Collective quantization of axially symmetric gravitating B=2 skyrmion

1.1 MB·English

by  Hisayuki Sato

Collective quantization of axially symmetric
gravitating B = 2 skyrmion
Hisayuki Sato1, Nobuyuki Sawado1, and Noriko Shiiki1,2
∗ † ‡
1Department of Physics, Faculty of Science and Technology,
Tokyo University of Science, Noda, Chiba 278-8510, Japan
2Department of Management, Atomi University, Niiza, Saitama 352-8501, Japan
(Dated: February 2, 2008)
In this paper we perform collective quantization of an axially symmetric skyrmion with baryon
number two. The rotational and isorotational modes are quantized to obtain the static properties
ofadeuteronandotherdibaryonicobjectssuchasmasses,chargedensities,magneticmoments. We
discuss how thegravity affects to those observables.
PACSnumbers: 12.39.Dc,21.10.-k,04.20.-q
7
0
I. INTRODUCTION studying the effects of gravityonsolitonspectra is inter-
0
esting itself. Nevertheless we further attempt to give an
2
The Skyrme model [1] is considered as an unified the- interpretation to the solution as a gravitating deuteron
n
or dibaryonic object and study their mass spectra and
a ory of hadrons by incorporating baryons as topological
other static observables.
J solitons of pion fields, called skyrmions. The topological

5 chargeis identified asthe baryonnumber B. Performing The first step towards the study of gravitational ef-
collective quantization for a B = 1 skyrmion, one can fects on the quantum spectra of skyrmions was taken in
4 obtain proton and neutron states within 30% error [2]. Ref.[10]byperformingcollectivequantizationofaB =1
v
Correspondingly multi-skyrmion solutions are ex- gravitatingskyrmion. It wasshownthere thatthe quali-
6
9 pected to represent nuclei [3, 4, 5]. The static proper- tative change in the mass difference, mean charge radius
1 ties of a B = 2 skyrmion such as mass spectra, mean andchargedensitiesunderthestronggravitationalinflu-
9 charge radius, baryon-number density, magnetic mo- ence confirms the attractive feature of gravity while the
0 ment, quadrupole moment and transition moment were reduction of the axial coupling and transition moments
6 studied in detail by Braaten and Carson upon collective by the strong gravity indicates the gravitational effects
0 zero mode quantization [6]. The results confirmed that as a stabilizer of baryons.
/
h the quantized B = 2 skyrmion can be interpreted as a
In this paper we extend the work [10] to the B = 2
t deuteron.
- axiallysymmetricgravitatingskyrmionandestimatethe
p TheEinstein-Skyrme(ES)modelinwhichtheSkyrme
e fields coupled to gravity was first considered by Luckock static properties of a deuteron or other dibaryonic ob-
h jects. We observe how strong gravity affects the baryon
andMoss[7]. Theyobtainedsphericallysymmetricblack
: observables e.g., mass spectra, mean charge radius,
v holesolutionswithSkyrmehair. Itisthefirstdiscovered
i counter example to the no-hair conjecture. Axially sym- baryon-number density, magnetic moment, quadrupole
X
moment and transition moment.
metric regular and black hole skyrmion solutions with
r
a B = 2 were constructed in [8]. Subsequently the model Although the Skyrme model describes nucleons with
was extended to the SU(3) and higher baryon number about30%error,thepossibilitythatitmayprovidequal-
with discrete symmetries to study gravitating skyrmion itatively correctdescriptionofthe interactionofbaryons
solutions [9]. with gravity can not be excluded. It is expected that in
In the Einstein-Skyrme theory, the Planck mass is re- theearlyuniverseorequivalenthighenergyexperiments,
latedtothepiondecayconstantfπ andcouplingconstant the gravitationalinteraction with baryons is not negligi-
α by Mpl = fπ 4π/α. To realize the realistic value of ble. We hope that our work could provide insight into
the Planck masps, the coupling constant should be ex- the observations in such situations.
