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Use the inverse Lorentz transformationAlong the particle's worldline $d{x}^{\prime }=u^{\prime }d{t}^{\prime }$, so the relativistic velocity-addition formula isFor subluminal collinear velocities, put $u/c=\tanh {\phi }_u$, $u^{\prime }/c=\tanh {\phi }_{u^{\prime }}$ and $v/c=\tanh {\phi }_v$. The hyperbolic tangent addition formula, obtained by dividing the supplied sine/cosine addition formulas, givesBecause hyperbolic tangent is injective on the real line,Thus rapidity is additive even though velocity is not.

Each positive increment in the instantaneous rest frame adds the same rapidity $\alpha =\mathrm{artanh}(1/2)=\frac{1}{2}\log 3$. Starting from zero rapidity, iterated collinear boosts with equal rapidity giveEvery finite number of increments leaves $u_n<c$; the speed of light is approached as $n\to \infty$.

For the simple symmetric random walk, independence and the Rademacher distribution give the moment-generating functionThe function ${\theta }^2/2-\log \cosh \theta$ is even; for $\theta \ge 0$ its derivative is $\theta -\tanh \theta \ge 0$, since the derivative of $\tanh \theta$ is at most one and $\tanh 0=0$. Here $\cosh$ and $\tanh$ are the hyperbolic cosine and hyperbolic tangent. Hence the exponential-moment estimate is

The supplied exponential maximal bound for a symmetric random walk now gives $\mathbb{P}({\max }_{k\le n}{S}_k\ge t\sqrt{n})\le \exp (-\theta t\sqrt{n}+n{\theta }^2/2)$. For $t>0$, choose $\theta =t/\sqrt{n}$; for $t=0$, use the elementary bound by one. Thus

Fix $a>1$, choose $0<\delta <a^2-1$, and set $n_m=\lfloor (1+\delta {)}^m\rfloor$. For large $m$ these form an increasing sequence. Apply the preceding exponential maximal bound for a symmetric random walk at $n_{m+1}$ with threshold $a\sqrt{2n_m\log \log n_m}$. The resulting probability is at mostWe have $n_m/n_{m+1}\to (1+\delta {)}^{-1}$ and $\log \log n_m=\log m+O(1)$. Choose $b$ strictly between $1$ and $a^2/(1+\delta )$; the displayed probabilities are $O(m^{-b})$. By the Borel-Cantelli first lemma, almost surely, eventually ${\max }_{k\le n_{m+1}}{S}_k<a\sqrt{2n_m\log \log n_m}$. For every $n_m\le n\le n_{m+1}$, monotonicity of $n\log \log n$ for large $n$ therefore givesIntersect the probability-one events for $a=1+1/r$ to obtain the upper law of the iterated logarithm:

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.

For a real steady amplitude, integrate the conserved-field equation twice. Periodicity forces the integration slope to vanish, and the zero mean fixes the remaining constant. HenceSubstituting this into the steady amplitude equation gives the nonlocal ordinary differential equationFor $A=R\mathrm{sech}(cX)$, use the hyperbolic secant identity $A^{\prime \prime }=c^{2}A-2c^2{A}^3/R^2$. Vanishing of the two independent coefficients givesChoose $c>0$ and $R>0$; an overall negative sign is another solution. Thus $\kappa >1$. The exact average of the proposed profile over the finite interval isFor $cL\gg 1$, the hyperbolic tangent differs from one by $O(e^{-cL})$, so $cL⟨A^2⟩\simeq 2R^2$. Eliminating $R^2=2c^2/(\kappa -1)$ gives the requested leading relationEquivalently $c+c^{-1}=K$, with $K=4\kappa /[L(\kappa -1)]$. Positive real roots exist precisely when $K\ge 2$; they are $(K\pm \sqrt{K^2-4})/2$ and are reciprocal. Each candidate must additionally satisfy $cL\gg 1$. For $L>2$, the leading necessary range is $1<\kappa \le L/(L-2)$, with equality at the coalescence $c=1$. Away from this leading-order fold, the inequality is strict.

The strict finite-window condition can also be seen by retaining the average exactly. Its consistency relation is $c+c^{-1}=K\tanh (cL/2)$. Since $\tanh (cL/2)<1$, it implies $K>2$, and henceThe strict endpoint distinction is exponentially small in the large-$cL$ approximation; the leading polynomial alone permits its double root. When $0<L\le 2$, this algebra imposes no finite upper bound on $\kappa >1$. The printed interval therefore needs $L>2$: for example $L=1$, $\kappa =1.1$ gives a large positive root $c\simeq 43.98$ of the leading equation and $cL\gg 1$, despite the printed upper endpoint being negative.

Finally, the single localized pulse of a conserved-field convection model is an approximate periodic solution. Its values agree at the interval ends, but its derivatives have opposite signs there; their magnitude is exponentially small, $O(cR{e}^{-cL/2})$. Periodic-tail corrections are needed for an exact smooth periodic boundary condition. The range derived from the truncated profile is a consistency condition, not a proof that every permitted parameter gives an exact stable periodic pulse.

The condition $L\ll na$ implies $|\tau a|\ll 1$. The small-argument expansion of the hyperbolic tangent gives $\tanh x=x+O(x^3)$, soUsing $f=k_{B}T\tau$ yields the entropic Hooke law for a one-dimensional chainThus the effective spring constant is $k_{B}T/(n{a}^2)$.

At fixed tension, the exact extension isIncreasing the temperature decreases the positive argument of the hyperbolic tangent, so the chain contracts. In the small-extension regime this becomeswhich is inversely proportional to $T$.