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In ideal magnetohydrodynamics, the first-order electric field is ${\mathbf{E}}_1=-\mathbf{v}\times B_0{\mathbf{e}}_z$. Retaining its product with both the background and first-order magnetic field gives the Poynting vectorThe first term is transverse and oscillatory, with zero period average. The second is quadratic and transports wave energy. A potential quadratic ideal electric-field term $-\mathbf{v}\times \mathbf{b}$ vanishes here because the real traveling-wave fields are parallel or antiparallel.

For the $+z$ branch, $\mathbf{b}=-\sqrt{{\mu }_0{\rho }_0}\mathbf{v}$, soFor the opposite branch, its sign reverses. The averaged Poynting flux is the total wave energy density, including equal kinetic and magnetic shares, multiplied by the group velocity. This agrees with Alfvén wave energy transport. Although the flux in this expression is electromagnetic, it supplies the transport of both components of the coupled fluid-field wave.

Use positive real local wavenumberswithin $|x|\le L$. If the exterior is vacuum, continue $k_1,k_3$ as $\omega /c$ outside the column. Define the positive phase integrals and barrier integralThe three WKB approximations, with the common $e^{-i\omega t}$ suppressed, areThe flux-normalized WKB amplitudes make squared travelling-wave coefficients proportional to directed Poynting flux. Since ${\Phi }^{\prime }=-k_1$ and ${\Psi }^{\prime }=k_3$, $A_{\mathrm{in}}$ and $A_{\mathrm{t}}$ travel to the right, while $A_{\mathrm{r}}$ and $A_{\mathrm{back}}$ travel to the left. The constants $\pi /4$ are a convenient Airy turning-point connection formula convention.

Let $J(x)={\int }_x^b\kappa (s)ds=\eta -I(x)$. The WKB approximation in the barrier can be rewritten asThe right Airy turning-point connection formula givesso $A_{\mathrm{t}}=D_R+i{G}_R/2$ and $A_{\mathrm{back}}=D_R-i{G}_R/2$. Impose no incident wave from the right, $A_{\mathrm{back}}=0$. ThenConsequently, within the leading Airy turning-point connection formula calculation,For an opaque barrier, $\eta \gg 1$,The transmission coefficient is a ratio of Poynting fluxes; the $k^{-1/2}$ normalization already includes their local wavenumber factors. The displayed denominator is useful for consistent matching and Wronskian flux conservation, but does not claim accuracy for subleading terms beyond the WKB approximation, or for merging wave turning points.

For real $k^2(x)$, the wave equation has the conserved Wronskian fluxFor this ordinary plasma wave, the time-averaged Poynting flux is $⟨S_x⟩=c^2\mathcal{J}/(8\pi \omega )$. With no incident wave from the right, evaluating the same Wronskian flux on both sides provesThe coefficients in f also satisfy this identity algebraically, before the opaque-barrier expansion.

A plasma wave resonance makes the ideal cold description singular and requires an appropriate causal regularization. A passive resonant layer can absorb wave energy into particle motion, or convert it into another branch. For the two measured outgoing channels one then haswith $A>0$ when resonant absorption or uncounted mode conversion occurs. A numerical value of $A$ requires the layer physics; a resonance by itself does not specify a universal loss fraction. Counting all converted and absorbed energy preserves conservation of energy.

Allow time dependence while keeping the constitutive relations linear, isotropic, nondispersive, and independent of time. The Maxwell equations in matter giveFor the Poynting vector $\mathcal{P}=\mathbf{E}\times \mathbf{H}$, the divergence identity yieldsThe first two terms are exact time derivatives. Taking the zero-field energy density as zero gives the electromagnetic energy densityConsequently the Poynting theorem is the continuity equationIn integral form,The left side is the rate of increase of stored electromagnetic energy plus outward Poynting flux. The right side accounts for work delivered to the free electric current; in an Ohmic medium ${\mathbf{J}}_f\cdot \mathbf{E}$ is the local rate of Joule heating.

Since ${\mathbf{e}}_{\rho }\times {\mathbf{e}}_{\phi }={\mathbf{e}}_z$, the Poynting vector is axial. The magnetic permeability cancels from $\mathbf{E}\times \mathbf{B}/\mu$ in each material layer:The signed Poynting flux through the plane with unit normal $+{\mathbf{e}}_z$ is its surface integral with area element $\rho d\rho d\phi$:ThusThe transmitted electromagnetic energy per unit time is the circuit's terminal power. For $VI>0$, the Poynting flux points in the direction of the inner electric current. The electromagnetic energy travels through the insulating gap and reaches a load at the end of the coaxial cable. The radii and material constants redistribute the electromagnetic energy density and the local Poynting vector, while their integrated Poynting flux depends only on the terminal electric potential difference and electric current.

