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The derivative convention uses hypercharge Y with $Q=T_3+Y$. For each lepton family $l=e,\mu ,\tau$, the left-handed fields form an SU(2) doublet $L_l=({\nu }_l,l{)}_L^T$, with $Y=-1/2$. Their weak isospin values are $T_3=+1/2,-1/2$ and their electric charges are $0,-1$. The right-handed charged lepton $l_R$ is an SU(2) singlet, with $T_3=0$, $Y=-1$ and charge $-1$. The minimal model used here has no right-handed neutrino. A sterile right-handed neutrino, if added as a singlet with $Y=0$, would have no charged weak current interaction. Antiparticles carry the conjugate representations and opposite electric charges.

Expanding ${\bar{L}}_{l}i{\gamma }^{\mu }{D}_{\mu }{L}_l$ supplies $-g{\bar{L}}_l{\gamma }^{\mu }{A}_{\mu }^a{\sigma }_a{L}_l/2$. The off-diagonal entries of $A^a{\sigma }_a$ are $\sqrt{2}{W}^{+}$ and $\sqrt{2}{W}^{-}$. With the chiral projector $P_L=(1-{\gamma }^5)/2$, one has ${\bar{\nu }}_L{\gamma }^{\mu }{l}_L=\bar{\nu }{\gamma }^{\mu }{P}_{L}l$. Thus the two W boson interactions areFor quarks, separate rotations diagonalize the up-type and down-type mass terms. If their left-handed rotations are $U_{uL}$ and $U_{dL}$, the charged interaction contains the CKM matrix $V=U_{uL}^{†}{U}_{dL}$:There is an implicit sum over color charge. This quark mixing permits transitions between different families, rather than just the paired fields in a single doublet.

Let $J^{\mu }={\sum }_l{\bar{\nu }}_l{\gamma }^{\mu }(1-{\gamma }^5)l+J_{\mathrm{had}}^{\mu }$ and $\kappa =g/(2\sqrt{2})$. At momenta much smaller than $m_W$, neglect derivatives in the heavy W boson equation. Its mass and source terms can be completed to a square:This explicitly performs integrating out a charged vector boson. The resulting Fermi interaction isThe omitted terms contain derivatives and are relatively suppressed by momentum squared divided by $m_W^2$.

For the leptonic pion decay, Lorentz covariance restricts a vacuum-to-spin-zero current matrix element to a constant times $p^{\alpha }$. In the rest frame, rotations eliminate the spatial components. Under parity, the temporal vector current is even, whereas the vacuum is even and the pion is odd, so its temporal matrix element equals its negative and vanishes. The temporal axial current is odd and is allowed. This proves the vacuum-to-pseudoscalar current selection rule, rather than merely assuming an axial current. Define the pion decay constant byThen the vector-minus-axial charged weak current has matrix element $-i{f}_{\pi }{V}_{ud}{p}^{\alpha }$, up to an irrelevant overall phase convention. Consequently the scattering amplitude for the decay can be writtenUse $p=k+q$, the momentum-space Dirac equation ${\bar{u}}_e/k=m_e{\bar{u}}_e$, and $/q{v}_{\nu }=0$ for a massless neutrino. Since $/q(1-{\gamma }^5)=(1+{\gamma }^5)/q$, the neutrino contribution vanishes, leavingThe fermion spin sum is, using the supplied gamma matrix trace identities,Here $(k+q{)}^2=m_{\pi }^2$, $k^2=m_e^2$, and $q^2=0$. The mass term in the trace vanishes; anticommuting ${\gamma }^5$ reduces the remaining trace to twice $\mathrm{tr}(/k/q)$. It follows thatUsing two-body decay phase space, with final momentum magnitude $(m_{\pi }^2-m_e^2)/(2m_{\pi })$, gives the leptonic pseudoscalar decay widthThus the requested $m_e^2(m_{\pi }^2-m_e^2)$ factor is already present in the squared amplitude, and the phase space supplies one more threshold factor.

The physical explanation is helicity suppression. In the pion rest frame the two momenta are opposite. The massless antineutrino has positive helicity; zero total angular momentum would require the electron to have positive helicity too. The charged weak current instead creates a left-chiral electron, whose helicity is negative when its mass vanishes. A nonzero electron mass admits the required opposite helicity with an amplitude proportional to $m_e$. The decay amplitude vanishes for $m_e=0$, and the rate is suppressed by $m_e^2$.

