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Use the same metric signature as above and define the Minkowski generating functional byHere $\hslash =1$. Integration by parts writes the quadratic action as $S_0=\frac{1}{2}\varphi K\varphi$ with $K=-\square -m^2+i0$; the infinitesimal imaginary part damps the oscillatory Gaussian functional integral and selects the vacuum Feynman propagator. Abbreviating repeated spacetime integrations, completing the square givesTranslation of the regulated integration variables therefore yieldsThe source-independent normalization contains the regulated functional determinant and its oscillatory phase; it cancels in $Z_0[J]/Z_0[0]$. Define ${\Delta }_F=i{K}^{-1}$. The Gaussian evaluation of a free scalar generating functional is equivalently $Z_0[J]/Z_0[0]=\exp [-J{\Delta }_{F}J/2]$.

With Fourier convention $e^{-ip\cdot x}$, $-\square$ acts as $p^2$. The inverse-kernel equation $K{K}^{-1}=I$ then givesThis is the time-ordered free two-point correlation function: differentiating twice with respect to $J$ and multiplying by $1/i^2$ recovers ${\Delta }_F$. It obeys $(\square +m^2){\Delta }_F=-i{\delta }^{(d)}$ in the distributional limit. The $i0$ is essential: the algebraic denominator alone would not distinguish Feynman, advanced and retarded Green functions.

Absorb the gauge coupling into the connection $A_{\mu }=A_{\mu }^a{T}_a$ and use $D_{\mu }={\partial }_{\mu }+A_{\mu }$. The anti-Hermitian generators make $U(x)=\exp [{\omega }^a(x)T_a]$ unitary for real parameters. With a matter field transforming as ${\psi }^{\prime }=U\psi$, require $D_{\mu }^{\prime }{\psi }^{\prime }=U{D}_{\mu }\psi$. Applying both sides to an arbitrary field givesExpanding $U=1+\omega +O({\omega }^2)$ and $U^{-1}=1-\omega +O({\omega }^2)$ yields the infinitesimal gauge transformation with an anti-Hermitian connectionThe adjoint covariant derivative is $D_{\mu }\omega ={\partial }_{\mu }\omega +[A_{\mu },\omega ]$. In components,The trace normalization makes the structure constants totally antisymmetric: $f_{abc}=-\mathrm{Tr}([T_a,T_b]T_c)$ and cyclicity of the trace supply antisymmetry in the remaining indices. Replacing $U=e^{\omega }$ by $e^{-\omega }$ reverses the infinitesimal parameter convention and gives $\delta A=D\omega$; mixing the two conventions would change the ghost operator incorrectly. The trace normalization also gives the invariant quadratic form $-\mathrm{Tr}(A_{\mu }{A}_{\nu })=A_{\mu }^a{A}_{\nu }^a$.

A gauge-invariant functional integral counts all representatives along each gauge orbit. The gauge condition $\mathcal{F}[A]=0$ selects a representative, and its delta functional enforces that selection. Assume a local patch where each orbit intersects the gauge slice once and the gauge-condition derivative has no residual zero modes. DefineChanging the orbit integration variable from $\omega$ to $\mathcal{F}[A^{\omega }]$ givesInsert this Faddeev-Popov gauge-orbit identity into the unfixed path integral. The action and measure are invariant under a change $A\mapsto A^{\omega }$, so integration over $\omega$ factors out as the gauge-group volume. Dividing by that redundant volume leaves the fixed integral and its Faddeev-Popov determinant. In a perturbative patch with fixed determinant sign or phase, the absolute value is replaced by the oriented determinant, up to a field-independent normalization:Here $D_{\mu }^{cb}={\delta }^{cb}{\partial }_{\mu }+f_{dbc}{A}_{\mu }^d$. This is a functional determinant of the gauge-condition Jacobian, not an arbitrary extra weight. The local invertibility assumption matters: residual gauge transformations or multiple orbit intersections, described by the Gribov ambiguity, prevent treating this derivation as a global unique gauge slice without further qualifications.

Both extra factors can be incorporated into an enlarged action. An auxiliary real field $b^a$ Fourier-represents the delta functional, while a Grassmann Gaussian integral over independent Faddeev-Popov ghost fields $c,\bar{c}$ represents the determinant:Thus the original fixed integral can be written with action $S+\int b\mathcal{F}+\int \bar{c}Mc$ and ordinary integration over the extra fields, with no explicit delta functional or determinant. The anticommuting fields give a determinant rather than the inverse determinant obtained from commuting Gaussian fields.

