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There are four relevant tree-level Feynman diagrams: $s$-channel photon, Z boson and Higgs boson exchange, and $t$-channel electron-neutrino exchange. The electron and Positron are incoming, and $W^{-}$ and $W^{+}$ outgoing. The Higgs diagram is normally tiny because its electron coupling is proportional to $m_e$; it should nevertheless be included in the requested list.

Let $v=2m_W/g$. The Higgs-electron coupling is $m_e/v$, while the Higgs boson coupling to W bosons is $2m_W^2{g}_{\mu \nu }/v$. Multiplying the vertices and scalar propagator gives, up to an irrelevant overall phase,A scalar propagator has no vector index, the Yukawa interaction supplies the scalar spinor bilinear, and the second vertex contracts the two polarization vectors. We retain the tree propagator and neglect widths, consistently with the stated leading-order calculation.

For unpolarized incoming beams the spin average is $1/4$, not a sum without normalization. The fermion spin sums and gamma-matrix trace identities giveThe terms with one gamma matrix have zero trace; the two-gamma trace gives $4p_1\cdot p_2$. Both final polarization states are specified as longitudinal, so no final polarization average is appropriate.

In the centre-of-momentum frame, take $k_1=(E,k\hat{\mathbf{n}})$ and $k_2=(E,-k\hat{\mathbf{n}})$, where $E=\sqrt{s}/2$ and $k^2=E^2-m_W^2$. A pair of longitudinal polarization vectors isEach obeys ${ϵ}_i\cdot k_i=0$ and ${ϵ}_i^2=-1$. Their contraction is angle independent:Consequently

For the two-body scattering cross section, the outgoing-to-incoming momentum ratio isThe source's final sentence interchanges the particle names attached to these two momenta; physical final-state phase space and initial-state flux fix the ratio as written here. Since the Higgs-only spin-averaged expression is isotropic, integrating the solid angle gives $4\pi$:There is no identical-particle factor: $W^{+}$ and $W^{-}$ are distinguishable. The threshold factor ${\beta }_W$ makes this contribution vanish when the outgoing momentum vanishes. This is the isolated squared Higgs amplitude for a specified longitudinal pair, not the complete physical cross section: the full result squares the coherent sum of all four amplitudes and includes their interference.

A heated cathode emits electrons, accelerated through a potential difference onto a metal target. Deceleration produces a bremsstrahlung continuum with ${\lambda }_{\min }=hc/(e{V}_{\mathrm{acc}})$; inner-shell ionization followed by electronic relaxation produces characteristic lines. The pictured instrument is consistent with the copper-target TEL-X-Ometer used for teaching diffraction. Its copper $K_{\alpha }$ radiation has wavelength about $0.154\mathrm{nm}$, comparable with atomic plane spacings. A nickel filter preferentially suppresses copper $K_{\beta }$, while slits collimate the beam; filtering does not remove every continuum photon.

A Geiger-Müller counter converts photon-induced gas ionization into countable electrical pulses. A scintillator with photomultiplier is another detector option. A detector-arm scan records count rate against scattering angle; coupled crystal rotation preserves the usual reflection geometry. These instrument features are documented by its manufacturer.^[ref]

For the numerical calculations below use $\lambda =0.154\mathrm{nm}$, explicitly identifying the characteristic line assumed. The photograph alone does not give a measured wavelength or angular travel limit; conclusions depending on these parameters are stated conditionally. A different selected line changes the angles but not a crystallographic extinction.

For the L12 structure, the crystal basis structure factor isEqual parity of $h,k,l$ makes all three phase factors $+1$, henceThe electron scattering factors are evaluated at the reflection's scattering vector.

The Boltzmann distribution for two atomic levels at thermodynamic equilibrium iswhere $\omega$ is the statistical weight of an atomic level. For a resolved level of total angular momentum $J$, $\omega =2J+1$ when magnetic substates are unresolved.

