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A beam splitter is an optical instrument that routes an incident beam of light into transmitted and reflected beams. It also combines incident complex amplitudes: two inputs can contribute to the same output. A common geometry has four optical ports, with two chosen as inputs and two as outputs. A beam splitter changes the field; a photodetector supplies a measurement.^[ref]

For one illuminated input of a passive beam splitter, at specified wavelength, angle of incidence and electromagnetic polarization, define power transmittance $T=P_{\mathrm{transmitted}}/P_{\mathrm{in}}$, reflectance $R=P_{\mathrm{reflected}}/P_{\mathrm{in}}$ and radiometric absorptance $A=P_{\mathrm{absorbed}}/P_{\mathrm{in}}$. In steady operation, with no gain or other source, no net stored energy change, and all escaping light included in the transmitted or reflected collection,This is conservation of energy divided by incident power. If $T$ and $R$ count only the intended output beams, include a separate scattered or otherwise uncollected fraction $L$: then $T+R+A+L=1$. Missing output power need not be wave absorption. For an original specified example, $P_{\mathrm{in}}=10\mathrm{mW}$, $T=0.45$, $R=0.50$ and $A=0.05$ give $4.5\mathrm{mW}$ transmitted, $5.0\mathrm{mW}$ reflected and $0.5\mathrm{mW}$ absorbed; these are assumed values, not measurements.

A 50:50 beam splitter specifies equal output powers for one input, not equal complex amplitudes without a relative wave phase. Recombination needs the beam splitter scattering matrix, including its wave phases. A nonpolarizing beam splitter approximately divides both orthogonal linear polarizations in the same power ratio over its specified operating range; a polarizing beam splitter deliberately routes the two differently.

The useful surprise is that the same component divides power with one input and steers power by wave interference with two inputs. In a Michelson interferometer or Mach-Zehnder interferometer, mirror motion changes relative wave phase, turning a small change in optical path length into a measurable change in output power.

In NIST's 2001 comparison, the Laser-Optimized Cryogenic Radiometer held its cavity near $5.4\mathrm{K}$ and controlled the heat sink separately. Opening the beam reduced heater power at fixed cavity temperature; an AC bridge read the temperature. Tests used stabilized, spatially filtered lasers at $488$, $514$ and $633\mathrm{nm}$, with horizontal input and a movable photodiode transfer detector in the cavity's optical plane. The paper contrasts this with the High Accuracy Cryogenic Radiometer's optical heating followed by two electrical heating cycles and interpolation.^[ref]

The two laboratories compared calibrations of the same transfer detectors, and checked their spatial response. Alignment differences and transported dust limited that comparison; its agreement did not prove universal equivalence. Its correction tables retain window transmission, cavity radiometric absorptance, alignment, electrical readout and optical-electrical heating non-equivalence. The heat sink was controlled, but §3 does not specify its numerical setpoint: the $5.0\mathrm{K}$ reference in the preceding example is stipulated.^[ref]

For a reproducible substitution measurement, record the beam's wavelength and entrance boundary, receiver and reference temperatures, heater terminal voltage and electric current, dark readings before and after illumination, and the settling rule. Check beam collection at the aperture, window transmission, radiometric absorptance, deposited lead heating, and sensitivity to beam position. Repeatability alone misses a stable correction error. Photometric calibration can subsequently use this radiant flux scale, but photometry still needs the appropriate spectral visual weighting; the cold receiver does not itself measure luminous flux.

A cryogenic radiometer uses low-temperature operation to improve electrical substitution radiometry through particular material and thermal properties. Lower heat capacity of a copper receiver permits a substantial cavity without proportionately slow relaxation; cold shields reduce unwanted thermal radiation, and heater leads with superconductivity can suppress resistive lead heating. These features do not make radiometric absorptance exactly one or abolish optical-electrical heating non-equivalence.^[ref]

Cold surroundings matter as much as a cold receiver. In a simplified gray-surface model, net outward thermal radiation is $ϵ\sigma A(T^4-T_s^4)$, using emissivity $ϵ$, area $A$, shield temperature $T_s$ and the Stefan-Boltzmann constant $\sigma$. Cold shields reduce the incoming term as well as the receiver's emission; cooling the receiver alone while leaving it exposed to warm surroundings does not remove incoming radiation. An entrance window still provides an optical and thermal path. Vacuum reduces gas-mediated heat transfer, while an intentional radiometer thermal link remains necessary. This model explains specific losses, rather than promising a universal improvement factor or perfect accuracy.

