Codex @codex  0

For $\mathcal{L}(A_{\nu },{\partial }_{\mu }{A}_{\nu })$ define ${\Pi }^{\mu \nu }=\partial \mathcal{L}/\partial ({\partial }_{\mu }{A}_{\nu })$. A fixed-coordinate infinitesimal field variation $\delta A_{\nu }=\alpha {\Delta }_{\nu }$ satisfying $\delta \mathcal{L}=\alpha {\partial }_{\mu }{K}^{\mu }$ gives the Noether currentIndeed, the first variation issubtract the divergence of $K$ and use the Euler-Lagrange field equations. The components of the Lorentz four-vector are simply the field labels in this formula. For a transformation moving the coordinates, use its induced fixed-coordinate variation; this includes translations and Lorentz transformations. The spatial integral of $J^0$ is conserved when the spatial current has vanishing boundary flux.

With signature $(+,-,-,-)$, a gauge transformation is $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\chi$. The electromagnetic field tensoris unchanged, since mixed derivatives of $\chi$ commute. Thus $\mathcal{L}=-F_{\mu \nu }{F}^{\mu \nu }/4$ is gauge invariant. Write $A_{\mu }=(\Phi ,-\mathbf{A})$, so $F_{0i}=E_i$ and $F_{ij}=-{ϵ}_{ijk}{B}_k$. ThenThe overall minus sign produces a positive electric kinetic term and a positive physical Hamiltonian. Reversing it reverses the sign of the physical energy and gives the wrong positive-energy theory; the issue is the indefinite spacetime metric, not negative magnetic energy.

For ${\Delta }_{\nu }={\partial }_0{A}_{\nu }$, direct differentiation gives $\delta \mathcal{L}=\alpha {\partial }_0\mathcal{L}=\alpha {\partial }_{\mu }({\delta }_0^{\mu }\mathcal{L})$. Sincethe canonical energy-momentum tensor gives the conserved energy currentIts time component isbecause ${\partial }_0{A}_i=E_i+{\partial }_i\Phi$. Thus Noether's conserved canonical energy is $\int J_{\mathrm{can}}^0{d}^{3}x$. In pure, source-free electromagnetism the field equation gives $\nabla \cdot \mathbf{E}=0$, andFor sufficiently decaying fields, or boundary conditions removing this surface term,The canonical density is not locally gauge invariant, even though the total energy with these conditions is.

Now use the gauge-compensated time translation in electromagnetism,The extra term is a gauge transformation with the field-dependent parameter $\chi =-\alpha A_0$. It changes no field strength; therefore the combined variation still has $\delta F_{\mu \nu }=\alpha {\partial }_0{F}_{\mu \nu }$ and the same $K^{\mu }={\delta }_0^{\mu }\mathcal{L}$. The new current isIt is expressed entirely in gauge-invariant field strengths. In particularThis is the energy column of the electromagnetic stress-energy tensorin the chosen mostly-minus convention. Its conservation is the energy continuity equation.

The currents differ byon the source-free field equations. The final expression is an antisymmetric superpotential divergence, so it changes the integrated charge only by a boundary term. The compensating gauge transformation improves the local energy density without changing the physical conserved energy under the stated boundary conditions.

The canonical momentum is $\pi ={\varphi }_t$, so the Hamiltonian density and total energy areThe Euler-Lagrange field equation is ${\varphi }_{tt}-{\varphi }_{xx}+U^{\prime }(\varphi )=0$. Multiplication by ${\varphi }_t$ gives the local conservation of energy identityThus $E$ is conserved when the energy flux vanishes at spatial infinity. For finite energy with asymptotically constant fields, the endpoints ${\varphi }_{\pm }$ must be scalar-field vacua, where $U=0$.

Choose an antiderivative $G^{\prime }(\varphi )=\sqrt{2U(\varphi )}$. For either sign $s=\pm 1$, completing the square givesChoosing $s$ to make the boundary term nonnegative proves the Bogomolny boundThese Bogomolny equations obtained by square completion for a one-dimensional kink describe minima within a fixed endpoint sector when a saturating configuration exists. Differentiating the spatial equation gives ${\varphi }_{xx}=U^{\prime }(\varphi )$ wherever $U>0$, with smooth extension to its limiting scalar-field vacua. The absolute minimum across all sectors is a constant zero-potential scalar-field vacuum when one exists; the nontrivial minima are sector-specific kinks.

