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Over $N$ half-lives, the duration is $T=N{t}_{1/2}=N\ln 2/\lambda$. The time-averaged thermal power isSince $\lambda T=N\ln 2$ and $e^{-\lambda T}=2^{-N}$,This is a time average, not the arithmetic mean of the initial and final powers. It follows directly from the assumed constant nuclear decay constant and constant recovered energy per decay.

The idealized isotope mass is $1.63\times 10^{-2}\mathrm{kg}$, about $16.3\mathrm{g}$. Both the alpha particle and daughter recoil energy are recovered in the stipulated heat balance. Thus the recovered energy per transformation isand the nuclear decay constant isWith $N_0=(m/M_{\mathrm{mol}})N_A$, thermal power is $P_0=\lambda N_0{E}_h$. ConsequentlyThe conversion of $210\mathrm{kg}\mathrm{kmo}{\mathrm{l}}^{-1}$ to $0.210\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$ is essential. This is an isotope-only energy-accounting result under the paper's prescribed branch and perfect heat-recovery assumptions; it is not the mass or design of a complete generator.

After three half-lives, $P(T)=P_0/8$. Therefore the minimum initial thermal power isEquality gives the stated terminal power exactly in the idealized exponential decay model. The associated duration is $415.2$ days. No thermal-to-electrical efficiency is needed because both specified powers are thermal.

Alpha decay releases energy in charged particles with short stopping ranges in matter, so much of their kinetic energy becomes local heat. An appropriate radioisotope thermoelectric generator can use a predictable radioactive decay heat source without requiring sunlight or a sustained nuclear chain reaction. Low penetrating photon and neutron output is desirable, but being an alpha emitter does not by itself establish the complete radiation spectrum.

The isotope's half-life must suit the mission. A short half-life gives high initial specific activity but a rapid decline in useful thermal power; a much longer one gives a steadier output but less activity per atom. Selection also depends on recoverable decay energy, specific power, daughter emissions, chemical and thermal stability, compatibility with the power-conversion system, availability and cost. Practical containment and radiological acceptability remain separate system-level requirements.

A thermoelectric generator converts a temperature difference into electrical output through the Seebeck effect; not all decay heat becomes electricity. NASA's operational account describes this heat-to-electricity distinction and the need to match source lifetime to long space missions.^[ref]

For the assumed three-half-life mission, the thermal power falls by a factor of eight. That large decline is the important disadvantage of this particular half-life even though the initial specific thermal power is high.

The first balance treats the fertile parent as being removed by neutron absorption. The second has a production contribution associated with capture in the parent and a removal contribution from absorption in the fissile population. Capture and total absorption are not interchangeable: in general absorption includes capture and fission, and a capture channel is the one that feeds the specified transmutation chain.

The lumped description assumes spatially averaged populations and flux, fixed effective reaction coefficients, no transport of material into or out of the system, and a shortened chain in which intermediate decay delays are neglected. It omits explicit precursor accumulation, competing reaction chains and the changing neutron spectrum. It also prescribes flux rather than solving a coupled power-and-composition feedback problem. Neglecting intermediate lifetimes requires them to be short relative to the variation being modelled; that assumption can fail during a rapid change in irradiation.

Put $g=\gamma P_0$. For a temporally uniform envelope, $A_z=igA$ and $A_{TT}=0$, so the propagation equation reduces to $-gA+\gamma P_{0}A=0$. Thus the power stays $P_0$ and the accumulated nonlinear wave phase is $gz$. The optical Kerr effect gives an intensity-dependent refractive index, and hence self-phase modulation. For the usual power normalization $\gamma \simeq n_2{\omega }_0/(c{A}_{\mathrm{eff}})$, with effective guided-mode area $A_{\mathrm{eff}}$. Losslessness preserves the background power; it does not remove the self-phase modulation. The carrier wave phase has already been factored out.

Write $A=(\sqrt{P_0}+ϵ)e^{igz}$. To first order,Cancelling the background and its phase derivative leavesThe last term couples a perturbation to its complex conjugate: the pump can transfer energy into a pair of sidebands. It is not an independent wave phase of each Fourier component. This sideband coupling is the linear origin of fibre modulational instability.

