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Area weighting of radiant exposure combines patch-average exposures in proportion to the receiving areas they represent. For disjoint patches with areas $A_j$ and mean radiant exposures $H_j$, the coefficients ${\alpha }_j=A_j/{\sum }_k{A}_k$ satisfy ${\sum }_j{\alpha }_j=1$, and the combined spatially averaged radiant exposure is ${\sum }_j{\alpha }_j{H}_j$. This follows by adding the received energies $A_j{H}_j$, rather than averaging a list of values without its geometry.^[ref]

An original coordinate example explains a calibration issue in image-based measurements. Let a rectangular receiving strip have length $\ell$ and constant width. Stipulate $H(x)=H_{0}x/\ell$ for $0\le x\le \ell$. Suppose a calibrated image coordinate $u\in [0,1]$ maps to physical position $x=\ell u^2$. Uniform sampling in $u$ is therefore nonuniform in physical area.

The physical area average isAn unweighted average over increasingly dense, uniformly spaced $u$ samples would instead converge to ${\int }_0^1{H}_0{u}^{2}du=H_0/3$. The factor $2u$ is the change of variables formula's area correction for the chosen mapping; the constant strip width and $\ell$ cancel in the normalized average.

Thus a Jacobian area correction for radiant-exposure maps can matter even if every local exposure value is correct. An optical camera measurement needs its spatial calibration and the receiving-surface map, not just pixel values. Overlapping patches must not be counted as disjoint areas, and local point samples need an explicit quadrature model before they can stand for patch averages.

A dipole anisotropy of a radiance field is a directional first-order departure from a constant radiance:Here $\mathbf{\omega }$ is a unit viewing direction, and the bound ensures nonnegative radiance. Writing $\mathbf{b}=ϵ\mathbf{a}$ with unit $\mathbf{a}$ gives the amplitude/direction convention in the title. The angular isotropy test with planar irradiance measurements derives its receiver response.

An original inversion uses opposite receiver orientations. Suppose measured irradiances are $E(+x)=3.4$, $E(-x)=2.6$, and all four $y,z$ readings equal $3.0\mathrm{W}{\mathrm{m}}^{-2}$. The pair means give $\pi L_0=3.0\mathrm{W}{\mathrm{m}}^{-2}$. Sincethe fitted vector is $\mathbf{b}=(0.20,0,0)$. Agreement of the three pair sums is an additional check on this restricted model and a stable measurement chain.

The first angular moment of a radiance field is $\int \mathbf{\omega }Ld\Omega =(4\pi /3)L_0\mathbf{b}$. By contrast, integrating $L$ over all directions gives $4\pi L_0$, independent of $\mathbf{b}$. Thus this example's projected-area-normalized spherical irradiance is $12\mathrm{W}{\mathrm{m}}^{-2}$, exactly the value for an isotropic field with the same $L_0$. One all-direction scalar cannot determine its dipole.^[ref] Receiver angular response must also be characterized rather than assumed ideal.

An energy-normalized pulse profile separates an optical pulse's temporal shape from its total laser pulse energy. For nonnegative radiant power $P(t)$ and finite $U=\int P(t)dt>0$, define $p(t)=P(t)/U$. Then $p$ has units inverse time and integrates to one. The radiant energy in a time interval is $U$ times the integral of $p$ over that interval.^[ref]

For an original two-lobe example, let $p=1/(8\mathrm{ns})$ during $0<t<2\mathrm{ns}$ and $p=3/(8\mathrm{ns})$ during $5<t<7\mathrm{ns}$, with zero values elsewhere. The two integrals are $1/4$ and $3/4$. Choosing $U=4\mathrm{nJ}$ gives powers $0.5$ and $1.5\mathrm{W}$, and energies $1$ and $3\mathrm{nJ}$. Multiplying $U$ by two doubles both powers without changing $p$.

The temporal energy centroid of an optical pulse is $\int tp(t)dt$ when this moment exists. Here it is $(1/4)(1)+(3/4)(6)=4.75\mathrm{ns}$, in the gap between lobes. An energy centroid need not be a location of maximum power.

