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A photon gas is a collection of photons described statistically as radiation. Here the equilibrium model is homogeneous, isotropic electromagnetic radiation in vacuum at temperature $T>0$: each photon has energy $E=c|\mathbf{p}|=h\nu$, with momentum $\mathbf{p}$ and frequency $\nu$, and there are two independent transverse states of electromagnetic polarization. An absorbing cavity with walls at a common temperature provides a way to establish this thermal equilibrium. The bulk formulas assume cavity dimensions much larger than the typical wavelength $hc/(k_{B}T)$, so the electromagnetic mode density in a box can replace individual cavity modes by a continuum. ^[ref]

The crucial constraint is whether photon emission and photon absorption, or other number-changing processes, can equilibrate the photon population. When they can, photon number is not an independently conserved quantity and the equilibrium photon chemical potential is ${\mu }_{\gamma }=0$. For each polarization and frequency, the mean occupation number then follows the Bose-Einstein distribution:If number-changing processes are ineffective but number-conserving interactions redistribute energy, a constrained Bose-Einstein distribution can instead have $f=[\exp ((h\nu -{\mu }_{\gamma })/(k_{B}T))-1{]}^{-1}$ with chemical potential ${\mu }_{\gamma }\le 0$ fixed by the number constraint. Thus thermal equilibrium with respect to energy exchange alone does not establish ${\mu }_{\gamma }=0$. The zero value can also persist in an already prepared Planck photon distribution during suitable subsequent evolution; continual photon emission is not required to preserve it. ^[ref]

Combining $f$ with the two-polarization density of states, $g_{\nu }/V=8\pi {\nu }^2/c^3$, gives the Planck photon distribution and the spectral photon energy density:Here $B_{\nu }$ is the spectral radiance in Planck's law; $u_{\nu }$ is stored energy density per unit frequency, not outgoing power per area. Their different units are connected by the full-sphere solid angle and the speed of light $c$. The Boltzmann constant is $k_B$, and $h$ is the Planck constant. ^[ref]

The equilibrium equation of state is particularly simple:The radiation constant $a_R$ is the same constant called $a_{\mathrm{rad}}$ in black-body radiation; $\sigma$ is the Stefan-Boltzmann constant. An outline of the energy density calculation substitutes $x=h\nu /(k_{B}T)$ and uses ${\int }_0^{\infty }{x}^3/(e^x-1)dx={\pi }^4/15$. The radiation pressure follows because the angular average of the squared directional component of photon momentum is $1/3$ for isotropy; the thermodynamic derivation of blackbody radiation pressure supplies an independent route. These formulas describe the radiation contribution, not the cavity walls. The isotropic massless relation $p=u/3$ survives a different isotropic spectrum, whereas the coefficient in $u=a_R{T}^4$ requires the zero-chemical potential Planck photon distribution.

For a computed example with stipulated $T=2.725\mathrm{K}$, the photon energy density is $u\simeq 4.17174\times 10^{-14}\mathrm{J}{\mathrm{m}}^{-3}$ and the radiation pressure is $p\simeq 1.39058\times 10^{-14}\mathrm{Pa}$. The photon number density in the International System of Units isHere $\zeta$ is the Riemann zeta function. The mean photon energy differs from the energy at the peak of a spectral curve: counting photons and weighting them by energy emphasize different frequencies. The calculations use the exact International System of Units values of the Planck constant, Boltzmann constant and speed of light. ^[ref]

A motivating observation is the cosmic microwave background spectrum measured by the Far Infrared Absolute Spectrophotometer (FIRAS) aboard COBE. Its apparatus used a polarizing Michelson interferometer, with one input viewing a roughly seven-degree sky patch and the other an internal blackbody. Scanning mirrors changed the optical path length difference; heat-sensitive detectors recorded the varying power, and Fourier transform spectroscopy recovered the spectral difference. Operation outside the terrestrial atmosphere removed atmospheric emission and absorption, while shielding and cooling the optics in a helium cryostat suppressed the instrument's own radiation. ^[ref]

