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The zeroth law of thermodynamics states the transitivity of thermal equilibrium: if two systems each equilibrate thermally with a third, they equilibrate with one another. A reproducible thermometric property can therefore label their temperature. Its empirical numerical scale is not itself the complete thermodynamic content; reversible heat-engine relations establish an absolute temperature scale.

The first law of thermodynamics expresses energy conservation. For a closed system, $dU=\delta Q-\delta W$ when $\delta W$ is work done by the system. internal energy is a state function, while heat and work depend on the path. In reversible simple compressible motion, $\delta W=pdV$ and $dU=TdS-pdV$; an open system additionally needs the energies carried by transferred matter. An ideal gas can expand freely with $Q=W=0$ and unchanged internal energy. The first law alone does not decide whether the reverse spontaneous process occurs.

The Second law of thermodynamics introduces entropy and restricts direction. For a reversible path, $dS=\delta Q_{\mathrm{rev}}/T$; for an irreversible process the entropy balance includes nonnegative production. The entropy of an isolated system cannot decrease. In a cycle, the Clausius inequality is $\oint \delta Q/T\le 0$, using the appropriate boundary/reservoir temperatures. The Kelvin–Planck statement forbids a cyclic device whose sole effect is extracting heat from one reservoir and turning all of it into work; the Clausius statement forbids moving heat from cold to hot with no compensating effect. A refrigerator is allowed because it consumes work. A two-reservoir engine has $\eta \le 1-T_c/T_h$, with equality in the reversible limit. Work may be fully dissipated as heat without contradicting this one-way restriction.

The third law of thermodynamics constrains low-temperature entropy. An equilibrium perfect crystal with a nondegenerate ground state approaches zero entropy as $T\to 0$, fixing the conventional absolute entropy reference. Degenerate ground states or frozen nonequilibrium disorder require care when discussing residual entropy. Under regular equilibrium conditions the heat capacity must decrease sufficiently for $\int C/TdT$ to remain finite; a constant positive classical heat capacity cannot be extrapolated to zero temperature. The unattainability formulation states that absolute zero cannot be reached by a finite sequence of admissible cooling operations. These statements concern the limiting thermodynamic regime, rather than predicting that every substance stays in one classical phase during cooling.

The laws work together: conservation sets a device's energy budget, entropy production restricts the attainable budget, and constitutive equations give the properties of its working material. For fixed temperature and volume, Helmholtz free energy decreases toward equilibrium; at fixed temperature and pressure the corresponding potential is Gibbs free energy. Stable equilibrium, phase coexistence and Maxwell relations follow from these potentials with appropriate constraints. Calorimetry, mechanical-work measurements, thermometry and reversible-path comparisons test different parts of this framework. None of the laws by itself supplies a particular gas equation of state or microscopic heat capacity.

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

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

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

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

An electric dipole has equal and opposite charges separated by a small vector from negative to positive, with electric dipole moment $\mathbf{p}=q\mathbf{d}$. More generally, a localized neutral distribution has $\mathbf{p}=\int {\mathbf{r}}^{\prime }\rho ({\mathbf{r}}^{\prime })d^3{r}^{\prime }$, independent of origin. At distances large compared with its size, the leading nonzero term in the multipole expansion isThe field points along $\mathbf{p}$ on either axial side and opposite $\mathbf{p}$ in the equatorial plane; its axial magnitude is twice the equatorial magnitude at equal radius. Near the charges the finite-charge expression must replace the far-field approximation.

A permanent dipole in an externally imposed slowly varying electrostatic field has interaction energy $-\mathbf{p}\cdot \mathbf{E}$, torque $\mathbf{\tau }=\mathbf{p}\times \mathbf{E}$ and force $\nabla (\mathbf{p}\cdot \mathbf{E})$ for fixed $\mathbf{p}$. A uniform field aligns it but gives no net force. For a linearly induced dipole $\mathbf{p}=\alpha \mathbf{E}$, the formation-inclusive energy is $-\alpha E^2/2$, not $-\alpha E^2$: the moment itself grows while the field is applied. Dipole density gives electric polarization, whose bound charges are ${\rho }_b=-\nabla \cdot \mathbf{P}$ and ${\sigma }_b=\mathbf{P}\cdot \hat{\mathbf{n}}$.

