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For a single chiral superfield, the supergravity F-term potential, with $\kappa =m_p^{-1}$ restored, isThe canonical Kähler potential has $K_{z\bar{z}}=K^{z\bar{z}}=1$, so the Kähler covariant derivative of a superpotential isConsequently the Polonyi model has the scalar potentialThe negative term is essential; retaining only the square of the supergravity auxiliary field would give the global F-term scalar potential rather than the supergravity F-term potential.

Set $x=\kappa z=a+iy$ and $b=\kappa \beta =2-\sqrt{3}$. In this notation $V=|\mu {|}^2{m}_p^2{e}^{|x{|}^2}U$, whereAt $a_0=\sqrt{3}-1$, $y=0$, one has $a_0+b=1$ and $1+a_0(a_0+b)=\sqrt{3}$, so $U=0$. To prove that this is a local minimum in both real directions, rather than checking only the real axis, put $u=a-a_0$. Direct expansion gives the stable zero-energy Polonyi vacuum identityEvery coefficient outside the square is strictly positive, and the exponential factor is positive. Thus $V\ge 0$ everywhere and equality requires $u=y=0$, provided $\mu \ne 0$. This proves the stronger conclusiona unique global minimum of this Polonyi model. If $\mu =0$, the scalar potential vanishes identically and this uniqueness and supersymmetry breaking conclusion do not hold.

For an independent local stability check, the Hessian matrix of $U$ at this global minimum is diagonal:Since $U$ and its first derivatives vanish there, differentiating the exponential introduces no additional Hessian matrix terms at the global minimum. For canonically normalized real fluctuations $z=z_0+(s+it)/\sqrt{2}$, the squared scalar masses are thereforewhich are both positive.

Although the vacuum energy is zero, the supergravity auxiliary field is not. In the convention $F^z=-e^{{\kappa }^{2}K/2}\overline{D_{z}W}$, its vacuum expectation value obeysThe matter fermion transformation contains a term proportional to $F^zϵ$, so no nonzero constant supersymmetry parameter leaves this vacuum invariant. Hence supersymmetry is spontaneously broken even though the vacuum energy vanishes: the positive auxiliary field contribution cancels the negative gravitational term. The matter fermion supplies the goldstino, which is absorbed by the gravitino in the super-Higgs mechanism.

Using the gravitino mass from a superpotential, and $z_0+\beta =m_p$, givesAn auxiliary field $F^z$ has mass dimension two. Defining the supersymmetry breaking scale by ${\Lambda }_{\mathrm{SUSY}}=\sqrt{|F^z|}$, one obtainsThus the gravitino mass is small compared with the supersymmetry breaking scale when that scale is sub-Planckian, as in the usual hierarchy $|\mu |\ll m_p$. This smallness is gravitational suppression; it is not a conclusion valid for arbitrary choices of the mass parameter.

Use the scalar components of the three chiral superfields, and set $a=|{\varphi }_0{|}^2$, $b=|{\varphi }_{+}{|}^2$, $c=|{\varphi }_{-}{|}^2$ and $\Sigma =a+b+c$. We assume the physical sign $g>0$ and a nonzero coupling $\lambda$. The coefficient $g$ below is exactly the coefficient in the supplied D-term scalar potential; it should not be silently replaced by a differently normalized squared gauge coupling. The canonical Kähler potential has identity Kähler metric, and the three Kähler covariant derivatives of a superpotential areSubstitution in the supergravity F-term potential givesIn particular, the subtraction cancels only half of the six ${\kappa }^{2}abc$ terms contributed by the three squares. The full scalar potential isIts nonnegativity follows term by term from $a,b,c\ge 0$ and $g>0$. This is a special property of this cubic superpotential, not of a general supergravity F-term potential.

