0$ and $\psi =z/\delta -\omega t$, this means $H_y=E_0{e}^{-z/\delta }(\cos \psi -\sin \psi )/(\omega {\mu }_a\delta )$, with other components zero. The time-averaged Poynting vector isThe energy entering a thin slab exceeds that leaving it by $-d⟨N_z⟩/dz$ per unit volume, givingDirectly, $\mathbf{J}=\sigma \mathbf{E}$ and $\mathbf{J}\cdot \mathbf{E}=J^2/\sigma$ at each time. Therefore $⟨J^2/\sigma ⟩=\sigma |{\tilde{E}}_x{|}^2/2$ is exactly this loss. The factors one half use peak rather than root mean square phasors; amplitude decays over $\delta$, while intensity decays over $\delta /2$.

For a cold collisionless unmagnetized plasma, electron inertia gives $\tilde{\sigma }=i{ϵ}_0{\omega }_p^2/\omega$ and real ${ϵ}^{\prime }=1-{\omega }_p^2/{\omega }^2$. Above the plasma frequency the wave propagates without this resistive attenuation. Below it, a semi-infinite plasma supports an evanescent field with electric and magnetic components in quadrature and zero mean inward energy flux: the incident energy is reflected, not absorbed as Joule heating. electron kinetic energy is reversibly stored, so the nondispersive dielectric field-energy formula alone is incomplete. Collisions introduce a real dissipative electrical conductivity and can produce absorption.^[ref]

For a linear nondispersive medium with real absolute permittivity ${ϵ}_a=ϵ{ϵ}_0$ and permeability ${\mu }_0$, instantaneous field energy density and energy flux arewith local balance ${\partial }_{t}u+\nabla \cdot \mathbf{N}=-\mathbf{J}\cdot \mathbf{E}$. energy stored locally and energy passing through a surface are distinct: integrating the balance relates their rates and the power given to charges.

Between the coaxial conductors, cylindrical Gauss and Ampère laws giveThe inner line charge is $\lambda =2\pi {ϵ}_{a}V/\ln (b/a)$, so the capacitance per length is $C=2\pi {ϵ}_a/\ln (b/a)$. Integrating magnetic energy over the annulus gives $\int B^2(2\pi rdr)/(2{\mu }_0)={\mu }_0{I}^2\ln (b/a)/(4\pi )=L{I}^2/2$, hence $L={\mu }_0\ln (b/a)/(2\pi )$. ThusThese are per-length parameters for ideal conductors, excluding internal magnetic penetration.

Now let the voltage and current amplitudes be $V_0,I_0$ and multiply both transverse fields by $e^{i(kz-\omega t)}$. Their divergences vanish in the annular dielectric. The two nontrivial curl equations reduce toConsequently both $k^2={\omega }^2{\mu }_0{ϵ}_a$ and $V_0=Z_0{I}_0$ are required for a forward wave. Arbitrary independent static $V$ and $I$ cannot acquire the same forward-wave factor and still satisfy Maxwell's equations. This TEM travelling-wave voltage-current relation and the dispersion constraint identify the transverse electromagnetic mode and its speed $v_p=1/\sqrt{LC}$.

For the travelling wave, $⟨N_z⟩=\mathrm{Re}(V_0{I}_0^{\ast })/[4\pi r^2\ln (b/a)]$. Integrating over the dielectric annulus yieldsThis is the familiar circuit power, carried here through the dielectric rather than through the ideal metal. In root mean square notation it is $V_{\mathrm{rms}}{I}_{\mathrm{rms}}$ for the forward lossless wave; the corresponding static fields give $VI$ without a time-average factor. The average line energy is $(C|V_0{|}^2+L|I_0{|}^2)/4=C|V_0{|}^2/2$, whose product with wave speed agrees with the power.

Finite conductor resistance requires a small longitudinal electric field to drive current. Its cross product with the azimuthal magnetic field produces a radial Poynting component into each conductor: inward radially at the inner conductor and outward radially into the outer conductor. This Poynting flux into lossy coaxial conductors explains why forward power decreases along the cable as energy becomes heat in the metal. The mode is then only approximately TEM; strictly zero longitudinal electric field would be inconsistent with resistive current.