For the usual Gaussian family of gauges, add $\xi b^a{b}^a/2$ and integrate out $b$. Completing the square producesThis is Gaussian gauge fixing with an auxiliary field; equivalently one averages the condition $\mathcal{F}=f$ with its Gaussian weight. Gauge-invariant observables are unchanged apart from normalization. For the Lorenz gauge ${\mathcal{F}}_a={\partial }^{\mu }{A}_{\mu }^a$, the chosen transformation convention gives $M=-{\partial }^{\mu }{D}_{\mu }$. The ghost action is $-\int \bar{c}{\partial }^{\mu }{D}_{\mu }c$, or $\int ({\partial }^{\mu }\bar{c})D_{\mu }c$ after integration by parts. The dependence of $D$ on $A$ accounts for the ghost interactions.

For a bosonic quadratic action $S_2=\frac{1}{2}\int \varphi K\varphi$, completing the square in the generating functional gives $Z_0[J]/Z_0[0]=\exp (-iJ{K}^{-1}J/2)$. Two functional derivatives therefore give the Feynman propagator $\Delta =i{K}^{-1}$, with boundary conditions and a pole prescription included in the inverse. Zero modes must be removed or fixed before an inverse exists.

Take Minkowski metric $\eta =\mathrm{diag}(1,-1,-1,-1)$, Fourier convention $e^{-ikx}$, and the Yang-Mills field strength $F_{\mu \nu }^a={\partial }_{\mu }{A}_{\nu }^a-{\partial }_{\nu }{A}_{\mu }^a+g{f}^{abc}{A}_{\mu }^b{A}_{\nu }^c$. The interaction terms do not contribute to the quadratic action. Integration by parts in $-F_{\mu \nu }^a{F}^{a\mu \nu }/4-(\partial \cdot A^a{)}^2/(2\alpha )$ yieldsThe color kernel is diagonal. For $k^2\ne 0$, use the projectors $P_L^{\mu \nu }=k^{\mu }{k}^{\nu }/k^2$ and $P_T^{\mu \nu }={\eta }^{\mu \nu }-P_L^{\mu \nu }$. They obey $P_L^2=P_L$, $P_T^2=P_T$, $P_L{P}_T=0$, soMultiplication explicitly gives the identity on both transverse and longitudinal components. Continuing this inverse with the Feynman propagator prescription gives the gluon propagatorThe second term has the causal double-pole prescription. Away from the pole this is the usual $-i{\delta }^{ab}[{\eta }_{\mu \nu }-(1-\alpha )k_{\mu }{k}_{\nu }/k^2]/(k^2+i0)$. For $\alpha =1$ it becomes the Feynman-gauge adjoint propagator; $\alpha \to 0$ gives a transverse covariant Landau gauge propagator.

The limit $\alpha \to \infty$ removes gauge fixing and has no finite inverse on the full field space. The longitudinal eigenvalue $-k^2/\alpha$ tends to zero and its inverse diverges. A contraction with conserved external currents kills the longitudinal term, but that does not define a gauge-unfixed Gaussian integral. This massless limit is not the massive-vector unitary-gauge limit: here it exposes the unfixed gauge orbit degeneracy.

For a compactly supported variation $\varphi \mapsto \varphi +\epsilon \eta$, the first variation of the energy functional isIntegration by parts has no endpoint term. Since $\eta$ is arbitrary, the static Euler-Lagrange field equation isIntegrating the specified polynomial derivative gives $U(\varphi )={\varphi }^2/2-{\varphi }^4+{\varphi }^6/2+C$. The polynomial factors as a nonnegative square times ${\varphi }^2$; its global minimum is zero when $C=0$. ThereforeThese are the three constant scalar-field vacua. For a localized finite-energy kink, the two ends approach vacua and the derivative tends to zero. Multiplying the field equation by ${\varphi }^{\prime }$ gives the first integralA nonconstant kink must stay within one interval between neighboring vacua. A proposed direct connection from $-1$ to $1$ would reach $\varphi =0$ at some finite point, where the first integral forces ${\varphi }^{\prime }=0$. Uniqueness for the smooth second-order differential equation then forces the identically zero solution, a contradiction. This is the intermediate-vacuum obstruction to a kink; two elementary kinks can span the outer vacua only in an infinite-separation limit, not as one finite-width static kink.

