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Let the equilibrium ideal gas be at rest, with uniform ${\rho }_0,p_0$ and ${\mathbf{B}}_0=B_0{\mathbf{e}}_z$. This local wave calculation neglects gravity, viscosity and magnetic diffusion. Define the adiabatic sound speed $c\equiv v_s=(\gamma p_0/{\rho }_0{)}^{1/2}$, the Alfvén speed $v_a=B_0/({\mu }_0{\rho }_0{)}^{1/2}$, and a wavevector $\mathbf{k}=k(\sin \theta ,0,\cos \theta )$, where $k=|\mathbf{k}|$ and $\theta$ is its angle to the equilibrium magnetic field. Perturbations have dependence $e^{i\mathbf{k}\cdot \mathbf{x}-i\omega t}$ and phase speed $v=\omega /k$.

The linearized ideal magnetohydrodynamic equations for mass conservation, pressure, induction and momentum giveThe relation between pressure and mass density follows from eliminating $\mathbf{k}\cdot {\mathbf{u}}^{\prime }$ between the two material evolution equations; it applies to the nonzero-frequency wave sector. The induction result automatically has $\mathbf{k}\cdot \mathbf{b}=0$.

The $y$ MHD wave polarization is perpendicular to both the field and wavevector. It has $\delta p=\delta \rho =0$ and obeysThus the transverse Alfvén wave hasIn the $x$-$z$ plane the two remaining velocity components instead obey the symmetric matrix eigenproblemTaking its determinant gives $v^4-(c^2+v_a^2)v^2+c^2{v}_a^2{\cos }^2\theta =0$. Solving this quadratic equation in $v^2$ derives both magnetoacoustic wave phase speeds,The plus sign is the fast magnetosonic wave, the minus sign the slow magnetosonic wave. The radicand is nonnegative, since it equals $[(c^2-v_a^2{)}^2+4c^2{v}_a^2{\sin }^2\theta ]/4$. Both squared speeds are nonnegative. Each nonzero mode has forward and backward propagation; an advected entropy disturbance is a separate zero-frequency mode in this static equilibrium.

Take $B_0=B_z>0$, $v_a=B_0/\sqrt{{\mu }_0{\rho }_0}$ and $\zeta =k(z-v_{a}t)$. The transverse ideal magnetohydrodynamic induction equation and magnetohydrodynamic momentum equation are satisfied by $u_x=-B_x/\sqrt{{\mu }_0{\rho }_0}$, $u_y=u_z=0$, with constant mass density. However, the longitudinal magnetohydrodynamic momentum equation also requiresHere $B_x^2=a^2{B}_0^2{\cos }^2\zeta$ is not constant. Balancing its gradient would requireThe adiabatic pressure equation instead gives ${\partial }_{t}p=0$, since this velocity is divergence-free and has no component along the spatial variation. No choice of $p_{\ast }(t)$ removes the spatially varying time derivative. Thus a nontrivial finite-amplitude linearly polarized sinusoidal Alfvén wave is not an exact pure traveling wave in a compressible ideal fluid. The magnetic-pressure obstruction to a linearly polarized Alfvén wave is quadratic in $a$, so the wave is a valid linear approximation and drives a compressive response at higher order. In an incompressible model pressure is a constraint variable; it is not governed by the same compressible adiabatic evolution law.

Allowing longitudinal flow and varying mass density does not rescue a regular one-dimensional periodic traveling solution with precisely this transverse magnetic field. To see this, let $U=u_z-v_a$ and use the traveling coordinate. Integrated continuity, induction and transverse momentum givewhere $m,A,D$ are constants. Eliminating $u_x$ givesAt any zero of $B_x$, finite mass density and velocity require the right side to vanish. Wherever $B_x\ne 0$, the coefficient then vanishes and fixes $\rho ={\mu }_0{m}^2/B_0^2$, constant; continuity fixes $U$, and adiabatic entropy transport fixes $p$. The longitudinal magnetohydrodynamic momentum equation again contradicts the varying $B_x^2$. The alternative $m=0$ makes transverse momentum force $B_x$ constant. Trivial cases such as $a=0$ or $k=0$ are excluded from the nontrivial wave conclusion.