The D-term vacuum branches of a cubic superpotential must be described as branches: there are not two isolated field-space minima. First suppose $\zeta >0$. Vanishing of $V_F$ requires $ab=ac=bc=0$, and vanishing of the D-term scalar potential requires $b-c=\zeta$. Together these conditions give the vacuum expectation valuesThe phase of ${\varphi }_{+}$ lies on a gauge orbit, and one may choose it positive and real. At this global minimum, all $D_{i}W$ and the real auxiliary field of the vector multiplet vanish. It is a supersymmetric vacuum, even though the internal gauge symmetry is spontaneously broken. For $\zeta <0$, interchange ${\varphi }_{+}$ and ${\varphi }_{-}$: the nonzero vacuum expectation value has magnitude $\sqrt{-\zeta }$. For $\zeta =0$, the supersymmetric vacuum is the entire family ${\varphi }_{+}={\varphi }_{-}=0$, with ${\varphi }_0$ arbitrary.

The other branch has ${\varphi }_{+}={\varphi }_{-}=0$ and arbitrary ${\varphi }_0$. Here $W=D_{i}W=0$ but the real auxiliary field is nonzero for $\zeta \ne 0$, andThus this branch has pure D-term supersymmetry breaking and a flat direction of a scalar potential in the neutral scalar. It is not stable at every neutral-field value. At a fixed $a=|{\varphi }_0{|}^2$, expansion in the charged scalars yieldsHence the two charged squared masses and the stability condition areAbove this threshold, both charged directions have positive mass squared. Continuity makes them uniformly positive in a sufficiently small neighborhood of the chosen neutral-field value, so the stationary branch is a genuine, non-strict local minimum valley. Its neutral direction stays flat. Below the threshold, a charged mass squared is negative and the branch is a saddle point. At equality it is a boundary of the stable valley, not a local minimum in the full field space: changing $a$ slightly downward and switching on the unstable charged field lowers $V$. For example, for $\zeta >0$, put $a=a_c-ϵ$, $b={ϵ}^2$, $c=0$. Since $M(a)=|\lambda {|}^{2}a{e}^{{\kappa }^{2}a}$ has $M^{\prime }(a_c)>0$, one finds $V-g{\zeta }^2=-M^{\prime }(a_c){ϵ}^3+O({ϵ}^4)<0$.

There are no further local-minimum branches under these generic assumptions. If $b+c>0$, the expression for $V_F$ is strictly increasing in $a$, so a stationary configuration with a charged condensate must have $a=0$. At $a=0$, an interior critical point with $b,c>0$ is impossible, becauseOn the boundary $c=0$, the relevant scalar potential is $g(b-\zeta {)}^2$; on $b=0$ it is $g(c+\zeta {)}^2$. Their nonnegative-domain minima give exactly the supersymmetric vacuum described above, or the origin of the charged-field branch. Thus, for nonzero $\zeta$, the two types are the zero-energy supersymmetric gauge orbit and the positive-energy D-breaking minimum valley above the threshold. If $\zeta =0$, the two types coalesce into the flat supersymmetric vacuum; the literal two-minimum assertion needs this parameter qualification.

In the global supersymmetry limit $\kappa \to 0$, the F-term scalar potential becomes $|\lambda {|}^2(ab+ac+bc)$ and the D-term scalar potential retains its form. The same two types of branches occur, with the threshold $|\lambda {|}^{2}a>2g|\zeta |$. Supergravity instead introduces the exponential and additional positive terms. The goldstino of global supersymmetry breaking is a physical fermion; in local supersymmetry it participates in the super-Higgs mechanism. A positive vacuum energy also gravitates in supergravity. In particular, the positive-energy branch here has $W=0$ and hence vanishing gravitino mass from a superpotential; it is not the tuned zero-energy broken vacuum of Question 3.