To extend this counting to the continuum, assume an ideal, nondegenerate plasma with classical free electrons and neglect the small difference between the translational masses of the two heavy ionic species. Let $z_e=e^{{\mu }_e/kT}$ be the electron fugacity in the Saha equation, with the electron chemical potential measured relative to its rest energy. The two electron spin states and the phase-space cell volume $h^3$ giveChemical equilibrium for capture equates the bound species' chemical potential to the sum for its parent ion and free electron. Thus applying Boltzmann factors to a bound level at energy $-I_n$ gives $N_n/N_{+}=({\omega }_n/{\omega }_{+})z_e{e}^{I_n/kT}$. Eliminating the electron fugacity proves the level-specific Saha ionization equation:The factor two is the electron spin multiplicity; the factor with power $3/2$ is the inverse free-electron translational state density. A total-ionization Saha equation sums the bound states into an internal partition function, whereas this expression refers to one specified level.

For dielectronic recombination, write the entrance ion as $X^{(z+1)+}$ in ground state $i$, and the intermediate state of $X^{z+}$ as $d=(j,n\ell )$. Capture excites the core from $i$ to $j$ while binding the incident electron. The resonance energy relative to the entrance continuum is $\bar{E}=E_{ij}-I_{n\ell }>0$. The doubly excited state can eject the electron by autoionization, or undergo radiative stabilization to a bound state.

Let $A_{di}^a$ be the partial autoionization rate back to the entrance ion, $A_a$ the total autoionization rate, $A_r$ the total radiative decay rate, and $A_s$ the rate of radiative decays that stabilize into bound levels. For an isolated narrow resonance, the Saha-Boltzmann equilibrium ratio has the bound-state energy replaced by the positive resonance energy:At equilibrium, inverse dielectronic capture and autoionization fluxes balance: $N_e{N}_i{C}_d=N_d^{\mathrm{eq}}{A}_{di}^a$. This establishes the capture coefficient even when the actual plasma is not in local thermodynamic equilibrium. In the low-density, weak-radiation recombining plasma, the intermediate level instead has the stationary balanceMultiplying its population by the successful stabilization rate gives the actual recombination coefficientHere the equilibrium population was used only to determine the capture rate by detailed balance; it was not assumed to equal the intermediate population when photons escape. This is precisely why the competing decay probabilities appear.

Use the supplied core-transition relation between the Einstein coefficients and atomic oscillator strength,Here $\alpha$ is the fine-structure constant and $a_0$ the Bohr radius, not a recombination coefficient. Factor out $A_{ji}$ and defineSubstitution yieldsThis definition accommodates several entrance and decay channels. In the common spectator-electron approximation with one autoionization channel and stabilization by the core transition, $A_s=A_r=A_{ji}$ and $A_{di}^a=A_a$. If all coupled substates of the captured $n\ell$ electron are summed, $g_d=2(2\ell +1)g_j$, givingFor an intermediate state resolved by fine structure, retain its actual $g_d$ instead of that summed weight. The stabilization probability itself is $A_s/(A_a+A_r)$; the requested ${\beta }_{j,n\ell }$ is not just this branching probability, because the prefactor has already factored out the core radiative rate and statistical weights of atomic levels. Radiative transitions to another still-autoionizing state require that state's eventual survival probability, rather than automatically counting every emitted photon as stabilization.

The total dielectronic recombination coefficient sums all accessible positive-energy resonances. In this isolated-resonance approximation it has the formwith atomic factors included in $c_d$. The resonance-temperature dependence of dielectronic recombination follows by differentiating a single term's logarithm:Thus low temperatures suppress a positive-energy resonance, although resonances very near threshold can still matter; at sufficiently high temperature a fixed set of resonances falls as $T^{-3/2}$. Several core-excitation series can produce several peaks. When autoionization is much faster than radiation, the capture-and-survival product is limited by the stabilizing radiative rate; when capture is weak, autoionization is the limiting inverse-capture rate. For a fixed Rydberg angular channel, autoionization typically decreases at high principal quantum number while the core radiative rate changes little. Capture into very high levels is vulnerable to subsequent collisional or field ionization, so finite-density survival can suppress the total rate. dielectronic recombination is resonant and strongly temperature-selective; its magnitude can be important in coronal ionization balance.

