AA higher-order Skyrme model
Sven Bjarke Gudnason , Muneto Nitta Institute of Modern Physics, Chinese Academy of Sciences, Lanzhou 730000, China Department of Physics, and Research and Education Center for Natural Sciences, Keio Univer-sity, Hiyoshi 4-1-1, Yokohama, Kanagawa 223-8521, Japan
E-mail: bjarke(at)impcas.ac.cn , nitta(at)phys-h.keio.ac.jp Abstract:
We propose a higher-order Skyrme model with derivative terms of eighth,tenth and twelfth order. Our construction yields simple and easy-to-interpret higher-orderLagrangians. We first show that a Skyrmion with higher-order terms proposed by Marleauhas an instability in the form of a baby-Skyrmion string, while the static energies of our con-struction are positive definite, implying stability against time-independent perturbations.However, we also find that the Hamiltonians of our construction possess two kinds of dy-namical instabilities, which may indicate the instability with respect to time-dependentperturbations. Different from the well-known Ostrogradsky instability, the instabilitiesthat we find are intrinsically of nonlinear nature and also due to the fact that even powersof the inverse metric gives a ghost-like higher-order kinetic-like term. The vacuum stateis, however, stable. Finally, we show that at sufficiently low energies, our Hamiltonians inthe simplest cases, are stable against time-dependent perturbations.
Keywords:
Skyrmions, higher-derivative terms, nonlinear instabilities a r X i v : . [ h e p - t h ] S e p ontents The Skyrme model [1, 2] is generally believed to describe low-energy QCD at large N c [3, 4]. It has also been derived directly from the QCD Lagrangian by means of partialbosonization [5]. As well known, it is not possible to perform a full bosonization in 3+1dimensions and hence the latter reference is bosonizing only the phases of the fermions.The Skyrme model has also been derived in the Sakai-Sugimoto model [6] by consideringthe effective action for the zero mode. All these derivations of the Skyrme model includea kinetic term as well as the Skyrme term, which is fourth order in derivatives. Skyrmeintroduced the term [1, 2] in order to stabilize the soliton – the Skyrmion – from collapse,– 1 –s otherwise is unavoidable due to Derrick’s theorem [7]. However, higher-order derivativecorrections higher than fourth order are generally expected.As expected in QCD and explicitly shown in the Sakai-Sugimoto model [6], an infinitetower of vector mesons exist as one goes up in energy scales. For each of these massivevector mesons, one can obtain effective operators in a pure pion theory by integrating outthe massive mesons. The interaction terms between the pions and the mesons yield newlow-energy effective operators. The first higher-derivative correction to the Skyrme modelis expected to be a sixth-order derivative term, see e.g. [8–20]. Physically, it corresponds tointegrating out the ω -meson [8, 9]; this can be seen from the phenomenological Lagrangianwith the interaction describing the decay ω → π + π − π .The sixth-order term – which we shall call the BPS-Skyrme term – recently caughtinterest due to its BPS properties when it is paired with a suitable potential [14–20].Here BPS simply means that the energy is proportional to the topological number – theSkyrmion number, B – of the model. This is a desired feature in nuclear physics wherebinding energies are very small.In principle, we expect infinitely many higher-derivative terms in the low-energy effec-tive action. However, as each term is larger in canonical dimension, it necessarily has tobe accompanied by a dimensional constant to the same power minus four. That constantis typically proportional to the mass of the state that was integrated out of the underlyingtheory. Therefore, as long as the energy scales being probed are much smaller than thelowest mass scale of a state that was integrated out, the higher-derivative expansion maymake sense and thus converge .Apart from a construction based on the hedgehog Ansatz by Marleau [10–13], no ex-tensive studies on higher-derivative terms in 3+1 dimensions, higher than sixth order, hasbeen carried out in the literature – to the best of our knowledge. Marleau considereda construction that yielded higher-order derivative corrections to the Skyrmion, but re-stricted in such a way as to give only a second-order equation of motion for the radialprofile (chiral angle function) [10–13]. When restricted to spherical symmetry, this con-struction gives stable profiles when certain stability criteria are satisfied [26]. Nevertheless,as we will show in Sec. 3, when relaxing the spherical symmetry, this construction becomesunstable. Longpr´e and Marleau later found that avoiding the instability was indeed dif-ficult [27, 28]; they proposed a stability criterion that, however, cannot be satisfied fora finite-order derivative Lagrangian without causing Derrick instability. We will proposeour interpretation of the instability as well as why it occurs and show that to finite order,it cannot be cured (stabilized). The instability occurs if perturbations are independentof one spatial direction. In particular, one can contemplate a perturbation in form of ababy-Skyrmion string which can trigger a run-away instability. The reason behind the For supersymmetrizations of the Skyrme model, see e.g. Refs. [21–24]. Mathematically, such series may not be well-defined or converge in any mathematical sense. We willnot dwell upon such obstacles here. Ref. [25] considered a higher-dimensional generalization of the Atiyah-Manton construction of Skyrm-ions using the holonomy of instantons; this reference considers eighth-order derivative terms in 7+1 dimen-sions. – 2 –nstability is basically the requirement of the radial direction to be special (that is, to obeyonly a second-order equation of motion, whereas the angular directions enjoy many morepowers of derivatives). This loss of isotropy brings about the latter mentioned instability.In this paper, we take the construction of higher-order derivative corrections to thenext level. The spirit of our construction is similar to that behind the Skyrme term andthe BPS-Skyrme term. Take the Skyrme term; it is fourth order in spacetime derivatives.The most general term with fourth-order derivatives will contain four time derivatives.The Skyrme term does not; it is constructed in such a way as to cancel the fourth-orderderivatives in the i -th space or time direction and contains four spacetime derivatives onlyas a product of second-order derivatives in two different space or time directions; e.g.( ∂ x φ ) ( ∂ y φ ) or ( ∂ t φ ) ( ∂ x φ ) . Note that this is the minimal number of derivatives in the i -th direction (we will denote this number by δ ). Two derivatives in the i -th direction is,however, only possible for terms up to and including sixth-order in derivatives in 3 spatialdimensions or eighth-order in derivatives in 4 spacetime dimensions. We prove, however,that the latter term vanishes identically in the Skyrme model ( S target space). For eighth-, tenth- and twelfth-order derivative terms, the smallest number of derivatives in the i -thdirection is four, i.e. δ = 4. That is, when we do not break isotropy.Our construction is straightforward and yields positive-definite static energies for thesystems. We find simple interpretations for the Lagrangians that we constructed. Theeighth-order Lagrangian can be understood as the sum of the Skyrme-term squared andthe kinetic term multiplied by the BPS-Skyrme term (the sixth-order term mentionedabove). The tenth-order Lagrangian can be interpreted as the Skyrme term multiplied bythe BPS-Skyrme term. Finally, the twelfth-order Lagrangian can be interpreted as theBPS-Skyrme term squared.We successfully achieve manifest stability for static energy associated with the higher-order Lagrangians. However, in order to check that time-dependent perturbations cannotspoil this stability, we construct the corresponding Hamiltonians. The Hamiltonians, aswell known, are important objects because they give rise to the Euler-Lagrange equationsof motion (as the Lagrangians do) and because we do not have any explicit time depen-dence, they are conserved and thus can be associated with the total energy. Althoughthe Hamiltonians do not suffer from the famous Ostrogradsky instability [29, 30] (see alsoRef. [31]), their highly nonlinear nature induces nonlinearities in the conjugate momentaand hence in the Hamiltonians themselves which potentially may destabilize the systemsand in turn their solitons. The dynamical instability we find is intrinsically different fromthe Ostrogradsky one, because we do not have two time derivatives acting on the samefield, but simply large powers of one time derivative acting on one field (see Appendix B).This implies that we only have a single conjugate momentum for each field (as opposedto several as in Ostrogradsky’s Lagrangian) and there is no run-away associated with alinear conjugate momentum in the Hamiltonian. Nevertheless, our construction yields anonlinear conjugate momentum which induces ghost-like kinetic terms. In particular, theterms containing fourth-order time derivatives are accompanied by two powers of the in-verse metric, which thus acquires the wrong sign – this term therefore remains negativein the Hamiltonian. The other effect we find is also related to the nonlinearities of the– 3 –igher-order derivative terms, namely, when a term has more than two time derivativesthe SO(3,1) symmetry of the Lorentz invariants is not simply transformed to SO(4) in-variants by the standard Legendre transform, but the latter SO(4) symmetry is broken.This breaking of the would-be SO(4) symmetry induces terms with both signs. This isalso related to our construction producing “minimal” Lagrangians, i.e. terms that are assimple as possible in terms of eigenvalues of the strain tensor. Although we find the abovedynamical instabilities in our Hamiltonians, we conjecture that the vacuum is stable.Finally, we argue that the Hamiltonian intrinsically knows that it is a low-energy ef-fective field theory and that the instabilities described above do not occur at leading orderfor time-dependent perturbations. We consider the simplest possible perturbation, i.e. ex-citing the translational zero mode, and associating the energy scale of said perturbationwith a velocity. We find exact conditions for when the instability sets in and estimate thevelocities for which the effective theory will break down. In all cases the critical velocitiesare of the order of about half the speed of light. Then we show that to leading order in thevelocity squared, there is no instability of the Hamiltonians of eighth and twelfth order.The paper is organized as follows. In section 2, we set up the formalism to constructthe higher-order derivative Lagrangians. In section 3, we review the Marleau constructionand show that it contains an instability already in the static energy. Section 4 presentsour construction of higher-order derivative Lagrangians with positive-definite static en-ergy. In section 5 the corresponding Hamiltonians are then constructed and dynamicalinstabilities are found and discussed. Section 6 then discusses the low-energy stability ofthe Hamiltonians. Section 7 then concludes with a discussion. Appendix A illustrates thebaby-Skyrmion string triggering a run-away perturbation found in the Marleau construc-tion while Appendix B provides a comparison of our dynamical instability with that ofOstrogradsky and the differences in their underlying Lagrangians. Traditionally, the Skyrme model is formulated in terms of left-invariant current L µ ≡ U † ∂ µ U (or equivalently the right-invariant current R µ ≡ ∂ µ U U † ), µ = 0 , , , U is the chiral Lagrangian field U = σ + iπ a τ a ∈ SU(2) , (2.1)with τ a the Pauli matrices, a = 1 , ,
3, and U obeys the nonlinear sigma-model constraintdet U = 1.The kinetic term is then simply given by L = 14 Tr ( L µ L µ ) , (2.2)and we are using the mostly-positive metric signature. Both the Skyrme term, which is offourth order in derivatives, and the BPS-Skyrme term [14, 15], which is of sixth order in– 4 –erivatives, is made out of antisymmetric combinations of L µ , L = 132 Tr [ L µ , L ν ][ L µ , L ν ] = −