For real $K,\Omega ,c_1,c_2$, put $d={\beta }_2{\Omega }^2/2$. Matching the two exponential coefficients givesA nonzero pair requires a zero determinant, henceFor unstable complex $K$, the real-coefficient assumption is too restrictive. The consistent pair is $ϵ=u{e}^{iKz-i\Omega T}+v^{\ast }{e}^{-i{K}^{\ast }z+i\Omega T}$, with complex $u,v$. Its equations give the same determinant and allow the growing and decaying branches. Merely inserting imaginary $K$ while treating the two original coefficients as real would not respect complex conjugation.

For $P_0>0$ and nonzero ${\beta }_2,\gamma$, exponential growth requires ${\beta }_2\gamma <0$ andFor ordinary positive Kerr nonlinearity this means anomalous group velocity dispersion, ${\beta }_2<0$. Define ${\Omega }_c=\sqrt{4g/|{\beta }_2|}$. The growing branch is $K=-iG$, with amplitude growth rateSideband power grows at rate $2G$, rather than $G$. The cutoff and zero-frequency modes have no exponential growth; in a degenerate zero-frequency sector a secular phase response can remain. Positive dispersion with positive $\gamma$ gives real $K$ and oscillatory perturbations.

Initially small ripples grow into pronounced intensity peaks and deplete the uniform background. Dispersion and self-phase modulation can then support localized bright pulses. A fundamental temporal bright soliton obeys $\gamma P_s{T}_0^2=|{\beta }_2|$, so its width is comparable to $1/{\Omega }_c$ when its peak power is comparable to the original power. The dominant ripple spacing is $2\pi /|\Omega {|}_{\max }$, not simply $1/{\Omega }_c$. Propagation distance $z$ is the evolution variable; $T$ measures time within the envelope. An ideal conservative fibre need not settle permanently into a unique pulse train: nonlinear compression, interactions and recurrence depend on the seed and boundary conditions.

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.

The wavelength is $\lambda =c/f\simeq 0.50\mathrm{m}$, and the aperture diameter is $D=1.0\mathrm{m}$. A uniformly illuminated circular aperture has normalized power pattern $[2J_1(x)/x{]}^2$, where $x=\pi D\sin \theta /\lambda$. Its first null is at $\sin \theta \simeq 1.22\lambda /D$, while its half-power point has $\sin \theta \simeq 0.514\lambda /D$. ThusThe first-null small-angle estimate is $0.61\mathrm{rad}$, or roughly $35^{\circ }$; using the sine gives about $38^{\circ }$. This is a broad beam because the diameter is only about two wavelengths. Real illumination and aperture efficiency alter the exact beam shape; a required allowable gain loss would specify the pointing tolerance more precisely.

For a matched, correctly polarized half-wave dipole, $A_{e,d}=G_d{\lambda }^2/(4\pi )$ with $G_d\simeq 1.64$. For an efficient dish, $A_{e,s}\simeq \eta \pi (D/2{)}^2$. Received available powers are proportional to these areas. With both voltages referred to the same matched receiver resistance,For ideal aperture efficiency this is about one fifth, or five times greater voltage from the dish. Terminal voltages without a common impedance or matching convention are not fixed by aperture alone: $V_{\mathrm{rms}}^2=P_{\mathrm{available}}{R}_L$ for the respective matched loads. A short-dipole rather than half-wave convention changes the order-one gain factor, not the main estimate.

