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A dispersion relation specifies which angular frequencies and wavevectors a linear wave model admits. Fix the plane wave conventionHere $a$ is a constant complex amplitude, $\mathbf{k}$ is the wavevector, and $\omega$ is the angular frequency. Substitution into a constant-coefficient linear partial differential equation replaces ${\partial }_t$ by $-i\omega$ and $\nabla$ by $i\mathbf{k}$, turning the wave problem into an algebraic condition. A system can have several roots; $\omega =\omega (\mathbf{k})$ means one chosen branch, rather than a promise that the full dispersion relation is single-valued. For real $\mathbf{k}$, $\mathrm{Im}\omega >0$ means temporal growth in this convention, while $\mathrm{Im}\omega <0$ means damping. The real-branch discussion of phase velocity and group velocity below assumes negligible loss.

For the one-dimensional wave equation $u_{tt}=c^2{u}_{xx}$, the dispersion relation is ${\omega }^2=c^2{k}^2$. The branch $\omega =ck$ carries right-moving travelling waves for $c>0$; both its phase velocity and group velocity are $c$. One may instead use the positive-angular frequency branch $\omega =c|k|$, with right-moving components at $k>0$ and left-moving components at $k<0$. A real wave pairs each $(k,\omega )$ with $(-k,-\omega )$; that conjugate pair represents the same travelling pattern. Branch and sign choices therefore matter when interpreting a dispersion diagram.

A particularly tangible dispersive example is a small-amplitude surface gravity wave on still water of uniform depth $h$. Neglect surface tension and viscosity, assume incompressible flow and irrotational flow, and impose a flat impermeable bottom. With $k>0$ and positive angular frequency,Here $g$ is gravitational acceleration and $\tanh$ is the hyperbolic tangent. Deep water means $kh\gg 1$: $\omega \simeq \sqrt{gk}$ and longer wavelengths have larger phase speed. Shallow water means $kh\ll 1$: $\omega \simeq \sqrt{gh}k$, an approximately nondispersive wave. Thus depth relative to wavelength, rather than depth alone, determines whether crests and wave packets travel together.^[ref]

The dispersion diagram below also includes the ideal deep-water capillary wave limit $\omega =\sqrt{\gamma /\rho }{k}^{3/2}$, where $\gamma$ is surface tension and $\rho$ is mass density. This limit requires the surface tension restoring term to dominate gravity; it reverses the trend of the surface gravity wave, giving larger phase speed at shorter wavelength. The combined gravity-capillary wave has ${\omega }^2=gk+(\gamma /\rho )k^3$.^[ref]

The curves use dimensionless normalizations chosen separately for each model; they are analytical diagrams, not experimental data. Their beauty is that a single dispersion relation encodes several observable behaviours: its value gives the angular frequency, its origin-to-point slope gives the one-dimensional phase velocity, its tangent slope gives the group velocity, and variation of that tangent slope predicts distortion of a wave packet. A straight line through the origin describes a nondispersive wave; an affine branch $\omega =ck+b$ with $b\ne 0$ has constant group velocity but phase velocity $c+b/k$, so those properties must be distinguished.

For $|\theta |\ll 1$ and $s=\delta Z\ll 1$, the Fay solution becomes $q\sim 4\delta \theta {\sum }_{j\ge 0}[((2j+1)s{)}^2+{\theta }^2{]}^{-1}$. The partial-fraction identity for the hyperbolic tangent sums this to the displayed smooth shock layer. It resolves the odd transition from $-\pi /Z$ to $+\pi /Z$ on angular thickness $O(\delta Z)$; at $\theta =0$ both sides vanish and the comparison is read through their derivatives.

Hyperbolic functions combine the exponential function to describe a hyperbola much as trigonometric functions describe a circle. Their two basic members are the hyperbolic cosine and hyperbolic sine, defined for every real number $t$ byTheir derivatives exchange the two functions:These formulas follow by applying differentiation to the exponential function definitions. ^[ref]

Adding and subtracting gives $\cosh t+\sinh t=e^t$ and $\cosh t-\sinh t=e^{-t}$. Multiplying these equalities proves the hyperbolic Pythagorean identity:Consequently the parametric equation $(x,y)=(\cosh t,\sinh t)$ traverses the entire right branch of the unit hyperbola $x^2-y^2=1$, where $x\ge 1$. For example, $t=0$ gives $(1,0)$, while $t=\pm 1$ gives approximately $(1.543,\pm 1.175)$. Compare $(\cos \theta ,\sin \theta )$ on the unit circle, where the squared coordinates add to one. The minus sign changes the geometry from a closed circle to an unbounded hyperbola.