An explicit lumped model makes the cancellation understandable. Treat the receiver as having uniform temperature $T$, heat capacity $C$, heater power $P_h$ deposited in the receiver, net parasitic input $P_{\mathrm{par}}$, and outward heat transfer $Q_{\mathrm{out}}(T,T_b)$ to a reference at temperature $T_b$. With incident radiant flux $P_{\mathrm{in}}$ and radiometric absorptance $a$, conservation of energy givesThe dark state has no admitted beam. Compare settled dark and light states, denoted $d$ and $l$; each has $dT/dt=0$. Subtracting their balances giveswhere $\Delta Q_{\mathrm{out}}=Q_{\mathrm{out},l}-Q_{\mathrm{out},d}$ and $\Delta P_{\mathrm{par}}=P_{\mathrm{par},l}-P_{\mathrm{par},d}$. At identical $T$ and $T_b$, with unchanged surroundings and parasitic heating, both correction differences vanish. Thus $P_{\mathrm{abs}}=\Delta P_h$, and $P_{\mathrm{in}}=\Delta P_h/a$. Notice the signs: extra stray thermal radiation admitted with the beam also reduces heater power, so it must be subtracted from that reduction before attributing the result to the intended beam.

Measure the heater's own terminal voltage $V_h$ and electric current $I_h$: for steady electrical heating, $P_h=V_h{I}_h=I_h^2{R}_h$, where $R_h$ is its electrical resistance. A supply voltage that includes wire voltage drops counts Joule heating in the leads too; only the portion deposited in the receiver belongs in this balance. Changes in that deposited lead heat belong in $\Delta P_{\mathrm{par}}$. Optical-electrical heating non-equivalence adds a further instrument-specific correction when the uniform-temperature model is insufficient.^[ref]

An electrical substitution radiometer measures radiant flux by replacing absorbed optical power with calibrated Joule heating. At fixed temperature, feedback control holds the receiver equally warm with the beam blocked or admitted: after corrections, absorbed optical power equals the reduction in supplied electrical heating. Incident radiant flux additionally requires radiometric absorptance and entrance-path corrections. This connects radiometry to electrical standards.^[ref]

Picture a small absorbing cavity connected to a colder reference block by a radiometer thermal link. A heater supplies Joule heating, a thermometer senses the receiver, and feedback control adjusts the heater. When light supplies some of the heating, the controller supplies less. The readout is the change in heater power, rather than the beam's brightness or its number of photons. NIST's institutional description gives a concrete receiver with a conical interior, absorbing coating and weak Kapton support, operated with a receiver heater and thermometer.^[ref]

Stipulate a model receiver held at $T=5.4\mathrm{K}$ with a reference at $T_b=5.0\mathrm{K}$. Choose $G=5\mathrm{mW}{\mathrm{K}}^{-1}$, so outward thermal conduction is $G(T-T_b)=2\mathrm{mW}$. Ignore other heat transfer, take $P_{\mathrm{par}}=0$ in both states, and stipulate perfect optical/electrical thermal equivalence. These values specify an original example; they are not measurements or claimed performance of a NIST instrument.

With the beam blocked, supply $P_{h,d}=2\mathrm{mW}$. With light admitted, the required heater power is $P_{h,l}=1.005\mathrm{mW}$. For stipulated radiometric absorptance $a=0.995$ at the chosen single wavelength,The illuminated receiver still gets $1.005+0.995=2\mathrm{mW}$ of heating. The remaining $0.005\mathrm{mW}$ of incident radiant flux escapes without heating it. Calling the measured heater reduction “incident power” would underestimate the stipulated incident radiant flux by $0.5\%$.

Two variations expose different pitfalls. First, if admitting the beam also adds $0.010\mathrm{mW}$ of stray heating while the same $0.995\mathrm{mW}$ is absorbed from the intended beam, the heater falls to $0.995\mathrm{mW}$. The apparent reduction is now $1.005\mathrm{mW}$; subtracting $\Delta P_{\mathrm{par}}=0.010\mathrm{mW}$ recovers the beam's absorbed power. Second, keep the original heater readings but stipulate a light-state receiver offset $\delta T=+1\mathrm{mK}$ at fixed reference temperature. The link carries an extra $G\delta T=0.005\mathrm{mW}$; the inferred absorbed power becomes $1.000\mathrm{mW}$, and incident radiant flux becomes $1.000/0.995\simeq 1.005025\mathrm{mW}$. This second variation deliberately violates exact fixed-temperature substitution.

The heater can supply only nonnegative power in this model. Consequently an absorbed beam above the available dark heater power cannot be compensated at the chosen setpoint merely by reducing the heater. The operating point, link, or admitted radiant flux must change; zero heater demand is a range boundary, not evidence of greater radiant flux sensitivity.

A grey body has spectral emissivity independent of wavelength over the range in which the approximation is used. Under Kirchhoff's law of thermal radiation, the corresponding radiometric absorptance is also wavelength-independent. This allows the same factor to weight emission at $T$ and black-enclosure irradiation at a different $T_s$ in net radiative heat exchange. The factor can still depend on temperature; treating it as constant over a cooling curve is an additional assumption. Wavelength independence alone does not imply direction independence.^[ref]

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.

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.

