Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 55 (Solitons and Instantons)

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.

Write $\Omega =e^{2\rho }$, $F={\partial }_x{A}_2-{\partial }_y{A}_1$, and $s=|\Phi {|}^2$, so the physical magnetic field is $B=F/\Omega$. Use finite-energy Abelian Higgs vortex boundary data: the field tends to unit magnitude at the ideal boundary, its phase winds $N$ times, and the boundary limits in the decomposition exist. This vacuum asymptotic behavior is the setting in which the stated degree gives an energy bound. We retain the current boundary term; degree alone must not be silently equated with magnetic flux on an open disk.

Define $j_i=\mathrm{Im}(\bar{\Phi }{D}_i\Phi )$. The gauge-covariant derivative of a charged scalar field satisfies $[D_x,D_y]\Phi =-iF\Phi$. Differentiating the current givesOn the other hand,Combining the two identities and completing the square in the magnetic and potential terms yields the conformal-surface vortex square completionAll integrations on the noncompact metric disk can first be made over $|z|<R<1$ and then passed to the limit. By the Stokes theorem, the two boundary contributions combine to $\frac{1}{2}\oint (A+j)$. At the vacuum boundary write $\Phi =e^{i\chi }$. Then $j=d\chi -A$, so this combined integral is $\frac{1}{2}\oint d\chi =\pi N$, even if the current term does not vanish separately. The usual additional condition of tangential covariant decay makes that current term vanish and yields magnetic flux quantization of an Abelian Higgs vortex, but is not needed if the combined boundary term has this phase limit. Thus for $N\ge 0$, $V\ge \pi N$. Reversing both signs in the square completion gives $V\ge -\pi N$ for $N\le 0$, henceFor positive degree, equality is attained precisely when the Bogomolny vortex equations areDistinguishing the physical field $B$ from the coordinate curvature coefficient $F$ prevents an erroneous conformal factor.

For the proposed field, set $q=x^2+y^2$ and choose the smooth connection one-formIndeed, with ${\partial }_{\bar{z}}=\frac{1}{2}({\partial }_x+i{\partial }_y)$ and $A_{\bar{z}}=\frac{1}{2}(A_1+i{A}_2)$,The second Bogomolny vortex equation follows fromThus both squares vanish, including at the simple zero $z=0$ by smoothness. At $r\to 1$, $\Phi \to e^{i\vartheta }$ and $A=2r^{2}d\vartheta /(1+r^2)\to d\vartheta$, proving degree one. Moreover, $j_{\vartheta }=4r^2(1-r^2)/(1+r^2{)}^3\to 0$, so the current boundary term really vanishes for this solution. The magnetic flux is explicitlyConsequentlyThis is the hyperbolic one-vortex at curvature minus one half; its finite energy follows directly from the vanishing squares and the finite flux.

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.

Now suppose the positive-sign Bogomolny equations $B_i=D_i\Phi$ hold. The gauge-theory Bianchi identity $D_i{F}_{jk}+D_j{F}_{ki}+D_k{F}_{ij}=0$ implies ${\sum }_i{D}_i{B}_i=0$, and therefore ${\sum }_i{D}_i{D}_i\Phi =0$. For the gauge equation,Consequently $(D\times B{)}_i+[\Phi ,D_i\Phi ]=[B_i,\Phi ]+[\Phi ,B_i]=0$. Thus the first-order Bogomolny equations imply both second-order field equations. The negative-sign convention works as well, with both occurrences of that sign tracked consistently.

Finally differentiate the norm twice, using the trace compatibility identity at each step:The second term vanishes, and $|B{|}^2=|D\Phi {|}^2$ for the Bogomolny equations. Thus the Higgs norm identity for a Bogomolny monopole givesThere is no extra factor 2 multiplying $v$: the printed energy density already contains both equal terms without an overall $1/2$.