132 Tr [ F µν F νµ ] , (2.3) L = 1144 η µµ (cid:48) (cid:15) µνρσ Tr [ L ν L ρ L σ ] (cid:15) µ (cid:48) ν (cid:48) ρ (cid:48) σ (cid:48) Tr [ L ν (cid:48) L ρ (cid:48) L σ (cid:48) ] = 196 Tr [ F νµ F ρν F µρ ] , (2.4)where we have defined F µν ≡ [ L µ , L ν ] , (2.5)and η µν is the flat-space Minkowski metric of mostly-positive signature. Proving that themiddle and right-hand side of Eq. (2.4) are identical is somewhat nontrivial; we will seethat it is indeed the case after we switch to the notation of eigenvalues, see below.Although one can construct higher-order terms with more than six derivatives using F µν (see Sec. 3), it is convenient to switch the notation to using invariants of O(4) instead n µ · n ν ≡ ∂ µ n · ∂ ν n , (2.6)where U = n + in a τ a , (2.7)and the boldface symbol denotes the four vector n ≡ ( n , n , n , n ) of unit length: n = 1.This tensor is the strain tensor.Since the Lagrangian is a Lorentz invariant, we can immediately see that the simplestinvariants of both O(4) and Lorentz symmetry we can write down, are given by (cid:104) r (cid:105) ≡ r (cid:89) p =1 η µ p +1 | r ν p n µ p · n ν p = ( − − r r (cid:89) p =1 η µ p +1 | r ν p Tr [ L µ p L ν p ] , (2.8)where the modulo function in the first index, p + 1 | r (meaning p + 1 mod r ), simplyensures that the index µ r +1 is just µ and η µν is the inverse of the flat Minkowski metricof mostly-positive signature.Another invariant of both SO(4) (which is a subgroup of O(4)) and of Lorentz sym-metry that we can construct is given by (cid:15) abcd (cid:15) µνρσ n aµ n bν n cρ n dσ , (2.9)which obviously vanishes for static fields. Therefore, we can safely ignore this invariant forthe static solitons.The most general static Lagrangian density with 2 n derivatives, can thus be writtenas −L n = (cid:88) r =1 ,...,n (cid:88) r = r ,...,n − r · · · (cid:88) r n = r n − ,...,n − (cid:80) p =1 ,..., ( n − r p a r n ,r n − ,...,r (cid:104) r n (cid:105)(cid:104) r n − (cid:105) · · · (cid:104) r (cid:105) , (2.10)– 5 –here it is understood that a factor of (cid:104) r p (cid:105) is only present when the index r p has a positiverange in the sum (including unity as its only possibility).The invariants (2.8) with the hedgehog Ansatz U = cos f ( ρ ) + ix a τ a ρ sin f ( ρ ) , (2.11)have an astonishingly simple form (cid:104) r (cid:105) = f rρ + 2 sin r fρ r , (2.12)where f is a profile function with the boundary conditions f ( ∞ ) = 0 and f (0) = π , f ρ ≡ ∂ ρ f and ρ = (cid:112) ( x ) + ( x ) + ( x ) is the radial coordinate.It is, however, not enough to work with a spherically symmetric Ansatz (i.e. the hedge-hog in Eq. (2.11)), as the system may have runaway directions when not restricting tospherical symmetry. It is clear that the static energy of the system is bounded from belowwhen all the coefficients a ≥ n .In this paper, our philosophy will be similar to the construction of the Skyrme term,namely we want to construct the higher-derivative terms with the minimal number ofderivatives in each spacetime direction. That choice, however, implies that some of thecoefficients a need to be negative. The prime example being the Skyrme term, for whichwe have a , = − a = 14 . (2.13)If we now consider d spatial dimensions, the smallest possible number of derivatives in the i -th direction (in the static case) is given by δ ≡ (cid:100) n/d (cid:101) , (2.14)where (cid:100) χ (cid:101) = ceil( χ ) rounds the real number χ up to its nearest integer. This of coursejust corresponds to distributing the derivatives symmetrically over all d spatial dimensions.This means that for d = 3, we can only have δ = 2 derivatives in the i -th direction for n ≤
3, i.e. at most six derivatives in total. We can also see that if we consider δ = 4derivatives in the i -th direction, then n = 4 , , ,
10, and 12 derivative terms.These are the terms we will focus on constructing in this paper.Since we now allow for some of the coefficients a to be negative, we have to find amethod to ensure the stability of the system or in other words positivity of the staticenergy of the system. For this purpose, it will prove convenient to use the formalism ofeigenvalues [32] of the strain tensor D ij ≡ −
12 Tr [ L i L j ] = n i · n j = V λ λ λ V T ij , (2.15)– 6 –hich we will denote as λ , λ , λ , (2.16) i, j = 1 , , V is an orthogonal matrix. It is now easy to prove that (cid:104) r (cid:105) = λ r + λ r + λ r . (2.17)This means that the invariant (cid:104) r (cid:105) has exactly the maximal number (i.e. 2 r ) of derivativesin one direction (and due to symmetry this term is summed over all spatial directions).Now our construction works as follows. We write down the most general Lagrangiandensity of order 2 n using Eq. (2.10). Then we calculate the number of derivatives of n in onedirection, say x . The general case has 2 n derivatives in the x -direction. Finding the linearcombinations with only δ (see Eq. (2.14)) derivatives in the x -direction is tantamount tosolving the constraints of setting the coefficients of the terms with 2 n , 2 n − · · · , δ + 2orders of derivatives in the x -direction equal to zero. The final step is to ensure that allterms provide positive semi-definite static energy when written in terms of the eigenvalues λ i , see Eq. (2.15). We will carry out the explicit calculation in Sec. 4. In this section we will review the construction of Marleau [10–13] for higher-order derivativeterms. The 2 n -th order Lagrangians are given by L = 14 η µµ (cid:48) Tr ( L µ L µ (cid:48) ) , (3.1) L = − η µµ (cid:48) η νν (cid:48) Tr ( F µν (cid:48) F νµ (cid:48) ) , (3.2) L = 1192 η µµ (cid:48) η νν (cid:48) η ρρ (cid:48) Tr ( F µν (cid:48) F νρ (cid:48) F ρµ (cid:48) ) , (3.3) L = − η µµ (cid:48) η νν (cid:48) η ρρ (cid:48) η σσ (cid:48) (cid:2) Tr ( F µν (cid:48) F νρ (cid:48) F ρσ (cid:48) F σµ (cid:48) ) − Tr (cid:0)(cid:8) F µν (cid:48) , F ρσ (cid:48) (cid:9) F νρ (cid:48) F σµ (cid:48) (cid:1)(cid:3) , (3.4) L = 0 , (3.5) L = 16144 η µµ (cid:48) η νν (cid:48) η ρρ (cid:48) η σσ (cid:48) η λλ (cid:48) η δδ (cid:48) (cid:20) Tr ( F µν (cid:48) F νρ (cid:48) F ρσ (cid:48) F σλ (cid:48) F λδ (cid:48) F δµ (cid:48) ) (3.6) −
92 Tr (cid:0)(cid:8) F µν (cid:48) , F ρσ (cid:48) (cid:9) F νρ (cid:48) F σλ (cid:48) F λδ (cid:48) F δµ (cid:48) (cid:1) + 72 Tr (cid:0)(cid:8) F µν (cid:48) , F ρσ (cid:48) (cid:9) (cid:8) F νρ (cid:48) , F λδ (cid:48) (cid:9) F σλ (cid:48) F δµ (cid:48) (cid:1) (cid:21) . The first three Lagrangians already have at most two derivatives in one direction δ = 2, aswe have seen in the previous section. Starting from the eight-order derivative term ( n = 4),the systematic construction works like this. Take n F -factors and contract their Lorentzindices as a matrix product and then subtract the following terms: the first one is madeby switching the second and the third F and then anti-commuting the first and the new There is a difference in a factor of two for these terms for n > – 7 –econd F (the old third F at position 2). The next term starts with the previous term andswitches the fourth and fifth F and then anti-commutes the third and new fourth F (theold fifth F at position 4). This continues as long as there are enough F factors to keep ongoing.Notice, that this construction cannot produce a tenth-order derivative term as it van-ishes identically.Although the Lagrangian densities (3.1-3.6) seem overly complicated in terms of theSkyrme term, Marleau found that for the hedgehog Ansatz, they simplify drastically [10–13]to L n = − sin n − ( f )2 ρ n − f ρ − − n n sin n ( f ) ρ n . (3.7)Notice, however, that for n > ρ ≥ f ≥ L = − (cid:104) (cid:105) , (3.8) L = 18 (cid:104) (cid:105) − (cid:104) (cid:105) , (3.9) L = − (cid:104) (cid:105) + 14 (cid:104) (cid:105)(cid:104) (cid:105) − (cid:104) (cid:105) , (3.10) L = 1316 (cid:104) (cid:105) − (cid:104) (cid:105)(cid:104) (cid:105) − (cid:104) (cid:105) + (cid:104) (cid:105)(cid:104) (cid:105) − (cid:104) (cid:105) , (3.11) L = 0 , (3.12) L = 5524 (cid:104) (cid:105) − (cid:104) (cid:105)(cid:104) (cid:105) − (cid:104) (cid:105)(cid:104) (cid:105) + 358 (cid:104) (cid:105)(cid:104) (cid:105) − (cid:104) (cid:105) + 298 (cid:104) (cid:105)(cid:104) (cid:105)(cid:104) (cid:105) − (cid:104) (cid:105)(cid:104) (cid:105) + 112 (cid:104) (cid:105) − (cid:104) (cid:105) (cid:104) (cid:105) + 12 (cid:104) (cid:105)(cid:104) (cid:105) , (3.13)where we have used F µν = − iX aµν τ a , X aµν = (cid:15) abc n bµ n cν + n µ n aν − n aµ n ν , (3.14) L µ = − iX aµ τ a , X aµ = (cid:15) abc n bµ n c + n µ n a − n aµ n , (3.15)and the contraction X aµν X aρσ = ( n µ · n ρ )( n ν · n σ ) − ( n µ · n σ )( n ν · n ρ ) + (cid:15) abcd n aµ n bν n cρ n dσ . (3.16)An easy check that one can make is to sum all the coefficients a in each Lagrangian densityand see that indeed the sum vanishes for all L n with n >
1. This simply means that thehighest power of derivatives vanishes for each of the higher-order Lagrangian densities.We can see from the reduced Lagrangian density (3.7), that for n >
3, correspondingto 8 or more derivatives, the non-radial derivative term (it is a combination of angularderivatives) acquires a negative sign. Since 0 ≤ sin f ≤ f inthe range f ∈ [0 , π ], there is no runaway asymptotically. Nevertheless, a negative sign– 8 –n the energy could signal some runaway instabilities that are just not allowed for by thespherically symmetric Ansatz (2.11). In fact, for the hedgehog Ansatz, Ref. [26] found astability criterion for the Marleau construction.In order to understand the instabilities in the Marleau construction, we take the La-grangian densities written in terms of the invariants, i.e. Eqs. (3.8-3.13) and plug in therelation (2.17) L = − (cid:0) λ + λ + λ (cid:1) , (3.17) L = − (cid:0) λ λ + λ λ + λ λ (cid:1) , (3.18) L = − λ λ λ , (3.19) L = − (cid:0) λ λ λ + 2 λ λ λ + 2 λ λ λ − λ λ − λ λ − λ λ (cid:1) , (3.20) L = 0 , (3.21) L = − (cid:0) λ λ λ + λ λ λ + λ λ λ + λ λ λ + λ λ λ + λ λ λ − λ λ λ − λ λ − λ λ − λ λ (cid:1) . (3.22)Clearly the construction yields non-manifestly positive terms for the eighth- and twelfth-order Lagrangians. We can see the trend that most terms that are products of derivatives inall three spatial dimensions are positive, whereas all terms that are products of derivativesin two spatial dimensions are negative. It is easy to construct a perturbation that can drive the system into a runaway direc-tion. Consider a perturbation that depends only on x , x but not on x , then it is clearthat for such perturbation the static energies for different Lagrangian densities become − L = 12 (cid:0) λ + λ (cid:1) , −L = 14 λ λ , −L = 0 , −L = − λ λ , −L = − λ λ . (3.23)For illustrative purposes, we will show an example of a run-away in Appendix A.Let us contemplate for a moment what the Marleau construction does. It is clear thatthe (cid:104) r (cid:105) -invariants themselves have a symmetric distribution of derivatives in all spatialdirections. There are therefore no preferred direction per se. Nevertheless, the Marleauconstruction is able to eliminate all terms with f pρ for p > n − n >
3. The way it works is to take the Lagrangian with2 n derivatives, L n , say using Eq. (2.10) and expand it in powers of f ρ . Then set thecombinations of the coefficients a to zero for all terms with higher powers of f ρ .One may ask whether the Marleau construction is unique and more importantlywhether there exists a construction for higher-derivative terms with more than six deriva- In Ref. [27, 28] a negative coefficient of the eighth-order Lagrangian was used to avoid the baby-Skyrmion string instability; that unfortunately yields a potential instability due to Derrick collapse of theentire soliton. – 9 –ives, that can provide at most two radial derivatives (i.e. at most f ρ ) and in the same timea positive-definite static energy. To answer this, let us count how many parameters are leftfree by the constraints setting terms with f kρ = 0 for k >
2. Table 1 lists the number offree parameters for a Lagrangian density with 2 n derivatives. We have used one parameterto normalize the second-order radial derivative term. Note that the number of invariants(2 n ) invariants constraints free parameters2 1 0 04 2 1 06 3 2 08 5 3 110 7 4 212 11 5 5 Table 1 . Number of derivatives, O(4) and Lorentz invariants, constraints and free parameters inthe Marleau construction. is indeed the partition function of n (in number theory). Notice however that the freeparameters merely allow one to write the same Lagrangian using different combinations ofinvariants (this should be straightforward from the point of view of group theory). Oncethe overall normalization is fixed, there are no free parameters left. In order to demonstratethis last point, let us construct the Lagrangians L n for n = 4 , , −L = a (cid:104) (cid:105) + a , (cid:104) (cid:105)(cid:104) (cid:105) − (cid:18) a + 316 a , (cid:19) (cid:104) (cid:105) − (cid:18) a + 98 a , (cid:19) (cid:104) (cid:105)(cid:104) (cid:105) + (cid:18) a + 516 a , (cid:19) (cid:104) (cid:105) , (3.24) −L = a (cid:104) (cid:105) + a , (cid:104) (cid:105)(cid:104) (cid:105) + a , (cid:104) (cid:105)(cid:104) (cid:105) + (cid:18) a − a , + 3 a , (cid:19) (cid:104) (cid:105)(cid:104) (cid:105) − (cid:18) a + 12 a , + 94 a , (cid:19) (cid:104) (cid:105) (cid:104) (cid:105) − (cid:18) a − a , + 52 a , (cid:19) (cid:104) (cid:105)(cid:104) (cid:105) + (cid:18) a − a , + 34 a , (cid:19) (cid:104) (cid:105) , (3.25)– 10 – L = a (cid:104) (cid:105) + a , (cid:104) (cid:105)(cid:104) (cid:105) + a , (cid:104) (cid:105)(cid:104) (cid:105) + a , , (cid:104) (cid:105)(cid:104) (cid:105) + a , (cid:104) (cid:105) + a , , (cid:104) (cid:105)(cid:104) (cid:105)(cid:104) (cid:105) + (cid:18) a − a , + 2827 a , − a , , + 2 a , + 79 a , , (cid:19) (cid:104) (cid:105)(cid:104) (cid:105) − (cid:18) a + 25288 a , + 2336 a , + 38 a , + 548 a , , (cid:19) (cid:104) (cid:105) − (cid:18) a + 365288 a , + 3736 a , + 12 a , , + 158 a , + 7348 a , , (cid:19) (cid:104) (cid:105) (cid:104) (cid:105) − (cid:18) a − a , + 712 a , − a , , + 98 a , + 516 a , , (cid:19) (cid:104) (cid:105)(cid:104) (cid:105) + (cid:18) a − a , + 23108 a , − a , , + 38 a , + 23144 a , , (cid:19) (cid:104) (cid:105) . (3.26)These are the most general Lagrangians with 1,2 and 5 free parameters, respectively, thatgive rise to the radial Lagrangian (3.7) with the coefficients c = 8 a + 6 a , , c = 25 a + 30 a , , c = 30 a + 15 a , + 24 a , + 36 a , + 18 a , , , (3.27)and with the characteristic of having only two radial derivatives (by construction of course).The Lagrangians in Eqs. (3.11) and (3.13) correspond to setting a = − / a , = 5 / a = − / a , = 11 / a , = 11 / a , , = − / a , = 13 / a , , = − / L = − a , (cid:18) (cid:104) (cid:105)(cid:104) (cid:105) − (cid:104) (cid:105) − (cid:104) (cid:105)(cid:104) (cid:105) + 516 (cid:104) (cid:105) (cid:19) , (3.28) L = − a , (cid:18) (cid:104) (cid:105)(cid:104) (cid:105) + 3 (cid:104) (cid:105)(cid:104) (cid:105) − (cid:104) (cid:105) (cid:104) (cid:105) − (cid:104) (cid:105)(cid:104) (cid:105) + 34 (cid:104) (cid:105) (cid:19) , (3.29) L = − a , , (cid:18) (cid:104) (cid:105)(cid:104) (cid:105)(cid:104) (cid:105) + 79 (cid:104) (cid:105)(cid:104) (cid:105) − (cid:104) (cid:105) − (cid:104) (cid:105) (cid:104) (cid:105) − (cid:104) (cid:105)(cid:104) (cid:105) + 23144 (cid:104) (cid:105) (cid:19) . (3.30)In order to normalize the above Lagrangian densities like Eq. (3.7), we need to set a , =1 / a , = 1 /
