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For an oriented piecewise smooth surface $S$ with boundary curve $C=\partial S$, and a continuously differentiable vector field $\mathbf{F}$ near $S$, Stokes theorem statesHere $\mathbf{n}$ is the chosen unit normal, $dS$ is positive surface area, and $d\mathbf{r}$ is the directed tangent line element. The positive traversal of $C$ is fixed by the right-hand rule relative to $\mathbf{n}$. Thus the flux of the curl through the surface equals the circulation along its oriented boundary.

Let $V\subset {\mathbb{R}}^3$ be a bounded region with piecewise smooth closed boundary $\partial V$, let $\mathbf{n}$ be its outward unit normal vector, and let $\mathbf{A}$ be a continuously differentiable vector field on a neighbourhood of its closure. The divergence theorem statesHere $dV$ is the positive volume measure, $dS$ is the surface area measure, $\nabla \cdot \mathbf{A}={\partial }_x{A}_x+{\partial }_y{A}_y+{\partial }_z{A}_z$, and the right side is the total outward flux. All boundary components must be included, with their normals pointing away from $V$.

For a nondegenerate surface take $R>0$, and abbreviate $C=\cosh R$, $H=\sinh R$. Close the upper oblate spheroid by its equatorial disk $D$, obtaining a half-ellipsoid $V$. The divergence isThe integral of $2y$ vanishes by reflection symmetry. At height $z$, the cross-section is a disk of radius $q=C\sqrt{1-z^2/H^2}$. Using planar polar coordinates,ConsequentlyThe outward unit normal vector on $D$ is $-{\mathbf{e}}_z$, so its flux is $-2\pi C^2$. Applying the divergence theorem to the closed surface and subtracting this disk contribution givesThis is a special case of polynomial flux through a half-ellipsoid. Including the closing disk is essential: the constant vertical component has zero divergence but nonzero flux through the open curved surface.

Since $\mathbf{F}=\nabla \Omega$, its divergence is the Laplacian of $\Omega$:The divergence theorem changes the outward flux through the whole closed boundary into $I={\int }_V(6x+4z)dV$. Rotational symmetry makes the $x$ term vanish. With $z=(u^2-v^2)/2$ and the Jacobian determinant from part (c),Both the curved face and the top disk are included in this surface integral through the divergence theorem.

The three-dimensional fundamental solution of the Laplace equation, with $\Delta F=+\delta$, isIt is harmonic away from the source, and its gradient has total outward flux one through a small source-centred sphere.

For $s=|{\mathbf{r}}^{\prime }|\in (0,1)$, place the image at ${\mathbf{r}}^{\ast }={\mathbf{r}}^{\prime }/s^2$, outside the unit ball. On $|\mathbf{r}|=1$, expansion of squared distances gives $|\mathbf{r}-{\mathbf{r}}^{\ast }|=|\mathbf{r}-{\mathbf{r}}^{\prime }|/s$. The method of images therefore gives the Dirichlet Green function inside a sphereThe boundary value is zero by the distance identity, and the image term is harmonic throughout the physical ball. Thus $\Delta G_0=\delta (\mathbf{r}-{\mathbf{r}}^{\prime })$ there. At the central source, take the finite limit of the weighted image term:

Let $a=\sinh 1$ and $c=\cosh 1$. The region is the first-octant portion of the ellipsoid $(x^2+y^2)/a^2+z^2/c^2\le 1$. In cylindrical coordinates, write $\rho =\sqrt{x^2+y^2}$ and $0\le \varphi \le \pi /2$. Away from the axis, the divergence isThe field is singular on the axis, so applying the classical divergence theorem there without qualification would be invalid. Instead remove $\rho <\epsilon$ and apply it to the remaining region. On the added cylindrical boundary the unit normal is $-{\mathbf{e}}_{\rho }$, andIts area element is $\epsilon d\varphi dz$, so its flux is $O(\epsilon |\log \epsilon |)\to 0$. The axial component of the field has zero dot product with this normal. The volume integral of the divergence is locally integrable, because its $1/\rho$ factor cancels the cylindrical measure. Thus removing a singular axis in a flux integral gives the desired improper surface integral with no extra axial term.