Absolute calibration came from moving an external blackbody into the sky aperture, filling the same beam used for the sky. This was an iron-loaded epoxy absorber shaped as a re-entrant cone. Repeated reflections made it a close approximation to a blackbody, while thermometers supplied its temperature. Mather and colleagues explicitly evaluated reflections, temperature gradients, radiation leaking through the material and leakage around its edge. These are limits on how faithfully the calibrator realizes thermal equilibrium radiation, rather than details that can be eliminated merely by calling it black. ^[ref]

During sky observations the internal blackbody temperature was adjusted to nearly null the spectral difference, reducing gain and drift errors. Fixsen and colleagues' 1996 analysis used calibrated spectra over wavenumbers $2$ to $21\mathrm{c}{\mathrm{m}}^{-1}$, about $60$ to $630\mathrm{GHz}$. They separated the sky's spatially uniform component, its dipole and Galactic emission, and varied sky cuts and foreground templates as controls. The resulting cosmic microwave background spectrum agreed with a single Planck function, with reported RMS deviations below 50 parts per million of its peak brightness. This is a residual statistic over the analyzed spectrum, not a pointwise relative error bound at every frequency. ^[ref]

The observation motivates using a photon gas as a physical model rather than merely a formal counting exercise. Its interpretation has a clear boundary: agreement with Planck's law constrains departures from an equilibrium spectral shape within the calibrated band and foreground assumptions; it does not directly measure radiation pressure or entropy density of a photon gas. Nor does it establish that today's cosmic microwave background is continually re-equilibrating with matter. A previously thermal spectrum can retain its Planck function shape as expansion of the universe lowers every photon frequency in the same proportion. The usefulness of the model is that one measured spectral temperature, together with its equilibrium assumptions, predicts energy density, radiation pressure, photon number density and entropy density of a photon gas consistently.

Planck's law gives the spectral radiance of an ideal blackbody in thermal equilibrium at absolute temperature $T>0$. It predicts how much power is carried in each small frequency or wavelength interval, per projected area and per solid angle. The whole curve is determined by $T$: no material-specific parameter enters this ideal Planck function.

Writing $\nu$ for frequency in hertz and $\lambda$ for wavelength in metres, the two forms of the Planck function areHere $h$ is the Planck constant, $k_B$ the Boltzmann constant, and $c$ the speed of light in vacuum. Their exact values in the International System of Units are $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$, $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$, and $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$. ^[ref]

The units matter: $B_{\nu }$ is measured in $\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}\mathrm{H}{\mathrm{z}}^{-1}$, whereas $B_{\lambda }$ is measured in $\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}{\mathrm{m}}^{-1}$. Both are spectral radiance, rather than radiant exitance or energy density. To report $B_{\lambda }$ per micrometre, multiply its per-metre value by $10^{-6}$; a micrometre-wide interval is $10^{-6}$ metres wide.

For a computed example, at $T=6000\mathrm{K}$ and $\lambda =0.500\mu \mathrm{m}$, the exponent is $hc/(\lambda k_{B}T)\approx 4.79592$. Substitution into Planck's law gives $B_{\lambda }\approx 3.17569\times 10^{13}\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}{\mathrm{m}}^{-1}$, or $3.17569\times 10^7\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}$ per micrometre. At the same wavelength, the illustrative temperatures $3000\mathrm{K}$ and $4500\mathrm{K}$ give $2.60268\times 10^5$ and $6.37793\times 10^6$ in the latter units. These are ideal predictions for stipulated temperatures.

The key change of variables is $\nu =c/\lambda$. Corresponding positive intervals carry the same power, so $B_{\lambda }d\lambda =B_{\nu }|d\nu |$ andEqual-width wavelength intervals have unequal widths in frequency; the factor $c/{\lambda }^2$ compensates for this. Merely replacing the horizontal-axis labels of $B_{\nu }$ by $c/\nu$ does not produce $B_{\lambda }$. This explains the frequency and wavelength peaks of Planck's law.