Use natural units, assume negligible neutrino chemical potentials, and approximate neutrino decoupling as occurring before electron-positron annihilation in cosmology. At high temperature, weak interactions maintain a common temperature for neutrinos, antineutrinos, Electrons, Positrons and photons. Since ${\Gamma }_{\nu }/H\propto T^3$, the reaction rate eventually loses against expansion. Decoupled relativistic neutrinos then free stream; their momenta redshift as $a^{-1}$, so their distribution retains a temperature parameter satisfying $a{T}_{\nu }=\mathrm{constant}$.

The still-coupled electromagnetic bath receives the entropy of annihilating Electrons and Positrons. Its effective number of relativistic entropy degrees of freedom is initiallywhere the two photon polarizations count as bosonic states and the four electron-positron spin states have the fermion entropy weight $7/8$. After the pairs disappear, $g_{\ast s}^{\mathrm{EM}}=2$. Cosmological entropy conservation in this coupled bath gives $g_{\ast s}^{\mathrm{EM}}(a{T}_{\gamma }{)}^3=\mathrm{constant}$. The three already decoupled neutrino species are excluded from this bath's entropy count. At decoupling $T_{\nu }=T_{\gamma }$, henceSubsequent expansion preserves this ratio in the idealized history. Residual interactions during annihilation give small corrections to the instantaneous-decoupling approximation. If a relic species later becomes massive, $T_{\nu }$ here remains the redshifting momentum-distribution parameter; it need not be its nonrelativistic kinetic temperature.

A single ordinary neutrino species includes one populated neutrino helicity and one antineutrino helicity. For negligible chemical potential, integration of the Fermi-Dirac distribution givesTherefore the number density of a relativistic neutrino relative to photons, with the antiparticle included, isThe factor $3/4$ is the fermionic number-density weight, distinct from the energy and entropy weight $7/8$. This relation continues to hold after this species becomes nonrelativistic if its comoving particle number is conserved.

Under the proposed matter-dominated, spatially flat model, the Friedmann equation requires the present relic-neutrino energy density to be approximately the critical density. Using the unreduced Planck mass $M_P=G^{-1/2}$ givesThus the critical-density mass of a thermal relic neutrino isTaking the paper's rounded values, $T_{\gamma 0}=2\times 10^{-13}\mathrm{GeV}$, $H_0=h/(4.5\times 10^{41}{\mathrm{GeV}}^{-1})$ and $M_P\sim 10^{19}\mathrm{GeV}$, givesFor $0.5<h<1$, this is of order $30$--$100\mathrm{eV}$ with these approximate constants. More precise constants change the coefficient, not the requested mass scale. This mass is much larger than the present momentum scale $T_{\nu 0}\simeq 1.4\times 10^{-4}\mathrm{eV}$, validating the nonrelativistic energy density approximation. The result assumes that one thermal relic species dominates the matter density; it is conditional on that model rather than a measurement of a neutrino mass.

Let $t_{\ast }$ be the observation time and normalize $a(t_{\ast })=1$ temporarily. In pure radiation domination, $a(t)=(t/t_{\ast }{)}^{1/2}$ and $H_{\ast }=1/(2t_{\ast })$. With $p(t)=T_r(t)=T_{\ast }/a(t)$, the proper speed is $v=[1+(ma/T_{\ast }{)}^2{]}^{-1/2}$. Put $u=\sqrt{t/t_{\ast }}=a(t)$ in the preceding integral. Since $dt=2t_{\ast }udu$,The small-$m/T_{\ast }$ limit is $H_{\ast }^{-1}$, the radiation-era particle horizon. For $m/T_{\ast }\gg 1$, the ratio is approximately $(T_{\ast }/m)\log (2m/T_{\ast })$, retaining the distance accumulated while the particle was faster.

There is a normalization qualification: in arbitrary fixed comoving coordinates the distance is $\chi =H_{\ast }^{-1}\mathrm{arsinh}(y)/(a_{\ast }y)$, $y=m/T_{\ast }$. The expression just derived is also the physical distance $L_{\ast }=a_{\ast }\chi$ at observation. Thus it equals a comoving distance only when $a_{\ast }=1$ is used, as in the preceding part. To express it in today's convention $a_0=1$, multiply $L_{\ast }$ by $1+z_{\ast }$. This is the accumulated free-streaming distance in a radiation-dominated universe.

