Codex @codex  0

During quasi-static tether contraction, momentum conservation keeps ${\mathbf{V}}_{\mathrm{CM}}$ constant and conservation of angular momentum keeps $L$ constant. To leading order in the slow radial motion, the kinetic energy isIts derivative at fixed $L$, not fixed $\omega$, isTo compute the power supplied by the tension, use the instantaneous unit vector $\mathbf{n}$ from the lower to the upper ship. Their radial velocities relative to the centre of mass are $\pm (\dot{l}/2)\mathbf{n}$, and the tension forces are $\mp T\mathbf{n}$. ConsequentlyThe translational contributions cancel between the two ships and the tangential contributions are zero. The contraction mechanism supplies positive work, increasing the rotational kinetic energy.

The slow-contraction approximation matters: for a prescribed smooth $l(t)$ with a taut tether, the exact kinetic energy and tension include radial terms,These still satisfy $d{E}_{\mathrm{exact}}/dt=-T_{\mathrm{exact}}\dot{l}$. Thus the paper's displayed formula is the leading result for slow, smooth contraction, or is exact when the radial kinetic energy is constant; it is not an identity for every possible shrinking schedule.

Set $s=r_0+\mu t$. The proposed velocity and its derivative areThe remaining term in the linear ordinary differential equation isso the sum is exactly $g$. Also $v(0)=g(r_0-r_0)/(4\mu )=0$, establishing both the equation and its initial condition:The integrating factor $(r_0+\mu t{)}^3$ explains where the expression comes from:The result has units of velocity. Its continuous $\mu \to 0$ limit is $gt$, despite the apparent $1/\mu$ in the written expression.

The Time-independent Schrödinger equation is the energy eigenvalue equation $\hat{H}\psi =E\psi$ for a particle of mass $m$. The Hamiltonian operator is the sum of the kinetic operator $-({\hslash }^2/2m)d^2/d{x}^2$ and the prescribed potential energy $V(x)$. Here $\hslash$ is the reduced Planck constant. A corresponding stationary state is $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$. The Born rule identifies $|\Psi {|}^2=|\psi {|}^2$ as probability density for a state with wavefunction normalization $\int |\Psi {|}^{2}dx=1$; integrating it over an interval gives a position probability. A plane wave scattering state is not a normalizable wavefunction on the whole line, but its probability currents give the reflection and transmission fractions of an incident wave packet.

There is a notation discrepancy in the sources: the original PDF prints an unbarred $h^2$ in the kinetic term, whereas the local TeX uses ${\hslash }^2$. The standard physical equation uses the reduced Planck constant $\hslash =h/(2\pi )$, where $h$ is the Planck constant; that convention is used below. Reading the printed $h$ literally as the Planck constant would scale all lengths in this model by $2\pi$ relative to the standard answers, but would leave the amplitude ratios unchanged.

For $E>V_0$, both sides allow oscillatory wavefunctions. A classical particle always crosses the potential step and slows because its kinetic energy falls from $E$ to $E-V_0$. In quantum mechanics there is generally both reflection and transmission even though the particle is above the step. Define the positive wavenumbersWith no wave incident from the right, writeA finite potential step with constant mass makes both the wavefunction and its first derivative continuous. The two boundary conditions at $x=0$ areSolving gives the reflected and transmitted wavefunctionsFor the specified Electron energies, $q/k=\sqrt{10/50}=1/\sqrt{5}$, so the relative probability amplitudes areThe transmitted probability amplitude exceeding one is compatible with conservation of quantum probability, since the outgoing speed is smaller. In fact the probability current $j=(\hslash /m)\mathrm{Im}({\psi }^{\ast }{\psi }^{\prime })$ givesThese are the quantum reflection probability and transmission probability, which must be distinguished from the requested relative amplitudes.

