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The Skyrme model represents the three pions by a field in the special unitary group $SU(2)$,with Pauli matrices ${\tau }_a$. Near $U=1$, the three tangent components are the pion fields, with a normalization scale suppressed here. Define $L_{\mu }=U^{†}{\partial }_{\mu }U$. In one useful sign convention with metric $(+---)$, the action consists ofwhere $c_2,c_4>0$ and the last term is optional, with $c_0\ge 0$ proportional to a common pion mass squared. The first term is a nonlinear sigma model kinetic term, and the second is the four-derivative term of the Skyrme model. For a static configuration of size $s$, the two derivative contributions scale as $E_2\propto s$ and $E_4\propto 1/s$, whereas a mass term scales as $s^3$. This Derrick scaling explains why the four-derivative term can stabilize a finite size instead of allowing collapse.

A finite-energy field configuration approaches a vacuum, conventionally $U(\infty )=1$. Compactifying space makes $U$ a map $S^3\to SU(2)\simeq S^3$. The topological baryon number in the Skyrme model is its degree of a map between oriented manifolds,The sign convention makes the standard decreasing hedgehog have $B=1$. Smooth evolution with the vacuum boundary condition preserves this integer. A Skyrmion is a localized soliton in such a sector; the unit soliton, after quantization, models a nucleon, and higher positive charges model multi-baryon systems. The topological conservation law is distinct from an ordinary Noether charge of isospin. Classical pion fields are bosonic, so obtaining fermionic nucleons also needs the quantum-statistics choice discussed below.

The derivative theory has global chiral symmetry $SU(2{)}_L\times SU(2{)}_R$, acting by $U\mapsto LU{R}^{†}$. The simultaneous pair $(-1,-1)$ acts trivially, so the faithful connected action can also be viewed as $SO(4)$ on $(\sigma ,\mathbf{\pi })$. In the massless theory the choice of vacuum breaks this to the vector subgroup. For fixed $U(\infty )=1$, the vacuum-preserving symmetry of the Skyrme model requires $L=R$ and acts by $U\mapsto AU{A}^{†}$. This is isospin, effectively $SO(3)$ because $A$ and $-A$ act the same way. A usual common pion-mass term explicitly preserves only this vector subgroup; full chiral symmetry is then an approximate massless-limit symmetry, not an exact symmetry of that term. Independent axial rotations change the vacuum and are not extra localized rigid-rotor coordinates in a sector with fixed boundary vacuum.

The space-time symmetry is the Poincaré group, comprising translations and Lorentz transformations, including spatial rotations. Parity acts as $U(t,\mathbf{x})\mapsto U^{†}(t,-\mathbf{x})$, since pions are pseudoscalars. For a static finite-energy solution, translations change its position; spatial rotations and isorotations change its orientation. These transformations generate collective coordinates, but a particular field can be unchanged by certain combined transformations.

For the usual low-charge minimum branches of the standard model, the relevant shapes and density symmetries are the following. These are not claims about every field of a given degree, all excited solutions or arbitrary modified pion potentials.

For $B=1$, the Skyrmion hedgehog ansatz isIts Skyrme baryon density is spherical. Indeed $B=-(2/\pi ){\int }_0^{\infty }{f}^{\prime }{\sin }^{2}fdr=1$. A spatial rotation rotates the pion direction, so the field itself is invariant under a compensating isorotation, not under every spatial rotation alone. Its proper combined stabilizer is a diagonal $SO(3)$.

For $B=2$, the toroidal two-Skyrmion has a ring-shaped density with an axial hole and full density symmetry $D_{\infty h}$. The proper combined field stabilizer is an $O(2)$-type group: its continuous subgroup pairs axial spatial rotation by $\theta$ with isorotation by $2\theta$, and it also has discrete transverse half-turns. The angular approximation $R(z)=z^2$ makes the axial factor two explicit. A static two-baryon minimum is consequently not simply two separate round unit lumps.

For $B=3$, the tetrahedral three-Skyrmion has a tetrahedral shell with four face-hole directions and full density group $T_d$. Its proper rotational group is $T\cong A_4$, the tetrahedral symmetry group of order 12. The actual field symmetries again pair these rotations with isorotations.

For $B=4$, the cubic four-Skyrmion has a cubic shell with six face-hole directions. Its full density group is $O_h$, the symmetry group of a cube, and its proper group is $O\cong S_4$, the rotational symmetry group of a cube of order 24. Cubic and octahedral symmetry name the same point group; this particular density shape is cubic. Reflections in these full density groups should not be confused with the proper rotation-isorotation group used for angular-momentum quantization.

