- Email: [email protected]
Gradient flow for well-separated Skyrmions
26 September 1996
PHYSICS EJZXiVlER
LETTERS
B
Physics Letters B 385 (1996) 187-192
Gradient flow for well-separated Skyrrnions P.W. Irwinl, Department
of Applied Mathematics
and Theoretical
N.S. Manton
Physics University of Cambridge,
Silver St., Cambridge
CB3 9EW UK
Received 22 June 1996 Editor: l?V. Landshoff
Abstract We discuss the gradient flow curves in the product ansatz in the Skyrme model. For m, = 0 we solve for the gradient flow trajectories. We also describe an algorithm to estimate the positions and relative orientation of two well-separated Skyrmions, given just the Skyrme field. PACS: 13.75.C~; ll.lO.Lm
1. Introduction In this letter we discuss some aspects of the B = 2 sector of the Skyrme model [ 11, in the region corresponding to two well-separated Skyrmions. It is well known that the product ansatz gives an accurate description of the Skyrme field configurations if the Skyrmions are widely separated [ 1,2]. Using this we obtain a collective coordinate approximation to twoSkyrmion dynamics accurate when they are widely separated. A Lagrangian describing this was derived in [ 31; a simplified version is used here, which ignores all velocity dependent interactions between the Skyrmions. We find the conserved quantities of this Lagrangian. They correspond to spin, isospin, and center of mass momentum. We study the gradient flow curves of the system. All conserved quantities are shown to be zero along these. For the case of massless pions ( mr = 0) , we obtain the trajectories of the gradient flow in closed ’ E-mail. [email protected]. 2 E-mail: [email protected]. 0370-2693/96/$12.00 Copyright PII SO370-2693(96>00867-2
form. We are unable to do this for 112, # 0 although the trajectories appear qualitatively the same. Finally, an algorithm is described for estimating the relative orientation and the centers of the Skyrmions given a Skyrme field which corresponds to two wellseparated Skyrmions but which is not explicitly in product form.
2. The Skyrme model The Skyrme model has the Lagrangian L =
J { d3x
--$Tr(P‘Rc)
+&Tr([~,,&l[~p,~“l) +~f~m~Tr(U+Ut
-2))
(2.1)
where R, = Ufd,U, U is the SU(2) valued Skyrme field, e and fr are free parameters of the model whose values are chosen to best fit experimental data, and mrr is the pion mass.
0 1996 Elsevier Science B.V. All rights reserved.
188
P.W. Irwin, N.S. Manton/Physics Letters I3 385 (1996) 187-192
The model has soliton solutions of finite energy. Finite energy implies that U tends to the identity at spatial infinity. Space is then compactified to S3 and thus each soliton solution has an associated integer, the degree, corresponding to the element of r3(S3) to which U belongs. As Skyrme argued, solitons of degree B may be interpreted as B nucleons [ 11. The symmetry group of the Skyrme Lagrangian is SO ( 3 ) isospin x Poincare group. For time-independent fields such as static solitons the symmetry group is reduced to SQ(3)isospin
x Euclidean
group of JR3.
(2.2)
Here, SO( 3)isospin acts via U(X) -+ AU(x) A+, where A E SU(2). The minimal energy B = 1 solution is the spherically symmetric “hedgehog” soliton, or Skyrmion, Uff(x)
= exp (if(r)2
97)
(2.3)
where Y = 1x1, 4 = 4 and 7 denotes the (vector of) Pauli matrices. f(r) is a function determined numerically, with f(0) = rr and f(r) -+ 0 as Y --f co. In the limit m, --+0,the function f(r) is asymptotically $ for large Y, where h is a constant [ 41. An SO( 3) subgroup of the symmetry group acts trivially on UH, since spatial and isospatial rotations are equivalent. But, acting with the rest of the symmetry group we obtain fields U(x)
= AUH(x
- X)A+
(2.4)
which are solutions to the field equation with the same energy. The moduli space approximation to single Skyrmion dynamics involves letting A, X become time-dependent and substituting (2.4) into (2.1). Physically, this means approximating the Skyrmion as a point particle with isospin degrees of freedom. One finds the Lagrangian
The moduli space approximation proceeds exactly as for the single Skyrmion. Let A, B, X, Y depend on time and substitute (2.6) into (2.1). The calculations are very lengthy however. They were presented in [ 31 for m, = 0 with the result L2 = -2M
+ ;Md
- $(3808
+ ;M?
+ ~A(cu2 +fi”)
- TrO) + Lint
(2.7)
where s=x-Y,s=(s(,
K=2R-h2f;, ;j-.T=B+&
;w=AtA,
The relative orientation of the two Skyrmions is given by 0 = A-'I?where d, B are the SU(3) matrices associated to A, B. Finally, the velocity dependent interaction terms are Z&t = /C{2V + (u2
’ O(V
X p)
+d)(v.