There are consequently four oriented elementary sectors: $-1\to 0$, $0\to -1$, $0\to 1$ and $1\to 0$. If opposite orientations are grouped together, there are two kink families and their antikinks. Spatial reflection reverses orientation, and $\varphi \mapsto -\varphi$ exchanges the two adjacent intervals; both preserve the energy. All four therefore have the same rest mass, even though the central and outer vacua have different fluctuation curvatures.

For the sector $0\to 1$, define $W(\varphi )={\varphi }^2/2-{\varphi }^4/4$, so $U=(W^{\prime }{)}^2/2$ and $W^{\prime }=\varphi (1-{\varphi }^2)$. The Bogomolny bound follows by completing the square:It is saturated by the Bogomolny equationLet $y={\varphi }^2$ on this positive branch. Then $y^{\prime }=2y(1-y)$, so $\log [y/(1-y)]=2(x-x_0)$. Thus the explicit kink in a phi-six model isIt has the required limits and satisfies the first integral. The translated center $x_0$ is arbitrary. All oriented elementary solutions areFor $\eta =1$ the endpoints are $0\to \sigma$, and for $\eta =-1$ they are $\sigma \to 0$. Direct evaluation ${\int }_0^1\sqrt{2U(\varphi )}d\varphi ={\int }_0^1\varphi (1-{\varphi }^2)d\varphi =1/4$ independently gives their common mass.

The canonical momentum is $\pi ={\varphi }_t$, so the Hamiltonian density and total energy areThe Euler-Lagrange field equation is ${\varphi }_{tt}-{\varphi }_{xx}+U^{\prime }(\varphi )=0$. Multiplication by ${\varphi }_t$ gives the local conservation of energy identityThus $E$ is conserved when the energy flux vanishes at spatial infinity. For finite energy with asymptotically constant fields, the endpoints ${\varphi }_{\pm }$ must be scalar-field vacua, where $U=0$.

Choose an antiderivative $G^{\prime }(\varphi )=\sqrt{2U(\varphi )}$. For either sign $s=\pm 1$, completing the square givesChoosing $s$ to make the boundary term nonnegative proves the Bogomolny boundThese Bogomolny equations obtained by square completion for a one-dimensional kink describe minima within a fixed endpoint sector when a saturating configuration exists. Differentiating the spatial equation gives ${\varphi }_{xx}=U^{\prime }(\varphi )$ wherever $U>0$, with smooth extension to its limiting scalar-field vacua. The absolute minimum across all sectors is a constant zero-potential scalar-field vacuum when one exists; the nontrivial minima are sector-specific kinks.

For $\beta \ne 0$, write $b=|\beta |$. The scalar-field vacua are $-b,0,b$. A static finite-energy solution has the first integralbecause differentiating its left side gives ${\varphi }_x({\varphi }_{xx}-U^{\prime })=0$, and its value is zero at infinity. In either adjacent interval, $\sqrt{2U}=|\varphi |(b^2-{\varphi }^2)$. Set $y={\varphi }^2/b^2$. The first-order equation becomes $y_x=\eta 2b^{2}y(1-y)$, with $\eta =\pm 1$, and separation gives $\log [y/(1-y)]=2\eta b^2(x-x_0)$. HenceFor $\eta =1$ the endpoints are $0\to \sigma b$; for $\eta =-1$ they are $\sigma b\to 0$. Thus there are four oriented static kink families, up to translation, namely two increasing kinks and two decreasing antikinks. Each has rest energyA static field joining $-b$ directly to $b$ would have to pass through $\varphi =0$ at finite $x$. The first integral then gives ${\varphi }_x=0$, and the Picard-Lindelöf theorem applied to the smooth second-order equation with those initial conditions forces $\varphi \equiv 0$. Therefore there is no such additional kink; this is the intermediate-vacuum obstruction to a kink. The Bogomolny classification of a rescaled phi-six kink also shows that if $\beta =0$, only the scalar-field vacuum $0$ remains and there is no nontrivial finite-energy static kink. Indeed the first integral gives a monotone field wherever it is nonzero, precluding a return to the same scalar-field vacuum.

