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Take the muon mass to be $m_{\mu }$, and use the quantum electrodynamics convention ${\mathcal{L}}_{\mathrm{int}}=-e\bar{\psi }{\gamma }^{\mu }\psi A_{\mu }$ with $e>0$. Let $p_1,p_2$ be the incoming muon and antimuon momenta and $k_1,k_2$ the outgoing photon momenta, so $p_1+p_2=k_1+k_2$. External particles are on shell: $p_1^2=p_2^2=m_{\mu }^2$ and $k_1^2=k_2^2=0$. The two tree-level Feynman diagrams are the two orders in which the photon legs attach to the muon line. There is no three-photon QED vertex, and therefore no single-photon annihilation diagram into two photons at this order.

The QED Feynman rules needed here are the vertex $-ie{\gamma }^{\mu }$, the Dirac propagatoran incoming muon spinor $u(p_1)$, an incoming antimuon adjoint $\bar{v}(p_2)$, and an outgoing photon polarization ${ϵ}_{\mu }^{\ast }(k)$. At a vertex enforce four-momentum conservation. Spinor products are ordered along the fermion line; exchanging the external photons introduces no relative minus sign. Define $R(q)=(/q+m_{\mu })/(q^2-m_{\mu }^2+i0)$ and $/ϵ={\gamma }^{\mu }{ϵ}_{\mu }$.

With the convention that a diagram contributes $i\mathcal{M}$, the two muon-antimuon annihilation amplitudes areTheir denominators are respectively $t-m_{\mu }^2+i0$ and $u-m_{\mu }^2+i0$, where $t=(p_1-k_1{)}^2$, $u=(p_1-k_2{)}^2$. A different consistent overall amplitude-phase convention has no physical effect.

A physical photon polarization vector obeys $k\cdot ϵ=0$, and represents an equivalence class $ϵ\sim ϵ+\alpha k$. For a real null momentum, adding $\alpha k$ preserves transversality. One can additionally choose transverse spatial vectors with ${ϵ}^0=0$ and ${ϵ}^{\ast }\cdot ϵ=-1$; there are two independent physical polarizations. The equivalence class removes the unphysical longitudinal direction, rather than imposing four independent physical polarization states.

The invariance of the total amplitude is a two-photon fermion Ward identity. Replace ${ϵ}_1^{\ast }$ by $k_1$ and put $q_t=p_1-k_1$, $q_u=p_1-k_2=k_1-p_2$. The external Dirac equations giveAway from propagator poles, $R(q)(/q-m_{\mu })=(/q-m_{\mu })R(q)=I$; the identity extends with the common Feynman prescription. ThereforeThe two terms in the bracket then give $-\bar{v}/{ϵ}_2^{\ast }u$ and $+\bar{v}/{ϵ}_2^{\ast }u$, which cancel. Exchanging photon labels proves the second Ward identity. Thus the sum is unchanged under ${ϵ}_i\mapsto {ϵ}_i+\alpha k_i$ for either photon. Neither diagram is generally gauge-invariant separately. Physically, longitudinal pure-gauge polarization does not couple to the observable scattering amplitude; only the two transverse photon degrees of freedom contribute.

For Compton scattering write the incoming momenta as $p,k$ and outgoing momenta as $p^{\prime },k^{\prime }$, with $p+k=p^{\prime }+k^{\prime }$. The outgoing muon contributes $\bar{u}(p^{\prime })$, the incoming photon contributes $ϵ(k)$ without conjugation, and the outgoing photon contributes ${ϵ}^{\prime \ast }(k^{\prime })$. The two tree-level Compton amplitudes areThese are the lower two diagrams. Their intermediate muon momenta are $p+k$ and $p-k^{\prime }$, with denominators $s-m_{\mu }^2+i0$ and $u-m_{\mu }^2+i0$. The first attaches the incoming photon before the outgoing photon along fermion flow; the second reverses that attachment order. Both contribute at order $e^2$ in the amplitude.

Compton scattering has one incoming and one outgoing muon and one photon on each side; annihilation has an incoming particle-antiparticle pair and two outgoing photons. Accordingly the external spinors/polarizations and physical Mandelstam channels differ, although both arise from the same two-vertex QED matrix element. Crossing symmetry relates them by continuing $p_2\mapsto -p^{\prime }$ and one outgoing photon momentum $k_1\mapsto -k$, with the corresponding external wavefunction replacements. The annihilation $t$ channel becomes the Compton $s$ channel and the other channel becomes its $u$ channel. Identical final photons require a factor $1/2!$ in an annihilation phase-space integral over an otherwise double-counted full final-state region; there is no analogous final-state identical-particle factor for the muon-photon Compton final state. Gauge cancellation holds for the summed Compton amplitude as well.