There is a further gauge-consistency qualification. In conventional two-derivative matter-coupled supergravity, a constant Fayet–Iliopoulos term gauges an R-symmetry and requires a corresponding nontrivial transformation of the superpotential. The apparent charges $0,+1,-1$ make the supplied cubic superpotential neutral, so a nonzero independent constant $\zeta$ does not by itself specify a consistent conventional supergravity action. The computed branches are those of the supplied scalar potential; a full embedding needs compatible charges or additional fields. This restriction is explained in Van Proeyen's derivation of the superpotential gauge-covariance condition.

More generally, the supergravity moment-map constraint on D-term breaking follows directly from gauge covariance. If $k^i{\partial }_{i}W=-{\kappa }^{2}rW$ and the moment map is $\mathcal{P}=i(k^i{K}_i-r)$, then at a point with $W\ne 0$,Consequently all chiral auxiliary fields vanishing implies the D-term also vanishes there. This relation has no analogue forcing that implication in an arbitrary global supersymmetry model. Its $W\ne 0$ hypothesis matters: the formal D-breaking branch above has $W=0$, so it does not contradict the identity. These conventional gauge-completion restrictions are additional to the elementary minimization calculation.

Let the basic temperature be $\overline{T}=T_{\ast }-\beta z$ with $\beta >0$, the density be ${\rho }_0$, gravitational acceleration be $-g\hat{\mathbf{z}}$, and thermal expansion coefficient be $a_T$. Denote kinematic viscosity, thermal diffusivity and magnetic diffusivity by $\nu ,\kappa ,\eta$. Write the dimensional magnetic perturbation as ${\mathbf{b}}_d$, the temperature perturbation as ${\theta }_d$, and use constant permeability ${\mu }_0$. Linearizing the Boussinesq equations and the resistive induction equation about the static state givesThe Lorentz force identity is $(\nabla \times {\mathbf{b}}_d)\times B_0\mathbf{c}=B_0(\mathbf{c}\cdot \nabla ){\mathbf{b}}_d-\nabla (B_0\mathbf{c}\cdot {\mathbf{b}}_d)$. Its gradient term is absorbed into $P_d$, the perturbation magnetohydrodynamic total pressure. The temperature source follows from ${\mathbf{u}}_d\cdot \nabla \overline{T}=-\beta u_{d,z}$.

Choose any reference length $d$, since no vertical boundaries are prescribed. Scale length by $d$, time by $d^2/\kappa$, velocity by $\kappa /d$, temperature perturbation by $\beta d$, magnetic perturbation by $B_0$, and total pressure by ${\rho }_0\nu \kappa /d^2$. The dimensionless parameters areThey are the Prandtl number, magnetic-to-thermal diffusivity ratio, Rayleigh number, and Chandrasekhar number, respectively. Division by the chosen scales givesIn particular $Q\zeta =B_0^2{d}^2/({\mu }_0{\rho }_0\nu \kappa )$, so the diffusivity convention agrees with the momentum coefficient.

For a plane wave put $\mathbf{l}=(k,0,m)$, $q=|\mathbf{l}{|}^2=k^2+m^2$, and $d_{\parallel }=\mathbf{c}\cdot \mathbf{l}=k\sin \alpha +m\cos \alpha$. The induction and temperature equations imply $\hat{\mathbf{b}}=i{d}_{\parallel }\hat{\mathbf{u}}/(s+\zeta q)$ and $\hat{\theta }=\hat{w}/(s+q)$. The projection $P=I-\mathbf{l}{\mathbf{l}}^{\mathsf{T}}/q$ removes pressure. Because $(P\hat{\mathbf{z}}{)}_z=k^2/q$, its vertical momentum equation isThus the oblique-field plane-wave magnetoconvection dispersion relation isIf out-of-plane velocity is also allowed, its decoupled magnetic branch satisfies $(s/\sigma +q)(s+\zeta q)+Q\zeta d_{\parallel }^2=0$; it has no buoyant driving. At stationary marginality, for $k\ne 0$,These are stationary thresholds; an oscillatory marginal mode, if present for particular diffusivity ratios, is a different onset calculation.

