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The statistical Hamiltonian in this question is already in thermal units, as indicated by $e^{-H}$ without an additional ${\beta }_T$. Define $W[h]=\log Z[h]$ and $F[h]=-W[h]$. The mean field and its linear response areThese are functional derivatives of the connected generating functional; the subtraction defines the connected correlation function. In the Landau approximation, neglect loop corrections and evaluate the field integral at a stable saddle $m_h$. It satisfies the Euler-Lagrange equationThis follows by varying the gradient term and integrating by parts, with periodic, decaying, or otherwise appropriate boundary conditions.

The two requested free energy functionals, following the source and Legendre conventions of the question, areIn the scalar-field source Legendre transform on the chosen stable branch, $h$ is chosen to produce $m$, and $\delta {\Gamma }_{\mathrm{L}}/\delta m=h$. Thus the imposed-source Helmholtz free energy and the fixed-order-parameter Gibbs free energy have the appropriate opposite source derivatives. These names are used in the question's magnetic-ensemble convention; the defining sign relation is what fixes the calculation. At leading Landau approximation there is no fluctuation-determinant term in $F_{\mathrm{L}}$.

Differentiate the saddle equation with respect to $h(y)$. The response obeysThereforeEquivalently, the inverse Hessian relation for a connected two-point function states that $G$ is the inverse kernel of ${\delta }^2{\Gamma }_{\mathrm{L}}/\delta m(x)\delta m(y)$. The factor $u_0/2$ follows from twice differentiating the quartic term $u_0{m}^4/4!$; it is not $u_0/6$. This tree-level connected response is obtained by varying the saddle. It does not require replacing the exact connected correlator by a product of the saddle values, which would incorrectly give zero.

For the requested single-momentum formula, assume a homogeneous source and a translationally invariant equilibrium phase, so $m_h(x)=m_0$ and $G(x,y)=G(x-y)$. SetThe Fourier transform with the printed positive sign sends $-{\nabla }^2$ to $q^2$, while the Dirac delta function transforms to one. Hence the Ornstein--Zernike correlation function hasThe inverse convention is $G(x)=\int d^{D}q(2\pi {)}^{-D}{e}^{-iq\cdot x}\tilde{G}(q)$. For general inhomogeneous $h(x)$, $K$ has nonconstant coefficients and $G$ depends separately on its two positions; the displayed momentum-diagonal formula then does not follow. The preceding differential equation still holds in that case.

At zero source and for $u_0>0$, the homogeneous saddle is $m_0=0$ for $r_0>0$, and $m_0^2=-6r_0/u_0$ on either selected stable ordered branch for $r_0<0$. Thus the Landau scalar correlation length isIn the ordered phase the negative bare quadratic coefficient is compensated by the positive curvature at the nonzero saddle. Retaining $m_0=0$ below the transition would instead give an unstable kernel and is not a physical correlation length. With the usual analytic thermal tuning $r_0=a_0(T-T_c)$, $a_0>0$, both branches diverge as $|T-T_c{|}^{-1/2}$, so the correlation-length critical exponent isThe high-temperature amplitude is $\sqrt{2}$ times the low-temperature amplitude for the same $a_0$. This statement is within Landau theory; fluctuations can change critical behavior outside the mean-field regime.

When $u_0=0$, the field theory is a Gaussian field theory, so integrating out short-wavelength modes can be done exactly rather than merely at a saddle. Introduce an ultraviolet momentum cutoff $\Lambda$ and a scale factor $b>1$, then split $\varphi ={\varphi }_{<}+{\varphi }_{>}$ into $|q|<\Lambda /b$ and $\Lambda /b<|q|<\Lambda$. Orthogonality of Fourier modes eliminates cross terms in the quadratic Hamiltonian. A constant source couples only to the zero-momentum mode and therefore does not couple to ${\varphi }_{>}$.

Before rescaling, the Gaussian functional integral factorizes:where $H_{<}$ has the original $r_0$ and $h$, restricted to low momenta. Up to a field-independent normalization, the eliminated modes contributeThe determinant affects the additive free energy, but generates neither a mass correction nor a source correction, since no interaction mixes low and high modes. This remains true beyond the Landau approximation: it is an exact integration of a Gaussian field theory, not neglect of the high-mode fluctuations.