-
2(U
X Ly ’ OV)
cm)
+(u~v)(u~ov)+(u~~)(v~c7”)}~ +K{(vxX~~(vXp)+(V~v)(Vx~.Ov) -
(u*V)(V.O(V
x p>>
-
(v~ov)(u.v)(u~v)}s
(2.8)
where u = 8, u = Y. As was discussed in [3], the omission of Lint causes no serious errors if s is large and cy, /3, U, v are small. Henceforth we drop Lint from (2.7)) leaving just the static interaction potential. We assume that for m, # 0,only the potential term in the Lagrangian is modified. So now LZ is L2 = -2Mf
;M,il+
;Mti
+ @(a2
+p”)
e--nb?s L1=-M+~M~+~Aa2.
(2.5)
where $Y .T = A+A with A, M constant. Let us now consider the two-Skyrmion sector. The product ansatz assumes that at large separations between two Skyrmions the field will be approximately of the product form U(x)
= AUH(x
- X)A+BU&x
- Y)B+.
(2.6)
-2/c(2c-
1)(0~00)----
s
(2.9)
with c M 0.96 for the physical value of m,, and c ---f 1 for mT -+ 0, the last term being the well known potential between two well-separated Skyrmions [ 3 I. The continuous symmetries of L2 imply that there exists conserved momentum P, angular momentum J, and isospin I.The symmetry group of Lz is SO (3) I x (S0(3)~ K R3), the same symmetry group (2.2) as
P.W. Irwin, N.S. Manton/Physics
for the full Skyrme equation. First introduce the center of mass R = i(X + Y) and rewrite (2.9) as L2 = -2M+M82+$Ms2+~A(Tr~~-‘+Tr~~-‘) -
2K(2C
-
1) {a(s)$0$
+ b(s)TrO}
(2.10)
where a(s) = b(s)
1!$+ y
+3
+ } 3m
=-{T$
g-s
>
$
e-md,
L2 is invariant under R” H R”+A”, implying P = 2MiZ is constant, i.e. total momentum is conserved. Next, L2 is invariant under equal left variations of A, Z3:
Letters B 385 (1996) 187-192
189
Skyrmions. Paths close to these will go out to a configuration of well-separated Skyrmions and then return to the toroidal solution. Therefore it is of interest to know the form of these curves for well-separated Skyrmions. The gradient flow curves may be computed numerically [6] using the full Skyrme fields. However this procedure is difficult to implement for well-separated Skyrmions. The product ansatz allows one to find analytic expressions for these curves. These may then be matched up with the curves obtained numerically which describe two-Skyrmion configurations with small separations, including the hedgehog and the toroidal solutions. Generally, if L is the Lagrangian for motion on a Riemannian configuration space M L = $gij(Cj)Q’Q’ - V(q)
A-+(l+e’IT”)A,
a-t(l+E”T”)Z?,
where T” are the generators implying -&4X1
+&a-‘)
then the equation for gradient flow is
of SO( 3)) (T”) i,i = -E,ij,
=o.
gij(q)$
(2.11)
d+A(1+8Ta),
l(Ta)usj.
g=O,
I(qi94i)
=
dL dh’,(q) aqi ds
J=$4sxS+A(cu+p).
.jdhf(q) s=o = gijq ~ ds
SO
’
(3.3)
JV dh:(q) to angular momentum
(3.2)
But if L is invariant under h,, then gi,j(q) and V(q) are separately invariant so
B+Z?(~+E”T’),
This corresponds
-$I.
We first note that any conserved quantity corresponding to an invariance of L is zero along a gradient flow curve. This may be seen as follows. If L is invariant under a l-parameter group of diffeomorphisms h,:M t-f M, s E Iw, then there exists the associated (Noether) momentum
(2.12)
L2 is also invariant under equal right variations of A, Z3 and a rotation of s,
si -+ si -
=
I
This can be rewritten as s(Acu + BP) = 0, where (Acv)~ = Jz”a,j. That is, there exists a conserved isospin Z=Acr+Bp.
(3.1)
conservation (2.13)
3. The gradient flow equations It has been proposed [ 51 that the union of gradient flow curves from the unstable B = 2 hedgehog solution is a candidate for the moduli space for two-Skyrmion dynamics in the Skyrme model. Practically all curves will end up at the minimal energy toroidal solution. A small family of curves will end at infinitely separated
dq”
ds
s=.