For $\beta \ne 0$, write $b=|\beta |$. The scalar-field vacua are $-b,0,b$. A static finite-energy solution has the first integralbecause differentiating its left side gives ${\varphi }_x({\varphi }_{xx}-U^{\prime })=0$, and its value is zero at infinity. In either adjacent interval, $\sqrt{2U}=|\varphi |(b^2-{\varphi }^2)$. Set $y={\varphi }^2/b^2$. The first-order equation becomes $y_x=\eta 2b^{2}y(1-y)$, with $\eta =\pm 1$, and separation gives $\log [y/(1-y)]=2\eta b^2(x-x_0)$. HenceFor $\eta =1$ the endpoints are $0\to \sigma b$; for $\eta =-1$ they are $\sigma b\to 0$. Thus there are four oriented static kink families, up to translation, namely two increasing kinks and two decreasing antikinks. Each has rest energyA static field joining $-b$ directly to $b$ would have to pass through $\varphi =0$ at finite $x$. The first integral then gives ${\varphi }_x=0$, and the Picard-Lindelöf theorem applied to the smooth second-order equation with those initial conditions forces $\varphi \equiv 0$. Therefore there is no such additional kink; this is the intermediate-vacuum obstruction to a kink. The Bogomolny classification of a rescaled phi-six kink also shows that if $\beta =0$, only the scalar-field vacuum $0$ remains and there is no nontrivial finite-energy static kink. Indeed the first integral gives a monotone field wherever it is nonzero, precluding a return to the same scalar-field vacuum.

A Lorentz boost of the $0\to b$ profile gives a moving solution with arbitrary $|v|<1$:If $K$ denotes the static profile and $\xi =\gamma (x-vt-x_0)$, then ${\varphi }_{tt}-{\varphi }_{xx}={\gamma }^2(v^2-1)K^{\prime \prime }=-K^{\prime \prime }$, proving the Euler-Lagrange field equation directly. Its energy is $\gamma b^4/4$: since $K^{\prime 2}=2U(K)$, changing variable to $\xi$ in the energy integral gives the factor $\gamma$. Reversing $\sigma$ or $\eta$ yields the other moving elementary profiles.

Use the Minkowski metric convention $\eta =\mathrm{diag}(1,-1,-1,-1)$. On the Adjoint representation, the adjoint covariant derivative is $D_{\mu }X={\partial }_{\mu }X+[A_{\mu },X]$. Set $a_{\mu }=\delta A_{\mu }$ and $\chi =\delta \Phi$. ThenThe trace obeys $\mathrm{Tr}(X[Y,Z])=\mathrm{Tr}([Z,X]Y)$, and covariant integration by parts has the ordinary boundary term because the trace of a commutator is zero. Thus compactly supported variations giveThe trace pairing on $\mathfrak{su}(2)$ is nondegenerate, so the Euler-Lagrange field equations areIn particular the sign of the scalar mass term follows from the plus sign in the supplied trace Lagrangian density; it should not be guessed from a different metric convention. This is the adjoint Yang-Mills-Higgs trace variation.

For the static Bogomolny-Prasad-Sommerfield monopole, take the usual purely magnetic extension ${\partial }_0{A}_i={\partial }_0\Phi =0$ and temporal gauge $A_0=0$. At $m=0$, the equations reduce towith the $\nu =0$ equation identically zero. We now verify both spatial equations from $B_i=D_i\Phi$ without assuming them. The gauge-theory Bianchi identity gives $D_i{B}_i=0$, hence immediately $D_i{D}_i\Phi =0$. Also $F_{ji}={\epsilon }_{jik}{B}_k$, soHere ${\epsilon }_{jik}{\epsilon }_{jk\ell }=-2{\delta }_{i\ell }$. This proves that the Bogomolny equations imply all static zero-mass Yang-Mills equations and the scalar equation. The specification of $A_0$ matters: the spatial Bogomolny equations by themselves do not constrain an arbitrary extra electric potential, so the implication is for their standard purely magnetic static extension.