A short derivation uses the Boltzmann distribution for a single electromagnetic mode with thermal energies $nh\nu$, $n=0,1,\dots$. Put $q=e^{-h\nu /(k_{B}T)}$. The geometric series gives $Z={\sum }_{n\ge 0}{q}^n=(1-q{)}^{-1}$ and mean thermal energy $h\nu q/(1-q)$. The electromagnetic mode density in a box, including both independent polarization of an electromagnetic wave states of a photon, is $8\pi {\nu }^2/c^3$ per volume per unit frequency. Multiplication gives spectral energy density $u_{\nu }=8\pi h{\nu }^3/[c^3(e^{h\nu /(k_{B}T)}-1)]$; isotropy gives $B_{\nu }=c{u}_{\nu }/(4\pi )$. This counts thermal excitation energy, excluding zero-point energy. ^[ref]

Planck's law connects quantum mechanics to a measurable radiation curve. Its small-$h\nu /(k_{B}T)$ limit is the Rayleigh-Jeans law, $B_{\nu }\approx 2k_{B}T{\nu }^2/c^2$; its large-$h\nu /(k_{B}T)$ limit is the Wien approximation, $B_{\nu }\approx (2h{\nu }^3/c^2)e^{-h\nu /(k_{B}T)}$. The suppression at high frequency makes the total energy density finite. Integrating the curve gives the Stefan–Boltzmann law, while locating its peak gives Wien's displacement law. Thus one formula predicts both the spectral shape and the total emission.

An ideal point emitter emits radiant flux from one location. It approximates a physical source when its emitting dimensions are small relative to the distances relevant to the calculation. With isotropy, its radiant intensity is $I=\Phi /(4\pi )$ for total radiant flux $\Phi$. An extended source generally requires integrating contributions from different emitting positions.

Pressure measures the compressive normal force per unit surface area. For a uniformly loaded surface of surface area $A$, the magnitude of the perpendicular force is $F_{\perp }=PA$; locally, $P=d{F}_{\perp }/dA$. Its SI unit is the pascal, $1\mathrm{Pa}=1\mathrm{N}{\mathrm{m}}^{-2}$. A force parallel to the surface contributes shear stress, rather than pressure. In a gas, the steady macroscopic force comes from many microscopic transfers of momentum; ideal gas pressure from wall collisions makes this connection explicit. ^[ref]

For a computed example, a uniform pressure of $100\mathrm{kPa}$ on a piston of surface area $10\mathrm{c}{\mathrm{m}}^2=10^{-3}{\mathrm{m}}^2$ produces a perpendicular force of $100\mathrm{N}$. If the opposite face is exposed to $80\mathrm{kPa}$, the net force due to the two pressures is $(100-80)\times 10^3\times 10^{-3}=20\mathrm{N}$, directed toward the lower-pressure side. The pressure on either face and their difference answer different questions; gauge pressure records a difference from a specified reference, often atmospheric pressure.

A scalar pressure describes a gas in local thermodynamic equilibrium with isotropic random particle velocities: every wall orientation has the same normal force per unit surface area. Without this isotropy, normal stresses can depend on direction and a pressure tensor is needed. Thus random motion alone is insufficient; a preferred direction matters.

In thermodynamics, pressure and volume are thermodynamic conjugate variables. For a simple compressible system, reversible expansion delivers work $\delta W=PdV$, and at fixed entropy $S$ and particle number $N$,where $U$ is internal energy. A large pressure therefore means both a large wall force per unit surface area and a large amount of expansion work per small increase in volume.

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 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]

Refraction changes the direction of a transmitted wave when it crosses between media with different phase velocities. For light at a planar interface, Snell's law relates the angles measured from the normal vector through the refractive indices. At normal incidence the direction remains unchanged even though the wavelength and phase velocity can change. The worked example under geometrical optics distinguishes transmission from total internal reflection.

For light, refraction explains why a submerged object looks nearer the surface and why an optical lens can form an image. A wavefront crossing a boundary must retain a consistent wave phase along that boundary: its normal spacing changes with phase velocity, and oblique light rays change direction. Normal incidence separates these two effects especially clearly.