tremely small with α O(10 39), which makes the ef-
−
∼
fectsofgravityontheskyrmionnegligible. We,therefore,
considerαasafreeparameterandstudy the strongcou-
pling limit to manifest the effects of gravityon skyrmion
spectra. The property of the skyrmion solution could
be drastically changed for large values of α. Certainly II. CLASSICAL GRAVITATING B =2
SKYRMIONS
Inthis section,wediscussB =2classicalregularsolu-
∗Electronicaddress: [email protected]
†Electronicaddress: [email protected] tions in the Einstein-Skyrme system (ES). The Skyrme
‡Electronicaddress: [email protected] Lagrangian coupled to gravity is defined by the La-
2
grangian whosezerothcomponentcorrespondstothebaryonnum-
ber density
= + , (1)
G S
LLG =L161πGLR, (2) B0 =−π2√1 g sin2F sinΘ(∂xF∂θΘ−∂θF∂xΘ). (10)
−
f2
= πgµνTr(U 1∂ UU 1∂ U) For the solutions to be regular at the origin x = 0
S − µ − ν
L 16
and to be asymptotically flat at infinity, the following
1
+ gµρgνσTr([U 1∂ U,U 1∂ U] boundary conditions must be imposed
32e2 − µ − ν
[U 1∂ U,U 1∂ U]), (3) ∂ f(0,θ)=∂ m(0,θ)=∂ l(0,θ)=0, (11)
− ρ − σ x x x
×
f( ,θ)=m( ,θ)=l( ,θ)=1. (12)
where f is pion decay constant and e is a dimensionless ∞ ∞ ∞
π
free parameter. U describes the SU(2) chiral fields. We
Fortheconfigurationtobeaxiallysymmetric,thefollow-
impose the axially symmetric ansatz on the chiral fields
ing boundary conditions must be imposed at θ = 0 and
as a possible candidate for the B = 2 minimal energy
π/2
configuration [6]
∂ f(x,0)=∂ m(x,0)=∂ l(x,0)=0, (13)
U =cosF(r,θ)+iτ n sinF(r,θ), (4) θ θ θ
R π π π
· ∂ f(x, )=∂ m(x, )=∂ l(x, )=0. (14)
θ θ θ
2 2 2
with
For the profile functions, the boundary conditions at the
n =(sinΘ(r,θ)cosnϕ,sinΘ(r,θ)sinnϕ,cosΘ(r,θ)),(5)
R x=0, are given by
∞
where n denotes the winding number of solitons, which
F(0,θ)=π, F( ,θ)=0, (15)
is equivalent to the baryon number B. Since we are in- ∞
∂ Θ(0,θ)=∂ Θ( ,θ)=0. (16)
terested in B = 2 skyrmions, we consider only n = 2. x x
∞
Correspondingly, the following axially symmetric ansatz
At θ =0 and π/2,
is imposed on the metric [11]
π
m l ∂ F(x,0)=∂ F(x, )=0, (17)
ds2 = fdt2+ (dr2+r2dθ2)+ r2sin2θdϕ2 (6) θ θ 2
− f f π π
Θ(x,0)=0 , Θ(x, )= . (18)
2 2
where the metric functions f , m and l are the function
of coordinates r and θ. This metric is symmetric with Sincethebaryonnumberisdefinedbythespatialintegral
respect to the z-axis (θ = 0). Substituting these ansatz of the zeroth component of the baryon current, we have
to the Lagrangian (3), one obtains the following static
energy density for the chiral fields
B= d3r√ gB0
Z −
√lsinθ
ε(x,θ)= x2 (∂ F)2+(∂ Θ)2sin2F 1 F1,Θ1
x x
8 = (2F sin2F)cosΘ . (19)
n (cid:0) (cid:1) 2π − (cid:12)
n2m (cid:12)F0,Θ0
+(∂ F)2+(∂ Θ)2sin2F + sin2F sin2Θ (cid:12)
θ θ lsinθ (cid:12)
o The inner and outer boundary conditions (F ,Θ ) =
0 0
+ √lsinθ f (∂ F∂ Θ ∂ F∂ Θ)2sin2F (π,0) and (F1,Θ1)=(0,π) yield B =2.
x θ θ x
2 (cid:20)m − By taking a variation of the static energy (7) with re-
n2f 1 spect to F and Θ, one obtains the equations of motion
+ lsin2θ sin2F sin2Θn(cid:0)(∂xF)2+ x2(∂θF)2(cid:1) froicrtfuhnecptrioonfisleffu,nmctioannsd. lTahreefideeldriveeqduaftrioomnstfhoertEhiensmteeitn-
1
+ (∂ Θ)2+ (∂ Θ)2 sin2F , (7) equations. WeshallshowtheirexplicitforminAppendix
x x2 θ o(cid:21) A.
(cid:0) (cid:1)
TheeffectivecouplingconstantoftheEinstein-Skyrme
where dimensionless variable x = ef r is introduced.