= 0,
(3.4)
hence I = 0 along gradient llow curves. We now wish to find the gradient flow curves for the product ansatz. First we rewrite A, g in ZQ in terms of the conserved I, and 0. Noting that TrAA-’ = 2iL2, T&&i =2p2, where it + p, we see that Trd&’
= 2( & -
&=AcY,
p=Z3p
p)” ,
and
Z=
(3.5)
so TrAA-’
+ Trg$-’
= $TrC?&’
+
Z2,
(3.6)
P.W. Irwin, N.S. Mation/Physics
190
and we may rewrite (2.10) L2 =
-
-2M
as
+ Mit2 + ;MS* + +I* + +,&8-t
2K(2C
-
1) {a(s)L?0$ + b(s)TrO}
(3.7)
It is an easy exercise to verify that J, I, k are zero along the gradient flow curves. Expressing (3.j in the form 0i.i = COS6)S,i -/- ( 1 - COS0) nini - Eijk nk Sin 8 with nini = 1, one finds that Trdd-’
=2d2+4(1
The interaction
-cos8)ii*.
potential
2K’{U(S)[COS0+
V
(3.8)
0 and S > 0, so again they repel to infinity. For trajectories near the neutral channel, initially S > 0 but again the Skyrmions rotate into the attractive channel. All trajectories approach the attractive channel except those with z = 1, in which case the Skyrmions separate to infinity. The case z = 1 includes the repulsive channel (8 = v), but if 6 < 7r then 6 < 0 so the Skyrmions asymptotically approach the neutral channel. For the case of massless pions the gradient flow equations simplify considerably and it is possible to explicitly solve for the trajectories of s, z, 8. Thus for the remainder of this section we assume that m, = 0. Eqs. (3.10), (3.11), (3.14) become
l&j 2
where~‘=~(2c-1). (3.9)
We are now in a position to calculate the gradient flow equations for the coordinates s, 8, n. Let z = $.n. Applying (3.2) to L2 we have
fU(s>(l-Z2)]
(3.12)
and (3.13)
z = 4K’a(s)
(3.16)
2(1 -c0se)
~=.E(z3_z)
;
s3
Ms2
+
(1 -COSe)
-$
[n-Z
may be combined
$f(e)=$fs*
(3.17)
(3.18)
&
(3.19)
(3.10)
cos 5 = pe 12A
(3.11)
on trajectories, for p some constant. g(z) = log Iz - z31, we have
$1.
(3.13)
to give
2(1 -c0se) (Z3 - Z>
’
Letting f( 6) = log 1cos ! 1, we have
(3.12)
Ati=--4K’.(,)Z[$-zn] -4K’U(s)
- 1)
ss
e
+(1+2CoSe)g}
;MS =
sin8(3z2
= -_2K
(3.15)
so
~MS=-~K’{[COS~+(~-COS~)Z*]~
ih8=2K’Sine[2b(s)
3z2 - 1
;MS=6K(1-cosB)7
is now
(1 -COS~)(FZ~~)*]
+b(s)(1+2cos8)}
Letters B 385 (1996) 187-192
;+
Ms2
’
MS
-$(z)=
A(1 -cosf3)
Integrating,
we find
z -zs=
qs*e%
+;
2
d
3
-g.
Also,
letting
(3.20)
(3.21)
(1 -p2e+$)i
(3.14) In the repulsive channel (z = 1, 0 = n-), 2 = 0, 6 = 0 and S > 0, so the Skyrmions separate to infinity. For trajectories that begin near the repulsive channel (z M l,e M n), initially S > 0, however the Skyrmions then twist towards the attractive channel (z = 0, 8 = r), so eventually they attract and s decreases. Similarly in the neutral channel (8 = 0)) 6 =
along trajectories, for q some constant. Note that the neutral channel (0 = 0) is a stationary point, since 4 = S = 0 (and z is undefined), unlike in the case m, # 0; also a = 0. Note also that f( 6) is not differentiable at B = 7~. So (3.18) is not valid at 8 = r, but this does not matter. One can see from (3.18) that s --f 0 as the attractive channel is approached (0 --+ Z->, but
P.W. Irwin, N.S. Munton/Physics
the attractive channel is never precisely reached in gradient flow, unless the initial field is in the attractive channel, moreover the validity of the product ansatz breaks down as s -+ 0. Again, g(z) is not differentiable at z = 0 or 1, hence (3.21) is not valid there. On a generic gradient flow curve z decreases monotonically with time, from 1 to 0. Therefore 3z2 - 1 changes sign as t increases. It follows from (3.15), (3.16) that s increases from zero to a finite value s,,-,~, which is attained when 3t2 - 1 = 0, and then decreases to zero. At the same time 8 decreases from n- to a positive value, and then returns to Z-. In fact, only the part of such a curve where s > 0 is relevant, as the product ansatz breaks down as s -+ 0. Knowing z (t) it is possible to find expressions for n(t), CC(t) as follows. From (3.12), (3.13)
iz.