Take stationary, transparent, homogeneous media with isotropy, linearity of their optical response, and real positive refractive indices, at one chosen frequency $f$. With $c$ the vacuum speed of light and ${\lambda }_0=c/f$ the vacuum wavelength, medium $j$ hasThe frequency is unchanged at a stationary boundary; the phase velocity, wavelength and wavenumber can change. Here $n_j$ is a phase refractive index, not the group refractive index. If there is optical dispersion, evaluate $n_j$ at this same frequency on both sides.^[ref]

For an original idealized air–water example, stipulate $n_{\mathrm{a}}=1$ and $n_{\mathrm{w}}=4/3$, with vacuum wavelength $600\mathrm{nm}$. Then $f=c/(600\mathrm{nm})\approx 4.99654\times 10^{14}\mathrm{Hz}$, the water phase velocity is $3c/4$, and its wavelength is $450\mathrm{nm}$. At an air angle of incidence $30^{\circ }$, Snell's law gives a water angle of refraction $\arcsin (3/8)\approx 22.0243^{\circ }$. These stipulated values illustrate the relations; actual refractive indices depend on material conditions and frequency. By reversibility of an optical ray, the same path emerging from water has outgoing air angle $30^{\circ }$.

The refractive index compares the phase velocity of light in a medium with the vacuum speed of light $c$. For a transparent, uniform medium with isotropy and a positive-index propagation branch,Here $\omega >0$ is the angular frequency, $k>0$ is the wavenumber, and $v_{\mathrm{ph}}=\omega /k$ is the speed of a surface of constant wave phase. The refractive index is a dimensionless quantity; it is a property of a specified material response at a specified frequency, rather than a universal number attached to the word “glass”. More general absorbing or negative-index media require the conventions discussed in refractive index and electromagnetic response. ^[ref]

For an original model example, choose $n=3/2$ at vacuum wavelength ${\lambda }_0=600\mathrm{nm}$. The phase velocity is $v_{\mathrm{ph}}=2c/3\simeq 1.999\times 10^{8}m/s$, the frequency is $f=c/{\lambda }_0\simeq 4.997\times 10^{14}\mathrm{Hz}$, and the medium wavelength is $400\mathrm{nm}$. These are stipulated model values, not measured properties of glass or water. A value at this single frequency does not determine a pulse's group velocity.

The usefulness of refractive index is that one measured response connects several different observations: Snell's law predicts refraction, optical path length predicts accumulated wave phase, and changes with frequency describe optical dispersion. Even at normal incidence, where a light ray does not bend, a changed refractive index changes wavelength and wave phase. The following comparison isolates that effect; the oblique geometry is developed in wavefront.

For a uniform linear medium with isotropy and local constitutive response, Maxwell equations give $k^2={\omega }^2\epsilon (\omega )\mu (\omega )$. Here $\epsilon$ is permittivity and $\mu$ is magnetic permeability. Since $c^{-2}={\epsilon }_0{\mu }_0$, the relative permittivity ${\epsilon }_r=\epsilon /{\epsilon }_0$ and the relative magnetic permeability ${\mu }_r=\mu /{\mu }_0$ obeyIf both responses are real and positive with negligible loss, and the ordinary positive-index branch is intended, then $n=+\sqrt{{\epsilon }_r{\mu }_r}$. If additionally ${\mu }_r\simeq 1$, then $n\simeq \sqrt{{\epsilon }_r}$. All these responses must be evaluated at the electromagnetic wave's frequency; a static relative permittivity cannot simply be substituted at an optical frequency. ^[ref]

With loss, use a complex index $\tilde{n}=n^{\prime }+i\kappa$ and explicitly fix the phase convention for complex amplitudes. For $E\propto \exp [i(kz-\omega t)]$ with $k=\tilde{n}\omega /c$, its wave amplitude has factor $\exp (-\kappa \omega z/c)$. A forward attenuating branch has $\kappa \ge 0$, while its wave phase is controlled by $n^{\prime }$. The complex relation ${\tilde{n}}^2={\epsilon }_r{\mu }_r$ alone leaves a choice of sign: the physical solution also needs causality and the appropriate radiation or decay condition. In negative-index media the physically selected real part can be negative. The positive square root is therefore not a universal rule for complex or negative-index media. Direction and polarization of an electromagnetic wave also matter in a plane wave in an anisotropic dielectric, for which one scalar index need not describe every propagation branch.

The relative refractive index of medium two with respect to medium one is $n_{21}=n_2/n_1$, using their absolute refractive indices at the same frequency. For transparent media with isotropy and positive refractive indices, Snell's law gives $n_{21}=\sin i/\sin r$, with the angles measured from the interface normal. Thus a refraction measurement in air directly determines an index relative to air. Converting it to an absolute refractive index requires the refractive index of that air; replacing it by one is an approximation.