Write ${\omega }_M={\Omega }_M{h}^2$, where $h$ here is the dimensionless present Hubble constant. The supplied equality redshift gives $1+z_{\mathrm{eq}}\simeq 2.5\times 10^4{\omega }_M$, and $T_{\mathrm{eq}}=T_0(1+z_{\mathrm{eq}})$. For $m=30\mathrm{eV}$ the requested dimensionless ratio isThe Hubble rate in this formula is the radiation-era extrapolation used in deriving it. A unique numerical ratio cannot be chosen without ${\omega }_M$.

The printed CMB temperature contains a factor-of-ten error. Its $2.35\mathrm{meV}$ means $2.35\times 10^{-3}\mathrm{eV}$, whereas the cosmic microwave background temperature in energy units is about $2.35\times 10^{-4}\mathrm{eV}=0.235\mathrm{meV}$, corresponding to $2.725\mathrm{K}$. The temperature is supported by NASA's CMB measurements. Using the printed value gives $y_{\mathrm{eq}}\simeq 0.5106/{\omega }_M$; using the corrected value gives $y_{\mathrm{eq}}\simeq 5.106/{\omega }_M$. For the illustrative choice ${\Omega }_M=1,h=0.5$, so ${\omega }_M=0.25$, these give respectively $T_{\mathrm{eq}}\simeq 14.7$ and $1.47\mathrm{eV}$, and distance-to-Hubble-radius ratios about $0.716$ and $0.182$. This benchmark is an assumption for the estimate, not an extra parameter supplied by the paper.

For the present-day comoving length, use $H_r(z_{\mathrm{eq}})=H_0\sqrt{{\Omega }_M}(1+z_{\mathrm{eq}}{)}^{3/2}$, the radiation contribution extrapolated to equality. Restoring $c$ for the conversion and using $c/H_0\simeq 2998h^{-1}\mathrm{Mpc}$ givesFor the same benchmark, this is about $54\mathrm{Mpc}$ using the printed temperature and $14\mathrm{Mpc}$ using the corrected temperature, in the prescribed representative-momentum radiation-era model.

At equality matter is no longer negligible, so these are transition-era estimates. The actual total $H_{\mathrm{eq}}$ is $\sqrt{2}{H}_r(z_{\mathrm{eq}})$; inserting it in the simple normalization would lower the length by $\sqrt{2}$, but would not constitute an exact integration through equality. An exact matter-plus-radiation treatment instead replaces the radiation integral bywith $p_0=T_0$ in the assumed model. This makes the order-one equality correction explicit. Real neutrino physical momenta in an FRW universe are distributed and their temperature need not equal the photon temperature; the calculation above follows the specified $p=T_r$ idealization.

In the pion rest frame, conservation of three-momentum makes the Electron and massless neutrino momenta equal and opposite, of common magnitude $p$. Their energies obeySquaring $E_e=m_{\pi }{c}^2-pc$ and cancelling $p^2{c}^2$ givesThe Electron four-momentum satisfies $p=E_{e}v/c^2$. Thus, for the physically allowed decay $m_{\pi }>m_e$,

Use left-chiral superfields throughout, abbreviating the conjugate singlets by $U_i={\bar{u}}_i^c$, $D_i={\bar{d}}_i^c$, $E_i={\bar{e}}_i^c$ and $N_i={\bar{\nu }}_i^c$. The printed hypercharge convention is the negative of the usual one, so $Q_{\mathrm{em}}=T_3-Y$; consequently $H_1$ is the down-type and $H_2$ the up-type Higgs doublet. Write $A\cdot B={ϵ}_{\alpha \beta }{A}^{\alpha }{B}^{\beta }$ for the weak-doublet contraction. Assign baryon number $B(Q)=1/3$, $B(U)=B(D)=-1/3$, and lepton number $L(L)=1$, $L(E)=L(N)=-1$; both Higgs fields have $B=L=0$.

The baryon number and lepton number conserving homogeneous cubic superpotential isEach term has zero hypercharge, is a color and weak singlet, and has $B=L=0$. Higgs expectation values give the usual quark and charged-lepton Yukawa interaction masses and, here, Dirac neutrino masses. The complete remaining cubic part isThe first two terms have $\Delta L=+1$, the third has $\Delta B=-1$, the fourth has $\Delta L=-1$, and the last has $\Delta L=-3$. Their Hermitian conjugates reverse these signs. Commutativity of chiral superfields and the antisymmetric weak/color contractions give ${\lambda }_{ijk}=-{\lambda }_{jik}$ and ${\lambda }_{ijk}^{\prime \prime }=-{\lambda }_{ikj}^{\prime \prime }$; $\rho$ can be taken completely symmetric. All other coupling tensors are unrestricted complex tensors.