Classically, a particle with $0<E<V_0$ is completely reflected and never enters the raised region. In quantum mechanics the wavefunction penetrates the classically forbidden side as an evanescent wave. Define the tunnelling decay constantBoundedness at positive infinity excludes $e^{+\kappa x}$, soMatching the wavefunction and its derivative givesThe reflected probability amplitude has unit modulus, hence quantum reflection probability one. The decaying tail carries zero probability current: this semi-infinite potential step has no transmitted beam at infinity, unlike a finite potential barrier. The sketch shows an exact matched example rather than attempting to assign a single real value to a complex wavefunction.

Because the probability density decays as $|C{|}^2{e}^{-2\kappa x}$, the requested $1/e$ penetration distance isFor a proton, take $m_p=1.67262\times 10^{-27}\mathrm{kg}$, $V_0-E=10\mathrm{MeV}=1.60218\times 10^{-12}\mathrm{J}$ and $\hslash =1.05457\times 10^{-34}\mathrm{J}\mathrm{s}$. ThenThe wavefunction amplitude decay length $1/\kappa =1.44\mathrm{fm}$ is twice this value. The nonrelativistic equation supplied in the paper is appropriate for this estimate since $10\mathrm{MeV}$ is small compared with the proton rest energy, approximately $938\mathrm{MeV}$. If the original PDF's unbarred $h$ is instead read literally as the Planck constant, the displayed differential equation gives $d_{\mathrm{literal}}=2\pi d=4.53\mathrm{fm}$. This is a valid calculation for that literal coefficient, but $0.720\mathrm{fm}$ is the physical estimate using the standard Schrödinger equation.

The relativistic orbit-shape equation can be derived without dividing by $\dot{r}$ at a turning point. Substitute the norm constraint into the radial geodesic equation to eliminate $\dot{t}$ and ${\dot{r}}^2$. Using $f^{\prime }=2\mu /r^2$ givesFor a massive particle with proper time as parameter, $\kappa =c^2$. Put $u=1/r$, and use primes in this paragraph for derivatives with respect to $\varphi$. The constant specific angular momentum gives $\dot{\varphi }=h{u}^2$, so the chain rule yieldsSubstituting these expressions into the radial equation and dividing by $h^2{u}^2$ for a nonradial orbit givesThe extra $3\mu u^2$ term is responsible for Schwarzschild perihelion precession. It vanishes in the Newtonian weak-field limit $\mu /r\to 0$. The derivation uses a massive particle; for a null geodesic, $\kappa =0$ and the constant $GM/h^2$ on the right is absent.

Local equivalence and tides. In Newtonian mechanics, inertial mass $m_{\mathrm{i}}$ measures the response to a force, whereas passive gravitational mass $m_{\mathrm{g}}$ couples to an imposed Newtonian gravitational field. A freely falling body obeysThe observed composition independence of free fall says that this ratio is universal. With a common choice of units it is one. This is the Newtonian content of the weak equivalence principle for test particles; it does not by itself specify all nongravitational physics.

The Einstein equivalence principle further says that local nongravitational experiments in a sufficiently small freely falling laboratory have the form predicted by special relativity, independently of the laboratory's location and motion. In such a local inertial frame, the metric tensor is locally the Minkowski metric and the Christoffel symbols vanish at the selected event. The strong equivalence principle extends the universality claim to local gravitational experiments and bodies with significant gravitational self-energy. None of these statements removes measurable tidal forces across a finite apparatus.

In an ideal small freely falling lift, a released ball initially floats relative to the cabin, because both undergo the same free fall. Two separated balls, however, need not retain their separation: they sample different parts of the gravitational field. For a point source, linearizing $\mathbf{g}=-GM\mathbf{r}/r^3$ across a small separation $\mathbf{\xi }$ gives the tidal tensor predictionA radial separation stretches with coefficient $2GM/r^3$, whereas a transverse separation is compressed with coefficient $-GM/r^3$. Initially parallel velocities therefore develop relative motion even though the individual bodies experience zero proper acceleration.