Consider a mechanically free isotropic model of a saturated ferromagnet, at fixed temperature and fixed magnitude of magnetization. Let $\mathbf{m}=\mathbf{M}/M_s$ be a unit vector, where $M_s$ is the saturation magnetization. The magnetostrictive strain is an infinitesimal strain tensor, measured relative to a reference with randomly oriented magnetic domains. This reference convention matters: it sets the orientational average of the preferred strain to zero. Define ${\lambda }_s$, the saturation magnetostriction, as the longitudinal relative length change in this model.

Rotational symmetry permits a symmetric second-rank tensor made from $\mathbf{m}$ of the form $a{\delta }_{ij}+b{m}_i{m}_j$, where ${\delta }_{ij}$ is the Kronecker delta. If we restrict attention to a change of shape with zero first-order volume change, its trace gives $3a+b=0$. Requiring the longitudinal component to equal ${\lambda }_s$ gives $a+b={\lambda }_s$. HenceThe tensor product $\mathbf{m}\otimes \mathbf{m}$ is symmetric and unchanged under $\mathbf{m}\mapsto -\mathbf{m}$. It uses two copies of the magnetic direction, so the third-rank isotropic tensor obstruction in isotropic linear magnetostriction vanishes does not apply. More generally, a small-magnetization expansion can contain $a{M}^2{\delta }_{ij}+b{M}_i{M}_j$; isotropy alone does not require its trace to vanish. The displayed saturated formula isolates the volume-preserving orientational contribution, rather than asserting that every material has zero volume magnetostriction.

Along a measuring unit vector $\mathbf{n}$, with $\mathbf{n}\cdot \mathbf{m}=\cos \theta$, the fractional length change isFor a stipulated ${\lambda }_s=20\times 10^{-6}$ and reference length $L=100\mathrm{mm}$, alignment gives $+2\mu \mathrm{m}$ and a perpendicular orientation gives $-1\mu \mathrm{m}$. Rotating from perpendicular to parallel therefore changes the length by $3\mu \mathrm{m}$, even though the saturated longitudinal value relative to the random reference is $2\mu \mathrm{m}$. A negative ${\lambda }_s$ reverses these elongation and contraction signs.

The eigenvalues of this infinitesimal strain tensor are ${\lambda }_s,-{\lambda }_s/2,-{\lambda }_s/2$, so $\Delta V/V=\mathrm{tr}{\epsilon }^{\mathrm{mag}}=0$ to first order. This is not exact finite incompressibility: using the illustrative deformation gradient $F=I+{\epsilon }^{\mathrm{mag}}$ givesA finite-deformation law would need higher-order corrections if exact volume conservation were required.

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.

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.

The Stefan-Boltzmann constant converts the fourth power of absolute temperature into blackbody radiant exitance:Here $k_B$ is the Boltzmann constant, $h$ the Planck constant and $c$ the speed of light. In the current International System of Units, these three constants have exact defining values, so $\sigma$ is exact as an expression in them and $\pi$. The displayed finite decimal is an approximation, not an exact decimal or a measured uncertainty. The 2022 recommended-constants table lists an ellipsis and an exact uncertainty for $\sigma$.^[ref]

A dimensional check uses $[k_B]=\mathrm{J}{\mathrm{K}}^{-1}$, $[h]=\mathrm{J}\mathrm{s}$ and $[c]=\mathrm{m}{\mathrm{s}}^{-1}$. Consequently $[k_B^4/(h^3{c}^2)]=\mathrm{J}{\mathrm{s}}^{-1}{\mathrm{m}}^{-2}{\mathrm{K}}^{-4}=\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-4}$, exactly the units needed in the Stefan–Boltzmann law.

For an equilibrium photon gas, the energy density is $u=a_{\mathrm{rad}}{T}^4$ with radiation constant $a_{\mathrm{rad}}=4\sigma /c$. The radiation pressure is $p_{\mathrm{rad}}=u/3=4\sigma T^4/(3c)$. These describe the radiation filling an equilibrium cavity; $u$ has units $\mathrm{J}{\mathrm{m}}^{-3}$, whereas the outward radiant exitance through a small opening is $cu/4$. The factor $1/4$ comes from averaging propagation directions and their projected area over a radiation field with isotropy, not from reducing the opening's emissivity.

