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Because $X_i=n_i/n_B$,Using $n_B=\eta n_{\gamma }$ and the given photon density in the deuterium equilibrium abundance formula givesAlthough the exponential favors deuterium once $k_{B}T<B_D$, the very small factor $\eta$ expresses the enormous excess of photons. The high-energy tail therefore photodissociates newly formed deuterium until the temperature is well below $B_D/k_B$; this is the deuterium bottleneck.

A larger $\eta$ increases the deuterium fraction at fixed temperature, so the bottleneck ends at a higher temperature and earlier time. With weak decoupling unchanged, fewer free neutrons then beta-decay before nuclear reactions bind them. Since primordial helium production is neutron-limited, the baryon-density effect on primordial helium makes $Y_{\mathrm{He}}$ larger than the standard value.

The number of momentum states in a shell isWith $E=p^2/(2m)$ this givesAt zero temperature all states up to $E_F$ are occupied, henceThe energy is $U=dN{E}_F/(d+2)$. Since a nonrelativistic ideal gas in $d$ dimensions obeys $pV=2U/d$,

Since $⟨z⟩=(\beta mg{)}^{-1}$,Using $\log N!\simeq N\log N-N$ and $S=k_B(\log Z_N+\beta ⟨E⟩)$ givesThis has the Sackur-Tetrode form with the effective volume $A⟨z⟩$, but its additive constant is $7/2$ rather than $5/2$. The extra unit arises because the atmosphere's vertical extent grows with temperature and its mean gravitational energy per atom is $k_{B}T$.

Increasing $\kappa$ makes ${\Gamma }_W/H$ smaller at every fixed temperature, so freeze-out occurs earlier, at a higher $T$. The equilibrium value $e^{-Q/(k_{B}T)}$ is then closer to one. Although some neutrons subsequently decay, the final neutron fraction is larger, andincreases, approaching unity if the surviving neutron and proton abundances become comparable.

Such a universe begins with less hydrogen and more helium. It has less fuel for long-lived hydrogen-burning stars; stellar evolution would tend to proceed through hotter, faster helium-burning stages, reducing typical stellar lifetimes. The shorter stable-energy window, altered heavy-element production, and lower primordial hydrogen abundance would make familiar water-rich, slowly evolving habitats less likely, although a detailed conclusion would require a stellar-evolution model.

At equal pressure and temperature, ideal gases have equal molecular number density. The mean air molecular mass is $.8(28)+.2(32)=28.8$, while helium has mass $4$, so

For $B=J=0$, $m=0$ remains stationary. At large $\beta K$, evaluating the free energy near a positive $m$ makes the negative cubic energy dominate the bounded entropy cost, so some $m>0$ has lower free energy than zero. At high temperature entropy uniquely favors zero. Continuity in temperature therefore forces a transition at an intermediate ${\beta }_{\ast }$.

Conservation of thermal energy and Fourier's law give the heat equationsThe density times the specific heat capacity, ${\rho }_i{c}_i$, is the volumetric heat capacity, and $k_i$ is the thermal conductivity.

The condition $T_1(-L,t)=0$ holds the left exterior face at a fixed zero reference temperature; $T_2(L,t)=0$ does the same for the right exterior face. The condition $T_1(0_{-},t)=T_2(0_{+},t)$ expresses perfect thermal contact, with no temperature jump at the interface. Finally,expresses continuity of the signed heat flux in the positive $x$ direction: the interface neither stores nor produces heat.

For $\omega \ne 0$, substitution of the separated solutions into the heat equations givesThe zero exterior boundary values are incorporated byTo obtain a eigenvalue equation for heat conduction in two joined slabs, write $u=\omega {\alpha }_{1}L$ and $v=\omega {\alpha }_{2}L$. Continuity of temperature and heat flux at zero gives the homogeneous linear systemA nonzero amplitude pair exists precisely when its determinant vanishes, equivalentlyIf $\cos u\cos v\ne 0$, division by $k_1{k}_2{\alpha }_1{\alpha }_2\cos u\cos v$ gives the printed transcendental equationThe printed necessity claim needs the qualification that both tangents are defined. If just one of $\cos u,\cos v$ vanishes, $D(\omega )\ne 0$. If both cosines vanish, however, $D(\omega )=0$ and legitimate normal modes remain. Their temperature at the thermal interface need not be zero; the temperature condition fixes the amplitude ratio, while each heat flux is zero there.

For a concrete counterexample, take identical slabs: ${\rho }_1={\rho }_2=\rho$, $c_1=c_2=c$, $k_1=k_2=k$, and $\alpha =\sqrt{\rho c/k}$. Thenwith each formula restricted to its own slab, satisfies both heat equations and all four boundary conditions. But $\omega \alpha L=\pi /2$, so neither tangent is defined. This directly disproves the unqualified printed assertion. The determinant condition retains these missing normal modes.

Finally, $\omega =0$ must be checked before dividing by it. Then $G_1=A(x+L)$, $G_2=B(L-x)$, and interface continuity gives $A=B$, while heat flux continuity gives $k_{1}A=-k_{2}B$. Positivity of the thermal conductivities forces $A=B=0$. Thus $\boxed{\omega =0\text{ gives only the trivial solution}.}$ In the determinant formula, its apparent zero at $\omega =0$ is spurious because the chosen sine basis degenerates there.

