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For $x>0$, differentiate the two explicit branches:The plus branch is strictly increasing and strictly convex for $x>0$. Its only zero is $x=0$, also its one-sided global minimum; it has no interior extremum or inflection point.

The minus branch rises initially, then has $y_{-}^{\prime }=0$ at $\sqrt{x}=4/5$, changing from positive to negative. Its second derivative changes sign at $\sqrt{x}=8/15$. Its zeros follow from $y_{-}=x^2(1-\sqrt{x})$. ConsequentlyThe origin is a one-sided local minimum of both graph branches, because their domain ends there; it is not an interior critical point. The minus branch has no global minimum since it tends to $-\infty$ as $x\to \infty$. Its maximum and inflection heights, useful for the sketch, are $256/3125$ and $28672/759375$, respectively.

The Cauchy-Riemann equations are $\boxed{u_x=v_y,u_y=-v_x}$. They give the gradient identityAt a regular intersection, where $f^{\prime }(z)\ne 0$, the two nonzero gradients are normal vectors to the two level sets, so their tangent directions are perpendicular. This is the orthogonality of regular real and imaginary level curves.

The regularity qualification is necessary: $|\nabla u|=|\nabla v|=|f^{\prime }|$, so at a critical point both normals vanish and the curves need not have unique tangents. For example $f=z^2$ gives diagonal lines for $u=0$ and coordinate axes for $v=0$, all meeting at the origin. Thus the conclusion is $\boxed{\text{right-angle intersection wherever }{f}^{\prime }(z)\ne 0}$, rather than an unrestricted claim at critical points.

For a holomorphic function, the derivative supplied by complex differentiability must have the same value along real and imaginary increments. Those two computations giveEquating real and imaginary parts proves the Cauchy-Riemann equations:Conversely, if the first partial derivatives are continuous near a point and obey these equations there, real differentiability giveswhich proves complex differentiability under this regularity assumption.

The gradients of the real and imaginary parts have dot productand $|\nabla u{|}^2=|\nabla v{|}^2=|f^{\prime }(z){|}^2$. At every point with $f^{\prime }(z)\ne 0$, the two level sets are regular curves, their gradients are nonzero normal vectors, and their tangent directions are therefore perpendicular.

The printed phrase "at every point" needs this regular-point qualification. At a critical point with $f^{\prime }(z)=0$, the level sets may cross or fail to have a unique tangent. For example, $f(z)=z^2$ gives $u=x^2-y^2$ and $v=2xy$: at the origin the zero level sets are respectively the diagonals and the coordinate axes. Thus the valid conclusion is $\boxed{\text{orthogonality at every regular intersection, where }{f}^{\prime }(z)\ne 0}$.

List the roots with their multiplicities. By the factor theorem, $Q(z)=a{\prod }_{k=1}^n(z-{\alpha }_k)$ with $a\ne 0$. The product rule, followed by division where $Q(z)\ne 0$, gives its logarithmic derivative:Let $Q^{\prime }(\beta )=0$. If $Q(\beta )=0$ too, then $\beta$ is already one of the roots, so either desired location conclusion follows immediately. Otherwise the displayed sum vanishes at $\beta$. If all roots have positive imaginary parts but $\mathrm{Im}\beta \le 0$, thenfor every $k$. Their sum cannot be zero. Thus $\mathrm{Im}\beta >0$.

For the unit disk conclusion, the same sum gives an especially transparent convex combination. Taking complex conjugates and setting $w_k=|\beta -{\alpha }_k{|}^{-2}>0$ givesTherefore, if every $|{\alpha }_k|<1$,The strict inequality follows because this is a finite weighted average of numbers strictly below one. This also proves the Gauss-Lucas theorem: every critical point of a nonconstant polynomial lies in the convex hull of its roots. The common-root case, handled before division, is essential when multiplicities occur.

At coexistence, the two phases have equal chemical potential. For a single component, $d\mu =-sdT+vdp$, with entropy and volume measured on the same molar or mass basis. Equating the phase derivatives along coexistence gives $\Delta vdp=\Delta sdT$. Reversible latent heat satisfies $L=T\Delta s$, henceHere $p$ and $T$ are coexistence pressure and absolute temperature, $L$ is latent heat for transforming phase 1 to phase 2, and $\Delta v=v_2-v_1$ uses the same amount of material as $L$.

