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In the Ffowcs Williams-Hawkings equation, the other source types are a surface acoustic monopole associated with acoustic thickness noise, and a volume acoustic quadrupole involving the Lighthill stress tensor. On an impermeable material surface, fluid and surface normal velocities agree. There is no through-surface mass-flux source; the remaining thickness source is $Q={\rho }_0{v}_n$. For a rigid body, the leading acoustic compact-source approximation to that source hasThus there is no leading net-volume acoustic monopole. To neglect thickness radiation beyond that leading cancellation, assume negligible volume displacement, as for ideal thin blades, or that its higher multipoles are small compared with the retained acoustic loading noise. Rigidity alone does not make a moving finite-volume body's local thickness source identically zero.

The volume acoustic quadrupole may be neglected for low Mach number motion when exterior turbulent or nonlinear stresses do not provide a competing strong source. We also assume small linear acoustics perturbations, a uniform reference sound speed, and negligible relevant viscous and entropy sources. These are source-strength approximations, particularly important if a loading contribution itself cancels by symmetry. Under them, the retained acoustic dipole is the force exerted by the object on the fluid, with the sign used in the previous solution.

Let $R=|x|$, $n=x/R$, and ${\tau }_0=t-R/c_0$. In the acoustic far field, $R$ is large compared with the object and $\omega R/c_0\gg 1$. For a source of size $\ell$ with $\omega \ell /c_0\ll 1$ and small surface Mach number, source-dependent delays and the Doppler factor can be neglected to leading order. The surface integral then contains just the total force $\mathcal{F}(\tau )={\int }_{S}FdS$. Differentiating its retarded time, rather than its $R^{-1}$ spreading factor, gives the radiating termThe sign follows from ${\partial }_{x_i}{\tau }_0=-n_i/c_0$. Differentiating $R^{-1}$ or the direction $n$ instead produces the lower-order near field. A constant total force does not radiate at this leading compact order.

The rotation introduces angular frequency $\Omega$ and, after combining the two blades, harmonics such as $2\Omega$. The acoustic compact-source approximation requires the propagation time $a/c_0$ to be small compared with ${\Omega }^{-1}$. ThusIt also bounds every blade element's Mach number by $\mu$. Consequently $|1-M_r|=1+O(\mu )$, and its leading value is one. The separate acoustic far field condition is $\Omega R/c_0\gg 1$.

Choose the positive rotation sense so that a first blade at phase $\alpha$ has radial and tangential unit vectorsIntegrating the given line force from $r=0$ to $a$ yields ${\mathcal{F}}_1=F{e}_{\varphi }(\Omega \tau )+D{e}_x$. Its axial component is constant, while ${\dot{\mathcal{F}}}_1=-F\Omega e_r$. Since $n=\cos \theta e_x+\sin \theta e_y$, the compact acoustic dipole sound from this blade isThe other blade has phase $\alpha +\pi$ and contributes the opposite rotating force. At a common compact retarded time, their total force is $2D{e}_x$, soThis cancellation calls for the next source-delay correction; it does not mean that the complete moving-source field vanishes.

Write $\alpha =\Omega {\tau }_0^{\ast }$ and $s=\sin \theta$, where ${\tau }_0^{\ast }=t-R/c_0$. Label the two arms by $\sigma =\pm 1$. Their positions are $y_{\sigma }=\sigma r{e}_r(\Omega {\tau }_{\sigma }^{\ast })$. To radiating far-field accuracy,The retarded time equation therefore gives, with ${\mu }_r=\Omega r/c_0$,For the first arm, $\sigma =1$, this is the requested phase expansion. The more explicit geometric error also makes the dimensional meaning of the printed $O(1/R)$ term clear.

Apply the Taylor theorem to the rotating line force. Since $d{e}_{\varphi }/d\alpha =-e_r$, its first two orders, expressed in a common reference frame, areAlso $v_{\sigma }=\sigma \Omega r{e}_{\varphi }(\Omega {\tau }_{\sigma }^{\ast })$, so the radial Mach number isThe absolute value causes no change of sign in this subsonic limit. Multiplying these two expansions before summing is essential: both the shifted force and the moving-surface retarded Jacobian contribute at the same order.