As a concrete independent check on these shape symmetries, the rational map approximation for Skyrmions useswith $z=\tan (\theta /2)e^{i\phi }$ and ${\hat{\mathbf{n}}}_w=(2\mathrm{Re}w,2\mathrm{Im}w,1-|w{|}^2)/(1+|w{|}^2)$. Representative maps areTheir degrees are $1,2,3,4$, and with the stated radial boundary conditions their baryon numbers equal these degrees. The angular Jacobian of a rational map is $J_R=[(1+|z{|}^2)|R^{\prime }|/(1+|R{|}^2){]}^2$. It vanishes at branch directions. $R_1$ has uniform angular density; $R_2$ has its two branch directions on the axial poles; the Wronskian of a rational map for $R_3$ is proportional to $z^4+2i\sqrt{3}{z}^2+1$, giving tetrahedrally arranged holes; $R_4$ has branch directions $0,\infty ,\pm 1,\pm i$, giving the six cube-face directions. Also $R_4(iz)=1/R_4(z)$ pairs a spatial quarter-turn with an isospin half-turn. These give illustrative approximate fields, not exact analytic solutions or a proof of global energy minimality.

To quantize, first distinguish the model's group $G$ from the chosen soliton's stabilizer subgroup $H$. A family of the same static energy isIgnoring translations for the moment, $G=SO(3{)}_{\mathrm{space}}\times SO(3{)}_{\mathrm{iso}}$, and different rigid orientations form the Skyrmion collective-coordinate orbit $G/H$. The connected stabilizer dimensions are $3,1,0,0$ for the four shapes, so their orientation-orbit dimensions are $3,5,6,6$; adding translations gives $6,8,9,9$. For the unit hedgehog, treating rotation and isorotation as six independent modes would double-count its locked orientation. For higher charges, arbitrary separation of constituent lumps is not an exact flat moduli space in this non-Bogomolny theory.

Let the orbit coordinates depend slowly on time. Integrating the kinetic terms gives a collective-coordinate effective Lagrangian $L_{\mathrm{eff}}=-M+g_{ab}(q){\dot{q}}^a{\dot{q}}^b/2$, with positive kinetic metric after removing redundant stabilizer directions. Its angular blocks are the spatial, isospin and mixed inertia tensors. Collective-coordinate quantization produces a rotor Hamiltonian on this orbit, with translations giving center-of-mass momentum. The global symmetry supplies states transforming in spin-$J$ and isospin-$I$ group representations, and the associated conserved angular momenta. Inertia tensors determine energy splittings; group symmetry alone does not specify those tensors or their numerical energies.

The topology and the static stabilizer supply further collective-rotation constraints for a Skyrmion. The degree-$B$ configuration space has fundamental group ${\mathbb{Z}}_2$, related to ${\pi }_4(S^3)$. Choosing the nontrivial Finkelstein-Rubinstein constraints character gives fermionic unit baryons. Wavefunctions live on the appropriate cover, and a lifted stabilizer operation obeysThe sign depends on whether the actual field-configuration loop is contractible, not just on whether the density looks symmetric. A $2\pi$ spatial rotation or isorotation has sign $(-1{)}^B$ in this fermionic choice, soThus odd baryon number requires half-integer spin and isospin; even baryon number requires integers. Discrete or continuous combined stabilizer symmetries impose additional restrictions on the allowed pairs and body-fixed rotor states. They must not simply be omitted, or replaced by trivial invariance under every density symmetry. The trivial character is a possible bosonic quantization but would not produce a fermionic nucleon. Improper field symmetries, implemented using the model parity transformation, can additionally constrain parity labels; scalar-density reflection symmetry alone is insufficient to infer those labels.

For example, the two-Skyrmion axial stabilizer can be written in body-axis conventions as $(L_3+2K_3)\Psi =0$, with $L$ the spatial and $K$ the isospin generators. A transverse spatial half-turn accompanied by an isospin half-turn has the nontrivial sign in the Finkelstein-Rubinstein constraints. The integer-spin scalar state $J=I=0$ is therefore excluded despite even $B$. A spin-one, isospin-zero state with zero axial body projection has the required minus sign under that half-turn; the spin-zero, isospin-one channel can also satisfy the constraints. Their relative energies require the inertia tensors.