The divergence theorem now reduces the calculation toThe integral over $s$ is the area of a quarter unit disk. All boundary fluxes are interpreted as convergent improper integrals at the axis.

For a continuously differentiable vector field $\mathbf{u}$ on a bounded region $V\subset {\mathbb{R}}^3$ with piecewise smooth closed boundary $S=\partial V$, the divergence theorem statesHere $\nabla \cdot \mathbf{u}$ is its divergence, $dV$ is the volume element, $dS$ is the surface area element, $\hat{\mathbf{n}}$ is the outward unit normal, and $d\mathbf{S}=\hat{\mathbf{n}}dS$. Thus a volume integral of the divergence equals the outward flux through the boundary.

For an admissible nonzero real trial function, integration by parts and the zero-flux boundary conditions reduce the Rayleigh quotient toA nonzero constant trial function attains zero. Positivity of $p$ also shows that equality requires $y^{\prime }=0$. HenceIf the conserved total amount is $M={\int }_{-L}^{L}cdx$, the normalized stationary solution isIts diffusive flux vanishes everywhere. The Boltzmann distribution fixes the spatial profile, while the diffusivity affects the relaxation toward it. Boundedness of $V$ makes the normalization finite on this finite interval.

Close the hyperboloid segment with the two disks at $z=\pm 1$, and let $V$ be its interior. The flux through either disk is zero because $A_z=0$. Thus the desired surface integral is the total outward flux through the closed boundary. Sincethe divergence theorem and cylindrical coordinates giveThe extra factor $r$ is the volume integral measure in cylindrical coordinates; omitting it would give the wrong flux.

Feynman diagrams organize the perturbative calculation of the matrix element in Fermi's golden rule. Each external line labels an incoming or outgoing particle state; each internal line contributes a propagator for a generally off-shell intermediate field; and each interaction vertex contributes a coupling and the appropriate spin, momentum and internal-symmetry structure. All indistinguishable allowed diagrams must be summed as probability amplitudes before taking the absolute square, so their interference contributes to the rate.

In the radiative pion decay diagram in part (b), solid arrowed lines are fermions; the wavy line joining the hadronic and leptonic vertices is a W boson propagator; and the wavy outgoing line is an external photon. Arrows track fermion number, so an outgoing antimuon has an arrow towards its production vertex. Each vertex conserves four-momentum; an internal propagator need not satisfy the on-shell energy–momentum relation. A closed loop Feynman diagram additionally requires an integration over an internal four-momentum.

The final Lorentz-invariant phase space, conservation Dirac delta functions, and the initial-state normalization or incident flux turn the squared scattering amplitude into a decay rate or cross section. A diagram supplies the dynamics; it is not itself the final density of states.

An accelerating point charge has a transverse radiation field in addition to its near field. At nonrelativistic speed and large distance $R$,The acceleration is evaluated at retarded time. Radiation fields decay as $1/R$, so their outward energy flux integrated over a sphere is independent of radius; the near electric field instead decays as $1/R^2$. The radiation vanishes along the acceleration and is strongest perpendicular to it.^[ref]

Using the Poynting vector givesThe integral uses $\int {\sin }^2\vartheta d\Omega =8\pi /3$. This is the Larmor formula. A harmonically accelerated charge with peak acceleration $a_0$ has mean power $q^2{a}_0^2/(12\pi {ϵ}_0{c}^3)$; external work supplies this radiated energy in steady operation. Uniform inertial motion does not radiate.

At relativistic speed the angular pattern becomes directionally concentrated and retarded-source factors matter. The total instantaneous radiated power generalizes to $q^2{\gamma }^6[a^2-|\mathbf{\beta }\times \mathbf{a}{|}^2]/(6\pi {ϵ}_0{c}^3)$, with $\mathbf{\beta }=\mathbf{v}/c$. The displayed simple ${\sin }^2\vartheta$ pattern is restricted to the nonrelativistic limit.