On one poloidal flux surface, the field-line angular velocity $\omega (\psi )$ is constant. In a frame rotating at that rate the relative velocity is $\mathbf{v}=\mathbf{u}-R\omega {\mathbf{e}}_{\varphi }=k\mathbf{B}/\rho$, parallel to the magnetic field. The Lorentz force is perpendicular to this motion, and the Coriolis force also does no work. Centrifugal acceleration comes from the effective potential $-R^2{\omega }^2/2$. Since entropy is constant along the surface, pressure work integrates to specific enthalpy $w$, with $dw=dp/\rho$ along the flow.

Therefore kinetic energy relative to the rotating poloidal flux surface, enthalpy and effective gravitational-plus-centrifugal potential have a constant sum:This is a corotating energy integral, not the laboratory energy alone: Maxwell stress tensor stresses can transfer energy and angular momentum in the laboratory frame. In fact the expression equals $|\mathbf{u}{|}^2/2+w+\Phi -\omega R{u}_{\varphi }$, namely the total magnetohydrodynamic Bernoulli invariant minus $\omega \ell$. Differential rotation between surfaces does not invalidate the argument, because the chosen angular velocity is fixed along the particular surface being followed.

At leading anisotropic order, perpendicular Alfvénic turbulence evolves independently of the compressive fluctuations. The slow modes and the entropy mode are transported by that turbulence. The fast magnetosonic frequency is of order $k_{\perp }(v_A^2+c_s^2{)}^{1/2}$, much larger than the reduced Alfvén frequency; the low-frequency reduced magnetohydrodynamics ordering filters out that fast-wave sector.

The passive evolution can be made explicit for a uniform ideal adiabatic plasma. Let $c_s^2=\gamma p_0/{\rho }_0$, $\delta =\delta \rho /{\rho }_0$, $b=\delta B_{\parallel }/B_0$, and define the normalized entropy perturbationUse $D={\partial }_t+{\mathbf{u}}_{\perp }\cdot {\nabla }_{\perp }$ and ${\nabla }_{\parallel }={\partial }_z+(\delta {\mathbf{B}}_{\perp }/B_0)\cdot {\nabla }_{\perp }$. Perpendicular force balance makes the leading total-pressure perturbation vanish:The ideal entropy equation gives $Ds=0$. The leading continuity and parallel-induction equations have the same compressive divergence term; subtracting them gives $D(\delta -b)=-{\nabla }_{\parallel }{u}_{\parallel }$. Parallel momentum gives $D{u}_{\parallel }=-{\nabla }_{\parallel }\delta p/{\rho }_0$. PutThe two compressive equations therefore becomeLet $c_T=v_A{c}_s/(v_A^2+c_s^2{)}^{1/2}$ be the tube speed and $h=c_s^{2}r/c_T$. ThenThus the slow and entropy fluctuations in reduced magnetohydrodynamics have characteristic amplitudes $f^{\pm }=u_{\parallel }\pm h$ obeyingThe two slow-wave populations propagate in opposite directions along the bent field lines while being mixed across them by ${\mathbf{u}}_{\perp }$. The entropy perturbation is carried by the fluid without a restoring-wave frequency. A pure entropy perturbation has $r=b=u_{\parallel }=0$ and $\delta =-s/\gamma$, so its density variation is accompanied by a temperature change with no leading pressure variation.

The coefficients and advecting fields in these three scalar equations come from the Alfvénic cascade; the scalars do not enter the leading perpendicular Alfvén equations. They therefore develop small perpendicular scales through passive mixing. If a scalar $f$ has constant variance flux ${\mathcal{F}}_f$ and is mixed on the Goldreich–Sridhar time ${\tau }_{\ell }\sim {\mathcal{E}}^{-1/3}{\ell }_{\perp }^{2/3}$, thenThe flux and amplitude need not equal the Alfvénic energy flux. This passive-spectrum statement uses the same locality and stationary inertial-range assumptions as part (a), and ignores additional damping. If a passive mode is absent initially and is not forced, its homogeneous leading equation keeps it absent; the Alfvénic cascade does not have to generate every compressive mode.