Denote the integrated numerator, including that Jacobian, by $\mathcal{B}$. Pairing the two blades cancels all terms odd in $\sigma$, including the first Doppler correction to the axial load. HenceThe factor $a/3$ comes from $a^{-2}{\int }_0^a{r}^{2}dr$. Only the time-dependent term radiates at order $R^{-1}$. Applying $-{\partial }_{x_i}$ to the integral now gives ${\rho }_{\mathrm{rad}}^{\prime }=(4\pi c_0^{3}R{)}^{-1}n\cdot {\partial }_t\mathcal{B}$, and thusThere are also nonradiating terms of order $R^{-2}$. Multiply by $c_0^2$ for the acoustic pressure. The stated coefficient uses the rotation convention fixed above; reversing the rotation reverses the corresponding signed load and phase convention.

This is a compact rotating two-blade loading source acting as an acoustic quadrupole. The compact total rotating force cancels, leaving the first spatial moment of the loading. Its two factors of the observer's projection into the rotor plane produce ${\sin }^2\theta$: there is no leading sound on the rotation axis and the density amplitude is maximal in the rotor plane. The configuration repeats after half a rotation, explaining frequency $2\Omega$. These statements concern amplitude; the corresponding acoustic intensity has a ${\sin }^4\theta$ factor.

If $F=0$, the displayed first-order contribution vanishes as well. For completeness, expanding the axial Jacobian to its next even order givesThe first remaining axial-load radiation is thenThus the constant axial total force does not produce the lower-order term, even though its moving spatial distribution can radiate at a higher order.

Ignoring magnetization, electric polarization charge transport gives bound current ${\mathbf{J}}_{\mathrm{b}}={\partial }_t\mathbf{P}$. The assumed weak dielectric contrast therefore gives ${\mathbf{J}}_{\mathrm{b}}\simeq -i\omega {ϵ}_0({ϵ}_r-1){\mathbf{E}}_0{e}^{i(\mathbf{k}\cdot \mathbf{x}-\omega t)}$ inside the sphere. This is the first Born approximation: the incident field supplies the electric polarization to leading order in ${ϵ}_r-1$.

For $\mathbf{x}=r\mathbf{n}$, the far-distance expansions are $|\mathbf{x}-{\mathbf{x}}^{\prime }|=r-\mathbf{n}\cdot {\mathbf{x}}^{\prime }+O(a^2/r)$ and $|\mathbf{x}-{\mathbf{x}}^{\prime }{|}^{-1}\simeq r^{-1}$. Substitute the retarded time into the phase in the Lorenz gauge vector potential. The phase becomes $kr-\omega t+(\mathbf{k}-k\mathbf{n})\cdot {\mathbf{x}}^{\prime }$, so, using ${\mu }_0{ϵ}_0=c^{-2}$,The assumptions $r\gg a$ and $k{a}^2/r\ll 2\pi$ control amplitude and phase errors respectively. In the radiation zone the supplied electric and magnetic fields satisfy ${\mathbf{B}}_{\mathrm{scatt}}=c^{-1}\mathbf{n}\times {\mathbf{E}}_{\mathrm{scatt}}$. Complex phasors therefore give the time-averaged radial Poynting flux $|{\mathbf{E}}_{\mathrm{scatt}}{|}^2/(2{\mu }_{0}c)$. The incident flux is $|{\mathbf{E}}_0{|}^2/(2{\mu }_{0}c)$, soOrient spherical coordinates along $\mathbf{q}$ to evaluate the form factor:ConsequentlyAt $q=0$ the bracket is interpreted by its limit $a^3/3$, not a singularity. The transverse factor accounts for polarization of an electromagnetic wave; for complex polarization of an electromagnetic wave its squared norm means the Hermitian norm.