Surface tension $\gamma$ is the tensile force per unit length across a line drawn in a liquid interface. For a clean interface with isotropy, it has the same value in every tangential direction: each side pulls the other along the surface, perpendicular to the line. A straight segment of length $\ell$ therefore transmits a force of magnitude $\gamma \ell$ from one interface. Here we assume constant $\gamma >0$, with fixed temperature and composition. The same coefficient measures the reversible work required to create surface area:The equality of units follows from $1\mathrm{J}=1\mathrm{N}\mathrm{m}$; the equality of the mechanical coefficients follows from the moving-rim argument below. ^{[ref] [ref]}

For a concrete example, take an ideal flat soap film spanning a rectangular wire frame with a straight sliding rim of length $\ell$. Pull the rim outwards in the plane of the film, perpendicular to the rim. Each of the film's two liquid–gas interfaces pulls the rim back towards the film with force $\gamma \ell$, so the outward holding force is $F=2\gamma \ell$. The two faces are two interfaces, each with its own surface tension. With the assumed toy value $\gamma =0.072\mathrm{N}{\mathrm{m}}^{-1}$ and $\ell =20\mathrm{mm}=0.020\mathrm{m}$, the required force is $2.88\mathrm{mN}$; one interface would contribute $1.44\mathrm{mN}$. This value of $\gamma$ is a stipulated model parameter, rather than a measured value for a particular soap solution.

Why does an interface resist being enlarged? Its molecules have a different environment from those in the bulk liquid; changing that environment has an energy cost. In the constant-surface tension model, reducing surface area lowers surface energy. This connects an easily measured rim force to the shape of a droplet, the Young–Laplace equation for its pressure, and the restoring mechanism of capillary waves. It is useful because one coefficient links forces, energy and geometry. ^[ref]

The moving-rim argument also explains the factor of two without memorizing a special formula. An outward displacement $dx$ creates surface area $\ell dx$ on each face, hence $d{A}_{\mathrm{total}}=2\ell dx$. Slow pulling without viscous dissipation requires work $Fdx=\gamma d{A}_{\mathrm{total}}$, giving $F=2\gamma \ell$. For one interface, the same argument gives $F=\gamma \ell$. A surface tension gradient instead produces Marangoni stress; the constant-$\gamma$ assumption excludes that additional effect.

Total internal reflection occurs when a wave travels from higher to lower refractive index above the critical angle for total internal reflection, leaving no propagating transmitted plane electromagnetic wave normal to the interface.

At an infinite planar interface, total internal reflection returns all incident normal power in the reflected plane electromagnetic wave. Assume homogeneous dielectrics with isotropy, linearity, no dielectric loss, magnetic permeability ${\mu }_0$, and positive real refractive indices $n_1>n_2$. Above the critical angle for total internal reflection, both TE polarization and TM polarization have unit optical reflectance despite the external evanescent wave.^[ref]

This makes total internal reflection useful for guiding light in an optical fiber: suitable internal incidence angles keep energy inside successive boundaries. Its more surprising feature is that perfect optical reflectance coexists with an external electric field. The evanescent field in total internal reflection can interact with a nearby object, as the frustrated total internal reflection experiment demonstrates.

For an original idealized example, choose $n_1=1.5$, $n_2=1$ at vacuum wavelength ${\lambda }_0=633\mathrm{nm}$. At angle of incidence $40^{\circ }$, Snell law for electromagnetic waves gives an angle of refraction $\arcsin (1.5\sin 40^{\circ })\approx 74.6186^{\circ }$, so a transmitted light ray still emerges. At ${\theta }_c\approx 41.8103^{\circ }$, its direction becomes tangent to the boundary. At $50^{\circ }$, the proposed sine of a real angle of refraction would be $1.5\sin 50^{\circ }\approx 1.14907>1$: the transmitted wavevector must instead have an imaginary normal component. These are predictions for the stipulated media, not measured properties of a particular block.

