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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.

A kaon is a pseudoscalar meson, with intrinsic parity $-1$. At rest, $P|\pm ⟩=-|\pm ⟩$. Combining this with the specified CP eigenvalues givesEquivalently, in the exchange-positive CP convention for neutral kaon mixing fixed by the question, $C|K^0⟩=-|{\bar{K}}^0⟩$ and $C|{\bar{K}}^0⟩=-|K^0⟩$. This sign is a convention for the flavor states; the specified CP eigenvalues and physical predictions must be kept consistent with it.

For two spin-zero pions, conservation of total angular momentum fixes the relative orbital angular momentum to zero. The resulting ${\pi }^{+}{\pi }^{-}$ or ${\pi }^0{\pi }^0$ state has CP $+1$. The lowest partial waves of ${\pi }^{+}{\pi }^{-}{\pi }^0$ and $3{\pi }^0$ have CP $-1$. Thus the dominant nonleptonic assignments in the CP symmetry approximation areThe two-pion mode has substantially more available kinetic energy and larger Lorentz-invariant phase space; the three-pion mode has an additional final particle and lies closer to its decay threshold. This helps explain why the CP-odd state has the longer mean lifetime. The actual mean lifetimes also depend on the decay amplitudes and on semileptonic channels, so counting final particles alone is not a quantitative prediction.

The middle panel of the figure in part (a) shows $\bar{s}\to \bar{u}{W}^{+}$, $W^{+}\to u\bar{d}$, with the $d$ as a spectator quark. The outgoing $d\bar{u}$ and $u\bar{d}$ form ${\pi }^{-}$ and ${\pi }^{+}$. In the bottom panel, an additional gluon produces $d\bar{d}$, supplying the neutral pion component for ${\pi }^{+}{\pi }^{-}{\pi }^0$. The charge-conjugate diagrams supply the other component of each of the kaon CP eigenstates. The drawn gluon splitting represents nonperturbative hadronization, rather than a quantitatively reliable perturbative prediction at the kaon scale.

These multiplicities describe the dominant low-orbital angular momentum modes, not an absolute rule for every three-pion configuration. In ${\pi }^{+}{\pi }^{-}{\pi }^0$, odd orbital angular momentum values of the charged pair and of its motion relative to ${\pi }^0$ can couple to total zero and give CP $+1$, but are strongly suppressed near decay threshold. Physical $K_L\to 2\pi$ also occurs through CP violation.

The observed dominant two-body transition is $\boxed{{\Sigma }^0\to {\Lambda }^0\gamma }$. Both baryons have valence quarks $uds$ and the same strangeness; an electromagnetic interaction transition is therefore possible. Their mass difference is only about $77\mathrm{MeV}/c^2$, below the lightest pion mass, so no additional on-shell strongly interacting particle fits. A photon is neutral and massless, and can carry the required angular momentum through a magnetic dipole transition between the two positive-parity spin-$1/2$ states. A single neutrino would violate lepton number and would not give the required fermion parity.

At rest, four-momentum conservation gives $p_{\Lambda }=p_{\Sigma }-k$ and $k^2=0$. ThusThe result is slightly below $(M_{\Sigma }-M_{\Lambda })c^2$ because the Lambda baryon recoils; rounded masses give about $74.5\mathrm{MeV}$.

The channels $n{\pi }^0$ and $p{\pi }^{-}$ are actually kinematically open: their total rest energies are about $1075$ and $1078\mathrm{MeV}$, below $M_{\Sigma }{c}^2\simeq 1193\mathrm{MeV}$. Their suppression is therefore not a decay threshold prohibition. Each changes strangeness from $-1$ to zero and so requires a weak interaction. The much faster allowed electromagnetic transition makes such weak branching fractions extremely small. “Not observed” should be understood as overwhelming electromagnetic competition, not an exact ban in the Standard Model. The photon identification likewise uses the known low-energy spectrum and dominant transition mechanism, not neutrality alone.

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.

For a holonomic conservative system the Lagrangian is $L(q,\dot{q},t)=T-V$, the difference between kinetic energy and potential energy. The Euler-Lagrange equations areIf $\partial L/\partial q_j=0$, this gives $d{p}_j/dt=0$ for the generalized momentum $p_j=\partial L/\partial {\dot{q}}_j$. Such a coordinate is a cyclic coordinate.