B.(nx$)=j.j.=O.
3) =h.n=O,
(nx
So if at t = 0, n = no, Z? = $0 then at all times n(t) , 3(t) lie in the no, $0 plane. Let
[email protected]= zo, n(t)
n(t)
. g(t) = z(t), i(t)
.no = a(t),
* $0 = p(t).
Then
z(t) = 4t) + P(t) + zo
(3.22)
Letters B 385 (1996) 187-192
191
t
=-
24~
M
J
dQ-[l
Z(T)[l-COSbJ(T>l
Z2(~W2 85(T)
.
0
(3.26) So given z (t) we may determine find n(t), i(t).
4. Relative orientation
a(t) , p(t)
and thus
and centers
In the previous section, expressions were given for gradient flow curves in the product ansatz corresponding to two well-separated Skyrmions. These curves may be matched up with the gradient flow curves obtained numerically starting with two Skyrmions at small separation, To do this one needs to identify from the Skyrme field U the coordinates used in the product ansatz; namefy the Skyrmions’ relative orientation and their centers X, Y. Here we sketch an algorithm to obtain the relative orientation and centers when U is precisely of product form. Then if U corresponds to two well-separated Skyrmions, U will be close to product form, and the same estimate should give the best approximation for the centers and relative orientation. Let the Skyrme field be expressed in the form U(x)
=(T(x)
+ir-m(x).
If U is of product form
and
U(x) n(t)
=nacosa(t)+nax(nox&~>sinrw(t)
s(t)
= sacosP(t)
(3.23)
- Y)B+
= AUH(.X - X)A+BU&
(4.1)
then +$a x (SO xita)
sinP(t)
(3.24) (+=cos.fl
and from (3.12),
cos f2 -sin
ft sin f2g,0R2,
(3.13) where
a(t)
=
I
O=A-‘B, fi =f(lxll>,
dr]A(r)l
0
x1
t
=-
12K
A
/T?(t) = J 0
J
&Z(7)[1-z*(7)11’*
0
dr(S(r)(
s3(r)
(3.25)
so
=
Ix - XI
)
x* =
f2=f(lx21),
Ix - YI,
(4.2)
192
P.W. Irwin, N.S. Manton/Physics
Our idea is just to use the function g to alternately estimate the centers X, Y and the relative orientation 0, until best estimates for both are obtained. If U is of product form then (+ M 1 in most of R3 with g M -1 in two regions approximately where the Skyrmions are centered. We assume that IT = -1 at two points only, and call these Xa and Ya. Our first estimates for X and Y are X0 and Ya. Now, using (7 and setting X = Xa and Y = Yo in (4.3), we evaluate the r.h.s. of (4.3) at enough points x to determine the quadratic form on the 1.h.s. of (4.3) and extract 0. Call this estimate 00. In practice, the points x should not be too close to X0 or Ya, nor too far away, to avoid large errors. The next step is to improve the estimates of the centers X, Y. This means using the formula (4.2) for u with 0 = 00, and seeking X, Y so that (T = -1 at Xa and Yo. A simple method is to evaluate (4.2) using (30, X0 and Ya as the product ansatz coordinates and then determine the points Xb, Yh say, where (+ = - 1. Then to compensate for the error, evaluate u again using 00, X0 - (XL - X0) and Ya - (YA - Ya) as coordinates. The points where F = - 1 should now be closer to X0, Ya, but any error can be compensated in a similar way. By one or more steps of this kind we find
Letters B 385 (1996) 187-192
a new estimate for the centers Xt, Yi. Given these, we can use (4.3) this time with X = Xl, Y = Y1, to obtain an improved estimate for 0, say 01. Then we can improve the estimate for the centers, and so on. It is not difficult to check that if the Skyrmion separation s is large, then this scheme should converge rapidly to give the centers and relative orientation.
Acknowledgements P.W.I. thanks PPARC for a research studentship.
References [ I] T.H.R. Skyrme, Proc. Roy. Sot. A 260 (1961) 127. [2] J.J.M. Verbaarschot, T.S. Walhout, J. Wambach and H.W. Wyld, Nucl. Phys. A 468 (1987) 520. [3] B.J. Schroers, Z. Phys. C 61 (1994) 479. [4] G.S. Adkins, C.R. Nappi and E. Witten, Nucl. Phys. B 228 (1983) 552; G.S. Adkins and CR. Nappi, Nucl. Phys. B 233 (1984) 109. [5] N.S. Manton, Phys. Rev. Lett. 60 (1988) 1916. [6] T. Waindzoch and J. Wambach, Nucl. Phys. A 602 (1996) 347.
