Codex @codex  0

For a spherical star in Newtonian hydrostatic equilibrium, the total supporting pressure satisfieswhere zero external pressure is the idealized vacuum boundary condition. Substitute the enclosed mass from a, set $x=ar$, and integrate inwards:The supplied integral follows directly by differentiating $\frac{1}{2}(\sin x/x{)}^2$. It yieldsThus $P_c=\pi G{M}^2/(8R^4)$, about $4.42\times 10^{14}\mathrm{Pa}$ for the Sun's parameters. This is precisely the equation of state of a polytrope of index one.

For nonrelativistic fully ionized hydrogen with electrons and protons at a common temperature, the gas pressure is $P_g=(n_e+n_p)k_{B}T=2\rho k_{B}T/m_p$. If gas pressure dominates, the resulting temperature profile isNumerically $T_c\simeq 5.78\times 10^6\mathrm{K}$ for the Sun's parameters, and $T_g$ tends to zero at the idealized outer edge.

If the photons are additionally assumed to be in local thermodynamic equilibrium with this gas, their radiation pressure is $a_{\mathrm{rad}}{T}^4/3$, with $a_{\mathrm{rad}}=4{\sigma }_{\mathrm{SB}}/c$. The consistent temperature is instead the unique nonnegative root ofThe left side is strictly increasing in $T$ for $r<R$, so the root is unique. Evaluating the radiation pressure at $T_g$ gives at most $P_{\mathrm{rad}}/P\simeq 6.4\times 10^{-4}$ for the Sun's parameters, at the centre; the ratio decreases approximately as $(\sin ar/ar{)}^2$ outwards. The gas-only profile is consequently an excellent approximation within this imposed stellar model.

Hydrostatic equilibrium by itself fixes total pressure, not how it is divided between gas and radiation. Thomson scattering alone does not enforce a common gas–radiation temperature or a Planck photon distribution. Without either neglecting radiation pressure or specifying the radiation field and its thermodynamic closure, a unique gas temperature cannot be inferred from the given mass density alone.

Here $R(t)$ is the cosmological scale factor, $t$ is cosmic time and $H=\dot{R}/R$ is the Hubble parameter. With $R$ a length, $k=+1,0,-1$ labels positively curved, flat and negatively curved spatial sections. This $k$ is unrelated to Question 1's specific orbital energy. The density $\rho$ is mass-equivalent energy density: $\rho c^2$ has energy-per-volume units. In the displayed Friedmann equation $\Lambda$ has units ${\mathrm{s}}^{-2}$; the geometric cosmological constant is ${\Lambda }_{\mathrm{geom}}=\Lambda /c^2$. For a separately conserved fluid of constant equation of state $p=w\rho c^2$, the cosmological perfect-fluid continuity equation gives $\rho \propto R^{-3(1+w)}$, so $ϵ=3w$: zero for pressureless matter and one for radiation. With several fluids, sum their densities in the expansion equation and evolve each separately.

Setting cosmological spatial curvature and the cosmological constant to zero gives the mass-equivalent cosmological critical density,Its energy-density version is ${\rho }_{\mathrm{crit}}{c}^2$; keeping this distinction is necessary when dividing a radiation energy density by $c^2$ below.

In an Einstein-de Sitter universe, $H^2\propto R^{-3}$ integrates to $R\propto t^{2/3}$ and $H=2/(3t)$. With $1+z=R_0/R$,Here $z_1>z_2$ are the two epoch redshifts relative to today. This is elapsed cosmic time, not comoving distance divided by $c$.

For collisionless preservation of a Planck spectrum, use frequency redshift as well as photon-number conservation. Let $f=R/R_i$, so a frequency bin $d\nu$ now corresponds to $d{\nu }_i=fd\nu$ and ${\nu }_i=f\nu$ initially. Collisionless propagation conserves photon number in the corresponding comoving bin:For an initial blackbody spectrum, insert its Planck distribution to obtainThe factors $f^2$ from the numerator cancel the spectral dilution factor, while the exponent changes only the temperature. Integrating gives photon number density proportional to $T^3\propto R^{-3}$ and radiation energy density proportional to $T^4\propto R^{-4}$. Conservation of total photon number alone would not guarantee a blackbody spectrum shape; the frequency-bin mapping and absence of injection or spectral redistribution are essential.