In cylindrical coordinates the squared speed is ${\dot{r}}^2+r^2{\dot{\theta }}^2+{\dot{z}}^2$. The holonomic constraint $z=r^2/a$ gives $\dot{z}=2r\dot{r}/a$, while gravity has potential energy $V=mgz$. ThereforeThere is no explicit $\theta$ dependence, so $\theta$ is a cyclic coordinate andThis is angular momentum about the vertical symmetry axis. At $r=0$ the polar angle is singular, but the Cartesian angular momentum $m(x\dot{y}-y\dot{x})$ remains well defined and has the same conserved value.

Time independence conserves the energy $E=T+mga\cos \theta$. The two cyclic coordinates giveThese are respectively angular momentum along the symmetry axis and angular momentum along the fixed upward vertical; $n$ is the axial spin rate. For $u=\cos \theta$ and $-1<u<1$, eliminate $\dot{\varphi }$ using $\dot{\varphi }=(h-Cnu)/[A(1-u^2)]$. The energy becomesMultiplying and rearranging givesTo justify the double-root condition, use the effective potentialEliminating the cyclic coordinates by Routh reduction gives the remaining equation $A\ddot{\theta }=-dU(\cos \theta )/d\theta$. A constant nonvertical angle therefore has $U^{\prime }(u_0)=0$ and $E=U(u_0)$. Since $P(u)=2A(E-U(u))(1-u^2)$, both $P(u_0)$ and $P^{\prime }(u_0)$ vanish. This argument uses the second-order equation; merely differentiating the energy identity at a stationary angle would not establish the derivative condition.

Thus steady nonvertical precession gives a common root in $(-1,1)$. The nonvertical qualification is implicit in the denominator used by the source. A top spinning exactly upright or inverted is also possible but corresponds to an endpoint $u=\pm 1$, not an interior root; the unrestricted literal claim would need that exception.

The distinction is captured by the double-root criterion for steady precession of a symmetric top.

Let $\mathbf{F}$ be the force on particle one, so the force on particle two is $-\mathbf{F}$, with $\mathbf{F}$ parallel to ${\mathbf{q}}_1-{\mathbf{q}}_2$. The combined torque isHence the combined angular momentum of particles one and two is conserved. Centrality of their interaction, in addition to equal and opposite forces, is what makes this torque sum zero.

The third particle is force-free, so its torque about the origin vanishes:Thus this angular momentum is conserved.

For the first particle, the derivative of its angular momentum about the origin is its torque:This need not vanish. For example take $f(r)=r^2/2$ and instantaneous positions ${\mathbf{q}}_1=(1,0,0)$, ${\mathbf{q}}_2=(0,1,0)$; the torque is $(0,0,1)$. Therefore it is not conserved in general, although it is in special configurations or for a constant potential.

For variations vanishing at the endpoints, the Euler-Lagrange equations areDifferentiate the stated energy expression using the chain rule. The acceleration terms cancel, leavingAlong a stationary trajectory the two brackets vanish. Thus $dH/dt=-F_t$, and in the absence of explicit time dependence $\boxed{H\text{ is constant}}$.

In plane polar coordinates, the squared speed is ${\dot{r}}^2+r^2{\dot{\theta }}^2$, so the kinetic energy is $T=\frac{1}{2}m({\dot{r}}^2+r^2{\dot{\theta }}^2)$. An inverse-square central gravitational force has potential energy proportional to $1/r$. With the sign convention printed in this paper, write $V=km/r$, givingFor attraction to a star of mass $M_{\star }$, this convention has $k=-G{M}_{\star }<0$. Keeping $k$ real and signed is essential; no sign is changed in differentiating the printed functional.

Since $F_r=mr{\dot{\theta }}^2+km/r^2$, $F_{\dot{r}}=m\dot{r}$, $F_{\theta }=0$ and $F_{\dot{\theta }}=m{r}^2\dot{\theta }$, the equations of motion areThe second equation gives the conserved angular momentum $J=m{r}^2\dot{\theta }$; equivalently $r\ddot{\theta }+2\dot{r}\dot{\theta }=0$. The autonomous-functional identity gives a second conserved quantity,This is the total mechanical energy; for attractive gravity the potential contribution is negative because $k<0$.