For vaporization, neglect liquid molar volume compared with ideal-vapour volume $RT/p$. The ideal-vapour Clausius-Clapeyron approximation becomesUsing the normal boiling reference $T_b=373.15\mathrm{K}$, $p_b=1.01325\times 10^5\mathrm{Pa}$ and the supplied constant $L_e$ givesThis deliberately constant-latent-heat extrapolation does not reproduce the supplied triple-point pressure exactly. It predicts the liquid-vapour branch, metastable against ice below the triple-point temperature; it is not a precision sublimation-pressure prediction.

The phase diagram has solid, liquid and vapour regions meeting at the triple point. The sublimation and vaporization boundaries have positive slopes, whereas the ordinary ice–water boundary slopes negatively because ice has larger specific volume. The liquid–vapour line terminates at a critical point; the original schematic below emphasizes topology and marks the supplied triple point.

For the melting-pressure estimate, Archimedes' principle gives ${\rho }_i=0.90{\rho }_w=900\mathrm{kg}{\mathrm{m}}^{-3}$, soConvert the molar evaporation heat using water molar mass $M\simeq 0.0180\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$, rounded from the tabulated value $18.0153\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$.^[ref] An enthalpy cycle through vapour then givesThis uses the question's temperature-independent evaporation heat all the way to the triple point; no replacement experimental fusion heat is inserted. The melting pressure inferred from flotation and latent heats, on holding $L_f$ and $\Delta v$ constant, isusing $T_t=273.16\mathrm{K}$ and $T=270.15\mathrm{K}$. Linearizing the logarithm about the triple point gives $53.4\mathrm{MPa}$ instead of $53.7\mathrm{MPa}$. Both are estimates within the stated constant-property approximation; the negative melting slope makes a positive pressure necessary to lower the melting temperature.

Notation. The source uses $p_s$ for both the hypothetical dry surface pressure and the actual surface pressure after evaporation. Distinguish them as $P_d$ and $P$, respectively, and interpret the specified pressure thickness $\Delta p$ as fixed. This conserves the original hydrogen inventory and permits $P$ to vary. Reading both occurrences as one fixed number would be inconsistent with adding water vapor to a closed atmospheric column mass budget.

Let $E(T)$ denote the saturation pressure of water, let $x$ be the mole fraction of water vapor in the mixed layer, and let $M_d,M_v$ be the molar masses of hydrogen and water. Surface phase equilibrium and the ideal gas law giveBecause the layer has constant temperature, its relative humidity isThus every level strictly above the surface is subsaturated: hydrostatic balance makes $p<P$, whereas the saturation pressure stays constant. The result does not require a dilute water vapor approximation.

The Clausius-Clapeyron relation, neglecting the specific volume of liquid water and treating the water vapor as an ideal gas, givesHere $T_t,E_t$ are the triple point values, $L_v$ is the specific latent heat, and $R$ is the molar gas constant. All temperatures in these expressions are absolute. As throughout the question, this is a constant-latent heat, ideal gas model, not an extrapolation valid through the real critical point of water.

For a well-mixed ideal gas mixture, its mean molar mass and water vapor mass fraction areBy hydrostatic balance, the mixed layer has total mass per unit area $\Delta p/g$, of which $w_v\Delta p/g$ is water vapor. The remaining atmosphere contains no water vapor. Consequently conservation of hydrogen mass givesEquivalently the requested quadratic equation isFor each $T_s$, calculate $E$, take the positive root of this quadratic equation, and retain it only if $0\le x<1$ and $P>\Delta p$. The last condition leaves a nonempty dry layer. In particular the dry constituent's surface partial pressure is not generally $P_d$: $P_d$ measures its column weight, and the hydrogen and water vapor have different molar masses.

A convenient parameter is the water vapor mole fraction itself. The mass conservation calculation yieldsBoth functions increase for $0\le x\le 1$. At the endpoint,Therefore the physically admissible hydrogen–water vapor mixture requiresBeyond this bound, the formal positive quadratic equation root requires $x>1$, equivalently a negative hydrogen partial pressure. It is the physical constraints, rather than loss of real roots, that end the solution branch. For a concrete example, put $P_d=\Delta p=1$ in consistent pressure units, $ϵ=9$, and $E=3$. The positive root of $P^2+23P-51=0$ is about $2.037$, giving $x\simeq 1.473$, which is impossible.

