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An antikink is a scalar-field kink with the opposite orientation of its vacuum boundary values. If ${\varphi }_K(x)$ joins vacua ${\varphi }_{-}$ to ${\varphi }_{+}$, then ${\varphi }_K(-x)$ joins ${\varphi }_{+}$ to ${\varphi }_{-}$. For an even double-well potential and an odd centered phi-four kink, the centered antikink is $-{\varphi }_K(x)$. With $Q=[\varphi (+\infty )-\varphi (-\infty )]/({\varphi }_{+}-{\varphi }_{-})$, a kink has $Q=1$ and an antikink has $Q=-1$. Their localized energies coincide by spatial reflection.

For the initial field $\varphi (x)=\tanh (x+c)-\tanh (x-c)-1$ and zero initial time derivative, evaluate the scalar-field momentum flux at the midpoint: ${\varphi }_x(0)=0$ and $\varphi (0)=2\tanh c-1$. The force on the left kink is the displayed positive value, attractive toward the right antikink. This exact initial expression has the large-separation asymptotic $32e^{-2d}$ for $d=2c$. If the initial velocity at the cut is $\psi (0)$, subtract $\psi (0{)}^2/2$.

For $V(\varphi )=\lambda (m^2-{\varphi }^2{)}^2$ with $\lambda ,m>0$, set $\kappa =\sqrt{2\lambda }m$. A well-separated kink–antikink pair at $-d/2,d/2$ is approximated by $\varphi =m\tanh \kappa (x+d/2)-m\tanh \kappa (x-d/2)-m$. At its midpoint, ${\varphi }_x=0$ and $\varphi =m-4m{e}^{-\kappa d}+O(e^{-2\kappa d})$. The static stress-energy tensor component $T^{11}={\varphi }_x^2/2-V$ is therefore negative there. Momentum conservation gives the force on the left half-line as $T^{11}(-\infty )-T^{11}(0)$, which is positive. Its leading magnitude is $F\sim 32m^2{\kappa }^2{e}^{-2\kappa d}$. Thus the pair attracts. This is a controlled leading-tail estimate at large separation, not an exact superposed solution or a theorem about all subsequent collision outcomes.

Use Minkowski spacetime signature $(+,-)$ and put $\kappa =|\lambda |>0$. The Euler-Lagrange field equation for the real scalar field isIndeed, differentiating the potential $U(\varphi )={\kappa }^2({\varphi }^2-1{)}^2/2$ gives $U^{\prime }=2{\kappa }^2\varphi ({\varphi }^2-1)$, while the kinetic term contributes ${\partial }_{\mu }{\partial }^{\mu }\varphi$.

Translation invariance gives the symmetric stress-energy tensorIts divergence is $({\partial }_{\mu }{\partial }^{\mu }\varphi +U^{\prime }){\partial }^{\nu }\varphi$, which vanishes by the Euler-Lagrange field equation. Consequently the conserved energy and physical spatial momentum areThe minus sign in $P=\int T^{01}dx$ follows from ${\partial }^1=-{\partial }_x$. Conservation assumes the associated stress-energy fluxes vanish at spatial infinity, as they do for the localized fields below.

A static finite-energy field configuration approaches vacua $\varphi =\pm 1$. Multiplying ${\varphi }^{\prime \prime }=U^{\prime }(\varphi )$ by ${\varphi }^{\prime }$ giveswhere the vacuum limits fix the constant. For a kink increasing from $-1$ to $1$, ${\varphi }^{\prime }=\kappa (1-{\varphi }^2)$. Integration gives $\mathrm{artanh}\varphi =\kappa (x-X)$ and henceIts negative is the antikink. The arbitrary center $X$ is a translational collective coordinate. The static kink mass is its rest energy; using ${\varphi }^{\prime 2}=2U$ and $y=\kappa (x-X)$ givesEquivalently, completing the static energy into a square gives $E=\frac{1}{2}\int [{\varphi }^{\prime }-\kappa (1-{\varphi }^2){]}^{2}dx+\kappa [\varphi -{\varphi }^3/3{]}_{-\infty }^{\infty }$, so this kink saturates its Bogomolny bound.

