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Photon absorption removes a photon and transfers its energy and momentum to matter. It contributes a negative photon-number collision source. It differs from Compton scattering, which redistributes a photon's energy and direction while preserving the count. The integrated source can vanish if photon emission and absorption balance.

An initially stationary isolated chemical atom of rest mass $M$ absorbing a single photon must recoil. If the internal excitation changes its rest mass by $\Delta M=\Delta E/c^2$, conservation of energy and conservation of momentum give $M{c}^2+h\nu =\sqrt{(M{c}^2+\Delta E{)}^2+(h\nu {)}^2}$ and hence the exact formula in the title. The additional positive energy supplies recoil. An external body sharing the momentum, or an initially moving chemical atom, changes this laboratory resonance condition.

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.

In vacuum, a photon of wave vector $\mathbf{k}$ carries momentum $\mathbf{p}=\hslash \mathbf{k}$ and energy $E=\hslash c|\mathbf{k}|$. In a direct optical interband transition this momentum is generally small compared with relevant crystal momenta, motivating the nearly vertical-transition approximation in wave vector space. Electromagnetic momentum in matter requires additional conventions.

Two photons meeting at angle $\alpha$ can produce an electron and a Positron when their combined four-momentum has sufficient invariant mass. For a head-on collision the threshold is $E_1{E}_2=m_e^2{c}^4$. Threshold means zero relative momentum in the centre-of-momentum frame, not necessarily in the laboratory.

Absorption of momentum $\mathbf{p}$ followed by emission of ${\mathbf{p}}^{\prime }$ changes a free atom's momentum by $\mathbf{p}-{\mathbf{p}}^{\prime }$. For isotropic emission of fixed magnitude, its mean vector is zero. The energy change then separates into a directional work term and two positive recoil terms. If absorption directions are also isotropic and uncorrelated with the initial momentum, the work term averages away. Equal incoming and outgoing photon magnitudes give two recoil energies per cycle.

A freely falling particle's physical momentum is measured in the local orthonormal frame of a comoving observer. With proper peculiar velocity $\mathbf{v}=ad\mathbf{x}/dt$ and Lorentz factor $\gamma =(1-v^2{)}^{-1/2}$, it is $m\gamma \mathbf{v}$. It differs by one factor of the scale factor from the conserved comoving momentum. For a nonrelativistic particle, its peculiar velocity therefore decays as $a^{-1}$.

Cecil Frank Powell's 1950 lecture illustrates the successive decays of pions and muons through their tracks and explains why the nearly fixed secondary muon range, allowing for range straggling, supports a single neutral decay partner. He describes its identification as a neutrino tentatively. The neutral particle itself leaves no recorded track. He contrasts this decay with the nuclear reactions caused by stopped negative pions, connecting the latter with the nuclear force.^[ref]

Here is a modern special relativity derivation for the dominant leptonic pion decay, assuming a free pion at rest, a massless muon neutrino, and no emitted photon. Write $m_{\pi }$ and $m_{\mu }$ for the pion and muon masses. By conservation of momentum the daughter momenta are opposite with equal magnitude $p$. By conservation of energy and the energy–momentum relation,Substitute $E_{\mu }=m_{\pi }{c}^2-pc$, square, and cancel $p^2{c}^2$. This givesThe resulting $p$ is fixed, although the muon's direction is not set by the incoming stopping track. Thus the kink angle relative to that earlier track is not the angle between the two daughters: in the pion rest frame, the daughters are back to back. A neutral-partner arrow in a diagram represents inferred momentum, not an observed string of grains.

For an original calculation, adopt rounded input values $m_{\pi }{c}^2=139.57\mathrm{MeV}$ and $m_{\mu }{c}^2=105.66\mathrm{MeV}$. They give $pc=29.79\mathrm{MeV}$, $T_{\mu }=4.119\mathrm{MeV}$ and $E_{\nu }=29.79\mathrm{MeV}$. Their sum satisfies $T_{\mu }+E_{\nu }=(m_{\pi }-m_{\mu })c^2=33.91\mathrm{MeV}$. These are rounded calculation inputs and outputs, not a fit to the 1947 plates.

