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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.