PHYSICS EJZXiVlER
LETTERS
B
Physics Letters B 385 (1996) 187-192
Gradient flow for well-separated Skyrrnions P.W. Irwinl, Department
of Applied Mathematics
and Theoretical
N.S. Manton
Physics University of Cambridge,
Silver St., Cambridge
CB3 9EW UK
Received 22 June 1996 Editor: l?V. Landshoff
Abstract We discuss the gradient flow curves in the product ansatz in the Skyrme model. For m, = 0 we solve for the gradient flow trajectories. We also describe an algorithm to estimate the positions and relative orientation of two well-separated Skyrmions, given just the Skyrme field. PACS: 13.75.C~; ll.lO.Lm
1. Introduction In this letter we discuss some aspects of the B = 2 sector of the Skyrme model [ 11, in the region corresponding to two well-separated Skyrmions. It is well known that the product ansatz gives an accurate description of the Skyrme field configurations if the Skyrmions are widely separated [ 1,2]. Using this we obtain a collective coordinate approximation to twoSkyrmion dynamics accurate when they are widely separated. A Lagrangian describing this was derived in [ 31; a simplified version is used here, which ignores all velocity dependent interactions between the Skyrmions. We find the conserved quantities of this Lagrangian. They correspond to spin, isospin, and center of mass momentum. We study the gradient flow curves of the system. All conserved quantities are shown to be zero along these. For the case of massless pions ( mr = 0) , we obtain the trajectories of the gradient flow in closed ’ E-mail. [email protected]. 2 E-mail: [email protected]. 0370-2693/96/$12.00 Copyright PII SO370-2693(96>00867-2
form. We are unable to do this for 112, # 0 although the trajectories appear qualitatively the same. Finally, an algorithm is described for estimating the relative orientation and the centers of the Skyrmions given a Skyrme field which corresponds to two wellseparated Skyrmions but which is not explicitly in product form.
2. The Skyrme model The Skyrme model has the Lagrangian L =
J { d3x
--$Tr(P‘Rc)
+&Tr([~,,&l[~p,~“l) +~f~m~Tr(U+Ut
-2))
(2.1)
where R, = Ufd,U, U is the SU(2) valued Skyrme field, e and fr are free parameters of the model whose values are chosen to best fit experimental data, and mrr is the pion mass.
0 1996 Elsevier Science B.V. All rights reserved.
188
P.W. Irwin, N.S. Manton/Physics Letters I3 385 (1996) 187-192
The model has soliton solutions of finite energy. Finite energy implies that U tends to the identity at spatial infinity. Space is then compactified to S3 and thus each soliton solution has an associated integer, the degree, corresponding to the element of r3(S3) to which U belongs. As Skyrme argued, solitons of degree B may be interpreted as B nucleons [ 11. The symmetry group of the Skyrme Lagrangian is SO ( 3 ) isospin x Poincare group. For time-independent fields such as static solitons the symmetry group is reduced to SQ(3)isospin
x Euclidean
group of JR3.
(2.2)
Here, SO( 3)isospin acts via U(X) -+ AU(x) A+, where A E SU(2). The minimal energy B = 1 solution is the spherically symmetric “hedgehog” soliton, or Skyrmion, Uff(x)
= exp (if(r)2
97)
(2.3)
where Y = 1x1, 4 = 4 and 7 denotes the (vector of) Pauli matrices. f(r) is a function determined numerically, with f(0) = rr and f(r) -+ 0 as Y --f co. In the limit m, --+0,the function f(r) is asymptotically $ for large Y, where h is a constant [ 41. An SO( 3) subgroup of the symmetry group acts trivially on UH, since spatial and isospatial rotations are equivalent. But, acting with the rest of the symmetry group we obtain fields U(x)
= AUH(x
- X)A+
(2.4)
which are solutions to the field equation with the same energy. The moduli space approximation to single Skyrmion dynamics involves letting A, X become time-dependent and substituting (2.4) into (2.1). Physically, this means approximating the Skyrmion as a point particle with isospin degrees of freedom. One finds the Lagrangian
The moduli space approximation proceeds exactly as for the single Skyrmion. Let A, B, X, Y depend on time and substitute (2.6) into (2.1). The calculations are very lengthy however. They were presented in [ 31 for m, = 0 with the result L2 = -2M
+ ;Md
- $(3808
+ ;M?
+ ~A(cu2 +fi”)
- TrO) + Lint
(2.7)
where s=x-Y,s=(s(,
K=2R-h2f;, ;j-.T=B+&
;w=AtA,
The relative orientation of the two Skyrmions is given by 0 = A-'I?where d, B are the SU(3) matrices associated to A, B. Finally, the velocity dependent interaction terms are Z&t = /C{2V + (u2
’ O(V
X p)
+d)(v.
