Near-BPS baby Skyrmions with Gaussian tails
IIFUP-TH-2021
Near-BPS baby Skyrmions with Gaussian tails
Sven Bjarke Gudnason , Marco Barsanti and Stefano Bolognesi Institute of Contemporary Mathematics, School of Mathematics and Statistics, Henan University,Kaifeng, Henan 475004, P. R. China Department of Physics “E. Fermi”, University of Pisa and INFN, Sezione di Pisa, Largo Pon-tecorvo, 3, Ed. C, 56127 Pisa, Italy
E-mail: gudnason(at)henu.edu.cn , marco.barsanti(at)phd.unipi.it , stefanobolo(at)gmail.com Abstract:
We consider the baby Skyrme model in a physically motivated limit of reach-ing the restricted or BPS baby Skyrme model, which is a model that enjoys area-preservingdiffeomorphism invariance. The perturbation consists of the kinetic Dirichlet term with asmall coefficient (cid:15) as well as the standard pion mass term, with coefficient (cid:15)m . The pionsremain lighter than the soliton for any (cid:15) and therefore the model is physically acceptable,even in the (cid:15) → (cid:15) → (cid:15) = 0 case, finding new nontrivial BPS solutions, for which we donot yet know the analytic form. a r X i v : . [ h e p - t h ] F e b ontents (cid:15) LO corrections 113.3 Axially symmetric solutions 20 ˜ m V + V The Skyrme model [1, 2] is an attractive model for a field-theoretic approach to nucleiand it is the low-energy effective theory of the Witten-Sakai-Sugimoto model [4]. Its sim-plest version overestimates the nuclear binding energies at the classical level by about anorder of magnitude compared with those observed in experiments. BPS solitons, on theother hand, saturate a Bogomol’nyi bound which implies that such solitons have vanishingbinding energies. A BPS Skyrme model was proposed in ref. [5] by Adam–Sanchez-Guillen–Wereszczynski (see also refs. [6–9]) which is a BPS theory that has BPS solutions for anyvalue of the topological charge or baryon number, B . Depending on the potential used inthis BPS theory, the solutions are often of a compacton type, that is, the solutions havesupport on a compact region of space and has a discontinuous derivative at the border of– 1 –he compacton region. This property can in part be traced back to the fact that the BPSSkyrme model does not have a kinetic term, but its only derivative term is of sixth orderand is the topological charge-density squared.Real-world nuclei have binding energies at the one-percent level of their rest mass, andit is expected that a small perturbation of a BPS model will yield multi-Skyrmion solutionswith low binding energies – at least at the classical level. Deforming the BPS Skyrmemodel is easy enough: a natural deformation is to try and revert back towards the chiralLagrangian and include the kinetic (Dirichlet) term with a small coefficient (cid:15) [10], see alsoref. [11] for a deformation in holography. Unfortunately, the numerical solutions becomeunprecedentedly difficult in the tiny (cid:15) limit, whereas for (cid:15) of order one, there are no severedifficulties with the model. This fact is due to several factors. The main reason is thatthe BPS model has an infinite moduli space, i.e. of volume-preserving diffeomorphisms;the theory describes the Skyrmions as an incompressible fluid [12], akin to the liquid-drop model description of nuclei. Indeed this fact about the model is very welcome forphenomenology, and especially for finite density applications [13–15]. However, this meansthat the (cid:15) = 0 theory has infinitely many solutions, whereas the 0 < (cid:15) (cid:28) S instead of S . A further similarity with the full Skyrme model, is that it also possessesa BPS version, which we shall call the BPS baby Skyrme model or restricted baby Skyrmemodel. This BPS version of the model consists, again analogously to the BPS Skyrmemodel in 3+1 dimensions, of the topological charge-density squared as well as a potential.Furthermore, the BPS baby Skyrme model also enjoys area-preserving diffeomorphisminvariance for static solutions [22], and it was in fact discovered earlier in this context, thanin the full 3+1 dimensional Skyrme model. In our previous paper [17], we chose to stickwith the most simplistic potential, namely the pion mass term. This had the consequencethat the BPS solutions in that model are of the compacton type [23–25]. Other potentials,however, will provide BPS solutions with either exponential (Gaussian) tails or power-lawtails [24]. Baby Skyrmions are also interesting in their own right, and in fact they are quitesimilar to the Skyrmions being studied heavily for the moment in magnetic materials, seeref. [26] for a review, and refs. [27–31] for some recent theoretical work.The BPS property originally appeared in theories where BPS solitons were solutionspreserving a fraction of supersymmetry. For the Skyrme model in 3+1 dimensions, this isnot the case simply because its target space is not K¨ahler [32]. The first attempts at su-persymmetrizing the Skyrme model ended up with a model with C P target space [33, 34],– 2 –hich is the right target space for the baby Skyrme model (but not the Skyrme model), al-though the bosonic Lagrangian was slightly different than that of the baby Skyrme model.The exact supersymmetric version of the baby Skyrme model was later constructed with N = 1 supersymmetry [35, 36], but it does not contain the bosonic Lagrangian of therestricted baby Skyrme model either. It turns out that the restricted baby Skyrme modelpossesses N = 2 supersymmetry [36, 37] and interestingly, supersymmetry forbids the pres-ence of the kinetic Dirichlet term, which exactly corresponds to the (cid:15) → , C ), and like the 2-dimensional case, su-persymmetry also forbids the presence of the kinetic Dirichlet term. This supersymmetricSkyrme model does possess soliton solutions, but they are not BPS states [41].In this paper, we construct a BPS sector which is composed by the Skyrme term(which is also the topological charge-density squared) as well as a potential that does notgive the pion a mass. Specifically, we will choose the pion mass term squared, which givesthe BPS solitons a Gaussian tail. The deformation of the BPS sector is given by thekinetic (Dirichlet) term, with coefficient (cid:15) , and the normal pion mass term, with coefficient (cid:15)m . This has the nice scaling with (cid:15) that keeps the mass of the perturbative pionsconstant and equal to m . Because of the deformation being in the form of both thekinetic (Dirichlet) term and a potential term (the standard pion mass term), the solutionof the deformation is not restricted harmonic, but some generalization which we will callgeneralized restricted harmonic (GRH). We perform large-scale high-definition brute-forcenumerical computations for the topological sectors Q = 2 and Q = 4 and are able to dial (cid:15) all the way to (cid:15) = 0 – finding new nontrivial BPS solutions, for which we do not knowan analytic expression. We know that for Q = 1, the restricted harmonic solution, i.e. thesolution for (cid:15) > Q = N = 2 and therefore it is expected that for large topological charge Q , the moststable solution is made of Q almost axially symmetric N = 2 components – stitchedtogether in some fashion. We then turn to calculating the bound states in the perturbativescheme, using the (cid:15) expansion up to O ( (cid:15) ) or next-to-next-to-leading order (N LO). Toleading order (LO), the correction to the energy is given by inserting the BPS solutioninto the deformation terms – this cannot explain a bound state, because in order to avoidovercounting, we simply cut off the tails of the BPS and LO energy densities, that otherwisewould overlap. This does explain an attractive force at LO. In order to understand theseparation distance of two N = 2 (or even two N = 1) constituents, we need also a repulsive In principle, the deformation is restricted harmonic, but not of the pure BPS solution, but of an (cid:15) -dependent BPS solution, where the total potential is used, including the (cid:15) -dependent pion mass term, seeapp. A. – 3 –orce at shorter distances. In our perturbative scheme, we do find such a force and it isdue to energy accumulation near the middle of the bound state, caused by our imposedgluing conditions. Unfortunately, this solution is based on the perturbative scheme, wherewe have linearized the perturbation field, and we believe that it misses other nontrivial(generalized) harmonic map solutions for bound states.The paper is organized as follows. In sec. 2, we introduce our version of the deformedrestricted baby Skyrme model, its BPS solutions, topological energy bounds and define aphysical length scale for the BPS solutions. In sec. 3, we review and modify the perturbative (cid:15) -expansion scheme introduced in ref. [17]. We then test the scheme on axially symmetricbaby Skyrmion solutions. In sec. 4, we perform full high-resolution numerical PDE com-putations and find nontrivial BPS solutions as well as near-BPS solutions. In sec. 5, wereview the long-range interactions of ref. [20] but with the notation of our model. In sec. 6,we attempt at calculating the binding energies perturbatively, but discover that we arefinding different bound states, than those we found using the full numerical computationsin sec. 4. We conclude the paper with a discussion in sec. 7. We have relegated somedetails of a(n (cid:15) -dependent) BPS solution that contains the potential of the deformationLagrangian to app. A. The model is based on the restricted (BPS) baby Skyrme model with a non-BPS defor-mation in the form of the kinetic term as well as the standard pion mass term, both withcoefficient (cid:15) , L [ φ ] = (cid:15) (cid:0) L [ φ ] − m V ( φ ) (cid:1) + L [ φ ] − V ( φ ) + 12 λ ( φ · φ − , (2.1)where φ = ( φ , φ , φ ) is a real 3-vector, the kinetic term and the Skyrme terms are givenby L [ φ ] = −
12 ( ∂ µ φ · ∂ µ φ ) , (2.2) L [ φ ] = −
14 ( ∂ µ φ · ∂ µ φ )( ∂ ν φ · ∂ ν φ ) + 14 ( ∂ µ φ · ∂ ν φ )( ∂ µ φ · ∂ ν φ ) , (2.3)respectively, and the two potentials are V p ( φ ) ≡ p (cid:0) − φ (cid:1) p , (2.4)with p = 1 ,
2. The nonlinear sigma model constraint φ · φ = 1 is enforced by means of theLagrange multiplier, λ . In this paper we will use the mostly positive metric signature.We have kept the pion mass, m as a free parameter, but in order to avoid clutter, wehave scaled away the coefficients in front of the BPS potential V ( φ ) and the Skyrme termby an appropriate choice of energy and length units (without loss of generality).The requirement of finite total energy effectively point compactifies 2-space from R to R ∪{∞} (cid:39) S , i.e. a 2-sphere. It would also induce a spontaneous symmetry breaking from– 4 –(3) to O(2), but we break this symmetry explicitly with the two potentials. Thereforethe target space is given by O(3) / O(2) (cid:39) S which is also a 2-sphere, hence allowing fortopologically nontrivial solutions with degree Q = 14 π (cid:90) d x Q = − π (cid:90) d x φ · ∂ φ × ∂ φ , (2.5)which is also equal to the number of baby Skyrmions in R . The solutions that minimizethe static energy E = − (cid:90) d x L (cid:12)(cid:12)(cid:12)(cid:12) ∂ t φ =0 , (2.6)are called stable baby Skyrmions, whereas local minima exists which we shall call meta-stable baby Skyrmions.In ref. [17] we have studied the case with only V as the potential, which in the limit (cid:15) → D ⊂ R , which could be of the shape of a disc, but not necessarily.This was the case because the potential V was absent. In fact, any potential with 0 < p < V gives rise to solutions with a Gaussiantail, even in the limit of (cid:15) →
0. We will see this explicitly in the next subsection.For the analytic calculations, it will be useful to use the following parametrization φ = 11 + | ω | ω + ¯ ω − i( ω − ¯ ω )1 − | ω | , (2.7)for which φ · φ = 1 is manifest. The Lagrangian components in terms of ω can be writtenas L [ ω, ¯ ω ] = − ∂ µ ω∂ µ ¯ ω (1 + | ω | ) , (2.8) L [ ω, ¯ ω ] = − ∂ µ ω∂ µ ¯ ω )( ∂ ν ω∂ ν ¯ ω ) − ( ∂ µ ω∂ ν ¯ ω )( ∂ µ ω∂ ν ¯ ω )(1 + | ω | ) , (2.9)the potentials as V p ( ω, ¯ ω ) = 1 p (cid:18) | ω | | ω | (cid:19) p , (2.10)with p = 1 ,