For a fixed circuit, Faraday's law of induction readsSurface orientation and contour direction obey the right-hand rule. The negative sign is Lenz's law: a passive induced current acts against the flux change, consistently with energy conservation. The induced electric field generally has nonzero curl and cannot be represented by one single-valued electrostatic potential.

For a material circuit moving with local velocity $\mathbf{u}$, the measured electromotive force includes the magnetic Lorentz force on its carriers:This distinguishes transformer induction in a stationary coil from motional induction in a changing circuit. A straight sliding rod of length $l$ moving perpendicular to itself and to uniform $B$ has $|\mathcal{E}|=Blv$. With circuit resistance $R$, the electrical power $(Blv{)}^2/R$ equals the external mechanical power needed to oppose magnetic drag. An ideal transformer gives $V_2/V_1=N_2/N_1$ in magnitude from common flux linkage; real leakage, resistance and core losses limit that simple result. pickup coils measure flux change rather than an absolute static magnetic field, so integration and a reference state are needed to recover a field history.

In charge- and current-free vacuum, Maxwell equations reduce toTaking a curl and using $\nabla \times \nabla \times \mathbf{E}=\nabla (\nabla \cdot \mathbf{E})-{\nabla }^2\mathbf{E}$ givesand the same wave equation for $\mathbf{B}$. For a plane wave $\mathbf{E}=\mathrm{Re}[{\mathbf{E}}_0{e}^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}]$, $\omega =ck$, $\mathbf{k}\cdot {\mathbf{E}}_0=\mathbf{k}\cdot {\mathbf{B}}_0=0$ and ${\mathbf{B}}_0=\hat{\mathbf{k}}\times {\mathbf{E}}_0/c$. Thus the fields are transverse, mutually perpendicular and in phase; polarization describes the electric-field trajectory at a fixed point.

The energy density and Poynting vector areFor a linearly polarized sinusoid with peak electric amplitude $E_0$, the mean intensity is ${ϵ}_{0}c{E}_0^2/2$. Electromagnetic momentum density is $\mathbf{S}/c^2$; normal absorption produces radiation pressure $⟨S⟩/c$, while ideal retroreflection doubles it. Linear superposition permits interference and standing waves, whose local electric and magnetic fields need not have the traveling-plane-wave equality point by point.

A fluid is described on a continuum scale by density, pressure and velocity fields. An Eulerian field is measured at fixed position; a material element changes at rate $D/Dt={\partial }_t+\mathbf{v}\cdot \nabla$. mass conservation requiresFor incompressible flow this becomes $\nabla \cdot \mathbf{v}=0$. A streamline is instantaneously tangent to velocity, whereas a pathline follows a particle through time; they coincide in a steady flow but need not in an unsteady one.

For an ideal inviscid fluid, pressure and body forces give the Euler equation. In the constant-density gravitational case,A stationary impermeable boundary requires zero normal velocity, but inviscid theory does not impose no slip. Taking its steady projection along a streamline gives $p/\rho +v^2/2+gz=\mathrm{constant}$. If the flow is also irrotational in the connected region, that constant is shared between streamlines. Inviscid does not mean irrotational: rotating ideal-fluid flows are possible. A velocity potential exists locally for an irrotational field, and in incompressible flow it obeys Laplace's equation. The Bernoulli assumptions must be checked before converting a pressure difference into a speed.^[ref]

Real fluids transmit shear stress. An incompressible Newtonian fluid with constant dynamic viscosity $\eta$ obeysThe constitutive shear stress is proportional to strain rate, and stationary solid boundaries usually impose no slip. viscosity diffuses vorticity and converts organized mechanical energy into heat; its local dissipation $2\eta e_{ij}{e}_{ij}$ is nonnegative. It cannot generally be included merely by changing the constant in a lossless Bernoulli equation.

