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Deflection changes one light ray's direction; focusing concerns relative directions of neighboring light rays and their bundle's cross section. To see the difference for a point-mass gravitational lens, send parallel incident light rays past the lens. A ray at impact parameter $b$ meets the optical axis approximately $f(b)=b/\hat{\alpha }=c^2{b}^2/(4GM)$ beyond the lens. Doubling $b$ quadruples this crossing distance. Thus the lens does not have a common focal length for all rays, although it can produce lensing magnification and an Einstein ring for a specified source-observer geometry.

The point-mass gravitational lens also illustrates why focusing must not be identified just with positive lensing convergence. Away from the central mass, differentiate the reduced deflection $\mathbf{\alpha }={\theta }_E^2\mathbf{\theta }/r^2$: its two-dimensional divergence is zero, so $\kappa =0$ there. Nevertheless its radial and tangential derivatives differ, giving lensing shear of magnitude ${\theta }_E^2/r^2$ and $\det A=1-{\theta }_E^4/r^4$. A vacuum ray bundle can be distorted and magnified. In the same ideal transparent geometrical optics model, surface brightness is preserved; the larger apparent solid angle supplies the increased received light.

Consider a parent atomic nucleus at rest. Write its signed nuclear polarization along a fixed axis $\hat{\mathbf{z}}$ as $P=⟨J_z⟩/(I\hslash )$, where $I\hslash$ is its maximal spin component. Let $\theta$ be the angle of the emitted electron momentum to that axis and let $\beta =v/c$. Use $\mathcal{P}$ for the parity operator, so it is not confused with $P$.

For an axially symmetric leading-dipole model, retain a constant and the term linear in the polarization vector $\mathbf{P}=P\hat{\mathbf{z}}$. Rotational symmetry makes the directional term proportional to $\mathbf{P}\cdot \hat{\mathbf{p}}$. Define its dimensionless coefficient as $A\beta$. At fixed electron energy the relative intensity and normalized probability density function per solid angle are thereforeThe factor $\beta$ is the conventional separation of the speed dependence; rotational symmetry alone does not determine that dependence or $A$. The coefficient $A$ contains the decay dynamics. This leading model omits higher angular terms and detailed recoil and radiative effects. A partially oriented higher-spin ensemble can also have alignment that is not specified by its vector nuclear polarization alone.

The unit-integral condition follows without any model of a detector:Thus $f$ has unit angular average, while the probability density function in the polar angle alone contains the Jacobian determinant factor:In particular, $f$ itself is neither a normalized probability density function in $\theta$ nor a detector count. Positivity on $-1\le u\le 1$ gives $|a|\le 1$.

Under full spatial reflection, $\mathbf{p}\mapsto -\mathbf{p}$ but $\mathbf{J}\mapsto \mathbf{J}$. The latter follows already for orbital angular momentum from $\mathbf{r}\times \mathbf{p}\mapsto (-\mathbf{r})\times (-\mathbf{p})=\mathbf{r}\times \mathbf{p}$; spin has the same axial vector transformation. Hence $\mathbf{P}\cdot \hat{\mathbf{p}}$ is a pseudoscalar, and $\cos \theta \mapsto -\cos \theta$. If parity conservation held for this model, equal probabilities for the inverted process would force $a=0$. Conversely, $a=0$ in an unpolarized sample proves nothing about $A$.

Integrating over the two hemispheres gives a useful original check:Choose $A=-0.8$, $P=0.5$, and $\beta =0.75$, so $a=-0.3$. Then the forward and backward hemisphere probabilities are $0.425$ and $0.575$, and their counting asymmetry is $-0.15$. These are chosen teaching values, not measured values from the Wu experiment. At a single energy, the angular shape determines the product $AP\beta$; it cannot separately determine $A$ without a polarization calibration.

The primary paper derives the geometric phase accompanying adiabatic transport around a closed parameter loop. Section 4 treats a spinning particle in a slowly changing magnetic field, with quantum phase proportional to its magnetic quantum number and the enclosed solid angle.