For $m=1$, write $R_{mathrmstat}=F(k)+QG(k)$, where $F=(1+k^2{)}^3/k^2$ and $G=(1+k^2)(\cos \alpha +k\sin \alpha {)}^2/k^2$. Direct differentiation gives $F^{\prime }(k)=-2k^{-3}+6k+4k^3$, so its two minima are $\pm 1/\sqrt{2}$, with $F=27/4$ and $F^{\prime \prime }=36$. For $Q>0$ the negative sign has the smaller magnetic penalty: at every positive $k$, $G(-k)<G(k)$ because both sine and cosine of the inclination are positive. The wavenumber selection in oblique-field magnetoconvection therefore starts from $a=-1/\sqrt{2}$. Expanding $F^{\prime }(k_c)+Q{G}^{\prime }(k_c)=0$ gives $k_c=a+C_{1}Q+O(Q^2)$, where $C_1=-G^{\prime }(a)/F^{\prime \prime }(a)$. Taylor expansion of the value then givesThis proves the small-field forms, including the order of the correction.

For large $Q$, $G$ is nonnegative and vanishes only at $b=-\cot \alpha$. The candidate $k=b$ has finite threshold $F(b)$, so a global minimizer must satisfy $QG(k_c)\le F(b)$ and hence tend to $b$. Here $G^{\prime \prime }(b)=2(1+b^2){\sin }^2\alpha /b^2>0$. Expanding the stationarity condition at $b$ yieldsThe value expansion isThe leading coefficient follows from $1+{\cot }^2\alpha ={\csc }^2\alpha$. The expansion holds at fixed $0<\alpha <\pi /2$.

Finally choose $\cot \alpha =1/\sqrt{2}$. Then the nonmagnetic minimizing wavevector also has $d_{\parallel }=0$, and its magnetic penalty vanishes. Since $F(k)\ge 27/4$ and $QG(k)\ge 0$, this is the global minimum for every nonnegative field strength:This field-independent stationary threshold in oblique magnetoconvection has a simple physical explanation: the disturbance is constant along the imposed field, so it does not bend field lines.

Let $x_{ij}\ge 0$ be shipments, with column sums eight and row sums at most the four capacities. Total demand is $32$, below total capacity $36$, so unused capacity is allowed. This is a transportation problem with capacity inequalities; equivalently add a dummy shop demanding four units with zero shipping cost.

Using the original PDF's cost $c_{12}=10$, take the feasible solutionIts row totals are $(7,8,8,9)$ and all column totals are eight. Four units of capacity remain unused at the second factory. Its shipping cost is $35+32+28+2+7+16=\boxed{120}$.

To prove that the cost is a global minimum, use the transportation dual certificate with capacity inequalities. If row potentials satisfy $u_i\le 0$ and $u_i+v_j\le c_{ij}$, every feasible solution satisfieswhere $s=(7,12,8,9)$. The second inequality uses the sign of $u_i$: multiplying a capacity upper bound by a nonpositive number reverses it. Choose $u=(0,0,-3,0)$ and $v=(4,7,2,5)$. The reduced costs areso every entry is nonnegative. The lower bound is $-24+8(4+7+2+5)=120$, equal to the displayed shipping cost. The minimum shipping cost is $120$.

For the production costs in a transportation problem, add the production cost of each factory to every real shipment from that factory. Unused capacity incurs no production cost, so the dummy-shop costs remain zero. The resulting real-shop cost matrix isA feasible solution isIts row totals are $(7,12,8,5)$ and column totals remain eight, leaving four units unused at the fourth factory. Its combined cost is $18+32+48+40+32+24+35=229$.

For a matching transportation dual certificate with capacity inequalities, take $u^{\prime }=(-1,-1,-3,0)$ and $v^{\prime }=(7,11,7,9)$. Their reduced costs areall nonnegative. The lower bound is $-7-12-24+8(7+11+7+9)=229$. Thus the minimum combined production and shipping cost is $229$, and the second allocation attains it.