A Lorentz boost of the $0\to b$ profile gives a moving solution with arbitrary $|v|<1$:If $K$ denotes the static profile and $\xi =\gamma (x-vt-x_0)$, then ${\varphi }_{tt}-{\varphi }_{xx}={\gamma }^2(v^2-1)K^{\prime \prime }=-K^{\prime \prime }$, proving the Euler-Lagrange field equation directly. Its energy is $\gamma b^4/4$: since $K^{\prime 2}=2U(K)$, changing variable to $\xi$ in the energy integral gives the factor $\gamma$. Reversing $\sigma$ or $\eta$ yields the other moving elementary profiles.

Work in units $\hslash =1$ and with the particle mass absorbed into the stated kinetic term. Divide the interval into $N$ pieces of length $\Delta =T/N$. For a suitable self-adjoint Hamiltonian operator, the Trotter product formula gives the regulated limitInsert $1=\int d{q}_j|q_j⟩⟨q_j|$ between neighboring factors. Because $V(\hat{q})$ is diagonal in the position operator basis, the given free-particle propagator yieldsThe exponent is the time-sliced action. Its continuum notation isThe measure is defined by the displayed finite-dimensional limits and their normalization, not by a translation-invariant flat measure on an infinite-dimensional space. The oscillatory limit needs a damping/analytic-continuation prescription. This is the time-sliced configuration-space path integral derived from the operator kernel.

For a quadratic potential energy, write $q=q_c+\eta$, where the classical path satisfies ${\ddot{q}}_c+V^{\prime }(q_c)=0$ and the endpoint conditions, and $\eta (0)=\eta (T)=0$. Expansion is exact:Integration by parts makes the linear term equal to $-[{\int }_0^T({\ddot{q}}_c+V^{\prime }(q_c))\eta dt]$, since the endpoint term vanishes. It is zero by the Euler-Lagrange equation. Also $V^{\prime \prime }$ is constant, independent of the endpoints. Thus classical-path factorization of a quadratic path integral is exact:Only the classical action carries the endpoint dependence.

Let $D={\partial }_t^2+V^{\prime \prime }$ on homogeneous Dirichlet boundary conditions. Expand $\eta$ in its normalized eigenfunctions, with eigenvalues ${\lambda }_n$. The fluctuation integral becomes a product of regulated integrals $\int d{a}_n\exp (-i{\lambda }_n{a}_n^2/2)$. Each is proportional to ${\lambda }_n^{-1/2}$, with a prescribed complex phase. Relative to $D_0={\partial }_t^2$, the normalization is thereforeThe subscript denotes the Dirichlet operator domain, and the square-root branch is continued from the free kernel with its convergence prescription. Equivalently one can use determinants of $-D$ and $-D_0$. For $V^{\prime \prime }={\omega }^2$, the Dirichlet oscillator determinant ratio isThis also checks the free limit. At conjugate times, the fluctuation operator has a zero eigenvalue and the ordinary finite-prefactor expression fails; the kernel is obtained by distributional continuation. Away from these times the classical boundary problem is unique and the stated factorization applies directly.

For the infinite-time oscillator with an external source, use vacuum boundary conditions and the Feynman i-epsilon prescription. For a smooth source of compact support, defineThe imaginary term damps the real-field integral. Integration by parts gives $S_{ϵ}[q]=\frac{1}{2}q{A}_{ϵ}q+Jq$, with the pairings understood as time integrals. Completing the square givesTranslate the integration variable. The source-independent Gaussian functional integral is $N$, soOn Fourier modes $e^{i\omega t}$, $A_{ϵ}$ has eigenvalue ${\omega }^2-m^2+iϵ$, henceFor $m>0$, closing the contour above for $t>0$ and below for $t<0$ gives $G(t)=-i{e}^{-im|t|}/(2m)$. The physical time-ordered oscillator quantum field theory propagator is $iG$, not $G$ itself. The shifted stationary path is $q_c=-GJ$ and obeys ${\ddot{q}}_c+m^2{q}_c=J$ with the same vacuum prescription. A retarded inverse would define a different boundary problem. Dividing by the zero-source integral removes $N$ and gives the vacuum source functional of a quantum oscillator. At zero mass an additional infrared prescription is needed for the inverse on constant modes.

Use the mostly-plus Minkowski metric, so $p^2=-p_0^2+{\mathbf{p}}^2$, and the source term $+\int J\varphi$. Define the vacuum-normalized generating functional, connected generating functional, and quantum effective action byThe Legendre transform is taken after expressing $J$ in terms of $\phi$, locally where this relation is invertible. Varying it gives $\delta \Gamma /\delta \phi =J$ and ${\Gamma }^{(2)}=(W^{(2)}{)}^{-1}$. The logarithm generates connected graphs; the Legendre transform removes graphs joined by a single propagator, leaving one-particle-irreducible vertices. This sign convention is important: its classical limit is $\Gamma =-S$, consistent with the positive vertices in this question.