π system is given by
The static (classical) energy is thus given by
α=4πGf2 (20)
fπ π
M =2π dxdθε(x,θ). (8)
e Z
which is the only free parameter.
We use the relaxationmethod to solve these nonlinear
The covarianttopological current is defined by
equations with the typical grid size 100 30. In Fig.1,
ǫµνρσ 1 the profile functions for α = 0,0.04,0.0×8,0.126 is pre-
Bµ = 24π2 √ gtr(U−1∇νUU−1∇ρUUU−1∇σU), (9) sented. No solutionexist for α&0.127. Also,the metric
−
3
functionsareshowninFigs.2 4. InFig.5,wedisplaythe Theseoperatorsarerelatedtothe usualcoordinate-fixed
−
metric functions at θ = 0,π/4,π/2 as a function of ra- isospin I and spin J by the orthogonal transformation,
i i
dial coordinate. Fig.6 shows α dependence of the static
energy M in unit of fπ/e. In Fig. 7 the energy densi- Ii=−Rij(A)Kj , Ji =−Rij(B)TLj. (29)
ties defined in Eq.(7) are plotted. The baryon densities
The commutation relations for these operators are
b=√ gB0 are shown in Fig. 8.
−
[K ,K ]=iǫ K , [L ,L ]=iǫ L ,
i j ijk k i j ijk k
[I ,I ]=iǫ I , [J ,J ]=iǫ J . (30)
i j ijk k i j ijk k
These operators satisfy the Casimir invariance by using
III. COLLECTIVE QUANTIZATION SCHEME
Eq.(29)
FOR THE B=2 SOLITON SOLUTION
K2=I2 , L2 =J2. (31)
The symmetry ofthe classicalsolitoninduces the follow-
In this section, let us try to give an interpretation to
ing conditions for the inertia tensors
the B =2 skyrmion solutionobtained in the last section
as a baryonic object by assigning quantum number of U =U , V =V , W =W =0,
11 22 11 22 11 22
spinandisospintoit. Weemploythesemi-classicalzero-
V =4U , W =2U . (32)
33 33 33 33
mode quantization following Ref. [6]. Let us introduce
dynamically rotated chiral fields Thus it is sufficient to calculate only U , U and V .
11 33 11
Inserting (4) into Eqs.(25) and (27) one obtains
Uˆ(r,t)=A(t)U(r )A (t), r =R(B(t))r, (21)
′ † ′
U = π dxdθ m√lx2sinθsin2F(1+cos2Θ)
and 11 fπe3 Z h 4f2
Rij(B)= 1Tr(τiBτjB†), (22) +√flsinθsin2F (1+cos2Θ)(x2F,2x+F,2θ)
2 +sin2F cos2Θ((cid:8)x2Θ2 +Θ2)
,x ,θ
where A(t) and B(t) are the time dependent SU(2) ma- + n2m sin2F sin2Θ , (33)
trices generating the isospin and the spatial rotations. lsin2θ
(cid:9)i
Substituting (21) into the Lagrangian(3), one finds
U = π dxdθ m√lx2sinθsin2Fsin2Θ
33 fπe3 Z h 2f2
1 1
L= M + aiUijaj aiWijbj + biVijbj, (23) +2√lsinθsin2F sin2Θ x2F2 +F2
− 2 − 2 f ,x ,θ
(cid:8)
where M is the static energy of the B = 2 skyrmion +sin2F(x2Θ2 +Θ2) , (34)
,x ,θ
given in Eq.(8). The Lagrangian is quadratic in time (cid:9)i
derivatives V = π dxdθ m√lx2sinθ(F2 +Θ2 sin2F
11 fπe3 Z h 4f2 ,x ,θ
a = iTrτ A A˙ , b =iTrτ B˙B . (24) +n2cot2θsin2F sin2Θ)
i i † i i †
−
+√lx2sinθsin2F (F Θ F Θ )2
The moments of inertia tensor Uij , Vij and Wij are ex- f ,x ,θ− ,θ ,x
pressed in terms of the chiral field U as +n2(F2(cid:8)+Θ2 sin2F)cot2θsin2Θ