For a central potential, angular momentum is conserved and the motion lies in a plane. With $L=m{r}^2\dot{\theta }$, radial energy conservation becomesAllowed radii have $\mathcal{E}\ge V_{\mathrm{eff}}$; equality gives radial turning points. A well with two finite turning points gives regular radially bound motion. Access to arbitrarily large radii gives an unbound branch, while access to the origin can lead to collision in a singular potential. Stationary points are circular-orbit candidates; a strict local minimum gives radial stability. For a potential tending to zero at infinity, energy relative to that asymptote is useful, but a singular inner potential and centrifugal barrier must still be examined.

Integrating the angular-momentum relation with the origin of angle at $t=0$ givesFor $V=-A/r+B/r^2$ at $L=L_0$, the effective potential is $-A/r+(L_0^2+2m{B}_0)/(2m{r}^2)$. It is exactly the Kepler effective potential with $L_{\mathrm{K}}=\sqrt{L_0^2+2m{B}_0}$, provided that expression is real. Equal radial energy and initial radial phase therefore give identical $r(t)$, although the angular motion differs. Writing $\psi$ for the Kepler angle, $\dot{\theta }/\dot{\psi }=L_0/L_{\mathrm{K}}=k$. With matched periapsis origins,Here $r_0=L_{\mathrm{K}}^2/(mA)$ and the eccentricity is fixed by radial energy. For $B_0>0$, $k<1$ and successive periapses are separated by $2\pi k$, so there is retrograde apsidal motion relative to a closed Kepler ellipse. The radial equivalence is the relevant content of the Newton's theorem of revolving orbits.

For the pure inverse-cube central force set $A=0$. The free-particle reference orbit at $B=0$ is a straight line, $r=d/\cos (\psi -{\psi }_0)$, not a finite-$A$ conic with its old $r_0$ held fixed. To cover attractive as well as repulsive cases completely, putwhere primes denote angle derivatives and $L\ne 0$. The inverse-cube central orbit classification gives the following physically positive-radius branches.

For $\lambda >0$, write $u=C\cos [\sqrt{\lambda }(\theta -{\theta }_0)]$ with $C>0$, so $r=C^{-1}\sec [\sqrt{\lambda }(\theta -{\theta }_0)]$. These are scattering orbits with positive energy and an inner turning radius $1/C$; the angular interval between the two infinite-radius ends is $\pi /\sqrt{\lambda }$. Repulsion ($B>0$) gives an interval smaller than $\pi$; weak attraction ($-L^2/(2m)<B<0$) gives a larger interval and can cause winding. The $B=0$ limit is the straight line.

For $\lambda =0$, $u=C+D(\theta -{\theta }_0)$. If $D\ne 0$, reciprocal-linear spirals connect an infinite-radius end to collision at infinite accumulated angle. If $D=0$ and $C>0$, every radius admits a circular solution at zero energy: attraction exactly balances centrifugal acceleration. Each is a marginal circular orbit under an inverse-cube force, with no radial restoring force.

For $\lambda <0$, put $\beta =\sqrt{-\lambda }$ and $u=C_{+}{e}^{\beta \theta }+C_{-}{e}^{-\beta \theta }$. Up to angle shifts and reversal, the three energy classes arewith positive-radius domains understood. The negative-energy branch is radially bounded but reaches the singular centre on either side of its outer turning point; it is not a regular closed orbit. Collision requires additional physics to continue through $r=0$. Zero angular momentum gives radial trajectories and is excluded from the angular parametrization above.

irrotational flow has $\nabla \times \mathbf{v}=0$, so locally $\mathbf{v}=\nabla \Phi$; the simply connected fluid domain here permits a globally single-valued velocity potential. Incompressibility gives $\nabla \cdot \mathbf{v}=0$, hence ${\nabla }^2\Phi =0$. In a multiply connected domain, zero local vorticity alone would not rule out circulation and a multivalued potential.