30 and a , , = 1 /
18, respectively.Note that the highest invariant we need to describe these higher-order Lagrangians isthe (cid:104) (cid:105) , which is the chain-contraction of the Lorentz indices of three O(4) invariants.In order to see whether the free parameters can change the Lagrangian densities, werewrite Eqs. (3.24-3.26) using the relation (2.17), obtaining L = − c (cid:0) λ λ λ + 2 λ λ λ + 2 λ λ λ − λ λ − λ λ − λ λ (cid:1) , (3.31) L = − c (cid:0) λ λ λ + 2 λ λ λ + 2 λ λ λ + λ λ λ + λ λ λ + λ λ λ − λ ( λ + λ ) − λ ( λ + λ ) − λ ( λ + λ ) (cid:1) , (3.32)– 11 – = − (cid:18) c −
23 ˜ c (cid:19) (cid:0) λ ( λ λ + λ λ ) + λ ( λ λ + λ λ ) + λ ( λ λ + λ λ ) − λ λ − λ λ − λ λ (cid:1) − (cid:18) c + 13 ˜ c (cid:19) (cid:0) λ λ λ + 2 λ λ λ + 2 λ λ λ − λ ( λ + λ ) − λ ( λ + λ ) − λ ( λ + λ ) (cid:1) + (cid:18) c − c (cid:19) λ λ λ , (3.33)where the coefficients c , , are given in Eq. (3.27) and we have defined˜ c ≡ a , + 13 a . (3.34)Notice that the eighth-order and tenth-order Lagrangians, L , depend on the combina-tions given in Eq. (3.27), which is just an overall normalization coefficient. The twelfth-order Lagrangian, on the other hand, has a residual free parameter, ˜ c . Say if we fix c of Eq. (3.27) to one, then we still have a one-parameter family of Lagrangians withdifferent eigenvalues λ i all giving rise to the reduced radial Lagrangian (3.7) upon usingthe hedgehog Ansatz (2.11).As for stability, it is clear that for L , all the free parameters just give rise to thesame Lagrangian with normalization c , and hence the negative terms cannot be elim-inated. For the twelfth-order Lagrangian, we have two parameters and two terms (thetwo first terms in Eq. (3.33)) that contain negative terms. However, eliminating both thefirst and the second term in the Lagrangian also kills the last term. Therefore for thesethree Lagrangian densities, there is no way of constructing stable static eighth-, tenth-,and twelfth-order Lagrangians with only second-order radial derivatives for the hedgehogAnsatz (2.11). By stable we mean that the static energy is bounded from below and henceis stable against non-baryonic perturbations, i.e. perturbations with vanishing baryon num-ber. In this section we will require positive-definite static energy and construct terms with eightand more derivatives. As shown in Eq. (2.14), the smallest possible number of derivatives inthe i -th direction is 4 for n = 4 , ,
6, corresponding to the eighth-, tenth-, and twelfth-orderLagrangians.
As a warm-up, let us rederive the kinetic, the Skyrme term and the BPS-Skyrme term.The difference for these terms with respect to the higher-order terms with 8, 10 and 12derivatives, is that δ = 2 for the second-, fourth- and sixth-order derivative term. Thismeans that we can consistently have only 2 derivatives in the i -th direction (whereas for8-12 derivatives, we need δ = 4). – 12 –irst, the kinetic term is trivial as it has only one possibility, i.e., −L = a (cid:104) (cid:105) = a ( λ + λ + λ ) . (4.1)Second, the Skyrme term is the simplest and first nontrivial example. We start bywriting −L = a (cid:104) (cid:105) + a , (cid:104) (cid:105) . (4.2)To eliminate the fourth-order derivatives in the i -th direction, we set − a = a , = c | , and arrive at −L = c | , −(cid:104) (cid:105) + (cid:104) (cid:105) ) = c | , ( λ λ + λ λ + λ λ ) . (4.3)Finally, let us rederive the BPS-Skyrme term. The most general form is −L = a (cid:104) (cid:105) + a , (cid:104) (cid:105)(cid:104) (cid:105) + a , , (cid:104) (cid:105) . (4.4)Eliminating the sixth-order derivatives in the i -th direction yields the constraint a + a , + a , , = 0 , (4.5)while eliminating the fourth-order yields a , + 3 a , , = 0 . (4.6)Their common solution is simply a , = − a = − c | , , and a , , = a = c | , , .Thus we obtain −L = c | , , (cid:18) (cid:104) (cid:105) − (cid:104) (cid:105)(cid:104) (cid:105) + 12 (cid:104) (cid:105) (cid:19) = c | , , λ λ λ . (4.7)We are now ready to move on to the more complicated higher-order derivative terms. Let us start with constructing the eighth-order Lagrangian. The most general static La-grangian can be written down as −L = a (cid:104) (cid:105) + a , (cid:104) (cid:105)(cid:104) (cid:105) + a , (cid:104) (cid:105) + a , , (cid:104) (cid:105)(cid:104) (cid:105) + a , , , (cid:104) (cid:105) . (4.8)This Lagrangian density, however contains generally eight derivatives in the same direc-tion; therefore, we will constrain the Lagrangian such that it has the minimal number ofderivatives in each direction; that is after constraining the above Lagrangian it will onlycontain terms with at most 4 derivatives in the i -th direction.Note that the above Lagrangian is constructed exactly as a sum over all possible Ferrersdiagrams in number theory or equivalently as a sum over all possible Young tableaux withthe total number of boxes equal to n (i.e. four here) [33]. Each row in the Young tableauis identified with the O(4) invariant. – 13 –e will hence, eliminate all terms with 8 and 6 derivatives in the (same) i -th direction;that is we allow for terms such as λ λ , but will eliminate terms such as λ or λ λ . Wechoose to solve these two constraints by eliminating a , , , and a , , , arriving at −L = c | , (cid:0) λ λ + λ λ + λ λ (cid:1) + c | , , (cid:0) λ λ λ + λ λ λ + λ λ λ (cid:1) , (4.9)where we have defined c | , ≡ a + 4 a , , c | , , ≡ a + 3 a , + 8 a , . (4.10)Thus the 3 free parameters a , a , and a , only appear in the above two combinations.Finally, in order to ensure stability of the static solutions, we require that both c | , > c | , , > c | , , = 2 c | , . The second termalso has a different interpretation than being the cross terms from the squared Skyrmeterm; it is simply the BPS-Skyrme term L multiplied by the Dirichlet term L . Since theBPS-Skyrme term is the baryon charge density squared, the latter term vanishes whereverthe baryon charge does.Writing the above Lagrangian in terms of the O(4) invariants, we get −L = a (cid:104) (cid:105) + a , (cid:104) (cid:105)(cid:104) (cid:105) + a , (cid:104) (cid:105) − (cid:18) a + 32 a , + 2 a , (cid:19) (cid:104) (cid:105)(cid:104) (cid:105) + (cid:18) a + 12 a , + a , (cid:19) (cid:104) (cid:105) . (4.11)As an example, we can set a = a , = 0 to get the following minimal Lagrangian −L min8 = a , (cid:0) (cid:104) (cid:105) − (cid:104) (cid:105) (cid:1) , (4.12)which yields a manifestly positive static energy for a , >
0. This minimal Lagrangian isof course nothing but the Skyrme term (Eq. (4.3)) squared. As another example, we set a = a , = 0 obtaining −L × = a , (cid:104) (cid:105) (cid:18) (cid:104) (cid:105) − (cid:104) (cid:105)(cid:104) (cid:105) + 12 (cid:104) (cid:105) (cid:19) , (4.13)which is clearly the Dirichlet term multiplied by the BPS-Skyrme term, see Eq. (4.7). Thiswas already clear from writing the Lagrangian in terms of the eigenvalues in Eq. (4.9). Wesee, however, also that if instead of setting a = a , = 0, we set a , = − a / −L × = a (cid:18) (cid:104) (cid:105) − (cid:104) (cid:105) − (cid:104) (cid:105)(cid:104) (cid:105) + 12 (cid:104) (cid:105) (cid:19) , (4.14)– 14 –hich is the same if we normalize a = a , .Plugging the hedgehog Ansatz (2.11) into Eq. (4.11), we get −L = c | , sin ( f ) ρ + 2 c | , , sin ( f ) ρ f ρ + ( c | , , + 2 c | , ) sin ( f ) ρ f ρ , (4.15)where the positive-definite coefficients are given in Eq. (4.10). We can again see the physicalinterpretation that the terms with coefficient c | , , are the BPS-Skyrme term multipliedby the kinetic (Dirichlet) term, while the two terms with coefficient c | , are the twoterms of the Skyrme term squared individually. We can also clearly see how the Marleauconstruction manages to cancel the third term in the above Lagrangian; setting c | , = − c | , , / i -th direction, that is, δ = 4. But it is also thefirst Lagrangian that has two physically independent terms, as shown in Eq. (4.9). We will now continue with the tenth-order Lagrangian. The most general static Lagrangianwith 10 derivatives can be written as −L = a (cid:104) (cid:105) + a , (cid:104) (cid:105)(cid:104) (cid:105) + a , (cid:104) (cid:105)(cid:104) (cid:105) + a , , (cid:104) (cid:105)(cid:104) (cid:105) + a , , (cid:104) (cid:105) (cid:104) (cid:105) + a , , , (cid:104) (cid:105)(cid:104) (cid:105) + a , , , , (cid:104) (cid:105) . (4.16)Using Eq. (2.14), we find that again in this case, we cannot have less than 4 derivatives inthe i -th direction. We will eliminate all terms with 10, 8 and 6 derivatives in the (same) i -th direction. Choosing to eliminate a , , , , , a , , , , a , , , and a , , , we get −L = c | , , (cid:0) λ λ λ + λ λ λ + λ λ λ (cid:1) , (4.17)where we have defined c | , , ≡ − a − a , . (4.18)Thus the 2 free parameters only appear in one combination which is fixed uniquely bynormalization. Finally, as usual in this construction, we require c | , , > −L = a (cid:104) (cid:105) + a , (cid:104) (cid:105)(cid:104) (cid:105) + a , (cid:104) (cid:105)(cid:104) (cid:105) − (cid:18) a + 43 a , + a , (cid:19) (cid:104) (cid:105)(cid:104) (cid:105) − (cid:18) a + 12 a , + 32 a , (cid:19) (cid:104) (cid:105) (cid:104) (cid:105) + (cid:18) a + a , + 2 a , (cid:19) (cid:104) (cid:105)(cid:104) (cid:105) − (cid:18) a + 16 a , + 12 a , (cid:19) (cid:104) (cid:105) . (4.19)Notice that although the coefficient a , appears in the above formulation of the Lagrangian,it does not influence the normalization coefficient c | , , given in Eq. (4.18). This isbecause when we write the invariants with the coefficient a , , (cid:104) (cid:105) (cid:18) (cid:104) (cid:105) − (cid:104) (cid:105)(cid:104) (cid:105) − (cid:104) (cid:105) + (cid:104) (cid:105)(cid:104) (cid:105) − (cid:104) (cid:105) (cid:19) = 0 , (4.20)in terms of the eigenvalues, λ i , we find that the above expression vanishes identically. Thisnontrivial relation among the invariants is in fact due to the observation we made in theprevious subsection for the eighth-order Lagrangian, namely that the Dirichlet term multi-plied by the BPS-Skyrme term can be written in two apparently different ways: Eq. (4.13)and Eq. (4.14). Thus the above relation can simply be written as (cid:18) (cid:104) (cid:105) − (cid:104) (cid:105) − (cid:104) (cid:105)(cid:104) (cid:105) + 12 (cid:104) (cid:105) (cid:19) − (cid:104) (cid:105) (cid:18) (cid:104) (cid:105) − (cid:104) (cid:105)(cid:104) (cid:105) + 12 (cid:104) (cid:105) (cid:19) = 0 , (4.21)where the two terms are equal as we found in the previous subsection and hence thenontrivial relation (4.20) follows.Hence, we can simplify the Lagrangian to −L = a (cid:104) (cid:105) + a , (cid:104) (cid:105)(cid:104) (cid:105) − (cid:18) a + a , (cid:19) (cid:104) (cid:105)(cid:104) (cid:105) − (cid:18) a + 32 a , (cid:19) (cid:104) (cid:105) (cid:104) (cid:105) + (cid:18) a + 2 a , (cid:19) (cid:104) (cid:105)(cid:104) (cid:105) − (cid:18) a + 12 a , (cid:19) (cid:104) (cid:105) . (4.22)As an example, we can set a = 0 and write the above Lagrangian as −L × = − a , (cid:0) −(cid:104) (cid:105) + (cid:104) (cid:105) (cid:1) (cid:18) (cid:104) (cid:105) − (cid:104) (cid:105)(cid:104) (cid:105) + 12 (cid:104) (cid:105) (cid:19) , (4.23)from which it is clear that this is simply the Skyrme term (Eq. (4.3)) multiplied by theBPS-Skyrme term (Eq. (4.7)). The static energy is positive definite because a , <
