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Bremsstrahlung is electromagnetic radiation emitted when a charged particle accelerates during interaction with another charged particle or field. The microscopic process $e^{-}+Z\to e^{-}+Z+\gamma$ creates a photon, while its inverse absorbs one. Thus it contributes to a photon-number collision source in the photon number current, unlike ordinary Compton scattering. Thermal bremsstrahlung averages this emission over a thermal distribution of charged particles.

An original toy model separates two ways to lose a central cathode ray signal. Let $x$ be forward depth into a uniform gas. Normalize the entering electron population to one. Write $U(x)$ for the fraction still in a narrow forward channel, and $D(x)$ for the fraction diverted into a wider collection channel. Assume constant removal rate $a$ per unit depth in both channels, and constant transfer rate $b$ from $U$ to $D$, with no return. Removal means that an electron stops or is trapped before reaching the collecting plane; it does not mean that its electric charge disappears. The wide ideal collector accepts all surviving members of both channels. This is a stipulated two-population differential equation model, not a general law of electron transport:The first equation gives exponential decay. Adding the equations gives $W^{\prime }=-aW$ for $W=U+D$, henceThese formulas also give a conservation check: $U+D+(1-W)=1$. The fraction $1-W$ has been removed before the plane, while $D$ remains present but misses the narrow collector. An exponential fit to the narrow signal therefore measures $a+b$, not the removal rate $a$ by itself. This is an identifiability issue: one declining signal cannot distinguish the two unknown rates.

For a worked example, choose $a=0.15\mathrm{c}{\mathrm{m}}^{-1}$ and $b=0.35\mathrm{c}{\mathrm{m}}^{-1}$. These are invented teaching inputs, not measured coefficients for air or Lenard's apparatus. At $x=2.00\mathrm{cm}$,Thus about $63.2\%$ misses the narrow collector, although only about $25.9\%$ has been removed in this model. Treating the whole narrow-signal deficit as stopping would confuse redistribution with removal.

Suppose a calibrated readout is proportional to arriving electron number with the same response for either channel. Choose an empty-path signal $R_0=1020$ arbitrary units, of which $B=20$ is background; the two collection geometries have the same calibrated reference. The calculated readings at $2\mathrm{cm}$ are $R_{\mathrm{n}}=20+1000e^{-1}\simeq 387.88$ and $R_{\mathrm{w}}=20+1000e^{-0.3}\simeq 760.82$. After background subtraction, define $T_j=(R_j-B)/(R_0-B)$ for each geometry. Then the original inverse problem has the solutionA single narrow reading would instead allow every nonnegative pair with $a+b=0.50\mathrm{c}{\mathrm{m}}^{-1}$. Using the raw ratios without subtracting $B$ gives an erroneous narrow rate of about $0.4834\mathrm{c}{\mathrm{m}}^{-1}$. In a real fluorescent screen, the assumed response must be tested: deposited energy and conversion to light need not remain constant as electrons slow. Repeating the comparison at several depths tests the constant slopes, although straight slopes alone do not establish the microscopic mechanism.

Actual electrons change kinetic energy and direction. Electron stopping power $S(K)=-dK/ds>0$ describes mean kinetic energy loss per travelled path length $s$; collisional losses include ionization, while radiative losses include bremsstrahlung. The continuous slowing-down approximation range integrates reciprocal total electron stopping power, neglecting fluctuations. It approximates travelled path length, not forward penetration depth.^[ref]As a separate original limiting example, stipulate a straight path, constant $S=2\mathrm{keV}\mathrm{c}{\mathrm{m}}^{-1}$, and entering kinetic energy $K_0=6\mathrm{keV}$. Then $K(s)=6\mathrm{keV}-Ss$ until stopping, and $R_{\mathrm{CSDA}}=3\mathrm{cm}$. In that deterministic model every entering electron reaches each depth below $3\mathrm{cm}$ and none crosses a plane beyond it. There is no exponential population tail. Real fluctuations and electron scattering make the situation more complicated, but the example already proves that continuous energy loss does not imply constant-rate removal.