In the ideal, leading anisotropic system, there are five separately conserved quadratic cascade quantities: the two oppositely propagating Alfvén populations, the two slow-wave populations, and the entropy fluctuation. With the amplitudes of part (b), one convenient normalization iswhere ${\mathbf{z}}_{\perp }^{\pm }={\mathbf{u}}_{\perp }\pm \delta {\mathbf{B}}_{\perp }/\sqrt{4\pi {\rho }_0}$. Constant background factors convert the velocity variances into modal energies and the entropy variance into an entropy-related free-energy contribution. These are the five quadratic cascade invariants of anisotropic magnetohydrodynamic turbulence.

To verify conservation rather than just count polarizations, the perpendicular Elsässer variables obeyDot each equation with $2{\mathbf{z}}_{\perp }^{\pm }$ and integrate. The propagation and nonlinear terms become surface fluxes of $|{\mathbf{z}}_{\perp }^{\pm }{|}^2$, and the pressure term becomes a surface flux because the corresponding field is solenoidal. For periodic boundaries, or boundaries with vanishing invariant flux, all these terms vanish and $d{I}_A^{\pm }/dt=0$. Their sum is proportional to perpendicular kinetic plus magnetic energy, while their difference is proportional to perpendicular cross-helicity; the two Elsässer energy invariants contain both pieces of information.

Likewise the slow-mode equations giveHere ${\mathbf{u}}_{\perp }$ is divergence-free and ${\nabla }_{\parallel }$ differentiates along the divergence-free vector $\hat{\mathbf{z}}+\delta {\mathbf{B}}_{\perp }/B_0$, so both integrals are boundary fluxes. Finally $Ds=0$ gives $d{I}_e/dt=-\int {\mathbf{u}}_{\perp }\cdot {\nabla }_{\perp }{s}^{2}dV=0$. This proves all five quadratic conservation laws under the same boundary assumptions.

The two Alfvén cascades interact dynamically: each population strains the other and is needed for its nonlinear transfer between scales. Their integrated invariants are nevertheless conserved separately, so this interaction does not transfer one population's invariant into the other. The slow and entropy cascades are driven by the Alfvénic mixing and field-line motion, with no feedback on the perpendicular cascade at the retained order. Their injection rates set independent fluxes; those fluxes need not be equal, even when the resulting perpendicular spectra share the $-5/3$ scaling.

The count is a count of independent quadratic modal cascade channels. Ideal passive advection also preserves higher moments, such as $\int F(s)dV$ for arbitrary smooth $F$, and analogous integrals for the slow characteristic scalars. These are not additional quadratic modal energies. Forcing and dissipative terms make the corresponding invariants source-and-sink balances; a stationary inertial range conserves their flux through scales rather than their total content in a continually forced domain.

Before electron-positron annihilation in cosmology, photons contribute two helicity states; electrons and positrons contribute two spin states each. Each of the three light neutrino flavours contributes one active neutrino helicity and one antineutrino helicity. At a common temperature the effective number of relativistic energy degrees of freedom is thereforeUsing $G_F=1.17\times 10^{-5}{\mathrm{GeV}}^{-2}$ and $M_{\mathrm{Pl}}\simeq 2.435\times 10^{18}\mathrm{GeV}$ in the scaling formula alone gives $T_F\sim 2\mathrm{MeV}$. Including the explicit expansion coefficient from (ii), but setting the unknown rate coefficient $A$ to one, gives $T_F\simeq 1.5\mathrm{MeV}$. Thus the warranted conclusion is a freeze-out temperature of order $1\mathrm{MeV}$, above the $0.511\mathrm{MeV}$ electron mass. The omitted weak-rate coefficient prevents a more precise prediction from this estimate.

