A NEW METHOD FOR A BABY-SKYRMION

AND A BABY-SKYRMION BRANE

TOKYO UNIVERSITY OF SCIENCE

MASAHIRO HAYASAKA

TÈRENCE DELSATE(UNIVERSITÈ DE MONS) , NOBUYUKI SAWADO(TUS)

ARX IV:1208.6341(HEP-TH)

A BABY-SKYRME BRANE MODEL

(IN 6DIMS.)

𝑆𝐸 = 𝑑6𝑥 −𝑔

1

2𝜒 6𝑅 − Λ 6 ,

𝑆𝐵 = 𝑑

6𝑥 −𝑔𝜅22𝜕𝑀𝒏 ∙ 𝜕

𝑀𝒏 −𝜅44𝜕𝑀𝒏 × 𝝏𝑁 𝒏

2 − 𝜅0 1 + 𝑛3 .

A baby-Skyrme action is used for the brane action.

The model is embedded a baby-skyrmion into the extraspace.

Generally speaking, there are 2 independent

methods to get a solution of a model.

The action in 6D is 𝑆𝑡𝑜𝑡𝑎𝑙 = 𝑆𝐸 + 𝑆𝐵,

variational method equation of motion

the Einstein-Hilbert action

not available available

(the Einstein eq.)

a baby-Skyrme action available

(for the energy)

available for only the axially

symmetric case

a Skyrme action available

(for the energy) available

(rational maps)

2 METHODS TO OBTAIN A SOLUTION

We solve the coupled system to obtain the baby-skyrmion brane solution.

THE METHOD FOR THE SKYRMION

Conor J. Houghton, Nicholas S. Manton & Paul M. Sutcliffe, Rational maps, monopoles and Skyrmions, Nucl.Phys. B510 (1998)507-537

𝑄 = 2 𝑄 = 3 𝑄 = 4

The well known method to obtain the minimal energy solutions with the platonic symmetry is so called the rational map ansatz. The rational function is determined in terms of the information by the analysis of a variational method. (Both results are coincide within a few percent of uncertainty.)

variational method equation of motion

the Einstein-Hilbert action

not available available

(the Einstein eq.)

a baby-Skyrme action available

(for the energy)

available for only the axially

symmetric case

a Skyrme action available

(for the energy) available

(rational maps)

2 METHODS TO OBTAIN A SOLUTION

We apply the procedure to a baby-Skyrmion.

Contrary to the case of the Skyrme model, the baby-Skyrme model has a difficulty to solve the EOM.

THE BABY-SKYRME MODEL(IN 2DIMS.)

The soliton is composed of a triplet scalar fields with unit length.

𝑛 𝑟, 𝜃 = 𝑛1, 𝑛2, 𝑛3 𝑛 ∙ 𝑛 = 1

𝐸𝑠𝑡𝑎𝑡𝑖𝑐 = 1

2𝜕𝑖𝑛 ∙ 𝜕

𝑖𝑛 +1

4𝜕𝑖𝑛 × 𝜕𝑗𝑛 ∙ 𝜕

𝑖𝑛 × 𝜕𝑗𝑛 + 𝜇𝑉 𝑛3 𝑑2𝑥

O(3)→O(2) symmetry

𝑄 =1

4𝜋 𝑛 ∙ 𝜕1𝑛 × 𝜕2𝑛 𝑑

2𝑥 ∶ the topological charge

A low dimensional mimic of the Skyrme model.

The notable difference from the Skyrme model is to need some potentials to stabilize a “baby-skyrmion” from a Virial theorem.

THE NORMAL BOUNDARY CONDITION

(IMPOSED AN AXIAL SYMMETRY)

3) 𝜕𝜃𝐹 𝑟, 𝜃 |𝜃=±𝜋2= 0 for 0 < 𝑟 < ∞,

4) Θ 𝑟,±𝜋

2= ±

3𝜋

2𝜋 for 0 < 𝑟 < ∞.

However, we could not obtain some ground states in this condition.

We need to improve it for the ground states.

1) 𝐹 0, 𝜃 = 𝜋, 𝜕𝑟Θ 𝑟, 𝜃 |𝑟=0 = 0 for −𝜋

2≤ 𝜃 ≤

𝜋

2,

2) 𝐹 ∞, 𝜃 = 0, Θ ∞, 𝜃 = 3𝜃 for −𝜋

2≤ 𝜃 ≤

𝜋

2,

THE SPATIAL SYMMETRY BREAKING

We extract a proper boundary conditions from the data of the variational method.

Marek Karliner & Italy Hen, Rotational symmetry breaking in baby Skyrme models, Nonlinearity 21 (2008)399-408

For , the charge 3 solutions exhibit a Z2 symmetry, while the potential has the O(2) symmetry.