In a long straight circular tube, steady fully developed laminar flow hasThis Hagen-Poiseuille equation follows by balancing the axial pressure gradient with viscous shear and setting $v_z(a)=0$. The parabolic profile, the fourth-power radius sensitivity and the pumping power $\Delta pQ$ are experimentally distinct consequences. Entrance regions, rough walls, non-Newtonian response or a changing temperature require additional modelling.

The Reynolds number $\mathrm{Re}=\rho UL/\eta$ compares inertia with viscosity. Small values favor creeping-flow balances; high values permit thin boundary layers, separation and turbulence, although transition depends on geometry and perturbations, not one universal critical number. At high Reynolds number an outer flow can be approximately inviscid while near-wall viscosity still determines drag. This resolves why a zero-drag ideal potential-flow model cannot predict the resistance of an ordinary moving body. In turbulent flow, fluctuating Reynolds stresses redistribute momentum and increase the required pressure drop; averaging the equations introduces closure requirements.

A controlled tube experiment measures pressure difference and collected volume per time while changing forcing. Use tracers or a velocity probe to test profile shape, allow sufficient entrance length, and monitor radius and temperature independently. Repeat with reversed flow and at several rates to test offsets and linearity. Dye mixing alone does not measure viscosity or establish a fully developed velocity profile. The original figure below separates a specified laminar velocity profile from an independent pressure/discharge measurement.

Use $c_p,c_V$ for the heat capacities of the entire $n$-mole sample. For a stable simple compressible gas, a Maxwell relation giveswhere ${\alpha }_V=V^{-1}(\partial V/\partial T{)}_p$ and ${\kappa }_T=-V^{-1}(\partial V/\partial p{)}_T>0$. This heat-capacity difference and mechanical stability explains the sign without assuming the internal energy depends only on temperature. It is strictly positive for an ordinary expanding gas; the unrestricted word “always” needs the qualification that equality is possible when thermal expansion vanishes.

For an ideal gas, thermodynamic consistency givesConsequently $U=U(T)$, while $H=U+pV=U+nRT$. Differentiating gives $c_p-c_V=nR$: at fixed pressure extra supplied heat permits expansion work in addition to the internal-energy increase.

Differentiate the supplied entropy at fixed $n$:Thus $U=5nRT/2$, consistently recovering $pV=nRT$, andThe coefficient in this entropy corresponds to five active quadratic thermal degrees of freedom per molecule. It should not be replaced by the monatomic value. Reference scales needed to make logarithms dimensionless are absorbed into the entropy constant and do not affect these derivatives.

With open windows the house equilibrates mechanically at approximately the same atmospheric pressure before and after heating. It is an open system: its mole number is not fixed. The isobaric ideal-gas internal-energy density isAt a fixed house volume, $n_{300}/n_{273}=273/300=0.91$: fewer molecules occupy the house, each with greater mean energy. Heating energy can enter while enthalpy leaves with escaping air. Treating the room as a sealed fixed-$n$ gas would give the wrong energy-density comparison.

Keep the problem's one-mole convention throughout. The excluded-volume parameter $b$ represents the loss of available translational volume due to short-range repulsion; it is an effective covolume, not simply the literal molecular material volume. The attraction parameter $a$ reduces the pressure below its purely repulsive value. In a mean-field picture, interaction energy is proportional to pair density, so the attractive free-energy contribution scales as $-a/V$ for one mole and the pressure correction as $-a/V^2$. Equivalently, molecular cohesive forces and the number striking a wall each supply a density factor. These parameters approximate microscopic correlations rather than measuring every pair potential exactly.