2, and the topological charge as Q = − i2 π (cid:90) d x (cid:15) ij ∂ i ω∂ j ¯ ω (1 + | ω | ) . (2.11)The ω parametrization will prove useful for finding the BPS solutions, which we will turnto next, but we will turn back to the φ parametrization for the numerical analysis.– 5 – .1 BPS solutions The model contains a BPS submodel, which is found by sending (cid:15) to zero ( (cid:15) →
0) in theLagrangian (2.1): L BPS [ φ ] = L [ φ ] − V ( φ ) + λ ( φ · φ − . (2.12)It contains also a different BPS submodel which includes the potential (cid:15)m V , but thesolutions are obviously dependent on (cid:15) , see app. A. The BPS equation can be found byperforming a so-called Bogomol’nyi trick (ignoring the Lagrange multiplier term) E BPS [ φ ] = 14 ( ∂ i φ · ∂ i φ )( ∂ j φ · ∂ j φ ) −
14 ( ∂ i φ · ∂ j φ )( ∂ i φ · ∂ j φ ) + 12 (1 − φ ) = 12 ( φ · ∂ φ × ∂ φ )( φ · ∂ φ × ∂ φ ) + 12 (1 − φ ) = 12 (cid:2) ( φ · ∂ φ × ∂ φ ) ∓ (1 − φ ) (cid:3) ± (1 − φ )( φ · ∂ φ × ∂ φ ) . (2.13)The BPS equation is thus Q = − φ · ∂ φ × ∂ φ = ∓ (1 − φ ) , (2.14)and when satisfied, the total energy is proportional to the topological charge or degree ofthe baby Skyrmion, Q , which is the minimum of the energy in the given topological sector(for given Q ).We will now change parametrization of φ to stereographic coordinates (2.7) and thenthe BPS equation reads ∂ ω∂ ¯ ω − ∂ ω∂ ¯ ω (1 + | ω | ) = ∂ r ω∂ θ ¯ ω − ∂ θ ω∂ r ¯ ωr (1 + | ω | ) = ∓ i | ω | | ω | . (2.15)Inserting the Ansatz ω = e i Nθ ζ ( r ), we obtain the equation ∂ r ζr = − N ζ (1 + ζ ) , (2.16)where we have chosen the lower sign. It will now prove convenient to change variables as[24] 1 + ζ = 11 − γ , (2.17) y = 12 r , (2.18)for which the differential equation becomesd γ d y = − N γ, (2.19)which has the solution [24] γ = e − ξ − κ , (2.20)– 6 –here ξ ≡ rR , the characteristic radius is R = √ N , (2.21)and κ is an integration constant. Changing variables back to ζ , we finally obtain thesolution ζ = 1 (cid:112) e ξ − , (2.22)where in order to ensure that the target space is fully covered, we must make sure that thefunction is singular at ξ = 0 and hence we have set κ = 0. For large values of ξ , we have ζ (cid:39) exp (cid:18) − ξ (cid:19) + O (cid:16) e − ξ (cid:17) . (2.23)It will prove useful to calculate the BPS mass M BPS = (cid:90) d x E BPS [ φ ]= ± i4 (cid:90) d x | ω | (1 + | ω | ) (cid:15) ij ∂ i ω∂ j ¯ ω = ± πN (cid:90) ∞ d r ζ (1 + ζ ) ∂ r ζ = − π | N | (cid:90) ∞ d ζ ζ (1 + ζ ) = 4 π | N | . (2.24)In the fourth line we have chosen the lower sign, corresponding to the above-found BPSsolution with the boundary conditions ζ (0) → ∞ and ζ ( ∞ ) = 0.The topological charge of such an axially symmetric configuration is thus given by Q = − N (cid:90) d r ζ∂ r ζ (1 + ζ ) = N. (2.25)We will call the topological charge N only for axially symmetric baby Skyrmions, whereasfor generically shaped multi-baby Skyrmions we will denote it by Q . In the previous section, the energy was shown to be bounded from below by the Bogomol’nyibound in the BPS limit, (cid:15) = 0, E ≥ M BPS = 4 π | Q | . (2.26)There is another BPS limit, which is the double limit (cid:15) → ∞ and m →
0. This limit is alsobounded from below due to the model becoming the pure O(3) sigma model (consisting– 7 –nly of the kinetic term) [36], E (cid:15) = 2 (cid:15) (cid:90) d x ∂ i ω∂ i ¯ ω (1 + | ω | ) ≥ (cid:15) (cid:12)(cid:12)(cid:12)(cid:12) − i (cid:90) d x (cid:15) ij ∂ i ω∂ j ¯ ω (1 + | ω | ) (cid:12)(cid:12)(cid:12)(cid:12) ≥ π(cid:15) | Q | = M lump , (2.27)where the topological solitons are instead called lumps.The total energy in the model (2.1) in the limit m → E (cid:15) ≥ M BPS + M lump ≥ π | Q | ( (cid:15) + 1) . (2.28)This bound is a composite of two limits and it is indeed only saturated in those two limits:i.e. in the limit of (cid:15) →
0, the model contains BPS baby Skyrmions and in the limit of (cid:15) → ∞ , m → m , for which the composite energy bound (2.28) no longercan be saturated. It is however possible to calculate a bound similar to that of eq. (2.26),but instead of including only V as the potential we include both the BPS potential andthe deformation potential, that is ˜ m V + V . Writing down the would-be BPS mass M would − be BPS = ± i4 (cid:90) d x | ω | (cid:112) ˜ m + (1 + ˜ m ) | ω | (1 + | ω | ) (cid:15) ij ∂ i ω∂ j ¯ ω = ± πN (cid:90) ∞ d r ζ (cid:112) ˜ m + (1 + ˜ m ) ζ (1 + ζ ) ∂ r ζ = − π | N | (cid:90) ∞ d ζ ζ (cid:112) ˜ m + (1 + ˜ m ) ζ (1 + ζ ) = 4 π | N | (cid:32) (cid:113) m (2 + ˜ m ) + ˜ m m − (cid:112) m m + 2 (cid:112) m (cid:33) , (2.29)which is the bound for the model (2.1) in the limit of (cid:15) → (cid:15)m = ˜ m fixed. Theexpression in the parenthesis on the last line of eq. (2.29) has the limiting value of unitywhen m → E (cid:15) (2+0 (cid:48) )+4+0 ≥ M would − be BPS + M lump ≥ π | Q | (cid:32) (cid:15) + 12 (cid:113) (cid:15)m (2 + (cid:15)m ) + (cid:15) m (cid:15)m − (cid:112) (cid:15)m (cid:15)m + 2 (cid:112) (cid:15)m (cid:33) . (2.30)– 8 –his energy bound can be Taylor expanded in (cid:15) as E (cid:15) (2+0 (cid:48) )+4+0 ≥ π | Q | (cid:20) m ) (cid:15) + m (cid:18) (cid:15)m (cid:19) (cid:15) − m (cid:15) O ( (cid:15) ) (cid:21) . (2.31)The BPS bound in the limit (cid:15) → (cid:15) , is due to the mass term and the lump mass. The higher-order corrections to theenergy bound come from the would-be BPS bound on the Skyrme term together with bothpotentials, i.e. ˜ m V + V . It will prove useful to know the length scale of the baby Skyrmion in the BPS limit, a.k.a. itsradius. The short answer is R = √ N , which depends only on N (since we have fixed thecoefficient of the BPS potential to be unity, which can always be done by a rescaling oflengths).For a more precise estimate of the radius, we can first calculate at which value of ζ thefraction β of the energy is contained. This in turn translates to a radius and hence yieldsthe coefficient multiplying R for the radius. The entire BPS energy is contained if ζ tendsto zero in the integral (2.24), − (cid:90) ∞ d ζ ζ (1 + ζ ) = 1 . (2.32)Therefore a fraction β of the BPS energy is contained by − (cid:90) ζ β ∞ d ζ ζ (1 + ζ ) = β, ⇒ ζ β = (cid:115) β + √ − ββ − , (2.33)with β ∈ (0 , ξ β = − log (cid:18) −
11 + ( ζ β ) (cid:19) = −
12 log(1 − β ) . (2.34)Using that ξ ≡ rR , we find the radius that contains the fraction β of the total BPS energyreads r β = (cid:114) −
12 log(1 − β ) R. (2.35)The radii corresponding to β = 0 . , . , .
99 are r . (cid:39) . R , r . (cid:39) . R and r . (cid:39) . R , respectively. (cid:15) We will now use the framework of perturbation theory around a back soliton solution, ϕ ,developed in ref. [17] where δ L BPS [ ϕ ] = 0 . (3.1)– 9 –he BPS solution in this model is given by eq. (2.22) in the ω parametrization. Since thenumerical calculations will be performed in the φ parametrization, it will be useful to writethe charge- N axially symmetric baby Skyrmion solution in that parametrization: ϕ = sin f ( r ) cos( N θ − α )sin f ( r ) sin( N θ − α )cos f ( r ) , f ( r ) = arccos (cid:16) − e − ξ (cid:17) , (3.2)with ξ ≡ r/R and R is the characteristic radius (2.21), x + i y = re i θ are the standard polarcoordinates in R and α is a phase modulus. The phase modulus is an internal parameterof the solution, but for a single solution is equivalent to a rotation in the plane by − α/N .Next we will consider the corrections to the energy order-by-order in (cid:15) , in the followingsections, starting with the leading order. The leading order (LO) correction is the first and linear order in (cid:15) , and in contradistinc-tion to the compacton case of ref. [17] comes from both the kinetic term as well as the(perturbative) pion mass term. Inserting the BPS solution into these terms thus gives (cid:15)M LO ( N, m ) = (cid:15) (cid:90) d x (cid:0) −L [ ϕ ] + m V ( ϕ ) (cid:1) = 4 π(cid:15) (cid:90) d r (cid:20) r ζ r + N ζ r (1 + ζ ) + rm ζ ζ (cid:21) = 4 π(cid:15) (cid:18) π
12 + N N m (cid:19) , (3.3)where in the following we will only consider N > m . The parenthesis on the last line takes the value 1 .
169 for N = 1 and m = 0, which is about 17% above the energy bound for the kinetic term. Turning on thepion mass, m = 0 .
5, increases the value of the parenthesis to 1 . E ( (cid:15), N, m ) = M BPS ( N ) + (cid:15)M LO ( N, m )= 4 πN + π (cid:15) π(cid:15)N log 2 + 4 πN (cid:15)m . (3.4)Note that this is strictly above the BPS bound (2.31) considered at the linear level in (cid:15) .Considering now the energy per N as a function of N , we can determine which solutionhas the lowest energy per baby Skyrmion (nucleon). We havedd N (cid:18) E ( (cid:15), N, m ) N (cid:19) = − π(cid:15) N (cid:0) π − N log 2 (cid:1) = 0 , (3.5)with solution N (cid:63) = π √ ≈ . , (3.6)– 10 – L O ( N ) / N N m = 0 m = 0.5 m = 1 12 14 16 18 20 22 24 26 28 30 32 34 1 2 3 4 Figure 1 . Leading-order energy per nucleon (baby Skyrmion) divided by (cid:15) : M LO N as a function of N for m = 0 , . , which is the N with the minimum energy per N . We notice that the formal minimum ofthe energy per N , to leading order in (cid:15) , does not depend on m .In fig. 1 is shown the leading-order energy correction, divided by (cid:15)N . This is the energycorrection per nucleon, to be multiplied by (cid:15) . We can see from the figure that the stableaxially symmetric baby Skyrmion, to leading order in (cid:15) , will have N = 2. For example, twoseparated 1-Skyrmions will have a higher energy than an axially symmetric 2-Skyrmion andan axially symmetric 4-Skyrmion will have higher energy than two separated 2-Skyrmions.Also an axially symmetric 3-Skyrmion will have higher energy than three 1-Skyrmions ora 1-Skyrmion and a 2-Skyrmion. LO corrections
The next corrections are the next-to-leading order (NLO) and next-to-next-to-leading order(N LO) corrections to the energy which are of order (cid:15) and (cid:15) , respectively, and they willbe calculated by introducing a linear perturbation around the background field φ = ϕ + δφ , (3.7)where ϕ = ( ϕ , ϕ , ϕ ) is the background solution around which δφ = ( δφ , δφ , δφ ) is asmall perturbation. We assume that δφ is of order (cid:15) , although this is not clear a priori ;we have to check it a posteriori .In order to capture the NLO and N LO corrections, we have to calculate the variation– 11 –f the model (2.1) up to third order in the fields (assuming that δφ = O ( (cid:15) )) L perturb [ ϕ , δφ ] = ∂ L ∂φ a (cid:12)(cid:12)(cid:12)(cid:12) δφ a + 12 ∂ L ∂φ a ∂φ b (cid:12)(cid:12)(cid:12)(cid:12) δφ a δφ b + ∂ L ∂∂ µ φ a (cid:12)(cid:12)(cid:12)(cid:12) ∂ µ δφ a + 12 ∂ L ∂∂ µ φ a ∂∂ ν φ b (cid:12)(cid:12)(cid:12)(cid:12) ∂ µ δφ a ∂ ν δφ b + 16 ∂ L ∂∂ µ φ a ∂∂ ν φ b ∂∂ ρ φ c (cid:12)(cid:12)(cid:12)(cid:12) ∂ µ δφ a ∂ ν δφ b ∂ ρ δφ c = (cid:15)m δφ + δλ ϕ · δφ − (cid:0) δφ (cid:1) + λ + δλ δφ − (cid:15)J µa ∂ µ δφ a − V µνab ∂ µ δφ a ∂ ν δφ b −