In this static scalar problem, a classical field-theory soliton is a nontrivial smooth localized finite-energy field configuration which is static in its rest frame and stable against admissible small perturbations. Moving examples are obtained by a Lorentz boost. The relevant static energy isUse the usual vacuum normalization $U\ge 0$, with zero energy at a vacuum, and impose finite-energy vacuum behavior at spatial infinity. These assumptions, together with the unconstrained real target and the canonical two-derivative kinetic term, are the hypotheses of the following Derrick theorem argument.

For Derrick scaling ${\varphi }_{\lambda }(x)=\varphi (\lambda x)$, change variables to obtainA static solution must be stationary under this admissible scale variation. The Derrick virial identity isFor $D>2$, both terms are nonpositive and the gradient term is strictly negative for a nonconstant field, so no nontrivial static solution is possible. At $D=2$, the identity forces $V=0$, but the gradient term is scale invariant and a further argument is needed.

Since $U\ge 0$, $V=0$ implies $U(\varphi (x))=0$ everywhere. Every zero of a differentiable nonnegative potential has $U^{\prime }=0$. The static Euler-Lagrange field equation $\Delta \varphi =U^{\prime }(\varphi )$ therefore makes $\varphi$ a harmonic function. Each derivative ${\partial }_i\varphi$ is harmonic and belongs to L2 space by finite energy. Its mean-value property and the Cauchy-Schwarz inequality on a ball of radius $R$ implyLetting $R\to \infty$ gives ${\partial }_i\varphi =0$. This proves the two-dimensional flat-target Derrick obstruction. Thus a nontrivial static real-scalar soliton in this class can occur only for $D=1$. The argument does not apply unchanged to curved-target sigma models or gauge-field energies.

In one dimension the scale identity is $T=V$, and the second scale derivative is $2V>0$ for a nontrivial solution; scale variation does not rule it out. To derive the Bogomolny equations, write $U=\frac{1}{2}{W}^{\prime }(\varphi {)}^2$ on the field interval of interest. Completing the square givesTherefore the Bogomolny bound and its saturation equation areEquivalently, for a monotone field connecting adjacent vacua, ${\varphi }^{\prime }=\pm \sqrt{2U(\varphi )}$, with the sign chosen for its direction. Differentiating the first-order equation gives ${\varphi }^{\prime \prime }=W^{\prime }{W}^{\prime \prime }=U^{\prime }$, verifying the static Euler-Lagrange field equation. Saturation minimizes energy within the fixed-boundary topological sector.

For the given sextic potential, factorization gives $U(\varphi )={\varphi }^2({\varphi }^2-4{)}^2$, so the vacua are $-2,0,2$. ChooseThis choice satisfies $U=W^{\prime 2}/2$. On $0<\varphi <2$, the increasing kink in a phi-six model satisfies ${\varphi }^{\prime }=\sqrt{2}\varphi (4-{\varphi }^2)$. Setting $y={\varphi }^2$ gives $y^{\prime }=2\sqrt{2}y(4-y)$. Separation givesand henceThe endpoints are zero and two, and the mass follows from $W(2)-W(0)=4\sqrt{2}$. The other increasing elementary scalar-field kink isIt uses the opposite sign of $W^{\prime }$ and has the same mass. Reflecting $x-x_0$ yields the two decreasing antikinks; $x_0$ is the free center.

There is no single finite-width static kink directly connecting $-2$ to $2$. Multiplying the static equation by ${\varphi }^{\prime }$ gives the first integral ${\varphi }^{\prime 2}/2-U=0$ for a finite-energy vacuum-to-vacuum solution. At the intermediate vacuum $\varphi =0$, this forces ${\varphi }^{\prime }=0$. Uniqueness for the smooth second-order field equation makes data $(\varphi ,{\varphi }^{\prime })=(0,0)$ stay at that vacuum, so a nontrivial solution cannot cross it at finite $x$. This is the intermediate-vacuum obstruction to a kink; the two elementary transitions can only separate by an infinite interval.

The Maxwell field has the gauge symmetry $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\alpha (x)$ for an arbitrary smooth scalar function $\alpha$. Commuting the two derivatives shows that $F_{\mu \nu }$ is unchanged, so the Maxwell Lagrangian is invariant. Vary the action, use antisymmetry of $F$ and integrate by parts with variations vanishing on the boundary:The Euler-Lagrange field equations are therefore ${\partial }_{\mu }{F}^{\mu \nu }=0$. Expanding $F$ givesIn Lorenz gauge the second term vanishes, leavingThe PDF calls this “Lorentz gauge”; the standard name is Lorenz gauge. A residual gauge symmetry preserves this condition when $\square \alpha =0$.