For the free real scalar field, integration by parts puts the action into the form $S=-\frac{1}{2}\varphi A\varphi$, where $A=-\square +m^2$ has Fourier symbol $p^2+m^2$. Include the Feynman i-epsilon prescription $A\mapsto A-i0$. Completing the square yieldsand consequentlyThe physical scalar propagator is $-i{A}^{-1}$, whereas the inverse quadratic kernel appearing in $W_0$ is $A^{-1}$. Differentiating this quadratic functional and Fourier transforming givesThe infinitesimal prescription specifies propagators and is dropped when quoting the local vertex kernel.

At the order represented by tree-level Feynman diagrams, stationary phase gives $W_{mtree}=S[\phi ]+\int J\phi$, with the classical field equation $\delta S/\delta \phi +J=0$. The Legendre transform cancels the source term, soThusOther connected tree graphs with several quartic vertices are one-particle reducible and belong to $W$, not to the proper vertices of $\Gamma$. This explains why no extra tree correction is hidden in these formulas.

The one-loop proper two-point graph is the tadpole diagram. The one-loop proper four-point graphs are the three bubble diagrams, with pairings $(12|34)$, $(13|24)$ and $(14|23)$ of the external legs. Each has Feynman-diagram symmetry factor $1/2$. The figure shows the topologies and every four-point channel; external momenta are incoming.

The tadpole factor can also be counted directly: there are $4\cdot 3=12$ assignments of two labeled external legs to one quartic vertex, while its $4!$ normalization leaves $12/24=1/2$. For a fixed four-point partition, the interchange of the two internal lines gives the same factor $1/2$; there are three distinct partitions. Graphs with an external self-energy insertion are one-particle reducible and are not additional contributions to ${\tau }_4$.

For the two-point divergence, perform a Wick rotation. The Euclidean tadpole contribution to the proper inverse kernel isAssume $m^2>0$ to separate the ultraviolet calculation from infrared issues. A Schwinger parameterization gives $1/(k_E^2+m^2)={\int }_0^{\infty }ds{e}^{-s(k_E^2+m^2)}$. Using the supplied Gaussian integral, initially in the convergence region, givesThe last expression provides the meromorphic dimensional regularization continuation. Set $d=4-ϵ$. Since $\Gamma (-1+ϵ/2)=-2/ϵ+O(1)$,If the coupling is written as ${\mu }^{ϵ}\lambda$, that factor multiplies the intermediate integral but does not change the displayed pole residue. The negative pole is a property of analytic continuation; it must not be inferred from the positive cutoff integrand without carrying out the continuation.

The added mass counterterm contributes $+B$ to ${\tau }_2$, because ${\Gamma }_{mtree}=-S$ and its Lagrangian term is $-B{\varphi }^2/2$. Hence cancellation requiresup to an arbitrary finite subtraction specifying the renormalization scheme. No external momentum occurs in the tadpole, so no one-loop wave-function renormalization is needed. At $m=0$ the unregulated tadpole is scaleless and vanishes in dimensional regularization; the massive calculation identifies the mass-dependent ultraviolet pole without conflating it with a massless infrared limit.

For completeness, the signs of the four-point graphs in this effective-action convention can be read from the Euclidean one-loop functional $\frac{1}{2}\mathrm{Tr}\log (A_E+\lambda {\phi }^2/2)$. Its expansion has terms $\lambda \mathrm{Tr}(A_E^{-1}{\phi }^2)/4$ and $-{\lambda }^2\mathrm{Tr}(A_E^{-1}{\phi }^2{A}_E^{-1}{\phi }^2)/16$. Differentiating four times givesThis is the Euclidean form, analytically continued to the physical external momenta. It confirms the three channel diagrams and their symmetry factors; evaluating its divergence is not needed for the requested mass counterterm.

Use complex Fourier amplitudes and let primes denote $z$ derivatives. Put $C=B^2/{\mu }_0$, $T=\gamma p+C$ and $d=i{k}_x{\xi }_x+i{k}_y{\xi }_y+{\xi }_z^{\prime }$. Static balance gives $(p+B^2/(2{\mu }_0){)}^{\prime }=-\rho g$. The perturbations from part (a) then implyFor the last identity, combine $\delta p$ with $B\delta B_y/{\mu }_0$ and use static balance; this is where the background magnetic-pressure gradient cancels.