16 Γ µνρabc ∂ µ δφ a ∂ ν δφ b ∂ ρ δφ c , (3.8)where the vertical bar “ | ” denotes evaluation on the background by setting φ = ϕ (tothe left of the bar) and we have defined the following quantities J µa ≡ − ∂ L ∂∂ µ φ a (cid:12)(cid:12)(cid:12)(cid:12) = ∂ µ ϕ a , (3.9) V µνab ≡ − ∂ L ∂∂ µ φ a ∂∂ ν φ b (cid:12)(cid:12)(cid:12)(cid:12) ≡ V µν ab + (cid:15)V µν ab ,V µν ab ≡ ( ∂ ρ ϕ · ∂ ρ ϕ ) η µν δ ab + 2 ∂ µ ϕ a ∂ ν ϕ b − ∂ µ ϕ · ∂ ν ϕ δ ab − ∂ ρ ϕ a ∂ ρ ϕ b η µν − ∂ µ ϕ b ∂ ν ϕ a ,V µν ab ≡ η µν δ ab , (3.10)Γ µνρabc ≡ − ∂ L ∂∂ µ φ a ∂∂ ν φ b ∂∂ ρ φ c (cid:12)(cid:12)(cid:12)(cid:12) = η µν (cid:0) ∂ ρ ϕ c δ ab − ∂ ρ ϕ a δ bc − ∂ ρ ϕ b δ ca (cid:1) + η νρ (cid:0) ∂ µ ϕ a δ bc − ∂ µ ϕ b δ ca − ∂ µ ϕ c δ ab (cid:1) + η µρ (cid:0) ∂ ν ϕ b δ ca − ∂ ν ϕ c δ ab − ∂ ν ϕ a δ bc (cid:1) , (3.11)and the Lagrange multiplier for the background fields, λ , is given by λ = − ( ϕ · ∂ ϕ )( ∂ µ ϕ · ∂ µ ϕ ) + ( ϕ · ∂ µ ∂ ν ϕ )( ∂ µ ϕ · ∂ ν ϕ ) − (1 − ϕ ) ϕ . (3.12)The Lagrange multiplier λ has been replaced by the expansion λ + δλ , where λ enforcesthe sigma model constraint for the background solution ϕ and δλ makes sure the total field φ = ϕ + δφ remains inside the O(4) group, i.e. it preserves the constraint φ · φ = 1 + O ( (cid:15) ).Varying δλ yields (cid:18) δφ + ϕ · δφ (cid:19) = 0 , (3.13)which has the solution [20, 17] δφ = ∆ × ϕ + 12 ∆ × ( ∆ × ϕ ) . (3.14)This form of the perturbation satisfies the constraint (3.13) up to O ( ∆ ) and since ∆ will turn out to be of order (cid:15) , that makes it of order O ( (cid:15) ). Notice that the perturbativecontribution to the energy from the terms multiplying δλ is exactly the bracket in eq. (3.13)and hence is of order O ( (cid:15) ), which we can safely disregard. Therefore, we do not need tofind the explicit expression for δλ to this order in perturbation theory; its job was to produceeq. (3.14). – 12 –e will now substitute the form of the variation (3.14) into the perturbation La-grangian (3.8): L perturb [ ϕ , ∆ ] = L perturb2 [ ϕ , ∆ ] + L perturb3 [ ϕ , ∆ ] + O ( (cid:15) ) , (3.15) L perturb3 [ ϕ , ∆ ] = L perturb3 , quad [ ϕ , ∆ ] + L perturb3 , cubic [ ϕ , ∆ ] , (3.16)where the NLO Lagrangian is given by L perturb2 [ ϕ , ∆ ] = (cid:15)m (cid:0) ∆ ϕ − ∆ ϕ (cid:1) − (cid:0) ∆ ϕ − ∆ ϕ (cid:1) + λ (cid:0) ∆ − ( ∆ · ϕ ) (cid:1) (3.17) − (cid:15) J µ · ∂ µ ∆ × ϕ − (cid:15) J µ · ∆ × ∂ µ ϕ − V µν ab ε acd ∂ µ (∆ c ϕ d ) ε bef ∂ ν (∆ e ϕ f ) , and the N LO Lagrangians read L perturb3 , quad [ ϕ , ∆ ] = 12 (cid:15)m (cid:0) ( ∆ · ϕ )∆ − ∆ ϕ (cid:1) − (cid:15) ϕ · ∆ )( J µ · ∂ µ ∆ ) − (cid:15) J µ · ∆ )( ∂ µ ϕ · ∆ ) − (cid:15) J µ · ∆ )( ϕ · ∂ µ ∆ )+ (cid:15) J µ · ∂ µ ϕ ) ∆ + (cid:15) ( J µ · ϕ )( ∆ · ∂ µ ∆ ) − (cid:15) V µν ab ε acd ∂ µ (∆ c ϕ d ) ε bef ∂ ν (∆ e ϕ f ) , (3.18) L perturb3 , cubic [ ϕ , ∆ ] = − (cid:0) ∆ ϕ − ∆ ϕ (cid:1) (cid:0) ( ∆ · ϕ )∆ − ∆ ϕ (cid:1) − V µν ab ε acd ∂ µ (∆ c ϕ d ) ∂ ν (cid:16) ∆ b ( ∆ · ϕ ) − ∆ ϕ b (cid:17) −
16 Γ µνρabc ε ade ∂ µ (∆ d ϕ e ) ε bfg ∂ ν (∆ f ϕ g ) ε chi ∂ ρ (∆ h ϕ i ) , (3.19)which is the complete perturbation Lagrangian to third order in (cid:15) .In principle, we should solve this nonlinear problem for ∆ , which however is almost asdifficult as the original problem, without introducing the perturbation theory on top of thesoliton background. Therefore, we will linearize the above Lagrangian, i.e. we will only usethe linear and quadratic parts in ∆ (i.e. eqs. (3.17) and (3.18)) to determine its equationof motion.One may then wonder why go to the third order in (cid:15) if it results in cubic terms in ∆ , that we anyway will discard once we turn to solving the equation of motion for theperturbation. The answer is that we need the last term of eq. (3.18), as it gives a term (cid:15)∂ in the equation of motion, whereas the last term in eq. (3.17) will give a term f ( ϕ, ∂ µ ϕ ) ∂ in the equation of motion for ∆ . Although we are preparing for the study of the limit ofvery small (cid:15) (say about 1 / (cid:15) ofcourse).The linearized static equation of motion for ∆ can thus be written down X ∆ aii + X ab ∆ bii + X abij ∆ bij + X abi ∆ bi + Λ∆ a + Λ ab ∆ b = − (cid:15)ε abc ϕ b ϕ cii − (cid:15)m ε ab ϕ b , (3.20)– 13 –here we have defined the quantities X ≡ (cid:15), (3.21) X ab ≡ − (cid:15)ϕ a ϕ b + ϕ aj ϕ bj , (3.22) X abij ≡ − ϕ ai ϕ bj , (3.23) X abi ≡ − (cid:15)ϕ ai ϕ b − ϕ aij ϕ bj + ϕ ajj ϕ bi − ϕ ai ϕ bjj + 2 ϕ aj ϕ bij + ( ϕ j · ϕ j )( ϕ a ϕ bi − ϕ ai ϕ b ) − ( ϕ i · ϕ j )( ϕ a ϕ bj − ϕ aj ϕ b ) , (3.24)Λ ≡ − (1 − ϕ ) ϕ − (cid:15)m ϕ , (3.25)Λ ab ≡ − (cid:15) ϕ aii ϕ b + (cid:15) ϕ a ϕ bii + (1 − ϕ ) ϕ ϕ a ϕ b − ( ϕ ij · ϕ j ) ϕ ai ϕ b + ( ϕ ii · ϕ j ) ϕ aj ϕ b + ( ϕ i · ϕ j ) ϕ aij ϕ b − ( ϕ j · ϕ j ) ϕ aii ϕ b + ( ϕ i · ϕ j ) ϕ a ϕ b − ( ϕ i · ϕ i ) ϕ a ϕ b − ε ac ε bd ϕ c ϕ d + (cid:15) m ( ϕ a δ b + δ a ϕ b ) . (3.26)It will now prove instructive to consider the equation of motion for the fluctuation ∆ at an asymptotic distance from the BPS baby Skyrmion background. In that case, thebackground solution (3.2) can be approximated as ϕ = e − ξ (cid:112) − e − ξ cos( N θ − α )2 e − ξ (cid:112) − e − ξ sin( N θ − α )1 − e − ξ (cid:39) e − ξ cos( N θ − α )2 e − ξ sin( N θ − α )1 − e − ξ + O ( e − ξ ) . (3.27)Inserting this approximation into eq. (3.20), we can write the equation of motion for thefluctuation, to second order in e − ξ as: (cid:15) (cid:0) ∂ ∆ a − ∂ ∆ δ a (cid:1) − (cid:15)m (cid:0) ∆ a − ∆ δ a (cid:1) − (cid:15)e − ξ (cid:16) w a δ b + δ a w b (cid:17) ∂ ∆ b + 4 (cid:15)R e − ξ w a x i ∂ i ∆ + 4 (cid:15)Nξ R e − ξ (cid:98) w a (cid:15) ij x j ∂ i ∆ + (cid:15)R (cid:18) ξ − N ξ − (cid:19) e − ξ (cid:16) − w a δ b + δ a w b (cid:17) ∆ b + (cid:15)m e − ξ (cid:16) w a δ b + δ a w b (cid:17) ∆ b − (cid:15)e − ξ ( w a w b − δ a δ b ) ∂ ∆ b − (cid:15)R e − ξ x i ∂ i ∆ δ a + 8 (cid:15)R e − ξ (cid:18) x i w a w b + Nξ ε ij x j (cid:98) w a w b (cid:19) ∂ i ∆ b + 2 (cid:15)m e − ξ (∆ a − ∆ δ a )+ 4 ξ R e − ξ w a w b ∂ ∆ b + 4 N R ξ (cid:98) w a (cid:98) w b e − ξ ∂ ∆ b − R e − ξ (cid:18) w a x i + Nξ (cid:98) w a (cid:15) ik x k (cid:19) (cid:18) w b x j + Nξ (cid:98) w b (cid:15) jl x l (cid:19) ∂ i ∂ j ∆ b − Nξ R e − ξ (cid:104) (3 ξ + 1) w a (cid:98) w b − (3 ξ − (cid:98) w a w b (cid:105) (cid:15) ij x j ∂ i ∆ b − R e − ξ (cid:20)(cid:18) N ξ (cid:19) w a w b + N (cid:18) ξ + 1 ξ (cid:19) (cid:98) w a (cid:98) w b (cid:21) x i ∂ i ∆ b − e − ξ (cid:98) w a (cid:98) w b ∆ b − e − ξ (∆ a − ∆ δ a )= − (cid:15)R (cid:18) ξ − N ξ − − m R (cid:19) e − ξ (cid:98) w a , (3.28)– 14 –here we have defined w = cos( N θ − α )sin( N θ − α )0 , (cid:98) w = − sin( N θ − α )cos( N θ − α )0 . (3.29)Clearly, only the first line of eq. (3.28) is not exponentially suppressed and therefore guar-antees the propagation of the fluctuation ∆ in the asymptotic regime (away from thebackground baby Skyrmion). Now, importantly, all the terms with a factor of (cid:15) on theleft-hand side of eq. (3.28) come from the O ( (cid:15) ) Lagrangian (3.18) and therefore had weonly gone to second order in the (cid:15) expansion, the linearized equation for the fluctuationwould have looked like this:4 ξ R e − ξ w a w b ∂ ∆ b − R e − ξ w a w b x i x j ∂ i ∂ j ∆ b − NR e − ξ (cid:104) w a (cid:98) w b − (cid:98) w a w b (cid:105) (cid:15) ij x j ∂ i ∆ b − R e − ξ w a w b x i ∂ i ∆ b − e − ξ (cid:98) w a (cid:98) w b ∆ b − e − ξ (∆ a − ∆ δ a ) = − (cid:15)R ξ e − ξ (cid:98) w a , (3.30)where we have kept only the terms with the highest powers of ξ on both sides. To thisorder, the equation of motion for the fluctuation is badly behaved, because it schematicallytakes the form O ab ∆ b = − (cid:15)R ξ e ξ (cid:98) w a , ξ ≡ rR , (3.31)whose right-hand side diverges exponentially. We expect of course that the fluctuationsgo to zero at asymptotic distances, and this will also be the case for the third-order in (cid:15) equation of motion (3.28), because of the terms (cid:15) ( ∂ − m )∆ a , which has the expectedexponential falloff compatible with the boundary conditions.We have assumed that the perturbation field is proportional to (cid:15) . In order to attemptto address this point, let us consider (for simplicity) a few of the terms of the third-orderin (cid:15) equation of motion (3.28) for the fluctuation: (cid:15) (cid:0) ∂ ∆ a − ∂ ∆ δ a (cid:1) − (cid:15)m (cid:0) ∆ a − ∆ δ a (cid:1) + 4 ξ R e − ξ w a w b ∂ ∆ b − e − ξ (cid:98) w a (cid:98) w b ∆ b − e − ξ (∆ a − ∆ δ a ) = 0 . (3.32)Since w and (cid:98) w are of order one, we can estimate at which distance from the backgroundbaby Skyrmion the third order terms become dominant: (cid:15) (cid:29) ξ R e − ξ , (3.33)where ξ ≡ rR and R = √ N . The smaller values of (cid:15) , the larger distances are neededbefore the third-order terms take over. Since the exponential of − r quickly becomesinfinitesimally small, the tail of the perturbation at asymptotic distances will take theform ∆ = c c e − m r , (3.34)– 15 –here c , are constants. Notice that this is seemingly independent of (cid:15) . All the (cid:15) -dependence is contained at distances (cid:15) (cid:46) ξ R e − ξ , (3.35)for which the (cid:15) -dependence becomes quite complicated. Indirectly, the dependence on (cid:15) ineq. (3.34) is possessed by c , by gluing it together with the solution are distances given byeq. (3.35). Importantly, in the limit of (cid:15) = 0, c , = 0 because there is no tail correction inthe BPS limit. To leading order, the coefficients must behave like c , ∝ (cid:15) p , with p ∈ Z > a positive integer. We have, however, not been able to prove rigorously that p = 1 (i.e. itcould be larger than one).It is worthwhile to compare the situation of the setting at hand – a baby Skyrmionhaving a Gaussian tail in the BPS limit and where the (cid:15) perturbation of the model entailsboth the Dirichlet kinetic term as well as the standard pion mass term, see eq. (2.1) – withthe compacton case studied in ref. [17].In the compacton case, as the name suggests, the BPS limit of the baby Skyrmionis a compacton (hence no tail in the BPS limit) and the (cid:15) perturbation in that case wasdone solely by adding the Dirichlet kinetic term (with coefficient (cid:15) ). Let us recapitulate thesituation in the compacton case of ref. [17]. The tail of the perturbation in that case becamenon-dynamic unless we went to the third order in (cid:15) (which would prevent the compactonsof knowing of one another and hence prevent the calculation of binding energies). Going tothe third order in the (cid:15) expansion gave a dynamic tail to the perturbation, which howeverwas nonanalytic in (cid:15) . The nonanalycity was due to the fact that the (inverse) propagatorhad the form (cid:15)∂ − m and the tail thus took the form e − mr √ (cid:15) , which when Taylor expandedin (cid:15) vanishes at any finite order.In this case of a baby Skyrmion with a Gaussian tail in the BPS limit, the situationdraws some similarities to the compacton case, but nevertheless is dramatically different.The similar property of the (cid:15) expansion, is that we still have to go to the third order in (cid:15) ,because otherwise – as demonstrated above – the equations of motion for the perturbationbecome ill-defined at asymptotic distances or at least incompatible with suitable boundaryconditions. A huge difference is that the baby Skyrmions, in the BPS limit, themselves haveGaussian tails, and therefore already “feel” each other when two or more of them are placedat a finite distance from each other. The BPS solution is thus nontrivial for any separationdistance. If we now turn on a small but finite (cid:15) according to the Lagrangian at hand (2.1),the perturbation field ∆ adds an exponential tail to the background solution. The fieldconfiguration now flows to the nearest GRH solution in field space. The solution to thegeneralized restricted harmonicity is, however, extremely difficult compared to the com-pacton case, which is nearly trivial. That is, axially symmetric baby Skyrmions placed atdistances such that their compacton regions do not overlap – for details, see ref. [17]. Now,since the background fields (meaning the fields of the background BPS baby Skyrmion)tend to the vacuum exponentially (or rather like the Gaussian), the governing equations ofmotion for the perturbation are simply (cid:15) (cid:0) ∂ ∆ a − ∂ ∆ δ a (cid:1) − (cid:15)m (cid:0) ∆ a − ∆ δ a (cid:1) = 0 , (3.36)– 16 –nd thus the tail of the perturbation (not the BPS solution) is exponential too, seeeq. (3.34), but in stark contradistinction to the compacton case, it is independent of (cid:15) (it is nevertheless “glued together” with a solution closer to the baby Skyrmion that is de-pendent on (cid:15) and hence a suppression of the tail is still in effect, but we shall check this aposteriori ). Of course, this is not an accident, but a consequence of the construction of thismore elaborate (and aimed at being more physical) model, compared with the compactoncase of ref. [17]. We include the Dirichlet kinetic term with a coefficient (cid:15) , because wewant small