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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.

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.

For completeness, allowing antikinks means ${\kappa }_i={\sigma }_i\cosh {\theta }_i$, ${\beta }_i={\sigma }_i\sinh {\theta }_i$, with ${\sigma }_i=\pm 1$. The velocities remain $\tanh {\theta }_i$. For opposite orientations, $|a_{12}|={\coth }^2[({\theta }_1-{\theta }_2)/2]$, and the general spatial shift isThis follows from the same two local limits; it makes explicit the orientation hypothesis behind the velocity-only all-kink answer.

Only the empty binary configuration contributes to $f$, and only the occupied configuration contributes to $g$. Thus the Hirota tau functions giveHere ${\beta }_1$ denotes the velocity parameter, while $\beta$ without a subscript remains the coupling of Sine-Gordon theory. For $\kappa >0$, the field approaches the adjacent scalar-field vacua $0$ and $2\pi$ at the two ends of space, so its topological charge is $1$. Its center is $X=0$, givingThe constraint ${\kappa }^2-{\beta }_1^2=1$ implies $\kappa =(1-v^2{)}^{-1/2}$. Hence this is precisely a Lorentz boost of the static Sine-Gordon kink, with the expected Lorentz contraction. As a direct check, if $F(X)=4\arctan e^X$, then $F^{\prime }=2\mathrm{sech}X$ and $F^{\prime \prime }=\sin F$, so $({\partial }_t^2-{\partial }_x^2)\varphi +\sin \varphi =({\beta }_1^2-{\kappa }^2)F^{\prime \prime }+\sin F=0$.

The real-parameter condition also permits $\kappa <0$. That choice reverses the topological charge and describes an antikink. The all-kink scattering formulas below use ${\kappa }_i>0$; the orientation dependence is stated explicitly at the end of the two-body calculation.

Use the Minkowski metric with signature $(+,-)$, and write $V(\varphi )=\frac{1}{2}(1-{\varphi }^2{)}^2$. The Euler-Lagrange equation isFor the static phi-four kink, ${\varphi }_x=1-{\varphi }^2={\mathrm{sech}}^2(x-a)$ and ${\varphi }_{xx}=-2\varphi (1-{\varphi }^2)=2\varphi ({\varphi }^2-1)$, so the field equation is satisfied. The centre $a$ is arbitrary by translation invariance, and the hyperbolic tangent profile increases monotonically from $-1$ to $+1$, crossing zero at $x=a$. The static kink and its endpoint topological charge are

Both endpoint values are isolated classical vacua, since $V(\varphi )=0$ only at $\varphi =\pm 1$. A continuous finite-energy deformation preserving the vacuum boundary conditions cannot change either endpoint to the other isolated classical vacuum. The topological charge is therefore unchanged: this kink cannot deform into a homogeneous classical vacuum, whose charge is zero. There is also a direct Bogomolny bound in this sector. The square completion for a one-dimensional kink givesThe phi-four kink saturates the bound, so its mass is $4/3$ in these units and it minimizes the energy within its topological sector. Its arbitrary position is a collective coordinate, not an instability. A kink and an antikink together have total charge zero and can annihilate without contradicting the protection of an isolated kink.

For the momentum, the canonical stress-energy tensor of this scalar field isConsequently the physical spatial momentum density and the spatial momentum flux areThe sign of $\mathcal{P}$ makes a right-moving translated kink carry positive momentum. Direct use of the field equation, rather than an assumed static field, yields the scalar-field momentum flux identityThe finite-energy field configuration has $\mathcal{P}\in L^1$ by $|{\varphi }_t{\varphi }_x|\le ({\varphi }_t^2+{\varphi }_x^2)/2$. Integrating the stress-energy conservation law over the left half-line gives the boundary forceUnder the usual vacuum falloff, the stress at the left endpoint is zero. More generally, smooth spatial cutoffs with derivative of order $1/R$ remove the left endpoint using the integrable energy density, so no pointwise limit of every derivative at infinity is needed. The identity expresses the force on the field to the left of $b$: positive force transfers momentum to the right. For well separated solitons, a cut between them measures the interaction force on the left soliton.

Take that cut at $b=0$. The specified symmetric pair has, at the initial time,The printed field profile does not itself specify the initial velocity. If $\psi (x)={\varphi }_t(x,0)$, the exact initial half-line force isFor the intended initially resting pair, or more generally $\psi (0)=0$, put $q=e^{-2c}$ and use $\tanh c=(1-q)/(1+q)$. The at-rest force for a symmetric phi-four pair isThe leading force is attractive, towards the antikink:The antikink feels the opposite force by the symmetry of the resting pair. This is an initial, large-separation interaction calculation, not a claim that the superposed profile is an exact static two-soliton solution. Without the initial-velocity condition, the additional momentum flux above prevents a unique force from being inferred from the printed profile alone.