The solar corona is a hot, dilute plasma in which collisional ionization mainly proceeds by direct electron impact and by excitation-autoionization. The principal inverse ion-changing processes are radiative recombination and dielectronic recombination. Three-body recombination is negligible at ordinary coronal densities. Photoionization, stimulated recombination and charge exchange are normally secondary in this fully ionized, collision-dominated regime, although they should be reconsidered near irradiated or neutral interfaces.

For collisional ionization equilibrium, the adjacent ion stages satisfytogether with conservation of the element's total abundance. Thus the low-density ion fraction depends primarily on $T_e$. This fixes the amount of Si X available for line emission, but not the populations of its excited levels: a coronal approximation does not impose a Boltzmann level distribution.

Within an ion, electrons collisionally excite, de-excite and redistribute levels. Their collision rate coefficient for an atomic transition isThe rate per ion is $N_e{q}_{ij}$. Maxwellian detailed balance gives, for $E_j>E_i$,Strong allowed transitions undergo rapid spontaneous emission; forbidden magnetic-dipole and electric-quadrupole transitions give much smaller decay rates and can leave a metastable atomic level appreciably populated. Radiative cascades feed levels from higher states. Absorption and stimulated emission add rates proportional to the local radiation field through the Einstein coefficients when significant. Proton collisions can also mix closely separated levels separated by fine structure and should be included when their rates matter.

Write $f_i=N_i/N(\mathrm{Si}\mathrm{X})$. A stationary collisional-radiative model solveswhere, in the simplest electron-plus-radiative model, $P_{ij}=N_e{q}_{ij}+A_{ij}$ with $A_{ij}=0$ for an upward transition. Add proton and radiation rates when appropriate. Include enough levels and radiative branches to capture population feeding. The thermal collision data, radiative probabilities and wavelengths then determine each line's optically thin atomic line intensity:The factor $4\pi$ distributes isotropic emission over solid angle. photon intensities omit $h\nu$. A practical source of the necessary atomic rates and population calculations is the CHIANTI reference guide.

Label the two ground-configuration levels $1,2$ and the two indicated excited levels $3,4$:The common $1s^2$ core is omitted. Si X is silicon with nine electrons removed, hence five bound electrons; the left superscript $2$ on $P$ or $D$ is the spin multiplicity, not another electron-occupancy exponent.

The observed lines correspond to decays $3\to 1$ and $4\to 2$. For approximately uniform plasma at the indicated $T_e\simeq 1.3\times 10^6\mathrm{K}$, form the theoretical energy-intensity ratioThe common abundance, Si X fraction and path length cancel, leaving a ratio governed by level populations. The frequency factor is ${\nu }_{42}/{\nu }_{31}=347/356$ for the rounded wavelengths. With observed photon counts, first apply the detector calibration and use the corresponding photon-ratio convention.

The electron-density diagnostic from metastable populations can be seen in a reduced rate model. The upper ground level $2$ has only a weak radiative decay to $1$. Ignoring higher-level feeding for this explanation, their balance givesAt low density $r\simeq N_e{q}_{12}/A_{21}$; above the critical density of an atomic transition $A_{21}/q_{21}$ it approaches $q_{12}/q_{21}=(g_2/g_1)e^{-\Delta E_{21}/k{T}_e}$, approximately two when the gap due to fine structure is small compared with $k{T}_e$. This saturation can occur while the strong upper-level transitions still remain radiatively dominated.

Let $A_3,A_4$ be the total radiative loss rates of levels $3,4$, and $b_{31}=A_{31}/A_3$, $b_{42}=A_{42}/A_4$ their line branching fractions. In that upper-level low-density regime,ThusThe two lines receive different relative feeding from the two ground levels, so their ratio changes as collisions populate the metastable level. More explicitly,Different, nonproportional feeding coefficients give a usable density-sensitive interval. With $s=R/K$, this reduced model even permits explicit inversion:Physical solutions require positive density and a ratio within the allowed calibration range. The actual Si X calibration should use the full multilevel population system, not discard cross-excitation, cascades or all competing branches simply to use this illustrative formula.