binding energies (that is a small perturbation added to the BPS model). Thenwe include the mass term, also with coefficient (cid:15) , in order to prevent that the pion massbecomes unrealistically large in the small- (cid:15) limit. Another feature – also by construction –is that the potential in the BPS sector ( V ) does not give rise to a pion mass and thereforethe coefficient of the potential is not restricted by the value of the pion mass.The picture that forms, which we will elaborate on in a later section, is that at asymp-totically large distances, the governing equation for the perturbation is eq. (3.36) and thetails of two or more baby Skyrmions attract each other (in the attractive channels) with aforce that is at least cubic in (cid:15) , but the mass of the fluctuations is independent of (cid:15) . Thepion mass can thus be set to any realistic value .A word of caution is that we calculate the perturbation using the linearized equation ofmotion that contains some (crucial) terms (3.18) at O ( (cid:15) ), but not the cubic terms (3.19) atthe same order. As for the calculation of the energy, we will compare the energy calculatedat NLO and N LO to the exact numerical solutions in the axially symmetric case in thenext section.The static energy of the perturbation is simply given by E perturb [ ϕ , ∆ ] = −L perturb [ ϕ , ∆ ] (cid:12)(cid:12)(cid:12) ∂ =0 . (3.37)A comment in store about the perturbation, ∆ , is that the entire static energy vanishesfor ∆ ∝ ϕ . However, since the expression (3.14) for δφ is nonlinear in ∆ , a few cross termssurvive if we take ∆ = δc ϕ + ∆ ⊥ , with ∆ ⊥ · ϕ = 0 a perpendicular perturbation. Thecross terms only give rise to a linear first-order PDE for δc , which must vanish once subjectto the boundary conditions lim | x |→∞ δc = 0 and δc (0) = 0, where the origin is at eachbackground baby Skyrmion center. This justifies setting ∆ = ∆ ⊥ . For more details, seeref. [17].It will now prove useful to specialize to the case of the background BPS baby Skyrmionsolution for ϕ , using transverse perturbations for ∆ = ∆ ⊥ and switching to polar coordi-nates in the plane, for which the static perturbation energy reads E perturb [ f, δf, δθ ] = E perturb2 [ f, δf, δθ ] + E perturb3 , quad [ f, δf, δθ ] + E perturb3 , cubic [ f, δf, δθ ] , (3.38) Of course, we are working with the 2-dimensional baby Skyrme model, but the aim is to test thisframework in two dimensions before attempting at addressing the 3-dimensional model. – 17 –ith E perturb2 [ f, δf, δθ ] = (cid:18) (cid:15)m sin f + (cid:15)N r sin(2 f ) (cid:19) δf + (cid:15)f r δf r + N r sin ( f ) δf r + f r r δθ θ + 12 (cid:18) cos f − cos(2 f ) + N r cos(2 f ) f r (cid:19) δf + (cid:18) cos f sin (cid:18) f (cid:19) − N r sin ( f ) f r (cid:19) δθ + Nr sin( f ) f r [2 δf r δθ θ − δf θ δθ r ] + N r sin(2 f ) f r [2 δf r δf + δθ r δθ ] + Nr cos( f ) f r δf δθ θ , (3.39)for the NLO terms, E perturb3 , quad [ f, δf, δθ ] = (cid:15) (cid:18) δf r + δf θ r (cid:19) + (cid:15) (cid:18) m cos f + N r cos(2 f ) (cid:19) δf + (cid:15) (cid:18) δθ r + δθ θ r (cid:19) + (cid:15) (cid:18) m cos f − f r + N r cos f (cid:19) δθ − (cid:15)Nr cos( f ) ( δf θ δθ − δf δθ θ ) , (3.40)for the N LO terms quadratic in ∆ ⊥ and E perturb3 , cubic [ f, δf, δθ ] = 12 r ( N sin( f ) δf r + f r δθ θ + N cos( f ) f r δf ) × (cid:2) N cos( f ) δf r δf + 2 δf r δθ θ + 2 N cos( f ) δθ r δθ − δf θ δθ r − N sin( f ) f r δf − N sin( f ) f r δθ (cid:3) + 14 sin 2 f (cid:0) δf + δθ (cid:1) δf, (3.41)for the N LO terms cubic in ∆ ⊥ , and we have defined the transverse perturbations ∆ ⊥ as ∆ ⊥ ≡ − sin N θ cos
N θ δf − cos f cos N θ cos f sin N θ − sin f δθ, (3.42)and the background BPS baby Skyrmion solution is given by f = f BPS = arccos (cid:18) − e − r R (cid:19) , (3.43)with R = √ N , (eq. (2.21)).The corresponding static equations of motion read X rr (cid:32) δf rr δθ rr (cid:33) + 1 r X r (cid:32) δf r δθ r (cid:33) + 1 r X θθ (cid:32) δf θθ δθ θθ (cid:33) + 1 r (cid:32) X δθθ X δfθ (cid:33) (cid:32) δf θ δθ θ (cid:33) + X rθ r (cid:32) (cid:33) (cid:32) δf rθ δθ rθ (cid:33) + (cid:32) Λ δf
00 Λ δθ (cid:33) (cid:32) δfδθ (cid:33) = (cid:32) − (cid:15) (cid:16) f rr + r f r − N r sin(2 f ) − m sin f (cid:17) (cid:33) , (3.44)– 18 –here we have defined the matrices X rr = (cid:32) (cid:15) + N r sin f (cid:15) (cid:33) , (3.45) X r = (cid:32) (cid:15) − N r sin f + N r sin(2 f ) f r (cid:15) (cid:33) , (3.46) X θθ = (cid:32) (cid:15) (cid:15) + f r (cid:33) , (3.47)as well as the functions X δfθ = 2 (cid:15)Nr cos f − Nr sin( f ) f rr + Nr sin( f ) f r , (3.48) X δθθ = − (cid:15)Nr cos f + 2 Nr sin( f ) f rr + Nr cos( f ) f r − Nr sin( f ) f r , (3.49) X rθ = Nr sin( f ) f r , (3.50)Λ δf = − (cid:15)m cos f − (cid:15)N r cos(2 f ) + N r sin(2 f ) f rr + N r cos(2 f ) f r − N r sin(2 f ) f r − cos f + cos 2 f, (3.51)Λ δθ = (cid:15)f r − (cid:15)N r cos ( f ) + N r sin(2 f ) f rr + N r cos ( f ) f r − N r sin(2 f ) f r − m cos f. (3.52)Firstly, the source term, i.e. the right-hand side of eq. (3.44) only exists for the upperequation, that is for the equation of motion for δf . Secondly, the mixing between theupper and the lower equations only appears in terms involving a θ derivative and a mixed r and θ derivative. Thus, δθ is a homogeneous source-free equation of motion, unless δf hasnontrivial θ dependence. Therefore, if we restrict to axially symmetric background BPSsolutions and turn on only axially symmetric perturbations, δf = δf ( r ), then δθ decouplesand is trivially satisfied ( δθ = 0 everywhere).The equation of motion for the perturbation of axially symmetric baby Skyrmions thusreduces to (cid:18) (cid:15) + N r sin f (cid:19) δf rr + 1 r (cid:18) (cid:15) − N r sin f + N r sin(2 f ) f r (cid:19) δf r + Λ δf δf = − (cid:15) (cid:18) f rr + 1 r f r − N r sin 2 f − m sin f (cid:19) , (3.53)and the corresponding static energy for the perturbation is E perturb2 [ f, δf ] = (cid:18) (cid:15)m sin f + (cid:15)N r sin(2 f ) (cid:19) δf + (cid:15)f r δf r + N r sin ( f ) δf r + 12 (cid:18) cos f − cos(2 f ) + N r cos(2 f ) f r (cid:19) δf + N r sin(2 f ) f r δf r δf, (3.54)for the NLO terms, E perturb3 , quad [ f, δf ] = (cid:15) δf r + (cid:15) (cid:18) m cos f + N r cos(2 f ) (cid:19) δf , (3.55)– 19 –or the N LO terms quadratic in δf and E perturb3 , cubic [ f, δf ] = N r sin(2 f ) δf r δf + N r (cid:18) −
32 sin f (cid:19) f r δf r δf − N r sin(2 f ) f r δf + 14 sin(2 f ) δf , (3.56)for the N LO terms cubic in δf . In order to verify the accuracy of our perturbative scheme, we start with axially symmetricbaby Skyrmions. As was shown in ref. [17], the perturbation can in this case be written as φ = ϕ + ∆ ⊥ × ϕ + 12 ∆ ⊥ × ( ∆ ⊥ × ϕ ) (cid:39) sin( f + δf ) cos N θ sin( f + δf ) sin N θ cos( f + δf ) + O ( δf ) , (3.57)where we have used eqs. (3.14), (3.42) and set δθ = 0. It is thus clear that δf is indeedan additive correction to the BPS background profile function f in the axially symmetriccase.Fig. 2 shows the profile function f = arccos( φ ) for the N = 1 , , (cid:15) = 0 .
01 and m = 0 . (cid:15) = 0) and (cid:15) expansion, denoted as “perturbation”. Inthe right-hand side panel, the BPS profile is subtracted off of all the profiles, so the differ-ences between the solutions are clearly visible. The characteristic radius, R = √ N (seeeq. (2.21)) of the baby Skyrmion is shown with a vertical dashed black line on each panelof the figure.Notice that the perturbation qualitatively captures the correct behavior of the exactsolution everywhere. For r (cid:46) . R the perturbative correction to the BPS solution matchesthe exact solution extremely well, see the right-hand side panels of fig. 2 (the characteristicradius R = √ N is marked with a vertical dashed black line on each graph) and this is theradius within which 99% of the energy of the BPS baby Skyrmion is contained. However,for r (cid:38) R the perturbative correction overshoots (undershoots) the exact solution for N = 1 , N = 4), which indicates the level of precision of the perturbative method at thelinear order in the perturbation field (corresponding to second and some third order termsin (cid:15) ).One could think that it would be necessary to include all terms at the third order inthe (cid:15) expansion, hence making the problem a nonlinear one. However, as the discrepancybetween the perturbative solution and the exact solution appears only at distances r (cid:38) R , the exponential suppression of the tail of the BPS background solution makes theperturbative contribution to the energy subdominant in that (asymptotic) region; in fact,less than 1% of the energy density of the BPS background baby Skyrmion resides beyond– 20 – r exact numericalBPSperturbation f − f B PS r exact numericalBPSperturbation −0.01 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0 2 4 6 8 10 12 14 f r exact numericalBPSperturbation f − f B PS r exact numericalBPSperturbation −0.005 0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0 2 4 6 8 10 12 14 f r exact numericalBPSperturbation f − f B PS r exact numericalBPSperturbation −0.02−0.015−0.01−0.005 0 0.005 0.01 0 2 4 6 8 10 12 14 Figure 2 . The profile function f = arccos( φ ) for N = 1 , , (cid:15) = 0 .
01 and m = 0 .
5. The BPS background profile function is shown as a red solid line. The perturbation(dark-green solid line) gets closer to the true solution (black solid line), obtained by numericalcalculation. (a) profile functions. (b) profile function with the BPS one subtracted off. The dashedblack vertical line marks R = √ N on each panel, which is the characteristic length scale of thebaby Skyrmion. – 21 –he distance of ∼ . R from the center of the soliton. In order to demonstrate that thisstatement is true, we now turn to calculating the perturbative corrections to the energy ofthe axially symmetric baby Skyrmions. N = 1 m = 0.5 m = 1 E ǫ BPSLONLO semianalyticNLO fitN LO semianalyticN LO fitexact N = 1 m = 0.5 E − M B PS ( ) − ǫ M L O ( , . ) ǫ NLO semianalyticNLO fitN LO semianalyticN LO fitexact −0.09−0.08−0.07−0.06−0.05−0.04−0.03−0.02−0.01 0 0.01 0.1 N = 1 m = 1 E − M B PS ( ) − ǫ M L O ( , ) ǫ NLO semianalyticNLO fitN LO semianalyticN LO fitexact −0.1−0.09−0.08−0.07−0.06−0.05−0.04−0.03−0.02−0.01 0 0.01 0.1 N = 2 m = 0.5 m = 1 E ǫ BPSLONLO semianalyticNLO fitN LO semianalyticN LO fitexact
25 26 27 28 29 30 31 32 0.01 0.1 N = 2 m = 0.5 E − M B PS ( ) − ǫ M L O ( , . ) ǫ NLO semianalyticNLO fitN LO semianalyticN LO fitexact −0.06−0.05−0.04−0.03−0.02−0.01 0 0.01 0.1 N = 2 m = 1 E − M B PS ( ) − ǫ M L O ( , ) ǫ NLO semianalyticNLO fitN LO semianalyticN LO fitexact −0.3−0.25−0.2−0.15−0.1−0.05 0 0.01 0.1 N = 4 m = 0.5 m = 1 E ǫ BPSLONLO semianalyticNLO fitN LO semianalyticN LO fitexact
50 52 54 56 58 60 62 64 66 0.01 0.1 N = 4 m = 0.5 E − M B PS ( ) − ǫ M L O ( , . ) ǫ NLO semianalyticNLO fitN LO semianalyticN LO fitexact −0.4−0.35−0.3−0.25−0.2−0.15−0.1−0.05 0 0.01 0.1 N = 4 m = 1 E − M B PS ( ) − ǫ M L O ( , ) ǫ NLO semianalyticNLO fitN LO semianalyticN LO fitexact −1−0.9−0.8−0.7−0.6−0.5−0.4−0.3−0.2−0.1 0 0.01 0.1
Figure 3 . The energy (mass) of the baby Skyrmions in the perturbative (cid:15) -expansion scheme,as a function of (cid:15) (on a log-scale): The red line shows the BPS bound, the orange line is theleading-order (LO) correction, the black pluses are the next-to-leading order (NLO) corrections in (cid:15) calculated using the linear perturbation, the green dashed line is a fit to the NLO points, thered crosses are the next-to-next-to-leading order (N LO) corrections in (cid:15) calculated using the samelinear perturbation, the dark-blue dashed line is a fit to the N LO points and the black line is theexact energy calculated using the full (nonlinear) equations of motion. The left columns show thetotal energy, E , the middle and right columns show the energy with the BPS and the LO correctionsubtracted off for m = 0 . m = 1, respectively, to better see the differences between theexact, NLO and N LO corrections.