Set $h=0$ in the even sextic Landau free energy. For $u>0$, near $r=0$ the nonzero stationary branch satisfies $M^2=-r/u+O(r^2)$ for $r<0$. The minimum value is $V_{\min }=-r^2/(4u)+O(r^3)$ below the transition and zero above. The order parameter is continuous, and if $r$ crosses zero linearly with temperature, the first temperature derivative of the minimized free-energy density is continuous too. There is no latent heat; a second derivative has a finite jump at this level.

For $u<0$, solve the two conditions $V^{\prime }(M)=0$ and $V(M)=V(0)$ with $M\ne 0$. Setting $q=M^2$ gives $r+uq+v{q}^2=0$ and $r/2+uq/4+v{q}^2/6=0$. Eliminating $r$ givesBoth the disordered and ordered minima have positive curvature there, but their values coincide. Crossing this phase coexistence boundary changes the global minimum discontinuously. The order parameter jumps; for $r=r(T)$ at fixed $u,v$, ${\partial }_T{V}_{\min }=r^{\prime }(T)M^2/2$ jumps as well, generically yielding a nonzero latent heat. Local extrema can persist past coexistence as metastable states, which explains hysteresis in a dynamically delayed transition. The spinodal points, where local curvature vanishes, do not locate the equilibrium transition.

Consider the $(u,r)$ plane at $h=0$ for $V=r{M}^2/2+u{M}^4/4+v{M}^6/6$, with fixed $v>0$. For $u>0$, the continuous boundary is $r=0$: above it the global minimum is zero, and below it there are two symmetry-related ordered pure thermodynamic phases. For $u<0$, the equilibrium boundary is the tricritical three-phase lineAbove this curve the disordered minimum wins; below it the two ordered minima win. On the curve,Its three zeros are exactly $M=0$ and $M=\pm \sqrt{-3u/(4v)}$, proving equality of three global minimum values. Spin inversion symmetry enforces the equality of the two ordered phases automatically, so tuning a single remaining coefficient can produce three-phase coexistence on a line in this symmetry plane. This does not contradict the generic phase-rule counting without symmetry restrictions.

The tricritical point is $(u,r)=(0,0)$, where the discontinuity vanishes, the first-order boundary meets the continuous boundary, and the positive sextic term becomes the leading stabilizer. The local metastability limits $r=0$ and $r=u^2/(4v)$, $u<0$, bracket the first-order curve but are not phase boundaries. In the full $(r,u,h)$ parameter space, the zero-field line is attached to tricritical wings; a nonzero field splits the equality of the two ordered phases.

For the stable branch found above, $t=x+\beta =1$ and $s=1$. Hence the vacuum expectation value isin the stated Planck units. To verify that it is a vacuum rather than merely a zero-energy stationary point, evaluate the Hessian matrix. At either zero-energy stationary branch,Because $H$ and its first derivatives vanish there, the Hessian matrix of $V$ is just $|m{|}^4{e}^{x^2}$ times this Hessian matrix. For $s=1$, both eigenvalues are positive. This proves a strict local minimum in both real scalar directions. For $s=-1$, $H_{xx}<0$, so the alternative $\beta =2+\sqrt{3}$, $⟨z⟩=-\sqrt{3}-1$ is a saddle point and is excluded from the stable zero-energy Polonyi vacuum.

There is also a useful global check. Set $u=x-(\sqrt{3}-1)$ on the stable branch. Directly completing squares givesBoth remaining coefficients are positive. Thus $H\ge 0$ everywhere, with equality only at $u=y=0$. The positive exponential prefactor proves that this is the unique global minimum, not just a metastable vacuum.