Operationally, evaluate the model ratio at the known temperature over a density grid and match the calibrated observed ratio to it, propagating line-flux and atomic-data uncertainties. No observed ratio or numerical atomic rate set is supplied here, so the result is this inference procedure rather than a numerical density. At low- or high-density saturation, the ratio constrains density poorly. Deconvolve blends or sum every unresolved transition in the theoretical feature; for example, the neighboring ${}^2{D}_{3/2}\to {}^2{P}_{3/2}$ branch can contribute near the 356-Angstrom feature when unresolved. Atomic line identifications and that nearby branch are tabulated in Del Zanna's coronal-spectroscopy thesis.

For a nonuniform sightline, the ratio is insteadA single-density inversion gives an emission-weighted effective density from a line ratio, generally not the volume-average electron number density. The density estimate is meaningful within its temperature, optical-depth, line-blending and spatial-uniformity assumptions. This is the precise interpretation of an average density obtained from the observed pair.

For ${\Omega }_m=1$ the preceding result reduces to $\kappa ={\Omega }_b{C}_H[(1+z_{\mathrm{re}}{)}^{3/2}-1]$. Using $C_H=0.032$ and ${\Omega }_b=0.05$ givesAt fixed $H_0$ and ${\Omega }_b$, a lower matter density lowers $H(z)$ relative to the flat dust value at positive redshift. The same electron density then persists for a longer scattering time, increasing the optical depth. Hence the reionization redshift giving unit depth is lower in an open universe.

In the high-redshift matter regime the scaling is $1+z_{\mathrm{re}}\propto {\Omega }_m^{1/3}$ at fixed optical depth; for ${\Omega }_m=0.3$ this suggests a redshift near fifty rather than seventy. This comparison holds the baryon density fixed, not the baryon fraction ${\Omega }_b/{\Omega }_m$, and is not an estimate of the actual observed reionization epoch.

The original PDF contains three algebraic misprints in this question. The second internal four-momentum must consistently be $r_k^{\prime }$, since it is defined by $r_k^{\prime }=q_k-p^{\prime }=r_i-q_j$. Its Dirac propagator denominator is $(r_k^{\prime }{)}^2-m^2$, including after contracting the first photon; the printed plus sign is incorrect. Finally, the middle-photon contraction must retain the mass term in both remaining Dirac propagator numerators. The following calculation keeps the electron mass and proves the corrected identities.

Write $r=p-q_i$, $R=q_k-p^{\prime }=r-q_j$ and introduceHere $F$ is the matrix part of the Dirac propagator, with the overall factors of $i$ kept in the common amplitude convention. The Clifford algebra proves $D(l)F(l)=F(l)D(l)=I$. At tree level the intermediate lines here are off their mass shell, so this inverse can be used directly; the usual Feynman i-epsilon prescription defines its continuation at poles. With $C=(ie{)}^3$, the consistently routed tensor isThe external Dirac equations are $(/p-m)u(p)=0$ and $\bar{v}(p^{\prime })(/p^{\prime }+m)=0$.

For the photon at the electron end, $/q_i=/p-/r$ and hence $/q_{i}u=-D(r)u$. Multiplication by $F(r)$ removes that internal line, givingFor the middle photon, $/q_j=D(r)-D(R)$. Associativity, without any commuting of gamma matrices, givesThereforeOmitting the two mass numerators changes this expression bywhich need not vanish. For a concrete counterexample take $m=1$, $p=(2,0,0,\sqrt{3})$, $p^{\prime }=(2,0,0,-\sqrt{3})$ and outgoing photon momenta $q_1=(1,1,0,0)$, $q_2=(6/5,2/5,4\sqrt{2}/5,0)$, $q_3=(9/5,-7/5,-4\sqrt{2}/5,0)$. All external momenta are on shell and conserve four-momentum. Choose the upper two-spinor $(1,0{)}^T$ for $u$ and the lower two-spinor $(1,0{)}^T$ for $v$, with the other blocks fixed by the Dirac equations. For $(i,j,k)=(1,2,3)$ and $(\mu ,\sigma )=(0,3)$, the correct middle contraction divided by $C$ is $1/3$, whereas the expression with the mass numerators omitted is $5/27$. Thus the printed middle identity is not valid for a massive electron.