An accelerated charge radiates. For ultrarelativistic circular motion, the emitted radiation is concentrated into a cone of angular width of order ${\gamma }^{-1}$ about the instantaneous velocity. As the cone sweeps past a distant observer, propagation-time compression produces a short pulse of width of order $\rho /({\gamma }^{3}c)$, where $\rho$ is orbit curvature radius. Its Fourier spectrum extends to high harmonics, with conventional critical angular frequencyThis is a spectral scale, not a sharp upper cutoff. For transverse acceleration and $v\simeq c$, the classical radiated power isThe radiation is linearly polarized for an observer in the orbit plane, with elliptical polarization away from it. The two sides of the orbit plane have opposite handedness. The strong forward preference and high-frequency spectrum follow from the relativistic radiation calculation.^[ref]

Magnetic bending fields steer the charge but do no work, so maintaining a storage-ring beam requires energy replacement by accelerating fields. Bright electron beams provide useful radiation from infrared to X-rays; stronger curvature or higher electron energy shifts the spectral scale. The classical formulas assume negligible quantum recoil.

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.

Synchrotron radiation is emitted by relativistic charges accelerated along curved paths. A magnetic field can supply the transverse acceleration while doing no work on the charge; the radiated energy must ultimately be replenished if the orbit energy is to remain constant. Relativistic beaming confines much of the radiation to angles of order $1/\gamma$ about the instantaneous velocity.

For circular motion of curvature radius $\rho$ and $\gamma \gg 1$,The exact classical circular-orbit power contains an additional factor ${\beta }^4$. A distant observer sees short pulses as the beamed direction sweeps past, producing a broad high-frequency spectrum. Radiation is predominantly linearly polarized in the orbital plane and becomes elliptically polarized off it. The critical frequency is a spectral scale, not a monochromatic emission frequency or a hard cutoff.

Storage rings provide intense radiation for spectroscopy and diffraction, while relativistic electrons in astrophysical magnetic fields also produce synchrotron emission. The classical theory relates total power, angular beaming and spectrum to the particle's trajectory; sufficiently strong quantum recoil requires corrections.^[ref]

Both Thomson scattering and Rayleigh scattering can be understood as radiation from a dipole driven by the incident electric field. Their different frequency dependences come from the response of a free charge compared with a bound charge.

For a free particle of charge $q$ and mass $m$, a harmonic field gives $m\ddot{\mathbf{x}}=q\mathbf{E}$, hence $\mathbf{p}=q\mathbf{x}=-q^2\mathbf{E}/(m{\omega }^2)$. The radiated field contains two time derivatives of $\mathbf{p}$, cancelling the response's ${\omega }^{-2}$. The Thomson cross section is consequently independent of frequency in the classical low-recoil regime:For incident linear polarization with unit polarization $\mathbf{ϵ}$, summing over outgoing polarizations gives $d\sigma /d\Omega =r_q^2[1-|\hat{\mathbf{n}}\cdot \mathbf{ϵ}{|}^2]$. Radiation vanishes along the acceleration and is strongest perpendicular to it. Averaging an incident unpolarized light beam giveswhere $\Theta$ is the scattering angle. Scattering through $90^{\circ }$ is fully linearly polarized perpendicular to the scattering plane for this ideal isotropic dipole model. The electron case dominates over equally charged heavier particles because the cross section scales as $m^{-2}$. recoil corrections become necessary when photon energy is no longer small compared with $m{c}^2$.

For a bound charge with natural frequency ${\omega }_0$ and damping rate ${\gamma }_d$,Far below resonance, ${\alpha }_{\mathrm{e}}$ is approximately constant, soThis is Rayleigh scattering. It requires a scatterer small compared with the wavelength and, for the simple expression, an isotropic linear dipole response. Atoms, molecules and small dielectric particles provide examples. The stronger scattering of shorter visible wavelengths explains blue skylight; absorption and the nonuniform solar spectrum prevent treating that simple scaling as a complete color calculation.^[ref]

For isotropic electric scatterers the normalized angular and polarization patterns are the same as in Thomson scattering. The frequency scaling distinguishes their limits: a free charge has a displacement proportional to ${\omega }^{-2}$, whereas a tightly bound charge has an approximately frequency-independent displacement. Near resonance, far outside the small-particle limit, or with anisotropic polarizability, the simple Rayleigh formula must be modified. In particular, the electric-only pattern must not silently be applied to a perfectly conducting sphere whose magnetic dipole response is also retained.