Mathematically, $p$ is a probability density function, but its weighting here is energy over time. It is not automatically a measured photon-arrival distribution, particularly when photon energies vary. A zero-energy trace cannot be normalized this way, and a noisy baseline-corrected trace with negative excursions is not automatically a nonnegative density. The Gaussian pulse energy and peak power article gives one normalized shape family.

An isotropic radiance field has the same radiance in every direction at the specified point: $L_{\mathrm{e}}(\mathbf{\omega })=L_0$. Angular isotropy is a local property; it does not by itself assert that $L_0$ is the same at every spatial position. Integrating this constant field gives the orientation-independent one-sided irradiance $\pi L_0$. The spherical irradiance article derives the corresponding all-direction value using its projected-area convention.^[ref]

An original counterexample shows why a few equal detector readings do not establish isotropy. Send six ideal narrow beams toward a point along the positive and negative Cartesian axes. Give every beam a normal-incidence irradiance $q=1\mathrm{W}{\mathrm{m}}^{-2}$. A one-sided planar receiver with unit normal $\mathbf{n}$ then recordsThe three axis-aligned receiver orientations all give $q$. However, a normal along $(1,1,1)/\sqrt{3}$ gives $\sqrt{3}q\approx 1.732\mathrm{W}{\mathrm{m}}^{-2}$. Three mutually perpendicular measurements therefore agree even though the idealized field is strongly directional. Finite narrow beams approximate this construction.

An angular isotropy test with planar irradiance measurements must sample enough orientations to test its stated angular model. A collimated light source and a rotating receiver can check angular response before such a test. Detector-area projection and the instrument's own angular sensitivity must be distinguished from variation in the incident field.

A Jacobian area correction for radiant-exposure maps converts a coordinate-space integral into one over the physical receiving surface. For a one-to-one planar calibration map $\mathbf{x}=f(u,v)$,The Jacobian determinant supplies physical area per coordinate-area increment. This change of variables formula preserves the radiant exposure definition as incident energy per receiving area.^[ref]

For an original two-dimensional example, let $0\le u,v\le 1$ and map $x=au$, $y=bv(1+u)$, with $a=2\mathrm{mm}$ and $b=3\mathrm{mm}$. Then $|\det Df|=ab(1+u)$. Stipulate $H(f(u,v))=H_0(1+u)$ with $H_0=2\mathrm{J}{\mathrm{m}}^{-2}$. The physical receiving area is $3ab/2=9\mathrm{m}{\mathrm{m}}^2$, andIts spatially averaged radiant exposure is $28/9\approx 3.111\mathrm{J}{\mathrm{m}}^{-2}$. Uniformly averaging the exposure values over $u,v$ without the determinant gives only $3\mathrm{J}{\mathrm{m}}^{-2}$.

The correction applies to exposure densities represented in distorted coordinates. If each detector sample already reports energy integrated over its physical patch, summing those calibrated energies requires no second multiplication by patch area. Otherwise the area is counted twice. An optical camera reduction must specify whether its values represent energy, irradiance, exposure or a further processed signal.

Optical energy outside a time gate is the radiant energy excluded by a specified ideal temporal acceptance set $G$. For a nonnegative radiant power profile of finite positive total energy,The time-integral definition is the radiometric one.^[ref] This observable needs a gate boundary but no fitted separation into a main field and an optical pulse pedestal.

For an original three-pulse example, let an optical prepulse, main pulse and optical postpulse carry $5$, $12$ and $3\mathrm{nJ}$ respectively, with disjoint supports. A gate containing only the main pulse admits $12\mathrm{nJ}$ of $20\mathrm{nJ}$, so $f_{\mathrm{out}}=0.40$. Widening that gate to include the postpulse reduces the fraction to $0.25$. The optical waveform has not changed; the admitted observable has.

For a translated interval $G_{\delta }=[a+\delta ,b+\delta ]$ with continuous boundary values,Unequal power at the boundaries produces first-order sensitivity to a timing shift. Equal boundary values eliminate this derivative but need not eliminate higher-order loss; quadratic energy loss from centred Gaussian gate timing error gives an existing example.