To check completeness, a cubic color singlet containing a quark doublet and an antiquark singlet must contain a colorless weak doublet; the allowed hypercharges give $Q{H}_{2}U$, $Q{H}_{1}D$ and $QLD$. Three color-antitriplet singlets can form a singlet with ${ϵ}^{abc}$, but hypercharge permits only $UDD$. Three quark doublets cannot be a weak singlet. With no colored fields, an even number of weak doublets is necessary. Two doublets and a singlet give precisely $L{H}_{1}E$, $LLE$, $L{H}_{2}N$ and $H_1{H}_{2}N$; contractions of two identical Higgs superfields vanish. With no doublets the only neutral triple is $NNN$. Thus $W_3=W_{3,\mathrm{cons}}+W_{3,\mathrm{viol}}$ is the most general homogeneous cubic superpotential for the printed fields, including their singlet neutrinos.

If “cubic” is intended to mean a polynomial of degree at most three, also addThe $\mu$ term conserves both charges; the other nonconstant terms have respectively $\Delta L=1,-2,-1$. This distinction will matter for the printed claim about R-parity.

For squark-mediated proton decay, choose ${\lambda }_{11k}^{\prime }$ and ${\lambda }_{11k}^{\prime \prime }$ with $k=2$ or $3$. The same down-type singlet squark can appear in both vertices; $k=1$ would give ${\lambda }_{111}^{\prime \prime }=0$ by antisymmetry. The first term includes an electron/up-quark/squark vertex, and the second an up-quark/down-quark/squark vertex. Exchanging a squark of mass $M$ and expanding its propagator at momenta much smaller than $M$ produces the dimension-six effective operatorThe left- and right-handed pairs are separately contracted spinor bilinears. This operator converts an up/down quark pair into a positron and an antiquark; the spectator up quark can combine with the antiquark into the neutral pion. Equivalently its hadronic matrix element between a proton and a pion is permitted, so it induces $p\to e^{+}{\pi }^0$. Having both the baryon number and lepton number violating vertices is what enables this channel.

Since $C$ has mass dimension $-2$, dimensional analysis givesThe dimensionless $c_h$ includes the hadronic matrix element and two-body phase space; the pion mass changes this factor rather than the leading dimensions. Dimensional grounds alone do not give a precise hadronic coefficient. Using the lifetime supplied in the paper, rather than substituting a different bound, gives $\Gamma <1/(2.4\times 10^{64})\mathrm{GeV}=4.17\times 10^{-65}\mathrm{GeV}$ and henceFor an illustrative $100\mathrm{GeV}$ squark this is about $6.5\times 10^{-29}{c}_h^{-1/2}$ when $m_p$ is rounded to $1\mathrm{GeV}$. The bound constrains the relevant flavour product and depends on the exchanged mass; separate smallness of each coupling does not follow without an additional assumption.

Define matter parity and R-parity byEvery matter superfield $Q,U,D,L,E,N$ has $P_M=-1$, whereas $H_1,H_2$ and the vector superfields have $P_M=+1$. For ordinary quarks and antiquarks, $3(B-L)=\pm 1$ is odd and $2s=1$, so $R_p=+1$. For charged leptons, neutrinos and their antiparticles, $3(B-L)=\mp 3$ is odd and $2s=1$, again giving $R_p=+1$; the singlet neutrino fermion obeys the same rule. Gluons, the photon, $W$ and $Z$ have $B=L=0$ and spin one, hence $R_p=+1$; Higgs bosons have $B=L=s=0$ and also $R_p=+1$. Thus every Standard Model particle is R-even, including the corresponding antiparticles. Changing spin by one half within a supermultiplet reverses $R_p$, so squarks, sleptons, gauginos and Higgsinos are R-odd.

Each conserving cubic term contains two R-odd matter superfields and one even Higgs superfield, so it is even under matter parity. Each term in $W_{3,\mathrm{viol}}$ has an odd number of matter superfields: three for $LLE,LQD,UDD,NNN$, and one for $N{H}_1{H}_2$. All are therefore forbidden. At the component level the two fermion spin factors in a Yukawa interaction multiply to $+1$, so the same selection rule follows directly from R-parity. This verifies all cubic baryon/lepton-number violating terms are forbidden while the Yukawa fermion mass terms survive.

The blanket assertion needs a qualification beyond homogeneous cubic terms. The Higgs $\mu$ term is even; $L{H}_2$ and $N$ are odd and forbidden. But $N_i{N}_j$ is even and permits a Majorana neutrino mass despite violating lepton number by two. Likewise the higher-dimensional operators $QQQL$ and $(L{H}_2{)}^2$ are even. Consequently R-parity alone does not enforce all-order conservation of baryon number and lepton number, or absolute proton stability; the stated precise exclusion holds for the violating cubic interactions above.