-
2(U
X Ly ’ OV)
cm)
+(u~v)(u~ov)+(u~~)(v~c7”)}~ +K{(vxX~~(vXp)+(V~v)(Vx~.Ov) -
(u*V)(V.O(V
x p>>
-
(v~ov)(u.v)(u~v)}s
(2.8)
where u = 8, u = Y. As was discussed in [3], the omission of Lint causes no serious errors if s is large and cy, /3, U, v are small. Henceforth we drop Lint from (2.7)) leaving just the static interaction potential. We assume that for m, # 0,only the potential term in the Lagrangian is modified. So now LZ is L2 = -2Mf
;M,il+
;Mti
+ @(a2
+p”)
e--nb?s L1=-M+~M~+~Aa2.
(2.5)
where $Y .T = A+A with A, M constant. Let us now consider the two-Skyrmion sector. The product ansatz assumes that at large separations between two Skyrmions the field will be approximately of the product form U(x)
= AUH(x
- X)A+BU&x
- Y)B+.
(2.6)
-2/c(2c-
1)(0~00)----
s
(2.9)
with c M 0.96 for the physical value of m,, and c ---f 1 for mT -+ 0, the last term being the well known potential between two well-separated Skyrmions [ 3 I. The continuous symmetries of L2 imply that there exists conserved momentum P, angular momentum J, and isospin I.The symmetry group of Lz is SO (3) I x (S0(3)~ K R3), the same symmetry group (2.2) as
P.W. Irwin, N.S. Manton/Physics
for the full Skyrme equation. First introduce the center of mass R = i(X + Y) and rewrite (2.9) as L2 = -2M+M82+$Ms2+~A(Tr~~-‘+Tr~~-‘) -
2K(2C
-
1) {a(s)$0$
+ b(s)TrO}
(2.10)
where a(s) = b(s)
1!$+ y
+3
+ } 3m
=-{T$
g-s
>
$
e-md,
L2 is invariant under R” H R”+A”, implying P = 2MiZ is constant, i.e. total momentum is conserved. Next, L2 is invariant under equal left variations of A, Z3:
Letters B 385 (1996) 187-192
189
Skyrmions. Paths close to these will go out to a configuration of well-separated Skyrmions and then return to the toroidal solution. Therefore it is of interest to know the form of these curves for well-separated Skyrmions. The gradient flow curves may be computed numerically [6] using the full Skyrme fields. However this procedure is difficult to implement for well-separated Skyrmions. The product ansatz allows one to find analytic expressions for these curves. These may then be matched up with the curves obtained numerically which describe two-Skyrmion configurations with small separations, including the hedgehog and the toroidal solutions. Generally, if L is the Lagrangian for motion on a Riemannian configuration space M L = $gij(Cj)Q’Q’ - V(q)
A-+(l+e’IT”)A,
a-t(l+E”T”)Z?,
where T” are the generators implying -&4X1
+&a-‘)
then the equation for gradient flow is
of SO( 3)) (T”) i,i = -E,ij,
=o.
gij(q)$
(2.11)
d+A(1+8Ta),
l(Ta)usj.
g=O,
I(qi94i)
=
dL dh’,(q) aqi ds
J=$4sxS+A(cu+p).
.jdhf(q) s=o = gijq ~ ds
SO
’
(3.3)
JV dh:(q) to angular momentum
(3.2)
But if L is invariant under h,, then gi,j(q) and V(q) are separately invariant so
B+Z?(~+E”T’),
This corresponds
-$I.
We first note that any conserved quantity corresponding to an invariance of L is zero along a gradient flow curve. This may be seen as follows. If L is invariant under a l-parameter group of diffeomorphisms h,:M t-f M, s E Iw, then there exists the associated (Noether) momentum
(2.12)
L2 is also invariant under equal right variations of A, Z3 and a rotation of s,
si -+ si -
=
I
This can be rewritten as s(Acu + BP) = 0, where (Acv)~ = Jz”a,j. That is, there exists a conserved isospin Z=Acr+Bp.
(3.1)
conservation (2.13)
3. The gradient flow equations It has been proposed [ 51 that the union of gradient flow curves from the unstable B = 2 hedgehog solution is a candidate for the moduli space for two-Skyrmion dynamics in the Skyrme model. Practically all curves will end up at the minimal energy toroidal solution. A small family of curves will end at infinitely separated
dq”
ds
s=.