,x ,x
+n2√l(cos2θ+ m) (cid:9)
Uij=8fπ1e3 Z d3xhmf√2lx2sinθ(cid:8)Tr(cid:0)U†[τ2i,U]U†(cid:2)τ2j,U(cid:3)(cid:1) fsinθ (F2 +Θ2l sin2F)sin2F sin2Θ . (35)
+gklTr U†∂kU,U†[τ2i,U] U†∂lU,U†[τ2j,U] ,(25) × ,θ ,θ i
Wij=Uij [(cid:2)τ2j,U]→i(x′time(cid:3)s(cid:2)∇)jU) , (cid:3)(cid:9)i (26) From Eqs.(28) and (32) we derive the constraint
Vij=Wij(cid:8)[τ2i,U]→i(x′times∇)iU)(cid:9). (27) (2K3+L3)|physi=0. (36)
(cid:8) (cid:9)
Body-fixed isospin operator K and angular momentum Insertingthebody-fixedoperator(28),Casimirinvariant
i
operator L are defined as a canonically conjugate to a (31)andconstraint(36)intotheLagrangian(23)onecan
i i
and b get the Hamiltonian operator as
i
I2 J2
∂L
K = =U a W b , H =M + +
i ∂ai ij j − ij j 2U11 2V11
∂L 1 1 4 K2
L = = WTa +V b . (28) + 3. (37)
i ∂bi − ij j ij j (cid:20)U33 − U11 − V11(cid:21) 2
4
α = 0.000 F (P,Z) 3 α = 0.000 Θ (P,Z) 1.4
3 22..74 1.5 1 . 21
F (P,Z) 12 ..255 2111....1852 Θ (P,Z) 1 0000....8642
0.9
1 0.6 0.5
0.5 0.3
0 0
8 8
7 7
6 6
5 5
4 4
Z 3 2 1 0 0 1 2 3 4 P 5 6 7 8 Z 3 2 1 0 0 1 2 3 4 P 5 6 7 8
α = 0.040 F (P,Z) 3 α = 0.040 Θ (P,Z) 1.4
3 22..74 1.5 1 . 21
F (P,Z) 12 ..255 2111....1852 Θ (P,Z) 1 0000....8642
0.9
1 0.6 0.5
0.5 0.3
0 0
8 8
7 7
6 6
5 5
4 4
Z 3 2 1 0 0 1 2 3 4 P 5 6 7 8 Z 3 2 1 0 0 1 2 3 4 P 5 6 7 8
α = 0.080 F (P,Z) 3 α = 0.080 Θ (P,Z) 1.4
3 22..74 1.5 1 . 21
F (P,Z) 12 ..255 2111....1852 Θ (P,Z) 1 0000....8642
0.9
1 0.6 0.5
0.5 0.3
0 0
8 8
7 7
6 6
5 5
4 4
Z 3 2 1 0 0 1 2 3 4 P 5 6 7 8 Z 3 2 1 0 0 1 2 3 4 P 5 6 7 8
α = 0.126 F (P,Z) 3 α = 0.126 Θ (P,Z) 1.4
3 22..74 1.5 1 . 21
F (P,Z) 12 ..255 2111....1852 Θ (P,Z) 1 0000....8642
0.9
1 0.6 0.5
0.5 0.3
0 0
8 8
7 7
6 6
5 5
4 4
Z 3 2 1 0 0 1 2 3 4 P 5 6 7 8 Z 3 2 1 0 0 1 2 3 4 P 5 6 7 8
FIG. 1: The profile function F and Θ in the cylindrical coordinate with α = 0.000,0.040,0.080,0.126 are shown. We use
dimensionless variables P =efπρ and Z =efπz. There exists no solution for α&0.127.
5
α = 0.040 f (ξ,ζ) 0.993 α = 0.040 m (ξ,ζ)
0.953 0.995
0.912 0.99
1 00..887333 1 0 .09.8958
0.792 0.95 0 .09.7957
ξζf (,) 0 .07.55 ξζm (,) 0 .008..589
0.75
0.25 0.7
1 1
0.8 0.8
0.6 0.6
ζ ζ
0.4 0.4
0.2 0.2
0 0 0.2 0.4 0.6 0.8 1 0 0 0.2 0.4 0.6 0.8 1
ξ ξ
α = 0.080 f (ξ,ζ) 0.945 α = 0.080 m (ξ,ζ)
0.885 0.995
0.825 0.98
1 00..776055 1 0 .09.6955
00..654855 0.95 0 .09.3952
ξζf (,) 0 .07.55 ξζm (,) 0 .008..589
0.75
0.25 0.7
1 1
0.8 0.8
0.6 0.6
ζ ζ
0.4 0.4
0.2 0.2
0 0 0.2 0.4 0.6 0.8 1 0 0 0.2 0.4 0.6 0.8 1
ξ ξ
α = 0.126 f (ξ,ζ) 0.855 α = 0.126 m (ξ,ζ) 0.96
0.705 0.92
0.555 0.88
1 00..420555 1 0 .08.48
0.95 0.76
ξζf (,) 0 .07.55 ξζm (,) 0 .008..589
0.75
0.25 0.7
1 1
0.8 0.8
0.6 0.6
ζ ζ
0.4 0.4
0.2 0.2
0 0 0.2 0.4 0.6 0.8 1 0 0 0.2 0.4 0.6 0.8 1
ξ ξ
FIG. 2: The metric function f in the cylindrical coordinate FIG. 3: Same as Fig.2 for themetric function m.