For , the solution keeps the O(2) symmetry.

𝑉 𝑛3 = 1 − 𝑛3

𝑉 𝑛3 = 1 − 𝑛32

𝑛 = sin 𝐹 𝑟, 𝜃 cos Θ 𝑟, 𝜃 , sin 𝐹 𝑟, 𝜃 sinΘ 𝑟, 𝜃 , cos 𝐹 𝑟, 𝜃

A RESULT OF THE VARIATIONAL METHOD

length of arrows direction of arrows

We plot the in this way.

This is the ground state for .

We extract a information of the direction of the arrows.

𝑉 𝑛3 = 𝜇 1 − 𝑛3

E density

𝑛

CARTOON FIGURE OF THE N VECTORS

IN THE GROUND STATE

We solve the upper half plane of polar coordinates.

CARTOON FIGURE OF THE N VECTORS

IN THE GROUND STATE

“phase flipping” points

1 parameter “r0”

We could extract the information of the boundary condition from the analysis.

THE NEW BOUNDARY CONDITION

(IMPOSED THE Z2 SYMMETRY)

1) 𝐹 0, 𝜃 = 𝜋, 𝜕𝑟Θ 𝑟, 𝜃 |𝑟=0 = 0 for −𝜋

2≤ 𝜃 ≤

𝜋

2,

2) 𝐹 ∞, 𝜃 = 0, Θ ∞, 𝜃 = 3𝜃 for −𝜋

2≤ 𝜃 ≤

𝜋

2,

3) 𝜕𝜃𝐹 𝑟, 𝜃 |𝜃=±𝜋2= 0 for 0 < 𝑟 < 𝑟0, 𝑟0 < 𝑟 < ∞,

𝐹 𝑟0, ±𝜋

2= 0

4) Θ 𝑟,±𝜋

2=

±𝜋

2 for 0 < 𝑟 < 𝑟0

±3

2𝜋 for 𝑟0 ≤ 𝑟 < ∞

.

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

18.06

18.08

18.10

18.12

18.14

18.16

E/B

r0

The ground state

DETERMINATION OF THE VARIABLE

It is possible to extract r0 directly from the variational method. We, however, determine r0 by solving the EOM.

r0

E/Q

r0

𝜇 = 0.08

COMPARISON OF THE BOTH RESULTS

variational method

EOM

𝜇 = 0.04 𝜇 = 0.06 𝜇 = 0.12 𝜇 = 0.64

Both r0 are perfectly coincide.

0

20

40

60

80

100

120

0.01 0.1 1 10

variational method

EOM

COMPARISON OF THE BOTH ENERGIES

small deformation large deformation 𝜇

𝐸

Both energies are perfectly coincide.

THE BABY-SKYRMION BRANE

THE BABY-SKYRME BRANE MODEL

(IN 6 DIMS.)

The action in 6D is

𝑆𝐸 = 𝑑6𝑥 −𝑔

1

2𝜒 6𝑅 − Λ 6 ,

𝑆𝐵 = 𝑑

6𝑥 −𝑔𝜅22𝜕𝑀𝒏 ∙ 𝜕

𝑀𝒏 −𝜅44𝜕𝑀𝒏 × 𝝏𝑁 𝒏

2 − 𝜅0 1 + 𝑛3 ,

𝑆𝑡𝑜𝑡𝑎𝑙 = 𝑆𝐸 + 𝑆𝐵,

𝑑𝑠(6)2 = 𝑀2 𝑟, 𝜃 𝑑𝑠(4)

2−𝐿2 𝑟, 𝜃

𝑀2 𝑟, 𝜃𝑑𝑟2 + 𝑟2𝑑𝜃2 ,

We derive the EOM of the actions and substitute those ansatz.

𝑑𝑠(4)2 = 𝑑𝑡2 − 𝛿𝑖𝑗𝑒

2𝐻0𝑡𝑑𝑥𝑖𝑑𝑥𝑗 . H0 is a Hubble parameter.

𝑛 = sin 𝐹 𝑟, 𝜃 cosΘ 𝑟, 𝜃 , sin 𝐹 𝑟, 𝜃 sinΘ 𝑟, 𝜃 , cos 𝐹 𝑟, 𝜃 .

We obtained the deformed baby-skyrmion brane!

The energy density The scalar curvature

THE RESULT

Unless the boundary condition of r0, the scalar curvature has distinct peaks located on the energy density peaks.

SUMMARY

We established the method how to get deformed solution of the baby-skyrmion by solving the EOM.

By using this method, we found the deformed baby-skyrmion brane.

WORK IN PROGRESS

Localization of the SM particles.

The solutions with various charges and various potentials.

Mechanism of the deformation.

Application of the method to other models.

Thank you.