In the supplied signed convention, and omitting pairing, the four contributions per nucleon areThe volume contribution is the constant $-15.7\mathrm{MeV}$. The positive surface correction decreases with $A$. At fixed proton fraction $Z/A$, the positive Coulomb correction grows as $A^{2/3}$, while the asymmetry correction is constant. Along the beta-stable valley the fraction $Z/A$ itself falls, and the asymmetry correction consequently grows.

Because the formula depends on both $A$ and $Z$, no unique plot against $A$ exists until a path $Z(A)$ is chosen. The original sketch below uses the continuous minimum of the nuclear Coulomb-plus-asymmetry terms,ignoring the small neutron-proton-electron rest-mass difference in locating that path. This is a stated smooth approximation to the valley of stability, not an assertion that nuclei have noninteger $Z$.

The four individual terms and their sum appear in one plot. A typical well-bound nucleus has $E_{\mathrm{bind}}/A\sim -8\mathrm{MeV}$ in the paper's convention, equivalently conventional positive binding energy per nucleon $B/A\sim +8\mathrm{MeV}$. If a positive-binding-energy plot is preferred, every plotted contribution must have its sign reversed. The competition between the surface and Coulomb penalties gives the familiar maximum of positive binding per nucleon near the iron-group nuclei.

There are three free-particle rest-energy terms and five nuclear corrections. In the paper's natural units, their meanings are:

The sign convention is significant: the quantity labeled $E_{\mathrm{bind}}$ in the supplied mass formula is added to the free-particle masses and is normally negative for a bound nucleus. It is the negative of the conventional positive nuclear binding energy $B$:The coefficients must therefore retain their printed signs; using positive $B$ while still adding it to the masses would be inconsistent.

Use a high-energy proton beam on a target to produce secondary pions, then select and focus the negative ones. A magnetic horn and a charge-selecting dipole system remove positive pions, limiting wrong-sign ${\nu }_{\mu }$ contamination. Select a narrow range of magnetic rigidity; if additional species rejection is available, velocity or mass discrimination reduces negative kaon contamination at the same momentum.

The selected ${\pi }^{-}$ travel through an evacuated decay tunnel and undergo ${\pi }^{-}\to {\mu }^{-}{\bar{\nu }}_{\mu }$. Their forward Lorentz boost concentrates the antineutrinos within an angle of order $1/{\gamma }_{\pi }$. Choose a tunnel long enough for substantial pion decay but short compared with the boosted muon decay length. Since ${\tau }_{\mu }\simeq 2.2\mu \mathrm{s}$ is much longer than ${\tau }_{\pi }\simeq 26\mathrm{ns}$, this reduces secondary ${\mu }^{-}\to e^{-}{\bar{\nu }}_e{\nu }_{\mu }$ contamination.

Downstream magnets sweep the remaining charged particles into absorbers, and shielding stops residual hadrons and muons. The antineutrino beam continues straight through. Parent-particle collimation, charge selection, and a forward detector improve purity; neutral-kaon decays, remaining wrong-sign parents, and muon decays prevent perfect purity. Neutrinos themselves cannot be focused or charge-selected by these magnets. Neutrinos produced after stopped-muon decay or capture are much less forward and can be reduced by directional and energy selection.

The weak charged current creates a negative charged lepton in a left-chiral state. For a massless particle, that would have negative helicity; part (e) instead requires positive helicity. A nonzero mass supplies the necessary wrong-helicity component. This helicity suppression is extremely strong for the fast electron, but much weaker for the slow muon.

For a positive-helicity charged lepton with spinors normalized to unit probability, the left-chiral weight is $(1-{\beta }_{\ell })/2$. With the box-state normalization of part (c), this gives the useful consistent rate ratioIndeed $1-{\beta }_{\ell }=2m_{\ell }^2/(m_{\pi }^2+m_{\ell }^2)$ cancels the extra energy-dependent factor in that density ratio, leaving the standard pion electron-to-muon decay ratio in inverted form:Using the rounded masses in this paper gives $R\simeq 8185$, close to the stated order of $8000$. Precision predictions require more accurate masses and electromagnetic radiative corrections. The essential mechanism is the charged-lepton-mass dependence of the decay amplitude; simply counting available final momenta misses it.