with α=0.040,0.080,0.126 is shown. Weuse ξ =P/(1+P)
and ζ = Z/(1+Z), instead of dimensionless variables P =
efπρ and Z =efπz.
tively. The parity is defined by the eigenstate of the
following operator
The corresponding energy (mass) eigenvalues are
P =eiπK3, (39)
i(i+1) j(j+1)
E =M + +
2U 2V
11 11
1 1 4 κ2 which means that the parity is (+) for even κ and ( )
−
+ (38) for odd κ.
(cid:20)U − U − V (cid:21) 2
33 11 11
Taking into account the Finkelstein-Rubinstein con-
where i(i+1),j(j+1) and κ are the eigenvalues of the straints for the axially symmetric soliton [6], one finds
Casimir operators (31), and the operator K , respec- that some combinations of (i,j) are not acceptable. The
3
6
α = 0.040 l (ξ,ζ) 0.998 1
0.995
0.993
0.99
1 0.988 0.9
0.985
0.95 0.983 α=0.040
0.8
ξζl (,) 0 .08.59 0.7
0.8
0.75 χθ,) 0.6 α=0.080
f (
0.5
1
0.8 0.4
ζ 0. 60.4 0.3 α=0.126 θθ== π /04
θ=π/2
0.2
0 0 0.2 0.4 ξ 0.6 0.8 1 0.2 0 0.2 0.4 χ 0.6 0.8 1
α = 0.080 l (ξ,ζ) 0.998
000...999876888 1 α=0.040
1 0.958
0.948
0.95
α=0.080
ξζl (,) 0 .08.59 0.95
0.8
θ)
0.75 χ, 0.9
l (
1
0.85
0.8
ζ 0. 60.4 α=0.126 θθθ=== ππ 0//42
0.2 0.8
0 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1
ξ χ
α = 0.126 l (ξ,ζ) 0.981
0.951 1
0.921 α=0.040
0.891
1 0.861
0.95 00..883011 0.95 α=0.080
ξζl (,) 0 .08.59
0.9
0.8
θ)
0.75 χ,
m ( 0.85
1 α=0.126
0.8 0.8
ζ 0. 60.4 θθ== π 0/4
0.2 0.75 θ=π/2
0 0 0.2 0.4 ξ 0.6 0.8 1 0 0.2 0.4 χ 0.6 0.8 1
FIG. 5: The metric function f, m and l for θ =
FIG. 4: Same as Fig.2 for themetric function l.
0,π/4,π/2 as a function of radial coordinate with α =
0.040,0.080,0.126 are shown. We use χ = x/(1+x),
allowed states are [6, 12] instead of dimensionless variable x=efπr.
ii 0 jj 0 ,provided i+j is odd,
3 3
| i| i
(for κ=0) of α in Table I. The mass difference from the classical
1 energy is shown as a function of α in Fig. 9. In Table I
[ii κ jj 2κ ( 1)i+j ii κ jj 2κ ].
√2 | 3 i| 3− i− − | 3− i| 3 i (and Fig.9), we show the results for i,j 3 with κ = 0,
≤
andalsosomeexcitedstatesforκ 2.Wepresentaspec-
(for κ=1, ,min i,[j/2] ) (40)
··· { } troscopic classification 2s+1L in T≤able I. For κ=0, the
j
Therefore κ>0 is allowed only if i 1 and j 2. system is in S-state, and then j equals to the intrinsic
≥ ≥
We show quantized energy spectra for various values spin s. For κ > 0 state, the orbital angular momentum
7
70 with the process γd 1S0.