Radiance resolves both position and direction: it is radiant flux per projected area and per solid angle. For a surface element of area $dA$, and a small outgoing cone $d\Omega$ at angle $0\le \theta <\pi /2$ to its outward unit normal,Here $\mathbf{x}$ locates the patch and $\mathbf{s}$ is a unit vector along the outgoing direction. The factor $dA\cos \theta$ is the patch's apparent area viewed along that direction. The same local definition applies to transmitted or received radiation, choosing the positive projected area for the crossing direction. ^[ref].

For a planar patch of area $A$ with spatially uniform, direction-independent radiance $L$, integrating over the patch givesTake $L=10\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}$ and $A=0.020{\mathrm{m}}^2$. The radiant intensities are $I(0)=0.20\mathrm{W}\mathrm{s}{\mathrm{r}}^{-1}$ and $I(60^{\circ })=0.10\mathrm{W}\mathrm{s}{\mathrm{r}}^{-1}$. A sufficiently narrow $0.010\mathrm{sr}$ cone therefore carries approximately $2.0\mathrm{mW}$ or $1.0\mathrm{mW}$, respectively. Both views have the same radiance $L$: the oblique view presents half the projected area. This is Lambert's cosine law, not a loss of radiance with viewing angle.

Radiance is useful because it separates a source's local directional output from the geometry of an observation. Integrating over outgoing directions gives radiant exitance; integrating incoming radiance at a receiver gives irradiance. In the stationary vacuum setting of vacuum conservation of specific intensity, a resolved uniform source retains its radiance along each light ray, while the radiant flux received from the whole source can decrease as its apparent solid angle shrinks.

Radiant exitance is total outward emitted radiant flux per actual emitting area, measured in $\mathrm{W}{\mathrm{m}}^{-2}$. Integrate radiance over the outward hemisphere, keeping the projected area factor:If the radiance is direction-independent, the spherical coordinates element is $d\Omega =\sin \theta d\theta d\phi$, soThe weighted solid angle is $\pi \mathrm{sr}$, although the whole hemisphere covers $2\pi \mathrm{sr}$. Directions near grazing contribute less because of the projected area. This geometric factor is the bridge between Lambert's cosine law and the total output of a diffuse patch. ^[ref].

For the uniform patch with $L=10\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}$, radiant exitance is $M=10\pi \mathrm{W}{\mathrm{m}}^{-2}$. Its area $0.020{\mathrm{m}}^2$ gives total radiant flux $\Phi =MA=0.20\pi \mathrm{W}\approx 0.628319\mathrm{W}$ over the outward hemisphere. This total differs from its radiant intensity $I(0)=0.20\mathrm{W}\mathrm{s}{\mathrm{r}}^{-1}$, which describes only one direction.

For an ideal blackbody, integrate Planck's law over frequency as well: $M=M_e=j^{\ast }=\sigma T^4$ by the Stefan–Boltzmann law. An ideal $6000\mathrm{K}$ emitting patch of area $1\mathrm{m}{\mathrm{m}}^2$ emits about $73.4881\mathrm{W}$ outward. Irradiance counts incoming radiation separately; for purely emitted output the net outward exchange per area is $M-E$. Reflected output must also be included if the surface reflects incident radiation.

Radiant flux is the power carried by electromagnetic radiation: radiation energy transferred per unit time, measured in watts, $\mathrm{W}=\mathrm{J}{\mathrm{s}}^{-1}$. Specify what is counted: all radiation emitted by a source, radiation within one beam, or radiation intercepted by a photodetector. Writing $Q(t)$ for the cumulative energy crossing the specified boundary givesThe first expression is an instantaneous derivative; the second is an average over the stated time interval. If $0.020\mathrm{J}$ arrives during $0.010\mathrm{s}$, the mean radiant flux is $2\mathrm{W}$. The photodetector's absorbed power equals this incident radiant flux only if it absorbs all of the incident radiation.

Keeping radiant flux fixed while widening a beam reduces its irradiance, the incident power per receiving area. Radiance additionally resolves the propagation direction and has units $\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}$; an optical camera can collect different radiant fluxes from sources with the same radiance but different apparent solid angles. Luminous flux, measured in lumens, weights the optical spectrum by the human eye's standardized response instead of counting energy equally at every wavelength. Equal radiant fluxes need not give equal luminous fluxes.^[ref]

Radiant power also differs from the optical power of an optical lens, which is reciprocal focal length in dioptres. Likewise, astronomical radiative flux is power per area. Stating the units prevents these shared words from hiding different physical quantities.