For the charged lepton, $p_{\ell }={\beta }_{\ell }{\gamma }_{\ell }{m}_{\ell }$ and $E_{\ell }={\gamma }_{\ell }{m}_{\ell }$. Their ratio is therefore its speed in natural units. Using part (a),Numerically,The electron is almost ultrarelativistic, while the muon is relatively slow. In ordinary units these are approximately $0.999974c$ and $0.271c$.

A spin-zero pion has no preferred emission direction. Choose the antineutrino momentum along $+z$. A massless antineutrino has positive helicity, so its spin points along $+z$. The muon moves along $-z$, and total $J_z=0$ requires its spin to point along $-z$:

The reversed configuration is obtained by rotating the entire event and is equally allowed. The two spins are opposite, while each is aligned with its own particle's momentum; both outgoing helicities are positive. A negative-helicity muon with this fixed antineutrino direction would give a nonzero total spin projection and is forbidden for this two-body decay of a spin-zero parent. The required positive-helicity charged lepton is the suppressed component of the weak charged current.

The charged pion has valence content ${\pi }^{-}=d\bar{u}$. Its constituents annihilate into a virtual W boson, followed by the lepton pair:

The quark vertex carries the $u$--$d$ Cabibbo-Kobayashi-Maskawa matrix element, and the leptonic vertex is the weak charged current. The pion's nonperturbative spatial structure is encoded in the pion decay constant when evaluating the decay amplitude.

In the simple box-state counting used here, the final three-momentum lies in a spherical shell: $dn=C{p}^{2}dp$, where the common constant $C$ contains the normalization volume and angular factor. Momentum conservation fixes the other final particle's momentum. Hence the density of states isCombining parts (a) and (b) givesWith the paper's rounded masses,By Fermi's golden rule, equal squared matrix elements make the rate ratio equal to this density ratio. Thus the naive constant-matrix-element estimate is $R\simeq 0.287$, which favours the electronic decay by about $3.49$ to one.

This is the particular noncovariant state-counting convention in the question. A calculation with the Lorentz-invariant phase-space measure places factors $1/(2E)$ in the measure and correspondingly normalizes its decay amplitude differently. Only the complete product of phase space and squared matrix element is convention-independent; mixing the two conventions would give a wrong rate.

The variable initial energy $E_0$ replaces $m_{\pi }$ in the kinematic expression before differentiation. The charged-lepton mass remains fixed:Thus the required Jacobian determinant in this one-dimensional change of variables isEquivalently, $E_0=\sqrt{p^2+m_{\ell }^2}+p$ gives $d{E}_0/dp=1+p/E_{\ell }$, so $dp/d{E}_0=E_{\ell }/(E_{\ell }+p)=E_{\ell }/m_{\pi }$, the same result. It is dimensionless in natural units.

In the pion rest frame, write its four-momentum as $P=(m_{\pi },0)$ and the charged lepton's as $p_{\ell }=(E_{\ell },\mathbf{p})$. By four-momentum conservation, the massless antineutrino has $k=P-p_{\ell }$, with $k^2=0$. ThereforeIts energy is $E_{\nu }=m_{\pi }-E_{\ell }$, and for a massless antineutrino, $E_{\nu }=|\mathbf{k}|$. The final momenta are equal and opposite, soThe expression is positive precisely when the two-body decay is energetically allowed, $m_{\pi }>m_{\ell }$, neglecting the neutrino mass.

Ignoring the requested final-mass difference and initially taking equal hadronic matrix elements, the common $W^{-}\to d\bar{u}$ factor cancels. A decay width is proportional to the squared decay amplitude, soWith the supplied Cabibbo angle,The quoted measurement, about $500\pm 10$, is larger by a factor $500/19.75\simeq 25.3$. If that whole difference were attributed to a ratio of other amplitudes, it would require a factor $\sqrt{25.3}\simeq 5.03$.

This does not invalidate quark mixing. The supplied Cabibbo-Kobayashi-Maskawa matrix is only a schematic hierarchy in powers of $\sin {\theta }_C$, particularly for $V_{ub}$; its omitted coefficients matter greatly after squaring. Equal phase space also does not imply equal hadronic decay amplitudes: the two transitions have different hadronic form factors and strong-interaction dynamics, and loop contributions can interfere with the charmless tree amplitude. Thus the calculation correctly predicts enhancement of the charmed channel, but the supplied CKM hierarchy alone does not quantitatively predict the measured ratio. It would be unjustified to treat $19.75$ as a precision prediction with only the quoted experimental error.