The total-energy denominator must cover the excluded light too. If the acquisition window misses an unknown tail, pulse-energy interval from incomplete tail measurements supplies conditional bounds rather than an exactly known fraction. A detection time gate on an electrical detector trace also includes the detector impulse response and is not automatically an ideal gate on incoming optical power. A characterized optical response and a stated detector baseline subtraction are needed to compare the two.

Radiant energy is the amount of energy emitted, transferred or received as electromagnetic radiation. It is measured in joules. For the specified beam or collection of rays crossing a boundary,where radiant flux ${\Phi }_{\mathrm{e}}$ is radiation power. The receiving geometry and time interval are part of the statement; dividing by a receiving area produces radiant exposure, rather than another total energy.^[ref]

For an original passive-optic example, stipulate $5.0\mathrm{mJ}$ incident during an interval, with no remaining change of stored energy afterward. Suppose $3.0\mathrm{mJ}$ is transmitted, $1.25\mathrm{mJ}$ reflected and $0.75\mathrm{mJ}$ absorbed. Their sum is the incoming $5.0\mathrm{mJ}$. A detector collecting only the transmitted path receives $3.0\mathrm{mJ}$, even though the source supplied more. This radiometric energy balance specifies the collection boundary instead of equating every energy readout in the apparatus.

A thermal detector converts absorbed radiation into an observable temperature response; a calibrated model relates that response to received energy. A documented NIST arrangement uses a slow thermal reference and calibrated beam splitting to establish pulse-energy measurements.^[ref] Absorbed energy, incident radiant energy and energy in an uncollected branch must remain distinct in that calibration.

Two sources can deliver the same radiant energy with different radiant flux histories or spectra. Their total energies agree only for the specified integration and collection; this alone does not determine peak power, radiant exposure at a point or detector response.

Radiant exposure measures the radiant energy received per unit surface area during a specified interval. At a fixed surface point $\mathbf{x}$, it is the time integral of incident irradiance $E_{\mathrm{e}}$:The area belongs to the receiving surface: its orientation and location matter. An irradiance in $\mathrm{W}{\mathrm{m}}^{-2}$ specifies a rate, while radiant exposure in $\mathrm{J}{\mathrm{m}}^{-2}$ specifies the accumulated arrival of energy.^[ref]

For a uniform rectangular pulse, $H_{\mathrm{e}}=E_0\tau$, with constant irradiance $E_0$ and duration $\tau$. Stipulate one pulse at $20\mathrm{W}{\mathrm{m}}^{-2}$ for $1\mathrm{s}$, and another at $80\mathrm{W}{\mathrm{m}}^{-2}$ for $0.25\mathrm{s}$. Both deliver $20\mathrm{J}{\mathrm{m}}^{-2}$, but their peak irradiances differ by a factor of four. The two radiant exposures agree only after each complete pulse has arrived. A shorter observation window can intercept different fractions of them, and a nonlinear receiver can respond differently even to their complete pulses.

Integrating over a receiving patch gives its incident radiant energy:Here ${\overline{H}}_A$ is the spatially averaged radiant exposure, rather than necessarily the local maximum. Uniform illumination is the special case where this spatially averaged radiant exposure equals the local value everywhere on the patch. In the rectangular-pulse example, a uniformly illuminated $1\mathrm{c}{\mathrm{m}}^2=10^{-4}{\mathrm{m}}^2$ patch receives $20\times 10^{-4}=2\mathrm{mJ}$.

For an original spatial example, consider a circular Gaussian beam at a fixed transverse plane, with constant Gaussian beam radius $w$ throughout the pulse. Let $P(t)$ be its total radiant power and $U=\int P(t)dt$ its laser pulse energy. Integrating its transverse irradiance in time gives the radiant exposure of a Gaussian beam:The normalization follows by integrating $H(r)2\pi rdr$ from $0$ to $\infty$. The disk of radius $w$ contains only $U(1-e^{-2})$, so its actual spatially averaged radiant exposure is $U(1-e^{-2})/(\pi w^2)$, not $U/(\pi w^2)$. With stipulated $U=1\mu \mathrm{J}$ and $w=1\mathrm{mm}$, the peak is $0.6366\mathrm{J}{\mathrm{m}}^{-2}$, while the average inside that disk is $0.2752\mathrm{J}{\mathrm{m}}^{-2}$. State the Gaussian beam radius convention and the receiving area with any reported value.