In geometrical optics, suppose radiation has a nonnegative radiance $L(\mathbf{x},\mathbf{s})$ whose directional contributions add as power. Here $\mathbf{s}$ is the actual propagation unit vector, $d\Omega$ is a solid angle element, and the collecting oriented surface $A$ has chosen unit normal $\mathbf{n}(\mathbf{x})$. Assume the following integrals exist. The outward and inward radiant fluxes areBoth quantities are nonnegative; their difference is the signed net radiant flux. The dot product supplies the cosine of the angle with the unit normal, hence the projected area. For incoming radiation the outward unit normal gives a negative dot product; the minus sign makes the collected power positive. Measuring the incoming angle from a source-facing unit normal instead gives the familiar positive $\cos \theta$. The units check is $\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}\times {\mathrm{m}}^2\times \mathrm{sr}=\mathrm{W}$.

Equivalently, the incoming surface integral is ${\Phi }_{-}={\int }_{A}EdA$ with local irradiance $E$; the emitted part of ${\Phi }_{+}$ is ${\int }_{A}MdA$ with radiant exitance $M$. For a small flat patch and a narrow accepted cone, nearly constant radiance and angle give $\Phi \simeq LA\cos \theta \Delta \Omega$. This approximation requires uniformity across the patch and cone. Direction-independent outgoing radiance over a whole hemisphere gives the different exact result ${\Phi }_{+}=\pi LA$, by Lambert's cosine law and the spherical coordinates calculation in radiant exitance. Replacing the cosine-weighted integral by the hemisphere's $2\pi$ solid angle would double the answer.^[ref]

These radiance formulas presume a description in additive directional power; they do not reconstruct arbitrary coherent wave interference from separate ray contributions. For general electromagnetic fields, the signed instantaneous power crossing an oriented surface is ${\int }_A\mathbf{S}\cdot \mathbf{n}dA$, using the Poynting vector $\mathbf{S}$; optical measurements commonly average over many wave periods. A zero net radiant flux can conceal equal nonzero incoming and outgoing radiant fluxes.

Radiant intensity describes how a source distributes its radiant flux over directions. A small cone of solid angle $\Delta \Omega$ carries approximately $I\Delta \Omega$ when $I$ varies little across it; integrating $I$ over all emitting directions recovers the total radiant flux. For a source with isotropy emitting $4\pi \mathrm{W}$ over a full sphere, $I=1\mathrm{W}\mathrm{s}{\mathrm{r}}^{-1}$ and a $0.01\mathrm{sr}$ cone carries $0.01\mathrm{W}$. A planar source satisfying Lambert's cosine law instead has direction-dependent radiant intensity, even though its radiance is direction-independent.

For an orbit of radius $b$ in the $xy$ plane, $\mathbf{a}=-b{\omega }^2(\cos \omega t,\sin \omega t,0)$. In spherical coordinates, at angle $\theta$ from the orbit normal, $⟨(\mathbf{n}\cdot \mathbf{a}{)}^2⟩=b^2{\omega }^4{\sin }^2\theta /2$. The acceleration covariance in nonrelativistic radiation therefore gives the displayed pattern. Its axial power per solid angle is twice its equatorial value; no observation direction has zero time-averaged power. Integrating gives $⟨P⟩=q^2{b}^2{\omega }^4/(6\pi {ϵ}_0{c}^3)$. The condition $b\omega \ll c$ is essential: relativistic synchrotron radiation has a different beamed pattern.

For radiation in vacuum with isotropy, energy is distributed equally per solid angle: write $u_{\Omega }=u/(4\pi )$ for energy density per unit solid angle. Let $\theta$ be the angle between a ray's path and the wall normal. Its normal velocity $c\cos \theta$ determines the arrival rate, while normal momentum per energy is $\cos \theta /c$. Their product is ${\cos }^2\theta$. For a perfectly reflecting cavity wall,The leading two is the normal momentum reversal; the incoming hemisphere contains half of the field's directional energy density. This is also the normal component of the bulk pressure tensor, $P_{nn}={\int }_{4\pi }{u}_{\Omega }{\cos }^2\theta d\Omega =u/3$. Isotropy makes all three diagonal components equal, while their sum is $u$ because the squared direction components sum to one. The factor one third therefore depends on three-dimensional angular geometry, not on the shape of the blackbody radiation spectrum.