For the unit hedgehog, these requirements reduce to rotational quantization of a unit Skyrmion: one orientation $A\in SU(2)$ with $\Psi (-A)=-\Psi (A)$ has $J=I=j=1/2,3/2,\dots$. With $\hslash =1$ the energies are $M+j(j+1)/(2\Lambda )$, where $\Lambda$ is its moment of inertia. Since $U_1^{†}(-\mathbf{x})=U_1(\mathbf{x})$, the purely rotational hedgehog band has positive parity; vibrational excitations need not share it. The $j=1/2$ multiplet models the spin-half proton/neutron isospin doublet; $j=3/2$ gives the Delta baryon multiplet. The familiar lowest rotor assignments for $B=2,3,4$ are respectively $(J,I)=(1,0),(1/2,1/2),(0,0)$. These assignments incorporate the corresponding field-stabilizer constraints and standard inertia ordering, not a prediction from density shape alone. Deformations, vibrational modes, radiation and binding dynamics lie beyond the rigid approximation. The two symmetry roles are therefore

A collective coordinate describes a position, orientation, or another parameter of a family of static solitons. For a family $\Phi (\mathbf{x};q)$, promote $q$ to a slowly varying $q(t)$ and substitute into the field action. In a scalar theory with unit kinetic coefficient, this gives the collective-coordinate effective LagrangianFor a gauge-theory soliton, one also solves the Gauss law constraint in gauge theory constraint and projects out pure gauge transformations; arbitrary variations of gauge representatives do not define the physical metric. Tangent vectors to an exactly equal-energy family are zero modes in field theory. If $V=E_0$ is constant, the Euler-Lagrange equations of this moduli-space approximation arethe geodesic equations of its Riemannian metric. The approximation neglects radiation and deformation modes; it describes motion sufficiently slow that these omitted degrees of freedom remain unexcited to the required accuracy.

In collective-coordinate quantization, take the wavefunction measure $\sqrt{\det g}{d}^{n}q$ and the minimal scalar Hamiltonian operatorThe Laplace-Beltrami operator supplies coordinate-invariant kinetic energy. Global identifications and statistics must be imposed on the wavefunctions; the classical metric alone does not choose them. Curvature-ordering terms and loop corrections are additional quantum input.

For the phi-four kink, use the normalization and profile of Question 1. Substituting $\varphi (x,t)={\varphi }_K(x-X(t))$ givesThe metric is constant because of translation invariance. This proves the translational dynamics of a phi-four kink: classically the centre moves at constant velocity, and quantum mechanicallyPlane waves label the continuous translational momentum, with no position-dependent potential. Uniform-motion Lorentz invariance upgrades the dispersion to $\sqrt{M^2+p^2}$; the displayed collective Lagrangian is its small-velocity expansion. Small perturbations also include an internal shape mode and continuum radiation, which this single collective coordinate omits. The fluctuation operator of a phi-four kink in this normalization isthe translational eigenfunction has ${\omega }^2=0$, the shape mode has ${\omega }^2=3c^2$, and continuum modes in the spectrum start at ${\omega }^2=4c^2$. Thus the free-coordinate states describe the kink's low translational energies, not its full excitation spectrum or quantum mass correction.

For two Abelian Higgs vortices at critical coupling, the static energy is $2M_v$ and the Abelian Higgs vortex moduli space has four real dimensions. Let $z_1,z_2$ be their positions, $Z=(z_1+z_2)/2$, and $q=z_1-z_2$. The centre of mass decouples; the relative metric is rotationally symmetric and can be writtenAt large separation, $F(\rho )\to M_v/2$, recovering two free particles. Although there is no static separation potential, the nonconstant metric produces velocity-dependent interaction. Coincidence is smooth in the relative coordinate for two identical vortices $w=q^2$, not in the double-valued $q$. Smoothness gives $F(\rho )\sim C{\rho }^2$ for $C>0$ near zero. A head-on geodesic continues through $w=0$ to the opposite real ray, so $q$ changes its line by $\pi /2$: the vortices scatter through a right angle. This geometric argument does not require an explicit formula for $F$.

With ordinary bosonic exchange statistics, relative wavefunctions are single-valued in $w$ and smooth at coincidence. In the separated polar coordinate they obey $\Psi (\rho ,\theta +\pi )=\Psi (\rho ,\theta )$, with even integer angular labels. Their kinetic operator isThe apparent singularity at $\rho =0$ must be resolved with the smooth $w$ coordinate and regularity, rather than arbitrary boundary conditions on a punctured cone. The free centre-of-mass motion and the asymptotically free relative geometry give quantum scattering states; a flat static energy does not imply that the metric is flat or that scattering is absent. This is not a prediction of a discrete family of static two-vortex bound separations. The smooth collision geometry is developed in David Tong's arxiv.org/abs/hep-th/0509216.

For a Skyrmion of baryon number one, the Skyrmion hedgehog ansatz iswhere $\mathbf{\sigma }$ are the Pauli matrices. Include a centre $\mathbf{X}$ and an orientation $A\in SU(2)$ through $U=A{U}_0(\mathbf{x}-\mathbf{X})A^{-1}$. Hedgehog symmetry identifies spatial rotations with opposite internal rotations, so there are three independent orientation coordinates, not six. Since $A$ and $-A$ give the same field, the physical orientation space is $SO(3)$, with SU(2) group as its double cover. Write $A^{-1}\dot{A}=i\mathbf{\omega }\cdot \mathbf{\sigma }/2$. The leading collective Lagrangian has the formwhere $\Lambda >0$ is the rotational moment of inertia obtained by integrating the profile's field kinetic energy.