At weak freeze-out, photons, electrons, positrons and all four neutrino flavors are relativistic and approximately at the same temperature. The effective number of relativistic energy degrees of freedom changes fromThe expansion rate at fixed temperature is larger by $\sqrt{50/43}\simeq 1.078$. Equating the supplied weak interaction time with the supplied age gives $T_f^3=\sqrt{g_{\ast }}$ when $T_f$ is measured in MeV. ThereforeEarlier cosmological weak freeze-out at a higher temperature leaves more neutrons. With negligible chemical potentials, the neutron-to-proton ratio in equilibrium is $r_f=\exp (-Q/T_f)$, where $Q=m_n-m_p\simeq 1.293\mathrm{MeV}$. If almost all surviving neutrons enter helium-4, the primordial helium mass fraction before neutron decay is $Y_f=2r_f/(1+r_f)$. Literal use of the given rate normalizations yieldsEquivalently, the leading response is $\delta Y_f/Y_f=[Q/(T_f(1+r_f))]\delta T_f/T_f$. This is a roughly one-percentage-point absolute increase in the helium fraction, rather than a one-percent relative increase.

The deuterium bottleneck delays nuclear burning, allowing free-neutron decay after freeze-out. If its end temperature is $T_N$, and ${\tau }_n$ is the neutron mean life, a sharper instantaneous-freeze-out estimate conserves total baryon number and givesThis formula evolves the neutron fraction, not just the neutron-to-proton ratio, because each neutron decay also creates a proton. After electron annihilation, let $r_T=(4/11{)}^{1/3}$. The radiation counts relevant to the age at fixed $T_N$ areConsequently $t_N^{(4)}/t_N^{(3)}\simeq \sqrt{3.36/3.82}\simeq 0.939$. Less time is available for neutron decay, adding to the helium increase. As an explicit illustrative completion, $T_N=0.07\mathrm{MeV}$ and ${\tau }_n\simeq 880\mathrm{s}$ give an increase of approximately $0.012$ in this toy model. The precise helium fraction requires the nuclear-burning temperature, baryon abundance and properly normalized weak rates; the question supplies only approximate time laws. In particular, the large no-decay fractions in the table must not be interpreted as precise predictions for the observed primordial abundance. The robust effect is extra relativistic energy raises the primordial helium fraction, with an absolute change of order $10^{-2}$ for one thermally populated flavor.

Before electron-positron annihilation in cosmology, the pair has two spin states, just as photons do, and the fermionic number integral gives $n_{\nu +\bar{\nu }}/n_{\gamma }=3/4$ at equal temperature. The factor follows by expanding $(e^x+1{)}^{-1}$ into an alternating exponential series: its $x^2$ integral is $3/4$ of the bosonic one.

After neutrino decoupling, cosmological entropy conservation in the photon-electron plasma changes its relativistic entropy count from $2+(7/8)4=11/2$ to $2$. The decoupled neutrinos do not share this heating. HenceThese counts include both the new neutrino and its antineutrino; another factor of two would double count them. This is the number density of a relativistic neutrino relative to photons, evaluated after photon heating.

They are nonrelativistic today, so the relic-neutrino energy density is dominated by rest mass:The number follows from the supplied critical density, without a Hubble-unit convention. It is conditional on the thermal history stated above; a sterile, colder or nonthermally produced state need not have this abundance.

Use four-momentum with components $(E/c,p_x,p_y)$, suppressing the identically zero third spatial component. The photon in the primed frame hasIts energy is $h{\nu }^{\prime }$ and its spatial momentum has magnitude $h{\nu }^{\prime }/c$. Apply the same Lorentz transformation as for the spacetime coordinates. With $\beta =v/c$,The time component gives the relativistic Doppler effectThe aberration of light follows by dividing spatial components by $E/c$:In particularwhere the tangent expression is understood only when its denominators exist. The sine and cosine equations determine the correct quadrant and also cover directions where the tangent/secant expression is undefined.

For the photon-number half-cone under relativistic aberration, the forward half of the primed emission angles is $-\pi /2\le {\theta }^{\prime }\le \pi /2$. The aberration of light is monotone on this interval, and its endpoints have $\cos \theta =\beta$. Thus exactly half of a uniform angular photon distribution lies within the lab-frame coneAs $v\to c^{-}$, $\alpha =\arcsin (1/\gamma )\sim 1/\gamma$, so the cone becomes narrow. Here “half” counts emitted photons or ray directions; it is not a claim that half of the observed energy or power lies there, since the relativistic Doppler effect changes the photon energies with angle.

Assume $\lambda >0$ and $v>0$. The vacuum manifold is the sphere $|\varphi |=v$. Choose ${\varphi }_0=v{e}_3$. For the given generators, $t^3{\varphi }_0=0$ whereas $t^1{\varphi }_0$ and $t^2{\varphi }_0$ are nonzero and independent. The Adjoint representation of SU(2) rotates this sphere, and the full connected stabilizer subgroup of $v{e}_3$ is generated by $t^3$. Hence the classical symmetry pattern isThe central element of SU(2) acts trivially on the triplet and belongs to this stabilizer; it does not add a separate broken direction. The gauge-field-free representative has zero gauge field strength and constant scalar magnitude. Locally near this nonzero classical vacuum, unitary gauge aligns the triplet along the third axis.

First make the gauge normalization explicit. Write the gauge kinetic term as $-\mathcal{N}{F}_{\mu \nu }^a{F}^{a\mu \nu }/4$, where $\mathrm{Tr}(t^a{t}^b)=\mathcal{N}{\delta }^{ab}$ if $F=F^a{t}^a$ is the matrix used inside the printed trace. Canonical normalization of a gauge kinetic term givesFor a conventionally normalized component trace, $\mathcal{N}=1$ and $g_c=g$. If the trace is the ordinary matrix trace in the displayed three-dimensional representation, $\mathrm{Tr}(t^a{t}^b)=2{\delta }^{ab}$ and $\mathcal{N}=2$. The PDF does not specify which trace convention is intended; both are covered by this formula.

In unitary gauge, the gauge covariant derivative isExpanding this kinetic term and the scalar potential gives the complete physical-field LagrangianHere a square of a vector means its contraction with the Minkowski metric, in signature $(+---)$. The labels 1 and 2 represent massive vectors; 3 represents the surviving Abelian vector.

To display their charge and all interactions more clearly, put $a_{\mu }={\mathcal{A}}_{\mu }^3$, $W_{\mu }^{\pm }=({\mathcal{A}}_{\mu }^1\mp i{\mathcal{A}}_{\mu }^2)/\sqrt{2}$, and defineWith the sign of gauge field strength printed in the paper, ${\mathcal{F}}^3=f+i{g}_{c}C$ and $G^{\pm }=({\mathcal{F}}^1\mp i{\mathcal{F}}^2)/\sqrt{2}$. Thus the same Lagrangian becomesThe physical masses areIn particular, literal adjoint matrix trace gives $m_W=gv/\sqrt{2}$; the standard canonical component convention gives $m_W=gv$. These are descriptions with differently normalized couplings, not different physical spectra.

The complex vector pair carries opposite charges under the unbroken U(1) gauge symmetry. The gauge field strength terms contain $a{W}^{+}{W}^{-}$ cubic interactions, $aa{W}^{+}{W}^{-}$ quartic interactions, and four-vector interactions involving the charged fields. There is no pure Abelian cubic or quartic self-interaction. The Higgs mode $\eta$ has cubic and quartic scalar interactions and couples through $2g_c^{2}v\eta W^{+}{W}^{-}+g_c^2{\eta }^2{W}^{+}{W}^{-}$. It is neutral and has no tree-level $\eta aa$ interaction. This is the physical charged-vector Lagrangian for an adjoint SU2 Higgs model.

If the angular fields ${\pi }_1$ and ${\pi }_2$ were retained, their vanishing potential masses would identify the two Goldstone bosons of the ungauged triplet. In the gauge theory they mix with the broken-direction gauge fields and can be removed by gauge fixing; the Higgs mechanism uses them as the longitudinal polarizations of the two massive vectors. They are not additional physical massless scalars. The physical degrees of freedom are conserved: $3+3\times 2=9$ before rearrangement, and $1+2\times 3+2=9$ afterwards.

This is not the Standard Model electroweak interaction. It has three original gauge bosons, leaving two massive charged vectors and one massless neutral vector, with no massive neutral Z boson. The Standard Model instead has $SU(2{)}_L\times U(1{)}_Y$, a complex Higgs doublet, and three massive vectors plus the photon. Its charge is $Q=T_3+Y$, not just the surviving $T_3$. Adding fermions cannot supply the missing gauge generator or turn the surviving neutral vector into both a photon and a Z boson. In particular, the usual right-handed fermions in the Standard Model are SU(2) singlets; they would have zero charge if only $T_3$ were available, instead of the charges produced by hypercharge.

