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A Ragone plot has specific energy $e_s=E/m$ on its horizontal logarithmic axis and specific power $p_s=P/m$ on its vertical logarithmic axis, normally in $\mathrm{Wh}\mathrm{k}{\mathrm{g}}^{-1}$ and $\mathrm{W}\mathrm{k}{\mathrm{g}}^{-1}$. The comparison must use the same mass convention: complete devices cannot fairly be compared with electrode-active-material values. At approximately constant delivered power, the discharge time isThus constant-time lines slope upward on the log–log plot. Ordinary capacitors lie toward high power and low energy; batteries toward higher energy and lower power; supercapacitors bridge the two. Technology and operating limits produce overlapping regions, not universal sharp boxes.

A physical bound-state wave function must be single-valued, finite, square-integrable and normalizable. Across a finite step in a regular potential, both $\psi$ and its first derivative are continuous when the particle mass is the same on either side; integrate the Schrödinger equation across an infinitesimal interval to obtain the derivative condition. The second derivative need not be continuous.

An infinite hard wall is an exception to derivative matching: the interior wave function vanishes at the boundary, but its derivative need not equal the exterior zero derivative. A delta-function potential also permits a prescribed derivative jump, and an effective-mass interface requires the appropriate mass-weighted matching condition. Thus ordinary finite-step conditions should not be asserted without qualification for every singular potential.

For longitudinal motion in transverse subband $i$,For a completely free particle take $E_i=0$ and use the full wave vector. In a semiconductor the parabolic-band mass is the appropriate effective mass rather than necessarily the bare electron mass.

The cobalt-60 mass is approximately $5.73\times 10^{-8}\mathrm{kg}$. The nuclear decay constant follows from the half-life, using a $365.25$-day year:Since radioactive activity is $A=\lambda N$, the parent population at certification isConvert the supplied atomic mass into molar mass, $M=0.05993382\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$, and use the Avogadro constant:This is about $57.3\mu \mathrm{g}$ of the specified isotope. It is not the total mass of the pellets or capsule, which can also contain other material.

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.

In the stated model, the limiting thicknesses are about $0.163\mathrm{m}$ of lead or $0.780\mathrm{m}$ of concrete. Use the unshielded equivalent dose rate from part (c), ${\dot{H}}_0=556.815\mu \mathrm{Sv}{\mathrm{h}}^{-1}$. Let $x=\mu t$, where $\mu$ is the supplied total linear attenuation coefficient. The stipulated radiation buildup factor gives the requirementThe left side decreases for $x>0$ because its derivative is $-x{e}^{-x}$. Solving the equality gives a unique threshold,Thus, with the source-to-observation distance kept at one metre,For example, rounding upward to $0.17\mathrm{m}$ of lead or $0.80\mathrm{m}$ of concrete satisfies the stipulated inequality. Equality at the threshold is not sufficient for the strictly smaller target.

A compact surrounding shield favours lead; concrete trades much greater thickness for a common structural material. The densities permit a useful planar comparison: the threshold areal masses $\rho t$ are about $1.84\times 10^3$ and $1.87\times 10^3\mathrm{kg}{\mathrm{m}}^{-2}$, respectively. The thinner shield is not proportionately lighter per equal projected area. Total enclosure mass also depends on geometry.

The idealized outline is continuous coverage around the source with the required material path length, including the closure; a direct unshielded line through a gap would invalidate this attenuation estimate. The calculation assumes the quoted observation distance, the fixed effective photon coefficients and the prescribed radiation buildup factor. It is not based on the more optimistic uncollided exponential attenuation alone.

Two mechanisms are a tau lepton pair with two leptonic decays, and a W boson pair with two leptonic decays. With equal beam energies, the laboratory is the centre-of-mass frame, and four-momentum conservation gives $E_{\mathrm{miss}}=\sqrt{s}-E_{e^{-}}-E_{{\mu }^{+}}$ in natural units.

For the first mechanism,There are four escaping neutrinos. Each tau lepton decay proceeds through a virtual W boson. The fixed-charge final state has branching fraction $B_{\mu }{B}_e$; the factor two for the inclusive $e^{\pm }{\mu }^{\mp }$ sample does not apply when the charges are specified.

For the second mechanism,There are two escaping neutrinos. The leading Feynman diagrams comprise both $s$-channel photon/Z boson exchange and $t$-channel electron neutrino exchange. Their scattering amplitudes interfere; the complete set is needed for a consistent high-energy prediction. In particular, the neutrino-exchange diagram alone does not define an independently observable production rate.