At the positron end, $q_k=R+p^{\prime }$. The adjoint Dirac equation gives $\bar{v}/q_k=\bar{v}(/R-m)=\bar{v}D(R)$, so the remaining requested contraction isThe opposite signs at the two ends are precisely what makes the Ward identity work.

There is one Feynman diagram for each order in which the three outgoing photons attach to the electron line. Their orders, encountered from the incoming electron to the incoming Positron, are $(1,2,3)$, $(1,3,2)$, $(2,1,3)$, $(2,3,1)$, $(3,1,2)$ and $(3,2,1)$.

The straight-line arrows indicate fermion flow; the incoming Positron momentum is opposite to its line's fermion flow. Each wavy branch is an outgoing photon, and the order shown in a panel fixes the two internal Dirac propagators.

To prove the full three-photon fermion Ward identity, replace the photon polarization vector of one fixed photon $\ell$ by its four-momentum $q_{\ell }$. Fix the order $(a,b)$ of the other two photons and write ${\Gamma }_a=/{ϵ}_a$, ${\Gamma }_b=/{ϵ}_b$. There are exactly three positions for photon $\ell$ relative to that order. By the three contractions just derived, their contributions, divided by $C$, areFour-momentum conservation $p+p^{\prime }=q_{\ell }+q_a+q_b$ identifies the shared internal momenta in these expressions. The three terms cancel exactly. Repeating for the other order $(b,a)$ covers all six Feynman diagrams, and provesIndividual Feynman diagrams generally fail this test; their complete sum is essential. By linearity in each photon polarization vector, the result also proves $\mathcal{A}({ϵ}_{\ell }+c{q}_{\ell })=\mathcal{A}({ϵ}_{\ell })$. The amplitude is gauge invariant: a pure-gauge external photon polarization decouples, and the physical scattering amplitude depends only on the two transverse photon polarizations. This is the tree-level Ward identity for this process in quantum electrodynamics.

With a vacuum annihilated by all $a_{\mathbf{p}}^s,b_{\mathbf{p}}^s$, define the particle and antiparticle occupation operators of a Dirac fieldEach counts its corresponding occupied modes. The total positive particle count isCommuting it through either creation operator gives $[N_{\mathrm{total}},a^{†}]=a^{†}$ and $[N_{\mathrm{total}},b^{†}]=b^{†}$, so both kinds of one-particle states have eigenvalue one. In a normalized discrete mode, $(a^{†}a{)}^2=a^{†}a$ by the anticommutation relations, so its occupation eigenvalues are $0,1$.

For a Dirac field there is also a distinct conserved signed fermion number. Inserting the mode expansion into the normal-ordered local density givesSpatial integration makes the same-sector momenta equal. The cross terms vanish by $u^{†}(\mathbf{p})v(-\mathbf{p})=0$; normal ordering of $b{b}^{†}$ supplies the minus sign for the antiparticles. Thus the density integral is not the total positive particle count. It generates the global phase symmetry, and for electron charge $-e$ the electric charge is $Q_{\mathrm{em}}=-e{Q}_D$.

The free normal-ordered Hamiltonian isso $N_e,N_{\bar{e}},N_{\mathrm{total}}$ and $Q_D$ are conserved in the free theory. In an interacting theory, pair creation can change the total count while preserving the signed charge. This distinguishes a positive occupation number operator from the charge behind Dirac fermion number conservation.

A Positron is the electron's antiparticle: it has the same mass and spin one-half, opposite electric charge $+e$, and fermionic statistics. A state $b_{\mathbf{p}}^{s†}|0⟩$ has positive energy $E_{\mathbf{p}}$ and physical momentum $\mathbf{p}$, as follows from the Hamiltonian and momentum operatorsThe charge commutator gives $[Q_{\mathrm{em}},b^{†}]=+e{b}^{†}$.