At fixed temperature, $dF=-pdV$. Substituting $p=RT/(V-b)-a/V^2$ and integrating gives the isothermal free-energy change of a Van der Waals gasThe arbitrary temperature-only term cancels. The internal-energy derivative is insteadExpansion raises the attractive internal energy toward zero. As a check, the entropy change is $R\ln [(V_2-b)/(V_1-b)]$, so $\Delta F=\Delta U-T\Delta S$. Constant $a,b$ and a homogeneous branch with $V>b$ are assumed. If a subcritical equilibrium path crosses coexistence, a Maxwell construction replaces the unstable homogeneous continuation; the displayed integral is the direct equation-of-state result, not a claim that such a continuation is physically stable.

At the Van der Waals critical point, the critical isotherm has a stationary inflection:Equating the resulting expressions for $R{T}_c$ gives $2V_c=3(V_c-b)$, henceIn reduced variables $x=V/V_c$, the critical curve is $p/p_c=8/(3x-1)-3/x^2$: it decreases through $(1,1)$ with zero first and second derivative, rather than having separate liquid/gas turning points at the critical temperature.

A microstate specifies the microscopic state: classically the positions and momenta of all particles, or quantum mechanically an allowed many-body state. A macrostate specifies a few coarse variables such as energy, volume and particle number. Many microstates can give the same macrostate. A statistical ensemble is a probability assignment over hypothetical copies prepared with the same macroscopic constraints; it does not require a sample to be literally divided into many physical copies. An observable's ensemble mean is $⟨A⟩={\sum }_i{p}_i{A}_i$, with an integral replacing the sum in classical phase space.

For an isolated system at fixed $U,V,N$, the microcanonical ensemble assigns equal probability to accessible microstates in a narrow energy shell. If their number is $\Omega (U,V,N)$, the Boltzmann entropy is $S=k_B\ln \Omega$. Thermal contact between two initially isolated systems allows energy to be shared while their sum stays fixed. The most probable division maximizes $S_1(U_1)+S_2(U-U_1)$, so equilibrium requires equal $\partial S/\partial U$. This gives the statistical temperature definitionFor positive temperature, energy tends to flow from lower entropy slope to higher entropy slope until those slopes agree. The narrow energy-shell width represents macroscopic resolution; it must not exclude the physically accessible states.

For a small subsystem exchanging energy with a much larger bath, the bath has energy $U_{\mathrm{tot}}-E_i$ when subsystem microstate $i$ has energy $E_i$. Expanding its entropy gives $S_b(U_{\mathrm{tot}}-E_i)=S_b(U_{\mathrm{tot}})-E_i/T+\cdots$. Counting compatible bath states therefore yields the Boltzmann distribution and canonical ensemble:The exponential weights individual microstates. An energy level of degeneracy $g_j$ has total probability $g_j{e}^{-\beta E_j}/Z$; a more degenerate excited level can outweigh one ground-state level even though each of its microstates is less probable. The Gibbs entropy $-k_B{\sum }_i{p}_i\ln p_i$ gives $S=k_B\ln Z+U/T$. The equilibrium energy variance is $⟨(\Delta E{)}^2⟩=k_B{T}^2{c}_V$, linking measurable fluctuations to a macroscopic response.

If particle exchange is also allowed, a reservoir fixes $T$ and chemical potential $\mu$. The grand canonical ensemble weights a state by $e^{-\beta (E_i-\mu N_i)}/\Xi$, with $\Xi ={\sum }_{N,i}{e}^{-\beta (E_i-\mu N)}$. Mean particle number is $k_{B}T\partial \ln \Xi /\partial \mu$, while volume remains a constraint in this version. Thus the ensemble follows the physical exchange conditions: fixed energy, thermal contact, or thermal and particle contact.

