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In the specified dust model $H_0=2/(3t_0)$, so ${\rho }_{\mathrm{crit}}=m_{\mathrm{pl}}^2/(6\pi t_0^2)$ and $n_B={\Omega }_{B0}{\rho }_{\mathrm{crit}}/m_B$. Photons and the six neutrino/antineutrino helicity states giveUsing the supplied GeV estimates literally givesThe rounded time/temperature conversions are unusually crude: converting $10$ Gyr to $4.8\times 10^{41}{\mathrm{GeV}}^{-1}$ and $3$ K to $2.6\times 10^{-13}$ GeV instead gives $Y_B\sim 10^{-10}$. Either set gives the same order-of-magnitude mass conclusion below, but the numerical abundance should identify which supplied units are used.

For a stable relic, the relevant comparison is energy density, not merely particle number. The matter density left over after the baryons is at most nine times the baryon density here. ConsequentlyPut $x=(m_{Y\mathrm{nr}}/m_Y{)}^{1/3}$ and use $Y_Y\simeq C_Y{x}^{3/2}{e}^{-x}$ with $C_Y\simeq 10^{-3}$. The condition becomesFor the literal supplied numbers the right side is about $30.5$, giving $x\gtrsim 25.6$ and $m_Y\lesssim 40$–$50$ GeV. With the physically consistent conversions the estimate is about $60$ GeV. The intended rough conclusion is therefore a sufficiently suppressed branch with $m_Y$ no larger than order $10^2$ GeV, within the toy cosmological particle freeze-out law. Comparing the two number yields alone gives a similar rough scale but misses the mass weighting. This is a thermal relic density bound for a tunable fermion.

For $m_Y\gtrsim m_{Y\mathrm{nr}}$, relativistic decoupling would instead leave $Y_Y\simeq 135\zeta (3)g_Y/(8{\pi }^4{g}_{\ast s})\simeq 4\times 10^{-3}$, incompatible with the matter-density bound for such heavy stable particles. These are conditional bounds on a populated, stable thermal relic. A short lifetime or an initially suppressed nonthermal population can change the conclusion; the supplied information alone does not constrain every possible production and decay history.

For small baryon fraction the CDM growing mode is ${\delta }_C\propto a$, and its gravity forcesFor an adiabatic cosmological perturbation to follow the unsuppressed growing mode, pressure must be below the gravitational forcing frequency: $c_s^2{k}^2\lesssim 4\pi G{a}^2{\bar{\rho }}_C$. With physical wavelength $\lambda =2\pi a/k$, this becomesThis is Jeans support versus forced baryon response. The wording “only grow” must refer to unsuppressed gravitational structure growth; it is not a theorem that all shorter-wavelength baryon perturbations are absent or constant. In the pressure-dominated regime a forced response ${\delta }_B\simeq (k_J/k{)}^2{\delta }_C$ remains. In particular, a falling sound speed can make this small response grow even before it catches up to CDM.

Before photon decoupling, the photon-baryon fluid has sound speed of order $c/\sqrt{3}$. Its Jeans scale is roughly a horizon scale, so smaller perturbations oscillate acoustically instead of freely following the growing CDM structures. After decoupling the baryon sound speed drops sharply. Above the new Jeans scale, pressure is negligible and subtraction of the CDM equation gives $({\delta }_B-{\delta }_C{)}^{\prime \prime }+\mathcal{H}({\delta }_B-{\delta }_C{)}^{\prime }=0$. For $a\propto {\tau }^2$, the difference is a constant plus a decaying $1/\tau$ term; its ratio to the growing CDM contrast tends to zero. Baryons therefore catch up. Under the stipulated post-decoupling $c_s\propto T\propto a^{-1}$, the physical Jeans length scales as $a^{1/2}$, the comoving scale as $a^{-1/2}$ and the corresponding Jeans mass as $a^{-3/2}$.

