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For $f_{\lambda }(x)=f(\lambda x)$ in $d$ space dimensions, write $T=\frac{1}{2}\int |\nabla f{|}^2$, $W=\int U(f)$ and $I=\int f^2$. Eliminating the rotation frequency by $\omega =Q/I$ gives the displayed fixed-charge scaling. The positive-power charge term can balance the gradient and potential terms even for $d\ge 3$. This evades the static Derrick theorem because the field is time-dependent and the relevant variations preserve its Noether charge.

For the internal rotation symmetry of two real scalar fields with $\delta {\varphi }_1=\alpha {\varphi }_2$ and $\delta {\varphi }_2=-\alpha {\varphi }_1$, the Noether charge is $Q=i\int (a_1^{†}{a}_2-a_2^{†}{a}_1)d^{3}p/(2\pi {)}^3$. Its commutators satisfy $[Q,b_{\pm }^{†}]=\pm b_{\pm }^{†}$, so these creation operators produce charge eigenstates in the one-particle state sector. Substitution gives $Q=N_{+}-N_{-}$, converting the rotation of real components into opposite charges in a complex basis.

An integral of the time component of a conserved current is a conserved charge when the spatial boundary flux vanishes. Charges need not come from an immediately apparent elementary symmetry; a field theory can have an infinite higher-derivative current hierarchy. A Noether charge is obtained from the symmetry current supplied by the Noether theorem.

For constant internal matrices acting on Dirac fields, the equal-time canonical anticommutation relations givewith the corresponding spatial delta functions for fields. The quartic terms cancel after fermion reordering. Integrating proves the displayed algebra for well-defined global Noether charges, with normal ordering and a regulator that preserves the nonanomalous internal symmetry. Orthogonal chiral projectors make the left/right mixed commutators zero. This reduces current-charge closure to a finite matrix calculation.

For ghost term $-({\partial }^{\mu }\bar{c})\cdot D_{\mu }c$ and the current convention $\delta S=-\int ({\partial }_{\mu }\theta )j_G^{\mu }$, localize $\delta c=\theta c$, $\delta \bar{c}=-\theta \bar{c}$. The two differentiated parameters give this current. Its Noether charge grades ghosts by $+1$ and antighosts by $-1$, up to the corresponding overall generator normalization.

For $\delta \varphi =\epsilon X$ and a variational symmetry of a Lagrangian density, write ${\Pi }^{\mu }=\partial \mathcal{L}/\partial ({\partial }_{\mu }\varphi )$ and $\mathcal{E}={\mathcal{L}}_{\varphi }-{\partial }_{\mu }{\Pi }^{\mu }$. The product rule gives $\delta \mathcal{L}/\epsilon =\mathcal{E}X+{\partial }_{\mu }({\Pi }^{\mu }X)$. Therefore $j^{\mu }={\Pi }^{\mu }X-K^{\mu }$ obeys ${\partial }_{\mu }{j}^{\mu }=-\mathcal{E}X$, so it is a conserved current on shell. Spatial integration gives a conserved Noether charge only when the boundary flux vanishes. The formula extends to several components by summation.

For $x^{\prime \mu }=x^{\mu }+\epsilon {\xi }^{\mu }$ and ${\varphi }_a^{\prime }(x^{\prime })={\varphi }_a(x)+\epsilon {\Delta }_a$, the fixed-coordinate field variation is $Q_a={\Delta }_a-{\xi }^{\nu }{\partial }_{\nu }{\varphi }_a$. Suppose action invariance is expressed as ${\delta }_0\mathcal{L}/\epsilon +{\partial }_{\mu }(\mathcal{L}{\xi }^{\mu })={\partial }_{\mu }{K}^{\mu }$. Writing ${\Pi }_a^{\mu }=\partial \mathcal{L}/\partial ({\partial }_{\mu }{\varphi }_a)$ and $E_a={\mathcal{L}}_{{\varphi }_a}-{\partial }_{\mu }{\Pi }_a^{\mu }$, the product rule gives ${\partial }_{\mu }{j}^{\mu }=-{\sum }_a{E}_a{Q}_a$. Thus $j^{\mu }$ is a conserved current on shell. A vanishing spatial boundary flux is additionally required for its Noether charge to be conserved. This formula includes internal transformations, translations and Lorentz transformations.