The perturbative corrections to the energy of the N = 1 , , m = 0 . , LO corrections (red crosses). Both the NLO (i.e. O ( (cid:15) )) corrections(3.54) and the N LO (i.e. O ( (cid:15) )) corrections (3.55)+(3.56) are calculated using the per-turbation, which is a solution to the linearized equation of motion (3.53), which containsall terms of the second-order-in- (cid:15) Lagrangian (3.17) and the quadratic terms of the third-– 22 –rder-in- (cid:15)
Lagrangian (3.18). The green dashed line shows a fit to the NLO correctionsand the dark-blue dashed line for the N LO corrections. The middle and right panels ofthe figure show the same information, for m = 0 . m = 1, respectively, but withthe BPS and LO corrections subtracted off, so as to better see the differences between theNLO corrections, the N LO corrections and the exact energies.First of all, we can see from the total energies, shown in the left-hand side panels offig. 3, that the precision of the perturbative scheme is extremely good. More precisely, theLO correction overshoots the exact result slightly, but the NLO correction is negative andundershoots the exact result. The N LO correction is then positive and comes extremelyclose to the exact energies. In fact, by a close inspection of the figures, it can be seen thatthe red crosses, corresponding to the perturbative N LO corrections, fit the exact energies(black solid lines) better than the N LO fits (dark-blue dashed lines). Although negligiblysmall, we expect the remaining discrepancy between the N LO corrected energies and theexact energies to be due to the approximation of linearizing the equation of motion for theperturbation and for truncating the (cid:15) expansion at the third order.The fits shown in fig. 3 contain no linear term in (cid:15) for the NLO corrections and nolinear and quadratic terms in (cid:15) for the N LO corrections. Nevertheless they describe theperturbative calculations very well and this is thus the a posteriori confirmation that δφ is of order (cid:15) (and possibly containing higher-order corrections too). It should be fairlyconvincing that the LO corrections, which are linear in (cid:15) are orders of magnitude largerthan the NLO and N LO corrections, that thus do not contain a linear term.Because we now have a dependence on m in this model, compared with the compactoncase of ref. [17], we have calculated the perturbative corrections for two different values of m , see fig. 3. Since the physical case has a nonvanishing pion mass, we have chosen twononvanishing values, namely m = 0 . m = 1.We will now use the fits to the NLO and N LO results in fig. 3 to write an approximateexpression for the N LO energy: E ( (cid:15), N, m ) = M BPS ( N ) + (cid:15)M LO ( N, m ) + (cid:15) M NLO ( N, m ) + (cid:15) M N LO ( N, m )+ (cid:15) M residual , ( N, m )= 4 πN + (cid:15) (cid:18) π πN log 2 (cid:19) + (cid:15)m (4 πN )+ (cid:15) (cid:0) − .
87 + 28 . N − . N (cid:1) + (cid:15) m (cid:0) . − . N − . N (cid:1) + (cid:15) (cid:0) . − . N + 20 . N (cid:1) + (cid:15) m (cid:0) − . . N + 26 . N (cid:1) + (cid:15) (cid:0) − . N − . N (cid:1) + (cid:15) m (cid:0) . − . N − . N (cid:1) , (3.58)where we have assumed that the dependence on m is quadratic. This fit should workreasonably well, at least for m ∈ [0 . , (cid:15) in the fit is to allow for some higher-order (residual) behavior of the perturbation. The– 23 –igher-order result for the energies in turn gives a correction to N (cid:63) of eq. (3.6), i.e.: N (cid:63) = 1 .
541 + (cid:15) ( − .
638 + 2 . m ) + (cid:15) (17 . − . m − . m )+ (cid:15) ( − .
18 + 22 . m + 0 . m + 1 . m ) + O ( (cid:15) ) . (3.59) m = 0.5 m = 1 N ★ ǫ (a) m = 0.5 m = 1 m = 0.5 ( E ( ǫ , N , m ) / N ) / ( E ( ǫ , , m ) / ) ǫ N = 1 N = 3 1 1.002 1.004 1.006 1.008 1.01 1.012 1.014 1.016 1.018 0.001 0.01 0.1 (b) Figure 4 . (a) The perturbatively corrected critical value, N (cid:63) , of the topological charge N of theaxially symmetric baby Skyrmion, as a function of (cid:15) . This critical value formally corresponds tothe smallest energy per N . (b) The ratio of the energy of the N -Skyrmion per N and the energyof the 2-Skyrmion divided by 2, for N = 1 , m = 0 . ,
1, as functions of (cid:15) . The perturbatively corrected N (cid:63) is shown in fig. 4(a) as a function of (cid:15) for m =0 . ,
1. It is seen from the figure that for (cid:15) sufficiently small (i.e. (cid:15) < .
1) the perturbativecorrections to N (cid:63) are too small to decisively make another topological charge sector thestable one. In order to check this explicitly, we plot in fig. 4(b) the ratio of the energy pertopological charge for N = 1 , N = 2; that is E ( (cid:15),N,m ) N E ( (cid:15), ,m ) ,for m = 0 . , (cid:15) . We can see that there is only a mild dependency on m and all values in the plot are greater than unity: Hence, the N = 2 axially symmetricbaby Skyrmion remains the stable solution. Increasing (cid:15) in the range shown in the figureactually makes the 2-Skyrmion more stable, as can be seen from fig. 4(b). We will now present numerical solutions to the full nonlinear equations of motion. It iscomputationally a very expensive task and because the static baby Skyrmions require only2-dimensional PDEs, we are able to use refined enough square grids for the computations.The numerical method used is the so-called arrested Newton flow on grids with about4096 lattice points and lattice spacing down to about ∼ . − or better. The code is an adaptation of the CUDA Ccode used in ref. [17], which is executed on a GPU cluster.– 24 –n this section, we explore numerical solutions in the topological sector Q = 4 and Q = 2 and include the axially symmetric solutions, which can be directly compared tothe very precise (exact) numerical ODE calculations. This gives another estimate on ournumerical precision. It also confirms in which part of the parameter space, the axiallysymmetric solutions exist and which type of solution has the lowest energy in the giventopological sector.Our first and main aim of this numerical exercise, is to confirm that the lowest energybaby Skyrmion solution – for small but finite 0 < (cid:15) (cid:28) N = 2 babySkyrmions sitting at some unknown separation distance from one another. This is sup-ported by the LO energy correction, which in the axially symmetric case is confirmed tobe the dominant correction to the energy in the small- (cid:15) limit.We begin with two N = 2 baby Skyrmions placed side-by-side, which thus is in thetopological charge sector Q = 4. The numerical results are shown in fig. 5 for representativevalues of (cid:15) . This figure and the following four are organized into four columns displaying thetopological charge density, the baby Skyrmion orientation in O(3) space, the total energydensity and finally the perturbation part of the energy density (cid:15) ( −L + m V ). The figuresin columns 1, 3 and 4 are shown with a color scheme normalized to the content on eachgraph. The figures in column 2 show the pion vector of the baby Skyrmion in the followingsense: φ = 1 corresponds to the vacuum and is shown with white, φ = − φ + i φ ) = 0 is shown with red, arg( φ + i φ ) = π is shown with green and finally arg( φ + i φ ) = π is shown with blue.The two N = 2 baby Skyrmions sitting side-by-side exist for all values of (cid:15) (in ourscanned range), even (cid:15) = 1 and (cid:15) = 0. Starting with (cid:15) = 1, the solution is made of tworing-like baby Skyrmions, that however are prolonged along the axis that joins them. Uponlowering (cid:15) , the greatest effect appears to be on the distribution of the energy and topo-logical charge within each (approximately) axially symmetric N = 2 baby Skyrmion. Theseparation distance between them is affected only mildly. More precisely, the separationdistance between the two N = 2 Skyrmions is seen to increase slightly for (cid:15) decreasingfrom (cid:15) = 1 to (cid:15) ∼ .
379 and below that, the separation distance appears to be constant inthe limit of (cid:15) →
0. This situation is quite different from the compacton case of ref. [17],where the limit of (cid:15) → (cid:15) → N = 2 Skyrmionsare axially symmetric. This is obviously because of their Gaussian tails in the BPS limit(i.e. in the limit of (cid:15) →
0) that makes it impossible to place two perfectly axially symmetricBPS solutions next to each other at any finite distance. Although we are able to lower (cid:15) allthe way down to (cid:15) = 0 with extreme resolution and in the same time taking into accountthe tails up to radii r ∼
15, the two N = 2 baby Skyrmions do not tend to perfectlyaxially symmetric solutions, see that last row of fig. 5 (i.e. for (cid:15) = 0). The BPS solutionshown in the last row of the figure is evidently the BPS solution closest to the limitingsequence of solutions for 0 < (cid:15) (cid:28)
1, but we should point out that the moduli space for– 25 – igure 5 . The Q = 2 + 2 baby Skyrmion for various values of (cid:15) ; this is stable Q = 4 babySkyrmion solution. The columns show the topological charge density, the pion vector orientationusing a color scheme described in the text, the total energy density and the perturbation energydensity (cid:15) ( −L + m V ). In this figure m = 0 . (cid:15) = 0 is infinite dimensional and hence this is just one solution out of infinitely many BPSsolutions that can take any shape as long as the volume is preserved. A slight polarizationof the constituent N = 2 Skyrmions – stretching the baby Skyrmions along the axis thatjoins them – persists in the limit of (cid:15) →
0, see fig. 5. This configuration in the topologicalcharge sector Q = 4, consisting of two N = 2 Skyrmions is the stable solution (i.e. withthe smallest energy per Q ) for all values of (cid:15) studied in this paper.Apart from the global minimum of the energy functional, which for Q = 4 sector For a definition of the volume of the soliton, see refs. [42, 43]. – 26 – igure 6 . The Q = N = 4 baby Skyrmion for various values of (cid:15) . The columns show thetopological charge density, the pion vector orientation using a color scheme described in the text,the total energy density and the perturbation energy density (cid:15) ( −L + m V ). The last row (i.e. for (cid:15) = 0 . m = 0 . we claim is attained by two N = 2 baby Skyrmions with a small separation distance,there exists also local minima – viz. metastable states. For this reason, we have triedwith several good guesses as initial conditions and observed what they flow to under thenumerical minimization of the energy by means of the arrested Newton flow algorithm.As N (cid:63) ∼ . N , it isexpected that the N = 4 axially symmetric baby Skyrmion is just a metastable state, ifit exists at all. An explicit check shows that the N = 4 solution exists for (cid:15) = 1, but it– 27 – igure 7 . The Q = 1 + 1 + 1 + 1 baby Skyrmion with (almost) tetrahedral symmetry, for variousvalues of (cid:15) . The columns show the topological charge density, the pion vector orientation using acolor scheme described in the text, the total energy density and the perturbation energy density (cid:15) ( −L + m V ). In this figure m = 0 . – 28 – igure 8 . The Q = 1 + 1 baby Skyrmion for various values of (cid:15) . The columns show the topologicalcharge density, the pion vector orientation using a color scheme described in the text, the totalenergy density and the perturbation energy density (cid:15) ( −L + m V ). The first row (i.e. for (cid:15) = 0 . N = 2 solution. In this figure m = 0 . decays to a(n almost) tetrahedrally symmetric arrangement of four N = 1 baby Skyrmionsfor (cid:15) ≈ .