For the fermionic quantization appropriate to baryons, the Finkelstein-Rubinstein constraints on the double cover impose $\Psi (-A)=-\Psi (A)$. In SU(2) representations, the central element $-1$ acts by $(-1{)}^{2j}$, so $j$ must be half-integer. Left and right group actions supply isospin and spin angular momentum; hedgehog symmetry makes their magnitudes equal. The rotational quantization of a unit Skyrmion therefore givesThe $j=1/2$ level has four spin-isospin states and models the nucleon doublet, proton and neutron, each with two spin states. The $j=3/2$ level has sixteen states and models the Delta baryon quartet, each with four spin states. The rotor predicts a splitting $3{\hslash }^2/(2\Lambda )$. Without the fermionic sign, single-valued functions on $SO(3)$ would instead allow integer $j$, which is a different quantization. High rotor levels can couple to deformation and pion radiation; this semiclassical approximation does not establish that its entire formal tower consists of stable particles.

nuclear radius scaling follows approximately $R=r_0{A}^{1/3}$ with $r_0\simeq 1.2\mathrm{fm}$ for a sharp matter-radius convention. Consequently the interior nucleon density is nearly independent of mass number: nuclear forces are short range and exhibit nuclear matter saturation. A diffuse nuclear charge distribution has a charge-weighted root mean square radius, not a hard surface. For a uniform sphere, $r_{\mathrm{ch}}=\sqrt{3/5}R$; charge radii and matter radii need not coincide, especially with neutron skins.

Elastic electron scattering measures the nuclear charge form factor. In a simple one-photon, spherical approximation, with $q$ a wave number,The scattering cross section relative to a point-charge prediction is proportional to $|F{|}^2$, so its low-$q$ slope constrains the radius and diffraction structure constrains the radial distribution. Finite momentum coverage and Coulomb distortion in electron scattering limit reconstruction. In Robert Hofstadter's experiment a magnetic spectrometer separated elastic and inelastic electron peaks; beam-charge normalization and closed-slit background checks prevented raw count structure from being mistaken for a nuclear charge form factor.^[ref]

A useful intrinsic-shape description is $R(\theta ,\varphi )=R_0[1+{\sum }_{\mu =-2}^2{\alpha }_{2\mu }{Y}_{2\mu }(\theta ,\varphi )+\cdots ]$. An intrinsic nuclear quadrupole moment $Q_0=\int {\rho }_{\mathrm{ch}}(3z^2-r^2)d^{3}r$ distinguishes prolate and oblate positive-charge distributions by positive and negative signs. The laboratory spectroscopic nuclear quadrupole moment is state dependent: an $I=0$ state has zero laboratory quadrupole moment even when its intrinsic shape is deformed. nuclear rotational bands approximately follow $E_I\propto I(I+1)$; harmonic nuclear quadrupole vibrations instead have equally spaced phonon energies and multiphonon spin multiplets. Measured electric quadrupole transitions test the collective interpretation in addition to level energies. Radius probes spatial extent; quadrupole observables and collective spectra probe shape.

A useful convention for the semi-empirical mass formula iswhere $B>0$ is the binding energy. The nuclear mass is $Z{m}_p+(A-Z)m_n-B/c^2$. Atomic-mass comparisons instead use hydrogen-atom masses and account for electron binding at the desired accuracy. No numerical fit coefficients are specified here.

The volume term represents short-range nuclear attraction with saturation: each nucleon has a limited number of close neighbours. The surface term subtracts the binding lost by nucleons at the boundary. Coulomb repulsion penalizes a large proton fraction. The asymmetry term penalizes unequal neutron and proton populations, reflecting the cost of filling different Fermi seas and nuclear interactions. Pairing favours even-even nuclei and penalizes odd-odd nuclei; odd-$A$ nuclei have zero pairing term in this approximation. For example, one may choose $\delta =\pm a_p{A}^{-1/2}$; an alternative fitted power requires a correspondingly different coefficient.

At fixed $A$, the smooth mass dependence on $Z$ is approximately parabolic. Neglecting the neutron-proton mass difference, pairing and small linear-in-$Z$ Coulomb corrections, minimization givesThus the preferred neutron excess grows with mass number. The surface penalty disfavors very light systems, while the growing Coulomb penalty reduces the binding per nucleon of very heavy systems; this explains broad fusion and fission energy trends. The formula is a bulk approximation: shell closures, deformation and small separation-energy differences can give important exceptions. A fitted average binding trend is not a reliable substitute for a measured decay threshold near a boundary of stability.