A non-topological soliton is a localized finite-energy field configuration whose existence or stability is not protected by a nontrivial topological charge. A conserved ordinary Noether charge can instead obstruct its decay. A Q-ball is a standard complex-scalar example: its profile is localized while its phase rotates in time.

Assume $\lambda >0$ and $v>0$. The minima of the scalar potential have $|\varphi |=v$. Choose the vacuum expectation value ${\varphi }_0=v{e}_3$. Its stabilizer subgroup consists exactly of matrices $\mathrm{diag}(R_2,1)$ with $R_2\in O(2)$, soThe continuous spontaneous symmetry breaking is $SO(3)\to SO(2)$ and has two broken generators. For the full orthogonal group, the cross-product notation requires the gauge connection to transform in the adjoint, as an axial vector: $R[A{]}_{\times }{R}^{-1}=[(\det R)RA{]}_{\times }$. The scalar field transforms as the ordinary vector $R\varphi$. This axial transformation of orthogonal gauge connections makes the stated covariant derivative transform correctly even for $\det R=-1$; assigning an ordinary-vector transformation to the connection would fail.

Near this nonzero vacuum, unitary gauge removes the two angular Goldstone bosons and gives $\varphi =(0,0,v+h)$. PutThese are the physical charged gauge bosons, neutral massless gauge boson, and radial scalar field. Directly substituting into the original gauge field strengths givesThe scalar field kinetic term has no radial-vector cross term, because ${\partial }_{\mu }\varphi$ and ${\mathbf{A}}_{\mu }\times \varphi$ lie in perpendicular internal directions. The complete physical gauge-scalar Lagrangian becomesFor each real vector the mass term is $m^2{A}_{\mu }{A}^{\mu }/2$, while for the complex pair it is $m_W^2{W}_{\mu }^{+}{W}^{-\mu }$. Hence the adjoint triplet Higgs spectrum isThe radial scalar also has $m_h^2=\lambda v^2$ in the question's potential normalization. The two lost scalar modes supply the longitudinal polarizations of the massive vectors through the Higgs mechanism.

For the lepton kinetic terms let $P_L=(1-{\gamma }^5)/2$, $P_R=(1+{\gamma }^5)/2$, and denote the left neutral triplet entry by $n_L=c_{\alpha }{\nu }_L+s_{\alpha }{N}_L$. The missing singlet is the perpendicular combinationIts overall sign is immaterial. The real orthogonal matrix $\left(\begin{array}{cc}{c}_{\alpha }& s_{\alpha }\\ -s_{\alpha }& c_{\alpha }\end{array}\right)$ is orthogonal, so the sum of the kinetic terms of $n_L$ and $L^S$ is ${\bar{\nu }}_{L}i/\partial {\nu }_L+{\bar{N}}_{L}i/\partial N_L$ with no mixed terms. Combining with the right triplet gives ordinary Dirac field kinetic terms for $E^{+},N,e$, and only a left-handed kinetic term for the massless neutrino. This is orthogonal completion of mixed neutral lepton kinetic terms. The heavy charged field is $E^{+}$, as printed in the PDF, not the TeX's $N^{+}$.

To compute the charge commutator, the canonical anticommutation relations implyThe quartic terms cancel on reordering the fermions. Thus well-defined normal-ordered integrated bilinears obey fermionic bilinear charge algebra, $[Q_C,Q_D]=Q_{[C,D]}$. Left/right mixed commutators vanish since $P_L{P}_R=0$.

In either triplet, writeThe displayed current has a factor two relative to the projected bilinears, since $1\mp {\gamma }^5=2P_{L,R}$. Consequently its charges are $T^{+}=2(Q_K^L+Q_K^R)$ and $T^{-}=(T^{+}{)}^{†}$. The neutral combinations have canonical anticommutators because $c_{\alpha }^2+s_{\alpha }^2=1$. Applying the matrix commutator giveswhere $Q_{\ell }=e\int :(E^{+†}{E}^{+}-e^{†}e):$ is the electromagnetic charge in the lepton sector. The neutrino, heavy neutral lepton and singlet contribute zero. This establishes triplet charged-current closure on electromagnetic charge, with the requested proportionality independent of normalization choices for the weak Noether charges. The PDF correctly writes $J^{-}=(J^{+}{)}^{†}$; the TeX repeats $J^{+}$ incorrectly. Equal-time algebra does not require conservation of the separate fermionic weak currents after symmetry breaking. For full gauge-theory charges, the corresponding charged-boson contributions must also be included.