= 0,
(3.4)
hence I = 0 along gradient llow curves. We now wish to find the gradient flow curves for the product ansatz. First we rewrite A, g in ZQ in terms of the conserved I, and 0. Noting that TrAA-’ = 2iL2, T&&i =2p2, where it + p, we see that Trd&’
= 2( & -
&=AcY,
p=Z3p
p)” ,
and
Z=
(3.5)
so TrAA-’
+ Trg$-’
= $TrC?&’
+
Z2,
(3.6)
P.W. Irwin, N.S. Mation/Physics
190
and we may rewrite (2.10) L2 =
-
-2M
as
+ Mit2 + ;MS* + +I* + +,&8-t
2K(2C
-
1) {a(s)L?0$ + b(s)TrO}
(3.7)
It is an easy exercise to verify that J, I, k are zero along the gradient flow curves. Expressing (3.j in the form 0i.i = COS6)S,i -/- ( 1 - COS0) nini - Eijk nk Sin 8 with nini = 1, one finds that Trdd-’
=2d2+4(1
The interaction
-cos8)ii*.
potential
2K’{U(S)[COS0+
V
(3.8)
0 and S > 0, so again they repel to infinity. For trajectories near the neutral channel, initially S > 0 but again the Skyrmions rotate into the attractive channel. All trajectories approach the attractive channel except those with z = 1, in which case the Skyrmions separate to infinity. The case z = 1 includes the repulsive channel (8 = v), but if 6 < 7r then 6 < 0 so the Skyrmions asymptotically approach the neutral channel. For the case of massless pions the gradient flow equations simplify considerably and it is possible to explicitly solve for the trajectories of s, z, 8. Thus for the remainder of this section we assume that m, = 0. Eqs. (3.10), (3.11), (3.14) become
l&j 2
where~‘=~(2c-1). (3.9)
We are now in a position to calculate the gradient flow equations for the coordinates s, 8, n. Let z = $.n. Applying (3.2) to L2 we have
fU(s>(l-Z2)]
(3.12)
and (3.13)
z = 4K’a(s)
(3.16)
2(1 -c0se)
~=.E(z3_z)
;
s3
Ms2
+
(1 -COSe)
-$
[n-Z
may be combined
$f(e)=$fs*
(3.17)
(3.18)
&
(3.19)
(3.10)
cos 5 = pe 12A
(3.11)
on trajectories, for p some constant. g(z) = log Iz - z31, we have
$1.
(3.13)
to give
2(1 -c0se) (Z3 - Z>
’
Letting f( 6) = log 1cos ! 1, we have
(3.12)
Ati=--4K’.(,)Z[$-zn] -4K’U(s)
- 1)
ss
e
+(1+2CoSe)g}
;MS =
sin8(3z2
= -_2K
(3.15)
so
~MS=-~K’{[COS~+(~-COS~)Z*]~
ih8=2K’Sine[2b(s)
3z2 - 1
;MS=6K(1-cosB)7
is now
(1 -COS~)(FZ~~)*]
+b(s)(1+2cos8)}
Letters B 385 (1996) 187-192
;+
Ms2
’
MS
-$(z)=
A(1 -cosf3)
Integrating,
we find
z -zs=
qs*e%
+;
2
d
3
-g.
Also,
letting
(3.20)
(3.21)
(1 -p2e+$)i
(3.14) In the repulsive channel (z = 1, 0 = n-), 2 = 0, 6 = 0 and S > 0, so the Skyrmions separate to infinity. For trajectories that begin near the repulsive channel (z M l,e M n), initially S > 0, however the Skyrmions then twist towards the attractive channel (z = 0, 8 = r), so eventually they attract and s decreases. Similarly in the neutral channel (8 = 0)) 6 =
along trajectories, for q some constant. Note that the neutral channel (0 = 0) is a stationary point, since 4 = S = 0 (and z is undefined), unlike in the case m, # 0; also a = 0. Note also that f( 6) is not differentiable at B = 7~. So (3.18) is not valid at 8 = r, but this does not matter. One can see from (3.18) that s --f 0 as the attractive channel is approached (0 --+ Z->, but
P.W. Irwin, N.S. Munton/Physics
the attractive channel is never precisely reached in gradient flow, unless the initial field is in the attractive channel, moreover the validity of the product ansatz breaks down as s -+ 0. Again, g(z) is not differentiable at z = 0 or 1, hence (3.21) is not valid there. On a generic gradient flow curve z decreases monotonically with time, from 1 to 0. Therefore 3z2 - 1 changes sign as t increases. It follows from (3.15), (3.16) that s increases from zero to a finite value s,,-,~, which is attained when 3t2 - 1 = 0, and then decreases to zero. At the same time 8 decreases from n- to a positive value, and then returns to Z-. In fact, only the part of such a curve where s > 0 is relevant, as the product ansatz breaks down as s -+ 0. Knowing z (t) it is possible to find expressions for n(t), CC(t) as follows. From (3.12), (3.13)
iz.