The antiquark $\bar{d}$ in the specified ${\overline{B}}_d^0=b\bar{d}$ is a spectator. For the charmed final state the leading external-W boson tree iswith $c\bar{d}=D^{+}$ and $d\bar{u}={\pi }^{-}$. For the charmless final state replace the active transition by $b\to u+W^{-}$, giving $u\bar{d}={\pi }^{+}$ and the same $d\bar{u}={\pi }^{-}$. Their weak amplitudes have factors $V_{cb}{V}_{ud}^{\ast }$ and $V_{ub}{V}_{ud}^{\ast }$, respectively, up to an irrelevant overall conjugation convention.

The diagrams determine the leading weak flavour factors. Strong-interaction matrix elements and, particularly for the charmless channel, additional loop diagrams can also contribute to physical branching fractions.

For both modes, a leading weak quark transition in $K^{+}=u\bar{s}$ is $\bar{s}\to \bar{u}+W^{+}$, followed by $W^{+}\to u\bar{d}$, with the original $u$ acting as a spectator quark. This produces the four quarks needed for ${\pi }^{+}{\pi }^0$, using the $u\bar{u}$ component of ${\pi }^0$. Producing three pions additionally requires a $d\bar{d}$ pair from strong hadronization. A representative quark-line diagram draws it as gluon emission and splitting.

The three-pion assignment is $u\bar{d}$, $u\bar{d}$, and $d\bar{u}$, matching ${\pi }^{+}{\pi }^{+}{\pi }^{-}$. The extra soft pair creation is a schematic representation of hadronization, not a controlled perturbative prediction of the relative decay rates. Both processes are first order in the effective weak interaction.

There is an explicit error or hypothetical assumption to address: the original PDF assigns $0^{+}$ to all these mesons. Taken literally, the two-pion final state has $L=0$, because all particles have zero spin and the initial $J$ is zero. Its parity is $(+1{)}^2(-1{)}^0=+1$. For three spinless particles, use two Jacobi orbital angular momentum quantum numbers $l,L$. Coupling to total zero requires $l=L$, so the orbital parity $(-1{)}^{l+L}$ is $+1$ and the three-particle parity is also $+1$. Thus under the printed $0^{+}$ assignment, both channels conserve parity.

The named physical pions and kaons are instead pseudoscalar mesons with $J^P=0^{-}$. With those physical assignments,ThereforeThese two opposite-parity final states from the same physical kaon are a classic illustration of parity violation in a weak decay. The physical conclusion requires correcting the printed parity assignment explicitly.

In the Wu experiment, Chien-Shiung Wu and collaborators studied beta-minus decay of polarized cobalt-60,A thin cobalt-60 layer on a cerium magnesium nitrate crystal was cooled by adiabatic demagnetization to a few millikelvin. A polarizing magnetic field produced nuclear polarization. An anthracene scintillation counter and photomultiplier detected the emitted electrons; angular gamma ray observations from the daughter de-excitation monitored whether the nuclei remained polarized.^[ref]

The electrons were emitted preferentially opposite the parent nuclear spin. Reversing the nuclear polarization reversed the asymmetry, while warming the sample removed both polarization and the effect. Those controls distinguished a spin-correlated asymmetry from a fixed detector bias.

The relevant angular correlation is a polarized beta-decay angular distribution of the formUnder parity, ${\mathbf{p}}_e\to -{\mathbf{p}}_e$ but $\mathbf{J}\to \mathbf{J}$, so the spin-momentum correlation changes sign. Parity conservation would require equal rates for these two directions relative to the same spin. The observed nonzero coefficient therefore demonstrated parity violation in beta decay. An unpolarized source would average this correlation to zero and could not provide the same test.

Parity is full spatial reflection, $\mathbf{x}\mapsto -\mathbf{x}$, represented by a parity operator $\mathcal{P}$. A polar vector, such as momentum, reverses, whereas an axial vector, such as spin, does not. A state of definite parity has $\mathcal{P}|\psi ⟩={\eta }_P|\psi ⟩$, with ${\eta }_P=\pm 1$ in the usual convention. A multiparticle state's parity includes both the product of its intrinsic parities and its orbital spatial factor, for example $(-1{)}^L$ for relative orbital angular momentum $L$.