A scattering measurement should subtract stray-light and detector backgrounds, control scatterer density and optical thickness, and calibrate detector response and incident power when comparing wavelengths. A polarization analyzer at $90^{\circ }$ tests the dipole pattern; multiple scattering or unresolved nonisotropic contributions can reduce its polarization.

antenna quantities. radiation resistance represents emitted power by $P_{\mathrm{rad}}=I_{\mathrm{rms}}^2{R}_{\mathrm{rad}}$ at the feed. It is not Joule heating in the antenna; a separate loss resistance accounts for that. antenna gain is the directional far-field intensity relative to an isotropic radiator supplied with the same input power:It includes radiation efficiency and equals antenna directivity only for a lossless antenna. antenna effective area is defined by the power available to a matched load, $P_{\mathrm{av}}=A_{\mathrm{eff}}{S}_{\mathrm{inc}}$, for incidence from a specified direction and matched polarization. It need not equal a literal geometric collecting area.

Gain–area relation. To derive the antenna gain-effective-area relation, take a passive reciprocal antenna. Its electromagnetic reciprocity makes the receive pattern proportional to the transmit pattern; write $A_{\mathrm{eff}}(\Omega )=KG(\Omega )$. The normalization follows from thermal equilibrium. In the Rayleigh-Jeans law limit, blackbody radiation radiance per unit frequency per unit solid angle is $2k_{B}T/{\lambda }^2$ for two polarizations. A single antenna polarization couples to one half, so the available incident noise power isThe open-circuit Johnson voltage spectrum of the antenna resistance is $4k_{B}T{R}_{\mathrm{rad}}d\nu$. A matched load receives half the open-circuit voltage, giving available power $k_{B}Td\nu$. In a lossless antenna at thermal equilibrium these powers agree; hence $\int A_{\mathrm{eff}}d\Omega ={\lambda }^2$. Since $\int Gd\Omega =4\pi$,The same efficiency factor multiplies receiving area and gain if reciprocal dissipative losses are included consistently. Impedance mismatch or polarization mismatch reduces delivered power further. The factor of two for thermal polarization modes and the factor of four between open-circuit voltage variance and matched-load power are separate effects.

Normalize the radar beam. Interpret the azimuth as signed about the beam centre at $\varphi =0$, so both sides of the narrow lobe are included. Put $u=\theta -\pi /2$. To leading order in the small angular width, $\sin \theta \simeq 1$ and the angular integrals can be extended over the real line:The Gaussian antenna beam solid angle gives the peak on the positive $x$-axis consequently hasHere $N$ is power per solid angle, equal to $r^2{S}_r$ in the far field. The leading neglected solid-angle correction is relative order ${\alpha }^2$.

Return using the supplied electric-dipole response. The electric-dipole radar cross section of a small conducting sphere determines this return. At target range $D$, the incident intensity is $S_t=C/D^2$. Write its root mean square electric field as $E_t$, with $S_t={ϵ}_{0}c{E}_t^2$. The induced electric dipole is $p_{\mathrm{rms}}=4\pi {ϵ}_0{a}^3{E}_t$. A dipole's angular radiation pattern, normalized by $\int {\sin }^2\chi d\Omega =8\pi /3$, gives $d{P}_{\mathrm{sc}}/d\Omega =(3P_{\mathrm{sc}}/8\pi ){\sin }^2\chi$. The backward direction is perpendicular to the induced dipole, so the supplied total dipole power givesThe received intensity is this angular power divided by $D^2$. Using the monostatic radar equation, or directly multiplying by the receive area, givesThe receiver threshold is a voltage across the matched load, so the detectable power is $V_m^2/R$, without another matching factor:This is the electric-dipole result following from the response data provided. It assumes $ka\ll 1$, far-field propagation on each leg, alignment at the beam maximum and no attenuation or additional receiver losses.