In the nuclear shell model, a level with total angular momentum $j$ has $2j+1$ substates for each nucleon species. The first three levels accommodate $2+4+2=8$ protons or eight neutrons. nuclear pairing makes a filled shell or a pair of like nucleons contribute $J=0$, and an orbital has spatial parity $(-1{)}^{\ell }$.

For oxygen-17, eight protons and eight neutrons form a closed core; the remaining neutron occupies $1d_{5/2}$, so $J^P=5/2^{+}$. For nitrogen-15, eight neutrons close their shell, while the seventh proton occupies $1p_{1/2}$. Equivalently it is a proton hole relative to the closed eight-proton shell, giving $J^P=1/2^{-}$. beryllium-10 is an even-even nucleus; pairing its four protons and six neutrons gives $J^P=0^{+}$.

boron-10 needs an additional qualification. The fifth proton and fifth neutron each occupy $1p_{3/2}$. addition of angular momentum permitsThe experimentally correct ground-state assignment is $3^{+}$, corresponding to aligned unpaired $j=3/2$ contributions in this simple description. However, one-particle level ordering alone does not select it over the other couplings: their splitting requires the nuclear residual interaction. Assuming a preference for the aligned proton-neutron coupling gives the intended prediction; without that assumption, the strictly deducible result is the allowed positive-parity multiplet. The evaluated ground state is $3^{+},T=0$, with a $1^{+},T=0$ level above it.^[ref]

The nuclear Fermi gas model treats protons and neutrons as two species of spin one-half fermions moving approximately independently in a confining nuclear mean field. At a temperature small compared with their Fermi temperature, the Pauli exclusion principle fills the lowest available momentum states. It does not allow all nucleons to sit at zero momentum.

For species $i=p,n$, two spin states occupy each momentum-space cell $(2\pi \hslash {)}^3/V$. Counting a filled sphere giveswhere $n_i=N_i/V$. The Fermi energy is measured from the mean-field well bottom; it need not equal the energy needed to remove a nucleon into vacuum. Occupied-state averaging givesEven at zero temperature there is kinetic energy and degeneracy pressure. As an explicitly illustrative input, total density $n=0.16{\mathrm{fm}}^{-3}$ and $N=Z$ give $p_{F}c\simeq 263\mathrm{MeV}$ and $E_F\simeq 37\mathrm{MeV}$. These numbers illustrate the formula; they are not additional data supplied by the paper.

Separate proton and neutron seas explain a kinetic contribution to nuclear symmetry energy. At fixed $n$, put $\delta =(N-Z)/A$ and define $E_F$ at $n_p=n_n=n/2$. ThenIncreasing one species' population raises its Fermi surface and costs energy, favoring similar populations before electrostatic and interaction terms are included.

pauli blocking suppresses collisions deep in a filled sea because scattering needs unoccupied final states. An energetic external probe can knock a nucleon into an empty state and reveal its initial Fermi motion, making the model useful for broad inclusive scattering distributions. A finite surface can be approximated using a position-dependent nuclear mean field.

A free gas cannot produce nuclear binding or saturation: those require an attractive interaction and its density dependence. The smooth gas picture also omits discrete nuclear shells, nuclear spin-orbit splitting, nuclear pairing, short-distance correlations and deformation. It cannot assign all ground-state spins from a momentum sphere. Its useful prediction is the scale and occupancy of nucleon motion, with a surrounding mean field supplied separately.

Smooth-model fission energy. Use natural units $c=1$, so the quoted nuclear masses are expressed as rest energies. Conservation of mass number and atomic number fixes the second fragment to $(A_2,Z_2)=((1-y)A,(1-y)Z)$. First omit nuclear pairing to compare a continuous range of $y$. Pairing can subsequently add ${\delta }_1+{\delta }_2-{\delta }_{\mathrm{parent}}$ for integer fragments; it has no smooth parity assignment for arbitrary real $y$.

In the parent-minus-daughters mass difference, the nucleon rest energy and volume terms cancel. The nuclear symmetry energy also cancels because both fragments have the same $Z/A$:The remaining liquid-drop fission energy isSplitting increases the surface energy cost, but reduces nuclear Coulomb energy. Both contributions are symmetric under $y\leftrightarrow 1-y$, so the symmetric split is stationary. Symmetry alone would not establish a maximum, since the two contributions compete.

For a full check, put $S=a_S{A}^{2/3}$, $x=a_C{Z}^2/(a_{S}A)$ and $t=y^{1/3}+(1-y{)}^{1/3}$. On $0\le y\le 1/2$, $t$ increases from $1$ to $2^{2/3}$. Expressing the two fractional-power sums through $t$ givesThe derivative can change sign only from negative to positive, so any interior stationary point in $t$ is a minimum. The largest value is therefore either the no-split endpoint $Q=0$ or the symmetric split. Thus symmetric fission maximizes the positive liquid-drop energy release: whenever this smooth model permits an exothermic split, its largest release occurs at $y=1/2$.