Take $x^{\mu }=(ct,x,y,z)$ and metric signature $(+,-,-,-)$. The four-current, four-potential and four-gradient areThe first two transform as contravariant four-vectors under an inertial-frame Lorentz transformation, within a consistently transformed electromagnetic gauge. The lowered-index gradient is a covector. Their contraction is a Lorentz scalar:Indeed, $J^{\prime \mu }={\Lambda }^{\mu }{}_{\nu }{J}^{\nu }$ and ${\partial }_{\mu }^{\prime }=({\Lambda }^{-1}{)}^{\alpha }{}_{\mu }{\partial }_{\alpha }$, so the two matrices cancel in the contraction. Local charge conservation in one inertial frame implies local charge conservation in every inertial frame. Integrating over a region gives the charge change equal to minus its outward current flux; global conserved charge requires no flux through the system boundary.

A Hertzian dipole is an ideal localized oscillating source much smaller than its radiation wavelength. Its electric version is an infinitesimal current element; its magnetic version is a small current loop with moment $\mathbf{m}=IA\mathbf{n}$. Such elements model the far-field radiation and radiation resistance of electrically small antennas. The approximation requires source dimensions much smaller than $c/\omega$.

The supplied magnetic field has a quasistatic $[m]/r^3$ term, an induction $[\dot{m}]/(c{r}^2)$ term and a transverse radiation $[\ddot{m}]/(c^{2}r)$ term. The electric field has induction and radiation terms. Brackets mean evaluation at the retarded time $t_r=t-r/c$, not a time average. Near terms mainly exchange stored energy; only the $r^{-1}$ fields give a nonvanishing outgoing power integral as $r\to \infty$.

For a fixed dipole axis the radiation fields satisfy $E_{\varphi }=-c{B}_{\theta }$. The radial Poynting flux is thereforeUsing $\int {\sin }^2\theta d\Omega =8\pi /3$ yieldsFor a general changing dipole direction, replace the squared scalar acceleration by $|\ddot{\mathbf{m}}{|}^2$.

Radiation resistance is the equivalent input resistance whose mean power equals radiated power, not ohmic heating. For a fixed small loop carrying $I(t)=I_0\cos \omega t$, $m=AI$, soThis uses peak current $I_0$; with rms current the power is $I_{\mathrm{rms}}^2{R}_{\mathrm{rad}}$.

The satellite loop needs its changing orientation as well as its induced current. Put the spin axis along z, the external field along x, and let $\mathbf{n}(t)=(\cos \omega t,\sin \omega t,0)$. With $A=\pi b^2$ and resistance $R=10^{-4}\Omega$, the resistance-only approximation givesWrite $m_0=B{A}^2\omega /R$. The full magnetic moment isIt has a static component and a rotating component at $2\omega$. Consequently $|\ddot{\mathbf{m}}{|}^2=4m_0^2{\omega }^4$, andHolding the dipole axis fixed while inserting the sinusoidal current would instead give $B^2{A}^4{\omega }^6/(12\pi {ϵ}_0{c}^5{R}^2)$, eight times smaller. That is the fixed-axis-loop result; it does not describe the stated rotating loop. This is frequency doubling of an induction-driven rotating loop; the changing vector direction is essential.

The resistance-only model is conditional: loop inductance is not supplied. If self-inductance L matters, solve $L\dot{I}+RI=BA\omega \sin \omega t$. In the steady state $I_0=BA\omega /\sqrt{R^2+{\omega }^2{L}^2}$. Its phase changes the static moment direction but not the twice-frequency component's magnitude, so the power becomesThe first expression is the requested resistive idealization and an upper estimate within this passive RL correction. Radiation reaction is negligible compared with the stipulated ohmic resistance. The radiated energy comes from rotational kinetic energy: the induced magnetic torque brakes the loop. Most lost rotational energy is Joule heating; a small part escapes as radiation. Maintaining a constant spin would require external mechanical power.