At fixed material and particle identity, a fixed initial kinetic energy suggests a narrow particle range distribution broadened by range straggling and measurement uncertainty. The synthetic example convolves a single chosen range with a normal distribution; a second example superposes two such chosen ranges. It explains why repeated comparable stopping distances can be informative, without pretending to extract a historical spectrum.

The decay equations do not identify the neutral partner as a neutrino on their own: a massless photon would satisfy the same energy and momentum equations. Nor does any isolated kink establish the pion's identity. The inference gains force from calibrated particle range, track grain density, repeated decay patterns and independent particle properties. Historical particle names should not obscure the distinction: a muon is a lepton, whereas a pion is a meson.

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.

Let $U=|\mathbf{U}|$, $\nu =\mu /\rho$, and $r=|\mathbf{x}|$. A decaying outer disturbance matched to the leading translating sphere in Stokes flow isHere $\Delta \varphi =0$ and $(\nu \Delta -\mathbf{U}\cdot \nabla )\chi =0$ away from zero. The substitution $\chi =h{e}^{\mathbf{U}\cdot \mathbf{x}/(2\nu )}$ reduces the latter equation to $(\Delta -U^2/(4{\nu }^2))h=0$, whose decaying radial solution is proportional to $e^{-Ur/(2\nu )}/r$. In the inner overlap, cancellation of the potential monopole gives $\mathbf{u}\sim -3a[\mathbf{U}+\mathbf{n}(\mathbf{U}\cdot \mathbf{n})]/(4r)$. The potential source's volume flux is $6\pi \nu a$, and its mass flux is $6\pi \mu a$. Downstream, the fluid wake width grows as $\sqrt{\nu x/U}$; its mass deficit balances this source, while its momentum deficit is $6\pi \mu aU$, the Stokes drag law.

Preparation uncertainty describes the outcome spread intrinsic to a specified quantum state preparation. Operationally, prepare many copies of the same quantum state, measure position on one group and momentum on another, and estimate the two variances. Ideal measurements still have the spreads predicted by the Born rule. No earlier position measurement need occur in the momentum group. The Heisenberg uncertainty relation constrains these two distributions belonging to the same preparation.

Measurement backaction concerns how a particular measurement in quantum mechanics changes a quantum state. Instrument resolution concerns how its reported outcomes relate to the intended observable. Those questions require a model of the instrument and its interaction with the system. Substituting an instrument error and an induced change for $\Delta x$ and $\Delta p$ in the Heisenberg uncertainty relation does not follow from its proof. The practical value of preparation uncertainty is that it bounds achievable input spreads before any such instrument model is chosen.

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.

A steady unidirectional incompressible viscous flow in a rigid channel, driven by $G=-dp/dz$, reduces the axial momentum equation to the displayed Poisson equation, with $w=0$ on the cross-sectional boundary. Circular-pipe flow is the Hagen-Poiseuille flow special case; an arbitrary cross section need not have that parabolic radial profile.

In the local rest frame of a gas, the pressure tensor is the directional momentum flux $P_{ij}=\int p_i{v}_{j}f(\mathbf{p})d^{3}p$, where $f$ is the phase-space distribution function per unit momentum volume and $\mathbf{v}$ is particle velocity. For an isotropic distribution, $P_{ij}=P{\delta }_{ij}$, recovering the kinetic pressure of an isotropic gas. An anisotropic distribution can exert different normal pressures along different directions.

For fixed channel width $b(z)=\beta z$, cross-sectional area is $A=\beta h^2/2$. With reduced gravity $G\varphi$, uniform cross-sectional fields and downward settling velocity magnitude $w$, volume conservation, momentum and particle deposition flux giveThe hydrostatic pressure force contains ${\int }_0^h(h-z)\beta zdz=\beta h^3/6$. No streamwise widening term occurs in a prismatic channel. Deposition uses the projected boundary width $\beta h$ under the vertical-settling model.