Take $\lambda >0$, $m>0$, use metric $(+,-)$ and set $\kappa =\sqrt{2\lambda }m$. The two scalar-field vacua are $\varphi =\pm m$; the positive kink joins $-m$ at the left end to $m$ at the right. The square completion for a one-dimensional kink givesThus its Bogomolny bound is $E\ge 4\sqrt{2\lambda }{m}^3/3$. Equality requires ${\varphi }_x=\sqrt{2\lambda }(m^2-{\varphi }^2)$; separating variables, or differentiating the resulting hyperbolic tangent, gives$X$ is the translational collective coordinate. The antikink is $-{\varphi }_K$ for the centered odd profile. Differentiating the first-order equation gives the static Euler-Lagrange field equation ${\varphi }_{xx}=4\lambda \varphi ({\varphi }^2-m^2)$. Hence the Bogomolny equation really does solve the second-order theory. The two distinct vacua and the above kink require the positive parameters; the degenerate case $m=0$, or the unstable negative-$\lambda$ potential, does not have this topological kink sector.

For a slowly moving kink, put $\varphi (x,t)={\varphi }_K(x-X(t))$. The saturated first-order equation implies $V({\varphi }_K)={\varphi }_K^{\prime 2}/2$, soThe stress-energy tensor has $T^{00}={\varphi }_t^2/2+{\varphi }_x^2/2+V$ and $T^{01}=-{\varphi }_t{\varphi }_x$. Using ${\varphi }_t=-v{\varphi }_K^{\prime }$ in the approximate translate therefore gives $E=M+M{v}^2/2$ and $P=Mv$. Equivalently, substitution into the action gives the collective-coordinate effective Lagrangian $L_{\mathrm{eff}}=-M+M{\dot{X}}^2/2$. These are the translational dynamics of a phi-four kink, withThe displayed remainder orders refer to the exact constant-velocity solution. The uncontracted translated profile is only an approximation: its error as a field starts at $O(v^2)$, but the static profile is an energy stationary point, so the corresponding static-energy error starts at $O(v^4)$ and does not change the displayed $v^2$ coefficient.

Since the relativistic action is invariant under Lorentz transformations, a Lorentz boost gives the exact moving kink, for $|v|<1$,Its conserved energy and momentum are $E={\gamma }_{v}M$ and $P={\gamma }_{v}Mv$, whose expansions agree with the preceding calculation. The width contraction is essential to solving the exact time-dependent equation.

Use the centered profile $K(x)=m\tanh (\kappa x)$. An appropriate approximate separated-lump initial condition isIt tends to $-m$ as $x\to -\infty$ and $m$ as $x\to +\infty$. Near $-a$ and $a$ it is a positive kink, and near zero it is an antikink; the other two tails cancel to exponentially small accuracy. The condition $\kappa a\gg 1$ makes these interpretations accurate. This sum is legitimate smooth initial data, not an exact static multi-kink solution. Its topological charge isand its energy is close to $3M$, with exponentially small interactions.

Adjacent kink–antikink pairs attract: the kink–antikink attraction from the stress tensor gives a negative midpoint pressure and hence an inward force on each outside kink. Direct overlap of the two outer kink tails is exponentially smaller than each adjacent kink–antikink overlap. The initial field is odd and its velocity zero, so the equation's reflection-plus-sign symmetry preserves $\varphi (-x,t)=-\varphi (x,t)$. The central zero remains at $x=0$ and the outer cores move in symmetrically. They collide with the central antikink. Excess energy can excite localized kink oscillations and outgoing radiation; the usual relaxational outcome is a remaining centered kink plus radiation after transient collisions or oscillations. On an infinite line radiation can carry energy away from the core even though total energy is conserved. Conservation of $Q=1$ prevents complete disappearance into vacuum, but it does not determine every bounce or exclude a temporarily re-emerging kink–antikink pair. The nonintegrable collision should not be described as exact elastic passage of all three solitons.

The static kink ${\varphi }_K(x)=4\arctan e^{x-a}$ joins adjacent scalar-field vacua $0$ and $2\pi$. It obeys ${\varphi }_K^{\prime }=2\mathrm{sech}(x-a)$ and has topological charge $Q=[\varphi (+\infty )-\varphi (-\infty )]/(2\pi )=1$. In the Sine-Gordon theory normalization with physical mass scale $m$ and coupling $\beta$, its classical mass is $8m/{\beta }^2$. A Lorentz boost gives ${\varphi }_K=4\arctan \exp [(x-vt-a)/\sqrt{1-v^2}]$. Spatial reflection gives an antikink.