The window also changes the entering population: source kinetic energy before the foil need not equal $K_0$ just outside it. For an electron initially at rest, an ideal static accelerating electric potential difference $V>0$ supplies electrical work $eV$, where $e$ is the elementary charge. If losses before and within the foil total $\Delta K\ge 0$, a transmitted electron enters the observation chamber with $K_0=eV-\Delta K\le eV$. A screen's brightness determines neither $V$ nor $\Delta K$. These formulas specify a modern ideal work-energy theorem argument; no numerical historical voltage or foil thickness is inferred from the diagram.

A Characteristic X-ray comes from an electronic transition filling an inner electron shell vacancy. The emitted photon carries the difference between atomic energies. Vacancy creation by direct electron impact requires kinetic energy above the shell's binding energy, exceeding the line energy. Higher tube voltage need not move the spectral line.^[ref]

An original stipulated shell model illustrates the distinction without using measured atomic data. Let positive binding energies be $B_K=9.0\mathrm{keV}$ and $B_L=1.2\mathrm{keV}$, ignoring relaxation corrections. Filling a K-shell vacancy from the L shell releasesAn $8.5\mathrm{kV}$ tube can energetically emit a $7.8\mathrm{keV}$ bremsstrahlung photon, yet cannot directly remove this K-shell electron. A $12\mathrm{kV}$ tube can create that vacancy by electron impact; if radiative filling occurs, the line is still $7.8\mathrm{keV}$, not $12\mathrm{keV}$. This argument concerns direct electron impact and an initially unexcited target; it does not describe every possible excitation route.

Appendix I, printed page I-2, distinguishes low-atomic-number beta shielding from an outer layer used to reduce secondary bremsstrahlung.

A crystal X-ray spectrometer selects a narrow wavelength interval by diffraction at known interplanar spacing $d$ and Bragg angle $\theta$. For fixed diffraction order $n$, a known $d$ lets an angular measurement determine $\lambda =2d\sin \theta /n$. This reverses the crystallography problem of finding $d$ from a known wavelength.

Siegbahn's laboratory developed X-ray tubes and evacuated spectrographs for longer wavelengths. Rotating the crystal selected reflections; photographic plates recorded positions. Opposite-side exposures measured angles on both sides of the direct beam. Later tube instruments mounted a beam-defining slit opposite the plate holder. A separate double instrument compared two reflected beams with two ionization chambers. These were different instruments, not one universal assembly.^[ref]

The Nobel biography describes constant-temperature spectrographs and a precision angular scale read by diametrically opposed microscopes.^[ref] Temperature control matters because thermal expansion changes $d$; an accurately read angle cannot compensate for an unknown interplanar spacing.

The schematic separates preparation, selection and readout. An X-ray tube target supplies Characteristic X-rays with a possible bremsstrahlung background. A collimator limits incident directions; evacuation reduces air X-ray attenuation. Rotating the crystal brings different wavelengths into reflection. A photographic plate stores line positions over an exposure, whereas an ionization chamber can measure collected electric charge at successive angles. Neither readout alone establishes absolute wavelength: an angular zero, angular-scale calibration, diffraction order, $d$ and any required refraction correction enter the inference. Narrower slits reduce accepted angular spread but also reduce signal; a weak line can become harder to locate despite a narrower instrumental response.

A control reported in Siegbahn's lecture photographed a line system with copper-sulphate and calcium-sulphate crystals. Irregular relative shifts around copper's characteristic frequency supported anomalous dispersion; the arrangement excluded differing absorption across that frequency as the explanation.^[ref] A finer angular scale alone cannot remove this physical refraction correction.