A perfectly absorbing wall exposed to this incoming hemisphere receives only $u/6$ from absorption. If it is in thermal equilibrium with the isotropic radiation, its outgoing emission supplies another $u/6$ of recoil, recovering $u/3$. A cold absorber before equilibrium is a different boundary condition. Similarly, the one-way incident energy flux is $cu/4$, whereas the net energy flux across a plane in the undisturbed isotropic field is zero. Dividing $cu/4$ by $c$ misses the extra normal-momentum projection and does not give the cavity pressure.

For a stipulated isotropic radiation energy density $u=6\mu \mathrm{J}{\mathrm{m}}^{-3}$, the bulk pressure is $2\mu \mathrm{Pa}$ regardless of its spectral shape. In thermal equilibrium, blackbody radiation supplies $u=a_{\mathrm{rad}}{T}^4$ with radiation constant $a_{\mathrm{rad}}=4\sigma /c$, so $p_{\mathrm{rad}}=a_{\mathrm{rad}}{T}^4/3$; Stefan–Boltzmann law defines $\sigma$. The equilibrium photon gas derives this same equation of state from its free energy.^[ref]

Radiometry measures the energy and power carried by electromagnetic radiation, including light. It asks how much radiation a source supplies, where it travels, and how much reaches a receiving surface. The measurements are physical rather than weighted by the sensitivity of the human eye: equal radiant fluxes at different wavelengths need not look equally bright. ^[ref].

The five quantities below answer different questions about the same radiation. The International System of Units makes the distinctions visible: a watt is a joule per second, while a steradian measures solid angle. ^[ref].

Resolving wavelength or frequency adds a further “per unit interval”, as in spectral radiance. The notation is consistent with NIST's table of spectral quantities. In particular, radiant flux is power, whereas the existing astronomical radiative flux is power per area. ^[ref].

This bookkeeping makes an optical camera's collected signal calculable from a source's radiance and the geometry of the receiving aperture. It also lets radiation measurements from different instruments be compared. For example, NIST describes separate calibrations of spectral radiance and spectral irradiance for instruments used to observe Earth. ^[ref].

For incident unpolarized light and the isotropic electric dipole response of Rayleigh scattering by a small sphere, let $\theta$ be the angle between the incident beam's propagation direction and the outgoing direction. The differential scattering cross-section and its relative angular pattern arewhere $d\Omega$ is an element of solid angle. Averaging over the two transverse directions of incident polarization of an electromagnetic wave gives this pattern. Forward scattering at $0^{\circ }$ and backward scattering at $180^{\circ }$ are equally strong: $F=1$. Sideways scattering at $90^{\circ }$ is weakest, but nonzero: $F=1/2$. ^[ref].

The normalization is consistent becauseThus integrating $d\sigma /d\Omega$ gives the total Rayleigh scattering cross-section $\sigma$. The plotted $F$ has maximum one; it is not itself normalized to unit solid angle integral.

For a single electric dipole driven by incident linear polarization, the angular distribution of electric dipole radiation instead has a ${\sin }^2\alpha$ factor, where $\alpha$ is measured from the electric dipole moment direction, parallel to the incident electric field. That direction is perpendicular to the incident beam. Confusing $\alpha$ with the beam angle $\theta$ would incorrectly predict zero forward scattering. Averaging the dipole patterns over incident unpolarized light yields the displayed $F(\theta )$.

For an incident beam uniform across its relevant impact parameters, an incident annulus has area $d\sigma =2\pi b|db|$. Rotational symmetry maps it to outgoing solid angle $d\Omega =2\pi \sin \theta d\theta$. The central-potential classical differential cross-section and the Rutherford impact-parameter relation therefore giveThis is a differential scattering cross-section, with units of area per steradian, conventionally ${\mathrm{m}}^2\mathrm{s}{\mathrm{r}}^{-1}$, not a detector count and not an area per degree. It applies for $0<\theta <\pi$, with a finite limiting value at $\pi$. Doubling incident kinetic energy divides it by four at fixed angle; doubling either electric charge number multiplies it by four.