Multiply the Fourier momentum equation by $-{\xi }^{\ast }$ and integrate from $a$ to $b$. Integrating the pressure-gradient term by parts gives $[{\xi }_z^{\ast }\delta \Pi {]}_a^b-\int d^{\ast }\delta \Pi dz$. The gravity term is $\int [-g{\rho }^{\prime }|{\xi }_z{|}^2-\rho g{\xi }_z^{\ast }d]dz$. The magnetic terms reduce toIndeed the term from $\delta B_z{B}^{\prime }{e}_y$ cancels exactly the $B^{\prime }$ term arising from $i{k}_{y}B\delta B$. This establishes, before completing the square,Now $\delta \Pi =A-Td$, soThus the magnetohydrodynamic energy principle has the requested quadratic formAll squares are squared complex moduli. With the prescribed vanishing surface term the quadratic form is real, as required for the self-adjoint stability problem.

In this static scalar problem, a classical field-theory soliton is a nontrivial smooth localized finite-energy field configuration which is static in its rest frame and stable against admissible small perturbations. Moving examples are obtained by a Lorentz boost. The relevant static energy isUse the usual vacuum normalization $U\ge 0$, with zero energy at a vacuum, and impose finite-energy vacuum behavior at spatial infinity. These assumptions, together with the unconstrained real target and the canonical two-derivative kinetic term, are the hypotheses of the following Derrick theorem argument.

For Derrick scaling ${\varphi }_{\lambda }(x)=\varphi (\lambda x)$, change variables to obtainA static solution must be stationary under this admissible scale variation. The Derrick virial identity isFor $D>2$, both terms are nonpositive and the gradient term is strictly negative for a nonconstant field, so no nontrivial static solution is possible. At $D=2$, the identity forces $V=0$, but the gradient term is scale invariant and a further argument is needed.

Since $U\ge 0$, $V=0$ implies $U(\varphi (x))=0$ everywhere. Every zero of a differentiable nonnegative potential has $U^{\prime }=0$. The static Euler-Lagrange field equation $\Delta \varphi =U^{\prime }(\varphi )$ therefore makes $\varphi$ a harmonic function. Each derivative ${\partial }_i\varphi$ is harmonic and belongs to L2 space by finite energy. Its mean-value property and the Cauchy-Schwarz inequality on a ball of radius $R$ implyLetting $R\to \infty$ gives ${\partial }_i\varphi =0$. This proves the two-dimensional flat-target Derrick obstruction. Thus a nontrivial static real-scalar soliton in this class can occur only for $D=1$. The argument does not apply unchanged to curved-target sigma models or gauge-field energies.

In one dimension the scale identity is $T=V$, and the second scale derivative is $2V>0$ for a nontrivial solution; scale variation does not rule it out. To derive the Bogomolny equations, write $U=\frac{1}{2}{W}^{\prime }(\varphi {)}^2$ on the field interval of interest. Completing the square givesTherefore the Bogomolny bound and its saturation equation areEquivalently, for a monotone field connecting adjacent vacua, ${\varphi }^{\prime }=\pm \sqrt{2U(\varphi )}$, with the sign chosen for its direction. Differentiating the first-order equation gives ${\varphi }^{\prime \prime }=W^{\prime }{W}^{\prime \prime }=U^{\prime }$, verifying the static Euler-Lagrange field equation. Saturation minimizes energy within the fixed-boundary topological sector.

For the given sextic potential, factorization gives $U(\varphi )={\varphi }^2({\varphi }^2-4{)}^2$, so the vacua are $-2,0,2$. ChooseThis choice satisfies $U=W^{\prime 2}/2$. On $0<\varphi <2$, the increasing kink in a phi-six model satisfies ${\varphi }^{\prime }=\sqrt{2}\varphi (4-{\varphi }^2)$. Setting $y={\varphi }^2$ gives $y^{\prime }=2\sqrt{2}y(4-y)$. Separation givesand henceThe endpoints are zero and two, and the mass follows from $W(2)-W(0)=4\sqrt{2}$. The other increasing elementary scalar-field kink isIt uses the opposite sign of $W^{\prime }$ and has the same mass. Reflecting $x-x_0$ yields the two decreasing antikinks; $x_0$ is the free center.