Away from a collision, let $r=q_1-q_2$ and $D=|r|>0$. The Lagrangian isDifferentiating $D^2=r\cdot r$ with respect to $x_1$ and $x_2$ gives $2D{\partial }_{x_1}D=2(x_1-x_2)$ and $2D{\partial }_{x_2}D=-2(x_1-x_2)$. ThereforeThe other coordinates give ${\nabla }_{q_1}D=r/D$ and ${\nabla }_{q_2}D=-r/D$. Substituting these into the Euler-Lagrange equations yields equal and opposite central forces:Adding the equations shows that the derivative of the total linear momentum is zero, so $P=m_1{\dot{q}}_1+m_2{\dot{q}}_2$ is constant. For the total angular momentum, the terms ${\dot{q}}_i\times {\dot{q}}_i$ vanish, givingThus both stated vectors are conserved quantities.

The total energy is also constant. Define $E=\frac{1}{2}{m}_1|{\dot{q}}_1{|}^2+\frac{1}{2}{m}_2|{\dot{q}}_2{|}^2+V(D)$. Since $\dot{D}=r\cdot ({\dot{q}}_1-{\dot{q}}_2)/D$,Hence $\boxed{E=T+V(D)\text{ is conserved}}$. The explicit cancellation establishes energy conservation for an arbitrary smooth separation-dependent potential, throughout any collision-free trajectory.

For an autonomous Lagrangian, the Euler-Lagrange equations implyIndeed, differentiating the expression cancels the acceleration terms, leaving ${\sum }_q\dot{q}(d{L}_{\dot{q}}/dt-L_q)-L_t$, which vanishes. Here $T$ is quadratic in the velocities, so $\sum \dot{q}{L}_{\dot{q}}=2T$. Thus the conserved energy isThe azimuth $\varphi$ is a cyclic coordinate. Its conjugate momentum is constant:Therefore $\boxed{{\sin }^2\theta \dot{\varphi }=h}$ is constant as well. Physically, $p_{\varphi }$ is the vertical component of angular momentum.

Differentiating orthonormality gives ${\dot{e}}_i\cdot e_j=-e_i\cdot {\dot{e}}_j$. With $\omega =\frac{1}{2}{\sum }_i{e}_i\times {\dot{e}}_i$, the vector triple product yieldsThe body coordinates $x_i$ of a fixed material point are constant. Differentiating its position $R_G+{\sum }_i{x}_i{e}_i$ therefore givesThis identifies $\omega$ as the angular velocity of the rigid body.

Because $G$ is the center of mass, $\int \rho rdV=0$, so the translational contribution to the angular momentum about $G$ vanishes. The remaining vector triple product givesThis symmetric matrix is the inertia tensor about $G$. The angular-momentum balance about the center of mass is $dL/dt={\tau }_{\mathrm{ext},G}$ in an inertial frame. With no external forces, that torque is zero, so the inertial vector $L$ is constant. Its components in rotating body axes need not individually be constant.

For the tool, let the disc lie at $z=0$ and the rod extend to $z=l=\sqrt{3}a$. The total mass is $3M$ and its center is at $z_G=l/3$. Axial symmetry makes the inertia tensor diagonal with $I_x=I_y$.

The standard disc moments are $M{R}^2/2$ about its normal and $M{R}^2/4$ about a diameter. A thin rod has moment $m{l}^2/12$ about a transverse axis through its center. The parallel axis theorem adds $m{d}^2$ for a parallel shift of distance $d$. ThereforeandThe rod contributes zero about its own longitudinal axis. Thus $\boxed{J=2M{a}^2{I}_3}$, an isotropic inertia tensor even though the shape is not spherical. This gives a nonspherical rigid body with isotropic inertia. Consequently $L=2M{a}^2\omega$. Conservation of $L$ in torque-free motion proves the angular velocity is constant for every initial orientation and spin.

beta decay is a weak charged current process. At quark level $d\to u{W}^{-\ast }\to u{e}^{-}{\bar{\nu }}_e$, with amplitude proportional to $G_F{V}_{ud}$ at low momentum transfer. A neutron in a nucleus can thereby become a proton:The second reaction denotes an energetically allowed nuclear conversion, not a decay of an isolated free proton. In a nuclear beta transition the mass number stays fixed and the atomic number changes by one. baryon number and electric charge are conserved; the neutrino or antineutrino supplies the missing energy, momentum and angular momentum. The three-body final state produces a continuous electron spectrum rather than a single electron energy. Neglecting neutrino mass and recoil, an allowed spectrum in electron total energy isThe Fermi function accounts for the electron's Coulomb interaction with the daughter; $E_0$ is the electron total-energy endpoint. forbidden beta decay adds momentum-dependent nuclear shape factors. The V-A interaction violates parity: it couples left-chiral fermion fields, not an equal mixture of left and right currents.