Free-free absorption is the inverse of bremsstrahlung: a free electron absorbs a photon during an encounter with an ion, which permits conservation of momentum. In local thermodynamic equilibrium its absorption coefficient and the thermal free-free emissivity obey Kirchhoff's law of thermal radiation, $j_{\nu }={\alpha }_{\nu }{B}_{\nu }(T)$. For a nonrelativistic plasma at $h\nu \ll k_{B}T$, ${\alpha }_{\nu }\propto n_e{\sum }_i{Z}_i^2{n}_i{T}^{-3/2}{\nu }^{-2}{g}_{\mathrm{ff}}$, including the stimulated-emission correction in the net coefficient.

Let the net cooling per unit mass of an ideal gas be $\mathcal{L}=A\rho T^{\alpha }-H$, with positive constants $A,H$. At fixed pressure and composition, $\rho T$ is constant, so $\mathcal{L}=C{T}^{\alpha -1}-H$. At a thermal equilibrium $T_0$, $C{T}_0^{\alpha -1}=H$. The specific heat capacity at constant pressure gives $c_{p}d(\delta T)/dt=-(\partial \mathcal{L}/\partial T{)}_p\delta T$. Thus $\delta T\propto e^{st}$ with the displayed growth rate. The equilibrium is stable for $\alpha >1$, unstable for $\alpha <1$, and marginal at linear order for $\alpha =1$. In particular, optically thin Bremsstrahlung with $\alpha =1/2$ is unstable under this heating law and the isobaric constraint. Fixed-density perturbations have a different criterion.

Take positive $\dot{Q}$ to mean net cooling per unit mass. At fixed pressure for an ideal gas of unchanged composition, $\rho \propto T^{-1}$, so $\dot{Q}=C{T}^{\alpha -1}-H$ with $C>0$. At thermal equilibrium, $C{T}_0^{\alpha -1}=H$. Because expansion work is included at fixed pressure, the perturbed temperature obeysThis isobaric stability criterion for density-dependent power-law cooling gives stability for $\alpha >1$ and instability for $\alpha <1$. For optically thin bremsstrahlung, $\alpha =1/2$, soA slightly cooler parcel becomes denser and cools faster; a slightly hotter parcel has reduced cooling and heats further. The conclusion assumes the sound-crossing time is short enough to maintain nearly common pressure, with conduction and other stabilizing processes omitted.

Take positive $\dot{Q}$ to mean net removal of thermal energy per unit mass. The isobaric thermal stability of a heated ideal gas depends on the density change accompanying a temperature disturbance. For an ideal gas of fixed composition, $p=\rho k_{B}T/\bar{m}$, so constant pressure gives $\rho T=\text{constant}$. Therefore the net cooling can be writtenwhere $C>0$ is constant for this pressure. At thermal equilibrium $T=T_0$, $C{T}_0^{\alpha -1}=H$.

The first law of thermodynamics at constant pressure includes the expansion work, so the relevant heat capacity is the specific heat capacity at constant pressure, $c_p$, rather than the constant-volume value. Linearizing its temperature equation givesHenceFor optically thin Bremsstrahlung, $\alpha =1/2$, soA hotter parcel expands and becomes less dense. Its cooling per unit mass, proportional to $T^{-1/2}$ at fixed pressure, decreases while its cosmic ray heating stays fixed, making it hotter still. A colder, denser parcel cools faster and runs away in the opposite direction. Here $H$ has units $\mathrm{W}\mathrm{k}{\mathrm{g}}^{-1}$ and $c_p{T}_0$ units $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$, so $s$ is an inverse time. The conclusion assumes the disturbance remains close to pressure equilibrium and neglects heat conduction and changes of composition. The relevant $\alpha$ is the temperature exponent in the supplied cooling law.

A planetary nebula combines a hot post-asymptotic-giant-branch stellar core, its ionized expelled envelope, neutral/molecular material and interstellar dust. Its ultraviolet-to-infrared spectral energy distribution is therefore not a single-temperature blackbody. The principal components are:

The observed mixture also contains attenuation by internal and foreground interstellar extinction. A spectrum containing all these components need not spatially coincide: stellar light, ionized gas, dusty clumps and the molecular envelope sample different regions.