To restore the original cutoff, use $x=b{x}^{\prime }$, $q^{\prime }=bq$, and choose the field rescalingSubstitution into the low-mode Hamiltonian gives the Gaussian momentum-shell scalingThe gradient coefficient is unchanged: the volume factor is $b^D$, the two derivatives contribute $b^{-2}$, and the two fields contribute $b^{2-D}$. The mass term gains $b^2$, while the source term gains $b^{(D+2)/2}$. ThusFor a nonconstant source the corresponding formula is $h^{\prime }(x^{\prime })=b^{(D+2)/2}h(b{x}^{\prime })$, after restricting its coupling to retained modes. The constant-source case avoids that extra source filtering.

Writing $b=e^{\ell }$, the exact Gaussian renormalization-group flow hasBoth perturbations are relevant at the massless zero-source Gaussian fixed point, with $y_t=2$ and $y_h=(D+2)/2$. There is no anomalous field rescaling here, so $\eta =0$ and the correlation-length critical exponent is again $\nu =1/y_t=1/2$. The coefficient changes arise from rescaling, not from an interaction-induced shift in $r_0$ during shell integration.

A stable real Gaussian functional integral requires $r_0>0$, or an infrared prescription that treats the zero mode at the massless point. With $u_0=0$ and $r_0<0$, the field energy is unbounded below; the Gaussian field theory alone cannot describe a stable ordered phase. The scaling laws are therefore interpreted about the Gaussian fixed point from the stable side, with finite-volume zero-mode regularization when needed.

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.

Take the couplings in the Hamiltonian to include inverse temperature, so the Boltzmann factor is $e^{-H}$. If physical energies are used instead, first replace $J,L,D,K$ by their products with $1/(k_{B}T)$. Split the on-site term equally between its two bonds. For spin states $s,t\in \{1,0,-1\}$ the spin-chain transfer matrix has entriesUsing the specified coupling coordinates, in the order $(1,0,-1)$ this isFor example, $e^{D/2}=x/z$ and $e^{-J+L+D}=y/z$. This is the nearest-neighbour Blume–Emery–Griffiths model with an additive constant and the paper's sign convention for $D$.

Summing the periodic spin chain gives the partition function $Z_N=\mathrm{tr}{W}^N={\sum }_{a=1}^3{\lambda }_a^N$. For finite real couplings all entries of $W$ are strictly positive; the Perron–Frobenius theorem gives a unique positive dominant eigenvalue with ${\lambda }_1>|{\lambda }_a|$ for $a\ne 1$. Since $W$ is a symmetric matrix, all its eigenvalues are real. Thus in the thermodynamic limit,The positivity condition is stronger and more useful than mere ordering by signed value: subdominant eigenvalues can be negative. Also, the printed strict ordering between the other two is not guaranteed for all couplings. For instance, $J=L=D=0$ makes $W$ the positive constant matrix $e^{K}11^T$ with two equal zero eigenvalues. Degeneracy there does not affect the largest-eigenvalue limit; no explicit generic eigenvalues are needed.

For spin magnetization, introduce a dimensionless field $h$ through $-h{\sum }_i{\sigma }_i$, or insert $S=\mathrm{diag}(1,0,-1)$ into the trace. Thenwhere $v_1$ is a normalized eigenvector of the largest eigenvalue and $W_{st}(h)=e^{h(s+t)/2}{W}_{st}(0)$. Spin inversion symmetry gives $v_{1,+}=v_{1,-}$, since the positive eigenvector of the largest eigenvalue is unique. Thereforeat zero field, both at finite $N$ and in the finite-coupling thermodynamic limit. The finite-$N$ result follows directly by pairing each configuration with its spin-reversed partner. Eigenvalues at a single fixed field do not determine a general observable: a field derivative of the largest eigenvalue, or its eigenvector, is required. Positivity and the real analytic dependence on the couplings exclude a finite-temperature spontaneous symmetry breaking transition in this one-dimensional finite-range chain; singular zero-temperature coupling limits require separate treatment.