0, seeEq. (4.18). Since the entire Lagrangian (4.18) is simply the Skyrme term multiplied by theBPS-Skyrme term, the complementary part of the Lagrangian (4.22) is also nontriviallyequal to the above expression. We can see how the other part looks like by setting a , = 0,getting −L × = a (cid:18) (cid:104) (cid:105) − (cid:104) (cid:105)(cid:104) (cid:105) − (cid:104) (cid:105) (cid:104) (cid:105) + 52 (cid:104) (cid:105)(cid:104) (cid:105) − (cid:104) (cid:105) (cid:19) . (4.24)– 16 –sing the eigenvalues, λ i , we find that this Lagrangian is equal to that of Eq. (4.23) for a = a , <
0. This is again a highly nontrivial relation between different O(4) invariants.Plugging the hedgehog Ansatz (2.11) into Eq. (4.19), we obtain −L = 2 c | , , (cid:18) sin ( f ) ρ f ρ + sin ( f )2 ρ (cid:19) sin ( f ) ρ f ρ , (4.25)where the positive-definite coefficient c | , , is given in Eq. (4.18). The physical interpre-tation is again very clear as the Skyrme term multiplied by the BPS-Skyrme term. It isalso clear from the above construction why the tenth-order term vanishes in the Marleauconstruction, because there is only one coefficient and there is no way to eliminate the f ρ term without setting the whole term to zero. The highest order in derivatives we will consider in this paper is twelve. The most generalstatic Lagrangian density with 12 derivatives can be written as −L = a (cid:104) (cid:105) + a , (cid:104) (cid:105)(cid:104) (cid:105) + a , (cid:104) (cid:105)(cid:104) (cid:105) + a , , (cid:104) (cid:105)(cid:104) (cid:105) + a , (cid:104) (cid:105) + a , , (cid:104) (cid:105)(cid:104) (cid:105)(cid:104) (cid:105) + a , , , (cid:104) (cid:105)(cid:104) (cid:105) + a , , (cid:104) (cid:105) + a , , , (cid:104) (cid:105) (cid:104) (cid:105) + a , , , , (cid:104) (cid:105)(cid:104) (cid:105) + a , , , , , (cid:104) (cid:105) . (4.26)Using Eq. (2.14), we find that this is the largest number of derivatives in a term whichcannot have less than δ = 4 derivatives in the i -th direction. Continuing along the linesof the previous subsections we eliminate all terms with 12, 10, 8 and 6 derivatives in the(same) i -th direction. Choosing to eliminate the coefficients a , , , , , , a , , , , , a , , , , a , , , , and a , , , we arrive at −L = c | , , λ λ λ , (4.27)where we have defined c | , , ≡ a + 9 a , . (4.28)Thus the 2 free parameters only appear in the above combination which is fixed once thenormalization of this Lagrangian is. As always in this construction, we require c | , , > −L = a (cid:104) (cid:105) + a , (cid:104) (cid:105)(cid:104) (cid:105) + a , (cid:104) (cid:105)(cid:104) (cid:105) + a , , (cid:104) (cid:105)(cid:104) (cid:105) + a , (cid:104) (cid:105) − (cid:18) a + 56 a , + 43 a , + 3 a , (cid:19) (cid:104) (cid:105)(cid:104) (cid:105)(cid:104) (cid:105) − (cid:18) a , + 43 a , , − a , (cid:19) (cid:104) (cid:105)(cid:104) (cid:105) − (cid:18) a + 12 a , (cid:19) (cid:104) (cid:105) + (cid:18) a + a , − a , , + 94 a , (cid:19) (cid:104) (cid:105) (cid:104) (cid:105) − (cid:18) a − a , + 16 a , − a , , + 32 a , (cid:19) (cid:104) (cid:105)(cid:104) (cid:105) − (cid:18) a , + 16 a , , − a , (cid:19) (cid:104) (cid:105) . (4.29)Notice that when the Lagrangian is written in terms of the eigenvalues, λ i , the only pa-rameter is the overall coefficient c | , , which is the combination (4.28) of a and a , .The 3 other parameters in the above Lagrangian thus have no influence on the physics andso we again expect nontrivial relations among the invariants. They are (cid:104) (cid:105) − (cid:104) (cid:105)(cid:104) (cid:105) − (cid:104) (cid:105)(cid:104) (cid:105) + 56 (cid:104) (cid:105)(cid:104) (cid:105) − (cid:104) (cid:105) = 0 , (4.30)and Eq. (4.20), where the first is the relation with coefficient a , and the latter appears withboth coefficients a , and a , , as well as a factor of (cid:104) (cid:105) and (cid:104) (cid:105) , respectively. The latterrelation was discussed already in the last subsection. The nontrivial relation Eq. (4.30)can be understood by writing it as (cid:18) (cid:104) (cid:105) − (cid:104) (cid:105)(cid:104) (cid:105) − (cid:104) (cid:105) (cid:104) (cid:105) + 52 (cid:104) (cid:105)(cid:104) (cid:105) − (cid:104) (cid:105) (cid:19) − (cid:0) (cid:104) (cid:105) − (cid:104) (cid:105) (cid:1) (cid:18) (cid:104) (cid:105) − (cid:104) (cid:105)(cid:104) (cid:105) + 12 (cid:104) (cid:105) (cid:19) = 0 , (4.31)which is exactly the two Lagrangians (4.23) and (4.24) with a , = a and the nontrivialrelation (4.30) follows.Hence, we can simplify the Lagrangian to −L = a (cid:104) (cid:105) + a , (cid:104) (cid:105) − (2 a + 3 a , ) (cid:104) (cid:105)(cid:104) (cid:105)(cid:104) (cid:105) + a , (cid:104) (cid:105)(cid:104) (cid:105) − a (cid:104) (cid:105) + (cid:18) a + 94 a , (cid:19) (cid:104) (cid:105) (cid:104) (cid:105) − (cid:18) a + 32 a , (cid:19) (cid:104) (cid:105)(cid:104) (cid:105) + 14 a , (cid:104) (cid:105) . (4.32)As an example, we can set a = 0 for which the above Lagrangian reads −L × = a , (cid:18) (cid:104) (cid:105) − (cid:104) (cid:105)(cid:104) (cid:105) + 12 (cid:104) (cid:105) (cid:19) , (4.33)which is clearly the BPS-Skyrme term (Eq. (4.7)) squared. Since the whole Lagrangian(4.32) is the BPS-Skyrme term squared, the complementary part – i.e. the part with co-efficient a – is also nontrivially the BPS-Skyrme term squared. We can write that partdown by setting a , = 0 in Eq. (4.32), yielding −L × = a (cid:18) (cid:104) (cid:105) − (cid:104) (cid:105)(cid:104) (cid:105)(cid:104) (cid:105) − (cid:104) (cid:105) + 32 (cid:104) (cid:105) (cid:104) (cid:105) − (cid:104) (cid:105)(cid:104) (cid:105) (cid:19) . (4.34)– 18 –sing the eigenvalues, λ i , we find that the above Lagrangian is exactly equal to that ofEq. (4.33) for a = 3 a , . This is the last nontrivial relation we find between different O(4)invariants. We expect the relation between these two formulations of the twelfth-orderLagrangian to play a role in the simplification of the fourteenth-order Lagrangian.Plugging the hedgehog Ansatz (2.11) into the Lagrangian (4.29), we get −L = c | , , sin ( f ) ρ f ρ , (4.35)where the positive-definite coefficient c | , , is given in Eq. (4.28). Again the physicalinterpretation is very clear as the above expression is simply the baryon-charge densityto the fourth power or equivalently the BPS-Skyrme term squared. In our construction,there is no term which is second order in f ρ and so there is no overlap here between thisconstruction and the Marleau construction at this order. It is clear why; in order to geta twelfth-order term with only two radial derivatives, we need either 6 derivatives in the θ direction and 4 derivatives in the φ direction or vice versa. Our construction eliminatessuch terms with 6 derivatives in the i -the direction and hence, this term is not present inour construction. In the last section we have constructed higher-order Lagrangians with positive-definitestatic energies. This together with the nontrivial topological charge π (cid:18) SU(2) × SU(2)SU(2) (cid:19) = Z , (5.1)guarantees time-independent stability of the Skyrmions (solitons). In this section, we willcheck that time-dependent perturbations are also under control. For this investigation, weneed to calculate the Hamiltonians corresponding to the Lagrangians obtained in the lastsection. The first step is to compose the O(4) and Lorentz invariants into time and spatial derivativeparts, respectively. We thus define (cid:104) r, (cid:105) ≡ r (cid:89) p =1 n i p · n i p +1 | r , (cid:104) r, (cid:105) ≡ ( n · n i )( n · n i r − ) r − (cid:89) p =1 n i p · n i p +1 . (5.2)The first index r in the brackets represents the number of spatial indices in the product ofinvariants while the second represents the number of time indices ( µ = 0). Notice that inthe above angular brackets all indices are lowered. Note also that there is no need for more– 19 –han one time index in the invariant, because two time indices always break the chain intotwo. Thus we can write the relevant Lorentz invariants as (cid:104) (cid:105) = −(cid:104) , (cid:105) + (cid:104) , (cid:105) , (5.3) (cid:104) (cid:105) = (cid:104) , (cid:105) − (cid:104) , (cid:105) + (cid:104) , (cid:105) , (5.4) (cid:104) (cid:105) = −(cid:104) , (cid:105) + 3 (cid:104) , (cid:105)(cid:104) , (cid:105) − (cid:104) , (cid:105) + (cid:104) , (cid:105) , (5.5) (cid:104) (cid:105) = (cid:104) , (cid:105) − (cid:104) , (cid:105) (cid:104) , (cid:105) + 4 (cid:104) , (cid:105)(cid:104) , (cid:105) + 2 (cid:104) , (cid:105) − (cid:104) , (cid:105) + (cid:104) , (cid:105) , (5.6) (cid:104) (cid:105) = −(cid:104) , (cid:105) + 5 (cid:104) , (cid:105) (cid:104) , (cid:105) − (cid:104) , (cid:105) (cid:104) , (cid:105) − (cid:104) , (cid:105)(cid:104) , (cid:105) + 5 (cid:104) , (cid:105)(cid:104) , (cid:105) + 5 (cid:104) , (cid:105)(cid:104) , (cid:105)− (cid:104) , (cid:105) + (cid:104) , (cid:105) . (5.7) (cid:104) (cid:105) = (cid:104) , (cid:105) − (cid:104) , (cid:105) (cid:104) , (cid:105) + 6 (cid:104) , (cid:105) (cid:104) , (cid:105) + 9 (cid:104) , (cid:105) (cid:104) , (cid:105) − (cid:104) , (cid:105) (cid:104) , (cid:105)− (cid:104) , (cid:105)(cid:104) , (cid:105)(cid:104) , (cid:105) − (cid:104) , (cid:105) + 6 (cid:104) , (cid:105)(cid:104) , (cid:105) + 6 (cid:104) , (cid:105)(cid:104) , (cid:105) + 3 (cid:104) , (cid:105) − (cid:104) , (cid:105) + (cid:104) , (cid:105) . (5.8) As a warm-up, let us first consider the Hamiltonians for the generalized Skyrme model,i.e. for the Lagrangian with the kinetic term, the Skyrme term and the BPS-Skyrme term[14–20]. Writing the Lagrangians in terms of the time-dependent brackets, we get L = c | ( (cid:104) , (cid:105) − (cid:104) , (cid:105) ) , (5.9) L = c | , (cid:0) (cid:104) , (cid:105)(cid:104) , (cid:105) − (cid:104) , (cid:105) + (cid:104) , (cid:105) − (cid:104) , (cid:105) (cid:1) , (5.10) L = c | , , (cid:18) − (cid:104) , (cid:105)(cid:104) , (cid:105) + 32 (cid:104) , (cid:105)(cid:104) , (cid:105) + 3 (cid:104) , (cid:105) − (cid:104) , (cid:105)(cid:104) , (cid:105)− (cid:104) , (cid:105) + 32 (cid:104) , (cid:105)(cid:104) , (cid:105) − (cid:104) , (cid:105) (cid:19) , (5.11)from which we can calculate the conjugate momenta π (2) · n = 2 c | (cid:104) , (cid:105) , (5.12) π (4) · n = 2 c | , ( (cid:104) , (cid:105)(cid:104) , (cid:105) − (cid:104) , (cid:105) ) , (5.13) π (6) · n = c | , , (cid:0) −(cid:104) , (cid:105)(cid:104) , (cid:105) + (cid:104) , (cid:105)(cid:104) , (cid:105) + 2 (cid:104) , (cid:105) − (cid:104) , (cid:105)(cid:104) , (cid:105) (cid:1) . (5.14)We can now write down the Hamiltonians in terms of the invariants with the time-dependent brackets H = c | ( (cid:104) , (cid:105) + (cid:104) , (cid:105) ) , (5.15) H = c | , (cid:0) (cid:104) , (cid:105)(cid:104) , (cid:105) − (cid:104) , (cid:105) − (cid:104) , (cid:105) + (cid:104) , (cid:105) (cid:1) , (5.16) H = c | , , (cid:18) − (cid:104) , (cid:105)(cid:104) , (cid:105) + 32 (cid:104) , (cid:105)(cid:104) , (cid:105) + 3 (cid:104) , (cid:105) − (cid:104) , (cid:105)(cid:104) , (cid:105) + (cid:104) , (cid:105) − (cid:104) , (cid:105)(cid:104) , (cid:105) + 12 (cid:104) , (cid:105) (cid:19) . (5.17)– 20 –rom the invariants, it is not clear whether the Hamiltonians are bounded from below ornot. Therefore, it is convenient to rewrite them in terms of the eigenvalues λ µ , H = c | (cid:0) λ + λ + λ + λ (cid:1) , (5.18) H = c | , (cid:0) λ (cid:0) λ + λ + λ (cid:1) + λ λ + λ λ + λ λ (cid:1) , (5.19) H = c | , , (cid:0) λ (cid:0) λ λ + λ λ + λ λ (cid:1) + λ λ λ (cid:1) , (5.20)where we have used the eigenvalues, λ µ , defined as (cid:101) D µν ≡ ( n µ · n ν ) = (cid:101) V λ λ λ λ (cid:101) V T µν , (cid:101) V ≡ (cid:32) σ w T u V (cid:33) , (5.21)where σ is a real scalar, v, w are real row-vectors of length 3 and V is a 3-by-3 real matrix. (cid:101) V (cid:101) V T = (cid:101) V T (cid:101) V = , which gives rise to the relations σ + w T w = σ + u T u = 1 , ww T + V V T = , u T V = − σw T . (5.23)We can clearly see that all the Hamiltonians (Eqs. (5.18-5.20)) are positive definiteeven when including time dependence.It is easy to show that the determinant of the matrix (cid:101) D µν vanishes. This can bechecked explicitly by using a parametrization of n with manifest unit length, e.g. n =(sin f sin g sin h, sin f sin g cos h, sin f cos g, cos f ). Alternatively this can be understood bynoting that the target space is three dimensional and there are no four independent tangentvectors n µ which in turn implies that the determinant of (cid:101) D vanishes (because one of thevectors must be linearly dependent on the others) . This has the following implication:one can always choose λ = 0. This simplifies the Hamiltonians to H = c | (cid:0) λ + λ + λ (cid:1) , (5.24) H = c | , (cid:0) λ λ + λ λ + λ λ (cid:1) , (5.25) H = c | , , λ λ λ . (5.26)Obviously, all three Hamiltonians are positive semi-definite.In the following subsections, we will check explicitly whether it also possible to establishpositivity also the higher-order Lagrangians constructed in the previous section. Actually the decomposition into temporal and spatial parts of (cid:101) D µν is not necessary when the Hamil-tonian only contains terms with 2 time derivatives, because in that case, one can form SO(4) singlets,e.g.2 (cid:104) , (cid:105)(cid:104) , (cid:105) − (cid:104) , (cid:105) − (cid:104) , (cid:105) + (cid:104) , (cid:105) = 2 (cid:101) D (cid:101) D ii − (cid:101) D i (cid:101) D i − (cid:101) D ij (cid:101) D ji + (cid:101) D ii = (cid:101) D µµ − (cid:101) D µν (cid:101) D νµ . (5.22)This will not be the case for more than two time derivatives, as we will see in the next subsection. We thank Martin Speight for pointing this out. – 21 – .3 8 derivatives