Let $s=q^2>0$, neglect $m_e$, and average over the two spin states of each incoming particle. Only the single-photon channel is being considered. The spectral-density symbol ${\rho }_h$ is not defined in the question, so specify its normalization before calculating. A useful photon-field convention isHere $d{\Pi }_X$ is the product of final-particle measures $d^{3}k/[(2\pi {)}^{3}2E]$, including the appropriate sums and symmetry factors. Lorentz invariance and current conservation give the transverse tensor form. In this convention ${\rho }_h$ is the hadronic photon spectral density, with mass dimension $-2$, rather than a density of a stationary time series.

The electron annihilation amplitude can be written $e\bar{v}(p_{+}){\gamma }_{\mu }u(p_{-})⟨X|A^{\mu }|0⟩$, up to an overall phase. The leptonic tensor after the initial spin average isIt obeys $q^{\mu }{L}_{\mu \nu }=0$ and $g^{\mu \nu }{L}_{\mu \nu }=-s$. In the center-of-momentum frame, the flux denominator in the hint is $2s$, since the relative speed is 2 and each beam energy is $\sqrt{s}/2$. Contracting the tensors gives the inclusive relativistic cross-sectionHere $\alpha$ is the fine-structure constant. The $1/4$ in the initial spin average is essential because the hint's particles were spinless.

For the other common convention, define the hadronic electromagnetic current with the coupling omitted, $J_h^{\mu }={\sum }_f{Q}_f{\bar{q}}_f{\gamma }^{\mu }{q}_f$, and write its inclusive tensor asEquivalently, for $i\int d^{4}x{e}^{iqx}⟨0|T{J}_h^{\mu }(x)J_h^{\nu }(0)|0⟩=(q^{\mu }{q}^{\nu }-q^2{g}^{\mu \nu }){\Pi }_J(q^2)$, one has ${\rho }_J=\mathrm{Im}{\Pi }_J/\pi$. At leading order in the electromagnetic interaction, $⟨X|A^{\mu }|0⟩$ is proportional to $e⟨X|J_h^{\mu }|0⟩/s$. ThusIf the symbol ${\rho }_h$ is instead used for this dimensionless hadronic electromagnetic-current spectral density, the last formula applies with ${\rho }_J$ renamed ${\rho }_h$. The normalization must not be silently switched between these formulas.

At $\sqrt{s}$ well above the strong-coupling scale of Quantum chromodynamics, asymptotic freedom makes production over distances of order $1/\sqrt{s}$ perturbative. The electromagnetic current initially creates a quark-antiquark pair. Subsequent strong interactions produce hadrons, but an inclusive sum over all hadronic final states is much less sensitive to this rearrangement than an exclusive channel. This is the regime in which quark-hadron duality motivates a leading parton model calculation, with radiative and power-suppressed corrections. It is not a pointwise theorem at individual resonances or near thresholds; the comparison is most reliable for sufficiently inclusive or suitably averaged high-energy observables. This argument remains restricted to photon exchange, even where additional electroweak channels could also contribute.

For one active massless quark flavor with charge $Q_{f}e$, the tree-level scattering amplitude isThe two gamma-matrix traces, the initial spin average, and the final color charge sum yieldwhere $t=-s(1-\cos \theta )/2$ and $u=-s(1+\cos \theta )/2$. The massless two-body Lorentz-invariant phase space then givesSumming the distinct final flavors, rather than interfering amplitudes for them, givesAccording to the permitted approximation in the paper, active flavors satisfy $m_f^2<s$ and are treated as massless; the others are omitted. This is the stipulated step approximation, not the exact pair-production threshold $s\ge 4m_f^2$.

The hadronic R ratio is $R=N_c\sum Q_f^2$, relative to the massless muon-pair relativistic cross-section $4\pi {\alpha }^2/(3s)$. If $N_u$ and $N_d$ active flavors have charges $2/3$ and $-1/3$, respectively,For $N_c=3$, the usual sets of three, four, five and six active flavors give $R=2,10/3,11/3,5$. In the two spectral conventions the same leading calculation gives

Use conformal time, $dt=ad\tau$, and let primes mean $d/d\tau$. Then $a^{\prime }=a\dot{a}$ and $a^{\prime \prime }=a{\dot{a}}^2+a^2\ddot{a}$. For $k=1$, $\Lambda =0$, the Friedmann equation and acceleration equation giveSubstitution yieldsUse the supplied solution without deriving it. Its first maximum has sine equal to one, so its amplitude must equal the physical turnaround scale in part (a):Choose the expanding branch with $B=0$ and the first zero at $\tau =0$.

