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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 the usual canonical scalar field theory, subtract the vacuum energy so that the potential $U$ is nonnegative, and write the static energy in $d$ space dimensions asAssume finite energy and sufficient decay to justify the scaling variation, with a fixed vacuum at infinity. Under Derrick scaling ${\varphi }_{\lambda }(x)=\varphi (\lambda x)$,A static solution must be stationary under this admissible variation. Thus the Derrick virial identity isFor $d\ge 3$ both terms are nonpositive, and the coefficient of $T$ is strictly negative. Therefore $T=W=0$: the field is a constant vacuum. This proves the Derrick theorem under its hypotheses. Negative potentials, additional derivative terms, gauge fields or time dependence can invalidate this particular obstruction; it is not a theorem about arbitrary energy functionals.

A non-topological soliton can instead be supported by a conserved ordinary Noether charge. For a complex scalar field with global $U(1)$ invariance, take $\varphi =e^{i\omega t}f(x)$ and let $I=\int f^2$. In the normalization with kinetic density $|\partial \varphi {|}^2/2$, its charge is $Q=\omega I$. At fixed charge, eliminating $\omega$ givesThe charge-constrained scalar-field scaling now readsThe positive scaling power of the charge term can balance the others even for $d\ge 3$. Such a Q-ball is time-dependent although its energy density is stationary, and the fixed-charge variational problem is different from the static one.

Suitable potentials permit this mechanism in any spatial dimension. For example, suppose $U(f)>0$ for $f\ne 0$ and the ratio below attains a strictly positive minimum:Such a potential can favor a large region of nearly constant nonzero amplitude. At its preferred amplitude, the bulk terms minimize at $I\approx |Q|/{\omega }_0$, with energy about ${\omega }_0|Q|$. A wall has surface energy of order $|Q{|}^{(d-1)/d}$, lower order than the bulk term for every fixed $d\ge 1$. For large charge the energy can therefore be below the free-particle threshold $m|Q|$. This explains how a localized fixed-charge solution may exist and resist dispersal. Existence and stability still require a suitable potential and charge range; not every complex-field theory or every member of a rotating family is a constrained minimum.

For the particular one-dimensional profile, substituting the rotating field into the equation givesMultiplying by $f^{\prime }$ shows thatBoth $f$ and $f^{\prime }$ vanish at infinity, so the integration constant is zero. HenceHere $U(f)=f^2/2-f^4/4$. This potential is not globally bounded below; the existence of the displayed localized solutions alone does not prove their nonlinear stability.

Use metric signature $(+,-)$ and setThe Lorentz boost of a Q-ball givesThe peak follows $x=x_0+vt$. Because the equation is Lorentz invariant and the field is a scalar, this is again a solution.

The canonical stress-energy tensor gives the energy and signed physical momentum at fixed time:The minus sign agrees with the scalar-field momentum flux in the chosen signature. For the boosted field,andThe first integral gives $U(f)=\frac{1}{2}[(f^{\prime }{)}^2+{\omega }^2{f}^2]$. Changing variable to $\xi$, so $dx=d\xi /\gamma$, yieldsThus the rest-energy normalization and the relativistic energy-momentum relation are

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.