Kink–antikink attraction from the stress tensor ($F\sim 32m^2{\kappa }^2{e}^{-2\kappa d}$)

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.

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 $\psi (0{)}^2/2$.