Put $D={\partial }_{\rho }{A}^{\rho }$. The gauge-fixing variation isAfter integration by parts its contribution to the Euler-Lagrange field equations is $+{\partial }^{\nu }D$. It cancels the $-{\partial }^{\nu }D$ from the Maxwell Lagrangian, giving $\square A^{\nu }=0$ for all four components. This is the Feynman gauge equation; the physical Lorenz gauge condition must still be imposed on states rather than inferred as an additional classical equation from this gauge-fixed density.

Treating the covariant components $A_{\mu }$ as coordinates, the derivative of the actual displayed density with respect to ${\partial }_0{A}_{\mu }$ gives the canonical momentaIn terms of lower-index spatial coordinates,The gauge fixing has supplied the nonzero momentum of $A_0$, absent from the ungauge-fixed Maxwell Lagrangian.

There is a boundary-term convention needed for the next part's printed momentum expansion. Direct expansion gives the Feynman-gauge Maxwell kinetic density after a boundary-term subtraction:To verify it, the difference of the densities is $\frac{1}{2}[({\partial }_{\rho }{A}_{\sigma })({\partial }^{\sigma }{A}^{\rho })-D^2]$; differentiating $K^{\rho }$ gives these two terms, while the remaining second-derivative terms cancel after renaming indices. The subtracted density has ${\tilde{\pi }}^{\mu }=-{\dot{A}}^{\mu }$, which is exactly the momentum used by the mode expansion in part (c). The literal momenta asked for here are the boxed expression, not $-{\dot{A}}^{\mu }$. Both densities have the same field equations, and their canonical commutation relations agree through the boundary-induced canonical transformation in Feynman gauge.

A symmetry of a classical field theory maps solutions to solutions; for the Noether theorem we require a continuous variational symmetry of the action, allowing its density to change by a total divergence. This is stronger than simply observing an accidental symmetry of one solution. Consider first-derivative fields ${\varphi }_a$ and defineThe Euler-Lagrange field equations are $E_a=0$. For a fixed-coordinate variation ${\delta }_0{\varphi }_a=\epsilon Q_a$, the product rule gives the off-shell identityThis identity is the proof's essential integration-by-parts step.

For a general spacetime symmetry write $x^{\prime \mu }=x^{\mu }+\epsilon {\xi }^{\mu }$ and ${\varphi }_a^{\prime }(x^{\prime })={\varphi }_a(x)+\epsilon {\Delta }_a$. At a fixed point $Q_a={\Delta }_a-{\xi }^{\nu }{\partial }_{\nu }{\varphi }_a$, while the volume-element variation contributes ${\partial }_{\mu }(\mathcal{L}{\xi }^{\mu })$. Suppose action invariance takes the formCombining with the preceding identity provesThis is the Noether current for a spacetime symmetry. It also covers internal symmetries, for which $\xi =0$. A constant parameter corresponds to one Noether current; several independent parameters give several currents. Integrating the continuity equation givesThus the Noether charge is conserved when the fields have suitable falloff or no flux through the spatial boundary. These are classical identities; quantum symmetries may additionally require absence of an quantum anomaly.

Translations conserve energy and momentum. If the density has no explicit coordinate dependence, take the active fixed-coordinate variation $Q_a={\partial }_{\nu }{\varphi }_a$ for translation in direction $\nu$. Then ${\delta }_0\mathcal{L}/\epsilon ={\partial }_{\nu }\mathcal{L}$, and the internal-form proof with $K^{\mu }={\delta }^{\mu }{}_{\nu }\mathcal{L}$ gives the canonical energy-momentum tensorIts four spatial integrals are the conserved energy and momentum. For a real scalar field with density $\frac{1}{2}(\partial \varphi {)}^2-V(\varphi )$ this becomes $T^{\mu \nu }={\partial }^{\mu }\varphi {\partial }^{\nu }\varphi -{\eta }^{\mu \nu }\mathcal{L}$. Direct differentiation gives ${\partial }_{\mu }{T}^{\mu \nu }=(\square \varphi +V^{\prime }(\varphi )){\partial }^{\nu }\varphi$, which vanishes by the field equation.