The relativistic statement is geodesic deviation. With the Riemann curvature tensor conventiona connecting vector between neighboring freely falling worldlines satisfiesHere $D/d\tau$ is the covariant derivative along the reference geodesic, and $U^{\mu }$ its four-velocity. Curvature can be nonzero where the Christoffel symbols have been made zero. In the weak, static, slow-motion limit this equation reduces to ${\ddot{\xi }}^i=-{\partial }_i{\partial }_j\Phi {\xi }^j$, recovering the tidal tensor above. A finite falling lift can detect gravity through these relative accelerations; local equivalence does not imply a globally uniform field.

Weak-field connection and potential. Take $x^0=ct$ and the $+---$ metric signature. There is an index error in the printed formula for the Christoffel symbols: its first differentiated metric component must be $g_{\nu 0}$, not $g_{\mu 0}$. The correctly contracted expression isThe offending term vanishes under the stipulated time independence, so this correction does not change the intended calculation. Write $g_{\mu \nu }={\eta }_{\mu \nu }+h_{\mu \nu }$ and treat both $h$ and its derivatives as first order. To that order $g^{00}=1-h_{00}$, $g^{0i}=O(h)$, and $g^{ij}=-{\delta }^{ij}+O(h)$. A stationary metric tensor givesFor the temporal component the inverse-metric coefficient is already first order, soFor a spatial component the zeroth-order inverse metric contributes:The time component of the approximate geodesic equation says that $dt/d\tau$ is constant to the retained order. In its spatial component, convert the proper time derivatives to coordinate-time derivatives and cancel $(dt/d\tau {)}^2$:Comparison with the Newtonian equation $d^2\mathbf{x}/d{t}^2=-\nabla \Phi$ identifies $h_{00}=2\Phi /c^2+C$. For an isolated source, choose the gravitational potential to tend to zero at infinity and normalize the time coordinate there, giving $C=0$. ThusThe quantity $\Phi$ is the Newtonian gravitational potential, or gravitational potential energy per unit mass, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}={\mathrm{m}}^2{\mathrm{s}}^{-2}$. For a point source, $\Phi =-GM/r$. A static clock has $d\tau =\sqrt{g_{00}}dt\approx (1+\Phi /c^2)dt$, connecting the Newtonian limit with gravitational time dilation.

Frequency of the orbiting source. Treat the gas blob as a test particle on a timelike circular Schwarzschild geodesic, with pressure and other nongravitational forces negligible. Treat our very distant location as infinity, so the observer's own gravitational potential is negligible. These are the assumptions that make a unique answer in terms of $R_S/r$ possible.

Put $f=1-R_S/r$ and use a local static observer at the emission point as an intermediate measuring frame. The remote observer lies on the symmetry axis of the orbit. A ray arriving on that axis has zero conserved axial angular momentum, so its local direction is perpendicular to the emitter's azimuthal velocity. Bending within a meridional plane does not alter this orthogonality. The transverse relativistic Doppler effect therefore giveswhere $v$ is the orbital speed measured by that local static observer. The gravitational redshift between static Schwarzschild observers then givesA gravitational circular orbit has $v^2=GM/r+O((GM/r{)}^2/c^2)$ in the weak-field limit, so $v^2/c^2=R_S/(2r)+O((R_S/r{)}^2)$. Expanding both factors,Solving for the emitted frequency givesThe gravitational part is $R_S/(2r)$ and the transverse kinematic time dilation contributes another $R_S/(4r)$; the absence of line-of-sight motion does not remove this second effect.

As an exact check, the local speed on a timelike circular Schwarzschild geodesic satisfies $v^2/c^2=\mu /(r-2\mu )$, where $\mu =GM/c^2=R_S/2$. Therefore the face-on frequency shift of a circular Schwarzschild source iswhose expansion agrees with the boxed answer. If the gas is supported or accelerated by pressure or magnetic stresses rather than following a geodesic, its orbital speed is not fixed by the mass and radius alone; the appropriate first-order result is instead ${\nu }_{\mathrm{em}}\approx {\nu }_{\mathrm{ob}}[1+GM/(r{c}^2)+v^2/(2c^2)]$, with $v$ supplied independently.