An energy readout alone therefore does not specify a surface's radiant exposure. In an ideal linear model for pulse energy measurement with a photodiode, dark-subtracted photocurrent $i_{\mathrm{ph}}(t)=R{P}_{\det }(t)$ has constant spectral responsivity $R$ at the chosen wavelength. Integrating the electric current gives signal electric charge $Q_{\mathrm{sig}}=R{U}_{\det }$. If a known patch of area $A$ is uniformly illuminated and all of its incident energy is collected, thenAs a separate stipulated example, $Q_{\mathrm{sig}}=0.40\mathrm{nC}$ and $R=0.40\mathrm{A}{\mathrm{W}}^{-1}$ correspond to $U_{\det }=1.0\mathrm{nJ}$. Uniform illumination over $A=1.0\mathrm{m}{\mathrm{m}}^2$ gives $H_{\mathrm{e}}=1.0\mathrm{mJ}{\mathrm{m}}^{-2}$. Background subtraction, collection geometry, detector linearity and an integration window covering the full response are part of this measurement model. Continuous-wave responsivity alone does not establish linearity for a short pulse: silicon-photodiode experiments have observed pulse-dependent saturation.^[ref]

A historical NIST measurement calibration separates these operations experimentally. A modulator produces pulses from a continuous-wave laser; a calibrated beam splitter sends one beam to a slow thermal reference and another to the pulse detector. The reference records the accumulated energy, while the pulse repetition rate and injection duration determine the mean energy per pulse. The detector's integrating electronics provide a pulse-energy readout. Converting such a readout to radiant exposure additionally requires the illuminated area and its spatial distribution.^[ref]

The beam-use term fluence denotes the same time integral for a directed beam normal to the chosen surface. For light arriving from many directions, specify the angular geometry: planar irradiance and spherical irradiance are different quantities, so a value defined using one cannot simply be substituted for the other.^[ref]

A radiometric energy balance accounts for radiant energy crossing the specified measurement boundary during a specified interval. For a passive optic, suppose the relevant transient has finished, no electromagnetic energy remains stored, and no additional energy source is present. Incident radiation is partitioned into all outgoing radiation and energy transferred to matter. Resolving outgoing paths givesThe channels must be disjoint: a scattering term cannot include energy already counted by a reflection or transmission collector. The definition of radiant energy concerns radiation emitted, transferred or received, so the location and collection geometry belong in each term.^[ref]

For an original example, inject $10\mu \mathrm{J}$ and measure $6\mu \mathrm{J}$ in the transmitted path plus $2\mu \mathrm{J}$ in the specular reflection. Calling the missing $2\mu \mathrm{J}$ absorbed is premature. If a separate collector finds $1.5\mu \mathrm{J}$ of diffuse outgoing radiation outside those channels, the inferred absorbed energy is only $0.5\mu \mathrm{J}$. The uncollected scattering in a radiometric energy balance would otherwise be misidentified as optical absorption.

A physical check compares calibrated incident and outgoing energy readouts while controlling apertures, backgrounds and the observation window. The scattered component may need a different collection geometry from the specular beam. Energy absorbed by matter can remain stored or later leave as heat or radiation; a more general control-volume balance must explicitly include storage, other heat transfer and work. It must not count absorbed energy twice by also treating the same material-energy increase as an independent loss.

Spatially averaged radiant exposure is the received radiant energy divided by the specified receiving area. For a patch $A$ and local radiant exposure $H(\mathbf{x})$,Both local and averaged values have units $\mathrm{J}{\mathrm{m}}^{-2}$, but the average need not equal the local maximum. The radiant exposure definition uses incident energy and the receiving surface's area.^[ref]

For an original unequal-area example, let one $3\mathrm{m}{\mathrm{m}}^2$ region receive a uniform $2\mathrm{J}{\mathrm{m}}^{-2}$, and a neighbouring $1\mathrm{m}{\mathrm{m}}^2$ region receive $10\mathrm{J}{\mathrm{m}}^{-2}$. They receive $6$ and $10\mu \mathrm{J}$ respectively. The average over the combined patch is thereforeThe unweighted average of the two listed exposure values would be $6\mathrm{J}{\mathrm{m}}^{-2}$, which is wrong for these unequal areas. The area weighting of radiant exposure gives the larger region three times the influence. The actual peak remains $10\mathrm{J}{\mathrm{m}}^{-2}$ in the stipulated piecewise-uniform model.