For the requested instantaneous mass estimates, define the baryonic Jeans mass as the baryon content of a sphere of radius ${\lambda }_J/2$:Using the supplied matter-era age gives $t_{\mathrm{dec}}=t_0(1+z_{\mathrm{dec}}{)}^{-3/2}\simeq 9.5\times 10^{13}$ s and ${\bar{\rho }}_C\simeq 1/(6\pi G{t}_{\mathrm{dec}}^2)\simeq 6\times 10^{-23}\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}$. ThenThe drop is by $(\sqrt{3}\times 10^{-5}{)}^3$. A definition using the total gravitating mass instead of baryon content gives values about ten times larger. The pre-decoupling number is an acoustic Jeans order estimate, not an accurate application of the nonrelativistic equations from part a when $c_s\sim c$.

Take a periodic comoving box of volume $L^3$, with $\mathbf{k}=2\pi \mathbf{n}/L$. A real field has the expansionwhere $[{\hat{a}}_{\mathbf{k}},{\hat{a}}_{\mathbf{q}}^{†}]={\delta }_{\mathbf{kq}}$ and the other ladder-operator commutators vanish. The coefficient of $e^{i\mathbf{k}\cdot \mathbf{x}}$ is therefore $w_k{\hat{a}}_{\mathbf{k}}+w_k^{\ast }{\hat{a}}_{-\mathbf{k}}^{†}$. This opposite momentum is required by reality; the single-oscillator shorthand with the same label on both operators must not be read literally as every traveling-wave Fourier coefficient.

The quadratic action has canonical momentum $\hat{\pi }=a^3\delta \dot{\hat{\varphi }}$. Its equal-time field commutator requiresChoose the vacuum annihilated by all $a_{\mathbf{k}}$ and the subhorizon positive-frequency condition. For constant $H$ and negligible effective mass, put $x=k/(aH)$, $C_k=L^{-3/2}H/\sqrt{2k^3}$. The candidate is $w_k=C_k(i+x)e^{ix}$. Since $\dot{x}=-Hx$,Direct substitution makes ${\ddot{w}}_k+3H{\dot{w}}_k+(k^2/a^2)w_k$ identically zero. Its Wronskian is $2iH{C}_k^2{x}^3=i/(L^3{a}^3)$, which has the required normalization. ThusThis is the existing Bunch-Davies mode normalization in a finite comoving volume. It is exact for a massless minimally coupled test field in de Sitter and the leading approximation for a light slowly evolving field. The homogeneous zero mode needs separate treatment.

For $k\gg aH$, the dominant term isThis is the positive-frequency harmonic-oscillator normalization with the physical box volume $a^3{L}^3$. Over a period much shorter than $H^{-1}$, the scale factor and physical frequency are effectively constant. Equivalently, the rescaled field $a{L}^{3/2}{w}_k$ approaches $e^{-ik\eta }/\sqrt{2k}$ exactly in the short-wavelength limit.

For $k\ll aH$, $w_k\to i{L}^{-3/2}H/\sqrt{2k^3}$ and ${\dot{w}}_k=O(x^2)$, so the growing field mode freezes. A single Fourier coefficient has variance proportional to $k^{-3}$, which by itself is not scale-independent. The number of modes in a logarithmic shell is proportional to $L^3{k}^3$. Their product givesThus equal logarithmic intervals have equal variance in exact de Sitter. Slow evolution of $H$ or nonnegligible effective mass produces a tilt. Integrating an exactly flat spectrum over arbitrarily many logarithmic intervals still requires infrared and ultraviolet cutoffs; the statement concerns power per interval, not a finite all-scale total variance.

For positive masses and distinct positions, Newtonian gravitational field givesInsert ${\mathbf{x}}_i=a(t){\mathbf{a}}_i$ with $a>0$. The separation numerator contributes a factor $a$ and the denominator $a^3$, soHence the time-dependent equations are satisfied wheneverThe fixed-shape condition is a Newtonian central configuration, and the resulting motion is a homothetic gravitational motion. It applies away from the collision scale $a=0$; a negative scale would require absolute values in the force scaling rather than the same positive-scale formula.