A real semicircular-block experiment isolates this threshold. The University of Maryland laboratory instructions place a laser, motor-driven rotation stage and photodetector together; an aperture limits the detector voltage. The semicircular block is centered on the rotation axis, with the photodetector aligned along its flat face. The instructions identify a strong grazing-beam signal and a weaker signal just before the critical angle for total internal reflection. The readout is detector voltage versus stage angle. Retracing the incident laser beam identifies normal incidence for alignment.^[ref]

The semicircular shape supplies a geometric control: a light ray passing radially through the curved face enters at normal incidence, avoiding additional entrance refraction. The weaker signal near the flat face locates a limiting angle of refraction; a vanishing remote-detector signal alone would not establish that the external electric field vanishes. Beam width and alignment limit angular resolution; no measured threshold or decay depth is supplied.^[ref]

Volume magnetostriction is the relative volume change associated with a magnetic-state change. To first order it is the trace of the magnetostrictive strain. Zero trace is an additional restriction, not a consequence of isotropy alone.

A wavefront connects points with the same wave phase at one specified time. It is a surface, rather than the trajectory of a particular piece of light. For a scalar monochromatic wave written as $u(\mathbf{r},t)=A(\mathbf{r})\cos \Phi (\mathbf{r},t)$, a wavefront is a connected level set of the phase function $\Phi$; different surfaces can share the same wave phase modulo $2\pi$. A plane wave has planar wavefronts. An outward wave from an ideal point source in a uniform medium has spheres as wavefronts, while an optical lens can turn nearly planar wavefronts into converging ones. The wavefront describes wave phase, not wave amplitude: drawing it does not say how bright the beam is. ^[ref]

The relation to light rays becomes precise in a stationary medium with isotropy, negligible loss and a real positive scalar refractive index $n(\mathbf{r},\omega )$. In the geometrical optics limit, spatial changes in the refractive index and wave amplitude are slow on the local wavelength scale. For a single smooth propagation branch, writeHere $\omega >0$ is the angular frequency, $c$ is the vacuum speed of light, $S$ has units of length, and the last relation is the optical eikonal equation. The local wavevector is $\mathbf{k}=\nabla \Phi =k_0\nabla S$. Its direction is normal to a smooth wavefront, since a gradient is normal to its level sets. A forward light ray has unit tangent vector $d\mathbf{r}/ds=\nabla S/n$, so it follows this normal under the stated assumptions. Along that light ray, $dS/ds=\nabla S\cdot (d\mathbf{r}/ds)=n$: the change in $S$ is the optical path length. This is the short geometric argument linking wavefronts, light rays and optical path length. ^[ref]

The diagram shows incident and transmitted branches separately, with the reflected branch omitted. Its stipulated refractive indices are $n_1=1$ and $n_2=3/2$, and its incident angle is $30^{\circ }$ to the interface normal. At a stationary interface, the frequency and tangential wavevector agree on the two sides: Snell's law gives $\sin {\theta }_2=1/3$, hence ${\theta }_2\approx 19.4712^{\circ }$. The wavefronts turn while their normal spacing decreases from ${\lambda }_0$ to $2{\lambda }_0/3$, where ${\lambda }_0$ is the vacuum wavelength. More explicitly, with $x$ normal to the interface and $y$ tangential, take $S_1=\sqrt{3}x/2+y/2$ for $x<0$ and $S_2=\sqrt{2}x+y/2$ for $x>0$. These functions agree at $x=0$; the drawn wavefronts are $S_j=m{\lambda }_0$ at $t=0$ with ${\varphi }_0=0$. This is an exact piecewise plane wave phase construction for the stipulated transparent media, not measured material data. ^[ref]

The normal-to-wavefront rule requires its assumptions. In a medium with anisotropy, the wavevector remains normal to a smooth wavefront, but the group velocity and energy flux need not follow it. ^[ref] Near a wave caustic, where multiple light rays meet, one smooth phase function may no longer describe the field; diffraction and the full wave amplitude matter. A wavefront is also not generally a surface of equal pulse arrival time. In a medium with optical dispersion, a monochromatic wave phase moves with the phase velocity, while a sufficiently narrow wave packet follows the group velocity. Even identical optical path lengths at one frequency can give different group delays. The usual equal-travel-time picture needs a common launch reference and a nondispersive ray model; an arbitrary broadband disturbance has no single monochromatic wavefront.^[ref]