→
Let us introduce the electromagnetic current in the
65 Skyrme model J(em)(x) which consists of the isoscalar
µ
and isovector part,
60
1
J(em)(x)= B (x)+I3(x), (41)
M 55 µ 2 µ µ
where B (x) is baryon current density given in Eq.(9)
µ
50 and I3(x) is the third component of the isospin current
µ
density, defined by
45
i
40 Iµa(x)=−efπ38 gµνTr[U−1[τ2i,U]U−1∂µU]
0 0.02 0.04 0.06α 0.08 0.1 0.12 +gµρgν(cid:0)σ Tr[U−1[τ2i,U]U−1∂νU]
FIG. 6: The coupling constant dependence of the classical [U−1∂µU,U−1∂νU] . (42)
×
B=2 skyrmion dimensionless mass derived from Eq.(8). (cid:1)
Inserting the dynamical field Uˆ in Eq.(21) into Eq.(41)
gives electromagnetic current operator Jˆ(em).
µ
ischosentobeitslowestvaluesothatitisconsistentwith To estimate the expectation value of these quantum
thequantumnumberj,theparityandthefactthats 3 operators, we describe the quantum spin states (40) in
≤
in the six-quark picture. It is seen that the mass differ- terms of the products of the rotation matrices
ence increases monotonically with increasing α. Fig. 1
1
(andalsoFigs.7,8)implies thatthe stronggravitymakes 2i+1 2
the sizeofthe skyrmionsmallerwhichmakesthe inertial hA|ii3k3i=(cid:18) 8π2 (cid:19) Di(iτ2A†)k3i3, (43)
moment smaller, resulting in increase in the mass differ-
1
ence. Inthecollectivequantization,theskyrmioncanbe 2j+1 2
B jj l = Dj(iτ2B ) . (44)
quantized as a slowly rotating rigid body and the mass h | 3 3i (cid:18) 8π2 (cid:19) † j3l3
differenceisinterpretedasaconsequenceoftherotational
kinetic energy. Thus the gravityworksfor increasingthe whereDi(A)m,m′ is wellknownWignerD function. The
kineticenergyoftheskyrmion. Inthenaivequarkmodel computation can be easily performed by the following
picture, the mass difference is ascribed to the hyperfine integration formula [14]
splittings. Theincreaseinthemassdifferencemayimply
tthhaetedffueecttsootfhethreedhuycptieornfinoeftshpelidttiisntagnscbeebceotmweeednoqmuianraknst, Z dB Di(B)∗m1m′1Dj(B)m2m′2Dk(B)m3m′3
by the gravity [13]. = 8π2 Ckm3 Ckm′3 , (45)
FromthedataoftheTableI,onecanstraightforwardly 2k+1 im1jm2 im′1jm′2
obtain the mass value in MeV unit, by multiplying f /e
π
for any parameter choice (f ,e). With f = 108 MeV, where Ckm3 is a Clebsch-Gordancoefficient.
π π im1jm2
e = 4.84, which is used in Ref. [6], our obtained mass The deuteron charge radius r2 1/2 is defined as the
value in the case of α = 0 is approximately 6% lower d
square root of (cid:10) (cid:11)
than the value obtained in Ref. [6]. There are two pos-
sible reasons for that. Firstly, the Lagrangianin Ref. [6]
r2 d d3rr2Jˆ(em)(r,t) d , (46)
includesthemasstermandhenceitslightlyenhancesthe d ≡ Z 0
mass value. Thus we also performed the same analysis (cid:10) (cid:11) (cid:10) (cid:12) (cid:12) (cid:11)
(cid:12) (cid:12)
including the mass term, which reduced the error from where d represents spin state of the deuteron. In this
| i
6%to0.5%inthemass. Secondly,thecoordinatesystem case, as only the isoscalar part of Jˆ(em) contributes to
0
theyusediscylindricalwhileweadoptedthesphericalco- the matrix element, the integration is straightforward.
ordinates. Sincethecylindricalcoordinaterequiresmuch
larger number of grid point to obtain torus-shape solu- π
r2 = dxdθ√ gx2B0(r,θ)
tionsthanthespherical,wesuspectthattheirresultsare (cid:10) (cid:11)d (efπ)2 Z −
not fully convergent,producing a slightly larger mass. 1
IV. ELECTROMAGNETIC PROPERTIES = dxdθx2sin2F sinΘ
−π(ef )2 Z
π
(∂ F∂ Θ ∂ F∂ Θ). (47)
x θ θ x
In this section we shall investigate the gravity effects × −
to the various observables, i.e. the mean charge radius Fig.10showstheαdependenceofthemeanchargeradius.
r2 1/2, magnetic moment µ , the quadrupole moment Duetotheattractiveeffectofthegravity,itdecreasewith
d d
(cid:10)Q,a(cid:11)ndthetransitionmomentµd np,whichisassociated increasing α.