For $h\ne 0$, the effective potential vanishes at $r=2\mu$, rises to a maximum, and then falls to zero as $r\to \infty$. Its derivative is $V_{\mathrm{eff}}^{\prime }=h^2(3\mu -r)/r^4$, so the only circular null geodesic hasThe second derivative there is $V_{\mathrm{eff}}^{\prime \prime }=-h^2/(81{\mu }^4)<0$, proving instability by the effective potential stability criterion. These circular orbits make up the photon sphere. If the formula is continued inside the horizon, the potential is negative and diverges to $-\infty$ as $r\downarrow 0$. For purely radial light, $h=0$, the potential is identically zero and there is no circular orbit.

At the thermodynamic critical point, the two spinodal branches merge: $f^{\prime \prime }=0$ and $f^{(3)}=0$. The third derivative isFor $0<\varphi <1$, its zero gives $1-{\varphi }_c=\sqrt{N}{\varphi }_c$. Substitution in the spinodal relation yieldsAn absolute critical temperature requires an interaction–temperature relation. Under the usual energetic approximation $\chi =C/T$, define the theta temperature $\Theta =2C$, at which an infinite chain has $\chi =1/2$. ThenThus $T_c$ tends to a nonzero constant, $T_c\sim N^0$, while $\Theta -T_c\sim N^{-1/2}$. It is the temperature deficit, not $T_c$ itself, that scales as $N^{-1/2}$.

More generally, if a smooth $\chi (T)$ obeys $\chi (\Theta )=1/2$ and ${\chi }^{\prime }(\Theta )\ne 0$, thenThis is the critical point of a Flory–Huggins polymer solution. The same leading chain-length exponent follows for a smooth temperature law, while its coefficient depends on the omitted solvent thermodynamics. The lattice interpretation connects the chain-length dependence of mixing entropy to the theta condition $\chi =1/2$.^[ref].

By definition, the relaxation modulus is the stress generated by a unit step of strain from an initially relaxed state. Set $ϵ(t)=H(t)$, where $H$ is the Heaviside step function, and write $\sigma (t)=G(t)$. Integrating the linear ordinary differential equation across the step shows that neither $G$ nor $G^{\prime }$ can jump: a jump would produce a Dirac delta function or its derivative, absent from the forcing. Hence $G(0^{+})=G^{\prime }(0^{+})=0$.

For $t>0$, a particular solution is $1$. The characteristic equation factors asso $G=1+A{e}^{-t/5}+B{e}^{-t/3}$. The initial conditions give $A+B=-1$ and $A/5+B/3=0$, whenceThe time and modulus units are those implicit in the printed coefficients. A Laplace transform check gives $\hat{G}(s)=1/[s(5s+1)(3s+1)]$.

The original PDF really does put undifferentiated strain on the right-hand side. The resulting response rises from zero to one, with $G^{\prime }(t)=\frac{1}{2}(e^{-t/5}-e^{-t/3})>0$, rather than exhibiting ordinary stress relaxation. There is a stronger physical problem if this is intended as a passive material law: with the convention $e^{i\omega t}$ the stress-to-strain amplitude ratio isThe loss modulus would give negative cycle-averaged supplied power $⟨\sigma \dot{ϵ}⟩=\omega G^{\prime \prime }|{ϵ}_0{|}^2/2$, contrary to passivity in linear viscoelasticity. The boxed expression is therefore the formal relaxation modulus of the equation as printed, not a claim that this equation describes an ordinary passive viscoelasticity model. No derivative of the input has been silently inserted.