For allowed beta decay the leading leptons have zero orbital angular momentum. Fermi transitions have $\Delta J=0$; Gamow-Teller transitions have $\Delta J=0,1$ with $0\to 0$ excluded. Both preserve nuclear parity. Higher lepton partial waves permit forbidden transitions with extra suppression. Nuclear matrix elements and phase space, as well as these selection rules, determine lifetimes. Particle-physics examples include free-neutron decay and pion beta decay, ${\pi }^{+}\to {\pi }^0{e}^{+}{\nu }_e$. semileptonic decays test quark mixing and the universality of the weak charged current. beta decay changes quark flavour through virtual W exchange; its continuous spectrum reflects an additional lepton in the final state.

parity reverses spatial coordinates and polar vectors such as momentum, while leaving axial vectors such as angular momentum unchanged. The correlation $\mathbf{J}\cdot {\mathbf{p}}_e$ is therefore parity odd. For polarized beta-emitting nuclei one can measurewhere $\theta$ is measured from nuclear polarization, $\mathcal{P}$ its magnitude and $A$ a decay-dependent coefficient. A nonzero spin–momentum asymmetry is incompatible with parity invariance.

The cobalt-60 experiment polarized nuclei at low temperature and compared beta emission along and opposite the polarization axis. Preferential emission on one side established weak-interaction parity violation. Reversing polarization tests whether the asymmetry follows nuclear orientation; warming the sample removes polarization, while gamma anisotropy monitors its loss. Detector acceptance, backgrounds and magnetic deflection of charged particles must be separated from a true emission asymmetry.^[ref]

In the modern weak charged current, the V-A interaction couples to left-chiral fermion fields. In the massless limit this selects left-helicity neutrinos and right-helicity antineutrinos. Massive charged leptons need not be pure-helicity states, consistent with the factor $v/c$. Strong and electromagnetic interactions conserve parity to the usual experimental approximation; parity violation alone should not be confused with CP violation or a failure of CPT.

The decay is energetically open, but two spin-zero pions have total angular momentum equal to their relative orbital angular momentum $L$. Conserving the parent spin requires $L=1$. Interchanging two identical neutral pions multiplies this orbital state by $(-1{)}^L=-1$, whereas their total state must be symmetric by Bose statistics. There is no different spin or species label to supply another minus sign. Thus the two-neutral-pion final state is forbidden. Its parity for $L=1$ would actually be $(-1{)}^2(-1{)}^1=-1$, matching the parent; the decisive obstruction is identical-boson symmetry, not the energy or the shared quark-wavefunction component.

An alpha emitter has an alpha-like cluster coupled to a daughter nucleus. Its positive decay energy is $Q=[M_P-M_D-M_{\alpha }]c^2$; atomic masses may be used consistently because the electron counts balance. The daughter and alpha share this energy according to momentum conservation. The attractive nuclear interior and repulsive exterior Coulomb potential produce a quasibound state with a barrier, so positive $Q$ does not imply rapid classical escape.

For daughter charge $Z_{d}e$, the exterior barrier is approximately $C/r$, $C=2Z_d{e}^2/(4\pi {ϵ}_0)$, with outer turning radius $b=C/Q$. For zero orbital angular momentum, the leading WKB probability isHere $R$ is a nuclear matching radius, $\mu$ the reduced mass, $\nu$ an interior assault frequency and $P_{\mathrm{pre}}$ a preformation factor. The barrier factor alone cannot determine the latter two. Nonzero emitted orbital angular momentum adds a centrifugal barrier and can strongly hinder a transition.

