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A standard thermal Cosmic neutrino background with one populated neutrino and one antineutrino helicity per species has $n_{\nu }+n_{\bar{\nu }}=3n_{\gamma }/11$ after electron-positron annihilation in cosmology. If this species is nonrelativistic today and dominates a spatially flat matter universe, equate $m_{\nu }(n_{\nu }+n_{\bar{\nu }})$ to the critical density $3H_0^2{M}_P^2/(8\pi )$, using the unreduced Planck mass. The photon number density gives the displayed conditional mass. Extra populated states or several massive species change the abundance count; vacuum-dominated cosmology changes the density available for matter.

Use natural units, the unreduced Planck mass $m_P=G^{-1/2}$, $V_0>0$ and $c\ne 0$. In a flat universe dominated by a canonical inflaton, $\varphi =(2m_P/c)\log (\beta t)$ and $a\propto t^p$ solve the inflaton equation of motion and Friedmann equation exactly when $p=16\pi /c^2$ and ${\beta }^2=V_0{c}^2/[2m_P^2(3p-1)]$. The positive-potential scaling solution requires $p>1/3$, while cosmic inflation requires $p>1$. Its constant equation-of-state parameter is $w=-1+2/(3p)$. In terms of the reduced Planck mass $M=m_P/\sqrt{8\pi }$, the exponential slope is $\lambda =c/\sqrt{8\pi }$ and the same exponent is $p=2/{\lambda }^2$.

The gravitino mass parameter in four-dimensional supergravity is the displayed expression. In a zero-energy supersymmetry breaking vacuum with only F-term breaking, $K_{i\bar{j}}{F}^i\overline{F^j}=3m_{3/2}^2/{\kappa }^2$. Writing this norm as ${\Lambda }_{\mathrm{SUSY}}^4$ gives $m_{3/2}={\Lambda }_{\mathrm{SUSY}}^2/(\sqrt{3}{m}_p)$. It is small relative to ${\Lambda }_{\mathrm{SUSY}}$ when the breaking scale is below the Planck mass. The super-Higgs mechanism supplies the longitudinal gravitino states.

Specify the gauge group, its gauge group representations and gaugings, a real Kähler potential $K$, a holomorphic superpotential $W$, and a symmetric holomorphic gauge kinetic function $f_{ab}$, with positive kinetic matrices. The latter includes the gauge couplings and theta angles; allowed Abelian Fayet–Iliopoulos terms require constants ${\xi }_a$. For local supersymmetry, also specify the Planck mass; constant Fayet–Iliopoulos terms in conventional N=1 supergravity require a compatible gauged R-symmetry and transformation of $W$. $W$ and $f$ are holomorphic; $K$ generally is not. Radiative corrections generally change $K$. The Wilsonian effective action obeys superpotential non-renormalization perturbatively, although physical couplings run through field normalization and nonperturbative corrections can change $W$. The gauge kinetic function does receive corrections while remaining holomorphic; its perturbative Wilsonian running has one-loop exactness of the holomorphic gauge kinetic function, not all-order constancy of the physical gauge coupling.

Use natural units with $c=\hslash =k_B=1$ and the unreduced Planck mass $M_{\mathrm{Pl}}=G^{-1/2}$. In a radiation-dominated universe, the effective number of relativistic energy degrees of freedom $g_{\ast }=\mathcal{N}$ givesThe ratio of weak conversion rate to expansion rate is proportional to $T^3$, so it decreases as the universe cools. Cosmological weak freeze-out occurs approximately at $\Gamma =H$, givingFor $g_{\ast }=10.75$, $G_F=10^{-5}{\mathrm{GeV}}^{-2}$ and $M_{\mathrm{Pl}}=1.22\times 10^{19}\mathrm{GeV}$ this estimate is $1.65\mathrm{MeV}$. Equivalently $g_{\ast }^{1/6}\simeq 1.5$ is only a mild multiplicative correction. The rate itself was given only to order of magnitude, so this is a decoupling scale, not a precise prediction for the weak freeze-out abundance.