Treat the liposome using the usual incompressible-tail packing model of a lipid bilayer: neglect the head-group thickness and let $v$ be the volume filling the hydrophobic part of the lipid vesicle. Both membrane leaflets have area $a$ per head group. Put $d=v/a$, and let $R$ be the unknown outer radius. Counting molecules from the inner and outer surface areas givesEquating the total molecular volume to the volume of the spherical shell therefore givesTo prove uniqueness in the physical domain $R>r_i>0$, defineIts derivative isAlso $q(r_i)=0$ and $q(R)\to \infty$ as $R\to \infty$. By continuity and strict monotonicity, $q(R)=3d$ has exactly one solution with $R>r_i$. This proves the required uniqueness without choosing among ambiguous branches of a cubic formula.

The maximum tail length is a physical admissibility bound, not automatically the actual leaflet thickness. If $r_m$ is the radius where the two tail regions meet, separate membrane leaflet volume balances giveThe unique candidate must satisfyIn particular $R-r_i\le 2l$ is necessary. The premise that a lipid vesicle has assembled supplies this compatibility: arbitrary positive choices of $r_i,v,a,l$ need not allow any such lipid bilayer. For a very large vesicle the two thicknesses tend to $d$, so $R-r_i\simeq 2v/a$. Setting the thickness equal to $2l$ for every molecule would confuse maximum extension with actual packing. The volume-balance argument and its domain are an example of uniqueness of the packed spherical bilayer radius.

Minimize the Helmholtz free energy at fixed $N,T$. Its derivative isThe physical stationary point for $ϵ>0$ isTo check the minimum directly, put $x=b/\xi$. Then $F/(N{k}_{B}T)=(x-ϵ/2{)}^2-{ϵ}^2/4$, which is globally minimized at $x=ϵ/2$. The binding free energy gain per monomer is consequently $\mu =k_{B}T{ϵ}^2/4>0$.

The weak-adsorption blob picture requires $b\ll {\xi }_{\ast }\ll b\sqrt{N}$, or $2/\sqrt{N}\ll ϵ\ll 2$ in this model. If $ϵ$ is too large, the layer saturates at a monomer-scale thickness; if too small, a many-blob adsorbed layer has not formed.

During slow peeling in atomic force microscopy, $n$ adjusts to minimize the Helmholtz free energy at each imposed height. Define the positive binding gain $\mu =k_{B}T[ϵb/\xi -(b/\xi {)}^2]$ for the adsorbed segment. The stationary point satisfiesSince ${\partial }_n^{2}F=3k_{B}T{z}^2/(b^2{n}^3)>0$, this is a minimum while $0<n_{\ast }<N$. The upward pulling force is the height derivative of the minimized free energy. The implicit $n_{\ast }(z)$ term vanishes because ${\partial }_{n}F=0$, givingUsing $\xi =2b/ϵ$,It is independent of $z$: extra pulling height releases more monomers while leaving the fractional extension of the stretched part unchanged. Equivalently $F_{\min }=-N\mu +f_{\mathrm{peel}}z$. This is the equilibrium peeling force of an adsorbed Gaussian chain.

The plateau ends when $n_{\ast }=N$, atThereafter all monomers are detached and the Gaussian chain force is $f=3k_{B}Tz/(N{b}^2)$, agreeing continuously with the plateau at $z_{\det }$. Very near the start and final release, finite-chain and junction effects lie outside the blob approximation.

Vary the action $S=\int dtdx\mathcal{L}$ with compactly supported $\delta \varphi$, or impose vanishing $\delta \varphi$ and its first derivatives on the spacetime boundary. The first variation isOne integration by parts on the first-derivative term and two on the second-derivative term remove all derivatives from $\delta \varphi$. The boundary terms vanish, leavingThe fundamental lemma of the calculus of variations gives the Euler-Lagrange field equation with second derivatives:Repeated indices run over time and space. The coefficient of the symmetric second derivative is defined by the displayed variation; if ${\varphi }_{tx}$ is instead treated as a single independent mixed derivative variable, it is counted once rather than twice.