Write the electromagnetic field tensor as $F^{\mu \nu }={\partial }^{\mu }{A}^{\nu }-{\partial }^{\nu }{A}^{\mu }$. Substitution into the Covariant Maxwell equation with the minus-plus-plus-plus metric givesChoose the Lorenz gauge ${\partial }_{\mu }{A}^{\mu }=0$. Since ${\partial }_{\mu }{\partial }^{\mu }={\nabla }^2-c^{-2}{\partial }_t^2$, the electromagnetic four-potential obeysConvolution with the outgoing retarded Green function gives the retarded electromagnetic potential

For the point particle, ${\tau }_{\ast }(x)$ is its unique retarded proper time, characterized byThus $y({\tau }_{\ast })$ is where the particle's worldline intersects the past light cone of the observation event $x$; the displayed expression is the covariant form of the Lienard-Wiechert potentials.

On the $z$-axis put $D=\sqrt{R^2+z^2}$. The retarded time is $t_{\ast }=t-D/c$. For the circular orbit, the spatial velocity at that time isBecause the displacement from the retarded source point to the observation point is orthogonal to this velocity, $|R^{\nu }{\dot{y}}_{\nu }|=\gamma cD$, while ${\dot{y}}^{\mu }=\gamma (c,\mathbf{v})$. The factors of $\gamma$ cancel, leaving

The outgoing solution obtained from the retarded fundamental solution of the wave equation isIn the radiation zone, put $r=|\mathbf{x}|$, $\mathbf{n}=\mathbf{x}/r$, and retain the leading $1/r$ term at retarded time $u=t-r$. The trace term disappears after applying the transverse-traceless projectorsoTwice using stress-energy conservation, ${\partial }_a{T}^{ab}=0$, and integrating by parts givesThe projector removes the trace, so in terms of the mass quadrupole momentthe far field isThe gravitational-wave energy flux isFor a trace-free symmetric tensor $A_{ij}$, the isotropic tensor integral over the observation direction givesConsequently the standard quadrupole formula, in the units $G=c=1$ used by the paper, isThus the printed coefficient $4\pi /5$ is inconsistent with the stated definition of $Q_{ij}$ and the standard wave-energy normalization; it appears to be a typographical error.

For the planet, choose its circular orbit in the $xy$-plane and write $\omega =2\pi /\tau$. With the star treated as fixed,and henceDirect differentiation givessoRestoring units multiplies this by $G/c^5$. If the printed $4\pi /5$ coefficient is followed literally, the answer is instead $4\pi$ times larger,For two bodies of comparable mass, $M$ is replaced by the reduced mass and $R$ by their separation.

The outgoing Green function of the electromagnetic wave equation has support on the past light cone. Convolution with that retarded Green function yields the retarded electromagnetic potentialwhere the source is evaluated at the retarded timeThis is the causal solution with no incoming radiation.

The electromagnetic four-potential is the field generated by the four-current $J^{\mu }$: every source volume element contributes with the inverse-distance Green-function factor. Electromagnetic influence propagates at speed $c$, so causality requires the source to be evaluated where its worldline intersects the field event's past light cone. ThusThe source time is exactly the retarded time $t_{\mathrm{ret}}=t-|x-x^{\prime }|/c$, rather than the observation time $t$.

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.

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.

Since $\mathbf{E}\cdot \mathbf{n}=0$, the Poynting vector isThe area on a distant sphere corresponding to solid angle $d\Omega$ is $r^{2}d\Omega$. ConsequentlyThe average is over the source motion at retarded time. At leading nonrelativistic order, differences between source-time and reception-time averaging are beyond the approximation. The units are power per solid angle, $\mathrm{W}\mathrm{s}{\mathrm{r}}^{-1}$.

The acceleration covariance in nonrelativistic radiation is another way to organize this result: with $C_{ij}=⟨a_i{a}_j⟩$, the angular factor is $\mathrm{tr}C-{\mathbf{n}}^{T}C\mathbf{n}$. Integrating over all observation directions uses $\int [|\mathbf{a}{|}^2-(\mathbf{n}\cdot \mathbf{a}{)}^2]d\Omega =(8\pi /3)|\mathbf{a}{|}^2$ and gives the averaged Larmor formula, $⟨P⟩=q^2⟨a^2⟩/(6\pi {ϵ}_0{c}^3)$.