Let us first rewrite the Lagrangian (4.11) in terms of the time-dependent brackets definedin Eq. (5.2) using Eqs. (5.3-5.6) L = 12 (4 a + 3 a , ) (cid:104) , (cid:105) (cid:104) , (cid:105) −
12 (8 a + 3 a , + 8 a , ) (cid:104) , (cid:105) (cid:104) , (cid:105) + a , (cid:104) , (cid:105)(cid:104) , (cid:105)− (4 a + 3 a , + 4 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105)(cid:104) , (cid:105) + (4 a + 2 a , + 4 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105) − (4 a + 3 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105) + (8 a + 3 a , + 8 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105)(cid:104) , (cid:105) + 4 a (cid:104) , (cid:105) + 3 a , (cid:104) , (cid:105)(cid:104) , (cid:105) − a + 2 a , ) (cid:104) , (cid:105) + 4 a , (cid:104) , (cid:105)(cid:104) , (cid:105)− (4 a + 3 a , + 4 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105) − a (cid:104) , (cid:105) − a , (cid:104) , (cid:105)(cid:104) , (cid:105) − a , (cid:104) , (cid:105) + 12 (4 a + 3 a , + 4 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105) −
12 (2 a + a , + 2 a , ) (cid:104) , (cid:105) . (5.27)The conjugate momentum can thus readily be obtained as12 π (8) · n = (4 a + 3 a , ) (cid:104) , (cid:105) (cid:104) , (cid:105) − (8 a + 3 a , + 8 a , ) (cid:104) , (cid:105) (cid:104) , (cid:105) + a , (cid:104) , (cid:105)(cid:104) , (cid:105)− (4 a + 3 a , + 4 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105)(cid:104) , (cid:105) + (4 a + 2 a , + 4 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105) − a + 3 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105) + 2(8 a + 3 a , + 8 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105)(cid:104) , (cid:105) + 4 a (cid:104) , (cid:105) + 3 a , (cid:104) , (cid:105)(cid:104) , (cid:105) − a + 2 a , ) (cid:104) , (cid:105) + 4 a , (cid:104) , (cid:105)(cid:104) , (cid:105)− (4 a + 3 a , + 4 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105) . (5.28)It is now straightforward to get the Hamiltonian H = 32 (4 a + 3 a , ) (cid:104) , (cid:105) (cid:104) , (cid:105) −
32 (8 a + 3 a , + 8 a , ) (cid:104) , (cid:105) (cid:104) , (cid:105) + a , (cid:104) , (cid:105)(cid:104) , (cid:105)− (4 a + 3 a , + 4 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105)(cid:104) , (cid:105) + (4 a + 2 a , + 4 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105) − a + 3 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105) + 3(8 a + 3 a , + 8 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105)(cid:104) , (cid:105) + 4 a (cid:104) , (cid:105) + 3 a , (cid:104) , (cid:105)(cid:104) , (cid:105) − a + 2 a , ) (cid:104) , (cid:105) + 4 a , (cid:104) , (cid:105)(cid:104) , (cid:105)− (4 a + 3 a , + 4 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105) + a (cid:104) , (cid:105) + a , (cid:104) , (cid:105)(cid:104) , (cid:105) + a , (cid:104) , (cid:105) −
12 (4 a + 3 a , + 4 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105) + 12 (2 a + a , + 2 a , ) (cid:104) , (cid:105) . (5.29)The final step is thus to rewrite the invariants in terms of the eigenvalues λ µ usingEq. (5.21), H = c | , (cid:0) λ λ + λ λ + λ λ (cid:1) + c | , , (cid:0) λ λ λ + λ λ λ + λ λ λ (cid:1) − c | , (cid:0) λ + λ + λ (cid:1) (cid:2) w T λ i w (cid:3) − c | , , (cid:0) λ λ + λ λ + λ λ (cid:1) (cid:2) w T λ i w (cid:3) + 4 c | , , (cid:0) λ + λ + λ (cid:1) (cid:2) w T ( λ i ) w (cid:3) (cid:2) w T λ i w (cid:3) − c | , (cid:2) w T ( λ i ) w (cid:3) + 4(2 c | , − c | , , ) (cid:2) w T ( λ i ) w (cid:3) (cid:2) w T λ i w (cid:3) , (5.30)where the coefficients c | , and c | , , are defined in Eq. (4.10) and λ i = diag( λ , λ , λ )is the 3-by-3 diagonal matrix of eigenvalues.– 22 –ote that the following inner products are positive semi-definite w T ( λ i ) p w ≥ , p ∈ Z > , (5.31)as are the eigenvalues themselves, λ µ ≥
0, with µ not summed over. Writing out explicitlythe above inner product, we get w T ( λ i ) p w = w λ p + w λ p + w λ p . (5.32)Plugging this into the Hamiltonian, we can write H = c | , (cid:0) − w + w ) (cid:1) λ λ + c | , (cid:0) − w + w ) (cid:1) λ λ + c | , (cid:0) − w + w ) (cid:1) λ λ + (cid:2) c | , , (cid:0) − w ( w T w ) (cid:1) − c | , w w (cid:3) λ λ λ + (cid:2) c | , , (cid:0) − w ( w T w ) (cid:1) − c | , w w (cid:3) λ λ λ + (cid:2) c | , , (cid:0) − w ( w T w ) (cid:1) − c | , w w (cid:3) λ λ λ , (5.33)from which it is easy to read off when the instability kicks in. Since σ + w T w = 1,the length of w cannot exceed 1, but that is not sufficient to establish stability of theHamiltonian.It is also clear from the above expression that as long as w is small enough, theHamiltonian is positive definite (for any values of λ i ).There are two sources of minus signs in the calculation of the Hamiltonian; one comesfrom the fact that the square of the time-time component of the inverse metric is notnegative. The second-order time derivatives in the Lagrangian density are accompanied by1 factor of the inverse metric giving exactly 1 minus sign and hence that term is positive inthe Lagrangian and also in the Hamiltonian. The fourth-order time derivatives, however,are accompanied by two factors of the inverse metric giving a plus and hence the termbecomes negative both in the Lagrangian and Hamiltonian. The other source of minussigns comes from our desire to eliminate higher powers of derivatives in the i -th direction.Throughout the paper, we have only used the invariants (2.8). However, we mentionedanother time-dependent invariant (2.9), which we neglected so far because it vanishesfor static configurations. Since we are considering time-dependent perturbations in thissection, we should consider including it. By construction it has 4 derivatives, but eachderivative appears only once in each direction. We choose to impose parity and time-reversion symmetry on the Lagrangian, which implies that the invariant (2.9) can onlyappear with even powers. Hence, the first Lagrangian where it can appear (squared) isthe eighth-order Lagrangian discussed in this section. Let us calculate its contributionexplicitly L (cid:48) = a (cid:15) (cid:16) (cid:15) abcd (cid:15) µνρσ n aµ n bν n cρ n dσ (cid:17) = a (cid:15) (cid:18) −(cid:104) (cid:105) + 43 (cid:104) (cid:105)(cid:104) (cid:105) + 12 (cid:104) (cid:105) − (cid:104) (cid:105)(cid:104) (cid:105) + 16 (cid:104) (cid:105) (cid:19) . (5.34) Recall that we use the mostly-positive metric signature. The conclusion remains the same by using themostly-negative metric signature, although the details change. – 23 –e claimed that the invariant vanishes for static contributions and so should its square;we can confirm this statement explicitly by observing that static part of the above La-grangian is exactly Eq. (4.20) and the claim follows. Turning on time-dependence, theabove Lagrangian can be written in terms of time-dependent brackets in Eq. (5.2) usingEqs. (5.3-5.6) as L (cid:48) = a (cid:15) (cid:18) − (cid:104) , (cid:105)(cid:104) , (cid:105) + 2 (cid:104) , (cid:105)(cid:104) , (cid:105)(cid:104) , (cid:105) − (cid:104) , (cid:105)(cid:104) , (cid:105) + 4 (cid:104) , (cid:105) − (cid:104) , (cid:105)(cid:104) , (cid:105)− (cid:104) , (cid:105)(cid:104) , (cid:105) + 2 (cid:104) , (cid:105)(cid:104) , (cid:105) − (cid:104) , (cid:105) + 43 (cid:104) , (cid:105)(cid:104) , (cid:105) + 12 (cid:104) , (cid:105) − (cid:104) , (cid:105)(cid:104) , (cid:105) + 16 (cid:104) , (cid:105) (cid:19) . (5.35)We now want to perform a Legendre transformation to get the corresponding Hamiltonian,starting with writing down the conjugate momenta12 π (8) (cid:48) · n = a (cid:15) (cid:18) − (cid:104) , (cid:105)(cid:104) , (cid:105) + 2 (cid:104) , (cid:105)(cid:104) , (cid:105)(cid:104) , (cid:105) − (cid:104) , (cid:105)(cid:104) , (cid:105) + 4 (cid:104) , (cid:105) − (cid:104) , (cid:105)(cid:104) , (cid:105)− (cid:104) , (cid:105)(cid:104) , (cid:105) + 2 (cid:104) , (cid:105)(cid:104) , (cid:105) (cid:19) , (5.36)and hence the Hamiltonian is simply H (cid:48) = a (cid:15) (cid:18) − (cid:104) , (cid:105)(cid:104) , (cid:105) + 2 (cid:104) , (cid:105)(cid:104) , (cid:105)(cid:104) , (cid:105) − (cid:104) , (cid:105)(cid:104) , (cid:105) + 4 (cid:104) , (cid:105) − (cid:104) , (cid:105)(cid:104) , (cid:105)− (cid:104) , (cid:105)(cid:104) , (cid:105) + 2 (cid:104) , (cid:105)(cid:104) , (cid:105) + (cid:104) , (cid:105) − (cid:104) , (cid:105)(cid:104) , (cid:105) − (cid:104) , (cid:105) + (cid:104) , (cid:105)(cid:104) , (cid:105) − (cid:104) , (cid:105) (cid:19) . (5.37)Rewriting it in terms of the eigenvalues λ µ using Eq. (5.21), we get H (cid:48) = − a (cid:15) λ λ λ λ = 0 . (5.38)As discussed in the previous subsection, one of the eigenvalues λ µ must vanish and we canalways choose it to be λ . In any case, the above contribution vanishes identically.We have seen in this subsection that although the static energy of the Lagrangian(4.11) is positive definite, the total energy obtained from the corresponding Hamiltonianis not. Thus the energy is not bounded from below and in principle the theory is unstable.Two comments are in store on this account. The dynamical instability encountered hereis not exactly due to Ostrogradsky’s theorem [30], because our Lagrangian by construction(by choice) does not contain (cid:3) n a , which requires a second conjugate momentum for thefield n a , see also App. B. To flesh this point out in more details, let us write the conjugate– 24 –omenta π (8) in details before dotting them onto n as π (8) a = 2 (4 a + 3 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105) n a − a + 3 a , + 8 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105) n a + 2 a , (cid:104) , (cid:105) n a − a + 3 a , + 4 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105) n a + 2(4 a + 2 a , + 4 a , ) (cid:104) , (cid:105) n a − a + 3 a , ) ( (cid:104) , (cid:105) n a + (cid:104) , (cid:105) n ai ( n i · n j )( n j · n ))+ 2(8 a + 3 a , + 8 a , ) ( (cid:104) , (cid:105)(cid:104) , (cid:105) n a + (cid:104) , (cid:105)(cid:104) , (cid:105) n ai ( n i · n ))+ 8 a n ai ( n i · n j )( n j · n k )( n k · n ) + 6 a , (cid:104) , (cid:105) n ai ( n i · n j )( n j · n ) − a + 2 a , ) (cid:104) , (cid:105) n ai ( n i · n ) + 8 a , (cid:104) , (cid:105) n ai ( n i · n ) − a + 3 a , + 4 a , ) (cid:104) , (cid:105) n ai ( n i · n ) . (5.39)Notice that we can write the conjugate momenta as π (8) a = 2 (cid:16) K ab + (cid:104) , (cid:105) K ab + (cid:104) , (cid:105) K ab + (cid:104) , (cid:105) K ab (cid:17) n b + 2 ( K + (cid:104) , (cid:105) K + K (cid:104) , (cid:105) ) δ ab n b . (5.40)In principle, now we would like to invert the equation to get an expression for n a in terms of π (8) a . The equation, however, is a cubic matrix equation; we will not attempt at finding theexplicit solution here. It is merely enough to notice that the inverse, which we assume toexist, is proportional to a cubic root involving π (8) a itself. Therefore, the Hamiltonian doesnot contain a term linear in π (which does not appear anywhere else in the Hamiltonian)and the Ostrogradsky theorem hence does not apply. The instability is thus much moreintricate and of nonlinear nature than the Ostrogradsky one.Our theory is a highly nonlinear field theory and the dynamical instability is rootedin this nonlinearity. In fact there are two different effects destabilizing the Hamiltonian athand. The first is due to the Lagrangian being composed of products of Lorentz invariants.When a term contains four time derivatives it is necessarily accompanied by two inversemetric factors, thus giving the same sign as for the potential part of the Lagrangian. Thisinduces a ghost-like kinetic (squared) term in the Hamiltonian, which thus is not boundedfrom below. Clearly this effect occurs for all even powers of squared time derivatives, butnot for odd powers (like 2,6,10 and so on). A different effect destabilizing the system isdue to higher powers (than two) of time derivatives giving larger factors in the conjugatemomentum (and also in the Euler-Lagrange equations of motion of course) and this inturn implies that the Hamiltonian does not recombine Lorentz SO(3,1) invariants as SO(4)invariants; this SO(4) invariance is broken and that is why w appears in the result (5.33).This yields mixed terms of both signs; of course the reason for the mixed terms of bothsign is that we used constraints to obtain a minimal δ = 4 Lagrangian. After breaking thewould-be SO(4) symmetry of the terms in the Hamiltonian, these constraints induce termsof both signs.Even though the Hamiltonian (5.33) is not positive definite, it clearly provides condi-tions for stability. If all factors in front of the λ s are positive, then the system is stableat the time-dependent level. This can be achieved in different ways; for instance, we couldchoose c | , = 0, c | , , > w i ( w T w ) < , ∀ i ∈ (1 , , . (5.41)– 25 –or c | , > w = 0 and so w is a vector that rotates the time-dependence of the strain tensor (cid:101) D µν intothe nonvanishing eigenvalues λ i .We will show this more explicitly with an example in the next section. In the followingsubsections, however, we will continue with the minimal δ = 4 Lagrangians and checkthat what we observed for the eighth-order Lagrangian is general and thus persists for thetenth-order and twelfth-order Lagrangians. We will now calculate the Hamiltonian corresponding to the Lagrangian (4.19) along thelines of the last subsection. Since the calculation is mostly mechanical and we showed theexplicit calculations for the eighth-order Lagrangian in the last subsection, we will not fleshout the steps here, but simply state the result H = (5 a + 4 a , ) (cid:104) , (cid:105) (cid:104) , (cid:105) − a + 2 a , + 3 a , ) (cid:104) , (cid:105) (cid:104) , (cid:105)(cid:104) , (cid:105) + (10 a + 2 a , + 9 a , ) (cid:104) , (cid:105) (cid:104) , (cid:105) + a , (cid:104) , (cid:105)(cid:104) , (cid:105)−