The term $b^{†}v(p)e^{ip\cdot x}$ in the field has negative frequency, but it creates this positive-energy Positron. Treating its coefficient as an annihilator of a physical negative-energy electron would give an unbounded energy interpretation. Fermionic quantization instead reorders that sector, giving positive antiparticle occupation energies and the opposite signed charge. The historical Dirac sea describes a Positron as a hole in filled negative-energy electron levels; the field-theory interpretation uses an antiparticle creation operator and normal ordering, without requiring a literal infinite material sea.

Under charge conjugation of a Dirac field, ${\psi }^C=C{\bar{\psi }}^T$, where $C^{-1}{\gamma }^{\mu }C=-({\gamma }^{\mu }{)}^T$, particle and antiparticle modes interchange. Electron–Positron pair creation and annihilation change total particle number while conserving electric charge. Positrons are positive-energy antiparticles, not physical negative-energy electrons.

Use signature $(+,-,-,-)$ and electron charge $q=-e$, with $e>0$. Then ${\mathcal{L}}_{\mathrm{int}}=-q\bar{\psi }{\gamma }^{\mu }\psi A_{\mu }=+e\bar{\psi }{\gamma }^{\mu }\psi A_{\mu }$. Write the process as $e^{-}(p,s)+\gamma (k,ϵ)\to e^{-}(p^{\prime },s^{\prime })+\gamma (k^{\prime },{ϵ}^{\prime })$. Physical external states satisfyand the photon polarization vectors obey $k\cdot ϵ=k^{\prime }\cdot {ϵ}^{\prime }=0$. Choose unit physical polarizations with ${ϵ}^{\ast }\cdot ϵ={ϵ}^{\prime \ast }\cdot {ϵ}^{\prime }=-1$; representatives differing by a multiple of the corresponding null momentum describe the same physical polarization. The external spinors obey $(/p-m)u(p)=0$, $\bar{u}(p^{\prime })(/p^{\prime }-m)=0$, with ${\bar{u}}_s{u}_r=2m{\delta }_{sr}$. Outgoing polarization appears complex conjugated, so this formula includes helicity states as well as real linear polarizations.

The needed QED Feynman rules are a vertex $-iq{\gamma }^{\mu }=+ie{\gamma }^{\mu }$, an internal Dirac propagator $i(/q+m)/(q^2-m^2+i0)$, the external spinors, and the incoming/outgoing photon polarization vectors. There is no internal photon at this order and no elementary two-photon–Dirac seagull vertex. The two photon orderings along the electron line give internal momenta $q_s=p+k$ and $q_u=p-k^{\prime }$. Defining the matrix element by $S_{fi}-{\delta }_{fi}=i(2\pi {)}^4{\delta }^4(p+k-p^{\prime }-k^{\prime })\mathcal{M}$, their sum isThe denominators are $2p\cdot k+i0$ and $-2p\cdot k^{\prime }+i0$. Both terms are essential; this is the tree-level Compton amplitude.

To prove the incoming Ward identity, replace $ϵ$ by $k$ and write $R(q)=(/q+m)/(q^2-m^2)$, with the Feynman boundary prescription understood. The physical nonzero-photon kinematics keeps these tree denominators off the electron pole. Sincethe external Dirac equations implyTherefore the two contracted terms are $\bar{u}(p^{\prime })/{ϵ}^{\prime \ast }u(p)$ and its negative, and cancel. By linearity,This is the two-photon fermion Ward identity. An analogous cancellation holds for the outgoing photon. Gauge-equivalent polarization representatives have the same scattering amplitude, so a longitudinal pure-gauge component is not an extra physical photon state. Individual diagrams do not generally satisfy this identity; their sum does.

For perturbative order, each elementary QED vertex has one photon leg. If $I_{\gamma }$ is the number of internal photon lines and $E_{\gamma }=2$, the vertex parity of a QED amplitude givesThus there is no third-order contribution. At fourth order, $V=4$ and, counting all three legs per vertex with four external legs, $I=(3V-4)/2=4$. A connected graph has loop number $I-V+1=1$.