A tau lepton pair has enhanced production at the Z boson particle resonance, $\sqrt{s}\simeq 91\mathrm{GeV}$. The four-neutrino missing energy has a broad continuous distribution. To derive a useful schematic, neglect daughter masses and parent spin correlations. In an unpolarized tau lepton rest frame, the Michel spectrum for $x=2E_{\ell }^{\ast }/m_{\tau }$ isA Lorentz boost gives $E_{\ell }/E_{\tau }=x(1+{\beta }_{\tau }u)/2$, with $u=\cos {\theta }^{\ast }$ uniform on $[-1,1]$ in this approximation. For ${\beta }_{\tau }\simeq 1$, the charged-daughter probability density function in $z=E_{\ell }/E_{\tau }$ isFor two independent decays, $y=E_{\mathrm{miss}}/\sqrt{s}=1-(z_1+z_2)/2$ has the convolutionand $⟨y⟩=0.65$. It vanishes at both endpoints and peaks on the larger-missing-energy side. This is a shape illustration, not the exact polarized Z boson prediction; spin correlations, acceptance, and radiative effects alter it.

The W boson process turns on near $\sqrt{s}=2m_W\simeq 161\mathrm{GeV}$. At exact on-shell production threshold, each neutrino has energy $m_W/2$, soThe cross section vanishes at the ideal two-body threshold, so this narrow limiting spectrum must not be called the maximum-rate spectrum. The large W boson-pair rate occurs above threshold, around $200\mathrm{GeV}$. There ${\beta }_W=\sqrt{1-4m_W^2/s}$ and each neutrino hasThusAn isotropic independent-decay schematic is triangular, centered at $E_{\mathrm{miss}}/\sqrt{s}=1/2$, with endpoints approximately $0.20$ and $0.80$ at $200\mathrm{GeV}$. Actual W boson particle polarizations and their correlations change this shape; their V-A interaction is not isotropic for a fixed helicity. The finite decay width, beam spread, and radiation also smear the edges.

The phrase “maximum rate” also needs a production regime: a Z boson peak is a resonant maximum, whereas pure photon-mediated tau lepton production has its continuum maximum near $4.2\mathrm{GeV}$. Indeed, ${\sigma }_{\gamma }\propto {\beta }_{\tau }(3-{\beta }_{\tau }^2)/s$ has $s=(1+\sqrt{21})m_{\tau }^2$. The displayed high-boost curve describes the resonant regime; at the continuum maximum use ${\beta }_{\tau }\simeq 0.532$ in the boost formula. An observed event-rate maximum additionally depends on collider luminosity and detection efficiency.

The ideal two-shower event fraction is $B_e^2$, approximately $4\%$ in simple channel counting or $3.18\%$ using the measured tau lepton electronic branching fraction. The selected particle decay chain isThe electron and Positron each make a contained electromagnetic shower; muons penetrate, and generic hadrons do not deposit all their energy as a contained electromagnetic shower.

To estimate $B_e$, use lepton universality and neglect final masses and Quantum chromodynamics corrections. The electronic and muonic decay widths each carry weight one. The accessible inclusive hadronic channels have weights $3|V_{ud}{|}^2$ and $3|V_{us}{|}^2$, where three is the color multiplicity in a decay width. The charm quark is inaccessible in physical tau lepton hadronic decays. With $|V_{ud}{|}^2+|V_{us}{|}^2\simeq 1$, the total weight is $1+1+3=5$, givingThe 2021 measured value $B_e=0.1782$ instead gives $B_e^2=0.0317552$.^[ref] The difference reflects effects omitted by the simple free-channel estimate, especially the hadronic contribution to the total decay width.

There are two tau leptons per produced pair and two per selected pair, so the fraction of all produced tau leptons belonging to these two-shower events is also $B_e^2$. The unrestricted fraction of individual tau leptons decaying electronically is $B_e$; these are different selections. A real high-energy shower cut gives an additional acceptance factor: $N_{\mathrm{selected}}/N_{\mathrm{pairs}}=B_e^2{\epsilon }_{ee}$, which cannot be computed without the cut and detector response. No factor of two is needed for the unique $e^{+}{e}^{-}$ assignment.

For a purely electromagnetic interaction continuum with five active quark flavors, the usual estimate is $N_{e^{+}{e}^{-}}/N_{\mathrm{had}}=3/11$. The unspecified particle resonance and the word “total” require qualifications before this can be treated as a unique physical prediction.