B.(nx$)=j.j.=O.
3) =h.n=O,
(nx
So if at t = 0, n = no, Z? = $0 then at all times n(t) , 3(t) lie in the no, $0 plane. Let
[email protected]= zo, n(t)
n(t)
. g(t) = z(t), i(t)
.no = a(t),
* $0 = p(t).
Then
z(t) = 4t) + P(t) + zo
(3.22)
Letters B 385 (1996) 187-192
191
t
=-
24~
M
J
dQ-[l
Z(T)[l-COSbJ(T>l
Z2(~W2 85(T)
.
0
(3.26) So given z (t) we may determine find n(t), i(t).
4. Relative orientation
a(t) , p(t)
and thus
and centers
In the previous section, expressions were given for gradient flow curves in the product ansatz corresponding to two well-separated Skyrmions. These curves may be matched up with the gradient flow curves obtained numerically starting with two Skyrmions at small separation, To do this one needs to identify from the Skyrme field U the coordinates used in the product ansatz; namefy the Skyrmions’ relative orientation and their centers X, Y. Here we sketch an algorithm to obtain the relative orientation and centers when U is precisely of product form. Then if U corresponds to two well-separated Skyrmions, U will be close to product form, and the same estimate should give the best approximation for the centers and relative orientation. Let the Skyrme field be expressed in the form U(x)
=(T(x)
+ir-m(x).
If U is of product form
and
U(x) n(t)
=nacosa(t)+nax(nox&~>sinrw(t)
s(t)
= sacosP(t)
(3.23)
- Y)B+
= AUH(.X - X)A+BU&
(4.1)
then +$a x (SO xita)
sinP(t)
(3.24) (+=cos.fl
and from (3.12),
cos f2 -sin
ft sin f2g,0R2,
(3.13) where
a(t)
=
I
O=A-‘B, fi =f(lxll>,
dr]A(r)l
0
x1
t
=-
12K
A
/T?(t) = J 0
J
&Z(7)[1-z*(7)11’*
0
dr(S(r)(
s3(r)
(3.25)
so
=
Ix - XI
)
x* =
f2=f(lx21),
Ix - YI,
(4.2)
192
P.W. Irwin, N.S. Manton/Physics
Our idea is just to use the function g to alternately estimate the centers X, Y and the relative orientation 0, until best estimates for both are obtained. If U is of product form then (+ M 1 in most of R3 with g M -1 in two regions approximately where the Skyrmions are centered. We assume that IT = -1 at two points only, and call these Xa and Ya. Our first estimates for X and Y are X0 and Ya. Now, using (7 and setting X = Xa and Y = Yo in (4.3), we evaluate the r.h.s. of (4.3) at enough points x to determine the quadratic form on the 1.h.s. of (4.3) and extract 0. Call this estimate 00. In practice, the points x should not be too close to X0 or Ya, nor too far away, to avoid large errors. The next step is to improve the estimates of the centers X, Y. This means using the formula (4.2) for u with 0 = 00, and seeking X, Y so that (T = -1 at Xa and Yo. A simple method is to evaluate (4.2) using (30, X0 and Ya as the product ansatz coordinates and then determine the points Xb, Yh say, where (+ = - 1. Then to compensate for the error, evaluate u again using 00, X0 - (XL - X0) and Ya - (YA - Ya) as coordinates. The points where F = - 1 should now be closer to X0, Ya, but any error can be compensated in a similar way. By one or more steps of this kind we find
Letters B 385 (1996) 187-192
a new estimate for the centers Xt, Yi. Given these, we can use (4.3) this time with X = Xl, Y = Y1, to obtain an improved estimate for 0, say 01. Then we can improve the estimate for the centers, and so on. It is not difficult to check that if the Skyrmion separation s is large, then this scheme should converge rapidly to give the centers and relative orientation.
Acknowledgements P.W.I. thanks PPARC for a research studentship.
References [ I] T.H.R. Skyrme, Proc. Roy. Sot. A 260 (1961) 127. [2] J.J.M. Verbaarschot, T.S. Walhout, J. Wambach and H.W. Wyld, Nucl. Phys. A 468 (1987) 520. [3] B.J. Schroers, Z. Phys. C 61 (1994) 479. [4] G.S. Adkins, C.R. Nappi and E. Witten, Nucl. Phys. B 228 (1983) 552; G.S. Adkins and CR. Nappi, Nucl. Phys. B 233 (1984) 109. [5] N.S. Manton, Phys. Rev. Lett. 60 (1988) 1916. [6] T. Waindzoch and J. Wambach, Nucl. Phys. A 602 (1996) 347.