A Lorentz boost with velocity $v$ givesWriting $z=\gamma (x-X-vt)$, the derivatives are ${\varphi }_t=-\gamma v{\varphi }_{\mathrm{K}}^{\prime }(z)$ and ${\varphi }_x=\gamma {\varphi }_{\mathrm{K}}^{\prime }(z)$. The momentum integral therefore becomesThe boosted energy isbecause ${\gamma }^{-1}=\gamma (1-v^2)$. Thus $E^2-P^2=M^2$ and $P/E=v$, exactly the energy–momentum relation of a relativistic particle of rest mass $4\kappa /3$. This is the relativistic energy and momentum of a phi-four kink. If $\lambda =0$, the double-well potential disappears and there is no localized static kink connecting these vacua.

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.

The Euler-Lagrange equation is obtained by varying the scalar field, integrating the derivative terms by parts, and requiring the coefficient of an arbitrary compactly supported variation to vanish:The canonical momentum is $\pi ={\varphi }_t$, so the Hamiltonian density and conserved energy areIndeed,Thus $dE/dt=[{\varphi }_t{\varphi }_x{]}_{-\infty }^{+\infty }=0$ for the finite-energy boundary conditions with no flux at infinity. The scalar-field vacua are the constant fields $2\pi m$, $m\in \mathbb{Z}$.

For a static field, multiply ${\varphi }^{\prime \prime }=\sin \varphi$ by ${\varphi }^{\prime }$ to obtain $\frac{1}{2}{\varphi }^{\prime 2}+\cos \varphi =C$. The vacuum limits fix $C=1$, so on the increasing interval $0<\varphi <2\pi$,ConsequentlyIts derivative is $2\mathrm{sech}(x-a)$, and its energy is $\int 4{\mathrm{sech}}^2(x-a)dx=8$. The increasing sign cannot change inside $(0,2\pi )$ because the derivative does not vanish there. This proves uniqueness up to translation for the prescribed boundary conditions.

There are continuously many translated static solutions with those same endpoints, but only one modulo translations. In the full theory, the adjacent-vacuum kinks and antikinks areTheir endpoints are $2\pi m$ and $2\pi (m+\sigma )$. Thus there are countably many adjacent-vacuum boundary sectors, each with a continuous center parameter; integer field shifts and reflection relate their shapes. A Lorentz boost gives moving kinks, replacing $x-a$ by $(x-vt-a)/\sqrt{1-v^2}$, with energy $8/\sqrt{1-v^2}$. There is no finite static kink joining nonadjacent scalar-field vacua: reaching an intermediate vacuum at finite $x$ would give both the equilibrium value and zero derivative, forcing a constant solution by uniqueness of the second-order initial value problem. Dynamical multisolitons are a different matter.

The printed Sine-Gordon Bäcklund transformation has a normalization error. With precisely the printed coordinates $\tau =x+t$, $\rho =x-t$,is the field equation. Put $s=({\varphi }_1+{\varphi }_0)/2$, $d=({\varphi }_1-{\varphi }_0)/2$. The printed relations give $d_{\rho }=b\sin s$ and $s_{\tau }=b^{-1}\sin d$, henceAdding and subtracting yieldsThese are the equations with mass squared four, not the unit-mass equation derived from the Lagrangian. A concrete counterexample is the seed ${\varphi }_0=0$, parameter $b=1$, and transformed field ${\varphi }_1=4\arctan e^{2x}$. It obeys both printed first-order relations but satisfies ${\varphi }_{1,xx}=4\sin {\varphi }_1$, so its unit-mass field-equation residual is $-3\sin {\varphi }_1$, which is not identically zero. The requested preservation claim is false with the printed normalization.

There are two equivalent repairs. Keep the printed first-order coefficients but take $\tau =(x+t)/2$, $\rho =(x-t)/2$; then ${\partial }_{\tau }{\partial }_{\rho }={\partial }_x^2-{\partial }_t^2$ and the preceding calculation proves preservation. Alternatively keep the printed coordinates and replace the coefficients $2b,2b^{-1}$ by $b,b^{-1}$. Then $d_{\rho }=(b/2)\sin s$, $s_{\tau }=(b^{-1}/2)\sin d$, and the differentiated equations carry a factor $1/4$, giving exactly ${\varphi }_{j,\tau \rho }=\frac{1}{4}\sin {\varphi }_j$. This is the light-cone normalization of the sine-Gordon Bäcklund transformation. For a zero seed it gives ${\varphi }_1=4\arctan \exp [(b\rho +b^{-1}\tau )/2]$; $b=1$ recovers the unit-width static kink.