In ordinary large short-range systems, relative fluctuations often decrease as $N^{-1/2}$ and the ensembles give the same bulk thermodynamics away from singular limits. Finite systems, critical fluctuations and long-range interactions require care. Interpreting a time average as an ensemble average additionally requires appropriate sampling or mixing dynamics; it is not guaranteed solely by naming an ensemble. thermal equilibrium probability assignments do not describe every driven or relaxing state.

Self-inductance relates current to its own magnetic flux linkage: $\Lambda =LI$, so the induced electromotive force is $-L\dot{I}$ for fixed geometry. A long solenoid has $B={\mu }_{0}NI/l$ and area $\pi a^2$. Its $N$ turns giveThe homopolar generator uses a disc rotating through this axial field; integrate the motional Lorentz force from shaft to rim:For a homopolar disc dynamo, choose the wiring and rotation sense giving positive magnetic feedback. Kirchhoff's voltage law isfor prescribed constant $\omega$. ThusA nonzero seed grows when $\omega >{\omega }_c$, decays when $\omega <{\omega }_c$, and is marginal at equality. This explains the printed $\omega \ge {\omega }_c$ as the threshold including its boundary; equality does not itself give exponential growth. An exactly zero initial current remains zero in the ideal homogeneous equation, and reversing the feedback sense prevents the instability.

In the torque-controlled homopolar dynamo equilibrium, a radial current element exerts a braking torque $d{G}_B=IBrdr$, so $G_B=IB{a}^2/2=C{I}^2$. For steady rotation and nonzero steady current, $G=C{I}^2$ and $\omega ={\omega }_c$. Hence the field intensity magnitude isThe two current signs give opposite magnetic polarities. Finally,The steady magnetic and rotational energies do not change. This is a steady-state balance, not an assertion that every dynamical trajectory must approach that state.

Insert $\nabla \times \mathbf{E}=-{\partial }_t\mathbf{B}$ and $\nabla \times \mathbf{H}=\mathbf{J}+{\partial }_t\mathbf{D}$ into the given identity. For a linear, time-independent, nondispersive medium, $\mathbf{D}=ϵ\mathbf{E}$ and $\mathbf{B}=\mu \mathbf{H}$, givingThe divergence theorem then yieldsThe left side is outward electromagnetic power flow. The current term is power delivered to matter, becoming Joule heating for an Ohmic conductor, and ${\partial }_{t}u$ is the change in stored electric plus magnetic energy. The minus sign says outward flux reduces stored energy unless supplied locally.

Follow the drawing's coordinates: lower strip at $z=0$, upper strip at $z=d$, current $I$ along $+x$ in the upper strip and back along the lower. The wording “both strips lie in the $xy$-plane” must mean parallel to that plane, since the drawing separates them in $z$. Neglect the side and end fringes using $l\gg a\gg d$. Then $I=V_0/(R+R_L)$ and the top-strip potential is $V(x)=V_0-IRx/l$, while the lower strip is equipotential at zero.

The harmonic function matching those two surfaces isIndeed ${\nabla }^2\varphi =0$, and it has the required potentials at $z=0,d$. ThusThe two current sheets reinforce their magnetic field between the strips; its direction follows the right-hand rule. The Poynting flow in resistive parallel strips follows from their cross product:Power flows along the gap towards the load and upwards into the resistive strip. Across a transverse area $ad$, $P_x(x)=IV(x)$ falls from $I{V}_0$ to $I^2{R}_L$. The upper surface receives ${\int }_0^{l}a{S}_z(z=d)dx=I^{2}R$, exactly the strip's heating; the ideal lower strip receives none. In the charge-free gap, ${\partial }_x{S}_x+{\partial }_z{S}_z=0$, consistent with a static field and no local dissipation. The small longitudinal electric field cannot be discarded if this upward energy transfer is to be represented.

Write $\mu =GM$ and $A=\mu m$. For a circular orbit of radius $r$, centripetal balance gives $v^2=\mu /r$. Thus $E=m{v}^2/2-A/r=-A/(2r)$ and $J^2=m^2{r}^2{v}^2=Amr$. Since $e=0$, $a=r_0=r$, these agree with the general expressions.