Multiplying the scalar equation by $\dot{a}$ givesThus its first integral is ${\dot{a}}^2/2-\Lambda /a=k/2$ for some real constant $k$.

In the pair potentialdifferentiating a pair with respect to ${\mathbf{a}}_i$ produces $G{m}_i{m}_j({\mathbf{a}}_i-{\mathbf{a}}_j)/|{\mathbf{a}}_i-{\mathbf{a}}_j{|}^3$. Every pair incident on $i$ contributes exactly once, so ${\nabla }_{i}W={\mathbf{G}}_i$. Moreover $W(\lambda {\mathbf{a}}_1,\dots ,\lambda {\mathbf{a}}_n)={\lambda }^{-1}W$ for $\lambda >0$. Differentiating with respect to $\lambda$ at one proves the relevant Euler homogeneous function theorem identity directly:Dot each fixed-shape equation with ${\mathbf{a}}_i$ and sum to obtainFor a nontrivial collision-free configuration with at least two positive masses, $W<0$ and $I>0$, and therefore $\Lambda >0$. Positivity requires an interacting pair; an isolated one-particle system is the degenerate zero-force exception rather than a positive-$\Lambda$ gravitational configuration. Pairwise cancellation also gives ${\sum }_i{\mathbf{G}}_i=0$, so the nondegenerate configuration has ${\sum }_i{m}_i{\mathbf{a}}_i=0$: its scaling origin is the centre of mass.

Finally, the system's kinetic energy is $I{\dot{a}}^2/2$ and its potential energy is $W/a$. ConsequentlyThis establishes the energy relation for every fixed configuration satisfying the algebraic condition, without assuming any particular particle arrangement.

The parallel axis theorem states $I_O=I_G+m{d}^2$ for two parallel axes, one through the centre of mass and the other a perpendicular distance $d$ away. To see this, expand the squared distance to the translated axis in the mass integral; its mixed term vanishes because the centroidal first moment is zero.

For a uniform solid hemisphere, the moment about a diameter in its flat face is half the corresponding full-sphere moment. With full-sphere mass $2m$, it is $\frac{1}{2}[\frac{2}{5}(2m)a^2]=2m{a}^2/5$. The original PDF gives the centroidal displacement $d=3a/8$; the TeX's $3a/5$ is incorrect. Moving the diameter axis to its parallel centroidal axis therefore yieldsThis is the moment of inertia of a uniform solid hemisphere, not that of a thin hemispherical shell.

Take a cross-section perpendicular to the rotation axis. Let $O$ be the underlying sphere centre, $G$ the centre of mass and $P$ the contact point. During rounded-surface contact, $OP=a$ vertically and $OG=d$ points toward the curved face, with downward component $d\cos \theta$. ThereforeFor rolling without slipping, $P$ is instantaneously at rest and the body rotates about it with angular speed $|\dot{\theta }|$. Thus the centroid speed is $b|\dot{\theta }|$. Translation and rotation contributeThe initial vertical base has $\theta =\pi /2$ and $K=0$, so conserved energy is $mga$. Equating the later $K+V$ to this value givesOn the initial roll toward the lowest centroid position, $\dot{\theta }<0$. Static contact friction does no work because the contacting material point is at rest; its coefficient must be large enough for the assumed no-slip motion. The formula describes the rounded-contact phase of the rolling uniform solid hemisphere and should not be continued through a change of contact geometry without another model.

Let $s=m_0/T$ be the constant fuel mass rate, and let $q(t)=M+m_0-st$ be the remaining rocket mass. Take positive motion opposite to the exhaust direction. Over a short interval $dt$, momentum conservation including the mass $sdt$ expelled at velocity $v-u$ givesConsequently, while the rocket slides forward,This is the constant-burn rocket with ground friction. Assuming forward sliding begins immediately and persists, integration from rest givesand henceA necessary physical assumption is missing from the printed request. With a common static friction and kinetic friction coefficient $\mu >0$, immediate motion requires $su\ge \mu g(M+m_0)$; equality allows motion as soon as fuel loss lowers the normal force. Then the acceleration is nonnegative initially and increases thereafter, so forward sliding is consistent. Without this assumption the displayed expression need not be the final speed and can even be negative.