→
8
α = 0.000 Energy density ε(P,Z) α = 0.040 Energy density ε(P,Z)
45 36
40 32
50 35 50 28
30 24
40 25 40 20
ε (P,Z) 2300 211 5050 ε (P,Z) 2300 11 6284
10 10
0 0
5 5
4 4
3 3
Z Z
2 2
1 1
0 0 1 2 3 4 5 6 7 0 0 1 2 3 4 5 6 7
P P
α = 0.080 Energy density ε(P,Z) α = 0.126 Energy density ε(P,Z)
30 20
27 18
50 24 50 16
21 14
40 18 40 12
ε (P,Z) 2300 11 9652 ε (P,Z) 2300 1 0864
3 2
10 10
0 0
5 5
4 4
3 3
Z Z
2 2
1 1
0 0 1 2 3 4 5 6 7 0 0 1 2 3 4 5 6 7
P P
FIG. 7: The energy density ε in the cylindrical coordinate with α = 0.000,0.040,0.080,0.126 are shown, in terms of the
dimensionless variables P =efπρ and Z =efπz.
TABLE I: The dimensionless quantized energy spectra up to i,j 3 for κ= 0, and also some excited states up to κ 2. In
≤ ≤
theparticle classification, we give thebaryonicdescriptions if available, otherwise we only show its spectroscopic classification
2s+1Lj.
Classification i j κ Parity α=0 α=0.040 α=0.080 α=0.126
classical skyrmion 69.7195 63.6722 56.9827 44.4862
Deuteron (3S1) 0 1 0 + 69.7227 63.6758 56.9866 44.4912
NN (1S0) 1 0 0 + 69.7241 63.6774 56.9886 44.4939
(3P2) 1 2 1 69.7294 63.6828 56.9943 44.5006
N∆ (5S2) 1 2 0 −+ 69.7339 63.6881 57.0003 44.5089
N∆ (3S2) 2 1 0 + 69.7368 63.6913 57.0041 44.5143
(3P2) 2 2 1 69.7387 63.6932 57.0059 44.5160
∆∆ (7S3) 0 3 0 −+ 69.7391 63.6935 57.0062 44.5162
(5P3) 1 3 1 69.7392 63.6935 57.0060 44.5156
∆∆ (1S0) 3 0 0 −+ 69.7475 63.7032 57.0176 44.5324
(5D4) 2 4 2 + 69.7479 63.7024 57.0153 44.5262
(5P3) 2 3 1 69.7485 63.7039 57.0177 44.5310
−
9
α = 0.000 Baryon density b(P,Z) α = 0.040 Baryon density b(P,Z) 0.12
0.11 0.11
0.1 0.1
0.09 0.09
0.2 0.08 0.2 0.08
0.07 0.07
P,Z) 0.15 000...000654 P,Z) 0.15 000...000654
b ( 0.1 00..0032 b ( 0.1 00..0032
0.05 0.01 0.05 0.01
0 0
4 4
3 3
Z 2 Z 2
1 1
0 0 1 2 3 4 5 0 0 1 2 3 4 5
P P
α = 0.080 Baryon density b(P,Z) α = 0.126 Baryon density b(P,Z)
0.145 0.21
0.13 0.19
0.115 0.17
0.2 0.1 0.2 0.15
0.085 0.13
0.07 0.11
P,Z) 0.15 0 .00.5054 P,Z) 0.15 00..0097
b ( 0.1 0 .00.2051 b ( 0.1 00..0053
0.05 0.05 0.01
0 0
4 4
3 3
Z 2 Z 2
1 1
0 0 1 2 3 4 5 0 0 1 2 3 4 5
P P
FIG. 8:
The baryon density b = √ gB0 in the cylindrical coordinate with α = 0.000,0.040,0.080,0.126 are shown, in terms of the
−
dimensionless variables P =efπρ and Z =efπz.