For a monatomic ideal gas, write the thermal energy density as $u=3n{k}_{B}T/2$. At low optical depth, radiative loss uses an astrophysical cooling function, often written $\mathcal{C}=n_e{n}_H\Lambda (T,Z)$, while $\mathcal{H}$ contains heating. At fixed density the net radiative gas cooling time is approximately $u/(\mathcal{C}-\mathcal{H})$ when the denominator is positive; a cooling flow closer to constant pressure must include enthalpy and compressional work. Comparing this time with the free-fall time of a uniform sphere determines whether pressure can be lost during gravitational contraction.

Hot ionized gas cools by bremsstrahlung and recombination; collisional excitation followed by photon escape supplies atomic and metal-line cooling at lower temperatures. molecular line cooling through rotational or vibrational lines and dust thermal emission become important in cold dense gas. Abundances, ionization state, density-dependent collisional de-excitation, and photon trapping all change the cooling function. Optically thick gas requires radiative transfer, not a simple local density-squared loss law.

Heating comes from photoionization, grain photoelectric heating driven by ultraviolet light, cosmic rays, shocks, turbulent dissipation and compression. photoionization heating deposits the photon energy above the ionization threshold as electron kinetic energy. A thermal equilibrium satisfies $\mathcal{H}=\mathcal{C}$; its stability depends on how the net loss changes under the relevant pressure or density constraint, not merely on equality. Different stable balances permit a multiphase interstellar medium. Spectral-line ratios constrain temperature and ionization, but interpreting them requires density diagnostics, interstellar extinction corrections and a radiation-field model.

An electron accelerated in an ion's Coulomb field emits bremsstrahlung. Both electron states are unbound, giving free-free radiation rather than a discrete bound-state line. Averaging over a Maxwellian electron distribution gives an optically thin spectral emissivity with dependenceHere $C_{\mathrm{ff}}$ depends on the emissivity convention, and the Gaunt factor supplies the quantum correction. No numerical emissivity normalization is needed for these diagnostics. For a mixture, sum $Z_i^2{n}_i$ over ion species. The bolometric emissivity is proportional to $n_e\sum Z_i^2{n}_i{T}^{1/2}$ with a frequency-averaged Gaunt factor. Thermal electron speeds control the spectrum while the encounter rate supplies its density-squared dependence.

temperature can be measured from the high-energy exponential cutoff at photon energy of order $k_{B}T$, provided the spectrum spans that turnover and foreground absorption and nonthermal contamination are modelled. At radio frequencies $h\nu \ll k_{B}T$, the exponential is nearly one, so the optically thin continuum is a poor thermometer by slope alone. Kirchhoff's law, ${\alpha }_{\nu }=j_{\nu }/B_{\nu }$, and the Rayleigh-Jeans law give ${\alpha }_{\nu }\propto n_e{n}_i{T}^{-3/2}{\nu }^{-2}{\bar{g}}_{\mathrm{ff}}$ for an isothermal thermal plasma. The transfer solution isin the radio limit. A resolved optically thick region therefore measures its electron temperature through brightness temperature, if it fills the beam. The turnover between thick and thin emission constrains the line-of-sight emission measure $\mathrm{EM}=\int n_e^{2}dl$ for a hydrogen-dominated fully ionized plasma. In the thin regime $T_b\simeq T{\tau }_{\nu }$, rather than $T_b=T$. The slowly varying Gaunt factor gives a nearly flat optically thin radio flux-density spectrum, in contrast to the ${\nu }^2$ thick limit at fixed projected area.^[ref]