The limiting temperature is obtained from $E(T_{\ast })=P_d+\Delta p$:This finite limit exists above $T_t$ if $E_t<P_d+\Delta p<E_t\exp [L_v/(R_v{T}_t)]$. The upper qualification matters mathematically: the stipulated constant-latent heat formula has a finite limiting saturation pressure as $T\to \infty$, so for an arbitrarily large unspecified $P_d+\Delta p$ the printed assertion of a finite breakdown temperature is not unconditional. For a concrete formal counterexample, choose $P_d+\Delta p=10^{15}\mathrm{Pa}$: with the usual triple-point constants the model has $E(\infty )\simeq 2.5\times 10^{11}\mathrm{Pa}$, so it never reaches the proposed inventory limit. This is a counterexample inside the stipulated equations, not a physically credible high-pressure ideal gas ocean. Actual liquid water also ceases to exist above its critical point, before such an extrapolation would make sense.

Physically, the imposed finite-pressure-thickness layer can hold no more than $\Delta p/g$ of water vapor per unit area. As $x\to 1$, it expels all its hydrogen into the dry upper layer. A still larger surface saturation pressure cannot coexist with this geometry. The layer must deepen, the temperature profile must change, or the assumed phase equilibrium must fail. If $P_d<\Delta p$, the additional restriction $P>\Delta p$ can also exclude the low-temperature end of the branch.

At a constrained critical point, the directional derivative of $f$ vanishes along every admissible tangent velocity:For a regular zero locus, $T_{a}M=\ker D{g}_a$. A linear functional vanishing on this kernel lies in the row space of $D{g}_a$. Equivalently there are Lagrange multipliers ${\lambda }_1,\dots ,{\lambda }_k$ such thatGeometrically, $\nabla f$ is orthogonal to the tangent space, so it lies in its orthogonal complement, spanned by the independent constraint gradients. Thus the level set of $f$ has no first-order change in a tangent direction. The equations provide candidates for constrained extrema; they must still be classified and compared, and boundary or singular cases must be treated separately. The condition is necessary for a regular constrained extremum, not a guarantee of a maximum.

A complex derivative computed along the real and imaginary directions gives $f^{\prime }=u_x+i{v}_x=v_y-i{u}_y$. Equality yields the Cauchy-Riemann equationsThe gradients satisfy $\nabla u\cdot \nabla v=u_x{v}_x+u_y{v}_y=0$. Since a gradient is normal to a regular level set, the two families meet at right angles by orthogonality wherever $f^{\prime }\ne 0$. At a critical point the regular-curve argument fails: for $f=z^2$ both gradients vanish at zero, so the assertion requires this qualification.

A complex derivative computed along the real and imaginary directions gives $f^{\prime }=u_x+i{v}_x=v_y-i{u}_y$. Equality yields the Cauchy-Riemann equationsThe gradients satisfy $\nabla u\cdot \nabla v=u_x{v}_x+u_y{v}_y=0$. Since a gradient is normal to a regular level set, the two families meet at right angles by orthogonality wherever $f^{\prime }\ne 0$. At a critical point the regular-curve argument fails: for $f=z^2$ both gradients vanish at zero, so the assertion requires this qualification.

The Joukowski map $f(z)=z+z^{-1}$ is holomorphic on $\mathbb{C}\setminus \{0\}$. A holomorphic function is locally conformal where its derivative is nonzero. Here $f^{\prime }(z)=1-z^{-2}$ vanishes at $z=\pm 1$. Thereforewhere the exclusion of zero is because the function is undefined there. Holomorphy alone does not supply conformality at the two critical points.

Put $z=1+\zeta$. Expanding the holomorphic function near its critical point givesFor a small ray $\zeta =r{e}^{i\theta }$, the leading image displacement is $r^2{e}^{2i\theta }$. Thus oriented ray angles are doubled modulo $2\pi$ at $z=1$. This is angle multiplication at a holomorphic critical point of multiplicity two; an unoriented angle between full tangent lines must also be interpreted modulo $\pi$. The map is not locally one-to-one there.

Choose a coarse-grained magnetization $m(\mathbf{x})$ as the order parameter: it vanishes in the paramagnetism phase and develops a spontaneous nonzero value in the ferromagnetism phase. In zero applied magnetic field, spin reversal sends $m\mapsto -m$, so the local Landau free energy contains only even powers. A minimal spatially inhomogeneous free-energy functional isThe positive gradient coefficient penalizes rapid spatial variations. Near an ordinary transition take $a(T)=a_0(T-T_0)$, $a_0>0$, and positive $b$; retain the sextic term when $b$ can approach or cross zero. The source $h$ is conjugate to the magnetization and explicitly breaks spin reversal.