Both neutrons and protons are nonrelativistic at this temperature. The nonrelativistic Maxwell--Boltzmann number density follows by the Gaussian momentum integral:Their equal spin degeneracies are $g_n=g_p=2$. Chemical equilibrium of the weak reaction gives ${\mu }_n+{\mu }_{{\nu }_e}={\mu }_p+{\mu }_e$; with negligible lepton asymmetry, ${\mu }_n\simeq {\mu }_p$. Taking the ratio and setting the nearly equal nucleon masses equal only in the prefactor gives the neutron-proton chemical equilibrium resultThe common baryon normalization cancels. Keeping the exact mass prefactor changes this by only order $Q/m_p$. Thus the printed simple exponential is the usual nonrelativistic equal-mass-prefactor approximation. Using $Q\simeq 1.29\mathrm{MeV}$ with the rough decoupling scale above gives a ratio of order one half; the rate's omitted numerical factors and gradual freeze-out must be included before using a more accurate neutron abundance.

Varying the canonical scalar action gives $\square \varphi +V_{,\varphi }=0$. In a homogeneous flat Friedmann-Lemaître-Robertson-Walker metric written in conformal time, $\sqrt{-g}=a^4$ and $g^{00}=a^{-2}$, soThe quadratic potential therefore givesEquivalently $\ddot{\varphi }+3H\dot{\varphi }+m^2\varphi =0$ in cosmic time. The intrinsic roll/oscillation time is $m^{-1}$ and the damping time is $H^{-1}$. If $m\gg H_0$, the scalar has long since begun rapid oscillations: averaging over a quadratic oscillation gives equal kinetic and potential energies, $⟨p⟩\simeq 0$, so it behaves as matter rather than accelerating dark energy. To remain nearly frozen or only begin rolling today requires the mass scale of quadratic quintessenceIf its energy is predominantly potential, equating $m^2{\varphi }_0^2/2$ to ${\rho }_{X0}=3M^2{H}_0^2{\Omega }_X$, with $M$ the reduced Planck mass, givesTaking the natural onset estimate $m\sim H_0$ and ${\Omega }_X\sim 0.7$ gives $|{\varphi }_0|\sim 2M\sim 5\times 10^{18}\mathrm{GeV}$, of order the Planck mass. More strongly frozen, negative-pressure evolution requires a smaller mass and correspondingly larger field. The density fixes the product $m|{\varphi }_0|$ at this accuracy, rather than a unique mass and field separately.

At neutrino decoupling, the interaction rate is comparable with the Hubble parameter. Equating the supplied rates givesHere the Planck mass is the unreduced $m_P=G^{-1/2}\simeq 1.22\times 10^{19}\mathrm{GeV}$ in natural units, $G_F\simeq 10^{-5}{\mathrm{GeV}}^{-2}$, and the effective number of relativistic energy degrees of freedom is $g_{\ast }=10.75$. The supplied rate gives an order-of-magnitude neutrino decoupling estimate; a precise neutron-proton conversion temperature uses its own weak-rate coefficients.

The chemical equilibrium condition for $n+{\nu }_e\leftrightarrow p+e^{-}$ is ${\mu }_n+{\mu }_{{\nu }_e}={\mu }_p+{\mu }_e$. With negligible lepton chemical potentials, ${\mu }_n={\mu }_p$. The nonrelativistic Maxwell--Boltzmann number density therefore givesbecause $g_n=g_p=2$ and the small mass difference can be neglected in the prefactor. This is the neutron-proton chemical equilibrium relation.

For the deuterium equilibrium abundance, ${\mu }_D={\mu }_n+{\mu }_p$ because photons have zero chemical potential. The sign convention in the paper is important: $B=m_D-m_n-m_p<0$, while the positive nuclear binding energy is ${\Delta }_D=-B$. Using spin multiplicities $g_D=3$, $g_n=g_p=2$ givesWith $n_B=\eta n_{\gamma }$, $m_n\simeq m_p\simeq m_N$, $m_D\simeq 2m_N$ and photon number density $n_{\gamma }=2\zeta (3)T^3/{\pi }^2$, this becomesHere $X_D=n_D/n_B$ is a number fraction; the fraction of baryons bound in deuterium is $2X_D$. The formula estimates the onset of the deuterium bottleneck opening while depletion of free neutrons and protons remains small. Because $\eta \ll 1$, stable deuterium becomes abundant only far below the positive nuclear binding energy, not simply at $T={\Delta }_D$.