For completeness, the physical rest/sliding solution with these equal coefficients is as follows. If $su\le \mu gM$, the rocket never starts before burnout and its final speed is zero. If $\mu gM<su<\mu g(M+m_0)$, static friction initially balances the thrust and release occurs atIntegrating only the sliding interval yieldsThis speed is positive because, with $y=q_s/M>1$, it equals $u[\log y-1+1/y]$, whose derivative with respect to $y$ is $(y-1)/y^2>0$. The $\mu =0$ case recovers the frictionless rocket equation. If static and kinetic coefficients differ, the release condition uses the static coefficient and the subsequent integral uses the kinetic coefficient; the prompt supplies no second coefficient.

For a compactly supported variation $\varphi \mapsto \varphi +\epsilon \eta$, the first variation of the energy functional isIntegration by parts has no endpoint term. Since $\eta$ is arbitrary, the static Euler-Lagrange field equation isIntegrating the specified polynomial derivative gives $U(\varphi )={\varphi }^2/2-{\varphi }^4+{\varphi }^6/2+C$. The polynomial factors as a nonnegative square times ${\varphi }^2$; its global minimum is zero when $C=0$. ThereforeThese are the three constant scalar-field vacua. For a localized finite-energy kink, the two ends approach vacua and the derivative tends to zero. Multiplying the field equation by ${\varphi }^{\prime }$ gives the first integralA nonconstant kink must stay within one interval between neighboring vacua. A proposed direct connection from $-1$ to $1$ would reach $\varphi =0$ at some finite point, where the first integral forces ${\varphi }^{\prime }=0$. Uniqueness for the smooth second-order differential equation then forces the identically zero solution, a contradiction. This is the intermediate-vacuum obstruction to a kink; two elementary kinks can span the outer vacua only in an infinite-separation limit, not as one finite-width static kink.

There are consequently four oriented elementary sectors: $-1\to 0$, $0\to -1$, $0\to 1$ and $1\to 0$. If opposite orientations are grouped together, there are two kink families and their antikinks. Spatial reflection reverses orientation, and $\varphi \mapsto -\varphi$ exchanges the two adjacent intervals; both preserve the energy. All four therefore have the same rest mass, even though the central and outer vacua have different fluctuation curvatures.

For the sector $0\to 1$, define $W(\varphi )={\varphi }^2/2-{\varphi }^4/4$, so $U=(W^{\prime }{)}^2/2$ and $W^{\prime }=\varphi (1-{\varphi }^2)$. The Bogomolny bound follows by completing the square:It is saturated by the Bogomolny equationLet $y={\varphi }^2$ on this positive branch. Then $y^{\prime }=2y(1-y)$, so $\log [y/(1-y)]=2(x-x_0)$. Thus the explicit kink in a phi-six model isIt has the required limits and satisfies the first integral. The translated center $x_0$ is arbitrary. All oriented elementary solutions areFor $\eta =1$ the endpoints are $0\to \sigma$, and for $\eta =-1$ they are $\sigma \to 0$. Direct evaluation ${\int }_0^1\sqrt{2U(\varphi )}d\varphi ={\int }_0^1\varphi (1-{\varphi }^2)d\varphi =1/4$ independently gives their common mass.

Let $y(x,t)$ be downward deflection, $0\le x\le \ell$, with $y(0,t)=y_x(0,t)=0$. Take a uniform circular elastic beam, $I=\pi r^4/4$, small slopes, and linear material response. Write $E$ for its axial elastic modulus and ${\eta }_s$ for its axial viscous coefficient; these are material quantities, not the solvent viscosity. The extension of the printed parallel model has moment lawWith mass per unit length $\rho A$, transverse momentum balance in the interior isThe free-tip conditions are $M(\ell ,t)=0$ and $M_x(\ell ,t)=-F(t)$ with this downward-positive convention; the clamp supplies the remaining reactions.