The other members are ratios and reciprocals: hyperbolic tangent is $\tanh t=\sinh t/\cosh t$, hyperbolic cotangent is $\coth t=\cosh t/\sinh t$, and hyperbolic secant is $\mathrm{sech}t=1/\cosh t$; denominators must be nonzero. The hyperbolic sine and hyperbolic cosine both solve the second-order linear differential equation $f^{\prime \prime }=f$. In fact $f(t)=A\cosh t+B\sinh t$ has $f(0)=A$ and $f^{\prime }(0)=B$, making this basis convenient for an initial value problem. Their geometry also makes hyperbolic substitution useful for integrals containing $\sqrt{1+u^2}$, and the hyperbolic cosine gives the physical catenary of a hanging chain.

Consider a stipulated dilute collection of independent two-level systems, each with longitudinal magnetic moment $s\mu$, $s=\pm 1$, and energy $E_s=-s\mu B_0$. The prescribed vacuum/applied magnetic flux density is $B_0={\mu }_0{H}_{\mathrm{app}}$. Ignore interactions, orbital magnetic response, and the small sample-field correction. This is the magnetic version of the existing independent spin-one-half two-level system, with alignment probabilities rather than an interacting Curie–Weiss model.

For temperature $T>0$, the canonical ensemble gives the one-moment partition function $Z_1=2\cosh x$, with $x=\mu B_0/(k_{B}T)$. Taking the expected value over its two Boltzmann weights givesHere $n$ is the number density, $k_B$ the Boltzmann constant, and $\tanh$ the hyperbolic tangent. For $|x|\ll 1$, its small-argument expansion of the hyperbolic tangent gives $M\simeq {\mu }_{0}n{\mu }^2{H}_{\mathrm{app}}/(k_{B}T)$, a Curie's law response in the applied-$H$ convention. For large $|x|$, $M$ approaches the saturation magnetization $n\mu$ with sign set by the field, and its differential magnetic susceptibility falls towards zero.

Both plots use stipulated $n=10^{26}{\mathrm{m}}^{-3}$, $\mu =9.2740101\times 10^{-24}\mathrm{J}{\mathrm{T}}^{-1}$, $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$, and ${\mu }_0=1.25663706\times 10^{-6}\mathrm{T}\mathrm{m}{\mathrm{A}}^{-1}$. The moment is approximately one Bohr magneton; the chosen vacuum permeability is a rounded model input. At $T=2\mathrm{K}$, $M_s=927.401\mathrm{A}{\mathrm{m}}^{-1}$ and ${\chi }_{\mathrm{model}}(0)=3.91410\times 10^{-4}$. The linear approximation overestimates $M$ by less than $3\%$ for $|x|\le 0.3$, corresponding to $|{\mu }_0{H}_{\mathrm{app}}|\le 0.89324\mathrm{T}$ or $|H_{\mathrm{app}}|\le 0.71082\mathrm{MA}{\mathrm{m}}^{-1}$. These parameters keep the modeled response weak; neglected magnetic feedback is of relative order ${\chi }_{\mathrm{model}}(0)$, rather than an explanation of a material's ordered phase.

At $B_0=0$, cooling increases the zero-field slope as $1/T$. At fixed $B_0\ne 0$, cooling instead gives ${\chi }_{\mathrm{model}}\sim 4{\mu }_{0}n{\mu }^2{e}^{-2\mu |B_0|/(k_{B}T)}/(k_{B}T)$: a saturated moment can hardly change further. The order of the zero-field and zero-temperature limits matters. This independent model has no finite-temperature collective phase transition; the diverging zero-field slope alone does not establish one.