Let $T_f$ denote cosmological weak freeze-out of the neutron-to-proton ratio, $r_f=e^{-Q/T_f}$, and let $t_N$ mark efficient nuclear burning. Free-neutron decay after freeze-out givesAlmost every surviving neutron enters helium-4, whose atomic nucleus contains two neutrons and four baryons. Thus neutron-limited helium synthesis givesA longer neutron half-life leaves more neutrons available, increasing the primordial helium mass fraction. If the longer lifetime reflects weaker weak interactions, the accompanying earlier cosmological weak freeze-out can reinforce that effect.

During radiation domination, $H\propto \sqrt{G{g}_{\ast }}{T}^2$ whereas the weak conversion rate is proportional to $T^5$. Hence $T_f\propto (G{g}_{\ast }{)}^{1/6}$. Increasing the effective number of relativistic energy degrees of freedom raises $T_f$, increasing $r_f$, and shortens the interval for free-neutron decay after freeze-out. Both increase the primordial helium mass fraction while nuclear burning remains efficient. The same reasoning gives the primordial helium response to a varying gravitational constant: larger $G$ over this interval produces more helium, with fixed microscopic rates and the approximate Friedmann equation unchanged. Smaller $G$ has the opposite effect. A particular varying-$G$ model must specify its background expansion and time history; the simple trend must not be extrapolated to changes so large that nuclear burning itself fails.

Cosmic inflation addresses the horizon problem by extending the causal past of the observed region, the flatness problem by damping the curvature contribution, and the cosmological monopole problem by diluting unwanted relics. A canonical inflaton can drive this expansion when its potential energy density exceeds twice its kinetic energy density.

Use the unreduced Planck mass $m_P=G^{-1/2}$, consistent with the coefficient in the Friedmann equation, and take $V_0>0$, $c\ne 0$. Seek the exponential-potential power-law inflation solution $\varphi =A\log (\beta t)$ and $a=a_{\ast }(t/t_{\ast }{)}^p$. The kinetic energy density is proportional to $t^{-2}$. The inflaton potential must have the same scaling, soSubstitution into the inflaton equation of motion givesThe Friedmann equation now reduces toConsequently the nontrivial positive-potential solution isA positive $V_0$ requires $p>1/3$. Accelerating expansion occurs exactly for $p>1$, givingThis is an exact solution, not a use of the slow-roll approximation. The first Hubble slow-roll parameter and equation-of-state parameter are constant:If the reduced Planck mass $M=m_P/\sqrt{8\pi }$ is used instead, the dimensionless exponential slope is $\lambda =c/\sqrt{8\pi }$ and the exponent becomes $p=2/{\lambda }^2$; these are the same solution.

For $p>1$, the comoving Hubble radius decreases as $(aH{)}^{-1}\propto t^{1-p}$. A region initially in causal contact can grow beyond that radius before the hot radiation-dominated universe begins. In the ideal power law extrapolated all the way to $t=0$, the particle horizon integral ${\int }_0^{t}d{t}^{\prime }/a(t^{\prime })$ diverges, so the solution has no finite past particle horizon. A finite, physically trustworthy inflationary interval instead requires a sufficiently large number of e-folds to enlarge an initially causal patch.

For a small curvature contribution, the Friedmann equation gives $|\Omega -1|=|k|/(a^2{H}^2)\propto t^{2-2p}\to 0$, solving the flatness problem. Conserved nonrelativistic magnetic monopoles have number density proportional to $a^{-3}$, while the inflaton energy density is proportional to $t^{-2}$; their energy-density ratio therefore falls as $t^{2-3p}$. This gives the required dilution, provided later reheating does not recreate them. Inflationary shear damping similarly suppresses anisotropic expansion.

The main failure is no automatic graceful exit or reheating. The first Hubble slow-roll parameter never reaches one, and the exponential inflaton potential has no finite minimum about which the field can oscillate. Additional dynamics are needed for a graceful exit from inflation and transfer to a hot radiation-dominated universe. The ideal solution also retains a curvature singularity at $t=0$; solving the horizon problem does not itself explain the beginning of the universe. For $c=0$, the constant inflaton potential instead permits constant-field de Sitter spacetime expansion; the logarithmic field solution is inapplicable.