4, see fig. 6. This is analogous to what happens in the compacton case [17].This solution is also a metastable state and in particular for large (cid:15) ∼
1, two N = 1 babySkyrmions are unstable and quickly combine into an axially symmetric N = 2 solution, seebelow. However, almost tetrahedrally symmetric configuration enjoys metastability due to– 29 – igure 9 . The Q = N = 2 baby Skyrmion for various values of (cid:15) . The columns show thetopological charge density, the pion vector orientation using a color scheme described in the text,the total energy density and the perturbation energy density (cid:15) ( −L + m V ). In this figure m = 0 . – 30 – = 4 E ǫ BPSLO (two N = 2 )Two numerical N = 2 Numerical tetrahedralNumerical N = 4 Exact N = 4 50 60 70 80 90 100 110 120 130 0.01 0.1 1 Q = 4 E − M B PS ( ) − ǫ M L O ( , . ) ǫ BPSLO (two N = 2 )Two numerical N = 2 Numerical tetrahedralNumerical N = 4 Exact N = 4−4−2 0 2 4 6 8 0.01 0.1 1 Figure 10 . Left panel: Energies of the baby Skyrmion solutions in the Q = 4 sector. The red lineshows the BPS bound, the orange line is the LO correction for two N = 2 baby Skyrmions. Thenumerical solutions are: the two N = 2 Skyrmions side-by-side (green crosses), the tetrahedrallysymmetric baby Skyrmion (blue dotted squares), and the axially symmetric N = 4 Skyrmion(magenta pluses). The numerical N = 4 solution is compared to the exact solution (using ODEcomputations) shown with a black solid line. The right-hand side panel shows the same energies,but with the BPS and the LO correction subtracted off. Around (cid:15) (cid:46) . N = 4 soliton becomes unstable and decays to the tetrahedrally symmetric one – which is justmetastable. Q = 2 E ǫ BPSLO (one N = 2 )Two numerical N = 1 Numerical N = 2 Exact N = 2 25 30 35 40 45 50 55 60 0.01 0.1 1 Q = 2 E − M B PS ( ) − ǫ M L O ( , . ) ǫ BPSLO (one N = 2 )Two numerical N = 1 Numerical N = 2 Exact N = 2−2−1.5−1−0.5 0 0.01 0.1 1 Figure 11 . Left panel: Energies of the baby Skyrmion solutions in the Q = 2 sector. The redline shows the BPS bound, the orange line is the LO correction for one N = 2 baby Skyrmion.The numerical solutions are: the two N = 1 Skyrmions side-by-side (green crosses), and the axiallysymmetric N = 2 Skyrmion (magenta pluses). The numerical N = 2 solution is compared to theexact solution (using ODE computations) shown with a black solid line. The right-hand side panelshows the same energies, but with the BPS and the LO correction subtracted off. Around (cid:15) (cid:38) . N = 1 baby Skyrmions side-by-side become unstable and decay to the axially symmetric N = 2 solution – which is the stable solution. the fact that the center N = 1 Skyrmion is confused by 3 attractive channels and cannotdecide which other N = 1 to combine into an N = 2 with.We increase (cid:15) for the (almost) tetrahedrally symmetric configuration of four N = 1– 31 –kyrmions from (cid:15) = 0 .
379 all the way up to (cid:15) = 1, see fig. 7. The arrangement, althoughmetastable, turns out to remain, even for (cid:15) = 1 and in fact it becomes perfectly tetra-hedrally symmetric for the larger values of (cid:15) due to larger attraction between the four N = 1 constituents. We also decrease (cid:15) all down to (cid:15) = 0 and find that it becomes lesstetrahedrally symmetric for the smaller values of (cid:15) – although we do not know exactlywhy. The attractive force between the N = 1’s become very weak and we can see from thesecond-last row of fig. 7 that it would be fairly easy to knock off the N = 1 baby Skyrmionon the right-hand side of the solution. In that sense, the model will be able to describeweakly bound nuclear clusters. Of course, in two dimensions it makes little sense to tryand compare with actual nuclear physics knowledge. In the last row of the figure, where (cid:15) = 0, the position of the constituents are just moduli and can be moved freely, of course.We could in principle search for further arrangements of four N = 1 baby Skyrmions,but we already know from the leading order calculation of the energy in (cid:15) , that the en-ergetically preferred constituents are the axially symmetric N = 2 baby Skyrmions. Wewill leave such a search for more clusters to future work as they will not be relevant in thesmall- (cid:15) limit.We will now determine the phase diagram in the Q = 4 topological charge sector. Tothis end, we plot the energies of the baby Skyrmions displayed in figs. 5, 6 and 7 in fig. 10(with exception of the (cid:15) = 0 ones, due to the logarithmic scale of the ordinate). Startingwith the least stable solution, the axially symmetric N = 4 baby Skyrmion (magentapluses) only exists for (cid:15) (cid:38) . N = 2 baby Skyrmions side-by-side. Indeed it isthe globally stable solution in the entire range of (cid:15) considered here. We also note that theLO energy is actually an extremely good approximation to that of the solutions made oftwo N = 2 Skyrmions or four N = 1 Skyrmions for (cid:15) (cid:46) . Q = 2 topological charge sector, for which the combinatoricpossibilities obviously are more limited. Of course, from sections 3.1 and 3.3 we know thatthe N = 1 baby Skyrmion has a larger mass than the N = 2 one, both of them with axialsymmetry. Nevertheless for a full understanding of the phase diagram in the Q = 2 sector,we need to study where the two N = 1 baby Skyrmions side-by-side are metastable andwhere they are unstable to collapse into an axially symmetric N = 2 baby Skyrmion, seefig. 8. With the intuition from the compacton case of ref. [17], we expect the two N = 1’sside-by-side to exist only for small values of (cid:15) . The critical value of (cid:15) above which thetwo N = 1’s side-by-side are unstable is roughly (cid:15) crit ≈ .
15. In the small- (cid:15) limit, thetwo N = 1 baby Skyrmions side-by-side appear to be two nearly axially symmetric babySkyrmions vaguely connected in the attractive channel. Because of the Gaussian tail inthe BPS limit, the two N = 1 baby Skyrmions cannot recover axial symmetry in the (cid:15) → Q = 2 sector) axially symmetric N = 2 baby Skyrmions. Nevertheless, we perform thesecalculations as a check of our numerical accuracy. The solutions are shown in fig. 9.– 32 –inally we plot the energies of the two types of solution in the topological charge Q = 2sector in fig. 11. In the left-hand side panel of the figure, we can see that both types ofsolution are very close in energy and are actually described quite well by the LO correctionto the energy for (cid:15) (cid:46) .
1. In the right-hand side panel, we show the same energies butwith the BPS and LO corrections subtracted off. First of all, we can see that the fullnumerical PDE solutions of the N = 2 baby Skyrmions (magenta pluses) match extremelywell with the exact energies from the ODE calculations (black solid line). Second of all,we confirm that the two N = 1 baby Skyrmions sitting side-by-side always have a slightlyhigher energy than the N = 2 one. The energy difference, however, is very small and hencethe metastability is possible even with a small energy barrier between the two kinds ofsolution. BPS Fig. 5 Fig. 7 Q = 4 50.2655 50.2683 50.2685BPS Fig. 8 Fig. 9 Q = 2 25.1327 25.1330 25.1357 Table 1 . Energies of BPS solutions (i.e. with (cid:15) = 0) compared with the theoretical BPS mass forgiven Q . In sharp contrast to the compacton case, we have in this model been able to send (cid:15) all the way to zero, obtaining numerical BPS solutions, 3 of which do not possess axialsymmetry, see the last row of figs. 5, 7, 8 and 9. Since (cid:15) = 0 cannot be plotted on thelogarithmic ordinate of figs. 10 and 11, and they are all of the same value (the BPS mass), wepresent the numerically calculated energies in tab. 1 as a handle on the numerical accuracyof our solutions. For the Q = 4 sector, the accuracy (discrepancy) of the numericallycalculated energy is about 6 × − , whereas for the Q = 2 sector it is 1 × − and1 × − , for the two solutions. E − E ǫ fit (a) Q = 4 E − E ǫ fit (b) Q = 2 Figure 12 . Binding energies for (a) the Q = 2 + 2 bound state solution of fig. 5 and (b) of the Q = 1 + 1 bound state solution of fig. 8. The fits are shown in the text. Here m = 0 . – 33 –e evaluate the binding energy BE Q ≡ QE − E Q , (4.1)of the Q = 2 + 2 bound state solution of fig. 5 in fig. 12(a) and of the Q = 1 + 1 boundstate solution of fig. 8 in fig. 12(b) and the fits displayed on the figures are given by BE (cid:39) . (cid:15) − . (cid:15) + 208 . (cid:15) − . (cid:15) , (4.2) BE (cid:39) . (cid:15) − . (cid:15) + 170 . (cid:15) − . (cid:15) , (4.3)and these binding energies are fitted for solutions with m = 0 .
5. The binding energiesare tiny and the slight oscillation of the numerically evaluated binding energies around thefit shows that we are close to the edge of our numerical precision. In principle, we couldalso numerically evaluate the energy of the bond between two axially symmetric N = 2Skyrmions, that is 2 E − E , but this quantity turns out to be at the level of 10 − andbeyond our numerical precision, hence we do not attempt at plotting it.We notice that there seems to be a linear contribution to the binding energy, in con-tradistinction with the compacton case of ref. [17]. We will discuss this point in latersections. In the case of the compactons, discussed in ref. [17], the restricted harmonic conditionwas not sufficient to fix the right background solution in the case of two interacting babySkyrmions. In particular, no restrictions in the choice of the relative orientation or sepa-ration distance between the non-overlapping compactons emerge from such a condition.In this paper, we have turned on a more physical potential and now have long-rangeforces. This means that at any separation distance, two baby Skyrmions feel each otherand the force between them is dependent on the relative orientation, as was shown byPiette-Schroers-Zakrzewski [20]. Since we are only interested in baby Skyrmion boundstates, among all the possible GRH maps we must choose among those that are in theattractive channel. This uniquely fixes the relative orientation and hence the choice of theGRH map. In the following, we briefly review the calculation of ref. [20], adapting it tothe case of our Lagrangian (2.1).Let u and v be baby Skyrmion solutions of charge N and M and let φ u and φ v betheir representation in 3-vector coordinates. Each field φ is a solution of the full equationsof motion: ∂ i j i = (cid:15)m n × φ + ( n × φ )(1 − n · φ ) , (5.1)where j i = (cid:15) φ × ∂ i φ + ∂ j φ ( ∂ j φ · φ × ∂ i φ ) . (5.2)When the two baby Skyrmions are well separated, the composite solution of total charge Q = N + M can be written as φ w ≡ φ u + v . In order to evaluate the interaction potential V between the two solitons, it is necessary to decompose the total energy in the form– 34 – [ φ w ] = E [ φ u ] + E [ φ v ] + V . To this end, it is useful to separate the coordinate space R into two regions such that φ u ≈ n is close to the vacuum n in region 2 and φ v ≈ n is closeto the vacuum n in region 1. We can this decompose the field into the vacuum n and aninfinitesimal correction: φ u ≈ n + δφ u + O ( δφ u · δφ u ) , (5.3)where δφ u · n = 0. Since φ u solves the Euler-Lagrange equation (5.1), δφ u satisfies thelinearized equation (∆ − m ) δφ u = 0 , (5.4)and we have that in region 1: φ w ≈ φ u + (cid:15) v × φ u + 12 (cid:15) v × ( (cid:15) v × φ u ) , (5.5)where (cid:15) v is linear in δφ v . The expansion for the field φ w in terms of δφ u in the region 1 isanalogous. The asymptotic result (5.4) is independent of (cid:15) as we expected and equivalentto that of ref. [20]. Once we have the form of φ w in the regions 1 and 2 (see eq. (5.5)), itis possible to evaluate the energy E [ φ w ] as E [ φ w ] ≈ (cid:90) d x E ( φ w ) + (cid:90) d x E ( φ w ) ≈ E [ φ u ] + E [ φ v ]+ (cid:90) d x (cid:2) j ui · ∂ i (cid:15) v + (cid:15)m ( (cid:15) v · n × φ u ) + ( (cid:15) v · n × φ u )(1 − n · φ u ) (cid:3) + (cid:90) d x (cid:2) j vi · ∂ i (cid:15) u + (cid:15)m ( (cid:15) u · n × φ v ) + ( (cid:15) u · n × φ v )(1 − n · φ v ) (cid:3) , (5.6)where (cid:15) u is linear in δφ u and is defined by (cid:15) u = 12 φ v × (cid:0) (1 + φ v · n ) δφ u − ( φ v · δφ u ) n (cid:1) , (5.7)and (cid:15) v is equivalent to eq. (5.7) with the superscripts u and v exchanged. In the end,using the equations of motion (5.1) and Gauss’s law, we can finally write the long-rangeinteracting potential V as V = (cid:15) (cid:90) Γ ( δφ v · ∂ i δφ u − δφ u · ∂ i δφ v ) d S i , (5.8)where Γ is a curve without self-intersections, separating region 1 from region 2 and d S i = ε ij ˙ γ j d t for any parametrization γ ( t ) of Γ. Given eq. (5.4), the form of this potential isequivalent to the one obtained in ref. [20] albeit with an overall factor of (cid:15) .With the result (5.8), it is now possible to calculate the long-range interaction fordifferent multisoliton configurations to determine the Skyrmions’ relative orientation max-imizing their attraction. For the multisoliton case of charge Q = 1 + 1, the potential takesthe form V ∝ (cid:15) cos( α − β ) e − m d √ m d , (5.9)– 35 –here d is the relative distance, α and β are the two respective phases (orientations) ofthe baby Skyrmions and we have assumed that δφ ∼ O ( (cid:15) ), see the discussion in sec. 3.2.From this expression, we recognize that the maximally attractive channel is obtained for α − β = π , i.e. when the two baby Skyrmions have opposite orientation. Following thisindication, we choose the GRH map of this topological sector with the solitons oppositeoriented. In the case of the baby Skyrmion pair of charge Q = 2 + 2, the asymptoticpotential is V ∝ − (cid:15) cos( α − β ) e − m d √ m d , (5.10)in which case the maximum of the attraction requires the solitons to have the same orien-tation, i.e. α = β and we have again assumed that δφ ∼ O ( (cid:15) ). Therefore, the proper GRHmap in this case is chosen to have the solitons equally oriented. Our choices, based hereon analytical considerations, are confirmed by the numerical calculations in figs. 5 and 8for small values of (cid:15) . xy } aRO r β Figure 13 . Sketch of the setup for calculating the binding energy between two axially symmetric N -Skyrmions. The radius containing most of the energy of the baby Skyrmion is r β R with asuitable value for β (i.e. the fraction of the BPS energy density contained within the radius); herewe shall use r . (cid:39) . √ N . The separation between the two baby Skyrmions is 2 a and thegluing boundary conditions in the problem are imposed at x = x = 0 (i.e. the y -axis in the figure). In the previous section, we have seen that two N -Skyrmions in the attractive channelattract each other at asymptotic distances. This attraction presumably breaks down dueto nonlinearities at some finite distance and for (cid:15) > N -Skyrmions separated by a finite distance 2 a , see fig. 13.In order to calculate the binding energies using the perturbative method of the (cid:15) -expansion that we have put forward in ref. [17] and this paper, we place two N -Skyrmionsside-by-side in the attractive channel, viz. with the pion field matching on the gluing line( x = 0), see fig. 13. In order to be able to glue the two baby Skyrmions together, we– 36 –eed appropriate boundary conditions at x = 0 (see fig. 13), which we shall call gluingconditions: ∂ x φ (0 , y ) = 0 ,φ (0 , y ) = 0 , (6.1) ∂ x φ (0 , y ) = 0 . The reason for the odd condition on φ , is that even boundary conditions on all φ a willturn the mirror Skyrmion into an anti-Skyrmion for odd N , which is not what we want toglue the solution with. The relatively simple-looking conditions above, become nonlinearRobin-type boundary conditions for the fluctuation field ∆ , once we use that φ = ϕ + δφ and δφ from eq. (3.14): ∂ x (cid:18) ϕ a + ε abc ∆ b ϕ c + 12 ( ϕ · ∆ )∆ a −
12 ( ∆ · ∆ ) ϕ a (cid:19) = 0 , a = 1 , , (6.2) ϕ a + ε abc ∆ b ϕ c + 12 ( ϕ · ∆ )∆ a −
12 ( ∆ · ∆ ) ϕ a = 0 , a = 2 . (6.3)It turns out to be a rather tricky boundary condition to implement numerically. For thisreason we will use the Ansatz (3.42) for ∆ which reduces the above nonlinear Robin gluingconditions to ∂ x (cid:18) δf cos f cos N θ + 12 sin f cos N θ (2 − δf − δθ ) − δθ sin N θ (cid:19) = 0 , (6.4) δθ cos N θ + δf cos f sin N θ + 12 sin f sin N θ (2 − δf − δθ ) = 0 . (6.5)We will now use a finite difference approximation for the x -derivative of δf and δθ : δf x = δf − − δf − + 3 δf h x , δθ x = δθ − − δθ − + 3 δθ h x , δf = δf , δθ = δθ , (6.6)where the subscript 0 refers to the x = 0 lattice point on the gluing line, − δf = δf + δδf , δθ = δθ + δδθ , (6.7)inserting these into the algebraic conditions and linearizing with respect to δδf and δδθ .This yields expressions for δδf and δδθ which are not particularly illuminating, so we willnot display them here. In principle the method is simple, δf is the previous value of δf and δf is updated by adding the solution for δδf to δf at each step of the algorithm (andsimilarly for δδθ ). Notice that δδf vanishes when δf is a solution to the full nonlinearRobin type gluing condition. Unfortunately, this simplest form of Newton iteration does Here for simplicity we use only a second-order formula, although in the numerical code we use a fourth-order formula. – 37 –ot have particularly good convergence properties [44]. For this reason we had to implementa line search algorithm following ref. [44], which updates δf and δθ using δf = δf + tδδf , δθ = δθ + tδδθ , t ∈ (0 , , (6.8)where t is a parameter that should be optimized for each step in the iteration. Note that t = 1 is just the simplest version of Newton iteration. We implement a rather crude linesearch algorithm that tries out 20 points of t in its interval and refines the search one time.It turns out that often the optimal value of t is around t ∼ . t ∼ .