The dominant action varies roughly as $Z_d/\sqrt{Q}$, explaining the Geiger-Nuttall relation and the enormous spread of alpha half-lives. angular momentum and parity constrain the emitted partial wave: a spin-zero positive-parity alpha requires $|J_P-J_D|\le L\le J_P+J_D$ and ${\pi }_P={\pi }_D(-1{)}^L$. Fine-structure alpha groups populate different daughter levels and therefore have different kinetic energies and penetrabilities. Shell structure and alpha clustering also change preformation probabilities, so equal $Q$ values do not guarantee equal lifetimes. Escape conserves the total energy throughout tunnelling; the particle does not borrow a measured energy from the barrier.

The symbols $EL$ and $ML$ mean electric and magnetic radiation multipoles of order $L$. The orders $L=1,2,3$ are called dipole, quadrupole and octupole. They describe the angular structure of charge/current radiation and the total angular momentum carried by the radiation; they do not change the photon's intrinsic spin from one. A real single photon has no monopole radiation mode.

For a single-photon transition, angular momentum addition requires $|J_i-J_f|\le L\le J_i+J_f$, $L\ge 1$, together with the electric/magnetic parity rules above. Thus a known final state and a measured multipole restrict both the spin difference and parity of the emitting level. For example, an E1 line reverses parity and allows $\Delta J=0,1$ except $0\to 0$; a mixed M1+E2 line preserves parity and must satisfy both angular triangles. An E2 line preserves parity and allows spin differences up to two.

Line energies identify connected levels, while calibrated intensities and coincidences test cascade flow and branching. If a proposed assignment permits a fast low-order transition to a lower state, an experimentally stringent absence of that line argues against the assignment. Nevertheless, an allowed transition can have a small or vanishing nuclear matrix element. Pure E2 radiation alone does not prove that M1 is forbidden by angular momentum. Detection sensitivity, efficiency and overlap must be considered, and in real nuclei internal conversion can remove photons; this question explicitly excludes that competing channel. Lifetimes or reduced-strength estimates make a missing-line argument substantially stronger than a bare selection-rule statement.

The common energetic principle behind nuclear fission and nuclear fusion is the nuclear binding energy. If the products are more tightly bound in total than the reactants, their smaller rest mass releasesThe two binding-energy expressions assume that the same numbers of protons and neutrons are counted on both sides; emitted free nucleons have zero nuclear binding. The energy appears mainly as product kinetic energy, with radiation and subsequent decays contributing according to the reaction.

The binding energy per nucleon rises rapidly through light nuclei, has a broad maximum in the iron/nickel region, and decreases more slowly for heavier nuclei. Combining suitable light nuclei can therefore increase their total binding, whereas splitting a sufficiently heavy nucleus into moderately heavy fragments can do the same. This is an energy comparison, not a rule that every fusion or fission channel is exothermic: the actual nuclear masses fix the sign of $Q$.

In nuclear fission, the liquid-drop model illustrates the competition between attractive short-range nuclear cohesion and long-range proton repulsive Coulomb interaction. Deformation increases surface area and costs surface energy, but separating the charge can reduce nuclear Coulomb energy. Their competition creates a fission barrier. An excited nucleus may pass over the barrier; spontaneous fission can occur through quantum tunnelling. The resulting fragments are often excited and neutron-rich, so they may emit prompt neutrons and photons and undergo later beta decay. The typical total energy released by a heavy-nucleus fission is of order a few hundred MeV, rather than an electronic-scale energy. Its size follows from an approximately MeV-per-nucleon binding gain summed over a heavy nucleus. A neutron from one fission can induce another; a nuclear chain reaction is the repeated physical process, distinct from the energy-release condition for one event.

In nuclear fusion, the initial charged nuclei repel one another. Exothermicity does not remove this entrance Coulomb barrier. A reaction requires sufficient overlap of their wavefunctions, achieved by energetic encounters and quantum tunnelling. For nonresonant low-energy charged-particle fusion, a useful separation isThe astrophysical S factor isolates the nuclear part from much of the rapidly varying Coulomb penetration probability. In a thermal population, the competition between the decreasing high-energy population and increasing penetration produces the Gamow peak. This explains how fusion can proceed below the classical barrier without requiring the mean thermal energy to equal its height.

Both processes conserve total energy, charge, baryon number and angular momentum. Their very different entrance barriers and mass ranges explain why a favorable $Q$ value by itself does not specify a reaction rate. Short-range attraction gives approximately saturated nuclear binding, while surface, Coulomb, symmetry and pairing terms account for the broad trend and its isotope-dependent departures.