For a homogeneous canonical inflaton, ${\rho }_{\varphi }={\dot{\varphi }}^2/2+V$ and $P_{\varphi }={\dot{\varphi }}^2/2-V$. Vacuum domination by a scalar field requires positive potential energy to dominate both kinetic energy and any competing material components. Then $P_{\varphi }\simeq -{\rho }_{\varphi }$ and the expansion is nearly vacuum-like. Exact scalar-driven acceleration requires ${\rho }_{\varphi }+3P_{\varphi }<0$, equivalently ${\dot{\varphi }}^2<V$; potential domination is a stronger condition.

The slow-roll approximation additionally neglects $\ddot{\varphi }$ compared with $3H\dot{\varphi }$ in the inflaton equation of motion. Define the unreduced Planck mass $m_{\mathrm{pl}}=G^{-1/2}$; it differs from the reduced Planck mass by $\sqrt{8\pi }$. Sufficient potential-flatness conditions aretogether with approach to the slow-roll attractor. The reduced equations are $H^2\simeq 8\pi V/(3m_{\mathrm{pl}}^2)$ and $3H\dot{\varphi }\simeq -V^{\prime }$.

For $V=m^2{\varphi }^2/2$, take $m>0$ and the positive-field rolling branch. ThenThe field velocity is constant to this order. Choosing the start at $t=0$ gives the quadratic-potential slow-roll solutionThe PDF puts $\pi$ inside this square root; the converted TeX incorrectly moves it outside. Next integrate $d\log a/d\varphi =H/\dot{\varphi }=-4\pi \varphi /m_{\mathrm{pl}}^2$:Thus the number of e-folds is the logarithmic expansion, not the expansion factor itself:Here ${ϵ}_V={\eta }_V=m_{\mathrm{pl}}^2/(4\pi {\varphi }^2)$. The usual quadratic-inflation endpoint estimate sets ${ϵ}_V\simeq 1$, givingThis is an endpoint estimate, not an exact result from the full equations. Exact acceleration ends at the Hubble slow-roll parameter ${ϵ}_H=1$, equivalently ${\dot{\varphi }}^2=V$. Inserting the slow-roll velocity into that last condition instead gives $\varphi \simeq m_{\mathrm{pl}}/\sqrt{6\pi }$; the different order-one coefficient illustrates the failing approximation at the endpoint. A precise value requires the full scalar and Friedmann evolution. The large initial-field contribution to $N$ is unaffected by that order-one uncertainty.

Finally $V({\varphi }_i)\sim m_{\mathrm{pl}}^4$ gives ${\varphi }_i^2\sim 2m_{\mathrm{pl}}^4/m^2$. The quadratic-inflation expansion from Planck density is thereforeFor $m\ll m_{\mathrm{pl}}$ this is enormous and the initial field is in the slow-roll regime. No unique numerical factor follows without specifying $m/m_{\mathrm{pl}}$. Extrapolating the classical model from Planck density is only a formal estimate; it does not establish control of quantum-gravity corrections at that starting scale.

Assume standard radiation-dominated expansion, negligible chemical potential, one neutrino helicity and one antineutrino helicity, and a stable species $X$ with no later entropy injection. Use natural units and unreduced Planck mass $m_{\mathrm{pl}}\simeq 1.22\times 10^{19}\mathrm{GeV}$. The relativistic energy density is ${\rho }_R={\pi }^2{g}_{\ast }{T}^4/30$. The Friedmann equation then givesFor thermal decoupling in cosmology, equate the interaction rate $G_A^2{T}_D^5$ to $H(T_D)$:This order-of-magnitude estimate assumes the supplied rate law remains applicable there.

A decoupled relativistic species cools as $T_X\propto a^{-1}$, while annihilations heat the coupled bath. Treat the supplied high-temperature count $100$ as the remaining bath's entropy count, approximately equal to its energy count. Cosmological entropy conservation down to ordinary neutrino decoupling givesSubsequent electron-positron annihilation in cosmology heats photons but neither neutrino population. Therefore the two-stage entropy dilution of an early-decoupled neutrino givesThe ordinary Cosmic neutrino background has $T_{\nu }/T_{\gamma }=(4/11{)}^{1/3}$. If the rounded count $100$ instead includes the new two helicities, the remaining bath count is $98.25$; this changes the answer only slightly, to about $0.341$. Keeping the extra species' entropy separate is essential in either convention.