The isoscalar part of the magnetic moments is ex- The R is the rotation matrix (22) and a and b are
ij i i
pressed in term of electromagnetic current as defined as Eq.(24). The anti-commutator relation in
Eq.(48) guarantees µˆ to be Hermitian operator. µ
1 i ij
µˆi = 2Z d3r√−gǫijkrjJˆ0(em)(r,t). (48) andµ′jk are diagonaland furthermore they satisfyµ11 =
µ = 0 , µ = µ and µ = 2µ . With these
22 ′11 ′22 ′33 − 33
Inserting the dynamical field Uˆ in Eq.(21) into Eq.(48), relations and using the body-fixed operator (28) , the
coordinate-fixed operator (29) , the relation of the mo-
one finds the magnetic momentum operator as
ment of inertia components (32) , and the constraint of
µˆ = 1 R (B)T,µ a +µ b , (49) Eq.(36) in Eq.(48), one can get
i I=0 2{ ij jk k ′jk k}
(cid:12) µ
(cid:12) µˆ = ′11J +terms proportional to K . (53)
where i I=0 −V l 3
(cid:12) 11
1 1 1 1 (cid:12)
µjk= (efπ)2 Z d3x2εjlmxl2Tr[U†[2τk,U]Cm], (50) Therefore weπonly use the component of µ′11,
µ′jk= (ef1)2 Z d3x21εjlmxl21Tr[U†(−ix×∇)kU µ′11 =−4(efπ)2 Z dxdθ√−gx2(cos2θ+1)B0(x,θ).(54)
π
C (x)],(51) And the V11 is moment of inertia given in Eq.(35). The
× m moments µ of deuteron is defined as the expectation
d
and Cm is value of the µˆ3 with j3 =1
Cm(x)= 8πi2εmnp(U†∂nU)(U†∂pU). (52) µd =(cid:10)µˆ3(cid:11)=−µV1′111. (55)
10
0.008
0.045 (5P )
∆∆((37PS33))10 0.007
(5P2)2
0.04 ∆∆(1S30)23 0.006
(5D )
42 µd 0.005
0.035
0.004
0.03 0.003
gy 0.002
ener 0.025 N∆(3S1)2 0 0.02 0.04 0.06α 0.08 0.1 0.12
n
o
xcitati N∆(5S2)1 FleIsGs.m1a1g:nTethieccmouompleinngtc(o5n5)s.tantdependenceofthedimension-
E 0.02
0.015 In Fig.11 α dependence of the µd is shown. It decreases
(3P ) withincreasingα,buttheeffectofgravityismoreevident
21
comparedwiththatofB =1[10]. Inthenon-relativistic
0.01 nuclear physics point of view, the magnetic moment of
thedeuteronconsistsofthesumofthemagneticmoment
NN(1S )
01 oftheprotonplusneutronandsomecorrectiontermsre-
0.005 lated to the D-state probability of the deuteron. This
Deuteron
(3S ) resultindicates thatthe assumptionthatthe deuteronis
10
almostS-state(contributionstotheotherstatesareonly
0
5%)maynolongervalidunderthestronggravitational
0 0.04 0.08 0.12
∼
α field. Therefore,wespeculatethattheeffectofgravityis
apparent on the change of such non-central components
FIG.9: Thecouplingconstantdependence ofthedimension- of the nuclear force. This will be more evident by exam-
lessmassdifferencefrom theclassical energyareshown. For ining the quadrupole moment. The quadrupole moment
α & 0.127, there exists no solution. We give the baryonic is given by
descriptions if available, otherwise we only show its spectro-
scopicclassification(2s+1Lj)i,whosesubscriptimeanseigen-
valueof theisospin operator (31). Qˆ = d3r√ g(3r r r2δ )Jˆ(em)(r,θ), (56)
ij Z − i j − ij 0
which is the same as the magnetic moment. Inserting
(21)intoEq.(56)oneobtainsthequadrupolemomentum
operator
3
2.8 Qˆ =R (B)TQ R (B), (57)
ij I=0 ia ab bj
2.6 (cid:12)
(cid:12)
where
2.4
1/2>d 2.2 Q = 1 d3x√ g(3x x x2δ )B (x,θ).(58)
2<r 2 ab 2(efπ)2 Z − a b− ab 0
1.8
Q satisfy Q = Q . And the symmetry relation for
ij 11 22
1.6
the quadrupole moments reduces the expression to
1.4
3 1
1.2 Qˆ =Q [ R (B)TR (B) δ ]. (59)
0 0.02 0.04 0.06α 0.08 0.1 0.12 ij I=0 33 2 i3 3j − 2 ij
(cid:12)
(cid:12)
Thus we only need the component of Q
FIG. 10: The coupling constant dependence of the dimen- 33
sionless mean charge radius (47).
π
Q = dxdθ√ gx2(3cos2θ 1)B0(x,θ).(60)
33 (efπ)2 Z − −