With zero quartic coupling, a proper one-loop vertex with two scalar and two labelled photon legs has two boxes, five triangles and two bubbles. A directed scalar chain connects the charged external ends; the internal photon joins its endpoints. Cubic vertices have one photon slot, seagulls two. All nine graphs have Feynman-diagram symmetry factor one with external labels fixed. At zero external momentum a transverse internal photon kills any graph with a cubic endpoint, leaving only the two seagull bubbles.

An ideal receding mirror reflecting a source beam of constant power $P$ receives force $(2P/c)(1-\beta )/(1+\beta )$ after including both Doppler-shifted reflected momentum and reduced impact rate. For fixed rest mass $M$, use $dt/d\tau =\gamma$ in the momentum equation. It reduces to $d\beta /d\tau =(1-\beta {)}^2/{\tau }_0$, where ${\tau }_0=M{c}^2/(2P)$. Starting from rest gives $\beta =\tau /(\tau +{\tau }_0)$. This continuum mirror model neglects individual-photon recoil and finite beam interception.

For a concrete independent example, stipulate perpendicular motion, $|q|=e$ with $e$ the elementary charge, uniform magnetic field $B=1.00\mathrm{T}$, incoming gyroradius $r_{\mathrm{in}}=0.100\mathrm{m}$ and outgoing gyroradius $r_{\mathrm{out}}=0.0500\mathrm{m}$. These are invented teaching inputs. Converting SI momentum to electronvolt units givesso $p_{\mathrm{in}}=29.9792458\mathrm{MeV}/c$ and $p_{\mathrm{out}}=14.9896229\mathrm{MeV}/c$. Both a Positron and a proton with these momenta trace the same ideal arcs. Their kinetic energies differ because the relativistic energy-momentum relation givesThe second expression avoids subtracting nearly equal quantities when $pc\ll m{c}^2$. For this calculation stipulate rounded rest energies $m_e{c}^2=0.511\mathrm{MeV}$ and $m_p{c}^2=938.272\mathrm{MeV}$. The Positron interpretation gives $K_{\mathrm{in}}\simeq 29.473\mathrm{MeV}$ and $K_{\mathrm{out}}\simeq 14.487\mathrm{MeV}$; the proton interpretation gives only $0.479\mathrm{MeV}$ and $0.120\mathrm{MeV}$. The corresponding proposed losses of kinetic energy are about $14.985\mathrm{MeV}$ and $0.359\mathrm{MeV}$.

The small proton kinetic energy follows from $K\simeq p^2/(2m_p)$, whereas here the Positron has $pc\gg m_e{c}^2$ and $K\simeq pc-m_e{c}^2$. Consequently magnetic bending alone cannot reject the proton. One needs an independent material constraint: could a proton at the proposed kinetic energy penetrate the plate and then travel at least the observed visible distance in the specified gas? A particle range shorter than that visible distance contradicts the hypothesis. The visible track need not include the stopping point: it already supplies a lower bound on particle range. No numerical particle range is assigned to this teaching example; that would require stopping data for the actual material and gas conditions. The experimental insight is to combine a geometric momentum measurement with an independent test of penetration.

For an admissible quantum state with zero position and momentum means, positivity of the squared norm $\Vert (p-isx)\psi {\Vert }^2$ for every real $s$ givesThe discriminant is nonpositive, hence $\Delta x\Delta p\ge \hslash /2$. Subtracting the expectation values from the position operator and momentum operator extends this outline to nonzero means.

The conjugate momentum to a periodic internal soliton angle generates shifts of that angle. Single-valued wavefunctions, or Bohr-Sommerfeld quantization with zero Maslov index, quantize this canonical momentum. A theta term can shift the relation between canonical momentum and mechanical charge without changing integer canonical labels.