Perfect-conductor qualification. perfect-conductor sphere dipole interference arises because the sphere also excludes the oscillating magnetic field. Solving the magnetic boundary problem gives $\mathbf{m}=-2\pi a^3{\mathbf{H}}_t$ in addition to the electric dipole. The magnetic polarizability of a perfectly conducting sphere follows from its induced surface currents.^[ref] In the backward direction the magnetic contribution has half the electric scattering amplitude and the same sign, so the returned intensity is $9/4$ of the electric-only value. Equivalently, the monostatic radar cross section is ${\sigma }_{\mathrm{back}}=9\pi k^4{a}^6$, rather than $4\pi k^4{a}^6$. Including both leading dipoles therefore givesBoth results have been distinguished because the supplied auxiliary data describe only the electric dipole, while the literal perfect-conductor boundary conditions also generate the magnetic one. The backward interference is a leading-order effect, not a higher-order correction in $ka$.^[ref]

Use axes adapted to the two orthogonal rotations in the diagram:For a linearly oscillating Hertzian dipole with real unit axis $\mathbf{d}$ and total time-averaged radiated power $P$, the angular distribution of electric dipole radiation isThe normalization follows from $\int {\sin }^2\vartheta d\Omega =8\pi /3$. Equal readings therefore implyHere $d_z\ne 0$, since otherwise all components would vanish. Reversing the axis is immaterial, so choose $d_z>0$. The complete family isDefine $a=\sqrt{2}-1$, $b=-\sqrt{2}-1$, and $D=1+(s_B\sqrt{2}-1{)}^2+(s_C\sqrt{2}-1{)}^2$. Then $d_z^2=1/D$, and the absolute intensity fixesFor the measured distance and intensity, the allowed axes and powers are:

The intensity data do not select a unique orientation or total power. This is the dipole-axis ambiguity from three intensity measurements: the readings constrain squared projections and lose their signs. For example, the first branch makes an angle $30.36^{\circ }$ with $A$, but the other branches are equally consistent with the three measurements. A further suitable observation direction or polarization measurement is needed to remove the ambiguity. The calculation uses radiation-zone intensities, as appropriate to the Hertzian dipole model.

A non-polarizing beam splitter divides the incident spectrum into a fixed arm and an arm with a translating mirror. Recombining their returned fields gives a detector signal as a function of mirror position $x$. The round-trip optical path difference is $D=2x$ relative to the equal-arm position, so the delay is $\tau =2x/c$ in air approximated as vacuum. Record both detector signal and calibrated position, subtract the constant background, and take a Fourier transform of the interferogram.

For matched arm polarizations and balanced throughput, write the ideal signal aswhere $S(\omega )$ is the incident spectral power density and $C$ is a constant set by the beam splitter and detector. This is the cosine-transform form of Fourier transform spectroscopy, or equivalently the Wiener-Khinchin theorem applied to delayed self-interference.

The delay-range resolution in Fourier transform spectroscopy follows because recording a finite delay range multiplies the correlation by a window. Its transform broadens each spectral line by an instrumental width of order the reciprocal delay range. A mirror travel range $L$ creates a delay range of order $2L/c$, soAt fixed wavelength, absolute wavelength resolution therefore improves inversely with travel range. The numerical coefficient depends on whether $L$ is a one-sided or total travel, on the window or apodization, and on whether resolution means first zero, half width or another criterion. The proportionality is the requested convention-independent result. Position sampling must also be fine enough to sample the fastest fringe variation without aliasing.

Antenna gain in direction $(\theta ,\varphi )$ compares the far-field power per solid angle with an isotropic radiator using the same accepted input power:It includes radiation efficiency; for the lossless dipole $P_{\mathrm{in}}=P_{\mathrm{rad}}$. The antenna effective area is defined for incidence from a specified direction with matched polarization and conjugate impedance matching:Its units are square metres and it need not equal the physical metal area.

For ordinary reciprocal antennas, electromagnetic reciprocity makes the coupling between two matched ports the same when transmitter and receiver are exchanged. Free-space transfer in one direction is proportional to $G_1{A}_2$, while the reverse transfer is proportional to $G_2{A}_1$. Thus $A/G$ must be a universal factor at that wavelength, independent of antenna or direction. The supplied relation fixes it asEquivalently, thermal detailed balance prevents an angular receiving pattern from collecting more radiation than its reciprocal emitting pattern would return in equilibrium. Reciprocity and the specified matching are the assumptions behind this statement; it is not a general theorem for active nonreciprocal devices.