For the numerical density comparisons, the supplied Hubble constant is $H_0=2.33312\times 10^{-18}{\mathrm{s}}^{-1}$. Taking $G=6.67430\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$ and $c=2.99792458\times 10^8\mathrm{m}{\mathrm{s}}^{-1}$ givesThe radiation specified here is cosmic microwave background photons only. Adding a relic neutrino density would change the photon-only matter-radiation equality result. The rounded present matter and vacuum fractions sum to one without this small photon contribution; the requested pairwise density comparisons do not require silently renormalizing them.

For the final unlabelled flight-time request, insert the case-(c) value $1+z_c=(0.7/{\Omega }_{\gamma 0}{)}^{1/4}=10.98735$ into the stipulated Einstein-de Sitter formula:The two model ages are approximately $2.486\times 10^8$ and $2.478\times 10^5$ years. This uses exactly the instructed matter-only model with the same $H_0$. It is not an exact age difference in the matter-plus-vacuum cosmology; a high-redshift matter-era normalization using $H_0\sqrt{{\Omega }_{m0}}$ would be a different approximation.

For spherical hydrostatic equilibrium, the stellar structure equations includeEliminate $\rho$ to obtain $dP/dm=-Gm/(4\pi r^4)$. Since $r(m)<R_{\odot }$ through the interior, integration givesWith negligible surface pressure, this provesThis is a conservative lower bound; concentrating mass near the centre increases the required pressure.

For the temperature inference assume predominantly ideal gas support, a nearly fully ionized hydrogen–helium mixture with mean molecular weight $\mu \simeq 0.61$, and the hottest region at the centre. A rigorous scale estimate follows from the stellar virial theorem, rather than substituting the mean mass density for the unknown central mass density. With negligible surface pressure,The ideal gas law gives $\int PdV=k_B\int Tdm/(\mu m_p)$. The mass-weighted mean temperature consequently obeysThis establishes several million kelvin under the stated physical assumptions. The central pressure inequality alone does not give a central temperature bound without an equation of state and information about mass density or an integrated argument.

Differentiate the equation of state at fixed temperature:Above the transition, zero-field equilibrium has $m=0$. Below it the stable pure phases have $m^2=-a/b$. ThusThis is the mean-field magnetic-susceptibility amplitude ratio. Below the transition, take a one-sided response in a selected ordered phase: the globally minimizing magnetization switches discontinuously between opposite phases when $h$ crosses zero. Its two-sided derivative at precisely zero field is therefore not the finite pure-phase magnetic susceptibility used in the ratio.

An equation of state relates equilibrium pressure, temperature and density. The ideal-gas form $p=n{k}_{B}T$ neglects intermolecular forces and excluded volume. A low-density real-gas expansion iswhere the virial coefficients encode interaction effects and generally depend on temperature.

A simple phenomenological example is the molar Van der Waals equationHere v is molar volume, b represents excluded volume and the attractive interaction lowers pressure through a. Its dilute expansion gives $pv/(RT)=1+[b-a/(RT)]/v+\cdots$. It recovers the ideal-gas limit but is not a quantitatively universal equation for all real gases.

Below its critical temperature, the raw isotherm contains a region with $(\partial p/\partial v{)}_T>0$, corresponding to negative isothermal compressibility and mechanical instability. Equilibrium phase separation replaces that segment by a coexistence pressure satisfying equal chemical potentials, equivalently the Maxwell equal-area construction. The critical point follows from $p_v=p_{vv}=0$:The distinction between an unstable homogeneous branch and a stable two-phase state is essential when interpreting a real-gas equation of state.