There is a qualification to the paper's unqualified wording. For arbitrary low $Z^2/A$, the symmetric split need not maximize $Q$; indeed it is a local minimum when $Z^2/A<(2/5)a_S/a_C=10$. The symmetric-maximum conclusion is valid in the heavy, fission-favorable regime relevant here, rather than a universal property of the formula.

At $y=1/2$,Requiring $Q_{\mathrm{sym}}>0$ gives

Barrier and fragment yields. Positive final-state Q value does not remove the fission barrier along a deformation path. Surface energy initially rises as the nucleus deforms; spontaneous fission then requires collective quantum tunnelling unless enough excitation is present. Its rate is exponentially sensitive to barrier height, width and collective inertia. Thus nuclei close to the exothermicity threshold can have immeasurably long spontaneous-fission mean lifetimes, while sufficiently heavier nuclei have smaller effective fission barriers or larger quantum tunnelling probabilities.

The fission-fragment mass distribution of low-excitation actinide fission is usually double peaked: complementary fragments have mass fractions roughly $0.4$ and $0.6$ rather than both exactly $1/2$. Nuclear shell model effects and the fission dynamics help favor asymmetric fragments. Maximizing the final smooth-model energy release is not a calculation of the fragment yield. The following yield curve is schematic, with no measured counts implied:

Capture excitation energies. Capturing a negligible-energy neutron forms a compound nucleus with the rest energy of the target plus the free neutron. Relative to the compound ground state, the neutron-capture excitation energy is thereforewhere $B$ is positive nuclear binding energy and $S_n$ the compound nucleus's neutron separation energy. The free neutron rest energy cancels in the binding-energy difference. With the supplied semi-empirical mass formula,At $Z=92$, uranium-235 has odd $A$, while uranium-236 is even-even. The pairing contribution is $+33.5(236{)}^{-3/4}=+0.55637\mathrm{MeV}$. By contrast uranium-238 is even-even and uranium-239 has odd $A$, giving $-33.5(238{)}^{-3/4}=-0.55286\mathrm{MeV}$. The individual contributions areThus the model predictsOf the $1.48305\mathrm{MeV}$ predicted difference, $1.10922\mathrm{MeV}$ comes from the change in nuclear pairing. It is the primary reason the even-even capture product is more strongly excited; the smooth terms account for the remaining $0.37383\mathrm{MeV}$.

Thermal versus fast-neutron fission. The last comparison uses the supplied observed capture excitations, $6.5$ and $4.8\mathrm{MeV}$, rather than replacing them by the preceding smooth-model estimates. Interpreting the quoted $6.2$ and $6.6\mathrm{MeV}$ activation energies as the barriers in the respective capture channels givesA simple neutron-induced fission threshold estimate for the second channel requires about $6.6-4.8=1.8\mathrm{MeV}$ additional centre-of-mass kinetic energy. Sub-barrier tunnelling is possible, but it does not give the large thermal-neutron fission probability of the first channel.

The barrier labels in the original PDF need care: it assigns them to ${}^{235}\mathrm{U}$ and ${}^{238}\mathrm{U}$, whereas the states formed after capture are ${}^{236}{\mathrm{U}}^{\ast }$ and ${}^{239}{\mathrm{U}}^{\ast }$. The comparison above uses the intended capture-channel interpretation. If the quoted numbers are literally barriers of the uncaptured target nuclei, the compound-nucleus barriers are missing and this numerical comparison does not follow from those data alone.

In a thermal reactor, neutron moderation slows fission neutrons so the large low-energy fission cross section of uranium-235 can support a nuclear chain reaction. Fuel composition, moderator choice and geometry must balance production against capture and leakage; capture resonances of uranium-238 during slowing are important losses from that neutron population. The reaction rate per target nucleus depends on the spectrum through $\int \varphi (E){\sigma }_f(E)dE$, where $\varphi (E)$ is differential neutron flux per unit energy. A fast spectrum can access the uranium-238 fission channel and changes this balance. Also, uranium-238 is a fertile material: capture followed by two beta-minus decays can produce plutonium-239, a fissile material, so capture can contribute to breeding even when it causes no immediate fission.

The original chart has $Z$ vertically and $N$ horizontally. Thus its region (a) is proton-rich relative to the beta-stable band, whereas (b) is neutron-rich.

Electron capture, $e^{-}+(Z,N)\to (Z-1,N+1)+{\nu }_e$, is the other proton-to-neutron weak process; it remains possible for some neutral atoms whose beta-plus decay lacks the $2m_e$ atomic-mass threshold.The emitted alpha particle contains two protons and two neutrons. Very heavy nuclei can also compete through nuclear fission; the chart alone does not determine each nuclide's branching fractions.