For even $N$, spin decimation on alternate sites sums the middle spin of each two-bond segment, hence the coarse spin-chain transfer matrix is $W^{\prime }=W^2$. DefineDirect multiplication gives $W^2=e^{2K}{z}^{-2}\left(\begin{array}{ccc}A& B& C\\ B& E& B\\ C& B& A\end{array}\right)$. Matching its entry ratios to the original parameterization yields the spin-1 chain decimation recursionThe remaining overall positive factor is absorbed into $K^{\prime }$. Keeping that factor preserves the free energy as well as normalized spin probabilities. Indeed $\mathrm{tr}(W^2{)}^{N/2}=\mathrm{tr}{W}^N$, exactly; on an odd ring an unmatched boundary segment needs separate handling rather than assuming a uniform two-site block decomposition.

Use a dimensionless Hamiltonian in the functional weight $e^{-H}$, assume $\alpha >0$, and let $\varphi (\mathbf{x})={\int }_{|\mathbf{p}|\le \Lambda }{d}^{D}p(2\pi {)}^{-D}{e}^{i\mathbf{p}\cdot \mathbf{x}}\tilde{\varphi }(\mathbf{p})$. Reality means $\tilde{\varphi }(-\mathbf{p})=\tilde{\varphi }(\mathbf{p}{)}^{\ast }$. At zero field, the quadratic form isIn the momentum-shell renormalization group, split into ${\varphi }_{<}+{\varphi }_{>}$ below and above $\Lambda /b$. Disjoint Fourier supports make the quadratic form split into $H_0[{\varphi }_{<}]+H_0[{\varphi }_{>}]$. Integration over the shell gives a source-independent Gaussian functional integral multiplying the partition function, or an additive constant in the effective free energy; it leaves the slow-mode quadratic coefficients unchanged before rescaling. A uniform field has support only at zero momentum, so it does not change this shell integration.

Restore the ultraviolet cutoff by ${\mathbf{p}}^{\prime }=b\mathbf{p}$, ${\mathbf{x}}^{\prime }=\mathbf{x}/b$, and chooseThe measure, two gradients and two fields have scale factors $b^D$, $b^{-2}$ and $b^{-(D-2)}$, whose product is one. The mass term has factor $b^D{b}^{-(D-2)}=b^2$, while the uniform source term has factor $b^D{b}^{-(D-2)/2}=b^{(D+2)/2}$. Hence the Gaussian momentum-shell scaling isFor a slowly varying nonuniform source the corresponding formula is $h^{\prime }({\mathbf{x}}^{\prime })=b^{(D+2)/2}h(b{\mathbf{x}}^{\prime })$. Here the nonuniform source on the right is its projection onto the retained Fourier modes; an eliminated source component contributes only to the field-independent Gaussian normalization. The field engineering dimension is $(D-2)/2$ and its Gaussian anomalous dimension is zero. The mass and uniform source are relevant perturbations of the Gaussian fixed point. Positivity of $\alpha$ is necessary for a stable kinetic term near this point. One can use a finite volume and positive mass as infrared regulators and then take the critical limit; at exactly zero mass the integral over the zero Fourier mode alone is not a normalized finite-volume Gaussian measure.

Let $X$ be a coarse-grained configuration, with restricted partition function $Z_X$ and free energy $F(X)=-k_{B}T\log Z_X$. Its Boltzmann distribution is $\pi (X)\propto e^{-\beta F(X)}$. Microscopic reversibility pairs every equilibrium microscopic history with a history read backwards, including reversal of all time-odd variables. Summing this identity over the microscopic histories representing a given coarse-grained path preserves it; a Markov assumption is not needed for this argument.

Write ${\mathbb{P}}_F$ for the probability density of a specified history conditional on its starting configuration $X_1$, and ${\mathbb{P}}_B$ for its physically reversed history conditional on starting at $\Theta X_2$, where $\Theta$ reverses time-odd variables. Equilibrium time-reversal symmetry givesSince $F(\Theta X)=F(X)$, the desired conditional path relation isFor a reversible Markov chain, this also follows by multiplying detailed balance ratios $P_{ij}/P_{ji}={\pi }_j/{\pi }_i$ along the history: the intermediate equilibrium weights telescope.