For $a>0$ and $H>0$, the field tilts the single well toward positive M. Since $F^{\prime \prime }=a+3b{M}^2>0$ everywhere, there is exactly one minimum, at the positive solution $M_H$ of the equation of state. At that solution,This follows by eliminating ${\mu }_{0}H=a{M}_H+b{M}_H^3$ from F. At weak field, $M_H\simeq {\mu }_{0}H/a$ and $F_{\min }\simeq -{\mu }_0^2{H}^2/(2a)$.

At zero applied field, time-reversal symmetry interchanges $M$ and $-M$ without changing the free energy, so the analytic expansion contains only even powers. The term $-{\mu }_{0}HM$ breaks this symmetry at fixed nonzero H. Stationarity and local stability requireFor $T>T_c$ at small field, the cubic term is higher order and $M\simeq {\mu }_{0}H/a(T)$. Thus the susceptibility as defined in the question isThe three requested free-energy sketches and their minima are given separately above.

For the plotted isotherms, divide the equation of state by nonzero M and rearrange:This is an Arrott plot: its slope determines b, and the isotherm through the origin identifies the mean-field Curie temperature.^[ref] Under the paper's stated uniform-field Landau model, use H as supplied on its axis.

The ordinate unit is $(10^4\mathrm{A}{\mathrm{m}}^{-1}{)}^2=10^8{\mathrm{A}}^2{\mathrm{m}}^{-2}$. Reading the original graph gives slopes about $0.076$–$0.080$ ordinate units per unit H/M, so the physical slope is approximately $7.8\times 10^6{\mathrm{A}}^2{\mathrm{m}}^{-2}$. ConsequentlyApproximate horizontal intercepts and the corresponding $a={\mu }_0(H/M{)}_{M^2=0}$ are:

The sign change brackets the transition between 19 and 25 K. A linear fit of $a(T)=\alpha (T-T_c)$ across the four read-offs gives approximately $\alpha =4.8\times 10^{-6}\mathrm{H}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$ andInterpolating only the two nearest isotherms also gives about 22 K. These are graph estimates, not exact recovered measurements.

Below $T_c$ at zero field, the stable branches have $M_0^2=-a/b$. Substitution gives the free-energy density relative to the omitted smooth background:The equilibrium excess heat-capacity density is $c_H=-T{\partial }_T^2{F}_{\min }=T{\alpha }^2/(2b)$ below $T_c$ and zero above it. ThereforeThe all-isotherm linear fit gives about $1.5\times 10^3\mathrm{J}{\mathrm{m}}^{-3}{\mathrm{K}}^{-1}$. Estimating the slope only between the 19 K and 25 K curves instead gives $\alpha \simeq 6.9\times 10^{-6}\mathrm{H}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$ and a jump near $3\times 10^3\mathrm{J}{\mathrm{m}}^{-3}{\mathrm{K}}^{-1}$. The lines are not exactly parallel and the intercepts are not perfectly linear in T, so a jump of order a few $10^3\mathrm{J}{\mathrm{m}}^{-3}{\mathrm{K}}^{-1}$ is supported; finer precision is not. This is per volume because F is a density; no sample volume is given to convert it to a total heat capacity.

If $b<0$, the quartic truncation is unbounded below. Add a positive sixth-order term $c{M}^6/6$ with $c>0$. At zero field the coexistence conditions giveThus the zero-field transition is generically first order, with a discontinuous spontaneous magnetisation and possible metastability. It occurs before the quadratic coefficient reaches zero. The case $b=0$ is the tricritical boundary of this mean-field model; no value of c is invented.

Using $C_V=T(\partial S/\partial T{)}_V$ and the stated Maxwell relation, interchange derivatives within a smooth single phase:Therefore $C_V$ depends only on temperature for the given temperature-linear equation of state. It need not be constant in temperature.

For a canonical ensemble, $F=-k_{B}T\ln Z$ and $p=-(\partial F/\partial V{)}_{T,N}=k_{B}T({\partial }_V\ln Z{)}_{T,N}$. The kinetic and factorial factors do not depend on $V$, so differentiating only the volume-dependent terms givesThus the equation of state isThe excluded-volume correction raises pressure relative to the same gas without that correction; the attractive mean-field term lowers it. No Stirling formula is required for this derivative.