A useful independent check integrates over all azimuths between two positive angles. For $0<{\theta }_1<{\theta }_2\le \pi$,This equals the incident annulus area $\pi (b_1^2-b_2^2)$ directly. A detector covering only a small patch at one angle does not collect this full azimuthal band.

A movable magnetic spectrometer selected scattered electrons by momentum and scattering angle. Magnetic fields dispersed momenta at the focal plane. With relativistic electrons, the inferred momentum determines outgoing energy $E^{\prime }$. Commissioning included steering the beam through the spectrometer at zero angle to check its optics and accepted solid angle. Counters sorted events into momentum and angle bins. The acceptance entered the conversion from counts to differential scattering cross-sections.^[ref]

Radiative corrections matter because photon emission can move events from an elastic scattering peak into lower outgoing-energy bins. The analyses subtracted the elastic radiative tail before unfolding inelastic radiation. Independent MIT and SLAC procedures provided a check. An uncorrected low-$E^{\prime }$ tail alone therefore cannot establish deep inelastic scattering.^[ref]

A solid angle measures how much of the surrounding directions a region occupies, as seen from a specified point $O$. Draw a sphere of radius $R$ centered at $O$ and mark where those directions meet its surface. If the marked patch has surface area $A_R$, thenA Euclidean similarity scales both $A_R$ and $R^2$ by the same factor, so the chosen radius does not affect the answer. On the unit two-sphere, the numerical value is simply the patch's surface area. Unlike an angle drawn in one plane, a solid angle measures coverage in two angular directions.

For example, the full sphere has surface area $4\pi R^2$ and hence solid angle $4\pi$ steradians. A hemisphere occupies $2\pi$ steradians. Three mutually perpendicular coordinate planes divide the directions into eight congruent pieces, so each octant occupies $4\pi /8=\pi /2$ steradians. A solid angle of one steradian therefore covers $1/(4\pi )\approx 7.96\%$ of all directions, rather than one full turn.

This makes a geometric question answer a collection question. For an ideal point emitter with isotropy, equal solid angles carry equal radiant flux. A detector accepting directions of total solid angle $\Omega$ intercepts the fraction $\Omega /(4\pi )$ of that emitted radiant flux, before losses. With direction-dependent radiant intensity $I$, the accepted radiant flux is instead $\int Id\Omega$. The solid angle of a disk below gives an exact example where the familiar inverse-square law is only a distant-source approximation.

A right circular cone with vertex $O$ selects all directions within a half-opening angle $\theta$ of its axis. Here $\theta$ is measured from the axis to a boundary direction, so the full opening in a meridian plane is $2\theta$. For $0\le \theta \le \pi$, interpret the selected directions as a spherical cap; for $\theta >\pi /2$, this is a region larger than a hemisphere.

The spherical coordinate system gives the area element $dA=R^2\sin \alpha d\alpha d\varphi$ on the sphere. Dividing by $R^2$ and integrating over the cap yieldsThe factor $\sin \alpha$ matters: a thin angular strip near a pole contains fewer directions than an equally wide strip near the equator. Treating equal intervals of both spherical coordinates as equal solid angles would overcount the polar directions.

At $\theta =45^{\circ }$, the solid angle is $\pi (2-\sqrt{2})\approx 1.8403$ steradians, or about $14.64\%$ of all directions. At $\theta =90^{\circ }$ it is $2\pi$ steradians, and at $180^{\circ }$ it is $4\pi$ steradians. For a small $\theta$ measured in radians, the Taylor series $\cos \theta =1-{\theta }^2/2+O({\theta }^4)$ gives the small-angle approximation $\Omega \approx \pi {\theta }^2$. Doubling a narrow half-opening angle therefore approximately quadruples its coverage.

Consider a flat circular disk of radius $a>0$, with its center a perpendicular distance $d>0$ from the viewpoint $O$. The line from $O$ to its center is normal to the disk. Its rim defines a right circular cone with $\tan \theta =a/d$ and $\cos \theta =d/\sqrt{d^2+a^2}$. The solid angle of a circular cone immediately givesFor $a=d$, this is the same $45^{\circ }$ half-opening example: about $1.8403$ steradians. Both the radius and distance enter; doubling both leaves the solid angle unchanged.