Consider an explicitly stipulated dilute independent spin-one-half two-level system with zero applied magnetic field. Each constituent has $J_z=\pm \hslash /2$ and $m_z=\gamma J_z$, with constant signed gyromagnetic ratio $\gamma$. Ignore interactions and the specimen's own magnetic field. At positive temperature $T$, thermal equilibrium with the rotating lattice gives energy levels $E_{\pm }=\mp \hslash \Omega /2$. Define the dimensionless variable $x=\hslash \Omega /(2k_{B}T)$, where $\hslash$ is the reduced Planck constant and $k_B$ is the Boltzmann constant. The partition function and Boltzmann distribution giveTaking the expected value giveswhere $n$ is number density. This is the rotational version of the independent-moment magnetic response. The angular momentum aligns with positive rotation even when the magnetic moment points oppositely. The probability ratio $p_{+}/p_{-}=e^{2x}$ provides a direct population check of the sign.

For $|x|\ll 1$, expand the hyperbolic tangent to obtainThe slope against $B_{\Omega }$ is positive for either sign of $\gamma$, while the slope against $\Omega$ has the sign of $\gamma$. At fixed $T>0$, reversing $\Omega$ reverses both means, and $\Omega =0$ gives zero. At fixed nonzero $\Omega$, the formal $T\to 0^{+}$ limit saturates $|⟨J_z⟩|$ at $\hslash /2$; the linear approximation must not be extrapolated there. At $T\to \infty$, the response vanishes. These limits describe the toy canonical ensemble, whose equilibration is assumed.

For a numerical example, stipulate $n=10^{28}{\mathrm{m}}^{-3}$, $T=300\mathrm{K}$, $\gamma =-1.76\times 10^{11}{\mathrm{s}}^{-1}{\mathrm{T}}^{-1}$, and rotation frequency $f=1000\mathrm{Hz}$. With $\Omega =2\pi f$, $\hslash =1.054571817\times 10^{-34}\mathrm{J}\mathrm{s}$, and $k_B=1.380649\times 10^{-23}J/K$, the model givesThese are teaching inputs and calculated outputs, not measured material values. The corresponding internal angular momentum density is $M_z/\gamma =+4.22\times 10^{-17}\mathrm{J}s/{\mathrm{m}}^3$. Keeping a rotor at prescribed $\Omega$ while its internal angular momentum changes requires the drive to supply the appropriate angular impulse. In an isolated body the mechanical part instead changes oppositely, as in magnetomechanical angular momentum balance. There is no creation of net angular momentum by choosing a rotating frame.

Choose six illustrative values $y=(-2.2,-1.8,-1,1,1.8,2.2)$, fixed mixture weights ${\pi }_1={\pi }_2=1/2$, known standard deviation $\sigma =1$, and initial means $({\mu }_1^{(0)},{\mu }_2^{(0)})=(-1,1)$. These are constructed data for a Gaussian mixture distribution, not measurements. With $\varphi$ the standard normal distribution probability density function, the fitted log-likelihood isAt the initial means, the component density ratio is $\varphi (y_i-1)/\varphi (y_i+1)=e^{2y_i}$. The first E-step therefore gives

The paired values $y$ and $-y$ have complementary mixture responsibilities, so $N_1^{(0)}=N_2^{(0)}=3$ exactly. The first M-step givesThe hyperbolic tangent expression follows by pairing the contributions at $y$ and $-y$. Assigning every negative value wholly to the first mixture component and every positive value wholly to the second would instead give means $\pm 5/3$. Those hard assignments discard the overlap: the expectation-maximization algorithm is not the same update as K-means clustering.

The observed-data log-likelihood rises from $-11.420338905$ initially to $-10.368745063$ after one update and $-10.340655142$ after two. Further updates approach means $\pm 1.638023833$, with log-likelihood approximately $-10.340509617$. These values describe this initialization; EM likelihood monotonicity supplies no certificate of a global maximum.

For stable numerical evaluation, put $a_{ij}=\log (1/2)-\frac{1}{2}\log (2\pi )-\frac{1}{2}(y_i-{\mu }_j{)}^2$ and $b_i={\max }_j{a}_{ij}$. The log-sum-exp function givesSubtracting $b_i$ keeps at least one exponential equal to one, avoiding the all-zero sum that direct evaluation of tiny component densities can produce.

The inequalities $0\le \log \cosh t\le t^2/2$ make this energy finite on $H^1$. Since the second derivative of its scalar potential is ${\mathrm{sech}}^{2}t\in (0,1]$, its Fréchet derivative is $DI(v)[h]=\int (\nabla v\cdot \nabla h+\tanh (v)h)$, with Taylor remainder between zero and half the squared $H^1$ norm of $h$. The derivative operator is strictly monotone because the hyperbolic tangent is strictly increasing.