Determine equilibrium by minimizing this free-energy functional, imposing the appropriate boundary conditions. Its Euler-Lagrange equation isFor a uniform system at $h=0$, compare the values of all stable stationary solutions, rather than accepting every solution of this equation. The selected global minimum gives the equilibrium magnetization; derivatives of its free energy give entropy, heat capacity and magnetic susceptibility. At an ordinary transition the two ordered minima emerge continuously from zero. Spatial dependence also permits domain walls and an estimate of the correlation length.

This is a mean-field approximation: it selects a minimizing profile and neglects statistical fluctuations around it. The symmetry constrains the expansion, but its numerical coefficients are phenomenological. Close enough to a transition, thermal fluctuations near a critical point can invalidate the resulting mean-field critical exponents.^[ref]

Landau theory expands a thermodynamic potential in an order parameter, constrained by symmetry. For a scalar with sign-reversal symmetry, a useful free-energy density isHere $h$ is the conjugate field. At $h=0$, minimizing rather than merely finding stationary points givesThe symmetry-related minima represent spontaneous symmetry breaking, as for a simple uniaxial ferromagnet with magnetization as order parameter. An infinitesimal field selects one sign. Above $T_c$, the susceptibility is $1/[a(T-T_c)]$; below it the stable-branch susceptibility is $1/[2a(T_c-T)]$. Minimization lowers $f$ by $-a^2(T-T_c{)}^2/(4b)$ below $T_c$, producing a heat-capacity density jump $T_c{a}^2/(2b)$ with continuous entropy.

A negative quartic coefficient requires a stabilizing $+c{\eta }^6/6$ with $c>0$. It can produce a first-order transition before the quadratic coefficient changes sign: coexistence occurs at $a(T-T_c)=3b^2/(16c)$ and ${\eta }^2=-3b/(4c)$. Odd terms are allowed when the assumed symmetry is absent. Landau theory predicts mean-field exponents; spatial fluctuations can change critical behaviour, especially close to a critical point in low dimensions. A gradient term extends the model to nonuniform order parameters but does not by itself supply dynamical equations.

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.

Use $v$ for molar volume to distinguish it from the fluctuating parcel volume below. The equation is $p=RT/(v-b)-a/v^2$. The parameter $b>0$ models excluded volume, while $a>0$ models the reduction in pressure from attractive interactions.

Below the critical temperature, a raw isotherm has a loop. The positive-slope segment is mechanically unstable and surrounding metastable portions do not give the stable equilibrium mixture. Replace the loop by a constant coexistence pressure $p_s$ satisfying the Maxwell equal-area condition,The endpoints are liquid and vapour molar volumes; intermediate mean volumes correspond to phase mixtures with changing proportions. Above the critical temperature, liquid-like and gas-like states connect continuously without a coexistence plateau.

In the pressure-temperature diagram, the liquid-vapour coexistence line ends at the critical point. Below it, the low-pressure side is vapour and the high-pressure side liquid. “Gas” conventionally denotes the region above the critical temperature, where compression alone cannot cross a liquid-vapour transition. The model has no solid phase.

The critical isotherm has an inflection: $p_v=p_{vv}=0$. Thuswhich giveFor fluctuations about a stable equilibrium value $x_0$, the system plus reservoir has probability proportional to $e^{-A(x)/(k_{B}T)}$. Expanding availability,gives a Gaussian distribution and, to this order, $⟨x⟩=x_0$ withThis requires positive curvature and sufficiently small fluctuations; it is not exact when higher-order terms matter.

The whole cylinder's volume is fixed. For the volume fluctuations of a small fluid parcel, interpret the small quantity as a material parcel containing $s$ moles, whose volume can fluctuate against the much larger remainder. Its mean volume at the critical density is $\bar{V}=s{v}_c=3sb$. At fixed temperature its availability is $A(V)=F_s(V,T)+p_{\mathrm{bath}}V$, soOn the critical isochore, approached from $T>T_c$,ConsequentlyEquivalently, $⟨\delta V^2⟩=k_{B}T{\kappa }_T\bar{V}$, with ${\kappa }_T=4b/[3R(T-T_c)]$. The mean-field variance diverges as $(T-T_c{)}^{-1}$, so its critical exponent is $1$; the root-mean-square fluctuation has exponent $1/2$. A fixed spatial observation box instead has number fluctuations, not fluctuating geometric volume. Finite size and non-Gaussian terms cut off the literal divergence; the approximation fails exactly at the critical point.