For the two relativistic fermion states, $g_{\ast s,X}=(7/8)2=7/4$. Photon and three standard neutrino species have present entropy degrees of freedom, referred to $T_{\gamma }$,Writing $\theta =T_X/T_{\gamma }$, the fractional entropy of a relativistically decoupled species isThus approximately 1.7% of the thermal particle entropy is in the new neutrinos and antineutrinos. If $100$ counts all states before the new decoupling, the fraction is $1.75\%$. These are the same rounded estimate. The question concerns the radiation/relic thermal entropy budget; it does not supply a model for entropy in later black holes or other gravitational structures.

If $X$ has a small mass but decoupled relativistically, its comoving particle number is still conserved. The relativistic-neutrino number-density ratio givesOnce nonrelativistic, ${\rho }_X\simeq m_X{n}_X$. The density bound on a relativistically decoupled massive relic is thereforeUsing the standard present photon abundance and critical-density normalization, this is approximately ${\Omega }_X{h}^2\simeq m_X/(9\times 10^2\mathrm{eV})$, where $h=H_0/(100\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1})$ is the present Hubble normalization, not a magnetic field. A stable relic must not exceed the permitted matter density; for example, if ${\Omega }_M{h}^2=0.15$, even assigning all matter to it gives only an order-$10^2\mathrm{eV}$ mass ceiling.

A stronger restriction can follow from neutrino free streaming and the observed formation of large-scale structure of the universe: a light thermal population acts as hot rather than cold dark matter, erasing structure below its streaming scale. If at most a fraction $f_{\max }$ of the matter may be in this population, replace the overclosure ceiling by $m_X\lesssim 9\times 10^2{f}_{\max }{\Omega }_M{h}^2\mathrm{eV}$. The allowed fraction is observational input and is not supplied, so the paper does not determine a unique tighter mass limit. A species still relativistic today is radiation rather than matter; its contribution to the effective number of neutrino species is $\Delta N_{\mathrm{eff}}=(T_X/T_{\nu }{)}^4\simeq 0.051$, and must instead satisfy the radiation-density constraints. These conclusions assume stable thermally populated states; decay or nonstandard entropy histories change the bounds.

Use natural units $\hslash =c=k_B=1$ and the unreduced Planck mass $M_P=G^{-1/2}$. Neutrino decoupling occurs when the weak interaction rate falls below the Hubble parameter. The weak-decoupling temperature estimate equates the given rate with the radiation-era expansion rate:Here $g_{\ast }$ counts the effective number of relativistic energy degrees of freedom, and the Fermi constant is $G_F\simeq 10^{-5}{\mathrm{GeV}}^{-2}$. The omitted reaction-rate coefficients make this an order-of-magnitude estimate: decoupling occurs at a temperature of order one MeV.

The charge-conserving channel is $n+{\nu }_e\leftrightarrow p+e^{-}$, together with $p+{\bar{\nu }}_e\leftrightarrow n+e^{+}$. The positron sign in the printed neutron-capture channel violates electric charge conservation. Chemical equilibrium requires ${\mu }_n+{\mu }_{{\nu }_e}={\mu }_p+{\mu }_e$. Negligible lepton chemical potentials therefore give ${\mu }_n\simeq {\mu }_p$. Since neutrons and protons are nonrelativistic particles, their Maxwell-Boltzmann distribution givesBoth nucleons have two spin states, and their masses agree closely in the prefactor. Thus neutron-proton chemical equilibrium impliesThe heavier neutron is suppressed by a Boltzmann factor.