In natural units, neglecting final masses, single-photon annihilation givesHere $\alpha$ is the fine-structure constant, $Q_f$ the quark electric charge in elementary-charge units, and the factor three counts orthogonal colour final states. Consequently the inverse hadronic R ratio isWith $u,d,s,c,b$ accessible, the denominator is $3[2(2/3{)}^2+3(1/3{)}^2]=11/3$. The corresponding values for $u,d,s$ and $u,d,s,c$ are $1/2$ and $3/10$. Inclusive Quantum chromodynamics corrections increase the hadronic denominator, approximately by $1+{\alpha }_s/\pi$ away from thresholds and particle resonances. These are annihilation ratios before selection and detection efficiency corrections, not ratios of uncorrected detector counts.

A future muon-collider Higgs resonance is a plausible reading of part (c), since its narrow decay width makes a detuning of a few $\mathrm{MeV}$ relevant. Even then, two further assumptions are necessary. First, being a few $\mathrm{MeV}$ below the Higgs boson peak does not eliminate it: for detuning $\Delta =\sqrt{s}-m_H$, the ideal narrow Breit-Wigner resonance cross section retains the relative factorFor ${\Gamma }_H=4\mathrm{MeV}$ and $\Delta =-4\mathrm{MeV}$ this is $1/5$, not zero. Resonant hadronic production can still change the requested denominator substantially.

Second, at $\sqrt{s}\simeq 125\mathrm{GeV}$ the Z boson continuum is not negligible. A more complete annihilation estimate including both tree-level Feynman diagrams usesHere $s_W^2$ is the squared sine of the weak mixing angle, $T_3^f$ is weak isospin, and $N_c^f$ is one for a lepton and three for a quark. Thus the continuum ratio is $F_e/[3(2F_u+3F_d)]$, rather than necessarily $3/11$. For example, $s_W^2=0.231$, $m_Z=91.1876\mathrm{GeV}$ and ${\Gamma }_Z=2.4952\mathrm{GeV}$ give approximately $0.102$ at $125\mathrm{GeV}$ before Quantum chromodynamics corrections.

If the unspecified particle resonance instead means the Z boson, detuning by a few $\mathrm{MeV}$ is tiny compared with its $2.5\mathrm{GeV}$ decay width. The resonant ratio remains approximately $\Gamma (Z\to e^{+}{e}^{-})/\Gamma (Z\to \mathrm{hadrons})=1/20.767\simeq 0.0482$.^[ref] This is a concrete counterexample to treating any slight downward detuning as permission to discard the particle resonance. The source supplies neither its identity nor the detuning and selection needed for a unique all-process ratio. The boxed continuum formula is the charge-counting answer under its stated approximation; a literal total must also include remaining resonant and other accepted hadronic channels.

In ordinary perturbative Standard Model reactions all three interactions conserve electric charge, four-momentum, total angular momentum, baryon number, and total lepton number. The last two are accidental conservation laws of these perturbative processes, not exact nonperturbative statements: electroweak sphalerons violate $B+L$ while preserving baryon-minus-lepton number. Total color charge is respected throughout.

The strong interaction and electromagnetic interaction also conserve each net quark flavor number, including strangeness, charm and bottom flavour. The weak charged current can change them through the Cabibbo-Kobayashi-Maskawa matrix. Isospin is an approximate strong interaction symmetry in the small light-quark mass difference limit; the different electric charges of $u$ and $d$ break it in the electromagnetic interaction. Approximate flavor symmetry is distinct from exact net-flavour-number conservation by those interactions.

Parity conservation and charge conjugation symmetry hold for the electromagnetic interaction and, assuming a negligible strong CP parameter, the strong interaction. The weak interaction violates both, and also violates CP symmetry in suitable processes. CPT symmetry holds under the usual local Lorentz invariant quantum field theory assumptions. Individual lepton family numbers are conserved in the conventional massless-neutrino approximation but not in neutrino oscillations; an assertion of exact family conservation in all observed reactions would therefore be false.

The magnetic rigidity gives the momentum, the bending direction gives the sign of electric charge, and the calorimetric energy then gives the mass. For planar tracks perpendicular to a uniform magnetic field, the Lorentz force and circular motion giveTrace each track away from its production or decay vertex. With the magnetic field into the page, a positive track bends counterclockwise and a negative track bends clockwise. For example, a positive particle initially travelling left accelerates downwards. The known incoming negative pion fixes the sign convention experimentally. A radius alone does not distinguish electric charge signs.