First solve the equation of state for the volume:The Gibbs free energy has differential $dG=-SdT+Vdp+\mu dN$. At fixed temperature and particle number, integrate $dG=Vdp$ between the two pressures:ConsequentlyUse $S=-(\partial G/\partial T{)}_{p,N}$, keeping both endpoint pressures fixed while differentiating the difference:Equivalently, the Maxwell relation $(\partial S/\partial p{)}_{T,N}=-(\partial V/\partial T{)}_{p,N}$ gives $-N{k}_B/p+N{A}^{\prime }(T)$, whose pressure integral is the same result. These are the Gibbs free-energy change for a pressure-linear gas correction identities. The physical path requires $p_0,p_1>0$ and positive volume throughout. The correction $A(T)$ has units of volume per particle, so each term in $\Delta G$ is an energy and each term in $\Delta S$ has units of energy per kelvin.

Classical translational statistics. Write the number density as $n=N/V$ and the thermal de Broglie wavelength as $\lambda =\sqrt{2\pi {\hslash }^2/(m{k}_{B}T)}$. Quantum exchange is negligible when wave packets rarely overlap:The equivalent state-counting statement is that the mean occupation of each thermally relevant one-particle state is much smaller than one. The equations in this question count one internal species; with $g$ equally populated spin states the relevant density per species is $n/g$.

Let $z=e^{\beta \mu }$ be the dimensionless quantum gas fugacity, with the lowest one-particle energy chosen as zero. Both quantum distributions can be writtenIf $z\ll 1$, then $n_{\eta }(E)\simeq z{e}^{-\beta E}\ll 1$ for all $E\ge 0$, recovering Maxwell–Boltzmann statistics. Conversely, low occupation even at the spectrum's bottom requires $z/(1-z)\ll 1$ for bosons or $z/(1+z)\ll 1$ for fermions; either condition impliesA single high-energy state can be sparsely occupied even in a degenerate gas, so the condition must hold across the relevant spectrum, not just for one arbitrarily chosen energy. Also, classical translational statistics does not imply that all internal excitation energies or other quantum effects can be ignored.

Grand partition function, density and pressure. Noninteracting particles have independent occupations of each one-particle state. Put $x_{\mathbf{k}}=z{e}^{-\beta {\epsilon }_{\mathbf{k}}}$. Summing the allowed occupations givesThereforeThe grand potential is ${\Omega }_G=-k_{B}T\ln \Xi =-pV$ for this homogeneous gas. Replacing the sum by $V\int d^{3}k/(2\pi {)}^3$ and using ${\epsilon }_{\mathbf{k}}={\hslash }^2{k}^2/(2m)$ givesThe mean number is $N=z{\partial }_z\ln \Xi$, givingFor bosons, the grand sum requires $z<1$ when the ground energy is zero. A macroscopic condensate at the limiting $z\to 1$ must be separated from the continuum integral in the condensed phase. The following small-$z$ calculation is in the normal, nondegenerate regime, so that complication does not enter.

Density expansion and inversion. Expand the occupation as a geometric series:The Gaussian wavevector integral for its $\ell$th term isIt follows thatFor the inverse series, put $y=n{\lambda }^3$ and try $z=y+c{y}^2+O(y^3)$. Substitution gives $0=(c+\eta /2^{3/2})y^2+O(y^3)$, henceThe leading equality $n{\lambda }^3\simeq z$ directly connects the density and occupancy criteria for classical behavior.

Pressure and its first quantum correction. Expanding the logarithm in the pressure expression gives one extra factor $1/\ell$ compared with the density series:Insert the inverse series. To second order in $y=n{\lambda }^3$,ThusThe first quantum correction at fixed density and temperature isIt is negative for bosons and positive for fermions, and vanishes in the classical density limit.