To obtain density, one also needs geometry. An angular extent and independent distance give a transverse size; an assumed shape provides a depth or volume. For approximately uniform emitting gas with astrophysical volume filling factor $f_V$, $L_{\mathrm{ff}}\propto n_e^2{f}_{V}V{T}^{1/2}$, so $n_e\propto [L_{\mathrm{ff}}/(f_{V}V{T}^{1/2}){]}^{1/2}$. Alternatively a measured column emission measure and assumed path length give an rms density. Clumping raises the mean squared density relative to the square of the mean: the inferred value is not automatically the mean gas density. Free-free brightness alone does not uniquely determine temperature, density and physical size. Resolved imaging, a distance, optical-depth information or an independent column-density diagnostic is needed to remove those degeneracies. Cluster SZ data are complementary because their leading signal depends on $\int n_e{T}_{e}dl$, whereas free-free emission depends on density squared. Metal lines, unresolved temperature structure, foreground absorption, synchrotron emission and background subtraction must be controlled before attributing the continuum to a single thermal component.

A heated cathode emits electrons, accelerated through a potential difference onto a metal target. Deceleration produces a bremsstrahlung continuum with ${\lambda }_{\min }=hc/(e{V}_{\mathrm{acc}})$; inner-shell ionization followed by electronic relaxation produces characteristic lines. The pictured instrument is consistent with the copper-target TEL-X-Ometer used for teaching diffraction. Its copper $K_{\alpha }$ radiation has wavelength about $0.154\mathrm{nm}$, comparable with atomic plane spacings. A nickel filter preferentially suppresses copper $K_{\beta }$, while slits collimate the beam; filtering does not remove every continuum photon.

A Geiger-Müller counter converts photon-induced gas ionization into countable electrical pulses. A scintillator with photomultiplier is another detector option. A detector-arm scan records count rate against scattering angle; coupled crystal rotation preserves the usual reflection geometry. These instrument features are documented by its manufacturer.^[ref]

For the numerical calculations below use $\lambda =0.154\mathrm{nm}$, explicitly identifying the characteristic line assumed. The photograph alone does not give a measured wavelength or angular travel limit; conclusions depending on these parameters are stated conditionally. A different selected line changes the angles but not a crystallographic extinction.

A low-frequency example is Galactic synchrotron emission: relativistic electrons spiral in the Galactic magnetic field and emit a broad nonthermal spectrum. In antenna-temperature units its approximate spectrum is $T_A\propto {\nu }^{-{\beta }_s}$ with ${\beta }_s$ around3, varying with the electron population and sky direction. It is a major, often dominant contamination at frequencies below roughly20–40GHz; radio point sources and bremsstrahlung can contribute in the same low-frequency region.

A high-frequency example is Galactic thermal dust emission. Grains heated by starlight have approximate modified-blackbody spectrum $I_{\nu }\propto {\nu }^{{\beta }_d}{B}_{\nu }(T_d)$, with grain temperatures of order tens of kelvin. Their emissivity-weighted emission rises rapidly through the microwave/submillimeter range and commonly dominates anisotropy contamination above roughly100–200GHz, especially at higher frequencies. Dusty extragalactic galaxies provide a related high-frequency foreground.

Synchrotron is the low-frequency contaminant and thermal dust the high-frequency contaminant. There is no universal crossover frequency at which either exceeds CMB anisotropy everywhere: sky masking, angular scale, polarization and source amplitude matter. Cleaner cosmological windows lie between these regimes, and multi-frequency observations distinguish the foreground spectra from a thermodynamic CMB temperature perturbation. The separation into low-frequency emission and high-frequency dust is also used in the Planck astrophysical-component products.

In the local orthonormal tetrad, the components of the photon number current areThey are respectively the photon number density and the photon number flux. Thus $J^{\mu }$ is the photon number four-current, not an energy current or an electric charge current. Photons carry no electric charge. For the isotropic background,Integrating the covariant Boltzmann equation over the invariant mass-shell measure, with vanishing momentum-boundary flux, giveswhere $C[f]$ is the collision operator in the covariant streaming equation. ThereforeCollisionless photon propagation satisfies this condition, as do Thomson scattering and ordinary Compton scattering when all participating photons are counted: each collision has one incoming and one outgoing photon. Photon emission, photon absorption, bremsstrahlung, double Compton scattering and electron-positron annihilation in cosmology or its inverse generally give a nonzero number source. A cancellation of net creation and destruction can nevertheless make the integrated source vanish even when individual number-changing reactions occur. Cosmological redshift itself changes photon energy, not photon number; in an expanding homogeneous universe a conserved photon number current gives ${\bar{n}}_{\gamma }\propto a^{-3}$.

