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For a lossless ideal Hertzian dipole driven by an incident wave, use its induced open-circuit voltage and its feed impedance $R_{\mathrm{rad}}+iX$. A conjugately matched load has impedance $R_{\mathrm{rad}}-iX$. The same induced current therefore delivers equal power to the load and to radiation resistance: $P_{\mathrm{sca}}=P_{\mathrm{load}}$. The scattered wave has the usual dipole angular pattern. In direction of gain $G$, its scattering cross-section is $G{A}_{\mathrm{eff}}$, since the load receives $S_{\mathrm{inc}}{A}_{\mathrm{eff}}$.

This ideal single-mode scattering is not the same as a signal that travels through an already matched antenna port, reflects from a mismatched terminating load, and returns. A zero port reflection coefficient removes that latter echo but does not make the physical dipole invisible to the incident field. Additional structural scattering, ohmic loss or a different load changes the result.

The operating voltage and electric current maximizing delivered photovoltaic power.

Adjustment of the electrical operating point of a photovoltaic device to maximize delivered power.

For a series Maxwell model with $\tau =\eta /k$, $E^{\prime \prime }=k\omega \tau /(1+{\omega }^2{\tau }^2)$. At fixed strain amplitude the energy dissipated per cycle is $\pi E^{\prime \prime }{\epsilon }_0^2$ and peaks at $\omega \tau =1$. At fixed stress amplitude it is instead $\pi {\sigma }_0^2/(\eta \omega )$, which has no finite-frequency maximum. Specifying the imposed variable and whether loss means energy per cycle or power is essential.

Keep the illumination and cell temperature approximately constant. Connect an ammeter and an adjustable resistor as a series circuit across the solar cell; connect a high-resistance voltmeter across its terminals. Vary the electrical resistance and record pairs of electric potential difference $V$ and delivered electric current $I$. This measures the illuminated current-voltage characteristic.

A load close to a short circuit estimates $I_{\mathrm{sc}}$ at $V\simeq 0$. With the load disconnected, an open circuit measurement using a high-resistance voltmeter estimates $V_{\mathrm{oc}}$ at $I\simeq 0$. Neither endpoint delivers useful power: $P=VI$ vanishes there. The intermediate readings locate the maximum of $VI$ and determine the solar-cell fill factor. Holding illumination fixed matters because the photogenerated electric current changes with light intensity; holding temperature fixed matters because the diode saturation current and junction response change with it.

This experiment distinguishes producing an electric potential difference from delivering energy to a load. To determine conversion efficiency as well, measure the incident power on the illuminated area, rather than equating it to a lamp's electrical rating.

Mechanical impedance is the ratio of a driving force complex amplitude to the resulting velocity complex amplitude at a specified point and in a specified direction. Its units are $\mathrm{N}\mathrm{s}{\mathrm{m}}^{-1}$. The direction of the force, the time convention and whether one is describing a travelling component or a total field must be stated. A positive real travelling-wave mechanical impedance transfers mean power $⟨P⟩=Z|v{|}^2/2$ when $|v|$ is the peak velocity wave amplitude.

Consider an original idealized narrow-band model for one matched spatial mode and one polarization of waves. In the classical Rayleigh-Jeans law regime, an available thermal power can be written $P=k_{B}T\Delta \nu$, where $\Delta \nu$ is the effective bandwidth and $T$ the equivalent input noise temperature. Assume a linear detector with stable positive gain and absorb its bandwidth, Boltzmann constant and response factor into $g$. Then the output is $S=g(T+T_r)$, with input-referred receiver noise temperature $T_r$. These are model assumptions, not a reconstruction of the historical electronics.

Two references with known noise temperatures $T_h\ne T_c$ giveso the two-load radiometer calibration yieldsSubtracting the reference equations eliminates the unknown receiver term; substituting back determines it. An arbitrary stable output offset can likewise be eliminated from the interpolation for $T_s$, although separating that offset from $T_r$ then requires additional information.

For deliberately chosen inputs, set $T_h=300\mathrm{K}$, $T_c=4.2\mathrm{K}$, $T_r=40\mathrm{K}$ and $g=2mV/K$. These are equivalent noise temperatures; physical load thermometry must be converted consistently at the observing frequency. The reference outputs are $S_h=680\mathrm{mV}$ and $S_c=88.4\mathrm{mV}$. A sky-input output $S_s=90\mathrm{mV}$ gives $T_s=5.0\mathrm{K}$. If independent estimates assign $1.4\mathrm{K}$ to atmosphere, $0.5\mathrm{K}$ to antenna loss and $0.1\mathrm{K}$ to ground pickup in this lumped additive model, the residual is $3.0\mathrm{K}$.