Neutrino decoupling occurs before most electron-positron pairs annihilate. Afterwards neutrinos redshift freely, $T_{\nu }a=\mathrm{constant}$, while the photons receive the pair entropy. For the still-coupled photon-pair gas, cosmological entropy conservation gives $g_{\ast s}{T}_{\gamma }^3{a}^3=\mathrm{constant}$. Its entropy degrees of freedom drop from $2+(7/8)4=11/2$ to $2$. Comparing this with the freely cooling neutrinos givesAfter annihilation, photon and neutrino temperatures differ. To write the total energy density as $({\pi }^2/30)g_{\ast }{T}_{\gamma }^4$, the neutrino contribution must include that temperature ratio:This is the paper's post-annihilation ${\mathcal{N}}_{\mathrm{eff}}$. Merely deleting the electron term while keeping all species at the same temperature would be incorrect. For comparison the entropy count, normalized to $T_{\gamma }^3$, is $g_{\ast s}=2+(7/8)6(4/11)=43/11\simeq 3.91$. These results assume instantaneous neutrino decoupling and relativistic neutrinos; the two counts represent different thermal moments.

Write the photon energy density as $ϵ=a_R{T}^4$, where $a_R=4\sigma /c$ by the Stefan–Boltzmann law. The photon gas has pressure $P=ϵ/3$ and zero photon chemical potential. The first law of thermodynamics therefore givesDividing by $T$ and integrating gives the entropy density of a photon gas:The integration constant is fixed to zero at zero temperature, consistently with the third law of thermodynamics.

The inequality $\Gamma \gg H$ says interactions act much faster than the expansion time, maintaining local thermal equilibrium. For adiabatic reversible expansion with no additional source of entropy, cosmological entropy conservation fixes the entropy in a comoving volume. Since that volume scales as $a^3$, and the relativistic degrees of freedom are fixed here, $T^3{a}^3$ is constant. ThusDuring radiation domination, the Friedmann equation gives $H^2\propto \rho \propto T^4$, and hence $\boxed{H\propto T^2}$. Fast interactions alone would not exclude an additional entropy-producing process; changing relativistic species likewise changes the simple temperature scaling.

At fixed particle number, expand the entropy differential in $(T,p)$. The definition of heat capacity at constant pressure and the Maxwell relation $(\partial S/\partial p{)}_T=-(\partial V/\partial T{)}_p$ giveSimilarly expansion in $(T,V)$, using heat capacity at constant volume and $(\partial S/\partial V{)}_T=(\partial p/\partial T{)}_V$, givesAt constant pressure, $dV=V\alpha dT$. Also $dV/V=\alpha dT-\kappa dp$, so $(\partial p/\partial T{)}_V=\alpha /\kappa$. Equating the two entropy expressions at constant pressure yieldsFor a mechanically stable system with positive isothermal compressibility, this also gives $C_p\ge C_V$.

Work in a gas region with $\rho ,p>0$ and physical adiabatic exponent $\gamma >1$. With $\xi =x/t$, each time derivative is $-\xi /t$ times differentiation in $\xi$, and each spatial derivative is $1/t$ times it. The continuity equation, Euler momentum equation and the entropy advection equation for pressure reduce toTheir coefficient determinant is $(u-\xi )[(u-\xi {)}^2-c^2]$, where the adiabatic sound speed is $c=\sqrt{\gamma p/\rho }$. If the determinant is nonzero, all derivatives vanish and the solution is uniform.

The apparent third characteristic $u=\xi$ does not give a smooth fan: on such an interval $u^{\prime }=1$, whereas continuity requires $\rho u^{\prime }=0$, impossible for $\rho >0$. It cannot support an isolated smooth entropy variation either. Subtracting $\gamma$ times the logarithmic density equation from the logarithmic pressure equation gives $(u-\xi )(\ln (p/{\rho }^{\gamma }){)}^{\prime }=0$. A nonzero continuous entropy derivative would force $u=\xi$ on an interval and give the same contradiction. Thus the specific entropy is constant throughout each smooth branch.

Every nonconstant branch therefore lies in one acoustic characteristic family. Let $\sigma =\pm 1$ denote it, with $\xi =u+\sigma c$, so $u-\xi =-\sigma c$. The constant entropy means $p=K{\rho }^{\gamma }$, $K>0$, and $c^{\prime }/c=(\gamma -1){\rho }^{\prime }/(2\rho )$. Continuity gives $u^{\prime }=\sigma c{\rho }^{\prime }/\rho =2\sigma c^{\prime }/(\gamma -1)$. Differentiating $\xi =u+\sigma c$ then provesIn particular, $u-2\sigma c/(\gamma -1)=J$ is a constant Riemann invariant. All nonconstant smooth local solutions have the explicit formon an interval where $c>0$. Along the left-going family ($\sigma =-1$) sound speed decreases with $\xi$; along the right-going family ($\sigma =+1$) it increases. These are the two self-similar ideal-gas rarefaction branches. Together with arbitrary uniform positive states they exhaust the smooth local solutions. A global solution can join them piecewise at fan edges, where derivatives generally jump, and can also have discontinuities; those do not constitute additional smooth branches.