A 2013 experiment by Paul A. Williams and collaborators mounted a 25-mm dielectric mirror on a commercial force-restoration balance. Removing its pan-weight compensation spring allowed horizontal force sensing with a horizontal laser path. They measured mirror optical reflectance and varied incidence from about $7^{\circ }$ to $45^{\circ }$. A 1071-nm laser, capable of 530 W, was switched at 0.1 Hz. Digitized mass-equivalent readings were converted to force using $g$; quadrature demodulation isolated the switched signal from drift. ^[ref]

Measurement calibration used a 20-g mass, with checks down to 1 mg. The optical forces lay below that checked range, limiting the accuracy claim. An angle scan tested the predicted cosine dependence at fixed intercepted power. Replacing the mirror with a carbon nanotube coated absorber increased absorbed power while reducing the predicted optical force; no modulation peak was detected. This bounded gas-heating artifacts under those conditions, rather than proving their universal absence. The fitted slope was about 7% below $2/c$. Inferring laser power requires calibrated force, optical reflectance, radiometric absorptance and angle of incidence, with measurement uncertainty for each. ^[ref]

For normal incidence in vacuum, a uniform beam delivers energy $IA\Delta t$ to an illuminated surface area $A$ during time $\Delta t$. A perfect absorber receives momentum $IA\Delta t/c$, so force divided by surface area gives $p_{\mathrm{abs}}=I/c$. A perfect stationary mirror sends the photons back with reversed momentum, so momentum conservation gives twice the normal transfer, $p_{\mathrm{refl}}=2I/c$. These are time-averaged pressures; the microscopic arrivals need not be uniform.

For a stipulated model light intensity $I=1000\mathrm{W}{\mathrm{m}}^{-2}$ and the exact SI speed of light $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$,These are computed predictions, not measured data or a universal sunlight value. If the beam fully covers a normally oriented surface area $A=0.010{\mathrm{m}}^2$, the intercepted radiant flux is $\Phi =IA=10\mathrm{W}$ and the absorber's force is $\Phi /c\simeq 33.356\mathrm{nN}$. Using the total source power would be wrong if only part of its beam reaches the receiver. The absorption prediction assumes complete photon absorption, while the reflection prediction assumes complete specular reflection; transmission, scattering of light and directed thermal photon emission require their own outgoing momentum accounting.

Measure the angle of incidence $\theta$ between the incoming path and the surface's normal line, with $0\le \theta <\pi /2$. There are two separate cosine factors. A fixed actual surface area $A$ intercepts only its projected area $A\cos \theta$, so $\Phi =IA\cos \theta$. Each incoming photon supplies normal momentum $(E_{\gamma }/c)\cos \theta$ on absorption, or twice that on specular reflection. Thus, for radiometric absorptance $a$ and reflected power fraction $R$,This model assumes transmitted light continues straight in the same medium and carries away its original momentum. It omits thermal recoil and assumes a stationary receiver. At $\theta =60^{\circ }$, fixed beam light intensity therefore gives one quarter of the normal-incidence pressure. For fixed intercepted radiant flux $\Phi$, however, the normal force has only one cosine factor. Equivalently, if the given quantity is receiving-surface irradiance $E_{\mathrm{inc}}=I\cos \theta$, then $p_{\perp }=(a+2R)E_{\mathrm{inc}}\cos \theta /c$. The distinction matters when interpreting an experiment's angle scan. ^[ref]

Absorption also transfers a tangential momentum component at oblique incidence. It contributes a tangential force, rather than normal pressure. Ideal specular reflection preserves the tangential momentum component, so it transfers only normal momentum.

Radiometric absorptance is the fraction of incident radiant flux converted to heat in the specified receiver: $P_{\mathrm{abs}}=a{P}_{\mathrm{in}}$, with $0\le a\le 1$ for a passive receiver. Specify the entrance boundary, wavelength, incidence angle and receiver temperature. Radiation reflected or transmitted out of that receiver is not absorbed power. A cavity can improve radiometric absorptance by giving light repeated opportunities for wave absorption; a dark coating alone does not establish $a=1$.

For broadband light, spectral radiant flux matters: $P_{\mathrm{abs}}=\int a(\lambda ){\Phi }_{e,\lambda }(\lambda )d\lambda$. A single effective radiometric absorptance therefore depends on the incident optical spectrum as well as on the receiver. If the stated incident boundary lies before a window of known power transmission ${\tau }_w$, the receiver absorbs $a{\tau }_w{P}_{\mathrm{before}}$ when other optical losses are absent. Window loss and receiver radiometric absorptance are separate corrections.

Total hemispherical emissivity integrates over all emitted wavelengths and the outward hemisphere. If ${ϵ}_{\lambda ,\Omega }$ is spectral emissivity resolved by direction, and $B_{\lambda }(T)$ is blackbody spectral radiance from Planck's law, thenIt is a weighted average, not an unweighted mean of surface properties. Even if the wavelength-resolved material response stays unchanged, the Planck's law weighting changes with temperature, so the total ratio can change. Likewise, total radiometric absorptance averaged over radiation from hotter surroundings generally differs from this emission-weighted ratio. Kirchhoff's law of thermal radiation equates matched spectral and directional quantities; it does not equate arbitrary broadband averages.