A receiver integrating all incident energy over the complete patch can determine $Q_A$; division by its known collecting area then gives the average. That measurement alone cannot identify the peak or reconstruct the spatial distribution. Subdivision with a characterized spatial response supplies additional information. For a finite aperture intercepting a Gaussian beam, use energy actually entering the aperture, as calculated in radiant exposure of a Gaussian beam, rather than the total energy of the untruncated beam.

Spherical irradiance counts incident radiance from all directions without selecting a receiving-plane normal. Using the projected-area convention specified in the CIE entry's note2,Here $d\Omega$ is an element of solid angle. An ideal small sphere of radius $a$ intercepts radiant flux ${\Phi }_{\mathrm{in}}=\pi a^2{E}_{\mathrm{e},\mathrm{o}}$. The denominator is its diametrical projected area, not its full surface area.^[ref]

For an original isotropic radiance field example, take $L_{\mathrm{e}}=L_0=2\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}$ in every direction. Then $E_{\mathrm{e},\mathrm{o}}=4\pi L_0=8\pi \mathrm{W}{\mathrm{m}}^{-2}$. A one-sided plane instead receivesThe factor four follows from the different angular geometries; it is not a unit conversion. Averaging the sphere's intercepted power over its actual $4\pi a^2$ surface also gives $E_{\mathrm{e},\mathrm{o}}/4$, rather than the projected-area-normalized quantity above.

An all-direction receiver must have a characterized angular response. A single flat irradiance detector inherently samples a hemisphere with projected-area weighting, so rotating it or enclosing it in a diffuser does not supply an all-direction value without an appropriate response model and measurement calibration. Report the angular and area conventions with the result; equal units do not make irradiance and spherical irradiance interchangeable.

The temporal energy centroid of an optical pulse is its energy-weighted mean time. For nonnegative radiant power $P(t)$ with finite positive energy $U$ and finite absolute first moment,The normalization uses radiant energy as integrated radiant flux.^[ref] The centroid is a moment of the energy-normalized pulse profile; it need not coincide with the largest power value, a threshold crossing or the middle of the FWHM interval.

For an original two-component example, let ninety percent of the energy belong to a symmetric narrow component centred at zero and ten percent to a symmetric component centred at $20\mathrm{ns}$. Linearity of the integral gives $t_E=0.9(0)+0.1(20)=2\mathrm{ns}$. Weak delayed energy can move the centroid even when the dominant peak remains near zero. Multiplying the whole power trace by a positive constant leaves $t_E$ unchanged; translating it by $t_0$ shifts its centroid by $t_0$.

A detector response can shift the measured centroid too. Suppose a normalized impulse response $h$ and normalized optical profile $p$ have finite absolute first moments. For their convolution $q=p\ast h$, direct integration with $t=s+u$ givesThus a causal exponential detector response of time constant $2\mathrm{ns}$ shifts the recorded centroid by $2\mathrm{ns}$ in this model. NIST's pulsed-radiometer characterization explicitly measures detector impulse responses and uses convolution to assess reproduced waveforms.^[ref] The centroid identity is the independent moment calculation above.

Baseline drift can cause a separate shift. Suppose a true pulse has $U=1\mathrm{nJ}$ and $t_E=0$, but a positive residual baseline of $1\mathrm{mW}$ remains across the analysis window from $-10$ to $90\mathrm{ns}$. Its added energy is $0.10\mathrm{nJ}$, centred at $40\mathrm{ns}$. The resulting centroid isA waveform measurement should therefore record the time window, baseline estimate and detector timing response with its reported centroid. A negative background-subtracted trace also needs care: it is not automatically a nonnegative energy distribution.