The main characteristic peaks identify nickel and phosphorus: the spectrum has a strong Ni $K\alpha$ peak near $7.48\mathrm{keV}$, the Ni $L$ region near $0.85\mathrm{keV}$ and P $K\alpha$ near $2.01\mathrm{keV}$. Weaker associated lines, including Ni $K\beta$ near $8.26\mathrm{keV}$, can accompany them. The continuous background is bremsstrahlung from decelerated electrons.

Small carbon and oxygen signals at low energy can arise from contamination, surface oxide or preparation materials; a copper contribution can arise from the adjacent substrate within the finite interaction volume. The Cu $L$ region lies close to Ni $L$, so overlap is particularly troublesome at low energy. A weak Cu $K\alpha$ contribution near $8.04\mathrm{keV}$ is better separated, though still close to Ni $K\beta$. At $30\mathrm{kV}$ the beam can excite these K lines.

Peak height is not an atomic fraction: ionization cross-sections, fluorescence yields, absorption, detector response, background and line overlap require correction. The large nickel peak therefore supports the nickel-rich assignment but does not directly give the nickel-to-phosphorus atomic ratio.

Reduce time, increase distance and use radiation-appropriate shielding. At fixed conditions, accumulated absorbed dose is proportional to exposure time. For an approximately isotropic point source, the inverse-square law reduces the uncollided radiation field as distance increases. Confinement of radioactive material prevents contamination; it addresses a different pathway from attenuation of an external field.

For alpha particles, short stopping ranges make light barriers sufficient against external particles, but confinement remains important. For beta particles, a low-atomic number shield such as plastic limits both penetration and secondary bremsstrahlung; high-atomic number material can then attenuate residual photons. For penetrating gamma rays, dense material such as lead, or a thicker bulk shield such as concrete, provides attenuation. One shield material is not equally appropriate for every radiation type.^[ref][ref]

The idealized external photon estimate is approximately $0.557\mathrm{mSv}{\mathrm{h}}^{-1}$ at one metre. First apply radioactive decay to the certified radioactive activity:Treat the source as isotropic and small compared with the observation distance. Neglect photon self-absorption, capsule attenuation and attenuation in air. The metal and capsule stop the primary beta particles, so the direct external contribution is taken to be the two gamma rays; secondary bremsstrahlung is neglected. Their total energy per transformation isThe incident photon energy flow per area at distance $r$ is thereforeInterpret the given $3{\mathrm{m}}^{-1}$ as the linear energy-absorption coefficient ${\mu }_{\mathrm{en}}$, rather than a total photon-removal coefficient. For an absorbing medium of mass density $\rho$, a thin volume $Sdx$ absorbs energy per time ${\Psi }_{E}S{\mu }_{\mathrm{en}}dx$ and has mass $\rho Sdx$. Its absorbed dose rate is consequentlyFor the estimate assume a water-like $\rho =10^3\mathrm{kg}{\mathrm{m}}^{-3}$ and local deposition of the secondary charged-particle energy. At $r=1\mathrm{m}$,For gamma rays, the radiation weighting factor is one, giving $\dot{H}=0.556815\mathrm{mSv}{\mathrm{h}}^{-1}$ under these assumptions.^[ref]

The phrase “equivalent absorbed dose” mixes two quantity names: absorbed dose is measured in gray, while equivalent dose is measured in sievert. Their rates are numerically equal here because the weighting factor is one. The estimate is a physical energy-deposition calculation; the assumed mass density is essential, and the stated inverse-length coefficient must not be used as if it already had area-per-mass units.