A single rarefaction wave enforces two matching equalities: equal specific entropy and one common Riemann invariant. The inequality $u_R>u_L$ does not enforce either. If the entropies differ, one smooth isentropic fan cannot connect them; even if they agree, arbitrary end velocities and pressures need not satisfy either acoustic-family invariant.

The general positive-gas Riemann problem has a left acoustic wave, a middle contact discontinuity, and a right acoustic wave. Across the contact the two intermediate states have the same velocity $u_{\ast }$ and pressure $p_{\ast }$, but can have different mass densities and entropies. Each acoustic wave is a rarefaction wave or a shock wave, depending on the required intermediate pressure relative to its adjacent initial state. Shocks satisfy the conservative jump conditions and increase entropy. A sufficiently large pressure contrast can produce a shock on one side even when the initial velocities obey $u_R>u_L$.

For sufficiently rapid separation, the two rarefactions instead exhaust the gas and create vacuum between them. Their vacuum-edge velocities are $u_L+2c_L/(\gamma -1)$ and $u_R-2c_R/(\gamma -1)$, so a finite vacuum gap occurs whenEquality is the limiting case in which the vacuum edges meet. Thus the additional features can be two acoustic waves, a contact carrying an entropy jump, shocks, or a vacuum region, rather than one fan joining the two original uniform states.

The initial-value problem is a Riemann problem. Both the one-dimensional Euler equations and the two constant initial states are unchanged under the simultaneous scaling $(x,t)\mapsto (\lambda x,\lambda t)$ for $\lambda >0$. There is no imposed length or time scale. Rescaling a physically admissible entropy solution therefore gives another solution with exactly the same initial data. Uniqueness of the physical Riemann solution makes the two coincide, soThe smooth pieces are consequently those in part (a), while shocks and contacts lie on rays $x=st$. Across shocks one uses the conservative mass, momentum and total-energy equations and the entropy admissibility condition, rather than extending the smooth entropy-advection equation through the discontinuity. Scale invariance by itself, without selecting the unique admissible solution, would not justify the uniqueness step.

For a relativistic species at zero chemical potential, $p=ϵ/3$ and $s=(ϵ+p)/T_i$, so its entropy density is proportional to $g_i{T}_i^3$. The Bose-Einstein distribution gives the bosonic normalization, while the fermionic thermal integral is $7/8$ of it. Species still in thermal equilibrium have $T_i=T$; freely streaming decoupled species can have another temperature and therefore contribute an extra factor $(T_i/T{)}^3$ when the total is expressed in units of $T^3$. This gives the two terms $N_{\ast }$ and $N_{SD}$ and their stated effective-degree-of-freedom weights.

After neutrino decoupling, the comoving entropy of the photon-electron-positron plasma is conserved separately from the neutrinos. Before electron-positron annihilation its effective entropy count is $2+(7/8)4=11/2$; afterwards it is $2$. Consequentlywhereas $a{T}_{\nu }$ stays equal to its pre-annihilation value. HenceThis is the instantaneous-decoupling, negligible residual-electron approximation used in the question. Heating of the still-coupled plasma, not cooling peculiar to neutrinos, creates the temperature difference.

Characteristics beginning at $\xi$ have $x=\xi +t{u}_I(\xi )$ and carry the constant value $u_I(\xi )$. In the sloping interval, $x=t+(1-t)\xi$. For $0\le t<1$ the resulting single-valued solution isIt is continuous and piecewise smooth, locally Lipschitz for $t<1$, but not a classical $C^1$ solution across the two derivative corners. Its weak derivatives satisfy the equation, since the function itself has no jumps there. The sloping gradient is $-1/(1-t)$ and blows up at $t=1$, when those characteristics coalesce.