The conditioning is essential. If the starting equilibrium weights are included in the path probabilities, the forward and reversed probabilities are equal, rather than differing by the displayed exponential. For example, a two-state continuous-time Markov chain with $F_1=0$, $F_2=\Delta >0$, rates $k_{12}=e^{-\beta \Delta }$ and $k_{21}=1$ has equal equilibrium-weighted probabilities for the two reversed single-jump histories. Their conditional rate ratio is instead $e^{-\beta \Delta }$. Normalized bridges conditioned on both endpoints also have a different normalization from the path weights used above. The PDF prints ${\mathbb{F}}_B$ in the denominator while its explanatory text defines ${\mathbb{P}}_B$; the latter is the intended notation. For the scalar order parameter used subsequently, $\Theta \varphi =\varphi$.

With periodic or sufficiently decaying boundary conditions, integration by parts givesAssume $\kappa \ge 0$ for a stable quadratic free energy; the diffusive smoothing discussed below requires $\kappa >0$. In a periodic box of volume $V$, use the unitary convention ${\varphi }_{\mathbf{q}}=V^{-1/2}{\int }_V\varphi (\mathbf{r})e^{-i\mathbf{q}\cdot \mathbf{r}}d\mathbf{r}$. A real field has ${\varphi }_{-\mathbf{q}}={\varphi }_{\mathbf{q}}^{\ast }$. Since the Fourier transform of a derivative converts ${\nabla }^2$ to $-q^2$, each Fourier mode obeysIt is an Ornstein-Uhlenbeck process. Multiplication by the integrating factor $e^{r(q)t}$ givesThe noise integral is understood as an Itô integral with a deterministic kernel; writing $\eta ds$ is shorthand for the corresponding Wiener increment. Direct differentiation in the mild stochastic sense checks both the equation and the initial value. The stated Kronecker-delta noise covariance corresponds to this finite-volume normalization. Unlike a conserved order parameter, the zero wavenumber mode decays at the nonzero rate $\Gamma a$.

The total charge of the widely separated $+1,-1$ pair is zero, so a uniform planar nematic director is compatible with the far boundary if it imposes no additional winding. Each integer defect is susceptible to energetic fission of an integer nematic disclination: the positive defect can split into two $+1/2$ defects and the negative defect into two $-1/2$ defects. Like-sign elementary defects repel, whereas opposite signs attract through their elastic distortion. This commonly produces subsequent encounters and annihilations of opposite half-integer defects.

The attraction can also be seen before fission. In the fixed-amplitude outer theory, a neutral pair of charges $q,-q$ separated by $R$ hasIts energy therefore falls as the separation decreases. In nematic defect-antidefect annihilation, opposite defects meet, the amplitude can vanish in their merging cores, and the local winding disappears; the remaining smooth director distortion relaxes towards the uniform energy minimum.

Direct approach and annihilation of the unsplit integer pair is also topologically allowed. Whether fission happens first, the paths of the four elementary defects, and the annihilation time require a specified dynamical law and core mobilities; the free energy alone does not fix them. Pinning, imposed boundary textures, or sustained thermal production of defect pairs can prevent a strictly defect-free final state. The stated final state presumes dissipative relaxation with no such obstruction.

Use the usual homogeneous, local, analytic Landau-Ginzburg theory; rotational symmetry and Z2 symmetry alone do not specify locality, analyticity, or the signs of the coefficients. The local derivative expansion contains every rotational scalar with an even total number of fields:The omitted terms include all allowed higher field powers and contracted spatial derivatives. Terms differing by a total derivative are equivalent for bulk behavior with suitable boundary conditions. The displayed gradient energy assumes spatial homogeneity. If only rotations about an origin are imposed, rotationally invariant position-dependent coefficients are also possible.

In the mean-field approximation, take a spatially uniform order parameter and minimize its Landau free energy. For $\gamma >0$, $u>0$ and $r=a(T-T_c)$ with $a>0$, keeping the quadratic and quartic terms givesThe single minimum splits into two minima as the temperature passes below the critical temperature. Choosing one minimum gives spontaneous symmetry breaking of the Z2 symmetry and a continuous phase transition. The order parameter is the expectation in a selected ordered state; averaging equally over both states gives zero. A Z2 symmetry does not guarantee a continuous phase transition: a negative quartic coefficient stabilized by a positive sextic coefficient can instead give a first-order phase transition.