Finally, at the stable vacuum $z+\beta =1$ and $D_{z}W=\sqrt{3}{m}^2$. Its supergravity auxiliary field hasThus the Minkowski vacuum breaks supersymmetry, even though its cosmological constant vanishes. If $m=0$, the potential is flat and the displayed tuned parameter and scalar value are not selected.

In the Landau theory approximation, choose the global minimum of the Landau free energy as a function of the magnetization. Since ${\mathcal{A}}_6>0$, the polynomial is a coercive function and a global minimum exists. The candidates are stationary points satisfyingCheck their values of the free-energy density; a stationary point or a metastable local minimum alone need not be the equilibrium state. Positive curvature ${\mathcal{A}}_2+3{\mathcal{A}}_4{m}^2+5{\mathcal{A}}_6{m}^4$ ensures local stability; when it vanishes, higher terms must be checked. The selected global minimum is the equilibrium magnetization within this approximation.

At zero magnetic field, equal opposite minima describe distinct phases exhibiting spontaneous symmetry breaking. A spontaneous magnetization is obtained by taking the thermodynamic limit before $h\to 0^{+}$ or $h\to 0^{-}$. A symmetric finite-volume average can instead have zero expectation even when the minima are nonzero. Thus one must specify the phase when interpreting the minimizer as the equilibrium expectation.

Discuss the zero-field phase boundaries, taking $h=0$ and writing $r={\mathcal{A}}_2$, $q={\mathcal{A}}_4$, $s={\mathcal{A}}_6>0$. This qualification matters because a conjugate field generally rounds the zero-field continuous transition. Nonzero stationary points satisfy $r+q{m}^2+s{m}^4=0$.

For $q>0$, the continuous phase transition occurs atOn the $r>0$ side the only stable phase is $m=0$; on the $r<0$ side the equilibrium global minima are $\pm m_0$ with $m_0^2\sim -r/q$. The magnetization therefore grows continuously from zero as $r$ changes sign.

For $q<0$, the first-order phase transition instead occurs while $r$ is positive. At phase coexistence, $m=0$ and $m=\pm m_0$ are equally deep global minimum points. With $a=m_0^2>0$, stationarity and equality of free energies giveEliminating $r$ gives $-q{a}^2/4-s{a}^3/3=0$, henceAt this coexistence value, the sextic even Landau potential factors as $f(m)=s{m}^2(m^2+3q/(4s){)}^2/6\ge 0$. This verifies that the three equal stationary values are global minima. Above this coexistence value of $r$ the disordered phase wins; below it the ordered phase wins, with a finite jump in the phase-selected magnetization. The values $r=0$ and $r=q^2/(4s)$ are the disordered and ordered spinodal points, respectively, not the coexistence boundary: the latter follows from the repeated root of $r+qa+s{a}^2=0$.

The two transition loci meet at the tricritical point,Two independent control parameters such as $T$ and $g$ can tune these two conditions. In a generic local phase diagram their coefficient map has a Jacobian matrix of rank two at the intersection; unspecified functions ${\mathcal{A}}_2(T,g)$ and ${\mathcal{A}}_4(T,g)$ do not by themselves determine the geometric orientation of the boundaries. The sextic term stabilizes this intersection and the negative-quartic first-order side.

These are mean-field critical exponents at a tricritical point. Hold the other control at its tricritical value and take the coefficients to be real analytic functions of temperature along a thermal approach with ${\mathcal{A}}_2=a(T-T_c)+o(T-T_c)$, $a>0$, ${\mathcal{A}}_6\to s>0$, and ${\mathcal{A}}_4=0$ or ${\mathcal{A}}_4=o(|{\mathcal{A}}_2{|}^{1/2})$. At fixed tricritical $g$, a generic real analytic temperature path with nonzero thermal slope has ${\mathcal{A}}_4=O(T-T_c)$ and meets the last condition. Paths tangent to a scaling direction can have different apparent temperature powers.