A representative Feynman diagram annihilates $u\bar{d}$ into $W^{+\ast }$, producing ${\nu }_{\mu }$ and an off-shell ${\mu }^{+}$; the charged antilepton emits a photon and becomes the outgoing ${\mu }^{+}$. The top panel below shows this contribution. A complete gauge-invariant decay amplitude includes radiation from the charged hadronic side, together with structure-dependent terms. The physical separation into inner bremsstrahlung, structure-dependent radiation, and their interference is explicit in the standard radiative pion decay description.^[ref]

Here is an explicit finite model of phase space alone, with its assumptions exposed. Neglect the neutrino mass, treat the photon as massless, and replace the spin-summed invariant decay amplitudes by constants over the available Lorentz-invariant phase space. Write $M=m_{\pi }$, $m=m_{\mu }$, $r=m^2/M^2$, and use natural units. The three-body phase space with one massive daughter follows by factoring through an intermediate invariant squared mass $s$:The two-body comparison has ${\Phi }_2(M;m,0)=(1-r)/(8\pi )$. Since ${\Gamma }_n=\overline{|{\mathcal{M}}_n{|}^2}{\Phi }_n/(2M)$, assuming $\overline{|{\mathcal{M}}_3{|}^2}/\overline{|{\mathcal{M}}_2{|}^2}=\kappa e^2/M^2$ givesHere $\kappa$ is a dimensionless constant encapsulating the assumed matrix element ratio, and ${\alpha }_{\mathrm{em}}$ is the fine-structure constant. If the two-body mode dominates, this ratio approximates the radiative branching fraction. Near $m=M$, $1-r^2+2r\ln r=(1-r{)}^3/3+O((1-r{)}^4)$, displaying the extra phase space suppression.

This model is not a prediction of a physical uncut radiative branching fraction. Actual inner bremsstrahlung has a soft-photon singularity, so a count of events with a separately observed photon requires a positive photon energy cut. To see both the missing input and the mass dependence explicitly, the leading soft-photon result for $E_{\mathrm{cut}}<E_{\gamma }<E_{\mathrm{soft}}\ll E_{\max }$ isThe coefficient follows from the radiation factor $e[p_{\mu }\cdot ϵ/(p_{\mu }\cdot k)-p_{\pi }\cdot ϵ/(p_{\pi }\cdot k)]$: angular integration gives ${\beta }^{-1}\ln [(1+\beta )/(1-\beta )]-2$, where $\beta =(1-r)/(1+r)$, and the soft energy integral is $\int d{E}_{\gamma }/E_{\gamma }$. Thus a finite physical number cannot be deduced from the two masses and phase space alone. The infrared divergence cancels only when the corresponding unresolved real radiation and virtual radiative corrections are combined in an inclusive observable.

The following are the leading quantum electrodynamics diagrams. Solid arrows indicate fermion-number flow, and wavy lines indicate photons. The heavy source line represents a charged nucleus; its recoil allows four-momentum conservation.

For Compton scattering, $e^{-}\gamma \to e^{-}\gamma$, there are two diagrams: absorption of the incident photon followed along the electron line by emission of the outgoing photon, and the opposite ordering. These are the $s$- and $u$-channel electron propagators. Both scattering amplitudes are order $e^2$.

For bremsstrahlung, $e^{-}Z\to e^{-}Z\gamma$, a virtual photon transfers momentum between the electron and the source. The real photon is emitted before or after this interaction along the electron line. These two dominant external-field diagrams have scattering amplitudes of order $Z{e}^3$. A full finite-mass charged-target calculation also includes radiation from the target; it is suppressed in the heavy stationary-source approximation used here.

For nuclear-field pair production, $\gamma Z\to e^{-}{e}^{+}Z$, crossing the preceding diagrams gives two order-$Z{e}^3$ contributions, with the incoming real photon and the exchanged virtual photon in either order along the pair's fermion line. An isolated real photon cannot produce a massive pair in vacuum: its invariant squared mass is zero, whereas the pair has invariant squared mass at least $4m_e^2$. The source absorbs recoil momentum.