Orient the unit normal $\mathbf{n}$ from $R_1$ to $R_2$ and write $\mathbf{H}$ for the magnetic field strength. For static electric fields, $\mathbf{E}=-\nabla \varphi$; the static Ampère's law gives $\nabla \times \mathbf{H}={\mathbf{J}}_f$. The curl product rule therefore gives the local Poynting flux and terminal power identityIntegrate over an oriented piece ${\Sigma }_R$ of the separating surface and apply Stokes theorem:Once all the wires are included, the first surface integral is ${\sum }_k{\varphi }_k{I}_k$: the electric potential is constant on each perfect conductor and the integral of its normal current density is its signed electric current. Equivalently, remove the wire cross-sections and apply Stokes theorem in the surrounding insulator. Each resulting hole has negative boundary orientation, and the magnetostatic Ampère law around it gives the positive contribution ${\varphi }_k{I}_k$.

For a closed separating surface the boundary term is absent. For a noncompact separating surface, take $R\to \infty$ with the usual isolated-circuit condition ${\oint }_{\partial {\Sigma }_R}\varphi \mathbf{H}\cdot d\mathbf{\ell }\to 0$. In either case,For a steady closed circuit the signed electric currents sum to zero, so shifting the reference of electric potential does not change the terminal power.

For a fully general noncompact surface, the hypotheses as printed need an additional condition at infinity. The exact result before imposing it isTime independence alone does not remove this boundary term or ensure a finite Poynting flux; for example, spatially constant crossed electric fields and magnetic fields carry a constant Poynting vector through an infinite plane. The stated terminal-power formula holds with the boundary condition specified above.

The leading electric dipole radiation fields areThe radiation electric field is transverse, $\mathbf{n}\cdot {\mathbf{E}}_{\mathrm{rad}}=0$. The vector triple product therefore gives the leading Poynting vectorIf $\theta$ is the angle between ${\mathbf{p}}^{\prime \prime }(t-r/c)$ and $\mathbf{n}$, thenUsing ${\mu }_0{ϵ}_0{c}^2=1$, the final result isThe leading Poynting flux is outward, vanishes along the instantaneous retarded ${\mathbf{p}}^{\prime \prime }$ axis, and is maximal in its perpendicular plane. If ${\mathbf{p}}^{\prime \prime }=0$, the radiation term is zero and no angle needs to be assigned. The large-$r$ expansion assumes bounded retarded electric dipole moment and time derivatives; it may also be read at fixed retarded time.

In SI units, Maxwell's equations areDotting the last equation with $\mathbf{E}/{\mu }_0$, and using the curl equation for $\mathbf{E}$, givesIntegrate over the fixed region and apply the divergence theorem. This proves Poynting's theorem: the rate of increase of electromagnetic energy stored inside, plus the outward Poynting flux, plus the rate of work on charges, is zero. The last term need not be positive for arbitrary externally driven currents; when positive it is power transferred from the field to matter.

For the travelling wave put $\phi =kz-\omega t$. A purely oscillatory magnetic field satisfying Faraday's law isBoth divergences vanish. The vacuum Ampère-Maxwell law then requires $k^2={\mu }_0{ϵ}_0{\omega }^2$, so with the positive constants in the question,A superposed static background is not part of the pure plane-wave solution. The electric and magnetic energy contributions are equal, givingConsequently $\boxed{|\mathbf{P}|=c\mathcal{U}}$ pointwise, including the nodes where both sides vanish.

The intensity of an isotropic radiator at distance $r$ is the radiated power divided by spherical area: $\mathcal{I}=P/(4\pi r^2)$. For a locally plane far-field electromagnetic wave in air, its mean Poynting flux is $\mathcal{I}=E_{\mathrm{rms}}^2/Z_0=E_{\mathrm{peak}}^2/(2Z_0)$, where the vacuum wave impedance is $Z_0\simeq 376.7\Omega$. ThusandBoth conventions are given because “field strength” alone does not specify RMS or peak. This estimate treats the stated transmitter power as radiated power and ignores ground reflection and absorption.