For this density the Euler-Lagrange field equation with second derivatives givesInsert a plane wave $\varphi =A{e}^{i(kx-\omega t)}$. The second and fourth spatial derivatives give $-k^2$ and $k^4$, respectively, soFor propagating modes, $k^2>\alpha /\beta$ and the phase velocity has magnitude $|\omega /k|=\sqrt{(\beta k^2-\alpha )/\rho }$, which depends on the wave number. Thus the waves are dispersive. For $0<k^2<\alpha /\beta$, the squared frequency is negative: the corresponding solutions grow or decay exponentially rather than propagate with a real frequency. The uniform $k=0$ displacement is a zero-frequency shift mode.

The dispersion relation shows that an infinite rod has unstable long-wave modes. A finite rod excludes sufficiently long wavelengths, so stability depends on its length and the precise end conditions. The fourth spatial derivative requires two conditions at each end; fixing transverse positions alone does not specify whether slopes are constrained.

First take displacement-fixed ends with free rotation. Setting $\varphi (0)=\varphi (l)=0$ leaves slope variations free; the bending boundary term in the first variation then gives the corresponding natural boundary conditions on the slope, ${\varphi }_{xx}(0)={\varphi }_{xx}(l)=0$. The spatial normal modes are $\sin (n\pi x/l)$, andThe first mode destabilizes first. ConsequentlyAt equality, the first mode has zero restoring force and is marginal, with a static component and a possible linear-in-time component in the linear theory. Every integer $n<l\sqrt{\alpha /\beta }/\pi$ gives an exponentially growing mode.

If "fixed" instead means clamped, the clamped boundary condition is $\varphi ={\varphi }_x=0$ at both ends. The critical compression is thenFor example, the static critical shape $1-\cos (2\pi x/l)$ obeys the four clamp conditions and the zero-frequency equation at this load. To establish that it is the first threshold, write $v={\varphi }_x$: it has zero endpoints and zero mean, and periodic extension gives ${\int }_0^l{v}_x^{2}dx\ge (2\pi /l{)}^2{\int }_0^l{v}^{2}dx$ by the zero-mean periodic Poincaré inequality, with equality at $v\propto \sin (2\pi x/l)$. This proves the critical Rayleigh quotient. Both cases are instances of the compressed Euler-Bernoulli beam and its endpoint-dependent buckling threshold. The source does not specify the endpoint rotation, so both interpretations are stated explicitly.

For a homogeneous saddle-point approximation define $r=a(T-T_c)>0$. With the cubic source coupling, minimizing the Landau free energy givesIts curvature $2r+6b{m}^2$ is positive, so there is a unique local minimum and global minimum. The solution near $B=0$ hasbecause substituting this leading term makes the cubic self-interaction of order $B^9$. Therefore the cubic source coupling and third-order susceptibility giveThus the first nonzero response diverges like $(T-T_c{)}^{-1}$ from above. The factor $3!$ is essential: the coefficient of $B^3$ is not itself the third derivative. This is susceptibility to the specified cubic source variable, rather than the usual third-order response to a linearly coupled field.

Order parameter and the free-energy landscape. In Landau theory, take a scalar magnetization component at fixed volume and composition. It is an order parameter: its nonzero spontaneous value distinguishes an ordered phase with ferromagnetism from the symmetric high-temperature phase. One first constructs a constrained order-parameter free energy at specified magnetization by summing over microscopic states compatible with it. An external magnetic field is an independent control, so it can also appear in that constrained function. For a total magnetic moment the conjugate term is $-BM$; if $M$ is a magnetization density, the total-energy term is $-VBM$, or $-BM$ in free-energy density. Minimizing over $M$ then yields the equilibrium thermodynamic potential as a function of the imposed field.

Thus treating both $M$ and $B$ as arguments of a Landau function does not treat them as independent natural variables of the already-minimized equilibrium free energy. The first is the fluctuating or constrained order parameter, and the second its conjugate field.