23 (5 a + 4 a , + 3 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105)(cid:104) , (cid:105) −
14 (5 a + 2 a , + 6 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105) + 12 (15 a + 6 a , + 12 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105)(cid:104) , (cid:105) −
512 (7 a + 2 a , + 6 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105) − a + 4 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105) + 3(10 a + 4 a , + 6 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105)(cid:104) , (cid:105) + 3(5 a + 2 a , + 3 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105)(cid:104) , (cid:105) − a + 2 a , + 9 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105)(cid:104) , (cid:105) + 5 a (cid:104) , (cid:105) + 4 a , (cid:104) , (cid:105)(cid:104) , (cid:105) − a + 6 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105) + 3 a , (cid:104) , (cid:105)(cid:104) , (cid:105)− (5 a + 4 a , + 3 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105) + 3(5 a + 6 a , ) (cid:104) , (cid:105) (cid:104) , (cid:105) + 2 a , (cid:104) , (cid:105)(cid:104) , (cid:105)− (5 a + 2 a , + 6 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105)(cid:104) , (cid:105) + (5 a + 2 a , + 4 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105) − a (cid:104) , (cid:105)− a , (cid:104) , (cid:105)(cid:104) , (cid:105) − a , (cid:104) , (cid:105)(cid:104) , (cid:105) + 13 ( a , + 4 a , + 5 a ) (cid:104) , (cid:105)(cid:104) , (cid:105) + 14 (6 a , + 2 a , + 5 a ) (cid:104) , (cid:105) (cid:104) , (cid:105) −
12 (4 a , − a , − a ) (cid:104) , (cid:105)(cid:104) , (cid:105) + 112 (6 a , + 2 a , + 7 a ) (cid:104) , (cid:105) . (5.42)Rewriting the Hamiltonian in terms of the eigenvalues, λ µ , using Eqs. (5.21) and (5.32),we get H = c | , , (cid:0) − w − w − w w − w w − w w (cid:1) λ λ λ + c | , , (cid:0) − w − w − w w − w w − w w (cid:1) λ λ λ + c | , , (cid:0) − w − w − w w − w w − w w (cid:1) λ λ λ . (5.43)Unfortunately, the Hamiltonian is not positive definite for arbitrary vectors w . The condi-tions for stability are clear however, viz. as long as w is small enough the Hamiltonian ispositive. – 26 – .5 12 derivatives We will now calculate the Hamiltonian corresponding to the Lagrangian (4.29) along thelines of the last subsection.A difference with respect to the other cases, however, is that some of the free parametersin the static energy give rise to terms with 6 time derivatives. To eliminate these we set a , , = − a − a , − a , , (5.44)which leaves us with four free parameters a , a , , a , and a , . The Hamiltonian in termsof the time-dependent brackets reads H = H a + H b + H c , (5.45)where we have defined H a ≡
34 (6 a + 5 a , ) (cid:104) , (cid:105) (cid:104) , (cid:105) − (12 a + 5 a , + 8 a , ) (cid:104) , (cid:105) (cid:104) , (cid:105)(cid:104) , (cid:105)−
38 (12 a + 5 a , + 18 a , ) (cid:104) , (cid:105) (cid:104) , (cid:105) + 34 (24 a + 5 a , + 16 a , + 18 a , ) (cid:104) , (cid:105) (cid:104) , (cid:105)(cid:104) , (cid:105) −
18 (48 a + 5 a , + 32 a , + 54 a , ) (cid:104) , (cid:105) (cid:104) , (cid:105) + a , (cid:104) , (cid:105)(cid:104) , (cid:105)−
12 (6 a + 5 a , + 4 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105)(cid:104) , (cid:105)−
16 (12 a + 5 a , + 4 a , + 3 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105)(cid:104) , (cid:105) + 12 (12 a + 5 a , + 8 a , + 6 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105)(cid:104) , (cid:105) + 14 (18 a + 5 a , + 12 a , + 18 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105) (cid:104) , (cid:105)−
13 (21 a + 5 a , + 14 a , + 18 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105)(cid:104) , (cid:105) + 14 (6 a + a , + 4 a , + 6 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105) , (5.46)– 27 – b ≡ − a + 5 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105) + 3(12 a + 5 a , + 8 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105)(cid:104) , (cid:105) + 32 (12 a + 5 a , + 18 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105)(cid:104) , (cid:105)−
32 (24 a + 5 a , + 16 a , + 18 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105)(cid:104) , (cid:105) + (12 a + 5 a , + 8 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105)(cid:104) , (cid:105)−
32 (24 a + 5 a , + 16 a , + 18 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105)(cid:104) , (cid:105)(cid:104) , (cid:105) + 12 (48 a + 5 a , + 32 a , + 54 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105)(cid:104) , (cid:105) + 6 a (cid:104) , (cid:105) + 5 a , (cid:104) , (cid:105)(cid:104) , (cid:105) + 4 a , (cid:104) , (cid:105)(cid:104) , (cid:105) − a + 4 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105) − (6 a + 5 a , + 4 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105) − a + 3 a , ) (cid:104) , (cid:105) + 6(6 a + 4 a , + 9 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105)(cid:104) , (cid:105) + 6 a , (cid:104) , (cid:105)(cid:104) , (cid:105)−
12 (12 a + 5 a , + 8 a , + 18 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105)(cid:104) , (cid:105) + 12 (12 a + 5 a , + 8 a , + 6 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105) + 3(3 a + 4 a , ) (cid:104) , (cid:105) (cid:104) , (cid:105)− a + 4 a , + 9 a , ) (cid:104) , (cid:105) (cid:104) , (cid:105) + 2 a , (cid:104) , (cid:105)(cid:104) , (cid:105)−
13 (12 a + 5 a , + 8 a , + 18 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105)(cid:104) , (cid:105) −
32 ( a + a , ) (cid:104) , (cid:105)(cid:104) , (cid:105) + 12 (18 a + 5 a , + 12 a , + 18 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105)(cid:104) , (cid:105) −
16 (21 a + 5 a , + 14 a , + 18 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105) , (5.47) H c ≡ a (cid:104) , (cid:105) + a , (cid:104) , (cid:105)(cid:104) , (cid:105) + a , (cid:104) , (cid:105)(cid:104) , (cid:105) −
14 (6 a + 5 a , + 4 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105) + a , (cid:104) , (cid:105) −
16 (12 a + 5 a , + 8 a , + 18 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105)(cid:104) , (cid:105) + 16 (12 a + 5 a , + 8 a , + 6 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105) −
14 ( a + 2 a , ) (cid:104) , (cid:105) + 18 (18 a + 5 a , + 12 a , + 18 a , ) (cid:104) , (cid:105) (cid:104) , (cid:105) −
112 (21 a + 5 a , + 14 a , + 18 a , ) (cid:104) , (cid:105)(cid:104) , (cid:105) + 124 (6 a + a , + 4 a , + 6 a , ) (cid:104) , (cid:105) . (5.48)Rewriting the Hamiltonian in terms of the eigenvalues, λ µ , using Eqs. (5.21) and (5.32),we get H = c | , , (cid:0) − w T w ) − w w + w w + w w ) (cid:1) λ λ λ . (5.49)Unfortunately, the Hamiltonian is not positive definite. The condition for stability isnevertheless clear; as long as w is small enough, the Hamiltonian is positive. In this section we argue that if the theory we constructed is regarded as a low-energyeffective theory, then not only the Skyrmions themselves can only be described at low– 28 –nergies, but perturbations of them also have to be below the scale of validity of theeffective theory. Let us first note what happens to the 3-vector w in the static limit. If we pick thetime-time component of the strain tensor and set λ = 0, we get n · n = w T ( λ i ) w, (6.1)which vanishes in the static limit and since λ i cannot vanish, then w = 0 must hold. Whenwe turn on time dependence, say by a boost, then what happens is that the 3 eigenvalues λ i receive corrections like λ i = ¯ λ i + v λ (cid:48) i + O ( v ) , ∀ i, (6.2)( i not summed over) where ¯ λ i is the static part of the eigenvalue and in order for the straintensor to receive a nonzero time-time component, w must be nonzero. We also know fromthe definition of the diagonalization matrices, that σ + w T w = 1 and it follows that thelength of w is smaller than or equal to unity: w T w ≤
1. The same thus holds for each ofthe components of w .It should now be clear from Eq. (6.1), that at small time derivatives corresponding tosmall velocities or equivalently to small energy scales of the perturbations, the componentsof w (cid:28)
1. This ameliorates the instability and if the perturbations are sufficiently small,then the instability does not occur. Nevertheless, the instability can happen at some criticalvalue of the derivatives, i.e. in the product of temporal and spatial derivatives. Althougha theory which is not manifestly stable is not particularly desirable, this is somewhatexpected, because the expansion in derivatives implicitly corresponds to a low-energy theorywhere high-energy states have been integrated out, leaving higher orders in derivatives aseffective operators in the low-energy effective theory. In particular, we expect the scale ofvalidity of the low-energy effective theory to be below the energy scale where the loweststate has been integrated out. For the Skyrme model with four derivative terms, thiscorresponds to the mass of the ρ meson, while for the generalized Skyrme model with onlythe sixth-order derivative term and the kinetic term, it corresponds instead to the mass ofthe ω meson.The simplest possible perturbations are just excitations of the lowest lying modes ofthe spectrum of the Skyrmions. The lowest modes are of course the zero modes, includingthe translational moduli (other are rotational modes etc.). Other low-lying modes includevibrational modes, see e.g. [35–37].Here we will consider the simplest possible mode to excite, namely the translationalzero mode. As it is a zero mode, the energy of the perturbation is simply given by therelativistic energy being γ ( v ) times the rest mass. Therefore the velocity v translates intoan energy scale. For other types of perturbations, their frequency translates into an energyscale. Let us take the direction of the motion as x for which the Lorentz boost becomes x − x → x − x − vt √ − v (cid:39) x − x − vt, (6.3) See also e.g. Ref. [34]. – 29 –here we have expanded the Lorentz boost in v so it is simply a Galilean boost. We willhence expand the Skyrmion fields in the velocity v as n ≡ ∂ n = ∂ i n ∂x i ∂t = n i δ i v = n v. (6.5)Since n is proportional to n it is clear that the determinant of the strain tensor (cid:101) D vanishesand hence that λ can be chosen to vanish. Although we chose the direction of the boostin this case, it is always possible to write n as a linear combination of the other three n i .Since we choose λ = 0 to be the vanishing eigenvalue, ( σ, u T ) T is the eigenvectorcorresponding to the zero eigenvalue. In this case of the translational zero modes, we knowthe form of the strain tensor (cid:101) D µν = v n · n v n · n v n · n v n · n v n · n n · n n · n n · n v n · n n · n n · n n · n v n · n n · n n · n n · n , (6.6)and so the eigenvector corresponding to the vanishing eigenvalue is σ = 1 √ v , u = − √ v v . (6.7)We need to estimate w . Although we cannot determine w exactly, we know that the lengthof w equals that of u and that it is related to σ and u via V as w = − σ − V T u : w T w = v v , w = v V V V = v √ v sin θ sin χ sin θ cos χ cos θ , (6.8)where θ and χ are functions of spacetime coordinates and possibly of velocity v .If we try to expand the Hamiltonians (5.33), (5.43) and (5.49) in small velocity v (cid:28) H = c | , (¯ λ ¯ λ + ¯ λ ¯ λ + ¯ λ ¯ λ ) + c | , , (¯ λ ¯ λ ¯ λ + ¯ λ ¯ λ ¯ λ + ¯ λ ¯ λ ¯ λ ) − c | , v (cid:2) sin (¯ θ )¯ λ ¯ λ + (cos ¯ θ + sin ¯ χ sin ¯ θ )¯ λ ¯ λ + (cos ¯ θ + cos ¯ χ sin ¯ θ )¯ λ ¯ λ (cid:3) + c | , v (cid:104) λ (cid:48) (¯ λ + ¯ λ ) + λ (cid:48) (¯ λ + ¯ λ ) + λ (cid:48) (¯ λ + ¯ λ ) (cid:105) + 2 c | , , v ( λ (cid:48) + λ (cid:48) + λ (cid:48) )¯ λ ¯ λ ¯ λ + c | , , v (cid:104) λ (cid:48) (¯ λ ¯ λ + ¯ λ ¯ λ ) + λ (cid:48) (¯ λ ¯ λ + ¯ λ ¯ λ ) + λ (cid:48) (¯ λ ¯ λ + ¯ λ ¯ λ ) (cid:105) + O ( v ) , (6.9) If one considers other modes than the translational zero modes, the expression below would insteadbecome of the form n a = iω a n a , (6.4) a not summed over; now the energy scale of the perturbation is directly set by ω a . – 30 – = c | , , (¯ λ ¯ λ ¯ λ + ¯ λ ¯ λ ¯ λ + ¯ λ ¯ λ ¯ λ ) − c | , , v (cid:104) sin (¯ θ )¯ λ ¯ λ ¯ λ + (cos ¯ θ + sin ¯ χ sin ¯ θ )¯ λ ¯ λ ¯ λ + (cos ¯ θ + cos ¯ χ sin ¯ θ )¯ λ ¯ λ ¯ λ (cid:105) + 2 c | , , v ¯ λ ¯ λ ¯ λ (cid:104) λ (cid:48) (¯ λ + ¯ λ ) + λ (cid:48) (¯ λ + ¯ λ ) + λ (cid:48) (¯ λ + ¯ λ ) (cid:105) + c | , , v (cid:104) λ (cid:48) ¯ λ ¯ λ + λ (cid:48) ¯ λ ¯ λ + λ (cid:48) ¯ λ ¯ λ (cid:105) + O ( v ) , (6.10) H = c | , , ¯ λ ¯ λ ¯ λ − c | , , v ¯ λ ¯ λ ¯ λ + 2 c | , , v ¯ λ ¯ λ ¯ λ (cid:104) λ (cid:48) ¯ λ ¯ λ + λ (cid:48) ¯ λ ¯ λ + λ (cid:48) ¯ λ ¯ λ (cid:105) + O ( v ) , (6.11)where the barred symbols stand for their static value. It is very difficult to prove positivityof the leading order terms in general (for a general perturbation) because they come withboth signs; note that λ (cid:48) i ∈ R is only real, but not necessarily positive in general. However,the full eigenvalues λ i > i . On physicalgrounds we expect the energy to increase by perturbing the system, so we expect theleading order correction to be positive.We can do a bit better by focusing on the translational zero mode. In order to estimatewhat happens to the eigenvalues for the translational zero mode, we expand the spacetimestrain tensor (cid:101) D µν to second order in v and find the eigenvalues are modified as λ i = V T n · n n · n n · n n · n n · n n · n n · n n · n n · n + v n · n n · n n · n n · n n · n V, (6.12)where ¯ λ i = lim v → λ i , v λ (cid:48) i = λ i − ¯ λ i . (6.13)If we now rescale the coordinate x → x (cid:48) = (1 + v ) − x and note that (1 + v ) (cid:39) v to second order in v , then we can write λ λ λ = V T n (cid:48) · n (cid:48) n (cid:48) · n n (cid:48) · n n · n (cid:48) n · n n · n n · n (cid:48) n · n n · n V. (6.14)Although we have written the diagonalization on the same form as for the static eigenvalues,it is quite nontrivial to estimate the change in the eigenvalues λ i ; in general the scaling weperformed will affect all eigenvalues and it is hard to even estimate the size of the change.In the case of the twelfth order Hamiltonian, H of Eq. (5.49), we know that eachterm has four derivatives with respect to x and hence it is easy to compare the energiesas follows. The static energy density of the non-boosted system is H = c | , , ¯ λ ¯ λ ¯ λ , (6.15)– 31 –hile for the Galilean boosted system, we have H boosted12 = c | , , (1 + v ) (1 − v )¯ λ ¯ λ ¯ λ = c | , , ¯ λ ¯ λ ¯ λ + O ( v ) , (6.16)that is, to leading order in v , the is no instability under the translational zero mode.In order to determine stability would require a next-to-leading order calculation which,however, is very difficult.For the tenth- and eighth-order Hamiltonians, the derivatives in the spatial directionsare not distributed symmetrically for all terms and therefore it is not possible to comparethe scaled system with the static one, because the scaling we performed breaks isotropy.The breaking of isotropy can also be seen from the appearance of sin θ and sin χ in theabove expressions; it corresponds to part of the diagonalization matrix V that rotates the3-dimensional strain tensor into a diagonal form. We note, however, that the eighth-orderHamiltonian, H of Eq. (5.33) is positive definite to leading order in v if we set c | , = 0and c | , , > v of the translationalzero mode, the eighth- and twelfth-order Hamiltonians are stable and so is the vacuum ofcourse. We expect the same to hold true for the tenth-order Hamiltonian, but it is notstraightforward to prove it.Although the proof of stability in the general case for general perturbations and to next-to-leading order turns out not to be straightforward, we would like to make the followingconjecture. Conjecture 1
The minimal Hamiltonians of orders 8, 10 and 12, in Eqs. (5.33) , (5.43) and (5.49) are stable to leading order in low-energy perturbations and in turn so is thevacuum. If instead we do not expand the Hamiltonians in v , but analyze the conditions for theHamiltonians to remain positive, we get H : 1 − w i + w j ) ≥ , for i (cid:54) = j,c | , , (cid:0) − w i ( w T w ) (cid:1) − c | , w j w k ≥ , for i (cid:54) = j (cid:54) = k, H : 1 − w T w − w i + w w + w w + w w + w j w k ) ≥ , for i (cid:54) = j (cid:54) = k, H : 1 − w T w + 2 w w + 2 w w + 2 w w ) ≥ . (6.17)Let us start with H ; if we choose to set c | , = 0, the problem of stability simplifies to1 − w i ( w T w ) ≥ , (6.18)which by the parametrization (6.8) can be written as1 + 2 v − v + 2 v (cos θ + cos(2 χ ) sin θ )(1 + v ) ≥ , v − v + 2 v cos(2 θ )(1 + v ) ≥ , (6.19)– 32 –f we take the approach of assuming θ and χ to be worst possible, meaning that their valueswill lead to the hardest possible constraint on v , then we get v <
1, but of course we shouldnot trust velocities close to 1 with a Galilean boost; therefore, reinstating the γ factor, weget v < √ . (6.20)A very rough estimate of the energy scale where the effective theory breaks down is then(1 + v )Λ (cid:39) . H ; the constraints for positivity with the parametrization (6.8) read8 − v + 4 v (4 + 5 v ) cos(2 θ ) + v (3 cos(4 θ ) + 8 cos(2 χ ) sin θ )8(1 + v ) ≥ , − θ ( v + v + v cos χ sin θ )(1 + v ) − v (1 + v + 2 v cos θ ) sin θ sin χ (1 + v ) − v sin θ sin (2 χ )(1 + v ) ≥ , − v cos χ sin θ v − θ ( v + v + 2 v cos χ sin θ )(1 + v ) − v (sin (2 θ ) sin χ + sin θ sin (2 χ ))(1 + v ) ≥ , (6.21)Taking again the approach of minimizing each constraint with respect to θ and χ to getthe hardest constraints on v , we arrive at1 − v − v (1 + v ) ≥ , − v − v (1 + v ) ≥ , (6.22)of which the first one gives the hardest constraint on v . Reinstating the relativistic γ factor,we get v < (cid:115) √ − (cid:39) . . (6.23)Considering finally H ; the constraints for positivity with the parametrization (6.8)read 8 − v − v + v (4 cos(2 θ ) + 7 cos(4 θ ) + 8 cos(4 χ ) sin θ )8(1 + v ) ≥ , (6.24)whose hardest constraint on v is 2 − v − v v ) ≥ . (6.25)– 33 –einstating the relativistic γ factor, we get the constraint v < (cid:115) √ − (cid:39) . . (6.26)We note that increasing the order of the Lagrangian, slightly reduces the maximalscale at which the theory will break down. This is somewhat counter intuitive, but weshould recall that we work at a fixed order of derivatives in the i -th direction and increasethe total number of derivatives.We have thus shown that relativistic speeds of the order of about half the speed oflight are necessary before the effective theory will break down. In this paper, we have constructed a formalism for higher-derivative theories based onO(4) invariants. We started with reviewing the construction made by Marleau, but foundthat it possesses an instability in the static energy for all the Lagrangians of higher thansixth order in derivatives. The instability can be triggered by a baby-Skyrmion string-likeperturbation that can then run away (see App. A). The problem of the latter constructionis the desire to limit the radial profile function to have a second-order radial equation ofmotion. This comes at the cost of the angular derivatives conspiring at large order inderivatives to create negative terms. This can be seen by writing the static Lagrangian interms of eigenvalues of the derivatives of the O(4) invariants. We cure the instability byconstructing an isotropic construction where no special direction (e.g. radial) is preferred tohave lower order in derivatives than others. This construction necessitates four derivativesin the i -th direction for the Lagrangians of order 8, 10 and 12.We successfully constructed positive definite static energies for the Lagrangians of or-der 8, 10 and 12 with very simple interpretations. The eighth-order Lagrangian can beinterpreted as the Skyrme term squared plus the Dirichlet energy (normal kinetic term)multiplied by the BPS-Skyrme term. The tenth-order Lagrangian instead can be inter-preted as the Skyrme term multiplied by the BPS-Skyrme term. Finally, the twelfth-orderLagrangian can simply be understood as the BPS-Skyrme term squared.Although our construction straightforwardly yields stable static energies for higher-order systems, constructing the full Hamiltonians revealed that time-dependent pertur-bations may potentially destabilize the system and in turn its solitons. The (dynamical)instability we found is intrinsically different from the Ostrogradsky instability as it is notrelated to the Hamiltonian phase space being enlarged, but just to the canonical momentabeing nonlinear and in turn inducing terms of both signs. The nonlinearity induces twoeffects that destabilize the Hamiltonian; one is simply the square of the inverse metric forfour time derivatives, which remains negative. The other effect is that under the Legen-dre transform from the Lagrangian to the Hamiltonian, nonlinearities break the normalwould-be SO(4) symmetry (which is basically the Wick rotated SO(3,1) Lorentz symme-try). Although this may not be problematic itself, it induces terms of both signs in ourconstruction. After reducing the expressions using the eigenvalue formalism, we obtain– 34 –lear-cut conditions for positivity of the Hamiltonian given in terms of one of the vectorsof the diagonalization matrix, which has the physical interpretation of a rotation of thestrain tensor into the time-direction. Further analysis may reveal whether this effect trulydestabilizes the Hamiltonian or not.Finally, we argued that to leading order in time-dependence of the perturbations underconsideration, our construction is stable. We checked this to leading order in velocityshowing that the Hamiltonians of eighth and twelfth order do not destabilize. In case ofthe tenth-order Hamiltonian, we have not been able to prove stability to leading orderalthough we expect the same to hold true also in this case. We conjectured that theHamiltonians corresponding to the minimal Lagrangians in our construction are stable toleading orders of general low-energy perturbations and in turn so is the vacuum.It will be interesting to study the dynamical instability that we encountered here morein detail and to see whether it is possible to cure it. One hope could be to use only oddpowers of squared time derivatives, giving always an odd number of inverse metric factors.This may, however, not be enough to construct a manifestly positive Hamiltonian due tothe second instability effect that we discussed above.Although it may be less likely in our case, it is possible that dynamical stability canbe achieved in some parts of parameter space, i.e. for certain values of the constants in theLagrangians. For a simpler dynamical system, namely the Pais-Uhlenbeck oscillator [31]islands of stability were found for several interacting systems and even bounded Hamilto-nians can be found in some cases [38–41]. In our Lagrangians it seems less likely to bepossible, because the instability that we found also manifests itself with just a single overallcoupling constant that can be scaled away.Another hope for a manifestly stable Hamiltonian could be some construction withinfinitely many derivative terms resummed in a clever fashion. One of our motivations to construct a higher-order Skyrme-like term was to probewhether black hole Skyrme hair is stable only for the Skyrme term or unstable only for theBPS-Skyrme term [44–47].In our construction with minimal number of derivatives in the i -th direction – whichwe call the minimal Lagrangians – all time derivatives are necessarily multiplied by spatialderivatives to leading order. Therefore if some instability occurs, it will be amplified bythe presence of solitons.Our higher-order terms if added to the Skyrme model will give corrections to the prop-erties of the Skyrmions, including the mass, size and binding energy. Not only Skyrmions,but also the Skyrme-instanton [48, 49] will receive corrections from the new higher-orderterms. It will be interesting to study such corrections in the future.Another interesting direction will be a supersymmetric extension of our discussion.While supersymmetric extensions of the Skyrme model (of the fourth order) were studied In Ref. [42], the Skyrme model was constructed as the low-energy effective theory on a domain wall upto the fourth-derivative order [42]. However, a non-Skyrme term containing four time derivatives also existsat this order [43]. The effective Lagrangian looks unstable at this order, but the domain wall itself must bestable for a topological reason. Probably, all terms with infinitely many derivatives cure this problem. – 35 –n Refs. [21–24], a discussion of topological solitons in supersymmetric theories with moregeneral higher-derivative terms can be found in e.g. Refs. [50–55].