Work in the locally planar shock frame, taking $v_u,v_d>0$ towards the surface. For a stationary normal shock wave with negligible radiative loss inside the discontinuity, the Rankine-Hugoniot conditions for a perfect gas express conservation of mass, conservation of momentum, and conservation of energy per unit area asHere $h$ is specific enthalpy, including the pressure work; using only internal energy would omit that work. The gravitational potential is continuous across an infinitesimally thin front. For cold upstream gas, $P_u$ and $h_u$ are negligible, while a monatomic gas has $\gamma =5/3$ and $h=5P/(2\rho )$.

The hydrodynamic equations apply to flow parallel to the guiding magnetic field when its stress contributions cancel across the front. An oblique magnetohydrodynamic shock requires magnetic stresses and Poynting flux; a radiation-mediated front or appreciable escaping radiation requires the corresponding radiative momentum and energy flux terms. These are additional physics, not terms hidden in $h$ here.

Corotation reaches the speed of light at $R_c=c/\Omega$. Extrapolating a dipole field gives $B_c\sim B_{\ast }(R_{\ast }/R_c{)}^3$ and magnetic energy density ${ϵ}_c\sim B_c^2/{\mu }_0$, suppressing order-one factors. An outward electromagnetic energy flux of scale ${ϵ}_{c}c$ through area of scale $R_c^2$ therefore givesThis assumes a dipole-dominated field tied to the rotating star and an effective outgoing Poynting flux near the light cylinder. The dipole extrapolation, angular pattern and surface-field convention affect the coefficient. A strictly aligned vacuum dipole would not radiate, whereas an oblique vacuum dipole does. A plasma-filled magnetosphere opens field lines and can spin down even an aligned rotator, so the light-cylinder estimate is a scaling argument, not an exact vacuum emission calculation or literal rigid motion at $c$. Force-free calculations explicitly demonstrate this distinction.^[ref]

The energy comes primarily from stellar rotation. With constant moment of inertia $I$, $E_{\mathrm{rot}}=I{\Omega }^2/2$, and conservation of energy gives $I\Omega \dot{\Omega }=-P$. ThusFor constant $K$, differentiating gives $\ddot{\Omega }=-3K{\Omega }^2\dot{\Omega }$, hence $n=\Omega \ddot{\Omega }/{\dot{\Omega }}^2=3$. Evolving field strength, inclination, moment of inertia or additional torques invalidate the constant-$K$ prediction; generally $n=3+(\Omega /\dot{\Omega })(\dot{K}/K)$ in this parametrization.

For radial inflow onto mass $M$, take $v_{\mathrm{ff}}\sim (GM/r{)}^{1/2}$ and $\rho \sim \dot{M}/(4\pi r^2{v}_{\mathrm{ff}})$. magnetic pressure decreases as $B_{\ast }^2{R}_{\ast }^6/({\mu }_0{r}^6)$, while ram pressure scales as $\rho v_{\mathrm{ff}}^2\propto \dot{M}(GM{)}^{1/2}{r}^{-5/2}$. Balancing them yieldsNumerical factors and disc geometry are not determined by this spherical estimate. The displayed proportionality holds $M$ fixed and requires a magnetospheric boundary outside the stellar surface.

In a disc, inflowing matter near $R_M$ has specific angular momentum approximately $\sqrt{GM{R}_M}$. If it is transferred to a slower star, the material torque is $N\sim \dot{M}\sqrt{GM{R}_M}$ and $I\dot{\Omega }\sim N>0$. The corotation radius is $R_{\mathrm{co}}=(GM/{\Omega }^2{)}^{1/3}$, distinct from the light-cylinder radius. As stellar spin approaches the inner disc's Keplerian frequency, net spin-up weakens. If $R_M$ lies outside corotation, the field rotates faster than the disc and can centrifugally expel matter in the propeller regime. A characteristic equilibrium isfor fixed mass and magnetic-coupling convention. A complete magnetic torque model can change the coefficient; a star also cannot rotate past its mass-shedding limit, of order $\sqrt{GM/R_{\ast }^3}$.

For opacity per unit mass $\kappa$, an isotropic radiative force balances gravity at $L_E=4\pi GMc/\kappa$. Efficient release of surface gravitational binding energy gives $L_{\mathrm{acc}}\sim GM\dot{M}/R_{\ast }$ and therefore the surface accretion Eddington rateThe radius dependence matters: it uses efficiency $\eta \sim GM/(R_{\ast }{c}^2)$, not an arbitrarily fixed reference efficiency. Substituting this maximal rate into the equilibrium period givesmass, opacity and geometry factors are held fixed. This is the classical radiatively efficient Eddington-limited scaling, not a universal upper bound on every mass supply rate: beaming, advection, outflows and opacity changes can alter it. For sufficiently small field, $R_M\lesssim R_{\ast }$ and the magnetic scaling must be replaced by surface-disc and mass-shedding physics.