For pressureless matter, $w=0$, the closed matter-dominated Friedmann solution isThe Big Crunch is at ${\tau }_c=2\pi$, or proper time $t_c=\pi A=\pi {\rho }_0/3$, measured from the Big Bang.

For radiation, $w=1/3$, the closed radiation-dominated Friedmann solution isThe Big Crunch is at ${\tau }_c=\pi$, or $t_c=2A$.

A radial null geodesic in FLRW spacetime obeys $d\chi /d\tau =\pm 1$. The spatial slice is a unit three-sphere times $a$, so a great circle has comoving circumference $2\pi$. The dust lifetime supplies conformal distance $2\pi$: one circumference, with the return occurring only in the crunch limit. Strictly before the singular endpoint no full return is completed. The radiation lifetime supplies conformal distance $\pi$: the photon reaches the antipode, half a great circle, in the crunch limit. These are limiting null rays from near the initial singularity, not photons at a regular event on the singular surface.

For the reaction $p+e\leftrightarrow H+\gamma$, chemical equilibrium requires ${\mu }_p+{\mu }_e={\mu }_H$ because thermal photons have zero chemical potential. Insert the nonrelativistic Maxwell-Boltzmann distribution number densities and use $m_H=m_p+m_e-\mathcal{B}$:For ground-state hydrogen, including its spin states, $g_e=g_p=2$, $g_H=4$; hence the degeneracy factor is one. Since $m_p/m_H\simeq 1$, the right side is $(m_{e}T/2\pi {)}^{3/2}{e}^{-\mathcal{B}/T}$.

Charge neutrality gives $n_p=n_e$, while baryon conservation gives $n_b=n_p+n_H$. Consequently $n_e=n_p=X_e{n}_b$ and $n_H=(1-X_e)n_b$. Inverting the preceding ratio and inserting the photon number density with $n_b=\eta n_{\gamma }$ gives the hydrogen-only Saha equationWriting $y=\mathcal{B}/T$, its coefficient is $(2\zeta (3)/{\pi }^2)\eta (2\pi \mathcal{B}/m_e{)}^{3/2}\simeq 3.2\times 10^{-16}$, consistent with the rounded $3\times 10^{-16}$. The negligible mass-ratio correction and ground-state approximation are the assumptions behind this form.

At $X_e=1/2$, the left side of the Saha equation is two. With its rounded coefficient, the recombination temperature therefore satisfiesNeglecting the logarithmic prefactor gives $y\sim 36$ andThis is the requested rough scale. Retaining the prefactor and the unrounded logarithm gives $y\simeq 42.0$ and $T_{\mathrm{rec}}\simeq 0.32$ eV; the displayed $0.4$ eV is not a high-accuracy numerical root of the Saha equation.

The small baryon-to-photon ratio means that radiation and the entropy of the free charged particles strongly favor ionization. A low mean photon energy does not eliminate the energetic tail, and ionized particles have a large phase-space advantage. The exponential binding factor must become very large before neutral atoms dominate this dilute gas. Accordingly $\mathcal{B}/T$ is of order forty rather than order one, so recombination occurs well below the hydrogen binding energy.

Photon decoupling is a dynamical loss of frequent scattering, distinct from the chemical conversion of charged particles into neutral atoms. Before recombination, Thomson scattering on free Electrons keeps photons coupled to the baryon plasma. The physical scattering rate isin units with $c=1$. Define the approximate decoupling temperature byEquivalently the photon mean scattering time becomes comparable to an expansion time. A refined last-scattering definition uses the optical depth and the maximum of the visibility function; it need not coincide exactly with the local rate criterion. As $X_e$ falls, scattering becomes inefficient, ordinarily after the half-ionized recombination stage, so $T_{\mathrm{dec}}<T_{\mathrm{rec}}$.

Residual electron freeze-out concerns the remaining ionized fraction. The Saha equation presumes sufficiently rapid forward and reverse reactions to maintain chemical equilibrium. Real recombination is slowed by radiative bottlenecks and expansion; the free-electron fraction can therefore exceed its equilibrium value. Eventually the effective recombination rate per free Electron, roughly ${\alpha }_{\mathrm{eff}}(T)n_e$, falls below $H$. The remaining Electrons cannot all find and recombine with protons within an expansion time. A small nonzero residual fraction survives while the Saha prediction decreases exponentially toward zero.

The sketch shows this lag and residual floor. Its kinetic curve and the position of the decoupling marker are schematic: the data supplied determine the equilibrium curve, but not a quantitative recombination history or an exact $T_{\mathrm{dec}}$. Determining those needs the expansion history and atomic transition rates. The horizontal axis decreases toward the right to follow cosmological cooling.