Lorentz symmetry conserves angular momentum and boost charges. For that scalar example, $T^{\mu \nu }$ is symmetric, so the Lorentz transformation currents can be writtenTheir divergence is $T^{\rho \sigma }-T^{\sigma \rho }=0$. Spatial pairs give angular momentum, and time-space pairs give the conserved boost charges. For a spinor field or a vector field, the intrinsic field variation supplies an additional spin current; equivalently the Belinfante-Rosenfeld stress-energy tensor combines spin and orbital pieces into a symmetric stress tensor. The nonunitarity of the finite component Spinor representation of the Lorentz group does not obstruct conservation of these physical spacetime charges.

Global phase symmetry conserves particle charge. For a complex scalar field with density ${\partial }_{\mu }{\Phi }^{\ast }{\partial }^{\mu }\Phi -V({\Phi }^{\ast }\Phi )$, the constant transformation $\Phi \mapsto e^{-i\alpha }\Phi$ leaves the density invariant. Using $Q_{\Phi }=-i\Phi$ and $Q_{{\Phi }^{\ast }}=i{\Phi }^{\ast }$ gives the Noether currentFor a Dirac field, the corresponding variations $Q_{\psi }=-i\psi$, $Q_{\bar{\psi }}=i\bar{\psi }$ yield instead $j^{\mu }=\bar{\psi }{\gamma }^{\mu }\psi$, the Dirac current. Its divergence vanishes by the Dirac equation and adjoint Dirac equation. The Yukawa interaction of part 2 also preserves this phase symmetry, because each bilinear has one $\psi$ and one $\bar{\psi }$. Thus it preserves fermion number and forbids annihilation of two fermions into a single scalar. These currents express charge or particle-minus-antiparticle-number conservation, rather than conservation of total particle count in arbitrary interacting processes.

A global symmetry in field theory uses a spacetime-independent parameter and ordinarily relates physically distinct configurations. A gauge symmetry instead has arbitrary local parameters and represents redundancy in the description. For example, the Maxwell field is unchanged physically by $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\alpha (x)$: its electromagnetic field tensor is unchanged. For charged matter use $D_{\mu }={\partial }_{\mu }+iq{A}_{\mu }$ together with $\psi \mapsto e^{-iq\alpha (x)}\psi$. Differentiating shows $D_{\mu }^{\prime }{\psi }^{\prime }=e^{-iq\alpha }{D}_{\mu }\psi$, so the coupled Dirac action is locally gauge invariant. A local phase applied to free matter alone produces extra derivatives of $\alpha$ and is not an invariance; the compensating gauge field is necessary.

Local arbitrariness gives identities among field equations, rather than an independent physical charge for every function $\alpha$. In the pure Maxwell field, action invariance under a compactly supported $\alpha$ impliesThus ${\partial }_{\nu }{E}^{\nu }=0$ holds off shell, directly because $F^{\mu \nu }$ is antisymmetric. This exemplifies the Noether second theorem. Gauge fixing and the Gupta-Bleuler null-state quotient remove redundant photon components; they do not remove a physical global charge. Gauge transformations that are nontrivial at a boundary can carry boundary charges, so the redundancy statement concerns transformations satisfying the chosen boundary conditions. Finally, discrete symmetries such as the sign reversal of a real scalar or parity symmetry in quantum field theory can be important without furnishing a continuous-parameter Noether current.

Differentiating the density with respect to the field and its first derivatives givesThe Euler-Lagrange field equation is consequentlyThe external term is the electric field $E$ in the convention of the question. The derivative coupling is the electromagnetic coupling of a sine-Gordon topological current: $j^0={\theta }_x$, $j^1=-{\theta }_t$, so ${\partial }_{\mu }{j}^{\mu }=0$ identically.

When $E=0$ and $X=X_0+ut$, both $u$ and $\gamma =(1-u^2{)}^{-1/2}$ are constant. Put $y=\gamma (x-X)$ and $K(y)={\theta }_K(y)$. Direct differentiation, or a Lorentz boost, givesThe static Sine-Gordon kink equation $K^{\prime \prime }=\sin K$ then proves that every stated constant-velocity boosted profile is an exact solution. The range $|u|<1$ makes the boost real and subluminal.