For scale only, take $B=50\mu \mathrm{T}$, a representative near-Earth field rather than a supplied orbital measurement. The terrestrial surface range reported by NOAA is roughly $25$–$65\mu \mathrm{T}$; the actual orbital value depends on position.^[ref] With $b=1\mathrm{m}$ and $\omega =2\pi (500{\mathrm{s}}^{-1})$, the resistance-only model givesInductance would reduce this power. The kilohertz signal is exceptionally weak and a metre-scale receiving antenna is electrically tiny. Ionospheric absorption/reflection and strong natural or instrumental low-frequency backgrounds make ordinary ground reception impractical. This is not a theorem that the ionosphere blocks every low-frequency mode: field-guided propagation and condition-dependent leakage are real, as the VIPER mission's measurements were designed to test.^[ref]

At a separation $r=100\mathrm{km}$, $k_{\mathrm{rad}}r\simeq 2.1$, so the observer is not securely in the far field. The radiation term alone has at most the characteristic magnetic amplitudein this optimistic resistance-only vacuum model. Near and induction fields must be retained for a full prediction; a simple far-field inverse-square power estimate is not an exact received signal here. Another satellite avoids the whole ground ionospheric path, but plasma propagation, receiving sensitivity, bandwidth, orientation, integration time and spacecraft noise remain unspecified. Detection is not forbidden in principle, but routine practical detection is implausible at this scale; a definitive instrument-level yes/no does not follow from the supplied data.

In the magnetic Aharonov-Bohm effect, coherent electron paths enclose inaccessible magnetic flux even though the field vanishes along the paths. The magnetic vector potential contributes relative phaseRecombination turns this phase into a fringe displacement. For an electron $q=-e$, changing flux by $h/e$ gives one full fringe period. Reversing the loop orientation reverses the phase sign. A gauge change modifies each path's phase convention but leaves the closed-loop phase and observed interference unchanged. The accessible region surrounds excluded flux, so a globally trivial gauge cannot remove that phase.

A shielded toroidal magnet provides a concrete test: electron holography compared waves passing through its hole and outside it. Superconducting and copper coatings confined the field and excluded electrons from the magnet; the measured relative phase tested enclosed flux rather than a classical Lorentz deflection along the paths.^[ref]

The differential vacuum Gauss's law isIntegrate over a volume $\Omega$ with closed boundary $S$. The divergence theorem converts the left side into outward electric flux:This is the integral form. Surface charge can be treated by a thin-layer limit or by the corresponding jump in the normal field; the enclosed charge then includes the charged sheet. The surface orientation must be outward, and the field entering the integral is the total field, not merely the field of the enclosed charges alone.

For an ideal inductor of constant self-inductance $L$, the passive-sign electric potential difference is $V_L=LdI/dt$. The supplied power is $V_{L}I$, so increasing the electric current quasistatically from zero storesAny electrical resistance would additionally dissipate energy as heat; it is not included in this stored magnetic energy.

For a long vacuum solenoid with $N$ turns, axial length $\ell$ and cross-sectional area $A$, Ampère's law gives $B={\mu }_0(N/\ell )I$. Its flux linkage is $N\Phi =NBA$, soThe nearly uniform field occupies volume $A\ell$, with negligible external-field energy in the long-solenoid approximation. Dividing its stored energy by this volume yieldsThe vacuum and linear-inductance assumptions matter; a nonlinear magnetic material requires the appropriate field-energy integral rather than simply substituting a varying inductance into $L{I}^2/2$.

Stokes theorem converts circulation around an oriented boundary into curl flux through a spanning surface. For a sufficiently smooth vector field on an oriented piecewise smooth surface $S$,The boundary orientation follows the right-hand rule relative to $\mathbf{n}$. The field must be smooth on the surface, not merely on its boundary; a singularity crossing the surface cannot be ignored. For a fixed loop, applying the theorem to Faraday's differential law gives the integral flux law of induction.