Let $z^{\mu }(\tau )$ be the point electric charge's world line, with proper time $\tau$ and four-velocity $U^{\mu }=d{z}^{\mu }/d\tau =\gamma (c,\mathbf{u})$. For the field event $x^{\mu }$, choose the retarded source event: $R^{\mu }=x^{\mu }-z^{\mu }({\tau }_{\mathrm{ret}})$ is future-directed and null. Write $R^{\mu }=(R,R\mathbf{n})$, where $R=|\mathbf{x}-\mathbf{z}(t_{\mathrm{ret}})|=c(t-t_{\mathrm{ret}})$. In the instantaneous rest inertial frame of that source event,Consequently the covariant expression matching this Coulomb potential isThe denominator is a Lorentz scalar, so the right-hand side is a four-vector with the required instantaneous-rest components. This describes the retarded potential in Lorenz gauge, with no added free electromagnetic wave. Lorentz covariance alone does not establish the boundary conditions or exclude additional terms depending on four-acceleration. The physical input is the retarded point-source solution of Maxwell equations: the potential at the retarded event depends on the instantaneous four-current, while its derivatives generate the acceleration field. Equivalently, evaluating the retarded source integral gives a Jacobian $1/(1-\mathbf{n}\cdot \mathbf{\beta })$ from the light-cone constraint; this fixes that input for arbitrary smooth subluminal motion.^[ref]

In any inertial frame, put $\mathbf{\beta }=\mathbf{u}/c$ and $\kappa =1-\mathbf{n}\cdot \mathbf{\beta }$. Then $-R\cdot U=\gamma cR\kappa$, and the Liénard–Wiechert potential isDifferentiate using $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$ and $\mathbf{B}=\nabla \times \mathbf{A}$, including the position and time dependence of the retarded time. For example, $\partial t_{\mathrm{ret}}/\partial t=1/\kappa$ and $\nabla t_{\mathrm{ret}}=-\mathbf{n}/(c\kappa )$. The resulting Liénard-Wiechert field isHere the dot denotes a derivative with respect to source coordinate time. The first term is the velocity or non-radiative field: it falls as $R^{-2}$ and exists even without acceleration. The second is the acceleration or radiation field, falling as $R^{-1}$ and transverse to $\mathbf{n}$. Its Poynting flux falls as $R^{-2}$, so its outward power integrated over a large sphere stays finite as its radius tends to infinity. The first term's squared-field flux instead vanishes in that limit. This large-distance distinction defines radiation; a moving near field can carry local energy without constituting radiation escaping to infinity.

Let starred quantities denote the capacitor's rest inertial frame, $u_0={ϵ}_0{E}^2/2$, $\beta =v/c$, and $\gamma =(1-{\beta }^2{)}^{-1/2}$. Work in the uniform interplate region of an ideal parallel-plate capacitor, neglecting its fringing field of a capacitor. To transform from that rest inertial frame to a laboratory in which the capacitor moves along $+x$, use the positive-sign Lorentz transformationFor a purely electric field, the spatial block of the electromagnetic stress-energy tensor isThis is the negative of the Maxwell stress tensor: it has $-u_0$ along the electric field and $+u_0$ perpendicular to it. The stress contribution to energy transport in a moving capacitor is essential to the Lorentz transformation of the energy density and flux.

Plates parallel to $xy$. Choose ${\mathbf{E}}_{\ast }=E\hat{\mathbf{z}}$ and ${\mathbf{B}}_{\ast }=0$. The electromagnetic field tensor and electromagnetic stress-energy tensor areThe transformed nonzero independent electromagnetic field tensor components are $F^{03}=\gamma E/c$ and $F^{13}=\gamma \beta E/c$. Thus $\mathbf{E}=\gamma E\hat{\mathbf{z}}$ and $\mathbf{B}=-\gamma vE\hat{\mathbf{y}}/c^2$, consistent with $\mathbf{B}=\mathbf{v}\times \mathbf{E}/c^2$. Transforming the electromagnetic stress-energy tensor givesIn particular, the laboratory Poynting vector is $\mathbf{S}=2u_0{\gamma }^{2}v\hat{\mathbf{x}}$, and the electromagnetic energy density is $u=u_0{\gamma }^2(1+{\beta }^2)$. The ratio $S_x/u=2v/(1+{\beta }^2)$ is subluminal for $v<c$, but differs from the capacitor's velocity. The difference from a simple convective flux isThe extra term is $v{T}_{\ast }^{xx}$, the rest-frame stress multiplied by the velocity. This shows why treating stored field energy as an unstressed parcel moving rigidly with the plates misses part of the energy transport.