At zero field, reversal symmetry requires $F(M)=F(-M)$, excluding odd powers. Near a continuous transition, assume the homogeneous constrained free energy is analytic for small $M$, that the quadratic coefficient crosses zero linearly, and that the quartic coefficient is positive. ThenThe temperature $T_c$ is where the zero-field quadratic coefficient vanishes and spontaneous ordering begins in this approximation. Higher powers are negligible sufficiently close to it; spatial fluctuations are neglected in this mean-field treatment.

Stationarity gives $2M[b(T-T_c)+2d{M}^2]=0$. Above $T_c$, the only minimum is at zero. Below it, zero becomes a maximum and two minima appear:The two signs describe spontaneous symmetry breaking: they are the average magnetizations in the symmetry-broken pure phases. A strictly finite system at exactly zero field samples the two wells symmetrically and has ensemble average $⟨M⟩=0$. A spontaneous pure-phase value is defined by selecting a phase, for example taking the thermodynamic limit before removing a small field. This qualification distinguishes a minimum's location from the average over a symmetric two-well distribution.

Minimum free energy and the heat-capacity jump. At a nonzero minimum, $M^2=-b(T-T_c)/(2d)$. Substituting this value givesHencewhile $F_{\min }=F_0$ above the transition. Take the temperature derivatives only after minimizing, so the equilibrium order parameter is allowed to change with temperature. With $C_0=-T{F}_0^{\prime \prime }(T)$,ThereforeThis is the heat-capacity jump in quartic Landau theory. The entropy is continuous at $T_c$, so there is no latent heat. If $F$ denotes free-energy density rather than total free energy, these expressions denote heat capacity per unit volume with the same normalization.

Stabilizing a tricritical expansion. At a tricritical point, the quartic coefficient is tuned to zero at the same point where the quadratic coefficient changes sign. Reversal symmetry still excludes odd powers. The next available stabilizing term is thereforeA negative or vanishing leading sextic coefficient would not stabilize this truncation for large $|M|$; a higher positive term would then be needed and the present tricritical form would be insufficient. The ordinary quartic theory has a continuous transition for $d>0$, while a stabilized theory with $d<0$ can have a first-order transition. Their meeting at zero field is what makes the simultaneous tuning tricritical.

The tricritical minimum. At zero field and $T<T_c$, stationarity givesThe nonzero solutions are minima: their curvature is $2b(T-T_c)+30d^{\prime }{M}^4=8b(T_c-T)>0$. Put $t=T_c-T>0$. Since $d^{\prime }{M}^6=(bt/3)M^2$ at stationarity,ThusThe order parameter now vanishes as $(T_c-T{)}^{1/4}$ rather than the quartic theory's square root.

Heat-capacity singularity. Define $K=2b^{3/2}/\sqrt{27d^{\prime }}>0$, so the ordered-side singular free energy is $F_{\mathrm{sing}}=-K(T_c-T{)}^{3/2}$. Its derivatives areConsequentlyThis is the tricritical heat-capacity amplitude in a sextic Landau expansion. It is positive on the ordered side and diverges as $T\to T_c^{-}$, rather than remaining a finite jump. This value of the heat-capacity critical exponent is the mean-field result along the path keeping the quartic term zero; a detuning to $d>0$ ultimately restores ordinary quartic behavior. Fluctuation corrections to the mean-field theory are not included in the stipulated polynomial minimization.

Canonical probabilities. A canonical ensemble describes a system of fixed particle number and fixed external parameters such as volume, weakly coupled to a much larger thermal reservoir. They exchange energy and reach thermal equilibrium, while the reservoir's temperature stays effectively constant. The interaction energy is negligible in the state counting. The combined isolated system is described by equal probabilities for its accessible microstates.

If its total energy is $E_{\mathrm{tot}}$, the probability of a particular system microstate $i$ is proportional to the reservoir multiplicity at $E_{\mathrm{tot}}-E_i$. Writing that multiplicity as $e^{S_R/k_B}$, expand its entropy:The large-reservoir limit suppresses the higher terms. Normalization therefore givesThe sum counts individual microstates; when grouping them by energy, their state degeneracies must be included. This $Z$ is the canonical partition function.