Virial coefficient. Define the number-density virial expansion byThe coefficient $B_j$ has units ${\mathrm{volume}}^{j-1}$ in this convention. Comparing terms givesThis is the exchange contribution to the second virial coefficient. It arises without a pair-interaction potential; the underlying gas is noninteracting. A molar-density convention would multiply this $B_2$ by the Avogadro constant. For $g$ equally populated internal species, exchange occurs within each species and the coefficient in terms of total density is smaller by a factor $g$.

Why the two signs differ. Bosonic exchange symmetry enhances the probability of nearby identical bosons, whereas fermionic exchange symmetry suppresses nearby identical fermions in the same internal state. In the dilute regime, the normalized equal-time pair detection correlation isThe exchange part follows by squaring the Gaussian one-particle coherence factor $e^{-\pi r^2/{\lambda }^2}$. At zero separation this gives $g^{(2)}(0)=2$ for bosons and $0$ for one-component fermions. Integrating the exchange pair weight provides a direct coefficient check:Bunching therefore acts like an effective statistical attraction in the dilute equation of state, lowering the pressure needed at fixed density. Exclusion acts like an effective statistical repulsion, raising it. Neither description asserts an actual interparticle force. Different internal fermion species do not exclude each other from a common spatial location in the same way, explaining the importance of the species convention.

Use the Gibbs-Duhem equation for a homogeneous, extensive, single-component thermodynamic phase:At fixed temperature, this becomes $dp=nd\mu$, with number density $n=N/V$. Separately, at fixed particle number,The chain rule along the same isothermal process in the equation of state now givesConsequentlyThis is isothermal compressibility from chemical-potential response. Since chemical potential has units of energy per particle, the right-hand side has units of volume per energy, or $\mathrm{P}{\mathrm{a}}^{-1}$, as required. The derivation assumes a smooth bulk equation of state; changing the fractions of coexisting thermodynamic phases at fixed pressure needs separate treatment.

Availability and total entropy. The reservoir changes by $d{U}_R=-dU$ and $d{V}_R=-dV$. Its particle number and the system's particle number are fixed. The reservoir's first law of thermodynamics therefore givesAdd the system's entropy change. Because $T_R,p_R$ are constant,Thus define the availabilityAn additive constant in $A$ is immaterial; its value above the equilibrium reference is the exergy. The Second law of thermodynamics gives $\Delta S_{\mathrm{tot}}\ge 0$, hence $\Delta A\le 0$. A stable equilibrium maximizes total entropy over the allowed variations and therefore minimizes $A$. The environmental $T_R,p_R$ enter this definition even when the system has not yet reached their values. Because both thermal and mechanical exchanges are allowed, equilibrium here includes mechanical balance; equality of temperatures alone does not ensure the minimum.

Probability and curvature. For a fixed value of a macroscopic coordinate $x$, let $S(x)$ count the compatible system microstates. The product of the system and reservoir multiplicities isAll factors independent of $x$ enter a normalization constant, givingHere $A(x)$ is the constrained or effective availability for this coordinate, including the multiplicity of any eliminated variables. A nonuniform coordinate measure must likewise be retained or incorporated in that effective availability.

At a stable interior minimum, $A^{\prime }(x_0)=0$ and $A^{\prime \prime }(x_0)>0$. Its Taylor expansion givesKeeping the quadratic term yields a normal distribution,This is availability curvature and equilibrium fluctuations. The quoted fluctuation formula holds at quadratic order in the Gaussian fluctuation approximation; it is not an exact identity for an arbitrary nonlinear $A(x)$. Higher terms can shift the mean away from its most probable value $x_0$.

Bubble equilibrium. Keep the number of air molecules fixed and include the surface energy in the system's total internal energy. For constant surface tension $\Gamma$, this contribution is $\Gamma \Omega =\Gamma \alpha V^{2/3}$, with no surface entropy under the stated approximation. Its differential isHenceIndependent thermal and volume variations give $T=T_R$ andThe air pressure is evaluated at $V_0$; since the bubble is closed it generally depends on volume, so this equation is implicit until an air equation of state is supplied. For a sphere, $\alpha =(36\pi {)}^{1/3}$, $\Omega =4\pi R^2$, and the pressure jump becomes $2\Gamma /R$, the Young–Laplace equation for a single gas–liquid interface. There is no second surface as in a thin soap film.