Neglecting inertia gives $M=F(t)(\ell -x)$. Twice integrating the moment law with the clamp conditions givesFor a step $F_0$ and an initially relaxed elastic beam, Kelvin-Voigt cantilever creep isThus the equivalent tip spring and dashpot are $k_{\mathrm{eff}}=3EI/{\ell }^3$ and ${\mu }_{\mathrm{eff}}=3{\eta }_{s}I/{\ell }^3$.

If the intended material is instead a true series Maxwell fluid, its moment satisfies $\dot{M}+M/\tau =EI{y}_{xxt}$. Together with momentum balance $\rho A{y}_{tt}+M_{xx}=0$, this gives the full interior equation $\rho A(y_{ttt}+y_{tt}/\tau )+EI{y}_{xxxxt}=0$, with the moment law retained for boundary and initial conditions. In the quasistatic limit,This Maxwell cantilever creep has an initial elastic jump followed by linear creep, not a finite plateau. The jump is the ideal massless step-load limit; inertia resolves the transient. A two-element series model has no nonzero relaxed modulus, so “Maxwell solid” is not an additional specification of a solid with finite long-time stiffness. Such a solid would need an additional spring. The small-deflection approximation requires $\delta \ll \ell$ and eventually fails for unbounded creep.

The moment of inertia of a uniform solid cylinder is taken about its symmetry axis. Write its mass density as ${\rho }_c=m/(\pi a^{2}L)$. A coaxial shell of radius $r$ and thickness $dr$ has mass $dm=2\pi {\rho }_{c}Lrdr$. ThusA hollow cylinder would have a different moment of inertia; uniform solid density is the intended assumption.

Let $s$ measure the centre's displacement uphill from its initial position, and choose signed angular velocity $\omega$ so that uphill rolling without slipping has $\dot{s}=a\omega$. Let $F$ be the signed contact force uphill. The translation and rotation equations areThe minus sign in the torque equation follows from the chosen rolling sense: uphill contact force opposes the positive spin. Under rolling without slipping, $\ddot{s}=a\dot{\omega }$. Eliminating $F$ givesUsing $s(0)=0$ and $\dot{s}(0)=V$, the no-slip motion and maximum vertical rise areIndeed the centre turns at $t=3V/(2g\sin \alpha )$, reaching $s_{\max }=3V^2/(4g\sin \alpha )$, whose vertical component is the boxed height. Conservation of energy independently gives the same answer from initial kinetic energy $m{V}^2/2+I(V/a{)}^2/2=3m{V}^2/4$.

The available static friction is at most $\mu mg\cos \alpha$. It can supply the required uphill force exactly when $mg\sin \alpha /3\le \mu mg\cos \alpha$. ThereforeInterpret the stated friction magnitude as limiting static friction and as kinetic friction during slip, with the same coefficient. Take $0<\alpha <\pi /2$, $\mu \ge 0$. In the slipping regime the contact point moves downhill relative to the slope, so kinetic friction acts uphill. To verify this direction rather than assume it unchecked, solve with $F=\mu mg\cos \alpha$. Define $b=g(\sin \alpha -\mu \cos \alpha )>0$. The result isThe centre's speed is $|V-bt|$, with the sign of $v$ specifying ascent or descent. The contact point's relative velocity isThus the assumed uphill kinetic friction is consistent throughout both ascent and descent, including when the centre's velocity changes sign. There is no later return to rolling without slipping before reaching the bottom. This is the characteristic slipping solid cylinder on an incline regime.

The centre turns at $t_{\ast }=V/b$, givingThe formulas apply up to the positive return time $t_b=2V/b$. Substitution into the signed velocity and angular velocity givesThe return spin can have either sign and vanishes at $\tan \alpha =5\mu$. It reverses when $3\mu <\tan \alpha <5\mu$, although the contact point still slips downhill. The unchanged centre speed does not mean that no energy was dissipated: the rotational kinetic energy has decreased. Indeed integrating the frictional power $Fw<0$ giveswhich agrees with the difference of the initial and final rotational kinetic energies.

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.