Choose the positive weightsThese are the volumetric heat capacities. The heat equations and integration by parts giveThe exterior boundary terms vanish because the temperatures there are zero. At the thermal interface, the remaining term isby continuity of temperature and heat flux. Hence the weighted energy estimate isBoth weights are positive, so $I\ge 0$. Zero initial temperatures give $I(0)=0$, forcing $I(t)=0$ for all $t>0$. Each temperature therefore vanishes throughout its slab. Applying this argument to a difference also proves uniqueness for arbitrary common initial data in this thermal interface problem.

Assume the rod has constant cross-sectional area $A$, no internal source of heat, no bulk motion, and no lateral heat flux. Assume also that each section has a uniform temperature, so that $T$ depends only on $x,t$, and that Fourier's law applies with the stated constant thermal conductivity $k$. The longitudinal heat flux density is then $q=-k{T}_x$.

By conservation of energy, the rate of increase of internal energy in a slice $[x,x+\Delta x]$ equals incoming minus outgoing heat:Divide by $A\Delta x$ and take the limit $\Delta x\to 0$. For a sufficiently differentiable temperature field this gives $\rho c{T}_t=-q_x=k{T}_{xx}$, and henceHere $\rho$ is mass density, $c$ is specific heat capacity, and $\kappa$ is thermal diffusivity.

A steady state has $T_s^{\prime \prime }=0$, so $T_s=a+bx$. The fixed temperature gives $T_s(0)=T_0$, and thermal insulation gives $T_s^{\prime }(L)=0$. ThusSubtracting this steady state produces the homogeneous mixed boundary conditionsThe partial differential equation and boundary conditions apply for $t>0$. The initial condition is attained as $t\downarrow 0$. Its slope at $x=L$ is $-T_0/L$, so it does not satisfy the insulated-end condition at the instant $t=0$. The resulting heat equation solution is smooth for $t>0$ and approaches the initial temperature continuously; its spatial derivative need not be continuous at $(L,0)$.

For separation of variables, put $U=X(x)H(t)$ and set $X^{\prime \prime }/X=H^{\prime }/(\kappa H)=-\lambda$. The boundary conditions are $X(0)=0$, $X^{\prime }(L)=0$. Multiplying $-X^{\prime \prime }=\lambda X$ by $X$ and using integration by parts givesA nonzero solution therefore has $\lambda >0$: the zero case forces $X$ constant, and then $X(0)=0$ forces $X=0$. Writing $\lambda ={\mu }^2$, the eigenfunctions and their temporal factors areThese sine functions are orthogonal on $[0,L]$, with squared norm $L/2$. Expanding the initial condition and multiplying by $\sin ({\mu }_{m}x)$ gives, using the permitted interchange of the sum and integral,Here $\cos ({\mu }_{n}L)=0$ and $\sin ({\mu }_{n}L)=(-1{)}^n$. Thus the Fourier series solution isThe exponential decay permits differentiation of the series for $t>0$, and each term satisfies both boundary conditions. At $t=0$ it is the mixed-end Fourier expansion of $-T_{0}x/L$.

The gained heat is the increase in internal energy. Since the paper specifies no cross-sectional area, first give the answer per unit cross-sectional area; multiply by $A$ for a rod of cross-sectional area $A$. The initial temperature has integral $T_{0}L/2$, whereas ${\int }_0^L\sin ({\mu }_{n}x)dx=1/{\mu }_n$. At $t_{\ast }=L^2/\kappa$ this givesNumerically $\Delta Q/A\simeq 0.45623855\rho c{T}_{0}L$. This is positive and less than the eventual gain $\rho c{T}_{0}L/2$, as expected from relaxation towards the uniform steady state.

The specific heat capacity $c$ is the energy needed per unit mass per unit increase in temperature. With $c$ constant, a small element of volume $dV$ has mass $\rho dV$ and thermal energy $\rho cTdV$, measured relative to the reference at which $T=0$. Summing over $\Omega$ givesChoosing a different reference temperature only adds a time-independent constant to this energy, so it does not affect the heat balance with volumetric heat generation.

Suppose $T^{(1)}$ and $T^{(2)}$ are two sufficiently regular steady state solutions with the same prescribed temperature on the boundary. Their difference $w=T^{(1)}-T^{(2)}$ satisfiesThe identical PDE source terms cancel. Apply the product rule and the divergence theorem to $wk\nabla w$; this is the weighted form of Green's first identity:Since $k>0$, the nonnegative integrand must vanish, so $\nabla w=0$. Hence $w$ is constant on each connected component of $R$, and its zero Dirichlet boundary condition makes each constant zero. Therefore $\boxed{T^{(1)}=T^{(2)}}$: the solution, if it exists, is unique. This proves Dirichlet uniqueness for variable-conductivity heat conduction; it does not require a constant thermal conductivity.