The quark transitions are $u\to d+W^{+}$ followed by $W^{+}\to e^{+}{\nu }_e$ in (a), and $d\to u+W^{-}$ followed by $W^{-}\to e^{-}{\bar{\nu }}_e$ in (b). The additional panel shows electron capture as $u+e^{-}\to d+{\nu }_e$ through W boson exchange. The remaining quarks in each nucleon are spectators.

Regions (c) and (d) lie beyond the proton and neutron binding limits, respectively. In (c), unbound protons can be emitted, $(Z,N)\to (Z-1,N)+p$; in (d), unbound neutrons can be emitted, $(Z,N)\to (Z,N-1)+n$. These are the proton drip line and neutron drip line limits, with one- or several-nucleon emission possible according to the nuclear separation energies. They are not merely more distant points on a beta-decay trajectory.

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.

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.

A ground-state baryon contains three quarks in a spatially symmetric $L=0$ wavefunction, giving positive parity. Each quark is a spin one-half fermion, so the complete wavefunction must be antisymmetric under interchange of its complete quark labels. A physical baryon is a colour singlet, with the antisymmetric colour factorConsequently the combined spin-and-flavour factor must be symmetric. The product notation in the question organizes the degrees of freedom; it need not mean that spin and flavour are separately unentangled. In particular the octet requires a correlated sum of products.

For the three light flavours, the flavour symmetry decomposition isThe subscripts describe permutation symmetry, with $M$ denoting mixed symmetry. The addition of angular momentum of three spin one-half particles givesThe symmetric spin quartet has total $J=3/2$; the two spin doublets have $J=1/2$.

For $J=3/2$, the spin factor is symmetric, so the flavour factor must also be symmetric: this gives the spin-$3/2$ baryon decuplet. Its flavour content is four $\Delta$ states, three ${\Sigma }^{\ast }$ states, two ${\Xi }^{\ast }$ states, and one $\Omega$ state.

For $J=1/2$, the mixed-symmetry spin functions combine with the mixed-symmetry flavour functions to make one fully symmetric spin-flavour factor for each octet weight: this gives the spin-$1/2$ baryon octet, containing two nucleons, three Sigma baryons, one Lambda baryon, and two Xi baryons. The two mixed-symmetry octets in the raw flavour product are not two distinct ground-state octets; their permutation partners are correlated with the spin factors. An antisymmetric flavour singlet would require a fully antisymmetric spin factor to make a symmetric product, but three spin states drawn from a two-dimensional single-quark spin space have no such factor.

Equivalently, the symmetric three-particle representation of spin-flavour symmetry has dimension $\left(\genfrac{}{}{0ex}{}{8}{3}\right)=56$ and decomposes under flavour and spin intoThese are approximate multiplets: the larger strange quark mass breaks exact flavour symmetry and splits their energies.

Gravitational evidence requires much more nonbaryonic matter than can be supplied by ordinary luminous material or by the baryon abundance permitted by nucleosynthesis. A stable, electrically neutral massive particle is a plausible cold component. Thermal production followed by cosmological particle freeze-out gives an additional motivation: an annihilation rate near a weak-interaction scale can leave an appreciable relic abundance. This is a mechanism and search hypothesis, not proof that the dark matter must be a WIMP or have any particular elastic-scattering cross section.

The plotted regions are conditional WIMP interpretations of different measurements. The experiments use different targets and distinguish signal from background differently:

The plot also includes CRESST experiment, whose cryogenic calcium-tungstate modules combine phonon and scintillation-light signals. Different nuclei contribute to its recoil interpretation. Its excess region is another conditional candidate region, not an extra confirmed detection.^[ref]

Around ten GeV, the shaded regions favour cross sections roughly $10^{-41}$–$10^{-40}\mathrm{c}{\mathrm{m}}^2$, while the XENON100 solid curve and the improved low-mass CDMS curve cut into or exclude much of those regions under the common elastic, spin-independent WIMP-nucleus scattering interpretation expressed per nucleon. There is tension within that shared model; the plot does not show mutually confirmed discoveries. electron backgrounds, threshold response, nuclear-recoil quenching factors, channeling assumptions and the halo velocity distribution can shift interpretations. Changing the interaction model can also alter target comparisons, but cannot establish a WIMP by assertion. The 100-live-day XENON100 result itself reported candidates compatible with estimated background.^[ref]

For a rough dimensional event estimate, $1\mathrm{c}{\mathrm{m}}^2=10^{-4}{\mathrm{m}}^2$ givesIf a collider production cross section had this value, that small yield, before acceptance and backgrounds, would not support a likely discovery. The plotted nuclear-scattering cross section is not the LHC production cross section, however. Converting between them needs a mediator, couplings, parton distributions and production kinematics; a light stable WIMP would normally be inferred from missing transverse momentum with a visible recoil object, not a visible WIMP track. The mass makes pair production kinematically accessible, but the plot and integrated collider luminosity alone do not decide discovery probability.