A non-ideal gas departs from $p=n{k}_{B}T$ because interactions and finite particle size affect its pressure. At sufficiently low number density $n=N/V$, the virial expansion separates these corrections:The second virial coefficient has dimensions of volume per particle, while $B_3$ has dimensions of volume squared. These are density-series coefficients, not the coefficients of a pressure-series expansion with the same names.

For classical particles interacting through a pair potential $u(r)$,Repulsion contributes positively and attraction negatively. A hard core of diameter $d$ gives $B_2=2\pi d^3/3$, four times the geometric volume of one sphere. The factor reflects the volume excluded to the centre of a second particle together with pair counting; it is not simply the occupied geometric volume.

Expanding the Van der Waals equation at $nb\ll 1$ givesThe virial coefficients of the van der Waals gas are therefore $B_2=b-a/(k_{B}T)$ and $B_3=b^2$. The Boyle temperature, $T_B=a/(k_{B}b)$, makes the leading correction vanish, but does not remove the higher coefficients or turn the gas exactly ideal. More realistic potentials give temperature-dependent higher coefficients that need not equal the van der Waals values.

A useful measurement is a sequence of dilute isothermal pressure–volume observations. Plotting $[p/(n{k}_{B}T)-1]/n$ against $n$ exposes $B_2$ as the zero-density intercept and $B_3$ as its initial slope. Control temperature and density calibration, check equilibration and adsorption on the container walls, and vary the fitted density interval to test truncation error. One measured virial coefficient cannot uniquely reconstruct the pair potential. Near coexistence or a critical point a short density series may fail, and at quantum-degenerate densities exchange statistics add corrections even without interactions.

Landau description. Introduce a macroscopic order parameter and expand an appropriate free energy analytically in that parameter, retaining terms permitted by the physical symmetries. Stable equilibrium minimizes this free energy globally; a local minimum with larger free energy is metastable. Coefficient changes with temperature can exchange the preferred minima. A continuous emergence from zero can produce a continuous transition, while competing separated minima can produce a first-order transition. Stability requires a positive highest retained even-power term. This local mean-field description neglects spatial fluctuations and nucleation dynamics unless gradient and dynamical terms are added; near a critical point those omissions can matter.

For this temperature-dependent cubic-quartic Landau free energy, abbreviate $B=a/(12\pi )>0$ and $A(T)=c(T^2-T_0^2)$, so $F=A{\varphi }^2/2-BT{\varphi }^3+\lambda {\varphi }^4/4$ and $\gamma =2B^2/(\lambda c)$.

Local behavior at the origin. To determine constrained order-parameter boundary stability, note that $F^{\prime }(0)=0$ and $F^{\prime \prime }(0)=c(T^2-T_0^2)$, the origin is a local minimum for $T>T_0$ and a local maximum for $0\le T<T_0$. At $T=T_0$, the quadratic term vanishes, but $F=-B{T}_0{\varphi }^3+O({\varphi }^4)<0$ for small positive $\varphi$. ThereforeThe equality case uses the stipulated domain $\varphi \ge 0$. On an unrestricted real line, the nonzero cubic term would instead make it a stationary inflection point.

Global stability in both inline cases. For $\varphi >0$, complete the square in $F/{\varphi }^2$:The quartic coefficient guarantees a bounded global minimum. A negative value exists precisely when the minimum of this quadratic factor is negative. Hence case a, $\gamma >1$, has a nonzero globally stable phase with $F<0$ for every physical temperature. Case b, $0<\gamma <1$, has such a phase exactly forAt $T=0$, the square's centre is at the boundary, but its negative constant still gives $F<0$ for small positive $\varphi$, so this endpoint is included.

At $T=T^{\ast }$,The two global minima $\varphi =0,{\varphi }^{\ast }$ have equal free energy, and $F^{\prime \prime }({\varphi }^{\ast })=\lambda ({\varphi }^{\ast }{)}^2/2>0$. The coexistence barrier of a cubic-quartic Landau transition has its maximum at ${\varphi }^{\ast }/2$, with barrier height $\lambda ({\varphi }^{\ast }{)}^4/64$.