Acknowledgments
We thank Martin Speight for useful discussions. S. B. G. thanks the Recruitment Programof High-end Foreign Experts for support. The work of S. B. G. was supported by the Na-tional Natural Science Foundation of China (Grant No. 11675223). The work of M. N. issupported in part by a Grant-in-Aid for Scientific Research on Innovative Areas “Topolog-ical Materials Science” (KAKENHI Grant No. 15H05855) from the the Ministry of Educa-tion, Culture, Sports, Science (MEXT) of Japan, by the Japan Society for the Promotionof Science (JSPS) Grant-in-Aid for Scientific Research (KAKENHI Grant No. 16H03984)and by the MEXT-Supported Program for the Strategic Research Foundation at PrivateUniversities “Topological Science” (Grant No. S1511006).
A Baby-Skyrmion string run-away perturbation in the Marleau con-struction
In this appendix, we will provide an example of the instability for illustrative purposes. Letus for concreteness limit the example to a system with a kinetic term and an eighth-orderMarleau Lagrangian of Eq. (3.28), L = L + L Marleau8 = −(cid:104) (cid:105) − (cid:104) (cid:105)(cid:104) (cid:105) + 316 (cid:104) (cid:105) + 98 (cid:104) (cid:105)(cid:104) (cid:105) − (cid:104) (cid:105) , (A.1)where we have set a = a , = 1 for simplicity (since there are only two constants, theycorrespond just to setting the length and energy units).Instead of evolving the full equations of motion, let us just make a simplified simulation,i.e. cooling the static equations of motion. That system can be written as − n bij (cid:88) r =1 ∂ (cid:104) r (cid:105) ∂n ai ∂n bj ∂ L ∂ (cid:104) r (cid:105) − n bij (cid:88) r,s =1 ∂ (cid:104) r (cid:105) ∂n ai ∂ (cid:104) s (cid:105) ∂n bj ∂ L ∂ (cid:104) r (cid:105) ∂ (cid:104) s (cid:105) = n a . (A.2)For illustrative purposes, we will choose a 1-Skyrmion and perturb the tale of it witha baby-Skyrmion string. The baby-Skyrmion string carries no baryon number and in thenormal Skyrme model it will just be some extra energy that can be radiated away to findjust the 1-Skyrmion being the minimum of the energy.In this example, however, we have switched the Skyrme term for the eighth-orderMarleau term and hence as shown in Eq. (3.23), the baby-Skyrmion string will give rise toa negative energy density that can cause a run-away.In Fig. 1 is shown the configuration at the initial time. The 1-Skyrmion is alreadythe minimum of the energy functional and its fields have been found using the hedgehogLagrangian (3.7) with n = 4. The baby-Skyrmion string is not a topological object, butjust a perturbation added to the configuration. In Fig. 2 is shown a series of three snapshots– 36 – igure 1 . The absolute value of the energy density |E| of the configuration containing a 1-Skyrmion(the colored sphere) and a baby-Skyrmion string (the black vertical string). Figure 2 . Cooling of the configuration shown in Fig. 1 at time τ = 0, τ = 60 and τ = 120,respectively. The figure shows the energy density in an xy -slice at fixed z = 0. It is seen fromthe figure that the 1-Skyrmion is unchanged as cooling time increases, but the energy of the baby-Skyrmion string (to the right) is growing negative. in cooling time τ = 0 , ,
120 of the configuration. The 1-Skyrmion is stable and remainsa solution, but the baby-Skyrmion string is seen to grow more and more negative. Finally,we show the peak energies of the two objects in Fig. 3. The 1-Skyrmion has positive peakenergy that remains stable, whereas the baby-Skyrmion string has a negative peak energythat grows rapidly more and more negative. This nicely illustrates the baby-Skyrmionstring instability found in the Marleau construction.– 37 – negative peak energy 10 100 1000 10000 100000 1x10
20 40 60 80 100 120 140 160 180
Figure 3 . Peak energies: positive for the 1-Skyrmion and negative for the baby-Skyrmion stringas functions of the cooling time τ . It is seen from the figure that the 1-Skyrmion is unchanged, butthe energy of the baby-Skyrmion string (to the right) is growing negative. B Difference from the Ostrogradsky instability
Let us compare the simplest possible term giving rise to four time derivatives in our theorieswith that of the Ostrogradsky-like theories, L = −(cid:104) (cid:105) = − ( ∂ µ n · ∂ µ n ) (B.1)which is just the kinetic term squared. Recall that the nonlinear sigma model constraint n · n = 1 implies 12 ∂ µ ( n · n ) = n · ∂ µ n = 0 . (B.2)Consider now integrating the Lagrangian (B.1) by parts as L = − ( ∂ µ n · ∂ µ n )( ∂ ν n · ∂ ν n )= − ∂ µ [( n · ∂ µ n )( ∂ ν n · ∂ ν n )] + ( n · ∂ µ ∂ µ n )( ∂ ν n · ∂ ν n ) + ( n · ∂ µ n ) ∂ µ ( ∂ ν n · ∂ ν n )= ( n · ∂ µ ∂ µ n )( ∂ ν n · ∂ ν n )= ∂ ν [( n · ∂ µ ∂ µ n )( n · ∂ ν n )] − [ ∂ ν ( n · ∂ µ ∂ µ n )] ( n · ∂ ν n ) − ( n · ∂ µ ∂ µ n )( n · ∂ ν ∂ ν n )= − ( n · ∂ µ ∂ µ n )( n · ∂ ν ∂ ν n ) , (B.3)which obviously differs from the Ostrogradsky-like Lagrangian [30] L = − ∂ µ ∂ µ n · ∂ ν ∂ ν n . (B.4)The reason why we do not have the Ostrogradsky instability, exactly, is because the propa-gator of Eq. (B.3) is not p ; it remains p and the term is still just a product of two kineticterms. – 38 –rying to formally manipulate the Ostrogradsky-like Lagrangian (B.4), we get L = − ∂ µ ∂ µ n · ∂ ν ∂ ν n = − ( ∂ µ ∂ µ n · ∂ ν ∂ ν n )( n · n )= − ∂ µ [( ∂ µ n · ∂ ν ∂ ν n )( n · n )] + ( ∂ µ n · ∂ µ ∂ ν ∂ ν n )( n · n ) + 2( ∂ µ n · ∂ ν ∂ ν n )( n · ∂ µ n )= − ∂ µ [( ∂ µ n · ∂ ν ∂ ν n )] + ( ∂ µ n · ∂ µ ∂ ν ∂ ν n ) + 2 ∂ ν [( ∂ µ n · ∂ ν n )( n · ∂ µ n )] − ∂ µ ∂ ν n · ∂ ν n )( n · ∂ µ n ) − ∂ µ n · ∂ ν n )( n · ∂ µ ∂ ν n ) − ∂ µ n · ∂ ν n )( ∂ ν n · ∂ µ n )= − ∂ µ [( ∂ µ n · ∂ ν ∂ ν n )] + ( ∂ µ n · ∂ µ ∂ ν ∂ ν n ) + 2 ∂ ν [( ∂ µ n · ∂ ν n )( n · ∂ µ n )] − ∂ µ [( ∂ ν n · ∂ ν n )( n · ∂ µ n )] + ( ∂ ν n · ∂ ν n )( ∂ µ n · ∂ µ n ) + ( ∂ ν n · ∂ ν n )( n · ∂ µ ∂ µ n ) − ∂ µ n · ∂ ν n )( n · ∂ µ ∂ ν n ) − ∂ µ n · ∂ ν n )( ∂ ν n · ∂ µ n )= − ∂ µ [( ∂ µ n · ∂ ν ∂ ν n )] + ( ∂ µ n · ∂ µ ∂ ν ∂ ν n ) + ( ∂ ν n · ∂ ν n )( ∂ µ n · ∂ µ n )+ ( ∂ ν n · ∂ ν n )( n · ∂ µ ∂ µ n ) − ∂ µ n · ∂ ν n )( n · ∂ µ ∂ ν n ) − ∂ µ n · ∂ ν n )( ∂ ν n · ∂ µ n ) , (B.5)where in the last equation we have used Eq. (B.2). We can identify the (cid:104) (cid:105) and the − (cid:104) (cid:105) terms in the last equation. However, deriving the constraint (B.2) once more, we get that ∂ ν n · ∂ µ n + n · ∂ µ ∂ ν n = 0 , (B.6)where ν can also be equal to µ and summed over; it is a general statement derived from thenonlinear sigma model constraint. Using this relation, however, we can simplify Eq. (B.5)to L = − ∂ µ [( ∂ µ n · ∂ ν ∂ ν n )] + ( ∂ µ n · ∂ µ ∂ ν ∂ ν n ) , (B.7)which is simply the Ostrogradsky-like Lagrangian that we started with. Therefore, we cansee that the Lagrangian (B.1) is not just the Ostrogradsky-like Lagrangian (B.4) up to atotal derivative.Nevertheless, this exercise should show that the Ostrogradsky system is intrinsicallydifferent from our Lagrangians and that we do not have p in the propagator, but just ahighly nonlinear theory. References [1] T. H. R. Skyrme, “A Unified Field Theory of Mesons and Baryons,” Nucl. Phys. , 556(1962). doi:10.1016/0029-5582(62)90775-7[2] T. H. R. Skyrme, “A Nonlinear field theory,” Proc. Roy. Soc. Lond. A , 127 (1961).doi:10.1098/rspa.1961.0018[3] E. Witten, “Global Aspects of Current Algebra,” Nucl. Phys. B , 422 (1983).doi:10.1016/0550-3213(83)90063-9[4] E. Witten, “Current Algebra, Baryons, and Quark Confinement,” Nucl. Phys. B , 433(1983). doi:10.1016/0550-3213(83)90064-0[5] A. Zaks, “Derivation of the Skyrme-witten Lagrangian From QCD,” Nucl. Phys. B , 241(1985). doi:10.1016/0550-3213(85)90321-9 – 39 –
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