The subtraction exposes what a noise budget must achieve. A $1\mathrm{K}$ error in the assigned atmospheric term becomes a $1\mathrm{K}$ error in the residual, even with exact detector calibration. Multiplicative attenuation requires a different correction: if an isothermal atmosphere has transmission $t$, its ideal contribution is $t{T}_{\mathrm{cosmic}}+(1-t)T_{\mathrm{atm}}$ in the same classical equivalent-temperature convention. Recovering $T_{\mathrm{cosmic}}$ then requires both subtracting atmospheric emission and dividing by $t$. The additive toy budget applies when its terms already refer to the same receiver input.

There is also a conditioning issue. If $S_h-S_c$ is small, a fixed output error strongly changes the inferred temperature. Widely separated reference temperatures improve the scale determination only if the receiver remains linear over that range; measuring the reference repeatedly tests stability rather than proving linearity. The experiment therefore motivates a model, a calibration and a set of independent checks, each with its own assumptions.

For a far-field free-space link with one reciprocal antenna transmitting and receiving at range $D$, the outward intensity is $P_{T}G/(4\pi D^2)$. The backward scattered intensity at the receiver is this times ${\sigma }_{\mathrm{back}}/(4\pi D^2)$. Multiplying by the matched antenna effective area $G{\lambda }^2/(4\pi )$ gives the displayed received power. Attenuation, misalignment and polarization mismatch require additional factors. The threshold for a root mean square voltage $V_m$ across a real matched load $R$ is $V_m^2/R$.

For a small, uniform surface at temperature $T$, enclosed by much larger black walls all at temperature $T_s$, the outward net heat transfer by thermal radiation isAssume an opaque grey body that is a diffuse thermal emitter, with the same emissivity $ϵ$ and radiometric absorptance at the surface's current temperature. Its emitting area $A$ sees only the enclosure: the view factor to the walls is one, with no self-viewing surface, opening or intervening obstruction. The intervening medium must neither absorb nor emit appreciable thermal radiation; a vacuum supplies a useful idealization. The enclosure is maintained at uniform $T_s$. The small-body assumption makes that reservoir description physically reasonable; genuinely black isothermal walls supply the incoming irradiance even without an infinite area ratio. Positive ${\dot{Q}}_{\mathrm{rad}}$ means net loss by the surface; negative means net gain. This expression counts thermal radiation alone, not all heat transfer.^[ref]

The reasoning is a two-way energy balance. Black walls supply irradiance $E=\sigma T_s^4$, so the surface absorbs $P_{\mathrm{abs}}=ϵAE$ and emits $P_{\mathrm{emit}}=ϵA\sigma T^4$. Subtract these to obtain ${\dot{Q}}_{\mathrm{rad}}$. For an opaque surface the remaining fraction $1-ϵ$ of incoming radiant flux is reflected; counting emitted plus reflected outgoing radiant flux minus total incoming radiant flux gives the same result. Kirchhoff's law of thermal radiation justifies equal spectral absorption and emission factors; the grey body assumption lets one factor be taken outside both spectral integrals.

Choose $ϵ=0.8$, $A=0.01{\mathrm{m}}^2$, $T=500\mathrm{K}$ and $T_s=300\mathrm{K}$. These are stipulated model inputs, not measured material properties. Using $\sigma \approx 5.670374419\times 10^{-8}\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-4}$ givesThe incoming irradiance is $459.300\mathrm{W}{\mathrm{m}}^{-2}$, corresponding to $4.5930\mathrm{W}$ incident on the surface; only $80\%$ is absorbed. At $T=T_s=300\mathrm{K}$, emitted and absorbed powers are both $3.6744\mathrm{W}$, so radiative equilibrium has zero net exchange despite continuing emission. Changing only emissivity to $0.2$ reduces both one-way powers and the net loss by a factor of four, giving $6.1694\mathrm{W}$ net. A low-emissivity coating therefore suppresses both radiative cooling to colder walls and radiative heating from hotter walls.

Near equal temperatures, write $T=T_s+\Delta T$. Taylor's theorem givesFor constant emissivity, the radiative heat transfer coefficient $h_{\mathrm{rad}}$ is approximately $4.8992\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-1}$ at $T_s=300\mathrm{K}$ and $ϵ=0.8$. This local linear approximation is useful for small temperature differences; the $200\mathrm{K}$ difference in the worked example needs the full fourth-power expression.

A cooling experiment requires a wider first law of thermodynamics balance. In a uniform-temperature model without a heater,where $C(T)$ is the sample's heat capacity and the last two terms are signed outward convection and thermal conduction losses. A vacuum suppresses convection but does not eliminate thermal conduction through supports or sensor leads. A cooling slope therefore cannot establish emissivity unless these paths and heat capacity are independently constrained. Multiple wall temperatures, incomplete view factors, wavelength-selective materials or appreciable intervening absorption require a more complete radiative transfer calculation.