Thus the continuous characteristic description has maximal interval $0\le t<1$. If “weak solution” is taken literally, it does not cease to exist then: the entropy continuation is a shock with left/right states 1 and 0 and Rankine-Hugoniot speed $1/2$, situated at $x=(t+1)/2$ for $t\ge 1$. It is a global entropy weak solution, with the step at $t=1$ as its limiting trace. This distinguishes breakdown of the characteristic parametrization from nonexistence of a weak continuation.

The exterior characteristics give $u=0$ for $x<0$ and $u=1$ for $x>t$. In the gap put $\xi =x/t$ and $u=f(\xi )$. The Inviscid Burgers equation becomes $(f-\xi )f^{\prime }=0$. Matching the increasing states selects the rarefaction wave $f(\xi )=\xi$, soIt is continuous and piecewise smooth at positive times and satisfies the entropy condition. A discontinuous shock joining the same increasing states would be nonentropic.

Introduce a constant filament bending modulus $B$, which sets the mechanical force scale. The filament has intrinsic curvature of an elastic filament ${\kappa }_0(s)$, and its actual signed curvature in a small-slope Monge representation is $\kappa \simeq {\zeta }_{xx}$. To quadratic order one may identify material arc length $s$ with $x$ in ${\kappa }_0$, whileAt fixed positive extension force, the mechanical potential is bending energy minus $FL$. Replacing $L$ by $\mathcal{L}-(\mathcal{L}-L)$ leaves, up to a force-dependent constant,Its first variation isThe higher-order Euler-Lagrange equation is thereforeFor a Fourier transform relation, take a periodic bulk segment, or a long-filament bulk approximation with negligible endpoint contributions; these are necessary boundary conditions because the question specifies no filament end conditions. Using $f(x)=\int \hat{f}(q)e^{iqx}dq/(2\pi )$, every nonzero wavenumber obeysA zero displacement mode is an arbitrary translation. In a periodic representation a constant intrinsic-curvature mode contributes a constant bending cost but no bulk slope, so it does not contribute to the following deficit. The Parseval identity givesFor clarity about finite normalization, on a periodic interval of length $L$ define ${\kappa }_n=L^{-1}{\int }_0^L{\kappa }_0(x)e^{-i{q}_{n}x}dx$, $q_n=2\pi n/L$. Then the equivalent exact quadratic-order formula isThis is force-extension of an intrinsically curved filament, determined by its particular fixed shape rather than by a thermal orientation entropy.

For an ensemble with quenched intrinsic curvature, first minimize each realization and then average the last expression, replacing $|{\kappa }_n{|}^2$ by its ensemble mean. For stationary bulk disorder, let $S_{\kappa }(q)$ be the power spectrum, normalized so that its correlation is $⟨{\kappa }_0(x){\kappa }_0(x+r)⟩=\int S_{\kappa }(q)e^{iqr}dq/(2\pi )$. The measurable mean deficit per unit projected length isIf the curvature is smooth enough that the second spectral moment is finite, one can take the large-force limit under this integral, obtaining the smooth-curvature high-force extension lawFor a stationary ensemble the integral is $L⟨({\kappa }_{0,x}{)}^2⟩$. Equivalently, the bulk slope is ${\zeta }_x\simeq -(B/F){\kappa }_{0,x}$; a plot of the mean deficit against $F^{-2}$ measures the mean squared intrinsic-curvature gradient, if $B$ is known. The $F^{-2}$ law is a smooth-curvature bulk result, not a universal law for every random filament. For example, ideal white intrinsic-curvature disorder with $S_{\kappa }(q)=S_0$ has an infinite second spectral moment; retaining the full integral instead gives $⟨\mathcal{L}-L⟩/L=S_0\sqrt{B}/(8\sqrt{F})$ in that ideal continuum model.

Endpoint conditions also matter. If each end is moment-free and has no transverse applied force, the natural conditions from the variation areat each end. These can produce a free-end curvature boundary layer of width $\ell =\sqrt{B/F}$. For a constant intrinsic filament curvature $\kappa$, integration of the higher-order Euler-Lagrange equation with zero transverse force gives $B({\zeta }_x{)}_{xx}-F{\zeta }_x=0$ and henceThe last formula follows by integrating ${\sinh }^2[(x-L/2)/\ell ]$, and tends to ${\kappa }^2{\ell }^3/2\propto F^{-3/2}$, despite the vanishing bulk curvature gradient. Thus a finite free-ended measurement must include or suppress endpoint bending before interpreting a bulk $F^{-2}$ fit.