An interaction conserves parity when its dynamics are invariant under this transformation, equivalently $[H,\mathcal{P}]=0$ for a time-independent Hamiltonian. Parity violation means that a process and its fully reflected counterpart need not have the same probability. It is more than a left-right asymmetry in an unreflected laboratory apparatus.

The electromagnetic and strong sectors of the Standard Model conserve parity to the accuracy relevant here, taking the strong-sector Yang-Mills theta term to vanish. The weak charged current selects left-chiral fermions and right-chiral antifermions, as defined by fermion chirality; the reflected right-chiral charged current is not equally present. The neutral weak current also generally has different left- and right-chiral couplings. Consequently the weak sector violates parity; this does not mean that every individual weak decay must have opposite initial and final parity. For a massive particle, helicity and fermion chirality should be distinguished.

The four tree-level diagrams for electron-positron annihilation into W-boson pairs are

The first two diagrams contain the electromagnetic interaction and weak interaction neutral electron currents and the gauge boson self-interaction vertices $\gamma W^{+}{W}^{-}$ and $Z{W}^{+}{W}^{-}$. The third has two weak charged current vertices joined by an electron neutrino propagator. In the fourth, the incoming electron-Positron pair couples to the Higgs boson, which then couples to $W^{+}{W}^{-}$.

The small factor is at the initial Higgs-electron vertex. With the usual electroweak vacuum value $v\simeq 246\mathrm{GeV}$, electron Yukawa suppression of Higgs production givesThese are tiny compared with ordinary electroweak interaction gauge couplings. The decay amplitude for the Higgs boson diagram carries this small coefficient, and its separate squared contribution carries its square. Thus the Higgs diagram is negligible because the electron Yukawa coupling is proportional to the electron mass. It vanishes in the massless-electron approximation, even though the Higgs boson-W boson coupling itself is not small. The remaining three gauge diagrams must be added coherently, including their interference.

For gamma decay, the nuclear gamma-ray multipole selection rules require the radiation's angular momentum $L$ to satisfy the triangle inequalityHere $J_i=J_f=5/2$, so $0\le L\le 5$. A real emitted photon has no monopole radiation mode, leaving $L=1,2,3,4,5$.

The two nuclear quantum states have the same parity. An electric multipole has parity factor $(-1{)}^L$, while a magnetic multipole has factor $(-1{)}^{L+1}$. Thus electric radiation requires even $L$ and magnetic radiation odd $L$:In particular, an $E0$ nuclear transition cannot emit the single real photon requested here; processes such as internal conversion are different decay channels.

For usual nuclear gamma ray energies, $kR\ll 1$, where $k=E_{\gamma }/(\hslash c)$ and $R$ is the nuclear size. Higher multipole radiation orders bring extra powers of this small parameter, so $M1$ and $E2$ are the likely important contributions, with $M3$, $E4$ and $M5$ generally much smaller. Often $M1$ is dominant, but an allowed $M1$ matrix element can be hindered and an $E2$ transition can be collectively enhanced. The quantum numbers alone do not fix a numerical relative contribution or an unconditional ordering of the $M1$ and $E2$ rates.

Treat the quoted maximum as the resonant Z boson contribution described by the Breit-Wigner resonance cross section. The electron and Positron each have spin $1/2$, while the Z boson has spin one, soNeglect the electron mass at this centre-of-mass energy. Each incident momentum magnitude is then $p_i=E_0/2$ in natural units. Write $B_e={\Gamma }_{ee}/\Gamma$. At the resonance peak the denominator is ${\Gamma }^2/4$, and the same partial decay width appears for production and decay:The peak resonance cross section and branching fractions relation contains $B_e^2$, because the electronic channel occurs at both ends of the resonant process.

For an area in ordinary units, restore $(\hslash c{)}^2$. The original PDF prints $\hslash c=197.5\mathrm{MeV}$, which is dimensionally incomplete: the intended standard conversion is $197.5\mathrm{MeV}\mathrm{fm}=0.1975\mathrm{GeV}\mathrm{fm}$. Also,ConsequentlyThis estimate uses the supplied resonance model. The full physical elastic electron-Positron cross section also contains nonresonant photon exchange and interference; an experimental use of the formula must isolate the resonant contribution.