For an outward coplanar Hohmann transfer, burn prograde at the inner circle to put its radius at the ellipse's pericentre, then burn prograde at apocentre to circularize at the station's radius. The station must be phased to meet the spacecraft there. With $r_p=R$, $r_a=\alpha R$,Energy at the first burn, with $v_0^2=\mu /R$, givesAngular momentum is conserved along the transfer ellipse, so $v_{2-}=v_1/\alpha$, while the outer circular speed is $v_{2+}=v_0/\sqrt{\alpha }$. For $\alpha =3$,The correct second impulse adds $\Delta v=v_{2+}-v_{2-}$. The reversed circularization burn in a Hohmann transfer, using the same impulse magnitude, instead leavesIts magnitude is below the circular speed at $3R$, so that point is the new apocentre. Define $q=r_a{v}_w^2/\mu =(\sqrt{2}-1{)}^2=3-2\sqrt{2}$. For an ellipse at apocentre, $q=1-e_w$, givingThe lunar radius is $R/3$. Since $r_{p,w}<R/3$, the osculating orbit intersects the Moon: the spacecraft crashes before reaching its formal pericentre. The calculation uses impulsive tangential burns and a spherical central gravitational field; it does not assume that a trajectory continues through the solid body.

The original plot uses $R$ as the length unit and $R/3$ as the lunar radius. The initial station position shown is phased for rendezvous: its initial angular lead is $\pi -\pi [(1+\alpha )/(2\alpha ){]}^{3/2}$, modulo $2\pi$.

In a charge-free region of homogeneous permittivity, Gauss's law and $\mathbf{E}=-\nabla V$ give Laplace's equation, ${\nabla }^{2}V=0$. Prescribed surface electric potentials specify a Dirichlet problem. Its Uniqueness of the Dirichlet problem permits replacing conductors by fictitious charges outside the physical solution region: a harmonic candidate with the same boundary conditions is the physical solution. For an exterior problem its behaviour at infinity must also be prescribed. For a proof, subtract two candidate potentials to obtain $u$ with zero boundary values. Green's first identity gives $\int |\nabla u{|}^2{d}^{3}x=0$, so $u$ is constant and its boundary value fixes it to zero. Conditions on a normal electric field instead determine a potential only up to an additive constant when the usual compatibility condition holds.

A cylindrical Gaussian surface around an infinite line charge gives $E_r=\lambda /(2\pi {ϵ}_{0}r)$. The electric potential difference is thereforeAn isolated infinite line charge has no finite potential zero at infinity. For the opposite pair, however, the logarithmic growth terms cancel and one can choose $V\to 0$ there. Define $r_b^2=(x-b{)}^2+y^2$ and $r_c^2=(x-c{)}^2+y^2$. superposition givesOn $x^2+y^2=bc$,Thus the circle, and its cylindrical extension, is an equipotential. This is an Apollonius circle: a constant ratio of distances gives a circle rather than a constant distance from either charge.

For the cylinder above the plane, call its centre height $H>a$ and set $s=\sqrt{H^2-a^2}$. The cylinder-plane image charges are $+\lambda$ at height $s$ and $-\lambda$ at height $-s$. Their distances are equal on the grounded plane, while their distance ratio is $(H+s)/a$ on the circular surface. The positive fictitious line lies inside the metal, so the exterior is charge-free and Uniqueness of the Dirichlet problem applies. A Gaussian surface enclosing the physical cylinder confirms that its charge per unit length is indeed $\lambda$. Its voltage isThe physical field energy per unit length at fixed cylinder charge is $U^{\prime }=\lambda V_c/2$. Differentiate with respect to height, keeping $a$ and $\lambda$ fixed:The electrostatic attraction per unit length of a cylinder above a plane is downward for either sign of charge. Placing the fictitious line at the cylinder centre would give the wrong answer at this separation. The long-cylinder approximation neglects the ends.