A noise temperature expresses received or internally added noise power on an equivalent temperature scale. For one matched receiving channel in the classical thermal limit, an ideal rectangular bandwidth $\Delta \nu$ gives $P=k_{\mathrm{B}}{T}_N\Delta \nu$, with Boltzmann constant $k_{\mathrm{B}}$. An equivalent noise temperature need not equal a component's physical temperature.

The distance for an order-one self-phase modulation at peak power $P$ in a lossless Kerr envelope model. With attenuation, the phase accumulates over an effective interaction length instead.

For lossless parallel Hertzian transmitting and receiving dipoles separated by distance $r$ at angle $\theta$ to their common axis, directional gains are each $G=3{\sin }^2\theta /2$. The emitted polarization matches the receiving axis's transverse projection. Matched-load power is therefore $P_L=P[3{\sin }^2\theta /2{]}^2[\lambda /(4\pi r){]}^2$. One angular factor describes transmission and the other reception. This assumes reciprocal ideal dipoles, conjugate matching, far-field separation and negligible dissipative loss.

For a steady point-like astronomical source with isotropy of emission, in transparent space with negligible relativistic or cosmological effects, conservation of energy spreads its luminosity over a sphere of surface area $4\pi d^2$. Its bolometric observed radiative flux isThis inverse-square law reflects geometric dilution: the same emitted power crosses increasingly large spheres. It does not require light to lose energy on the journey. For a fixed luminosity, doubling Euclidean distance quarters the observed radiative flux. ^[ref].

With $L_0=10^{26}\mathrm{W}$ and $d_0=20$ parsecs, the result is $F_0\simeq 2.09\times 10^{-11}\mathrm{W}{\mathrm{m}}^{-2}$. At $40$ parsecs, the same source gives $F_0/4\simeq 5.22\times 10^{-12}\mathrm{W}{\mathrm{m}}^{-2}$. An ideal collecting area of $1{\mathrm{m}}^2$ at $20$ parsecs intercepts approximately $2.09\times 10^{-11}\mathrm{W}$, before detector losses.

Ohm's law states that an Ohmic conductor has electric current proportional to its applied electric potential difference, with constant electrical resistance under fixed material conditions:Here $V=\varphi (A)-\varphi (B)$ is the terminal electric potential difference, and positive $I$ enters at $A$ and exits at $B$. This sign convention makes $VI$ the electrical power absorbed. The straight-line relation through the origin lets one constant predict the response at many applied electric potential differences. Merely defining $R=V/I$ at one nonzero operating point does not establish Ohm's law: a component may have a ratio that varies with applied electric potential difference. Material and temperature must be specified and held fixed; many components have a nonlinear response. ^[ref].

For the ideal resistor model $R=2\Omega$ at electric potential difference $V=5\mathrm{V}$,This power is converted to internal energy by Joule heating; a watt is a joule per second. If the model holds for time $10\mathrm{s}$, the transferred energy is $125\mathrm{J}$. Doubling electric potential difference doubles electric current and quadruples power for the same fixed $R$. The model keeps material temperature fixed even while calculating Joule heating; appreciable changes in temperature may require a changing electrical resistance. ^[ref].

In 1827, Georg Ohm published his mathematical account of electrical transport in Die galvanische Kette, mathematisch bearbeitet. On printed page 36, his relation $S=A/L$ expresses electric current in terms of total driving tension and a reduced length incorporating electrical conductivity and geometry. This is historical notation: his $A$ is not the wire cross-sectional area used here, and his $L$ is not simply the wire's physical length. The striking idea is that material response and geometry can be combined into one quantity governing an electrical circuit. ^[ref].

Physical Review 75, 1912–1925 (1949). Primary classical calculation of total radiated power and the angular and spectral properties of accelerated energetic electrons.

Optical-electrical heating non-equivalence means that equal absorbed optical and electrical powers do not necessarily give identical thermometer responses. Light and the heater may deposit heat in different places, producing different temperature gradients and different routes of thermal conduction to the support. Holding one thermometer reading fixed need not hold every part of the cavity at the same temperature. This differs from incomplete radiometric absorptance: one concerns where absorbed heat goes, the other how much incident light becomes absorbed heat.

A useful accounting convention iswhere the signed correction ${\delta }_{\mathrm{ne}}$ is determined by characterization of the instrument and represents the residual departure from the lumped electrical substitution heat balance. It is not a universal extra source of energy. Varying the beam location or heater location, comparing equivalent temperature excursions, and checking repeatable dark returns can test the assumptions; no single small correction proves that every other correction vanishes.