For a calibrated total energy $E$, the relativistic energy-momentum relation givesIf the calorimeter instead supplies only kinetic energy $T$, solve $(T+m{c}^2{)}^2=p^2{c}^2+m^2{c}^4$ to obtainCompare with candidate pion, kaon, and proton masses, together with the electric charge sign and shower behavior. An electron makes an electromagnetic shower; a muon is penetrating; a hadron generally requires a hadronic calorimeter and its own calibration. If either energy escapes or the response is uncalibrated, simply substituting a deposited signal for $E$ need not identify the particle correctly. The absent figure prevents attaching identities to individual drawn tracks, but the requested identification procedure does not require its geometry.

The conversion from a measured beam peak to a particle resonance mass is fixed, but the missing graph supplies neither the peak position nor its width. Let $T_R$ be the incoming pion kinetic energy at the peak and $\Delta T$ the full width at half maximum after allowing for instrumental resolution. With a stationary proton and natural units,Thus the peak invariant mass isThe width must also be transformed to the centre-of-mass frame. To first order about the peak,For a broad feature, transform the two half-height positions separately: if they are $T_{-}$ and $T_{+}$, then the corresponding width in invariant mass is $\sqrt{(m_p+m_{\pi }{)}^2+2m_p{T}_{+}}-\sqrt{(m_p+m_{\pi }{)}^2+2m_p{T}_{-}}$. An energy-dependent numerator or width can also displace the observed maximum from the resonance pole parameter.

The low-energy ${\pi }^{+}p$ channel is normally dominated by ${\Delta }^{++}(1232)$, a Delta baryon with $Q=2$, $B=1$ and $I=3/2$. As an external reference example, not a reading of the absent plot, $M=1232\mathrm{MeV}$ and $\Gamma =117\mathrm{MeV}$ give $T_R\simeq 190\mathrm{MeV}$, $\Delta T\simeq 154\mathrm{MeV}$ and $\tau \simeq 5.63\times 10^{-24}\mathrm{s}$.^[ref] The actual graphical estimates remain unavailable. In particular, using the laboratory width directly in $\tau =\hslash /\Delta T$ misses the factor $m_p/M$.

Put $M_f=m_p+m_{{\pi }^{+}}+m_{\Lambda }+m_{K^0}$. In natural units, a stationary target proton and a beam proton of total energy $E_b$ have the Mandelstam invariantAt fixed-target production threshold the final particles have zero relative momentum in their centre-of-mass frame, so $s=M_f^2$. Restoring $c$, the total beam energy and its kinetic energy areUsing rounded masses in $\mathrm{MeV}/c^2$, $(938.27,139.57,1115.68,497.61)$, gives $E_{b,\mathrm{thr}}\simeq 2.921\mathrm{GeV}$ and $T_{b,\mathrm{thr}}\simeq 1.983\mathrm{GeV}$.

One possible strong interaction quark-line flavour rearrangement isIt needs newly created $d\bar{d}$ and $s\bar{s}$ pairs, with no net change of any quark flavor. For example, keep the first $uud$ as the outgoing proton; allocate the second incoming $u,d$ and the new $s$ to ${\Lambda }^0$, the remaining incoming $u$ and new $\bar{d}$ to ${\pi }^{+}$, and the new $d,\bar{s}$ to $K^0$. Two gluon splittings illustrate this flow. This is a possible hadronization topology, not a unique perturbative description of the proton collision.

After several short mean lifetimes, the surviving coherent beam is approximately $|K_L⟩=(|K^0⟩-|{\bar{K}}^0⟩)/\sqrt{2}$. In ordinary matter the forward scattering amplitudes of $K^0$ and ${\bar{K}}^0$ differ. For example, the $s$ in a neutral antikaon can enter a strange baryon through strong absorption; the opposite strangeness assignment has different available channels. If their forward transmission amplitudes through a thin block are $t_0$ and ${\bar{t}}_0$, respectively, thenThus the transmitted forward beam contains a regenerated short-lived component, with prompt two-pion decays after the block. Differential absorption changes the intensities as well as the relative phase. Scattered particles outside the forward beam can provide an incoherent component. This neutral kaon regeneration does not require CP violation: the material itself distinguishes matter and antimatter.

For a narrow unstable state, the effective Hamiltonian has the complex eigenvalue $m_j-i{\Gamma }_j/2$. With natural units and ${\Gamma }_j=1/{\tau }_j$, its Schrödinger equation givesThe first factor is the ordinary energy phase; squaring the second gives the exponential decay of the survival probability, $e^{-t/{\tau }_j}$. This assumes the approximately constant decay rate regime; exact very early and very late behavior need not be exponential.