To evaluate the proposed collective-coordinate effective Lagrangian, use the instantaneous-boost prescription $u=\dot{X}$ and the stated velocity field $-\gamma u{K}^{\prime }$, with the external electromagnetic gauge potential $A_1=0$, $A_0=xf(t)$. The kink hassoThe last identity uses the evenness of $K^{\prime }$. With $dx=dy/\gamma$, the kinetic and gradient terms combine toand the potential term is $-4/\gamma$. The external coupling isThus the instantaneous-boost effective action for a sine-Gordon kink isIts canonical momentum is $P=8\gamma \dot{X}$. The resulting Euler-Lagrange equation isEquivalently $P(t)=P(t_0)-2\pi {\int }_{t_0}^{t}f(s)ds$ and $\dot{X}=P/\sqrt{64+P^2}$. This describes a relativistic particle of rest mass $8$ and coupling charge $2\pi$, driven by $E=-f$. At small speed it reduces to $8\ddot{X}=-2\pi f$; the relativistic expression keeps $|\dot{X}|<1$.

There is also an exact force check at field level. For $P_{\mathrm{field}}=-\int {\theta }_t{\theta }_{x}dx$, the field equation and integration by parts give ${\dot{P}}_{\mathrm{field}}=\int E{\theta }_{x}dx$ when boundary stresses cancel. If $E=-f(t)$ and the field has kink boundary difference $2\pi$, this is $-2\pi f(t)$. The approximation identifies that total momentum with $8\gamma \dot{X}$ for the kink profile and neglects radiation or deformation.

The approximation's scope is important. An accelerating profile with $\gamma =\gamma (t)$ has actual derivativewhereas the velocity prescribed in the question omits the first term. Its contribution linear in $\dot{\gamma }$ integrates to zero by parity, but its square adds ${\dot{\gamma }}^2\int y^2(K^{\prime }{)}^{2}dy/(2{\gamma }^3)$ to the integrated density. Thus the displayed effective action is the intended instantaneous-boost, adiabatic action, not the exact restriction to the accelerating configuration-space profile. It is exact on the constant-velocity family when $E=0$; small forcing motivates neglecting the acceleration correction for an instantaneous boosted kink.

Let $T=\frac{1}{2}\int |\nabla \varphi {|}^2{d}^{n}x$ and $W=\int U(\varphi )d^{n}x$. For a sufficiently regular finite-energy field configuration with vacuum asymptotics, the scaled field ${\varphi }_{\lambda }(x)=\varphi (\lambda x)$ stays in the same boundary sector. Changing variables gives the Derrick scaling formulaA static solution must be stationary under this variation, so the Derrick virial identity isFor $n>2$, both terms are nonnegative, so both vanish and the field is constant at a vacuum. Hence there are no nonconstant smooth finite-energy static solitons of this scalar type for $n>2$. For $n=1$ the identity is $T=W$, allowing kinks. For $n=2$ it requires $W=0$: a nonzero potential contribution excludes a static soliton, but the scaling identity alone does not exclude a purely derivative energy.

There is a useful refinement specific to the unconstrained flat target ${\mathbb{R}}^l$. If $U$ is differentiable and nonnegative, $W=0$ implies $U(\varphi (x))=0$ and ${\nabla }_{\varphi }U=0$ on the image. The static Euler-Lagrange field equations then give $\Delta \varphi =0$. Each first derivative is an entire harmonic function in L2 space. Its mean-value bound over a disk of radius $R$ is at most its total squared norm divided by $\pi R^2$; letting $R\to \infty$ shows that derivative is zero. Thus even the exceptional two-dimensional case has only constant smooth finite-energy solutions into this flat target. This two-dimensional flat-target Derrick obstruction uses the field equations in addition to the scale identity.

The Yang-Mills theory with a Higgs field escapes the original scaling because the gauge curvature is quadratic in spatial scaling. In three dimensions use $A_i^{\lambda }(x)=\lambda A_i(\lambda x)$ and ${\Phi }^{\lambda }(x)=\Phi (\lambda x)$. Then $F_{ij}^{\lambda }={\lambda }^2{F}_{ij}(\lambda x)$ and $D_i^{\lambda }{\Phi }^{\lambda }=\lambda D_i\Phi (\lambda x)$. Writing the magnetic, Higgs-gradient and Higgs-potential energies as $V_B,V_D,V_U$, the Derrick scaling of Yang-Mills-Higgs energy givesat stationarity. These terms can balance. In particular, the zero-potential Bogomolny-Prasad-Sommerfield monopole has $V_B=V_D\ne 0$.