For the relic estimate, define $M=m{c}^2$ and $\theta =k_{B}T$. The sharp-freeze-out relic dilution model givesAssume stable particles, no subsequent annihilation, and $RT=\mathrm{constant}$. Conserved comoving particle number then givesThe present rest-energy density and its critical fraction areThus the supplied toy model yields a cosmologically substantial density, of the right broad scale to motivate a dark-matter candidate. It is not an independent precision prediction: $P$ is imposed and the abundance depends exponentially on it. If entropy degrees of freedom change, entropy conservation instead gives an extra factor $g_{\ast s,0}/g_{\ast s,f}$ in $n_0$ when using the photon temperature. The species content needed to evaluate that correction is not supplied. A realistic calculation also follows residual annihilations with the abundance Boltzmann equation rather than stopping exactly at equilibrium freeze-out.

The tau-neutrino charged-current threshold requires specifying the stationary target and the lightest allowed final state. For a free neutron, the lowest charged current channel is ${\nu }_{\tau }n\to {\tau }^{-}p$. At threshold both final particles are at rest relative to one another in the centre-of-momentum frame, soThe equal-nucleon-mass approximation gives $m_{\tau }+m_{\tau }^2/(2m_N)\simeq 3.46\mathrm{GeV}$. More generally $E_{\nu }^{\mathrm{thr}}=[(m_{\tau }+M_X{)}^2-M_T^2]/(2M_T)$ for a target of mass $M_T$ and a recoil system with minimum allowed invariant mass $M_X$. A stationary electron in ${\nu }_{\tau }{e}^{-}\to {\tau }^{-}{\nu }_e$ instead gives $(m_{\tau }^2-m_e^2)/(2m_e)\simeq 3.09\mathrm{TeV}$. The customary $3.5\mathrm{GeV}$ answer therefore presumes a nucleon target; nuclear thresholds and practical deep inelastic scattering thresholds differ.

Treat a nucleon as incoherent collinear spin-half quarks and antiquarks carrying fractions $x$ of its momentum. Neglect their masses and transverse motion, use a charged-current four-fermion interaction, and take $Q^2\ll m_W^2$. The subprocess energy is $\hat{s}\simeq xs$. Antineutrino scattering on a quark has the same selected-current helicity factor $(1-y{)}^2$ as in part a; crossing to an antiquark changes the allowed spin pairing and gives a flat distribution. Multiplying the elementary strength by the respective parton densities givesHere $q,\bar{q}$ are the effective weak flavour combinations with the isoscalar proton-neutron average and CKM weights absorbed into their definitions. With only light flavours and charge symmetry, the relevant up and down combinations coincide on an isoscalar target. Rescaling those density definitions requires rescaling the prefactor consistently.

The equation is a leading-order scaling result: it neglects logarithmic QCD evolution, heavy-quark thresholds, higher-twist and target-mass corrections, nuclear modifications, and weak-propagator suppression. The current produces ${\mu }^{+}$ for an incident muon antineutrino. At higher momentum transfer the weak propagator supplies a factor approximately $[m_W^2/(m_W^2+Q^2){]}^2$, and at low energy resonance and quasielastic scattering invalidate a pure DIS description.

For a nucleon at rest and beam energy well above its mass scales, $s\simeq 2m_{N}E$. If the leading parton distributions are approximately independent of beam energy, integration over $x,y$ therefore makes both total cross sections approximately proportional to $E$. The plotted plateaus are evidence for scattering on approximately pointlike constituents in this range, with the departures expected from QCD evolution and threshold effects.

Define the momentum-weighted contents $Q={\int }_0^{1}xq(x)dx$ and $\bar{Q}={\int }_0^{1}x\bar{q}(x)dx$. Since ${\int }_0^1(1-y{)}^{2}dy=1/3$, the two cross sections obeyRead the original high-energy plateaus as approximately $0.67$ and $0.33$ in the plotted $10^{-38}\mathrm{c}{\mathrm{m}}^2/GeV$ units. With $R={\sigma }_{\bar{\nu }}/{\sigma }_{\nu }\simeq 1/2$ and $r=\bar{Q}/Q$,Using the literal readings $0.33/0.67$ gives $r=0.1905$, so twenty percent is an appropriate graphical estimate, not a precision fit. A purely valence target would have $R=1/3$, while a fully symmetric sea would give $R=1$.