For locating the extrema away from coexistence,Retain only real positive roots; when both are positive, ${\varphi }_{-}$ is a local maximum and ${\varphi }_{+}$ a local minimum. For $T<T_0$, only ${\varphi }_{+}$ is positive. If $\gamma <8/9$, the pair first appears on cooling at $T_{\mathrm{sp}}=T_0/\sqrt{1-9\gamma /8}>T^{\ast }$; if $8/9\le \gamma <1$, a higher-free-energy nonzero minimum already exists at arbitrarily high temperature. These local-stability limits must not be mistaken for the global coexistence condition. The boundary $\gamma =1$ also favors a nonzero minimum at every finite temperature, but has no finite $T^{\ast }$.

Cooling and transition temperature. For $\gamma <1$, equilibrium starts at the zero minimum at high temperature and switches to the nonzero minimum at $T^{\ast }$. The jump is finite, $0\to {\varphi }^{\ast }>0$, and there is a barrier between the coexisting states. This is a first-order transition. The latent heat of a temperature-dependent cubic-quartic Landau transition follows from differentiating the minimized free energy: the heat released on cooling isIf $F$ is an energy density rather than total energy, $L$ is likewise a density. An actual system can supercool below $T^{\ast }$ before nucleation, so $T^{\ast }$ is the equilibrium transition temperature.

For thermal symmetry nonrestoration in a cubic-quartic model, with $\gamma >1$, the equilibrium global minimum is nonzero already at arbitrarily high temperature. Following it does not force a transition. A system initially prepared in the metastable zero minimum can still leave it by nucleation while $T>T_0$; that minimum loses local stability at $T_0$. This is the sense in which a change is possible rather than inevitable. There is no equilibrium coexistence temperature for $\gamma >1$ in this model. A kinetic transition can depend on preparation and nucleation dynamics; $T_0$ is the limiting loss of local stability of the zero state, not a second coexistence temperature. Continuing $T^{\ast }$ to an imaginary value does not define a physical transition temperature.

Mass window in the specified electroweak phase transition model. For the mass window in a cubic electroweak transition model, put $x=m_H^2/m_0^2>0$. The supplied ratios implyThe two strict conditions therefore requireSince $x(A+x)$ is increasing for $x>0$, the finite window isUsing the specified $A$ and $m_0$ givesAt $m_H=134$ u, $x=0.3581801$, $\gamma \simeq 0.0030033<1$ but ${\varphi }^{\ast }/T^{\ast }\simeq 0.1481<1$. Thus 134 u lies outside the required window, failing the order-parameter-strength condition. This does not remove the polynomial model's finite first-order jump; it fails the additional requirement ${\varphi }^{\ast }>T^{\ast }$. The result is a bound within the specified simplified model, not a model-independent Higgs-mass bound.

Landau theory expands the free energy in a small order parameter and keeps terms allowed by the system's symmetries. For a scalar order parameter $q$ with sign-reversal symmetry in Landau theory $q\mapsto -q$, a useful free-energy density isAt zero conjugate field $h$, the quartic Landau equilibrium branches obtained by minimizing areThe single minimum becomes two symmetry-related minima: the order parameter turns on continuously and a chosen equilibrium phase spontaneously breaks the symmetry. At the critical temperature the quadratic restoring term vanishes. The mean-field exponent is $q\propto (T_c-T{)}^{1/2}$; the susceptibility above $T_c$ is $\partial q/\partial h=1/[a(T-T_c)]$.

More general allowed terms describe other transitions. A negative quartic term, stabilized by a positive sixth-order term, can give a first-order transition with a discontinuous order parameter and metastable minima. Without the sign-reversal symmetry, a cubic term may also be allowed. The expansion is phenomenological: its coefficients depend on microscopic physics. Minimizing a uniform polynomial neglects important spatial fluctuations, particularly near a critical point, so its critical exponents need not be exact.^[ref]

A critical point requires $f_x=f_y=0$. Since $y^4+y^2+1>0$ for every real $y$, the first partial derivative forces $x=0$. The second then reduces to $3y^2-3=0$, so the only critical points are $(0,1)$ and $(0,-1)$.

At these points the Hessian matrix is diagonal:The first is a positive-definite matrix, giving a strict local minimum; the second is an indefinite matrix, giving a saddle point. Their values are $f(0,1)=-1$ and $f(0,-1)=3$. HenceThe minimum is local: along $x=0$, $f(0,y)=y^3-3y+1$ is unbounded below as $y\to -\infty$.