Writing the surviving state as $A_S(t)|K_S⟩+A_L(t)|K_L⟩$, invert the two kaon CP eigenstates definitions:Here $t$ is the proper time. For a relativistic beam of fixed Lorentz factor $\gamma$, substitute $t=t_{\mathrm{lab}}/\gamma$; in ordinary units the mass phase is $m_j{c}^{2}t/\hslash$.

Normalize the incident intensity to one. Since $|K^0⟩=(|K_S⟩+|K_L⟩)/\sqrt{2}$, the initial probability amplitudes are $A_S(0)=A_L(0)=1/\sqrt{2}$. Define $g_j=e^{-i{m}_{j}t}{e}^{-t/(2{\tau }_j)}$. The flavor probability amplitudes are $(g_S+g_L)/2$ and $(g_S-g_L)/2$. Their absolute squares arewhere $\Delta m=m_L-m_S$ in natural units. Thus $A=B=1/4$ for both, $\varphi =\Delta mt$, andMultiplying all three coefficients by the incident intensity gives any other normalization. Crucially, $C$ is time dependent. The printed template cannot hold with all of $A,B,C$ constant. For example, a nonzero constant $C$ would leave an undamped oscillation at arbitrarily late times, even after both unstable components have decayed.

In seconds, $\varphi =\Delta m{c}^{2}t/\hslash$. With $x=t/{\tau }_S$, the supplied mass difference and mean lifetimes giveInitially the intensities are one and zero. Their sum is $\frac{1}{2}(e^{-x}+e^{-r_{\tau }x})$, so these are unconditioned survival intensities, not probabilities conditioned on survival. The curves first meet at $x=\pi /(2\times 0.474975)\simeq 3.307$. Thereafter the neutral antikaon component briefly exceeds the $K^0$ component, before the interference dies away and both approach the surviving long-lived intensity $e^{-r_{\tau }x}/4$.

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.

For a three-body leptonic decay through the Fermi interaction, neglecting daughter masses and radiative corrections, lepton universality givesin natural units. Therefore scaling the muon decay rate by $m_{\tau }^5/m_{\mu }^5$ predicts only one leptonic channel, not the total tau lepton decay rate.

The tau lepton has two allowed charged-lepton channels, $e^{-}{\bar{\nu }}_e{\nu }_{\tau }$ and ${\mu }^{-}{\bar{\nu }}_{\mu }{\nu }_{\tau }$, as well as hadronic channels from $W^{-\ast }\to d\bar{u},s\bar{u}$. Each allowed quark channel has three color charge choices and its Cabibbo-Kobayashi-Maskawa matrix factor. Charm production is kinematically inaccessible in a physical hadronic final state at this mass. Treating the inclusive hadronic channels by a simple quark-hadron duality estimate,Thus, using $m_{\mu }\simeq 105.7\mathrm{MeV}/c^2$ and $m_{\tau }\simeq 1777\mathrm{MeV}/c^2$,Finite daughter masses, hadronic decay thresholds and resonances, Quantum chromodynamics radiative corrections, and electroweak radiative corrections refine this estimate. The threefold color multiplicity in a decay width and extra decay channels are the main correction to naive fifth-power scaling.

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.

This is the charge-and-parity-conjugate reaction of part (i). There are $t$-channel Z boson and crossed W boson contributions, with all fermion arrows reversed relative to part (i). For matched unpolarized kinematics and negligible CP violation,Combining all four results in the regime specified in part (i),In particular, conjugating both initial particles restores the part-(i) rate; conjugating the neutrino alone does not. Initial spin polarization and energies near a weak-boson mass require a separate comparison.

Let $M_P$ and $M_D$ denote neutral atomic rest masses for the appropriate parent and daughter, and initially neglect electronic binding energy corrections and daughter excitation. The Q values areFor beta-minus decay, the daughter nucleus has one extra positive charge. Its neutral atom contains one more electron than the neutral parent, cancelling the emitted electron's rest mass in the atomic accounting. For beta-plus decay, the neutral parent contains one more electron than the neutral daughter; together with the emitted Positron, this gives the $2m_e$ subtraction. Thus beta-plus decay requires an atomic mass difference exceeding $2m_e{c}^2$, whereas electron capture need not.

For electron capture, the captured bound electron is already included in $M_P$. Capture from a shell leaving an atomic hole of excitation energy $E_{\mathrm{hole}}$, and a nuclear excited state of energy $E_x$, gives approximatelyThe hole's energy later appears as atomic radiation or Auger electrons. Subtract $E_x$ for either beta decay as well. Exact endpoints also depend on daughter ionization and small atomic binding energy changes; the boxed expressions are the conventional leading atomic-mass formulas.