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.

Use $\hslash =c=1$ in the field-theory calculations. Work with signature $(+,-)$ and $a\ne 0$. Since $x=(x_{+}+x_{-})$ and $t=(x_{+}-x_{-})$, ${\partial }_{+}={\partial }_x+{\partial }_t$, ${\partial }_{-}={\partial }_x-{\partial }_t$, and the Sine-Gordon equation is ${\varphi }_{+-}=\sin \varphi$. Introduce $w=(\rho +\kappa )/2$ and $d=(\rho -\kappa )/2$. The specified Sine-Gordon Bäcklund transformation becomes $d_{+}=a\sin w$, $w_{-}=a^{-1}\sin d$. Its compatibility givesAdding and subtracting these equations establishesThus a compatible transformed field satisfies the same Sine-Gordon equation. Smoothness is needed to interchange mixed derivatives. The statement cannot literally include $a=0$, since one defining equation contains $1/a$.

Generating the kink. With zero seed, the two first-order equations give ${\rho }_{+}=2a\sin (\rho /2)$ and ${\rho }_{-}=2a^{-1}\sin (\rho /2)$. On a nonconstant branch put $p=\tan (\rho /4)$; then $p_{+}=ap$, $p_{-}=a^{-1}p$. Integration givesFor $a>0,C>0$, this is the Sine-Gordon kink, after absorbing $C$ in the center position. Its speed and Lorentz factor obeyChanging the sign of $a$ or of the exponential coefficient selects the corresponding antikink orientation; $|v|<1$ whenever $a$ is finite and nonzero. Constant vacuum branches, omitted by division by $\sin (\rho /2)$, can be added separately.

The continuously parametrized Bäcklund transformation does more than produce this one solution. Successive compatible transformations and Bianchi permutability for sine-Gordon Bäcklund transformations construct multisoliton solutions; expansion of the generating relations gives the local conserved-charge hierarchy of sine-Gordon theory. Together with its Lax pair and inverse scattering transform, this is the structure behind classical integrability, elastic scattering and the absence of generic radiative energy loss in soliton collisions. Existence of a transformation alone is not a proof that every generated conserved charge is independent and in involution; those are additional properties of the integrable system.

Transforming the static kink at $a=1$. The seed obeys ${\kappa }_x=2\mathrm{sech}x$, $\sin (\kappa /2)=\mathrm{sech}x$, $\cos (\kappa /2)=-\tanh x$. Adding and subtracting the two Sine-Gordon Bäcklund transformation equations yieldsAgain set $p=\tan (\rho /4)$. These simplify to $p_x=-p\tanh x$, $p_t=\mathrm{sech}x$. The first gives $p=f(t)/\cosh x$, and the second gives $f^{\prime }=1$. ThereforeContinuous inverse-tangent branches may be chosen without creating artificial jumps. At spatial infinity the field approaches the same vacuum, so its net topological charge is zero. For large $|t-t_0|$ there are two well-separated transitions near $x_{\pm }=\pm \mathrm{arcosh}|t-t_0|$, with speeds tending to zero. The two transitions exchange their kink/antikink orientations as they pass through the collision, while the field remains smooth.

This is the Sine-Gordon threshold kink-antikink solution, a separatrix between finite-speed scattering and the bounded Sine-Gordon breather motion, rather than a finite-period breather. In fact it is the $v\to 0$ limit of the scattering field in question 3. At $t=t_0$, $\rho =0$ and ${\rho }_t=4\mathrm{sech}x$, so the reduced energy is $\int {\rho }_t^{2}dx/2=16$, exactly twice the reduced single-kink energy $8$.