Use Minkowski spacetime with $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$ and units $\hslash =c=1$. The transformation rotates the two components in their internal plane. In particular,Thus both terms in the Lagrangian density are invariant: this is an internal rotation symmetry of two real scalar fields, rather than a transformation of spacetime. The equal masses are essential to this symmetry.

The Noether theorem gives the Noether currentIndeed, the cross terms in its divergence cancel, and the two Klein-Gordon equations implyFor fields decaying sufficiently at spatial infinity, there is no outward current flux. The conserved Noether charge is therefore

In canonical quantization, the canonical momenta are ${\pi }_k={\dot{\varphi }}_k$, with equal-time canonical commutation relationsConsequently $Q=\int d^{3}x({\pi }_1{\varphi }_2-{\pi }_2{\varphi }_1)$. Operators of different components commute, so this expression is Hermitian without an ordering correction. Write ${\int }_{\mathbf{p}}=\int d^{3}p/(2\pi {)}^3$, $E_{\mathbf{p}}=\sqrt{{\mathbf{p}}^2+m^2}$ and $a_{k\mathbf{p}}=a_{\mathbf{p}}^k$. Substituting the oscillator expansions and using the Dirac delta function from the spatial integral givesThe terms with two annihilation operators cancel after $\mathbf{p}\mapsto -\mathbf{p}$; the terms with two creation operators cancel in the same way. In the remaining terms useThere is no commutator constant between the distinct components. The result is already in normal ordering:

For the charged oscillator basis of a scalar doublet, computeIt follows that $[Q,b_{\pm \mathbf{p}}^{†}]=\pm b_{\pm \mathbf{p}}^{†}$. The Fock vacuum satisfies $Q|0⟩=0$, so $b_{+\mathbf{p}}^{†}|0⟩$ has charge $+1$, and $b_{-\mathbf{p}}^{†}|0⟩$ has charge $-1$. To obtain a normalized one-particle state, replace the sharp momentum by ${\int }_{\mathbf{p}}f(\mathbf{p})b_{\pm \mathbf{p}}^{†}|0⟩$, where ${\int }_{\mathbf{p}}|f(\mathbf{p}){|}^2=1$. The charge is unchanged. Equivalently,The generator convention is $\delta {\varphi }_k=i\alpha [Q,{\varphi }_k]$, which reproduces the original rotation signs.

For stability of a two-scalar quartic potential, the relevant potential energy density isAlong either axis boundedness requires $\lambda \ge 0$; along ${\varphi }_1={\varphi }_2$ it requires $\lambda +\mu \ge 0$. Conversely the displayed sum is nonnegative whenever both conditions hold. With $m^2>0$ the origin is then the unique global minimum, including the equality cases. Hence the stable-vacuum conditions areStrict positivity of the quartic term away from the origin would instead require $\lambda >0$ and $\mu >-\lambda$. That stronger condition is unnecessary for a positive mass term. If $m=0$, the same non-strict conditions give boundedness, but the equality cases have flat directions and do not give an isolated classical vacuum.

Finally, the interaction changes the variation of the potential energy byThe interacting Euler-Lagrange equations consequently giveThus the same Noether charge is conserved for all field configurations precisely when $\mu =\lambda$. Then $V_4=\lambda ({\varphi }_1^2+{\varphi }_2^2{)}^2$ preserves the internal rotation symmetry of two real scalar fields; imposing stability additionally requires $\lambda \ge 0$.

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.

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.