Let $A$ be the constant cross-sectional area, and suppose that the temperature is uniform on each cross-section. Consider a slice $[a,b]$. Its stored thermal energy is $\rho cA{\int }_a^{b}T(x,t)dx$. Since the lateral surface is insulated and there is no internal heat generation, conservation of energy giveswhere $q$ is the heat flux in the positive $x$ direction. By Fourier's law, $q=-k{T}_x$, soThe slice is arbitrary; therefore the local heat equation isThe constant $\kappa$ is the thermal diffusivity. The cancellation of $A$ uses a rod of uniform cross-section; the one-dimensional model assumes no transverse temperature variation.

Subtract the constant steady state profile: set $v=T-T_0$. ThenThe last equation follows from Fourier's law and zero heat flux at the insulated end. For a nonzero separated solution $v=X(x)H(t)$, separation of variables givesTo see that $\mu >0$ for a nonzero eigenfunction, multiply the spatial equation by $X$ and use integration by parts:Both boundary terms vanish. If the right side were zero, $X$ would be constant and $X(0)=0$ would make it zero. Thus write $\mu ={\lambda }^2$ with $\lambda >0$. Solving gives $X=A\sin (\lambda x)$, and $X^{\prime }(L)=0$ requires $\cos (\lambda L)=0$. The mixed Dirichlet-Neumann modes on an interval therefore haveThe eigenvalues of $-d^2/d{x}^2$ are ${\mu }_n={\lambda }_n^2$; ${\lambda }_n$ are their positive square roots. By superposition, with coefficients for which the resulting series and its required derivatives converge at positive times,The eigenfunctions are orthogonal on $[0,L]$ and have squared norm $L/2$. Expanding the initial excess temperature $v(x,0)=T_{0}x/L$ givesThus the requested temperature isAt the insulated endpoint $\sin ({\lambda }_{n}L)=(-1{)}^{n-1}$ cancels the alternating coefficient sign, leavingEvery summand is positive. The endpoint temperature never drops below $T_0$; it decreases towards $T_0$ as $t\to \infty$ and starts at $T(L,0)=2T_0$.

For completeness, the Sturm-Liouville eigenfunction expansion gives the initial profile in the $L^2$ sense. Here the coefficients are absolutely summable, so the Weierstrass M-test also gives uniform convergence on $[0,L]\times [0,\infty )$, and the initial values are recovered everywhere. For positive times the exponential factors justify all required termwise derivatives, giving a classical solution. There is, however, an initial incompatibility with a Neumann boundary condition at $(L,0)$: the initial slope is $T_0/L$, while the insulated Neumann boundary condition requires zero slope for $t>0$. The temperature is continuous to the initial edge, but $T_x$ cannot extend continuously to that corner. One must not differentiate the initial series term by term at $x=L$.

If $Pdx+Qdy=dF$ for a twice continuously differentiable potential $F$, then $P=F_x$ and $Q=F_y$. Equality of mixed partial derivatives gives the necessary condition for an exact differential:This condition alone does not ensure global exactness on a domain with holes.

For the Helmholtz free energy $A=U-TS$, use the stated identity for internal energy and the product rule:Thus $A_T{|}_V=-S$ and $A_V{|}_T=-p$. Equality of mixed partial derivatives yields the Maxwell relationThe temperature and volume are the independent variables here, with composition and amount of gas fixed.

Next express $S=S(T,V)$ and substitute its total differential into $dU=TdS-pdV$:Matching the coefficient of $dV$ givesThis is the thermodynamic internal pressure identity: an equation of state determines how internal energy changes under an isothermal change of volume.

For the Dieterici equation, set $c=an/(VR)$ while differentiating at fixed $V$. ThenSubstitute into the thermodynamic internal pressure identity. The term proportional to $T$ cancels $p$, leavingThe physical gas domain has $T>0$ and $V>nb$.

For the ideal gas instead, $p=nRT/V$ gives $T(\partial p/\partial T{)}_V-p=0$. Therefore $U$ is independent of volume at fixed amount of gas. Integrating the remaining condition $(\partial U/\partial T{)}_V=C_V$, with constant heat capacity at constant volume, giveswhere $U_0$ is independent of both $T$ and $V$ and fixes the reference for internal energy.

Insert the linear initial profile into the transformed integral from (b):The constant part integrates to $ax$. The other part is an absolutely integrable odd function, so its integral vanishes. HenceAs a direct check, $u_t=u_{xx}=0$ and $u(x,0)=ax$. Thus a linear temperature profile is stationary on the whole line. More generally, every affine function is preserved by this heat-kernel solution.

Along a change in volume at fixed temperature, the first law of thermodynamics givesUse the Helmholtz Maxwell relation derived in (a) to replace the entropy derivative. This provesFor the stated Van der Waals equation, temperature-independent $a,b$ give $P_T{|}_V=R/(V-b)$. Hence $U_V{|}_T=a/V^2$. Integrating with respect to $V$ yieldsThe dilute-limit condition holds at every fixed $T$ and forces $h(T)=cT$, soThe result is the internal energy of a van der Waals gas in the fixed-amount convention of the question. Its volume derivative is positive even though the attractive energy correction itself is negative.