For mass comparisons below, the original PDF really prints $m^2/\beta$. Restoring physical coordinates $X=x/m$, $T=t/m$ gives the angular-field action ${\beta }^{-1}\int dtdx[\frac{1}{2}(\partial \varphi {)}^2+\cos \varphi -1]$ and literal kink mass $8m/\beta$, for $\beta >0$. The usual Sine-Gordon theory instead has prefactor $m^2/{\beta }^2$ and mass $8m/{\beta }^2$. The equation of motion and every field above are unchanged by this overall normalization; the quantum scattering comparison in question 3 is not.

Take $0<v<1$ and $\Gamma =(1-v^2{)}^{-1/2}$. The Sine-Gordon kink-antikink scattering solution tends to the same vacuum at both spatial ends and hence has zero net topological charge. At large $|t|$, the two transition centers, found by setting the magnitude of the inverse-tangent argument to one, obeyThey are a kink and an antikink moving with asymptotic speeds $\pm v$, with no outgoing radiation. Their left-right orientations interchange at the collision. To define the soliton time delay, label a transmitted trajectory by its preserved rapidity rather than its left-right position. The right-moving asymptotes areAfter restoring physical lengths, the forward shift is $\Delta X=-2\log v/(m\Gamma )>0$. The arrival-time difference is thereforeThis is a time advance relative to the extrapolated incoming free motion.

The semiclassical scattering phase. For an outgoing energy wave packet with full S-matrix phase $S=e^{i\delta (E)}$, stationary phase in $e^{-iET+i\delta (E)}$ shifts its arrival time by $d\delta /dE$. The full S-matrix phase from a classical soliton delay uses the paper's $\delta$, twice the phase in a convention $S=e^{2i{\delta }_{\mathrm{pw}}}$. In the center-of-momentum frame, $E=2{\mathcal{M}}_K\cosh (\theta /2)$, so $dE/d\theta ={\mathcal{M}}_K\sinh (\theta /2)$. It follows thatThe literal PDF normalization gives ${\mathcal{M}}_K=8m/\beta$ and consequentlyThe integrable logarithmic singularity at zero causes no divergence of this phase difference. With the intended standard angular-field prefactor $m^2/{\beta }^2$, the result is insteadThe soliton time delay determines only phase differences, not an energy-independent constant or a choice of $2\pi$ branch. Keeping this distinction avoids an arbitrary high-energy subtraction.

Bound-state poles. Set $a_j=\pi j/N$. Each factor of the exact amplitude can be written $\cosh [(\theta -i{a}_j)/2]/\cosh [(\theta +i{a}_j)/2]$. Its denominator vanishes in the physical rapidity strip atThe corresponding numerator is nonzero, and no other numerator cancels the pole. In the direct kink–antikink fusion interpretation, analytically continue the constituent rapidities to $\pm i{u}_n/2$. Their momenta sum to $(2{\mathcal{M}}_K\cos (u_n/2),0)$, so the relativistic bound-state mass from a rapidity pole givesThese neutral particles form the Sine-Gordon breather spectrum at reflectionless couplings: there are $N-1$ breathers, none for $N=1$, and the putative $n=N$ state is at the unbound $2{\mathcal{M}}_K$ threshold. The amplitude fixes these ratios to the physical kink mass; its rapidity dependence alone cannot fix the overall mass scale or a mass-renormalization prescription relating that mass to $m$.

It is important to distinguish direct and crossed interpretations rather than count every occurrence of a pole twice. Crossing symmetry sends $i{u}_n$ to $i(\pi -u_n)=i{u}_{N-n}$. At the original angle the momentum-difference invariant is $4{\mathcal{M}}_K^2{\sin }^2(u_n/2)={\mathcal{M}}_{N-n}^2$, so the crossed interpretation exchanges the complementary member of the same tower. In particular the transmission residue alternates sign: direct evaluation of the remaining factors gives ${\mathrm{Res}}_{i{u}_n}{S}_T=i(-1{)}^n{R}_n$, $R_n>0$. Thus one must retain charge-channel/crossed-channel information, not reject every negative-imaginary transmission residue or declare an additional particle for a crossed pole. Direct and crossed locations coincide in the reflectionless pole set. The familiar $N=2$ case has one breather of mass $\sqrt{2}{\mathcal{M}}_K$ despite the negative-imaginary transmission residue; this diagonal reflectionless example is also displayed in Castro-Alvaredo, Chen, Doyon and Hoogeveen, section 4.2.