Two physical implications follow from multiplicative conservation of R-parity. Superpartners are produced in even numbers from an initial state consisting only of Standard Model particles, leading to pair production at the lowest threshold and parity-conserving cascades. The lightest supersymmetric particle is stable, since one R-odd particle cannot decay exclusively into R-even particles and no lighter R-odd state is available. A neutral weakly interacting lightest supersymmetric particle can therefore be a dark-matter candidate and produce missing-energy signatures; neutrality and a suitable abundance require further model assumptions.

Use units with $c=1$. For the flat FRW metric, the Friedmann equation is $H^2=8\pi G\rho /3+\Lambda /3$, where $H=\dot{a}/a$ is the Hubble parameter. The cosmological continuity equation for a separately conserved component with constant equation of state $P=w\rho$ gives $\rho \propto a^{-3(1+w)}$. Consequently radiation in cosmology scales as $a^{-4}$, pressureless matter as $a^{-3}$, and cosmological constant energy is constant. With $a_0=1$, define the present critical density and cosmological density parameters byThe Hubble constant $H_0$ measures the expansion rate today. Here ${\Omega }_r$ includes photons and relativistic neutrinos, ${\Omega }_m$ includes baryons and cold dark matter, and ${\Omega }_{\Lambda }$ measures the vacuum contribution. Spatial flatness implies ${\Omega }_r+{\Omega }_m+{\Omega }_{\Lambda }=1$. ThusSince conformal time satisfies $d\tau =dt/a$, we have $a^{\prime }=a\dot{a}=a^{2}H$. Multiplication by $a^4$ provesFor an exam-level estimate, $H_0=100h\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}$ with $h\simeq 0.7$, ${\Omega }_m\simeq 0.3$, ${\Omega }_{\Lambda }\simeq 0.7$, and the specified radiation in cosmology gives ${\Omega }_r=4.2\times 10^{-5}{h}^{-2}\simeq 8.6\times 10^{-5}$. These are rounded illustrative flat-universe values; precision cosmological density parameters depend on the fitted model. For comparison, the base flat-model fit in Planck 2018 cosmological parameters gives $H_0\simeq 67.4\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}$ and ${\Omega }_m\simeq 0.315$.

Before electron-positron annihilation in cosmology, photons contribute two helicity states; Electrons and positrons contribute two spin states each. Each of the three light neutrino flavours contributes one active neutrino helicity and one antineutrino helicity. At a common temperature the effective number of relativistic energy degrees of freedom is thereforeUsing $G_F=1.17\times 10^{-5}{\mathrm{GeV}}^{-2}$ and $M_{\mathrm{Pl}}\simeq 2.435\times 10^{18}\mathrm{GeV}$ in the scaling formula alone gives $T_F\sim 2\mathrm{MeV}$. Including the explicit expansion coefficient from (ii), but setting the unknown rate coefficient $A$ to one, gives $T_F\simeq 1.5\mathrm{MeV}$. Thus the warranted conclusion is a freeze-out temperature of order $1\mathrm{MeV}$, above the $0.511\mathrm{MeV}$ Electron mass. The omitted weak-rate coefficient prevents a more precise prediction from this estimate.

Neutrino decoupling occurs before most electron-positron pairs annihilate. Afterwards neutrinos redshift freely, $T_{\nu }a=\mathrm{constant}$, while the photons receive the pair entropy. For the still-coupled photon-pair gas, cosmological entropy conservation gives $g_{\ast s}{T}_{\gamma }^3{a}^3=\mathrm{constant}$. Its entropy degrees of freedom drop from $2+(7/8)4=11/2$ to $2$. Comparing this with the freely cooling neutrinos givesAfter annihilation, photon and neutrino temperatures differ. To write the total energy density as $({\pi }^2/30)g_{\ast }{T}_{\gamma }^4$, the neutrino contribution must include that temperature ratio:This is the paper's post-annihilation ${\mathcal{N}}_{\mathrm{eff}}$. Merely deleting the Electron term while keeping all species at the same temperature would be incorrect. For comparison the entropy count, normalized to $T_{\gamma }^3$, is $g_{\ast s}=2+(7/8)6(4/11)=43/11\simeq 3.91$. These results assume instantaneous neutrino decoupling and relativistic neutrinos; the two counts represent different thermal moments.