Integrating the ${\sin }^2\theta$ radiation pattern uses $\int {\sin }^2\theta d\Omega =8\pi /3$, so its lossless gain isThe receiver lies on the $x$ axis, broadside to both $z$-directed dipoles. Thus $G_t=G_r=3/2$ and their linear polarizations match. At $f=100\mathrm{MHz}$, $\lambda =c/f\simeq 3.00\mathrm{m}$, givingAt separation $r=100\mathrm{m}$, the incident flux is $P_0{G}_t/(4\pi r^2)$. The Friis transmission equation therefore gives the matched-load powerThe far-field and weak-coupling approximations are well satisfied at a distance of about $33$ wavelengths for an electrically short dipole. Matching refers to the available received power, not interception of all power in a geometric area equal to the wire's physical size.

The outgoing radiation field has magnitude $|E_{\theta }|=|\ddot{p}(t_r)|\sin \theta /(4\pi {ϵ}_0{c}^{2}r)$ and $B_{\varphi }=E_{\theta }/c$. The radial Poynting vector, averaged over one cycle, givesThus the $(x,z)$ cut has a two-lobed ${\sin }^2\theta$ shape, with zeros along $\pm z$ and maxima along $\pm x$. In the $(x,y)$ plane, $\theta =\pi /2$ throughout and the cut is a circle: there is no azimuthal dependence. At fixed direction the flux falls as $1/r^2$, while power per solid angle is independent of observation radius.

As a receiver for propagation along $x$, the antenna responds to the incident electric field component along $z$. For linear polarization at angle $\psi$ to $z$, the available load power is proportional to ${\cos }^2\psi$: it is maximal for $z$ polarization and zero for $y$ polarization. Equal incident intensity in circular $yz$ polarization gives half the matched-linear power. This polarization response follows from the induced voltage's projection along the dipole axis.

The receiving dipole remains matched. To model its actual reradiation, let its feed impedance be $R_{\mathrm{rad}}+iX$ with no ohmic loss, and its matched load be $R_{\mathrm{rad}}-iX$. The induced current dissipates equal powers in the load and in radiation resistance. Thus matched-load scattering of a Hertzian dipole has $P_{\mathrm{sca}}=P_2$, even though its port is matched.

The scattered wave has the same dipole gain toward the original antenna. Applying the reciprocal link factor a second time gives the reciprocal Hertzian-dipole echo:Each leg supplies an inverse-square spreading factor. This is the carrier-averaged power during a returned pulse with the stated transmitted power during its pulse. The pulse-train average additionally depends on duty cycle, which is not given. Treating the pulses as narrowband keeps the same gain and wavelength approximation.

A matched feed does not make a physical dipole a perfect absorber with zero external scattering. A model retaining only an antenna load-mismatch echo would set its reflected travelling-port wave to zero and would not determine this reradiation. Conversely, an arbitrary antenna structure can add scattering not fixed by gain and effective area alone. The numerical answer uses the ideal lossless Hertzian-dipole scattering model just stated.

Denote the orbit radius by $b$ here to distinguish it from observation distance; it is the quantity called $R$ in this part of the paper. WriteFor observation at spherical coordinates $(\theta ,\varphi )$, measured from the orbit normal,The acceleration magnitude is constant, $|\mathbf{a}|=b{\omega }^2$. Therefore the radiation pattern of a circularly moving nonrelativistic charge isTime averaging removes the azimuthal dependence. The axial intensity is twice the equatorial intensity, and no direction has zero mean intensity: the instantaneous dipole axis continually rotates. This is also the sum of the two orthogonal linear-oscillation patterns with their relative phase quadrature. At equal radius and linear-oscillation amplitude, the total mean power is twice the linear result. The calculation requires $b\omega \ll c$; relativistic synchrotron radiation is outside this approximation.