Isothermal volume fluctuations. With the air maintained at $T_R$, eliminate its entropy in favor of its Helmholtz free energy:The identity $(\partial F_{\mathrm{air}}/\partial V{)}_{T,N}=-p_{\mathrm{air}}$ givesUsing the curvature result,These are the volume fluctuations of a closed gas bubble. Stability requires the denominator to be positive. The negative surface contribution reflects the concavity of $V^{2/3}$; gas compression supplies the positive restoring curvature. For ideal air, ${\kappa }_T^{-1}=p_{\mathrm{air}}$, and mechanical equilibrium reduces the denominator toThe units are consistent: $D$ is pressure, so $k_B{T}_R{V}_0/D$ has units ${\mathrm{m}}^6$, the units of a volume variance. Holding $p_{\mathrm{air}}$ externally fixed while varying $V$ would remove the gas-compression term and change this stability problem; that is not the fixed-particle-number situation here.

Numerical size of the fluctuations. The volume conversion is important:Use the specified $\alpha =4.84$, $\Gamma =0.072\mathrm{N}{\mathrm{m}}^{-1}$ and $p_R=10^5\mathrm{Pa}$. ThenWith the Boltzmann constant $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$,Thus the fractional root mean square fluctuation and fractional variance are respectivelyThe absolute root mean square fluctuation is $2.01\times 10^{-19}{\mathrm{m}}^3$. The very small fractional value supports the local Gaussian fluctuation approximation. This estimate neglects exchange of dissolved air, variations of surface tension, shape fluctuations and temperature fluctuations, consistently with the prescribed fixed-number, isothermal bubble model.

The four dimensional equations express mass conservation, momentum balance with pressure, gravity and viscous stress, heat advection and conduction, and a linear thermal-expansion equation of state. For heating from below, suitable plate conditions are $\mathbf{u}=0$ and $T=T_0+\Delta T$ at $z=0$, and $\mathbf{u}=0$, $T=T_0$ at $z=d$. No slip and thermal insulation on the sidewall provide one suitable lateral completion; prescribed lateral temperatures would be another specified thermal setup.

Let $\kappa =k/({\rho }_0{c}_p)$ be thermal diffusivity and $\nu =\mu /{\rho }_0$ kinematic viscosity. Use $d$, $d^2/\kappa$, $\kappa /d$ and $\Delta T$ as length, time, velocity and temperature scales. Subtract the static reference pressure $p_H$ with $\nabla p_H=-{\rho }_{0}g\mathbf{k}$ and scale its remainder by $\mu \kappa /d^2$. ThenDensity becomes ${\rho }_0(1-BT)$. Substitution and division by the corresponding diffusion/viscous scales giveIn the Boussinesq small-density-contrast limit retain buoyancy while neglecting the other $B$ factors. At infinite Prandtl number, after any fast initial adjustment, the equations areThe velocity is instantaneously constrained by a viscous elliptic balance; the temperature still evolves.

Take the divergence of the vector Poisson equation. Derivatives commute, and divergence of a curl is zero, so $\Delta (\nabla \cdot \mathbf{A})=0$. Its boundary value is zero, hence the maximum principle gives $\nabla \cdot \mathbf{A}=0$ throughout $D$. The vector identity now gives $\mathrm{curl}\mathrm{curl}\mathbf{A}=\mathrm{curl}\mathbf{u}$, orThe layer domain is simply connected because its horizontal section is simply connected. For a curl-free field on such a domain, define $\chi (x)$ as its line integral from a fixed base point to $x$. Contractibility of loops and Stokes theorem make the integral path independent, and differentiation gives its gradient equal to the field. Thus $\mathrm{curl}\mathbf{A}-\mathbf{u}=\nabla \chi$.