At zero field on the ordered side with ${\mathcal{A}}_4=0$, stationarity gives $m^4=-{\mathcal{A}}_2/s$; under the more general approach ${\mathcal{A}}_4=o(|{\mathcal{A}}_2{|}^{1/2})$, this relation holds to leading order. Thus the order-parameter critical exponent isFor the zero-quartic representative, putting this minimum into the Landau free energy gives the singular order-parameter contribution (and the leading asymptotic contribution under the stated negligible-quartic approach)Using $C_{\mathrm{s}}=-T{\partial }_T^2{f}_{\mathrm{s}}$ at fixed other control yields $C_{\mathrm{s}}\sim (T_c-T{)}^{-1/2}$, specifically $T{a}^{3/2}/(4\sqrt{s(T_c-T)})$ to leading order. HenceThis heat-capacity critical exponent refers to the nonzero ordered-side singular amplitude; in this mean-field approximation to Landau theory, the disordered global minimum contributes zero singular free energy, while any regular background remains. Fluctuation corrections lie outside the present approximation.

In the disordered phase, differentiating the equation of state at $m=h=0$ gives the magnetic susceptibility $\chi =1/{\mathcal{A}}_2$. On the ordered branch its curvature is ${\mathcal{A}}_2+5s{m}^4=4|{\mathcal{A}}_2|$, so $\chi =1/(4|{\mathcal{A}}_2|)$. Both give the magnetic-susceptibility critical exponentAt the tricritical temperature with both lower coefficients zero, $h=s{m}^5$, so $m=\mathrm{sgn}(h)(|h|/s{)}^{1/5}$ and the critical-isotherm exponent isThe Widom scaling relation $\gamma =\beta (\delta -1)$ and Rushbrooke scaling relation $\alpha +2\beta +\gamma =2$ both check these powers. Their validity here is a mean-field conclusion, not a claim that fluctuations in every spatial dimension preserve them.

The equilibrium magnetization is a global minimum of the Landau free energy. Its candidates are the stationary pointsThus $m=0$, or $x=m^2\ge 0$ is a nonnegative root ofOne retains candidates with nonnegative curvature $f^{\prime \prime }(m)$ and compares their free energies; the candidate with the smallest value is the equilibrium state. Equal global minima describe phase coexistence.

The equilibrium magnetization is a global minimum of the Landau free energy. Its stationary values solve the polynomial equationand a local minimum must satisfyOne compares the value of $f$ at every such local minimum and chooses the smallest. Since $a_6>0$, the polynomial tends to positive infinity as $|m|$ tends to infinity, so a global minimum exists.

For the Landau free energy $G(\varphi )={\varphi }^6/6+g{\varphi }^4/4+\epsilon {\varphi }^2/2$, positive $g$ gives a continuous phase transition at $\epsilon =0$. For negative $g$, phase coexistence between the disordered global minimum $\varphi =0$ and the ordered global minima occurs on $\epsilon =3g^2/16$, with ${\varphi }^2=-3g/4$. Indeed, on that locus the polynomial is $G(\varphi )={\varphi }^2({\varphi }^2+3g/4{)}^2/6\ge 0$, so these three zeros are global minima. For $g<0$, the disordered and ordered spinodal points are respectively $\epsilon =0$ and $\epsilon =g^2/4$; equality of stationary free energies determines phase coexistence, while loss of a local minimum determines a spinodal point.

For an ideal soap film with equal constant surface tension $\gamma$ on both faces, the effective tension is $\tau =2\gamma$. Choose a unit normal $N$ from the minus side to the plus side, and $\Delta p=p_{-}-p_{+}$. Translating the Young–Laplace equation to $S=-dN$ givesA single interface uses $\tau =\gamma$. An outward-oriented spherical soap bubble has $H=-1/R$, giving $\Delta p=4\gamma /R$. The positive-convex curvature sum in the Young–Laplace equation article is $-2H$. ^[ref]