The information in the original question is insufficient to determine a unique spin-induced change of charge density. It specifies ideal dielectric filaments with a prescribed excess linear charge density, but not the charge carrier species, their spin state, charge carrier mobility, binding potential or relaxation mechanism. A classical line charge has no spin variable at all. The following conditional conclusions are supported by the fields already calculated.

For a carrier moving with its filament, the local magnetic field in its instantaneous rest frame is, for the transverse fields here,where $\mathbf{u}$ is that carrier's laboratory velocity. Exclude its filament's singular self-field and first retain only the prescribed straight convective motion.

In Case 1 there is a common rest frame in which both filaments and the excess charges are stationary and the interfilament magnetic field vanishes. Equivalently, in the laboratory ${\mathbf{B}}_A=\mathbf{v}\times {\mathbf{E}}_A/c^2$, so B's co-moving charges have ${\mathbf{B}}_{A,\mathrm{rest}}=0$. Thus uniform common translation alone creates no extra mutual rest-field spin splitting. This does not exclude intrinsic spin-orbit coupling of electrons moving within their atomic binding potentials; that material physics was never specified.

In Case 2 there is relative motion and no frame in which both source distributions are stationary. Put $E_{\ast }=\eta /(2\pi {ϵ}_{0}d)$ as in part (c). At the respective filament axes, the external fields experienced in each carrier distribution's own rest frame areFor the first expression B is at rest in $O$; for the second A is at rest in $O^{\prime }$ and the field comes from B moving at $-v\hat{\mathbf{x}}$. For a carrier with magnetic dipole moment ${\mathbf{\mu }}_s$, the interaction $H_s=-{\mathbf{\mu }}_s\cdot {\mathbf{B}}_{\mathrm{rest}}$ distinguishes opposite transverse spin orientations. A field gradient could exert a spin-dependent force $\nabla ({\mathbf{\mu }}_s\cdot {\mathbf{B}}_{\mathrm{rest}})$ on mobile carriers; spin relaxation could instead produce a spin polarization. These are possible responses, not consequences of charge density alone. In B's frame the displayed interaction is directly a magnetic-moment coupling to the moving source's field; treating the moving carrier's electrostatic orbital term in another frame also requires consistent binding and acceleration dynamics, including any applicable Thomas precession.

For example, if $\eta >0$ and mobile electron spins can relax, the lower magnetic-energy orientation has the electron's magnetic moment toward $+\hat{\mathbf{y}}$ and its spin toward $-\hat{\mathbf{y}}$. Oppositely oriented moments would experience opposite gradient forces. The sign changes if the source charge changes sign. None of these statements fixes a net charge redistribution without a confinement and response model, and the stipulated zero transverse size supplies no resolved “sides” on which charge could accumulate.

Therefore Case 1 has no new mutual spin effect from common bulk motion alone; Case 2 permits motion-induced spin-dependent interactions, but neither case determines an additional charge-density profile from the supplied data. In particular, the factor $\gamma$ relating proper and laboratory linear charge densities in part (b) follows from length contraction and is not an effect of spin-orbit coupling. If the charges remain pinned in an ideal dielectric, their imposed density remains the prescribed density in either case.

Spin-orbit coupling is a relativistic interaction between a particle's spin and its orbital angular momentum or, more generally, its motion through an electric field. A moving electron experiences a transformed magnetic field, which couples to its magnetic dipole moment. For an electron bound in a central electrostatic potential energy $U(r)$, the leading interaction, including the Thomas precession correction, isHere $\mathbf{L}$ is orbital angular momentum and $\mathbf{S}$ is spin. The factor of one-half relative to the simplest rest-field argument comes from Thomas precession of the accelerated electron frame.^[ref]

For example, in the hydrogen atom, the $2p$ states with total angular momentum $j=1/2$ and $j=3/2$ have different $\mathbf{L}\cdot \mathbf{S}$ eigenvalues. Their splitting is part of the atomic fine structure. The interaction redistributes and splits spin-dependent energies; it is not the same physical effect as the spin-independent convective magnetic force between charged filaments.