Because $S[x_0]=0$ at $x=0$, the factor $Z$ cancels from the ratio. Write $z=\beta \hslash \omega$ and $q=\sqrt{m\omega /\hslash }x$. The exact answer isAt high temperature, $z\ll 1$. Expanding the hyperbolic tangent givesThe displayed leading expression is the ratio of classical Boltzmann factors for potential energy $V(x)=m{\omega }^2{x}^2/2$ and $V(0)=0$. Indeed the classical canonical ensemble givesThe momentum Gaussian integral is independent of $x$ and cancels in the ratio. This is the classical phase-space limit of the thermal kernel. Here $\beta =1/(k_{B}T)$, with $k_B$ the Boltzmann constant; the associated classical position variance is $1/(\beta m{\omega }^2)$. At fixed $x$, $R$ ultimately tends to one as temperature tends to infinity, while its width grows.

At low temperature, $z\gg 1$. Now $\tanh (z/2)=1-2e^{-z}+O(e^{-2z})$, so at fixed $q$Unlike the classical high-temperature expression, this limiting width remains nonzero at zero temperature, reflecting the ground state of the quantum harmonic oscillator.

Let $a_j=\nu {\kappa }_j^{2}t+{\kappa }_{j}x$. Differentiating the logarithm givesFor definiteness order ${\kappa }_1<{\kappa }_2$. Then $e^{a_1}$ dominates as $x\to -\infty$, while $e^{a_2}$ dominates as $x\to +\infty$. The limiting states areWithout this ordering convention, $u(-\infty )=-2\nu \min ({\kappa }_1,{\kappa }_2)$ and $u(+\infty )=-2\nu \max ({\kappa }_1,{\kappa }_2)$. Thus the states $u_{1,2}$ correspond to $-2\nu {\kappa }_{1,2}$ after matching each label to its dominant exponential; the propagation speed does not depend on the naming of the two states.

Their exponent difference isConsequently, the Viscous Burgers travelling shock moves atSet $\Delta u=u_L-u_R=2\nu ({\kappa }_2-{\kappa }_1)>0$ and $\xi =x-Vt$. Factoring out the mean exponential gives the hyperbolic tangent profileA precise measure of the width must be stated. The natural length in the hyperbolic tangent argument isThus the layer has thickness of order $\nu /\Delta u$. A different common convention, jump divided by maximum gradient, gives $\delta =\Delta u/|u_x(0)|=8\nu /\Delta u=2\ell$; the far-tail exponential e-folding length is $\ell /2$. These differ only by their stated definitions, rather than by different solutions.

Two distinct limits require ${\kappa }_1\ne {\kappa }_2$. The source permits arbitrary real constants, but in the degenerate case ${\kappa }_1={\kappa }_2$ the solution is uniform, $u=-2\nu {\kappa }_1$, and there is no shock wave or finite shock width.

Put $w=\mathrm{artanh}z$. The hyperbolic tangent identity $z=(e^{2w}-1)/(e^{2w}+1)$ gives $e^{2w}=(1+z)/(1-z)$. Taking a compatible branch of the complex logarithm givesAs an identity for a single-valued inverse hyperbolic tangent, this requires consistent branch cuts; one cannot choose unrelated logarithm branches on the two sides. Near $z=0$ the branch with value zero is fixed by $\mathrm{artanh}0=0$. The full inverse relation also allows $w\mapsto w+i\pi k$, $k\in \mathbb{Z}$, and $z=\pm 1$ are singular points.

For $|y|>1$, the separated integral isInverting givesThe upper branch has $x+C>0$, where $y>1$ decreases towards $1$ as $x\to \infty$. The lower branch has $x+C<0$, where $y<-1$ decreases to $-\infty$ as $x\uparrow -C$. These are separate maximal intervals: the pole is not part of a solution. For the domain $x\ge 0$, a lower-branch initial value below $-1$ necessarily reaches that pole in finite positive $x$. The direction field sketch above displays both exterior branches and a central hyperbolic tangent branch.