Matching the exact phase. For the unwrapped reflectionless sine-Gordon transmission phase, at real $\theta \ge 0$ choose the continuous unwrapped branch with ${\delta }_N(0)=\pi N$. ThenEvery factor has unit modulus, consistent with purely transmitting elastic scattering. Differentiating is simpler than integrating its complex logarithm:For fixed $\theta >0$ the sum becomes a Riemann sum, and the elementary integral yieldsSince ${\beta }^2=8\pi /(N+1)$ and $\gamma =8\pi /N$, replacing $\gamma$ by ${\beta }^2$ changes this only at subleading order. The leading phase difference agrees with the standard semiclassical result. The derivative approximation is not uniform as $\theta \to 0$ at fixed $N$; integrating its logarithm gives a finite leading phase difference nevertheless. A compatible constant is the given branch $\pi N$, whose choice is supplied by the exact amplitude rather than the classical trajectory.

As a useful endpoint check, ${\int }_0^{\infty }\log \tanh (u/2)du=-{\pi }^2/4$, obtained by expanding the two logarithms $\log (1-e^{-u})-\log (1+e^{-u})$. Thus the leading phase with coefficient $2N/\pi$ tends to $\pi N/2$. The exact unwrapped endpoints arewhere the extra $\pi /2$ is subleading. Neither should be replaced by zero by an unannounced branch convention. Finally, the literal $m^2/\beta$ action gives a phase derivative of order $\sqrt{N}$, whereas the printed exact amplitude gives order $N$. The requested exact/semiclassical matching requires correcting the overall action normalization to $m^2/{\beta }^2$. This is an actual inconsistency between pages 2 and 4, not a change to the sine-Gordon equation or the classical scattering field.

Choose the positive-exponential branches from part (i) and set their additive constants to zero. First take $0<b<1$, and defineThen $\tan ({\varphi }_b/4)=e^{\gamma (x-vt)}$ and $\tan ({\varphi }_a/4)=e^{-\gamma (x+vt)}$. The tangent subtraction formula givesConsequently the allowed Sine-Gordon superposition formula produces the smooth fieldThis is the negative of the Sine-Gordon two-kink solution, and hence a two-antikink configuration. The auxiliary seeds have opposite topological charges, but their charges cannot simply be added to infer the charge of the nonlinear two-step Bäcklund transformation. Indeed, the displayed final field tends to $2\pi$ at the left spatial end and $-2\pi$ at the right, so its total topological charge is $-2$.

Let $T=|t|$ become large. Near the right transition, $x=vT+O(1)$, the tangent argument has the asymptotic formso the local field is $-4\arctan e^{\gamma (x-vT)+\log v}$, a single antikink. Near the left transition, the local field is $4\arctan e^{-\gamma (x+vT)+\log v}+o(1)$, again a decreasing antikink. The resulting asymptotic center lines areThus two incoming antikinks with topological charges $(-1,-1)$ and velocities $(+v,-v)$ separate again with exactly the same topological charges and velocities. There is no radiative tail in these asymptotic profiles. Labeling the outgoing objects by their preserved rapidities makes this elastic soliton scattering; labeling the left and right lumps instead describes reflection with exchanged velocities.

For the right-moving soliton, its incoming intercept is ${\gamma }^{-1}\log v$ and its outgoing intercept is $-{\gamma }^{-1}\log v$. The spatial shifts are therefore $\Delta x_{+}=-2{\gamma }^{-1}\log v$ and $\Delta x_{-}=2{\gamma }^{-1}\log v$. Define the soliton time delay as the change in arrival time at a fixed distant spatial point relative to continuation of the incoming straight line, so $\Delta t=-\Delta x/v_{\mathrm{particle}}$. Both objects have the same signed soliton time delay, which is an advance:This is the Sine-Gordon two-kink time advance. In physical coordinates $T_{\mathrm{phys}}=t/m$, the time shift is $2\log v/(m\gamma v)$. The explicit intercepts fix the sign convention unambiguously.