Relaxing the Z2 symmetry permits odd terms. A nonzero linear source generally rounds this ordinary continuous phase transition. A cubic term can instead give a first-order phase transition: for $V(\varphi )=r{\varphi }^2/2+w{\varphi }^3/3+u{\varphi }^4/4$, $u>0$ and $w\ne 0$, equal minima occur atThe order parameter jumps from zero to ${\varphi }_{\ast }$. A continuous phase transition without an exact Z2 symmetry is still possible with additional tuning; odd terms do not logically exclude it.

For the multicritical even Landau potential, set $r={\mu }^2=a(T-T_c)$ and assume an integer $n\ge 2$, ${\lambda }_n>0$ and $\gamma >0$. For $n>2$, all lower even interactions must be absent or tuned to zero to obtain these mean-field critical exponents. The stationary equation isIts nonzero solutions are minima because $V^{\prime \prime }({\varphi }_0)=-(2n-2)r>0$. Thus the order-parameter critical exponent isSubstituting into the minimum free energy density givesThe singular heat capacity per unit volume is $c_s=-T{\partial }_T^2{f}_{\min }$. Smooth variation of the other coefficients does not change the leading power. ConsequentlyFor $n=2$, this means a finite heat capacity jump, with ${\alpha }_{\mathrm{MF}}=0$; for $n=3$, the tricritical point has ${\beta }_{\mathrm{MF}}=1/4$ and ${\alpha }_{\mathrm{MF}}=1/2$. The case $n=1$ would contain only a quadratic potential and would not stabilize an ordered phase.

The Gaussian fluctuation correction near a critical point is more singular than the mean-field approximation contribution whenThis is the upper critical dimension of an even scalar interaction. Equivalently, its coupling has engineering dimension $2n-(n-1)d$, positive below $d_c$. Dominant fluctuations signal breakdown of the mean-field approximation; the Gaussian expression is not then a calculation of the exact interacting heat-capacity critical exponent. At $d=d_c$, equality of the powers allows logarithmic corrections.

Absorb $k_{B}T$ into the free energy, so the statistical weight is $e^{-F}$. Under the Gaussian fixed point rescaling, invariance of the gradient energy gives the engineering dimensionThe engineering dimensions of ${\varphi }^2,{\varphi }^4,{\varphi }^6$ are respectively $1,2,3$. Their coupling eigenvalues are $d-[{\varphi }^{2m}]$, hence $2,1,0$. The quadratic and quartic interactions are relevant operators; the sextic interaction is a marginal operator at this order.

The three steps of the momentum-shell renormalization group are:

Here $b=e^s>1$, so increasing $s$ means flowing toward longer distances. At leading order, the fast field has a Gaussian distribution with covarianceThis perturbative integration requires $k^2+{\mu }_0^2>0$ on the shell. By the Wick theorem, odd fast-field moments vanish and $⟨{\eta }^4⟩=3I_b^2$, $⟨{\eta }^6⟩=15I_b^3$. Therefore the first term of the cumulant expansion of a coarse-grained free energy isThis is the quartic interaction generated by a sextic interaction. Including the rescaling, the requested flow isIn spherical coordinates, $I_b=(2{\pi }^2{)}^{-1}{\int }_{\Lambda /b}^{\Lambda }{k}^2/(k^2+{\mu }_0^2)dk$. In particular, for ${\mu }_0^2=0$,The same integration shifts the quadratic coefficient by $90{\lambda }_0{I}_b^2$ before rescaling and adds an irrelevant constant to the free energy. Thus setting the bare quartic coupling to zero does not place the theory on the tricritical critical surface.