Use peak-amplitude phasors with time factor $e^{-i\omega t}$ and $\mu ={\mu }_0$. Faraday's law of induction and the Ampere-Maxwell law giveIn the good-conductor limit $\sigma \gg \omega ϵ$, take the root with positive attenuation into the slab:The real physical field is $\mathrm{Re}[{\tilde{E}}_y{e}^{-i\omega t}]$. A thick slab excludes a returning wave from its rear face. The skin depth is the electric-field amplitude decay length; the intensity decay length is $\delta /2$.

Faraday's law of induction fixes ${\tilde{H}}_z=k{\tilde{E}}_y/({\mu }_0\omega )$. The surface impedance of a good conductor and the complete transmitted magnetic field are consequentlyThe sign of the imaginary part belongs to the stated time convention. Average the Poynting vector using complex conjugation and the factor one half appropriate to peak phasors:As an independent energy check, $-{\partial }_x⟨N_x⟩=\sigma |E_0{|}^2{e}^{-2x/\delta }/2$, the mean Joule heating per unit volume.

Ohm's law gives ${\tilde{J}}_y=\sigma {\tilde{E}}_y$ and ${\tilde{B}}_z={\mu }_0{\tilde{H}}_z$. Hence the averaged Lorentz force per unit volume isIntegrating through the illuminated half-space gives the positive inward radiation pressureFinally let $Z_0=\sqrt{{\mu }_0/{ϵ}_0}$ be the vacuum wave impedance. Tangential-field boundary conditions at normal incidence give the electric-amplitude reflection coefficient $r=(Z-Z_0)/(Z+Z_0)$. Since $|Z|\ll Z_0$,Here $N_0=(1-R)I_{\mathrm{inc}}$ by surface energy balance. Combining the leading formulas gives $p\simeq 2I_{\mathrm{inc}}/c$, the nearly perfect-reflector limit. The displayed reflection expansion also assumes a good conductor relative to the incident vacuum wave, $\omega {ϵ}_0/\sigma \ll 1$.

For an attractive inverse-square force, write $V(r)=-\kappa /r$ with $\kappa >0$, and take nonzero conserved angular momentum $J$. The conic section orbit hasThe physically allowed cases are ellipse for $-m{\kappa }^2/(2J^2)\le E<0$, parabola for $E=0$, and hyperbola for $E>0$. The minimum allowed energy gives the circular special case $e=0$. Here $E$ is measured relative to infinity. Zero angular momentum instead gives radial collision or escape trajectories, interpreted as degenerate limits rather than regular closed ellipses.

Measure $\theta$ from the upward vertical, let ${\mathbf{e}}_3$ point from the support along the symmetry axis, and use the support as the reference point for all moments of inertia. Choose ${\mathbf{e}}_2=(\hat{\mathbf{z}}-\cos \theta {\mathbf{e}}_3)/\sin \theta$ and ${\mathbf{e}}_1={\mathbf{e}}_2\times {\mathbf{e}}_3$. These form an instantaneous principal-axis frame; its transverse axes need not spin with marks painted on the body. The three infinitesimal Euler angles rotations add as vectors:Resolving the vertical direction givesA convention matters here: in genuinely material-fixed transverse axes, the first two components are rotated by the spin angle $\chi$. For one conventional choice they are $\dot{\theta }\cos \chi +\dot{\varphi }\sin \theta \sin \chi$ and $-\dot{\theta }\sin \chi +\dot{\varphi }\sin \theta \cos \chi$. Since $I_1=I_2$, that rotation changes neither kinetic energy nor the axial and vertical conserved momenta. The simplified displayed components use the instantaneous transverse frame just defined.^[ref]