6. Although thisalgorithm is rather computationally expensive, ref. [44] showed the line search is one ofthe cheapest algorithms with improved convergence properties for this type of algebraicRiccati equation.
Figure 14 . Numerical solution for ∆ on the background of two N = 2 baby Skyrmions placednext to each other with separation distance 2 a for (cid:15) = 0 .
01. The gluing conditions described in thetext are imposed in the center ( x = 0) of each panel. The panels show (from left to right) ∆ , ∆ ,∆ , E BPS+LO , E NLO and E N LO . The panels are cropped so as to render the content clearly visibleand do not represent the borders of the computation. We now turn to solving the coupled PDEs (3.20) with the boundary conditionslim | x |→∞ ∆ = , ∆ ( x ) = , (6.9)for all x being centers of N -Skyrmions, as well as the gluing conditions described above.The “boundary condition” at all x is to prevent the perturbation from trying to unwrap– 38 – igure 15 . Numerical solution for ∆ on the background of two N = 2 baby Skyrmions placednext to each other with separation distance 2 a for (cid:15) = 0 . x = 0) of each panel. The panels show (from left to right) ∆ ,∆ , ∆ , E BPS+LO , E NLO and E N LO . The panels are cropped so as to render the content clearlyvisible and do not represent the borders of the computation. the soliton solution. Since the calculation is computationally very expensive, we haveimplemented the code in CUDA C and run it on an NVIDIA GPU cluster.Figs. 14 and 15 show the solution for the perturbation, ∆ , on the background of two N = 2 baby Skyrmions sitting side-by-side, for (cid:15) = 0 . , . , ∆ , ∆ , E BPS+LO , E NLO and E N LO , respectively,while the rows correspond to increasing values of (half of) the separation distance a . TheBPS+LO, NLO and N LO energy densities are defined as E BPS+LO = −L [ ϕ ] + V ( ϕ ) + (cid:15) (cid:0) −L [ ϕ ] + m V ( ϕ ) (cid:1) , (6.10) E NLO = E perturb2 = −L perturb2 [ ϕ , ∆ ] , (6.11) E N LO = E perturb2 + E perturb3 = −L perturb2 [ ϕ , ∆ ] − L perturb3 [ ϕ , ∆ ] . (6.12)It is clearly visible from the first two columns of the figures, that each baby Skyrmion isan N = 2 Skyrmion, since the angular modulation of the perturbation field ∆ possessesthe characteristic sin(2 θ ) and cos(2 θ ) behavior.– 39 –he gluing conditions (6.1) reflect the fact that φ , are even in x , whereas φ is odd.This fact translates to ∆ being even and ∆ , being odd in x . For this reason, we can seediscontinuities in the field in the first and third columns of figs. 14 and 15. Of course, dueto the symmetry of the problem, the calculation has only been carried out on the left-handside of the panels (i.e. for x ≤ x ≥
0) are simply (flipped)mirrors of the solution.For large separation distances, 2 a , the solution for the perturbation field ∆ is almostidentical to the axially symmetric solution (3.57) presented in fig. 2. This is because whenthe background solution ϕ ≈ ϕ vac = (0 , , a , on the other hand, the background soliton solution is not yet that close to thevacuum point, and furthermore, the intrinsic perturbation solution around the backgroundsoliton is also far from vanishing.The above-mentioned discontinuities occur when ϕ is nonvanishing at x = 0, becausethe second condition of eq. (6.1) demands that φ = 0 and thus the perturbation mustaccount for this by compensating with ∆ , (cid:54) = 0. Since ∆ , are odd fields in x , the gluingconditions necessarily induce discontinuities in the perturbation at the gluing line ( x = 0).The energy density is seen from the figures (14 and 15) to be concentrated near the gluingline for small a , whereas for large a the energy density is just equal to that of the axiallysymmetric energy. Finally, we can see that the perturbation field becomes less dominantnear the Skyrmion solution for (cid:15) = 0 .
01 compared with the (cid:15) = 0 . (cid:15) tends to zero.In order to calculate the binding energy of two charge- N baby Skyrmions, we first needto determine their optimal separation distance; that is the distance where the total energyof the boundary state is at a local or global minimum and then read off the energy at thatpoint.In an attempt to understand which separation distance (2 a ) is preferred by two N = 2baby Skyrmions in the attractive channel, we calculate the energy density as a function of a for various values of (cid:15) . The result is shown for (cid:15) = 0 . , . , . , . (cid:15) expansion. At the leading order (LO),there is no fluctuation field and the BPS background solution ϕ is used to calculate theenergy from the Lagrangian (2.1), including the BPS energy. Of course, we have stitchedtogether two baby Skyrmions side-by-side, so the LO energy is evaluated without overlap:this means we integrate the energy over the background BPS solution up to the gluing line( x = 0) and then multiply by two. Clearly, this order is flawed in the following sense: Thefurther we push the two baby Skyrmions together, the less the LO energy is. This cannot bethe correct physical picture. For that, the perturbation field ∆ is necessary. In particular,the gluing conditions are crucial as should become clear momentarily. For completeness,we show both the energy calculated to order (cid:15) (NLO) and to order (cid:15) (N LO). As we cansee from the figure, what happens is that the LO energy goes down as a decreases, and sodoes the NLO energy, but the N LO energy goes up. This is simply the gluing conditiontwitching the field at the gluing line and building up energy localized around the gluing– 40 – = 0.01 E a N LO semianalyticNLO semianalyticLO semianalytic (a) ǫ = 0.0207 E a N LO semianalyticNLO semianalyticLO semianalytic (b) ǫ = 0.0428 E a N LO semianalyticNLO semianalyticLO semianalytic (c) ǫ = 0.0886 E a N LO semianalyticNLO semianalyticLO semianalytic (d)
Figure 16 . The LO (blue squares), NLO (green crosses) and N LO corrections (red crosses) to theenergy of two N = 2 baby Skyrmions as functions of the separation distance 2 a , see fig. 13. Thepanels show different values of (cid:15) : (a) (cid:15) = 0 .
01, (b) (cid:15) = 0 . (cid:15) = 0 . (cid:15) = 0 . x = 0) (and multiplying by 2). A large black square shows the minimum of theN LO energy in panels (c) and (d). The large gray and black squares on the N LO curves of panels(a) and (c) correspond to the solutions depicted in figs. 14 and 15, respectively. – 41 –ondition, see also figs. 14 and 15. x } aR r β R r β } a (a) xaR r (cid:0) R r (cid:0) R compacton R compacton a (b) Figure 17 . (a) A nontrivial nonlinear GRH map solution with two charge- N baby Skyrmionsnext to each other. The cores of the baby Skyrmions (with r < r β are only slightly deformed, butthe tail is nontrivially connecting the two solitons with very short separation distance 2 a . (b) Atrivial (linear) GRH map solution with two charge- N baby Skyrmions next to each other. Theyare separated by the would-be compacton radius of a limit described in the text. Here the tails areeasily and linearly connected, but the separation distance is much larger. First of all, we notice that there is no minimum of the energy as function of a for (cid:15) (cid:46) .
03. Second of all, the minimum appears at a very large separation distance, comparedto those observed in the full brute-force numerical computations made in sec. 4. We willnow claim we are actually finding a bound state solution with the semianalytic calculationusing the (cid:15) expansion, it is just not the same solution that we found by full numerical brute-force computations in sec. 4. Instead, the solution that we seem to be pushed towards inthis perturbative scheme of the (cid:15) expansion, is another solution of the form that we willexplain now. The full numerical solutions found in sec. 4 are nontrivial nonlinear GRHmaps, whereas the perturbative solutions that we find in this section are trivial GRH maps,see fig. 17. The sense in which the latter are “trivial” GRH maps is the following. If weconsider taking the limit (cid:15) → √ (cid:15)m =: ˜ m held fixed. What this limit amounts to,is to kill off the kinetic term ( L ) while keeping the pion mass term V . In this limit, thereis a BPS solution with compacton radius R compacton given by eq. (A.10). The second andtrivial solution is flowing to this trivial GRH map solution for small (cid:15) , with R compacton ofeq. (A.10).In order to provide some evidence for this wild claim of a second “trivial” and “lin-ear” GRH map solution, we plot a corresponding to the minimum of the N LO energy asa function of (cid:15) in fig. 18. The figure compares the N LO calculation (red crosses) withthe full numerical brute-force computation of fig. 5 (green crosses) and the match is quitepoor as anticipated above by eye-balling the numbers. However, the figure also shows thecompacton radius of eq. (A.10) as a brown dashed line and the match with the N LOresults for intermediate values of (cid:15) is quite reasonable. We should mention that the semi-analytic method we are using is failing to find a minimum for (cid:15) (cid:46) . = 4 a ǫ N LOnumerical(A.10)
Figure 18 . Separation distance 2 a between two N = 2 baby Skyrmions as a function of (cid:15) . Thesemianalytic method to order N LO is shown with red crosses and the full numerical brute-forcecomputation of fig. 5 is shown with green crosses. For comparison, the compacton radius R compacton of eq. (A.10) is shown with a brown dashed line. computation.A comment in store is about the matching of the N LO calculation of the energy forthe bound state and the compacton radius mentioned above. Of course, we cannot expectfull agreement between these two quantities, because our kinetic term is not eliminated –it has coefficient (cid:15) (which however is very small). The point is that we consider similarbehavior and order of magnitude of the two quantities as evidence that we are flowing tothe trivial GRH map solution using the semianalytic approach ( (cid:15) expansion) and not tothe same solution found in sec. 4.So how come we do not find the nontrivial GRH map solution (with the semianalyticapproach), that we found numerically in sec. 4? We note from fig. 16 that the semianalyticapproach of calculating the bound state (and thereby the binding energy) did not find aminimum of the energy at neither NLO nor N LO around a ∼ .
8, for any value of (cid:15) . Weattribute this failure of the semianalytic approach to the fact that we have linearized theequation of motion for the perturbation field ∆ .Since we do not find the same types of GRH map solutions with the semianalyticmethod (this section) and with the full numerical brute-force computations (sec. 4), itmakes no sense comparing them. In fact, we cannot even be sure that the binding energiesof these two types of solution scale in the same way with (cid:15) .We now turn to the bound state of two N = 1 baby Skyrmions, which are onlymetastable (see fig. 1), since the axially symmetric N = 2 solution has lower energy thanthe bound state of two N = 1 Skyrmions. The bound state does, nevertheless, exist for (cid:15) (cid:46) .