For the freely jointed chain, rigid links have independent orientations, so the restoring force is entropic. Let $T$ now denote absolute temperature, and $\beta =1/(k_{B}T)$ with Boltzmann constant $k_B$. In the usual three-dimensional chain, a segment at polar angle $\theta$ has force energy $-Fb\cos \theta$. Its angular canonical partition function and the full chain partition function areDifferentiating $\log Z_N$ with respect to $\beta F$ yields the exact mean projected displacementHere $\mathcal{L}$ is the Langevin function; this is force-extension of a three-dimensional freely jointed chain, in the fixed-force canonical ensemble. At high extension, $\coth \xi =1+2e^{-2\xi }+\cdots$, soThus the smooth quenched-curvature bulk deficit falls as $F^{-2}$, whereas the three-dimensional freely jointed chain deficit falls as $F^{-1}$ and depends explicitly on temperature. The distinction is mechanical straightening of a fixed intrinsic shape versus orientational entropy; thermal bending fluctuations of a stiff filament have not been included in the first calculation.

If the comparison chain is also restricted to two dimensions, it is a planar freely jointed chain rather than the three-dimensional model. In this case,The last identity follows directly by differentiating the integral defining $I_0$; $I_0$ and $I_1$ are Modified Bessel functions of the first kind. At large $\xi$ the angular weight is proportional to $e^{-\xi {\theta }^2/2}$, giving $⟨\cos \theta ⟩\simeq 1-⟨{\theta }^2⟩/2=1-1/(2\xi )$. Consequently $Nb-⟨L⟩\sim N{k}_{B}T/(2F)$: the same inverse-force exponent, with half the three-dimensional coefficient. Both versions assume inextensible links and forces below the regime where the links themselves stretch.

The photon number density is obtained by integrating the Planck photon distribution. With $x=h\nu /(kT)$,The integral converges, since its integrand is of order $x$ near zero and decays exponentially at infinity. Hence $n=\alpha T^3$, with $\alpha$ independent of temperature.

For equilibrium radiation undergoing adiabatic expansion, the total entropy in a comoving volume is constant. Since the entropy density is $\beta T^3$, this means $\beta T^{3}V$ is constant. A comoving volume scales as the cube of the scale factor, soThis agrees with the cosmological redshift of each photon frequency. The conclusion uses entropy conservation during the expansion.

A first-order phase transition exchanges two distinct global minima of the Landau free energy. The equilibrium free energy remains continuous, but a first derivative such as the magnetization or entropy jumps. A temperature-driven transition generically has latent heat; crossing $h=0$ between symmetry-related ordered phases instead has a magnetization jump without requiring latent heat. A continuous phase transition has no such order-parameter jump and usually has a divergent correlation length and nonanalytic higher responses.

For $u>0$, $h=0$ and $r$ crossing zero, minimizing $V=r{M}^2/2+u{M}^4/4$ gives $M=0$ for $r>0$ and $M=\pm \sqrt{-r/u}$ for $r<0$. The minima merge continuously at zero. The curvature vanishes there, so the long-distance response becomes singular.

For $u<0<v$, put $z=M^2$. A nonzero stationary minimum satisfies $r+uz+v{z}^2=0$. Equality with the $M=0$ free energy givesThe finite separation demonstrates the first-order phase transition. Metastability occurs where a nonglobal local minimum persists; its disappearance at a spinodal point is not the equilibrium coexistence criterion.