An open film with the same resting gas on both sides has zero pressure jump. Neglecting film weight, soap film pressure balance gives $H=0$: a minimal surface. A pressurized bubble instead has nonzero mean curvature. These statements assume equilibrium, constant surface tension, and negligible normal viscous stress. Surfactant gradients or motion require the fuller normal stress balance. ^[ref]

The variational outline is short. With fixed boundary, a normal displacement $fN$ gives $\delta A=-2{\int }_{M}HfdA$ by the first variation of area formula. For a closed interface oriented outward, $\delta V={\int }_{M}fdA$, so stationarity of $\tau A-\Delta pV$ gives $-2\tau H-\Delta p=0$. This is a first variation calculation; testing for a stable minimal surface needs the second variation, and deciding a global minimum needs a comparison of competitors.

A stable minimal surface has nonnegative second variation of surface area for every admissible normal variation $fN$ fixing the boundary. For a two-sided minimal surface in three-dimensional Euclidean space, the quadratic form isHere ${\nabla }_{M}f$ is the gradient for the induced Riemannian metric. The gradient term penalizes rapid changes in the displacement; the principal curvature term can make the second variation negative. A plane has only the nonnegative gradient term. Nonnegativity alone, especially with zero modes, does not prove a strict local minimum or a global minimum.

For a finite catenoid between fixed rings, the answer depends on the ring separation, waist radius, and allowed variations. The axisymmetric second variation of a catenoid tests a specified restricted class. Comparing with two disconnected disks tests a different class of competitors, including a change in topology. Consequently, a stationary catenoid can have greater surface area than two disks without immediately collapsing under every small perturbation.

For $m\ne 0$ and positive $\beta$, the Polonyi model has a unique zero-energy global minimum at $\beta =2-\sqrt{3}$, $⟨z⟩=\sqrt{3}-1$. Writing $u=\mathrm{Re}z-(\sqrt{3}-1)$ and $y=\mathrm{Im}z$, its potential bracket is $(u^2+y^2+\sqrt{3}u{)}^2+(2\sqrt{3}-3)u^2+(4-2\sqrt{3})y^2$, a sum of nonnegative terms. The real and imaginary curvatures are $4\sqrt{3}$ and $8-4\sqrt{3}$. The other zero-energy stationary point with $\beta =2+\sqrt{3}$ and $z=-\sqrt{3}-1$ is a saddle point. The stable vacuum has nonzero supergravity auxiliary field, so it has supersymmetry breaking even with zero cosmological constant. The $m=0$ branch does not fix these values.

A free energy is a thermodynamic potential that compares possible states under specified environmental constraints. For a closed system of fixed composition, treated as a simple compressible system with only pressure-volume work, stable thermodynamic equilibrium minimizes Helmholtz free energy at fixed temperature and volume, or Gibbs free energy at fixed temperature and pressure. The choice follows the experiment: a rigid sealed vessel and a sealed cylinder with a freely moving loaded piston impose different constraints. Both may exchange heat with a thermal reservoir. See W. C. Carter, MIT thermodynamics notes, sections 2-1.14 and 2-1.19.

The useful insight is that internal energy alone cannot rank these alternatives: exchanging heat also changes the surroundings' entropy. Subtracting $TS$ packages that accounting into a property of the chosen system. Adding $pV$ additionally accounts for displacing the surroundings at fixed pressure. A free energy is a state function; supplied heat describes a transfer during a process. Under the stated constraints, supplied heat is $Q=\Delta U$ at fixed volume or $Q=\Delta H$ at fixed pressure, whereas neither $\Delta F$ nor $\Delta G$ generally equals $Q$.

A lower free energy identifies a thermodynamic preference, without specifying a timescale. A barrier can trap a system in a local minimum, producing metastability although a lower global minimum exists. The enzyme example illustrates the corresponding distinction for chemical reactions: changing the route can change the reaction rate while preserving the endpoint Gibbs free energy difference and chemical equilibrium.