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For $U={\varphi }^2({\varphi }^2-b^2{)}^2/2$ and $b>0$, the three scalar-field vacua are $-b,0,b$. The static first integral is ${\varphi }_x^2=2U$. In either adjacent interval, $y={\varphi }^2/b^2$ obeys $y_x=\pm 2b^{2}y(1-y)$, giving the displayed four oriented kink profiles with $\sigma ,\eta \in \{1,-1\}$. Each has rest energy $b^4/4$. The intermediate-vacuum obstruction to a kink excludes a direct static connection from $-b$ to $b$: reaching zero would give zero derivative, and uniqueness of the smooth second-order field equation forces the zero solution. A Lorentz boost replaces $x-x_0$ by $(x-vt-x_0)/\sqrt{1-v^2}$.

// Destination: relativistic-quantum-field.bigb

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.

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.

There is no single finite-width static kink directly connecting $-2$ to $2$. Multiplying the static equation by ${\varphi }^{\prime }$ gives the first integral ${\varphi }^{\prime 2}/2-U=0$ for a finite-energy vacuum-to-vacuum solution. At the intermediate vacuum $\varphi =0$, this forces ${\varphi }^{\prime }=0$. Uniqueness for the smooth second-order field equation makes data $(\varphi ,{\varphi }^{\prime })=(0,0)$ stay at that vacuum, so a nontrivial solution cannot cross it at finite $x$. This is the intermediate-vacuum obstruction to a kink; the two elementary transitions can only separate by an infinite interval.

The potential factorizes asIt is even and nonnegative, with three degenerate vacua at $\varphi =-2,0,2$. Its two intervening maxima occur at $\varphi =\pm 2/\sqrt{3}$ and have height $256/27$.

A finite-energy static solution in one dimension satisfiesMultiplication by ${\varphi }^{\prime }$ and use of the vacuum boundary conditions gives the first integralAny path from a negative vacuum to a positive vacuum must pass through the intermediate vacuum $\varphi =0$. There both $U$ and ${\varphi }^{\prime }$ vanish. The Picard-Lindelöf theorem then forces a solution reaching $(\varphi ,{\varphi }^{\prime })=(0,0)$ at finite $x$ to remain there. Equivalently, the first-order orbit approaches $\varphi =0$ only as $|x|\to \infty$. This intermediate-vacuum obstruction to a kink means that a single kink cannot connect $-2$ to $2$; it splits into two elementary kinks at infinite separation.