Use a selected translationally invariant pure thermodynamic phase and define the connected lattice correlation functionAway from criticality its long-distance tail has an exponential factor $e^{-r/\xi }$, possibly multiplied by a power. Equivalently ${\xi }^{-1}=-{\lim }_{r\to \infty }{r}^{-1}\log |G(r)|$ when this limit exists; different directions can have different lengths on an anisotropic lattice. At a thermodynamic critical point the correlation length diverges and the tail becomes algebraic. Differentiate $M=N^{-1}⟨{\sum }_n{\sigma }_n⟩$ with respect to the physical field appearing in the energy Statistical Hamiltonian. With ${\beta }_{\mathrm{th}}=1/(k_{B}T)$, the correlation-function susceptibility sum rule isIn physical continuum coordinates the sum becomes $a^{-D}\int d^{D}rG(r)$. Differentiating with respect to the dimensionless source ${\beta }_{\mathrm{th}}h$ instead removes the explicit inverse-temperature factor. Below the transition, take a one-sided pure-phase limit rather than a macroscopic symmetric mixture.

A normalized blocking kernel $B_b({\sigma }^{\prime },\sigma )$ assigns a probability or a delta constraint to a coarse configuration for each microscopic one, with ${\sum }_{{\sigma }^{\prime }}{B}_b({\sigma }^{\prime },\sigma )=1$. Define the blocked Boltzmann weight bySumming over ${\sigma }^{\prime }$ proves exact equality of the partition functions. Repeated real-space renormalization group steps compose the kernels and integrate out shorter-scale variables. To retain equality one must keep all generated interactions and the field-independent constant; a finite truncation is an approximation. Long-distance observables are mapped through the corresponding coarse observables, not identified blindly with microscopic spins. In physical units,A subsequent coordinate rescaling restores the reference lattice spacing while expressing couplings in cutoff units.

Let $F(u,C)=-({\beta }_{\mathrm{th}}N{)}^{-1}\log Z=C+f(u)$ be free energy per microscopic site. Invariance of $Z$ givesThe constant transforms as $C_1=b^D{C}_0+c(u_0)$. Therefore, with $g(u)=b^{-D}c(u)$, the identity-operator contribution to renormalization-group free energy givesThe second equality follows by substitution and induction. The inhomogeneous term records eliminated-mode entropy, determinants, and normalization factors, all multiplying the identity operator; it is not an interaction with the retained spins. Strictly, this $f$ can contain regular background terms. After an analytic subtraction of an inhomogeneous renormalization recursion, the genuinely singular part satisfies the homogeneous scaling equation, except for possible logarithmic resonances.

A renormalization-group fixed point satisfies $R_b(u_{\ast })=u_{\ast }$. Linearize in scaling coordinates $s_i$: $s_i^{\prime }=b^{{\lambda }_i}{s}_i$. A relevant operator has ${\lambda }_i>0$, an irrelevant operator has ${\lambda }_i<0$, and a marginal operator has ${\lambda }_i=0$ and needs nonlinear analysis. These ${\lambda }_i$ are logarithmic scaling eigenvalues; the discrete Jacobian eigenvalues are $b^{{\lambda }_i}$, not ${\lambda }_i$. The critical surface is the stable manifold obtained by tuning all relevant coordinates to zero. A repulsive renormalization-group trajectory leaves the fixed point along a relevant direction as the coarse-graining length grows. Stable and unstable manifolds become curved away from linear order.

The left section has $h=0$, one irrelevant coordinate $w$ and a relevant thermal coordinate $t$. Its critical surface is $t=0$ in that section. With two relevant variables, the full critical surface has codimension two: the second section shows repulsion in the $(t,h)$ plane at $w=0$.

Assume an isolated fixed point with two relevant scaling fields, analytic changes from physical controls to $(t,h)$, short-range isotropic scaling, no dangerously irrelevant variable in the thermodynamic scaling function, and no marginal or resonant logarithm. Let ${\lambda }_t,{\lambda }_h>0$ denote their scaling eigenvalues. The homogeneous singular free-energy density obeysChoose a stopping scale $L=b^p\simeq |t{|}^{-1/{\lambda }_t}$ so the renormalized thermal field is order one. Irrelevant fields vanish at this scale. With $s=D/{\lambda }_t$ and $\Delta ={\lambda }_h/{\lambda }_t$,Continuous scale interpolation absorbs the immaterial integer stopping-scale convention. Nonuniversal thermal, field and free-energy metric factors can be included explicitly; the normalized functions depend only on the universality class. The $+$ and $-$ labels refer to $t>0$ and $t<0$, which flow to different sides of the critical surface.