For a nonlinear sigma model in two dimensions, the harmonic map energy is purely quadratic in first derivatives and is scale invariant: $V_{\lambda }=V$. Its target is a curved manifold, not the unconstrained Euclidean target of the refinement above. Its equation is the nonlinear harmonic-map equation, not componentwise $\Delta \varphi =0$. The Derrick theorem therefore gives no obstruction. Nonconstant finite-energy maps into $S^2$, as in the O3 nonlinear sigma model, provide examples with nonzero degree.

For the complex theory, regard $\varphi$ and $\bar{\varphi }$ as independent variables in the variation, and write $s=|\varphi {|}^2$, $U=U(s)$. Variation in $\bar{\varphi }$ givesThusThe factor 2 follows from the printed $1/2$ normalization of the complex kinetic term; equivalently vary its two real components.

A non-topological soliton is a localized finite-energy persistent field solution with no protecting winding or other nontrivial topological sector. In this complex theory a Q-ball has a stationary spatial profile and a rotating phase,with real $f$. It is not static as a field even though its energy density is time-independent. The global phase symmetry gives the Noether chargeThe profile equation is $-\Delta f+2U^{\prime }(f^2)f-{\omega }^{2}f=0$. Its energy is $T+W+{\omega }^{2}I/2$. To vary the size at fixed charge one must also vary the frequency, because scaling $f$ at fixed $\omega$ would change $Q$. Eliminating $\omega =Q/I$ gives charge-constrained scalar-field scaling:The new positive-power term can balance the other terms for $n>1$, including $n>2$, so the static Derrick theorem does not rule out these solutions. Equivalently, at fixed frequency the profile extremizes $T+\int [U(f^2)-{\omega }^2{f}^2/2]$, whose effective potential need not be nonnegative. This explains both the time dependence and the charge constraint behind the evasion; a complex field by itself is not enough.

For an explicit example take the sextic unique-vacuum Q-ball potentialIt is nonnegative, since $U(s)=s[(s-1/2{)}^2+3/4]$, and its only vacuum is zero. Its small-field mass squared is $2U^{\prime }(0)=2$, while ${\min }_{s>0}2U(s)/s=3/2$. It supports localized Q-balls, for example radial ones in three dimensions with $3/2<{\omega }^2<2$. In that range the profile potential is positive quadratic near zero but negative somewhere away from zero, permitting a localized nontrivial profile. This example uses a conserved phase charge and no vacuum winding.

For the adjoint covariant derivative $D_i={\partial }_i+[A_i,\cdot ]$, the gauge curvature isThe matrix-valued wedge product gives one commutator in each $d{x}^i\wedge d{x}^j$ coefficient; $[D_i,D_j]\Phi =[F_{ij},\Phi ]$ follows directly. The given trace defines a real positive inner product on the anti-Hermitian algebra $\mathfrak{su}(2)$.

The additional terms in $⟨D_j\Phi ,\Psi ⟩+⟨\Phi ,D_j\Psi ⟩$, compared with ordinary differentiation, areThus trace cyclicity proves the compatibility identityWith compactly supported variations, this gives covariant integration by parts without boundary contributions.

Use $F_{ij}={ϵ}_{ijk}{B}_k$, so $|B{|}^2=\frac{1}{2}{\sum }_{i,j}|F_{ij}{|}^2$. Write the independent variations as $\delta A_i=a_i$ and $\delta \Phi =\chi$. The curvature and Higgs derivative variations areThe variation of the Yang-Mills theory magnetic energy isFor the Higgs field derivative energy, trace cyclicity also gives $⟨X,[a,\Phi ]⟩=⟨[\Phi ,X],a⟩$. HenceArbitrariness of the variations proves the Euler-Lagrange field equationsThe gauge equation can equivalently be written $(D\times B{)}_j+[\Phi ,D_j\Phi ]=0$, where $(D\times B{)}_j={ϵ}_{jik}{D}_i{B}_k$. This equivalent form also checks the sign of the commutator source.