For the deuterium equilibrium abundance, retain the paper's convention $B=m_D-m_n-m_p<0$; the positive nuclear binding energy is ${\epsilon }_D=-B\simeq 2.22\mathrm{MeV}$. Chemical equilibrium of $n+p\leftrightarrow D+\gamma$ gives ${\mu }_D={\mu }_n+{\mu }_p$ because the equilibrium photon chemical potential vanishes. Using the nonrelativistic Maxwell--Boltzmann number density for each species and dividing givesHere $g_D=3$ and $g_n=g_p=2$. With the baryon-to-photon ratio $\eta =n_B/n_{\gamma }$, the photon number density $n_{\gamma }=2\zeta (3)T^3/{\pi }^2$, and $m_D\simeq 2m_N$, $m_n\simeq m_p\simeq m_N$, this becomesIn this convention $X_D=n_D/n_B$ counts nuclei, so its contribution to the fraction of baryons in deuterium is $2X_D$.

The deuterium bottleneck explains why helium-4 formation waits until $T$ is far below ${\epsilon }_D$. At higher temperature, there are vastly more photons than baryons; even the dilute energetic tail can destroy newly formed deuterium. A crude estimate balances ${\eta }^{-1}{e}^{-{\epsilon }_D/T}$ against one, giving $T\sim {\epsilon }_D/\log ({\eta }^{-1})\sim 0.1\mathrm{MeV}$ for $\eta$ of order $10^{-9}$. The power-law factors in the deuterium equilibrium abundance and the nuclear reaction rates refine this estimate. Once stable deuterium accumulates, further reactions can efficiently place nearly all surviving neutrons into helium-4.

Write $r_N=n_n/n_p$ at the onset of nuclear burning. Each helium-4 nucleus requires two neutrons and contains four baryons. Neutron-limited helium synthesis therefore gives the primordial helium mass fractionFor a realistic rough cosmological weak freeze-out estimate, $r_f$ is about $1/6$ and free-neutron decay after freeze-out reduces it to about $1/7$, giving roughly one quarter of the baryonic mass in helium-4. The survival relation that conserves baryon number is $X_n(T_N)=r_f(1+r_f{)}^{-1}{e}^{-\Delta t/{\tau }_n}$; the corresponding $r_N$ is $X_n/(1-X_n)$.

The rough neutrino decoupling rate by itself does not determine that one-quarter value. Substituting $T_d=1.8\mathrm{MeV}$ literally into $e^{-Q/T_d}$ would give $r\simeq 0.49$ and $Y_4\simeq 0.65$ before neutron decay. Neutron-proton chemical equilibrium persists to a lower temperature, approximately $0.7$--$0.8\mathrm{MeV}$ once the actual conversion rates are included. The rate scaling fixes the MeV scale; accurate abundances need the conversion coefficients and the delay through the deuterium bottleneck. This sequence is described in David Tong's discussion of nucleosynthesis.

Increasing $Q$ reduces the equilibrium neutron-to-proton ratio, so the primordial helium abundance decreases. Holding the freeze-out temperature $T_f$ and the decay survival factor fixed, the primordial helium response to the neutron-proton mass difference isFor $Q\simeq 1.29\mathrm{MeV}$ and $T_f=0.7$--$0.8\mathrm{MeV}$ this is about $0.85$--$0.87$, a reduction of roughly fifteen percent. A larger $Q$ also changes neutron decay and detailed weak rates; their quantitative effects require more microscopic information. The sign from the Boltzmann factor is already clear.

Use natural units and the unreduced Planck mass $M_P=G^{-1/2}$; it is $\sqrt{8\pi }$ times the reduced Planck mass. A homogeneous canonical inflaton has ${\rho }_{\varphi }={\dot{\varphi }}^2/2+V$ and $P_{\varphi }={\dot{\varphi }}^2/2-V$. Vacuum domination by a scalar field requires ${\dot{\varphi }}_i^2/2\ll V({\varphi }_i)$, with other matter and the curvature term negligible in the Friedmann equation. The equation-of-state parameter is then near $-1$, and the Friedmann acceleration equation gives accelerating expansion.

For sustained slow-roll inflation, Hubble friction must dominate the inflaton acceleration: $|\ddot{\varphi }|\ll 3H|\dot{\varphi }|$, with the scalar potential changing only slightly in one Hubble time. In terms of the potential slow-roll parameter and second potential slow-roll parameter, this requiresFor the quartic scalar potential,Thus $|{\varphi }_i|\gg M_P$ supplies a broad slow-roll range when $\lambda \ll 1$. At an initial density of order the Planck density, potential domination givesso both potential slow-roll parameters are small for small $\lambda$. Small initial kinetic energy is a separate condition; the height of the scalar potential alone does not enforce it. These classical equations describe the post-Planck evolution under the model assumptions, rather than resolving quantum gravity at the initial endpoint.