Use target signature $(-,+,\dots ,+)$ and worldsheet signature $(-,+)$. The Polyakov action has local worldsheet diffeomorphism and Weyl invariance, together with global target-space Poincaré symmetry. Under a worldsheet coordinate change $\sigma \mapsto {\sigma }^{\prime }$, the embedding coordinates are scalars and the worldsheet metric is a tensor:A Weyl transformation acts as $g_{ab}\mapsto e^{2\rho (\sigma )}{g}_{ab}$ and leaves $X$ unchanged. In two dimensions $\sqrt{-g}$ scales by $e^{2\rho }$ and $g^{ab}$ by $e^{-2\rho }$, so the action is invariant. Target translations act by $X^{\mu }\mapsto X^{\mu }+a^{\mu }$, and Lorentz transformations by $X^{\mu }\mapsto {\Lambda }^{\mu }{}_{\nu }{X}^{\nu }$ with ${\Lambda }^{\mathsf{T}}\eta \Lambda =\eta$; they leave $g$ unchanged. Infinitesimally the latter have $\delta X^{\mu }={\omega }^{\mu }{}_{\nu }{X}^{\nu }$, where ${\omega }_{\mu \nu }=-{\omega }_{\nu \mu }$.

Locally, the two diffeomorphism functions put the metric into the form $g_{ab}=e^{2\rho }{\eta }_{ab}$, and Weyl invariance removes the conformal factor. In this conformal gauge, the embedding coordinates are free scalar fields and satisfyThe metric equation must still be imposed after gauge fixing. Varying the ungauged metric gives the worldsheet stress-energy tensor constraintWith a dot and prime denoting the two coordinate derivatives, the independent Virasoro constraints areequivalently $(\dot{X}\pm X^{\prime }{)}^2=0$. Local conformal gauge leaves residual conformal transformations; on general worldsheet topologies it also leaves global moduli, rather than removing all metric information.

For target translations, differentiation of the Lagrangian with respect to ${\partial }_a{X}_{\mu }$ gives the Noether currentThe embedding equation is ${\partial }_a{j}^{a\mu }=0$. In conformal gauge, $j^{\tau \mu }={\dot{X}}^{\mu }/(2\pi {\alpha }^{\prime })$, so integration over the closed string givesPeriodic boundary conditions make its time derivative vanish. For an infinitesimal Lorentz transformation, the antisymmetric coefficient of ${\omega }_{\mu \nu }$ gives the current $j^{a,\mu \nu }=X^{\mu }{j}^{a\nu }-X^{\nu }{j}^{a\mu }$. Its divergence vanishes because the remaining product of embedding derivatives is symmetric in $\mu ,\nu$. Thus the Noether charge isThese are the target-space Noether charges of a string.

For the circular solution take $R>0$, and let ${\mathbf{e}}_r=(\cos \sigma ,\sin \sigma )$ and ${\mathbf{e}}_{\sigma }=(-\sin \sigma ,\cos \sigma )$. Its derivatives areEach spatial coordinate has equal second derivatives in $\tau$ and $\sigma$, while the time coordinate has both second derivatives zero. Furthermoreso both the field equations and the Virasoro constraints hold, including the collapse instants.

At target time $t=X^0=R\tau$, the string is a circle in the $X^1{X}^2$ plane of radius $R|\cos (t/R)|$. It contracts from radius $R$ to zero and re-expands; the negative cosine reverses material labels by a half-turn rather than creating a negative radius. The spatial configuration repeats after target time $\pi R$. This is a pulsating circular string. Its induced metric degenerates at collapse, but the conformal-gauge embedding and constraints give its continuation.

The integrals of $\sin \sigma$ and $\cos \sigma$ vanish, so the momentum isThe spatial velocity is radial, giving zero angular momentum density in the plane; all boost-charge terms integrate to zero as well. Hence $J^{\mu \nu }=0$ for every pair of indices.

At maximum expansion, all string elements are instantaneously at rest, and the length is $L_0=2\pi R$. Thus the rest mass per unit length, equal to the string tension, isAt other times the instantaneous circumference is $L=2\pi R|\cos \tau |$, and the elements have speed $|\sin \tau |$. Their Lorentz factor is $1/|\cos \tau |$, so the energy remains $TL/\sqrt{1-v^2}=2\pi TR$. The rest-energy density is $T$; the total rest-frame energy divided by the changing circumference is not constant. This distinguishes the requested intrinsic mass density from the larger energy density of moving string elements.

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