The planar form is Green theorem. For a region $D$ with positively oriented boundary and sufficiently smooth $P,Q$,The outer boundary is counterclockwise and any holes have clockwise boundaries. For instance $P=-y/2,Q=x/2$ makes the area of $D$ equal to half the line integral of $xdy-ydx$; the unit circle gives area $\pi$.

A closely related use of “Green's theorems” in field theory is Green's identities, obtained by applying the divergence theorem to $\varphi \nabla \psi$:and, after subtracting the expression with $\varphi ,\psi$ exchanged,These relate interior sources to boundary data in electrostatics and other boundary-value problems. For example, a harmonic potential difference vanishing on a Dirichlet boundary condition has zero integral of its squared gradient by the first identity, establishing uniqueness. Both the planar circulation theorem and the field identities are included here to make the convention clear.

On the curved wire surface, the outward normal is $\hat{\mathbf{r}}$ and the Poynting vector has the constant radial value $N_r(a)=-I^2/(2\sigma {\pi }^2{a}^3)$. Its outward flux isThe Poynting vector is radial, so it has zero normal component on the two end faces in this approximation. This is therefore also the outward flux through the complete wire surface. Equivalently, positive electromagnetic power $P_{\mathrm{loss}}$ flows inward.

The fields are steady, so the stored electromagnetic energy does not change. Poynting's theorem consequently requires the incoming power to equal the Ohmic heating in part (e), exactly as found. The energy reaches the wire through the surrounding electromagnetic field. A local check is $\nabla \cdot \mathbf{N}=r^{-1}{\partial }_r(r{N}_r)=-I^2/(\sigma {\pi }^2{a}^4)=-\mathbf{E}\cdot \mathbf{J}$.

Let the approximately conserved magnetic flux be $\Phi$. ThenThe length $L$ shown in the original diagram is the untapered core arc; the two tapers are separate lengths $s$ each. In the untapered core, flux conservation givesFor one taper use distance $u$ from its narrow end. The straight taper in the diagram has local radiusand magnetic field intensityIts contribution to the line integral in the magnetostatic Ampère law is found explicitly:Adding the gap, the untapered arc and both tapers givesThereforeThis uses a linear radius profile, approximately uniform magnetic flux density over each section, a linear constant magnetic permeability, and negligible leakage and magnetic fringing. For a non-linear taper profile $a(u)$ the honest generalization isThus an arbitrary smooth taper with the same end radii and length need not give the printed formula; the straight-sided geometry supplies the additional information.

The closed surface integral sums the outward electric flux: $\mathbf{E}$ is the electric field, $d\mathbf{S}$ its outward vector area, $Q$ the total enclosed charge and ${ϵ}_0$ the vacuum permittivity. Gauss's law relates that flux to enclosed charge for any closed surface. High symmetry can make the field's magnitude constant on a suitable Gaussian surface and fix its direction; without symmetry, flux alone does not determine the local field.

For an isolated conducting sphere, charges reside at its surface and the exterior field is radial. ThusThe conductor has zero interior field. Integrating exterior field energy givesLet the initial charge per drop be $q=4\pi {ϵ}_{0}a{V}_1$. After disconnecting the charging supply, charge is conserved. Volume conservation gives final radius $A=2^{1/3}a$, total charge $2q$ andInitial mutual energy is negligible at the stipulated large separation. HenceThe electrostatic energy increase on equal-drop coalescence is positive. It can be supplied by the decrease in surface energy during coalescence and/or by external mechanical work bringing like charges together. Remaining energy can become motion and heat. If the charge is too large, surface-energy release alone need not suffice to make merging spontaneous; conservation does not assert that it will. Keeping the drops connected to a voltage supply would instead be a different charge-balance problem.