A useful local version applies to a small surface patch of area $dA$, at distance $r$ from $O$. If $\psi$ is the angle between its unit normal and the line to $O$, its unsigned geometric contribution isThe numerator is the projected area. For a visible patch, perpendicular projection followed by radial projection onto a small piece of a sphere gives this formula. Integrating counts the directions once provided no direction encounters multiple counted patches. For an opaque detector, use its unoccluded face. At grazing view, the projected area tends to zero.

For this on-axis disk, a ring of radius $\rho$ has $r=\sqrt{d^2+{\rho }^2}$ and $|\cos \psi |=d/r$. Its polar coordinates gives $dA=\rho d\rho d\varphi$, so the same answer follows directly from the surface integral:The changing distance and obliquity are both already accounted for; an extra detector-wide cosine factor would count the projection twice.

For $d\gg a$, the Taylor series of $(1+a^2/d^2{)}^{-1/2}$ givesThus the accepted fraction of an ideal point emitter's isotropic output is approximately $A/(4\pi d^2)$, where $A=\pi a^2$. This is the geometry behind the inverse-square law for a small detector. At fixed $a$, however, taking $d\to 0^{+}$ gives $\Omega \to 2\pi$, so the accepted fraction approaches $1/2$ rather than diverging.

For a chosen numerical example, let $a=1\mathrm{cm}$ and $d=5\mathrm{cm}$. The exact solid angle is approximately $0.122015$ steradians, giving an acceptance fraction $0.00970966$. An ideal point emitter with isotropy and total radiant flux $1\mathrm{mW}$ would therefore send about $9.71\mu \mathrm{W}$ onto the disk. The distant-source approximation gives $10.0\mu \mathrm{W}$, about $3.0\%$ high. This calculation assumes unobstructed propagation and counts incident radiant flux; photon absorption, detector response, and any optical attenuator require separate factors.

Spectral radiance is radiance per unit frequency or wavelength interval. Its astronomical synonym specific intensity is usually written $I_{\nu }=L_{\nu }$ or $I_{\lambda }=L_{\lambda }$; this $I_{\nu }$ is a spectral radiance, distinct from the radiant intensity $I$ in radiometry. For positive small intervals,The two expressions refer to the same spectral slice, with the positive widths related through frequency $\nu$ and wavelength $\lambda$. For frequency measured in hertz, $\mathrm{Hz}={\mathrm{s}}^{-1}$, the units are $[L_{\nu }]=\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}\mathrm{H}{\mathrm{z}}^{-1}$. For wavelength measured in metres, $[L_{\lambda }]=\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}{\mathrm{m}}^{-1}$. Integrating over a spectral band gives its radiance, and then multiplying by the collecting projected area and accepted solid angle gives a small uniform beam's radiant flux. A spectral measurement therefore retains information that the band-integrated power alone loses.

For an ideal blackbody, Planck's law gives $L_{\nu }=B_{\nu }(T)$ and $L_{\lambda }=B_{\lambda }(T)$ at temperature $T$, independent of outgoing direction. Its existing frequency and wavelength peaks of Planck's law illustrate why the coordinate matters. The simpler flat-band calculation below isolates that same change of variables without the thermal physics.

A steradian is the solid angle subtended at a sphere's center by a patch of surface area $R^2$. The whole sphere subtends $4\pi \mathrm{sr}$. In the SI, the steradian has unit one; retaining its symbol identifies a solid angle, just as radians identify an angle. A dimensionless quantity still needs its physical meaning specified.^[ref]

The emission coefficient per volume, frequency and solid angle of a Maxwellian plasma has $j_{\nu }\propto n_e{\sum }_i{Z}_i^2{n}_i{T}^{-1/2}{e}^{-h\nu /(k_{B}T)}{\bar{g}}_{\mathrm{ff}}$. Define the angular convention before supplying a numerical coefficient. The high-frequency cutoff measures temperature, while the normalization measures a density-squared emission integral. The radio spectrum is almost flat when optically thin and rises as ${\nu }^2$ when optically thick at fixed projected area.