To calculate the sextic renormalization-group beta function to second order, let $V={\lambda }_0\int d^{3}x(\phi +\eta {)}^6$. The required cumulant expansion isThe subtraction removes disconnected contributions. The term $20{\lambda }_0{\phi }^3{\eta }^3$ at each vertex supplies the primitive sextic correction. The Wick contractions giveThe first term contains three propagators connecting the two vertices; its factor $6$ is the number of bijections between the three fields at each vertex. The second contains a local tadpole diagram at each vertex and one connecting propagator. The primitive contribution to the free energy is thereforeExpanding the slow fields about a common point gives a negative local sextic correction proportional towhere ${\chi }_b$ restricts a momentum to the shell. This connected Feynman diagram has two independent loop momenta. Other Wick contractions also generate lower interactions and tadpole diagrams; they must be included or subtracted consistently when fixing the tricritical critical surface.

To extract the logarithmic tricritical sextic beta function, use normal ordering to remove local tadpoles and tune the quadratic and quartic relevant operators. The massless three-dimensional propagator is $C(z)=1/(4\pi |z|)$. The same three-line contraction over short separations $a<|z|<ba$ givesIts positive coefficient and the negative sign of the second cumulant expansion term establish the sign of the flow. Matching this logarithm to a renormalized coupling, or performing consistent iterated shell integration, givesThe precise normalization of $C$ is not needed. In a sharp momentum cutoff scheme, simply forcing all three internal momenta into one infinitesimal shell does not extract this two-loop logarithm: the shell restrictions remove its linear-in-$ds$ phase space. The finite-shell cumulant and the logarithmic renormalized calculation above are distinct stages of the computation. Field normalization corrections at order ${\lambda }^2$ enter this renormalization-group beta function only at order ${\lambda }^3$.

A positive, stabilizing sextic coupling is marginally irrelevant in three dimensions on the tricritical critical surface. Reversing the convention for $s$ reverses the sign of the renormalization-group beta function. A negative sextic coupling without higher stabilizing powers makes the potential unbounded; moreover, the untuned bare choice $g=0$ generally flows away through its generated relevant operator.

Let $h_i=\delta F/\delta p_i$ be the polar molecular field, with the positive functional derivative convention used in the paper. During a pure advective displacement $u_j$, the polar order parameter changes byThe free energy change is $\delta F=\int h_i\delta p_i{d}^{d}r$. Assume periodic boundaries, or boundary conditions that eliminate the surface work, and incompressibility ${\partial }_j{u}_j=0$. Integration by parts in the advective term givesRelabelling the dummy indices in the rotational and flow alignment of a polar order parameter terms givesTherefore $\delta F=\int {\Sigma }_{ij}^p{\partial }_i{u}_j{d}^{d}r$, with the reversible stress of a polar liquid crystalFor a local free-energy density $f(\mathbf{p},\nabla \mathbf{p})$, an explicit choice isIts divergence is exactly $-p_k{\partial }_j{h}_k$, by the Euler-Lagrange equation for $h_k$. A pressure term can be reassigned in an incompressible flow; the chosen representative realizes the printed divergence without that ambiguity. The force density ${\partial }_i{\Sigma }_{ij}^p$ has mechanical power $-\int {\Sigma }_{ij}^p{\partial }_i{v}_j$, the negative of the advective free energy rate, which checks the stress sign.

For the stated uniaxial nematic order normalization, the nematic director is an eigenvector of $Q$ with eigenvalue $\lambda$, and the two perpendicular eigenvalues are $-\lambda /2$. ThereforeUsing precisely the Landau-de Gennes free energy convention in the preceding solution givesIf the quartic invariant were instead normalized as $b\mathrm{Tr}{Q}^4/4$, its coefficient would be $\bar{b}=9b/32$; the physical predictions are unchanged after redefining $b$.

For a nonzero global minimizer of the free energy, compare opposite values of the scalar nematic order parameter:The lower one has $\bar{c}\lambda <0$, soThe claim concerns the stable ordered phase, not every metastable stationary point. For $\bar{c}=0$ the two signs are degenerate. For $\bar{c}\ne 0$, equality of the ordered and isotropic free energies, together with stationarity, gives ${\lambda }_t=-\bar{c}/(2\bar{b})$ and ${\bar{a}}_t={\bar{c}}^2/(4\bar{b})>0$. The finite jump ${\lambda }_t$ exhibits the first-order phase transition driven by the cubic invariant.