Take $k>0$ so the disturbance decays at depth. Differentiating the given candidate potential yieldsThe $x$ and $z$ second derivatives of $\Phi$ cancel in the Laplacian. At $z=0$, $v_z=h_0\omega \cos (\omega t)\sin (kx)={\partial }_{t}Z$, so the linearized kinematic free-surface condition holds. This uses $k{h}_0\ll 1$; advection of the displaced surface is higher order.

Since ${\cos }^2(kx)+{\sin }^2(kx)=1$, the local squared speed is $h_0^2{\omega }^2{\cos }^2(\omega t)e^{2kz}$. The depth integral is thereforeIt is independent of horizontal position; no further spatial average is needed for this kinetic energy. Replacing the upper boundary $Z$ by zero changes it only beyond quadratic order in amplitude.

Relative to the flat surface, the gravitational energy at each horizontal position changes by $\rho g{\int }_0^{Z}zdz=\rho g{Z}^2/2$. Averaging over one wavelength givesThe energy of a deep-water standing gravity wave, averaged over wavelength, is $T+⟨U{⟩}_x$ and must be constant for a free inviscid standing wave, which requires equal coefficients of the complementary squared sine and cosine:As an independent check, the linearized dynamic boundary condition ${\partial }_t\Phi +gZ=0$ at $z=0$ gives exactly $g-{\omega }^2/k=0$. Infinite depth, negligible surface tension and no dissipative forcing are part of this gravity-wave model.

At fixed temperature, equilibrium blackbody radiation has fixed energy density $u$, with radiation pressure $p=u/3$. Doubling the volume therefore doubles its internal energy, so $\Delta U=U_0$. The work done by the radiation isThe first law of thermodynamics, with positive work done by the radiation, givesThus $1.00\mathrm{mJ}$ is absorbed from the walls. The walls must replace the energy used to do work as well as supply the extra stored energy; using only $\Delta U=750\mu \mathrm{J}$ would omit that work. This is a reversible isothermal expansion of a photon gas, not an adiabatic expansion.

Mutual inductance expresses how a changing electric current in one circuit induces an electromotive force in another. Choose winding orientations and define their linked magnetic fluxes byFor a reciprocal linear magnetic system, the same $M$ appears in both expressions. Faraday's law of induction gives ${\mathcal{E}}_2=-Md{I}_1/dt$ when the secondary electric current does not itself change. The sign depends on winding orientation and is consistent with Lenz's law.

The stored magnetic energy isIts nonnegativity for arbitrary electric currents requires $M^2\le L_1{L}_2$. Strong coupling makes $|M|$ approach $\sqrt{L_1{L}_2}$; flux leakage makes it smaller.

An ideal electrical transformer has two windings linked by the same time-varying core magnetic flux. The induced winding voltages are proportional to their turn counts. For appropriately chosen terminal polarities,The electric current convention in the power relation takes primary power into the device and secondary power out to the load. Consequently a load electrical impedance $Z_L$ appears at the primary as $Z_{\mathrm{in}}=(N_1/N_2{)}^2{Z}_L$. A larger secondary voltage therefore comes with a smaller available secondary electric current, rather than extra energy.

A steady direct electric current produces no sustained transformer action after its switching transient. Practical departures from the ideal include winding electrical resistance, leakage magnetic flux, finite magnetizing electric current, core saturation, eddy currents and magnetic hysteresis losses.

Free charge is charge assigned to externally supplied or mobile carriers in a macroscopic description. Bound charge describes the net charge associated with a material's polarization density $\mathbf{P}$. Its bulk and surface densities arewhere the normal points out of the polarized material. Uniform bulk polarization density has no bulk bound charge, but it can still have surface bound charge.

The electric field is sourced by the total electric charge. Introducing the electric displacement field separates the free source in the macroscopic field equation:This does not mean that bound charges are physically absent from $\mathbf{D}$: they are represented through $\mathbf{P}$. In a linear isotropic dielectric, $\mathbf{P}={ϵ}_0{\chi }_e\mathbf{E}$ and $\mathbf{D}=ϵ\mathbf{E}$, with $ϵ={ϵ}_0(1+{\chi }_e)$.