Plates parallel to $yz$. Now ${\mathbf{E}}_{\ast }=E\hat{\mathbf{x}}$. The rest tensors areThe electric field is parallel to the Lorentz boost, so $F^{01}=E/c$ remains unchanged and $\mathbf{B}=0$. The cancellation between $T_{\ast }^{00}$ and $T_{\ast }^{11}$ givesThere is no Poynting flux in the ideal interior in this orientation, although the region containing the electric field moves. The longitudinal negative stress contribution cancels the convective term: $S_x={\gamma }^{2}v(u_0+T_{\ast }^{xx})=0$. Field energy is removed at the receding boundary and supplied at the advancing boundary by work on the charged plates; it need not flow through the interior to accompany the moving region. At edges, omitted fringing fields provide additional local structure.

For a finite idealized interplate rest volume $V_0$, length contraction gives $V=V_0/\gamma$. The field-only integrated laboratory energies would consequently be $U_{xy}=\gamma (1+{\beta }^2)U_0$ and $U_{yz}=U_0/\gamma$, with $U_0=u_0{V}_0$. Their different transformation laws are not paradoxical: the field alone is not a closed system. Include the plates and the supports holding them apart, whose mechanical stresses balance the Maxwell stress tensor. The total closed-system four-momentum then has the ordinary Lorentz transformation law. This is the same distinction between a subsystem and a conserved whole that appears in b.

The Poynting flux through an oriented surface is the normal component of the electromagnetic energy-flow vector. Integrating it over area gives outward electromagnetic power. A scale estimate of the form energy density times propagation speed must still specify the field and geometric regime.

For static electric fields and magnetic fields, Ampère's law and $\mathbf{E}=-\nabla \varphi$ give $\mathcal{P}=\mathbf{E}\times \mathbf{H}=\varphi {\mathbf{J}}_f-\nabla \times (\varphi \mathbf{H})$. Stokes theorem on an oriented separating surface then giveswhere each perfect conductor crossing it has constant electric potential ${\varphi }_k$ and signed electric current $I_k$. The line term is absent for a closed surface and must vanish in the limiting sense at infinity for a noncompact surface. Under that condition the incoming Poynting flux equals terminal power. If the signed electric currents sum to zero, the result is independent of the reference for electric potential.

For a static rest-frame electromagnetic stress-energy tensor with no Poynting flux, a Lorentz transformation to a laboratory where it moves along $+x$ gives $u={\gamma }^2(u_0+{\beta }^2{T}_{\ast }^{xx})$ and $S_x={\gamma }^{2}v(u_0+T_{\ast }^{xx})$. Thus energy flux includes stress as well as rest energy density. In an ideal parallel-plate capacitor, $T_{\ast }^{xx}=+u_0$ for motion perpendicular to the electric field, but $T_{\ast }^{xx}=-u_0$ for motion parallel to it. The former has $S_x=2{\gamma }^{2}v{u}_0$; the latter has zero interior flux, with work at the moving plates transferring energy at the boundaries. The field subsystem need not have an integrated four-momentum independent of simultaneity surface; include material support stresses for a closed system.

For a complex solution of $E^{\prime \prime }+k^2(x)E=0$ with real $k^2$, the Wronskian fluxis constant: differentiation leaves $\mathrm{Im}(\overline{E}{E}^{\prime \prime })=-k^2\mathrm{Im}|E{|}^2=0$. The directed wave flux is therefore conserved even when the local wavenumber varies. In a one-dimensional ordinary plasma wave, the mean Poynting flux is $c^2\mathcal{J}/(8\pi \omega )$ in Gaussian units.

For constant positive asymptotic wavenumbers $k_{-},k_{+}$, write the left field as $e^{i{k}_{-}x}+r{e}^{-i{k}_{-}x}$ and the transmitted right field as $t{e}^{i{k}_{+}x}$. Their fluxes are $k_{-}(1-|r{|}^2)$ and $k_{+}|t{|}^2$. Conservation givesHere the conserved quantities are flux fractions. The complex amplitude coefficients themselves need not sum to1, and a transmission fraction need not equal $|t{|}^2$ when the channel normalizations differ.