Take ${\varphi }_i>0$; the negative branch follows by the symmetry $\varphi \mapsto -\varphi$. Dropping kinetic energy from the Friedmann equation and acceleration from the inflaton equation of motion givesWriting $b=M_P\sqrt{\lambda /(6\pi )}$, the quartic-potential slow-roll solution isIndeed, $\dot{\varphi }\simeq -b\varphi$, and integrating $H=\dot{a}/a$ gives $\log (a/a_i)=\pi ({\varphi }_i^2-{\varphi }^2)/M_P^2$. These expressions apply during slow-roll inflation; their formal late-time saturation must not be used beyond the end of that approximation.

The exact end of accelerating expansion is ${ϵ}_H=-\dot{H}/H^2=1$, equivalently ${\dot{\varphi }}^2=V$ for the scalar-dominated flat model. The quartic-inflation endpoint estimate uses the leading potential condition ${ϵ}_V\simeq 1$, givingThe numerical coefficient is an endpoint estimate: the slow-roll approximation breaks down there. Imposing $|{\eta }_V|=1$ estimates when slow roll loses accuracy and gives a different order-one coefficient, not a different exact acceleration condition.

The slow-roll e-fold count isThe quartic-inflation expansion from Planck density combines this initial estimate with the conventional endpoint to giveFor small $\lambda$ the dominant term is enormous; the order-one endpoint uncertainty does not affect that conclusion.

To estimate the reheating temperature, assume prompt, efficient reheating converts the end-of-inflation energy into a thermal radiation bath. With the same endpoint convention, $V_e\simeq \lambda M_P^4/(4{\pi }^2)$, and equating this to the radiation energy density givesThe quartic-inflation prompt-reheating estimate fixes the scaling; the endpoint kinetic energy changes the coefficient by order one. Without an inflaton coupling and its thermalization history, $V$ alone cannot specify the actual reheating temperature.

Finally, during cosmic inflation the comoving Hubble radius decreases:An initially causally connected patch is enlarged enough that the present observable region can lie within it. Its portions may subsequently be far outside one another's Hubble radii while retaining their common earlier thermal history. This solves the horizon problem if the number of e-folds is sufficiently large, conventionally of order sixty for the relevant thermal history; $N\sim 2\pi /\sqrt{\lambda }$ can readily exceed that. The necessary count depends on reheating and later expansion. The causal mechanism and the conversion back to a hot universe are discussed in David Tong's account of inflation.

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.

Use Planck units for the following formulas. The black-hole area theorem is a classical statement with hypotheses. Assume the Einstein field equations, the null energy condition and appropriate global future regularity/predictability. For a clean version of the focusing sketch, take the future horizon generators to remain on a regular horizon and to be future complete in affine parameter. The usual strong asymptotic predictability formulation supplies the global causal control needed to exclude the same focusing pathology; an arbitrary spacetime with naked future breakdown is not covered by the theorem.

Let ${\ell }^a$ be an affinely parametrized generator of a smooth portion of the event horizon, and let $\theta =d\log dA/d\lambda$ be its null expansion. The horizon is a null hypersurface, so its generators have zero null twist. In four dimensions the Null Raychaudhuri equation isThe Einstein field equations give $R_{ab}{\ell }^a{\ell }^b=8\pi T_{ab}{\ell }^a{\ell }^b\ge 0$, since the metric terms vanish when contracted with a null vector. The screen-space null shear term is also nonnegative, so ${\theta }^{\prime }\le -{\theta }^2/2$. If ${\theta }_0<0$ at ${\lambda }_0$, integrating this inequality while the congruence remains regular givesIt focuses within affine distance at most $2/|{\theta }_0|$. The vanishing area element is a focal point to an earlier spacelike horizon cut. Past such a point the null geodesic can no longer generate an achronal boundary, but the event horizon is precisely such a boundary. Future completeness and the global hypotheses exclude its escaping this contradiction by ending prematurely at a pathology. Therefore $\theta \ge 0$ everywhere the smooth horizon description applies.