The remaining real parameter choices are covered without changing the calculation. For any $b\ne 0$ with $b^2\ne 1$, put $v_b=(1-b^2)/(1+b^2)$, $u=|v_b|$, $\gamma =(|b|+|b{|}^{-1})/2$ and $ϵ=\mathrm{sgn}(b{v}_b)$. The same choice of zero additive constants givesEach scattered object's topological charge is $-ϵ$, the velocities are $\pm u$, and the signed soliton time delay is $2\log u/(\gamma u)$. If $b=\pm 1$, the superposition coefficient vanishes and this representative is the vacuum; there is no pair of separated moving solitons and no scattering delay to assign. Thus the scattering conclusion requires the nondegenerate case $0<u<1$.

With zero seed, both Bäcklund transformation equations concern ${\varphi }_a/2$. On a nonconstant branch, separation of variables uses $d\log |\tan ({\varphi }_a/4)|=d{\varphi }_a/[2\sin ({\varphi }_a/2)]$, givingFor $C>0$, absorb its magnitude into an additive constant $c=\log C$ in the exponent. A convenient smooth representative isThe constant-phase condition for this traveling profile determines its velocity:The profile is a traveling soliton with velocity $v_a=(1-a^2)/(1+a^2)$, strictly between $-1$ and $1$. The scalar-field vacua on the two sides differ by $2\pi$. With the topological charge convention $Q=[\varphi (+\infty )-\varphi (-\infty )]/(2\pi )$, this branch has $Q=\mathrm{sgn}a$: it is a Sine-Gordon kink for $a>0$ and an antikink for $a<0$. Its width is proportional to ${\gamma }_a^{-1}$, and its derivative decays exponentially away from its center, giving a finite-energy field configuration.

For completeness, the classical rest mass in this paper's coupling convention is $8m/\beta$, as derived in Question 2; a Lorentz boost gives energy $8m{\gamma }_a/\beta$. This localized, topologically protected traveling field is the required classical field-theory soliton. A negative $C$ reverses the field and hence the topological charge; $C=0$ gives the vacuum rather than a soliton. Vacuum shifts by $4\pi$ can be made without changing the displayed Bäcklund transformation equations.

Write the rapidity parameters as ${\kappa }_i=\cosh {\theta }_i$, ${\beta }_i=\sinh {\theta }_i$, and define $E_i=e^{X_i}$. The signed coefficient in the Sine-Gordon multisoliton tau representation isFor distinct rapidities, $0<{\alpha }_{12}<1$. In particular, $B_{12}$ cannot be taken as a real logarithm of a positive coefficient. The finite sums defining the Hirota tau functions can instead be evaluated directly with the real, negative $a_{12}$. They giveThe physical field is a continuous branch of a multivalued function, equivalently $\varphi =4\arg (f+ig)$ with the argument followed continuously. The principal inverse tangent alone jumps when $f$ changes sign.

Follow the first kink with $x=v_{1}t+O(1)$. Then $X_1=O(1)$ and $X_2={\kappa }_2(v_1-v_2)t+O(1)$. The two possible local limits arewhere the second field is written on the continuous kink branch. Thus both limits are single Sine-Gordon kinks of the same width and velocity, but their centers obey $X_1=0$ or $X_1+\log {\alpha }_{12}=0$. Following the second kink gives the same conclusion with labels exchanged. The incoming and outgoing velocities are thereforeThere is no change in the asymptotic rapidities or kink profiles.

Define the spatial shift as the outgoing center intercept minus the incoming center intercept. Since the large-$E_2$ limit occurs afterwards when $v_1>v_2$, and beforehand when $v_1<v_2$, the soliton time delay isThe time formula uses $\Delta t=-\Delta x/v_1$ and requires $v_1\ne 0$. Its dependence on the velocities is explicit on substitutingFor a faster right-moving kink, $\Delta x_{1|2}>0$ and $\Delta t<0$: it arrives earlier than its freely continued incoming trajectory. If $v_1=0$, report the finite spatial shift; a fixed-position arrival-time delay for a stationary kink is undefined. Coincident velocities are excluded from a separated collision asymptotic.