The nonzero one-loop classes on the open electron line are an internal-electron self-energy insertion, a vertex correction at either external-photon vertex, and a virtual photon spanning both photon vertices, giving the four-point box class. The first figure sketches each class for one photon ordering; interchange the two external photon attachments for its crossed partner. This yields two internal-self-energy graphs, four vertex graphs and two spanning-box graphs. They use the usual photon propagator, for example $-i{g}_{\mu \nu }/({\ell }^2+i0)$ in Feynman gauge.

For completeness, one-loop diagram classes for Compton scattering also depend on whether one is drawing an unamputated correlation function or a physical amplitude. Electron self-energies can be attached to either external electron leg, and photon vacuum polarization can be attached to either external photon leg, for both photon orders. In an on-shell renormalized S-matrix, these are accounted for by LSZ reduction and external-field residues rather than counted again as independent amputated graphs. Photon-leg corrections are not vacuum-polarization insertions on an internal Born photon: there is no such Born line.

At the same fourth order, a renormalized calculation has one mass or electron-field counterterm on the internal electron line, or one vertex counterterm at either Born vertex; the crossed partners are included. Each counterterm is a relative order-$e^2$ correction to the order-$e^2$ tree amplitude. External-field normalization must follow the same convention.

A possible closed electron loop with three photon vertices, linked to the external electron line by a virtual photon, cancels between its two orientations by Furry's theorem. Transposing the loop and using $C^{-1}{\gamma }^{\mu }C=-({\gamma }^{\mu }{)}^T$ gives an extra sign $(-1{)}^3$, so its orientation-reversed partner is its negative. A one-photon tadpole likewise vanishes. Disconnected vacuum bubbles are removed by vacuum normalization. The first genuine radiative corrections are fourth order, with the internal self-energy, vertex and spanning-box classes and their consistently included renormalization terms.

After complete ionization, each hydrogen nucleus supplies one ion and one electron, while each helium-4 nucleus supplies one ion and two electrons. Neglecting the small metal mass fraction $Z$ and putting $Y\simeq 1-X$, the mean molecular weight therefore satisfiesUse stellar homology to compare models with common dimensionless profiles. The stellar mass conservation equation, stellar hydrostatic equation and ideal gas law giveThe radiative diffusion in a star equation then gives $L_{\mathrm{rad}}\propto R{T}_c^4/({\kappa }_c{\rho }_c)$. Inserting the specified opacity yields $L_{\mathrm{rad}}\propto R{T}_c^7/(Z{\rho }_c^2)\propto {\mu }^7{M}^5/Z$. Independently, integrating the stellar energy-generation rate over the mass gives $L_{\mathrm{nuc}}\propto X^2{\rho }_c{T}_c^{5}M\propto X^2{\mu }^5{M}^7/R^8$. Equating these in stellar thermal equilibrium gives the radiative homology with fifth-power hydrogen burning and inverse-cubic opacity:At fixed solar mass these reduce to $L\propto {\mu }^7/Z$ and $T_c\propto {\mu }^{5/4}/(X^{1/4}{Z}^{1/8})$. The stellar homology approximation, rather than merely the local power laws, is what permits these relations between whole stars.

Equal luminosities imply ${\mu }_2/{\mu }_1=(Z_2/Z_1{)}^{1/7}=2^{-1/7}$. Thusso $X_2=0.8$ to one significant figure. Retaining the unrounded hydrogen mass fraction when comparing central temperatures givesThe first model has the higher central temperature.

For homogeneous hydrogen burning at fixed $M,Z$, fuel conservation gives $M{E}_0\dot{X}=-L$. Since $d\mu /dX=-5{\mu }^2/4$ and $L=L_0(\mu /{\mu }_0{)}^7$, the homogeneous fuel-depletion luminosity feedback obeysIntegration with the initial mean molecular weight ${\mu }_0$ givesThis idealized stellar evolution law stops when $X=0$; it does not predict a physically attainable infinite luminosity. Indeed $\mu =4/3$ at fuel exhaustion, giving $t_H=M{E}_0[1-(3{\mu }_0/4{)}^8]/(10{\mu }_0{L}_0)$, strictly before the formal singularity.