For a frictionless fixed support, the support does no work and gravity is conservative, so the total energy is constant. The heavy symmetric top Lagrangian isThe two cyclic coordinates $\chi ,\varphi$ giveEquivalently, gravitational torque has no component along the vertical or symmetry axis. Eliminate $\dot{\varphi }=(J_z-J_3\cos \theta )/(I_1{\sin }^2\theta )$ and the axial spin from $E=T+V$:This is the heavy symmetric top reduction at fixed $J_3,J_z$. A constant inclination must satisfy the reduced equation of motion $I_1\ddot{\theta }=-U_{\mathrm{eff}}^{\prime }$, not merely have zero instantaneous velocity. Differentiation at fixed conserved momenta gives, for $0<\theta <\pi$ and $\Omega =\dot{\varphi }$,Therefore uniform precession of a heavy symmetric top requiresFor a prescribed upward inclination $0<\theta <\pi /2$, real precession rates exist only if the quadratic discriminant is nonnegative:For positive spin the absolute-value sign can be dropped. The two roots are the slow and fast branches and coalesce at the threshold. The familiar upright-limit critical magnitude is $2\sqrt{I_{1}mgh}$; strict inequality gives gyroscopic stabilization of an inverted symmetric top. This limiting stability condition should not be confused with a positive minimum over every possible inclination. As $\theta$ approaches $\pi /2$ from below, the fixed-inclination threshold approaches zero. At the horizontal angle $\Omega =mgh/J_3$ for $J_3\ne 0$; below the horizontal the quadratic admits real solutions even at $J_3=0$. The stated minimum therefore needs the inclination convention made explicit.

Use the Euler-Bernoulli beam equation for a uniform slender elastic beam, small deflection and small slope, neglecting axial tension, shear deformation, rotational inertia of a cross-section and damping. Let $E$ be Young's modulus, $I={\int }_A{z}^{2}dA$ the second moment of area about the neutral axis, and $\rho$ the mass per unit length. If density instead denotes mass per volume, replace this $\rho$ by ${\rho }_{\mathrm{bulk}}A$. The bending stiffness is $EI$; the small-slope curvature is $y_{xx}$. The bending and kinetic energies areTwice integrating the variation of $U$ by parts gives the restoring load $-EI{y}_{xxxx}$ per unit length. Newton's second law therefore givesAt a pivot, displacement is fixed but slope is free; zero end bending moment gives $EI{y}_{xx}=0$. This also follows from the natural boundary term $EI{y}_{xx}\delta y_x$ in the variation.

With separation of variables, $y=XT$ givesChoose the time normalization $T(0)=1$. Initial rest removes the sine time term, so $T(t)=\cos \omega t$. The spatial characteristic roots are $\pm ik,\pm k$, givingThe two conditions at $x=0$ give $B+D=0$ and $-B+D=0$, hence $B=D=0$. At $L$, addition and subtraction of $X(L)=X^{\prime \prime }(L)=0$ give $C\sinh kL=0$ and $A\sin kL=0$. Nontrivial positive-frequency modes therefore have $C=0$ and $k_n=n\pi /L$. Thus the simply supported beam normal modes areThe zero-frequency cubic solution also vanishes under these four endpoint conditions, so $n=0$ is not an additional mode.

For the specified release, small slope requires $|a|L\ll 1$. Expand the initial deflection in the Fourier sine series. Its coefficients areWriting ${\omega }_1={\pi }^2\sqrt{EI/\rho }/L^2$, the resulting motion isOnly odd normal modes occur because the initial profile is symmetric about the midpoint, and their amplitudes decrease as $n^{-3}$. The frequencies grow as $n^2$, so the motion is dispersive and is not a single shape multiplied by one cosine. Nevertheless the ideal undamped solution is exactly periodic with period $2\pi /{\omega }_1$. At $t=\pi /(2{\omega }_1)$ all odd-mode cosines vanish, so the bar is instantaneously straight; at $t=\pi /{\omega }_1$ it has the opposite initial deflection. The first mode dominates displacement, with third-mode amplitude only $1/27$ of the first.