Optical fiber attenuation is the reduction of transmitted optical power along an optical fiber. Define a positive loss in decibels by comparing optical power $P(0)$ already launched into the guide with surviving optical power $P(z)$:The factor ten applies to a power ratio; the factor twenty applies to a wave amplitude ratio only when power is proportional to its square under the same conditions.^[ref]

For a uniform illustrative model with constant loss coefficient ${\alpha }_{\mathrm{dB}}$ in $dB/km$ and distance $z$ in kilometres,For ${\alpha }_{\mathrm{dB}}=0.2dB/km$ over $10\mathrm{km}$, the loss is $2\mathrm{dB}$ andThus about $63.1\%$ of the launched optical power survives and $36.9\%$ is lost. This is an illustrative coefficient, not a property of all optical fibers, and is independent of the earlier illustrative refractive indices.

The exponential decay follows from assuming each equal distance interval transmits the same fraction. For two such intervals, power ratios multiply; their logarithms add, so the losses in decibels add. Two $5\mathrm{km}$ sections each transmit $10^{-0.1}\simeq 0.794328$; together they transmit $(0.794328{)}^2\simeq 0.630957$. The loss curve is straight in decibels and curved on a linear power scale.

Wave absorption and scattering of light can remove guided optical power even in a straight fibre, while bends can let it escape. Entrance losses, joints and connectors contribute additional losses; they belong separately from a uniform distance coefficient. This is why a measured end-to-end power ratio need not equal the straight-fibre model.^[ref]

For the damped harmonic oscillator, the mechanical energy is the sum of kinetic energy and spring potential energy:Thus $b{\dot{x}}^2$ is the power transferred out of the oscillator's mechanical energy. For $b>0$, the rate vanishes instantaneously at a turning point, despite nonzero stored potential energy; it is not a constant loss per unit time. In eddy-current damping the loss becomes Joule heating. ^[ref]

For an underdamped response, advancing one oscillation period multiplies both displacement and velocity by $e^{-\zeta {\omega }_0{T}_{\mathrm{d}}}$, so $E(t+T_{\mathrm{d}})=e^{-2\zeta {\omega }_0{T}_{\mathrm{d}}}E(t)$. This exact cycle relation does not imply that instantaneous energy is a pure exponential function throughout each cycle.

An outgoing radiation pulse carries a finite amount of electromagnetic energy through successively larger enclosing spheres. In vacuum, far from a localized source and without absorption or additional reflections, let $P_{\mathrm{emit}}(t)$ be the emitted power. The retarded time at radius $r$ is $s=t-r/c$. Define ${\overline{S}}_r$ as the angular average of the radial Poynting vector over the sphere, not a cycle average:Here $\overline{u}$ is the angularly averaged energy density in the leading radiation zone approximation. Multiplying by shell area removes geometric dilution: the energy per unit radial thickness translates outward without losing its shape. These angular averages do not assume that an actual electric dipole emits equally in all directions. Nor do they specify an exact spherically symmetric transverse solution of the Maxwell equations.

For an explicitly stipulated model, emitAfter emission ends, $t>\tau$, the pulse lies between $r=c(t-\tau )$ and $r=ct$, provided this shell is in the radiation zone. Its radial thickness remains $c\tau$. Substituting $s=t-r/c$ in the shell integral $\int 4\pi r^2\overline{u}dr$ gives $U=P_0\tau /2$ at every such time. The same substitution in the time integral of $4\pi r^2{\overline{S}}_r$ at any enclosing sphere gives that same total crossing energy. This is conservation of energy in two different views.

Along any fixed direction with nonzero outgoing power, compare the same part of the pulse at two radii, hence at different arrival times. The leading electric field amplitude falls as $1/r$, while its squared wave intensity falls as $1/r^2$. At $2r$ the field is one half and the energy flux one quarter; the signal arrives $r/c$ later. A small fixed area receiver therefore intercepts less energy at greater distance even though every enclosing sphere receives the same total pulse energy. The inverse-square law expresses spreading over area, rather than destruction of energy. Close to the source, the neglected near and far field terms prevent extrapolating these formulas down to $r=0$.

The passive sign convention defines terminal electric potential difference $V=\varphi (A)-\varphi (B)$ and positive electric current entering at $A$. Then $VI$ is absorbed electrical power: positive means the component receives energy, negative means it supplies energy. Reversing both reference directions leaves the product unchanged. For a positive electrical resistance obeying Ohm's law, both $V$ and $I$ reverse together and Joule heating remains $I^{2}R\ge 0$.