Use retarded time $t_r$ and reserve $r$ for the large observation distance. The source velocity is nonrelativistic, $|\mathbf{u}|/c\ll 1$. There are two errors in the original PDF that must be corrected before the radiation zone calculation: the vector-potential prefactor contains an extraneous $1/r$, and the displayed magnetic radiation field has the wrong sign for the stated outward unit vector $\mathbf{n}$. The extra distance factor is dimensionally wrong; the magnetic sign would make the Poynting vector point inward. The following derivation uses dimensionally consistent retarded potentials and outgoing radiation.

After field electron emission, let $h$ denote the plate separation of the parallel-plate capacitor, which the question does not specify. The uniform capacitor electric field has magnitude $E_c=|V|/h$, and a nonrelativistic electron of mass $m_e$ has constant acceleration magnitude $a=e{E}_c/m_e$. The sign of its charge affects the direction of the radiation field but not its radiated power.

In the supplied nonrelativistic Liénard-Wiechert field, $\dot{\mathbf{\beta }}=\mathbf{a}/c$. At polar angle $\theta$ to the acceleration, $|\hat{\mathbf{n}}\times (\hat{\mathbf{n}}\times \mathbf{a})|=a\sin \theta$, henceThe outgoing Poynting vector has magnitude $S={ϵ}_{0}c{E}_{\mathrm{rad}}^2$. Its radiated power per solid angle is thereforeUsing $\int {\sin }^2\theta d\Omega =8\pi /3$ gives the Larmor formula and the requested constant power during the uniform-field part of the flight:The brackets in the given field require the acceleration at retarded time; the radiation reaches the far observer after the propagation delay. This derivation assumes that radiation reaction is a small correction to the acceleration and that the flight is nonrelativistic, for example $e|V|\ll m_e{c}^2$ for release from rest. Entry, emission and impact transients are not described by the uniform-field model. A result solely in terms of $V$ cannot be obtained without a plate spacing or a specified internal field.

Only the $1/r$ fields transport a nonzero outgoing power to infinity. Retain those terms from the Hertzian dipole fields:The brackets mean evaluation at the retarded time,for a point source at the origin in vacuum. They are not a time average. The radial Poynting vector in the radiation zone isIntegrating over a sphere usesThereforeNear-field terms either give inverse powers of $r$ after the surface integral or describe reversible exchange with the near field. The emitted power labelled by source time $t_{\mathrm{src}}$ is ${\mu }_0\ddot{p}(t_{\mathrm{src}}{)}^2/(6\pi c)$; the enclosing surface receives it one light-travel time later. For a changing dipole orientation, the general form is $P={\mu }_0|\ddot{\mathbf{p}}{|}^2/(6\pi c)$.

Let $t$ label the reception event in the source reference frame. Its distance from the source and the relevant retarded time areThe received phase is $2\pi f_0{t}_e$. Because the observer's proper time obeys $d\tau =dt/\gamma$, its measured frequency isEquivalently, write $\beta =v/c$ and $s(t)=vt/r(t)$. The incoming Doppler factor is $D_{\mathrm{in}}=\gamma (1-\beta s)$, obtained also by contracting the photon four-momentum with the observer's four-velocity. The frequency decreases monotonically, withIn particular, the instant of closest approach is not the zero-shift point: a moving receiver looking perpendicular to its motion sees a factor $\gamma$, not $1/\gamma$. The latter factor applies to a different moving-emitter geometry. The graph uses source-frame reception time, with observer proper time $\tau =t/\gamma$.

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.

An ideal electric dipole at the origin with smooth electric dipole moment $\mathbf{p}(t)$ has Lorenz gauge electric potential and magnetic vector potential, for $r>0$,For a short proof, place electric charges $sq$ at $s\mathbf{p}/(2q)$, $s=\pm 1$, and let $q\to \infty$ with $\mathbf{p}$ fixed. The corresponding retarded times are $t_0+s\mathbf{n}\cdot \mathbf{p}(t_0)/(2qc)+O(q^{-2})$. Each Liénard–Wiechert potential has denominator $r-s(\mathbf{n}\cdot \mathbf{p}+r\mathbf{n}\cdot {\mathbf{p}}^{\prime }/c)/(2q)+O(q^{-2})$. The opposite $q/r$ terms cancel, leaving the displayed electric potential. The products of signed electric charge and velocity add, giving the displayed magnetic vector potential. Bounded time derivatives on the relevant interval justify the Taylor expansions.