For a standard collinear Lorentz boost, put $\beta =v/c$ and $\Gamma =(1-{\beta }^2{)}^{-1/2}$. Its equations are $x^{\prime }=\Gamma (x-\beta ct)$ and $c{t}^{\prime }=\Gamma (ct-\beta x)$. Define the rapidity $\varphi =\mathrm{arctanh}\beta$. Since ${\cosh }^2\varphi -{\sinh }^2\varphi =1$ and $\cosh \varphi >0$, $\cosh \varphi =\Gamma$ and $\sinh \varphi =\Gamma \beta$. ThereforeAdding and subtracting, and using $\cosh \varphi \mp \sinh \varphi =e^{\mp \varphi }$, givesA second boost multiplies these factors by $e^{\mp {\varphi }^{\prime }}$, so the combined rapidity is $\varphi +{\varphi }^{\prime }$. The hyperbolic tangent addition formula then gives the relativistic velocity addition lawVelocities are signed along the common axis, so this formula also covers oppositely directed boosts.

Put $s=\delta Z>0$. For the Fay solution, $1/\sinh (ns)=2e^{-ns}[1+O(e^{-2ns})]$. At $s\gg 1$ the first harmonic therefore giveswhich agrees with the late-time limit in (i).

For the opposite limit, expand the reciprocal hyperbolic sine as a geometric series and interchange absolutely convergent sums at every $s>0$:This exact positive-denominator form is useful near the shock. For small $s$ and $|\theta |$, expand the denominator of the terms with small $(2j+1)s$ and use $\sin \theta \sim \theta$. The leading sum isFor the shock-layer scaling $\theta =s\eta$, the passage to this sum follows from dominated convergence: $\cosh z-1\ge z^2/2$ and $1-\cos \theta \ge c{\theta }^2$ for small $|\theta |$ provide a summable bound proportional to $[(2j+1{)}^2+{\eta }^2{]}^{-1}$. The same leading expression matches the outer small-angle range $s\ll |\theta |\ll 1$. More generally, split the exact sum at $(2j+1)s=b$ with $\max (s,|\theta |)\ll b\ll 1$: the low terms have vanishing relative Taylor error, while the discarded exact and approximate tails are smaller than the leading expression by $O(\max (s,|\theta |)/b)$.

Now apply the partial-fraction identity for the hyperbolic tangent with $\eta =\theta /s$:The Fay shock-layer asymptotics are thereforeThe angular shock thickness is $O(\delta Z)$, and its matched states are $\pm \pi /Z$. At $\theta =0$, both formulas vanish; the limiting slopes agree. The denominator throughout this derivation is $\sinh (n\delta Z)$ as printed in the PDF, not the erroneous $\sin (n\delta Z)$ in the converted TeX.

An odd number of inhibitory connections gives $\Gamma <0$. Write $r=|\Gamma {|}^{1/3}>0$. The cube roots of $-r^3$ have arguments $\pi ,\pi /3,-\pi /3$, soThe real mode is always damped. The pair has negative real part precisely when $r\cos (\pi /3)<1$, givingfor strict linear asymptotic stability. For $r>2$ the origin is unstable to an oscillatory mode.

At $r=2=1/\cos (\pi /3)$ the eigenvalues are $-3/\tau$ and $\pm i\sqrt{3}/\tau$. Thus the linear oscillation frequency is $\boxed{{\omega }_H=\sqrt{3}/\tau }$ and its period is $2\pi \tau /\sqrt{3}$. Varying $r$ crosses the imaginary axis transversely, since $d\mathrm{Re}{\lambda }_{+}/dr=1/(2\tau )>0$.

The hyperbolic tangent has a negative cubic saturation and no quadratic term. To check the nonlinear onset, let $v$ be a critical complex eigenvector of $M=\beta W$ and $p$ its normalized adjoint eigenvector, with $p^{\ast }v=1$. For this weighted cycle, ${\overline{p}}_i{v}_i=1/3$. Since $Mv=(1+i\sqrt{3})v$, the cubic derivative evaluated on $(v,v,\overline{v})$ has projectionIts real part is strictly negative. With no quadratic contributions, this gives a negative first Lyapunov coefficient: the generic onset is a supercritical Hopf bifurcation, with a small attracting periodic orbit just beyond the threshold. Exactly at threshold the linear system has undamped oscillations, but nonlinear cubic damping makes sufficiently small oscillations decay; one must not infer a nonzero sustained cycle at the critical parameter solely from the imaginary eigenvalues.

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.