15, see fig. 8. Figs. 19 and 20 display the solution for the perturbation field, ∆ ,on the background of two N = 1 baby Skyrmions sitting side-by-side for (cid:15) = 0 .
01 and (cid:15) = 0 . , ∆ , ∆ , E BPS+LO , E NLO and E N LO , respectively, while the rows correspond again to increasingvalues of a , just like in the two N = 2 bound state, described above. The gluing conditions– 43 – igure 19 . Numerical solution for ∆ on the background of two N = 1 baby Skyrmions placednext to each other with separation distance 2 a for (cid:15) = 0 .
01. The gluing conditions described in thetext are imposed in the center ( x = 0) of each panel. The panels show (from left to right) ∆ , ∆ ,∆ , E BPS+LO , E NLO and E N LO . The panels are cropped so as to render the content clearly visibleand do not represent the borders of the computation. are similar in this case as in the previous case, making the fields ∆ , discontinuous. It isclear from the oscillatory behavior of the perturbation fields with the angle around eachbaby Skyrmion, that this is indeed an N = 1 constituent, since it behaves like cos θ or sin θ .One thing is clear already from glancing at the figures: Although we are able to push thebaby Skyrmions slightly closer to each other, compared with the previous case, it is stillnot close enough to be compared with the solution of fig. 8 in sec. 4.In order to attempt at calculating the binding energy of the bound state in this case,we again perform a large number of PDE calculations in this semianalytic approach formany values of the separation distance 2 a , and (cid:15) = 0 . , . , . , . LO for all values of (cid:15) . The minimum, however, againappears at quite large values of a and we thus again think that the semianalytic approachis finding the trivial GRH map solution, mentioned above.In fig. 22 we compare (half of) the separation distance a of the minimum of the energyat N LO of fig. 21 as a function of (cid:15) (red crosses) with that extracted from the full numerical– 44 – igure 20 . Numerical solution for ∆ on the background of two N = 1 baby Skyrmions placednext to each other with separation distance 2 a for (cid:15) = 0 . x = 0) of each panel. The panels show (from left to right) ∆ ,∆ , ∆ , E BPS+LO , E NLO and E N LO . The panels are cropped so as to render the content clearlyvisible and do not represent the borders of the computation. brute-force calculations of fig. 8 (green crosses) and the match is quite bad. For startersis the N LO value for a positive, whereas the numerically extracted one is negative. Wewill now compare the N LO result of a to the would-be compacton radius of eq. (A.10),which corresponds to the limit of eliminating L but keeping V with coefficient (cid:15)m =: ˜ m .The match is similar to the previous case of two N = 2 baby Skyrmions. Of course, asmentioned above, the would-be compacton radius assumes the absence of the kinetic term( L ), which is a bold approximation, even for the range of values of (cid:15) shown in fig. 22. Wethink that also for this bound state, we have found the trivial GRH map solution usingthe semianalytic approach. The missing minimum for (cid:15) > . (cid:15) . Nevertheless, looking at fig. 21(d), it isclear that the minimum of the N LO energy is nearby and will continue the curve drawnin fig. 22.Again in this bound state case, we will not compare the binding energy found herewith that calculated in sec. 4, because we believe they are in fact two different GRH mapsolutions. – 45 – = 0.01 E a N LO semianalyticNLO semianalyticLO semianalytic (a) ǫ = 0.0207 E a N LO semianalyticNLO semianalyticLO semianalytic (b) ǫ = 0.0428 E a N LO semianalyticNLO semianalyticLO semianalytic (c) ǫ = 0.0886 E a N LO semianalyticNLO semianalyticLO semianalytic (d)
Figure 21 . The LO (blue squares), NLO (green crosses) and N LO corrections (red crosses) to theenergy of two N = 1 baby Skyrmions as functions of the separation distance 2 a , see fig. 13. Thepanels show different values of (cid:15) : (a) (cid:15) = 0 .
01, (b) (cid:15) = 0 . (cid:15) = 0 . (cid:15) = 0 . x = 0) (and multiplying by 2). A large black square shows the minimum ofthe N LO energy. The large gray and black squares on the N LO curves of panels (a) and (c)correspond to the solutions depicted in figs. 19 and 20, respectively. – 46 – = 2 a ǫ N LOnumerical(A.10) −1.5−1−0.5 0 0.5 1 1.5 2 2.5 3 0.01 0.1
Figure 22 . Separation distance 2 a between two N = 1 baby Skyrmions as a function of (cid:15) . Thesemianalytic method to order N LO is shown with red crosses and the full numerical brute-forcecomputation of fig. 8 is shown with green crosses. For comparison, the compacton radius R compacton of eq. (A.10) is shown with a brown dashed line. E − E ǫ fit (a) Q = 4 E − E ǫ fit (b) Q = 2 Figure 23 . Binding energies for (a) the Q = 2 + 2 bound state solution splitting up into two N = 2Skyrmions and (b) of the Q = 1 + 1 bound state solution splitting up into two N = 1 Skyrmions.The fits are shown in the text. Here m = 0 . – 47 –t will nevertheless prove useful to plot the binding energies of the bound state solutionscalculated perturbatively in this section. Fig. 23 shows the binding energies of the Q = 4and Q = 2 sectors. The fits are given by BE (cid:39) − . (cid:15) + 0 . (cid:15) − . (cid:15) + 62 . (cid:15) , (6.13) BE (cid:39) . (cid:15) + 2 . (cid:15) − . (cid:15) + 142 . (cid:15) , (6.14)and these binding energies are fitted for solutions with m = 0 . BE NQ ≡ QN E N − E Q . (6.15)The binding energies are even more tiny than those calculated numerically in sec. 4. Wenotice that for these “trivial” GRH solutions found in the perturbative scheme, the bindingenergies are smaller than the nontrivial ones found in sec. 4. This suggests that thesesolutions are only metastable.We recall that the binding energy of splitting two N = 2 Skyrmions up into their N = 2constituents was tiny and too small to calculate accurately in sec. 4. Indeed, it turns outto be of the same order, i.e. O (10 − ), also in the perturbative scheme of the (cid:15) expansion.Nevertheless, the binding energy of the two N = 1 Skyrmions is found to be an orderof magnitude smaller than what was calculated numerically in sec. 4. This supports ourclaim that the bound states found in this section with the perturbative method are differentsolutions, which we call “trivial” GRH map solutions, compared with the nontrivial onesof sec. 4.A comment in store is about the order of (cid:15) that the binding energy is captured by.Both the fits (6.13)-(6.14) and (4.2)-(4.3) seem to have a linear term in (cid:15) . This is not incontradiction with the calculation of sec. 5, where we stated that the interaction potentialis of order (cid:15) . The asymptotic region, where we can calculate the Skyrmion-Skyrmioninteraction potential with a linear approximation yields a contribution to the binding energyof order (cid:15) . Nevertheless, there may well be contributions to the binding energy linear in (cid:15) at short distances, where the linear approximation breaks down, see eq. (3.33). We endthis discussion with recalling that compactons had a binding energy of order (cid:15) [17]. In this paper, we have studied the near-BPS regime of the baby Skyrme model with aphysical pion mass (i.e. one that does not diverge in the BPS limit) and a potential that isnecessary for the BPS solution. The latter potential is chosen such that the BPS solutionshave Gaussian tails (i.e. e − r ), which means that we avoid a typical complication of theBPS solutions being compactons – as was the case studied in ref. [17]. The problem withthe compactons, is that the fields deviate from the vacuum only on a compact region ofspace and the derivatives exhibit a jump from a negative value to zero at the border of saidregion. The true solution in the near-BPS regime does not possess such a discontinuity inthe first derivative of the fields and therefore we had invented in ref. [17] a cusp condition– 48 –o make the total field smooth, but inducing a counter-cusp to the perturbation field. Thishas all been avoided in the present paper. The motivation as stated above, is to get apion mass that is of the same order as the kinetic term and both are proportional to asmall parameter, (cid:15) , hence giving rise to physical pions in the theory and also to avoid thetechnical difficulties coming along with the mentioned cusps. The leading order correctionto the energy in (cid:15) of an axially symmetric charge- N baby Skyrmion is also in this caseobtained by evaluating the Dirichlet (kinetic) energy of the BPS solution, which luckily isfinite. This correction is linear in (cid:15) . Bound states, however, cannot be studied unless wego to a higher order in (cid:15) and introduce a perturbation field along with suitable boundaryconditions. This is because the tails of the BPS solutions fall off to spatial infinity andwould just overlap each other if nothing extra is introduced in the theory. The key to theperturbation field was again a transverse field used in ref. [17], inspired by the work inref. [20]. The perturbation field in this paper does not require a so-called cusp condition,mentioned above, which is a relieving simplification. First we studied the energies of thebaby Skyrmions in the axially symmetric case up to order O ( (cid:15) ) which we call N LOand find excellent agreement between the perturbative scheme and full numerical ODEcalculations.Then we move on to performing large-scale full numerical brute-force computationsof Skyrmions with charges Q = 4 and Q = 2, which include the simplest bound statesof the most stable baby Skyrmions in the theory. For the Q = 4 sector, the most stablesolution is a bound state of two N = 2 approximately axially symmetric baby Skyrmionsquite well separated, but still having their tales intertwined in a bound state. The axiallysymmetric Q = N = 4 solution is unstable for small (cid:15) , but another metastable solutioncomposed of four N = 1 baby Skyrmion in a nearly tetrahedral arrangement was found.Interestingly, this solution was completely tetrahedrally symmetric in the compacton casestudied in ref. [17], whereas in this case two of the solitons move close together and repelthe other two, one more than the other. This is probably explained by a quadrupole force,which is incompatible with the tetrahedral symmetry. In the Q = 2 topological sector, themost stable solution is simply the Q = N = 2 axially symmetric solution, but for small (cid:15) ,there is also a bound state of two N = 1 baby Skyrmions. Finally, we would like to stressthat we have been able to find numerical BPS solutions for all the mentioned cases with (cid:15) = 0 set strictly to zero. That is, these solutions are nontrivial BPS solutions for whichwe do not yet know a suitable analytic Ansatz.The last part of the paper is an attempt at calculating the binding energies perturba-tively in our semianalytic (cid:15) expansion scheme, which was very successful for the compactoncase of ref. [17]. Unfortunately, it turns out that this scheme is finding another type of GRHmap solution, which is in some sense trivial and more similar to the case of two compactonsplaced near each other. Because the two solutions (i.e. the trivial perturbative GRH mapsolutions and the nontrivial full numerical solutions mentioned above) show evidence ofbeing quite different in nature, we have not attempted at comparing their (binding) en-ergies. We have, however, fitted the binding energies in both the perturbative scheme ofthe (cid:15) expansion as well as in the full numerical computations. These fits also reveal thatwe are finding different solutions with the two different methods. Interestingly, it appears– 49 –hat the binding energy in this model, where the BPS solutions have a Gaussian tail, has aleading term which is linear in (cid:15) – in stark contradistinction to the case of the compactonsof ref. [17], where the binding energy appears only at the order O ( (cid:15) ).Although the compactons require painful cusp conditions for allowing one to studytheir near-BPS limit, the solitons with tails turn out to be even more difficult and canprobably be well understood only by using nonlinear techniques or simply full numericalcomputations.It would be interesting to consider other potentials and as we found in this paper,compared with ref. [17], everything depends strongly on the choice of the potential. Poten-tials in the BPS sector, giving rise to a power-law tail [24], would probably provide verydifferent properties with respect to the (cid:15) expansion as well as to the solutions in general.We will leave the investigation of such cases to future studies. Acknowledgments
S. B. G. thanks the Outstanding Talent Program of Henan University for partial support.The work of S. B. G. is supported by the National Natural Science Foundation of China(Grants No. 11675223 and No. 12071111). The work of M. B. and S. B. is supported bythe INFN special project grant “GAST (Gauge and String Theories)”.
A BPS solution with potential ˜ m V + V Writing the BPS equation for the Skyrme term, but including both ˜ m V + V instead ofonly V (as in eq. (2.13)), we have Q = − φ · ∂ φ × ∂ φ = ∓ (cid:113) (1 − φ ) + 2 ˜ m (1 − φ ) . (A.1)Switching to stereographic coordinates yields (cid:15) ij ∂ i ω∂ j ¯ ω (1 + | ω | ) = ∓ i | ω | (cid:112) ˜ m + (1 + ˜ m ) | ω | | ω | . (A.2)Inserting the axially symmetric Ansatz ω = e i Nθ ζ ( r ), we obtain ∂ r ζr = − (1 + ζ ) (cid:112) ˜ m + (1 + ˜ m ) ζ N , (A.3)where we have chosen the lower sign. Switching to the γ (2.17) and y variables (2.18), thedifferential equation reduces to d γ d y = − (cid:112) γ ( ˜ m + γ ) N . (A.4)Neatly, this differential equation reduces exactly to eq. (2.19) in the limit of ˜ m → (cid:18) √ γ + (cid:113) γ + ˜ m (cid:19) = − yN − κ, (A.5)– 50 –here κ is an integration constant. Solving for γ yields γ = 14 e − yN − κ (cid:16) − ˜ m e yN + κ (cid:17) . (A.6)The boundary condition corresponding to ω being singular at y → γ = 1 at y = 0,which determines κ : κ = log 2 + ˜ m − (cid:112) m ˜ m . (A.7)Inserting the integration constant κ into the solution, we find γ = e − yN (cid:16) ˜ m − e yN (cid:16) m − (cid:112) − ˜ m (cid:17)(cid:17) (cid:16) m − (cid:112) m (cid:17) . (A.8)In terms of ζ and r , we have ζ = (cid:118)(cid:117)(cid:117)(cid:116)
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