Table Of ContentAre There Static Textures?
Lukas Lichtensteiger1 and Ruth Durrer2
1AILab, Computer Science Department, Universit¨at Zu¨rich, Winterthurerstrasse
190, CH-8057 Zu¨rich, Switzerland
9 2D´epartement de Physique Th´eorique,Universit´e de Gen`eve, Quai Ernest
9
Ansermet 24, CH-1211 Gen`eve, Switzerland
9
1
n
a
J
4
Abstract
1
v WeconsiderharmonicmapsfromMinkowskispaceintothethreesphere.
4 Weare especially interested in solutions which are asymptotically constant,
2 i.e. converge to the same value in all directions of spatial infinity. Physical
0 3-space can then be compactified and can be identified topologically (but
1 notmetrically!)withathreesphere.Therefore,atfixedtime,thewindingof
0
themap is defined. Weinvestigate whether static solutions with non-trivial
9
winding number exist. The answer which we can proof here is only partial:
9
Weshow that within a certain family of maps no static solutions with non-
/
h zero winding number exist. We discuss the existing static solutions in our
p family of maps. An extension to other maps or a proof that our family of
- maps is sufficiently general remains an open problem.
o
r
t
s 1 Introduction
a
:
v
Manyphysicalproblemscanbedescribedbyscalarfieldsϕwithtopologicallynon-
i
X trivial targetspaces. The equation of motion for ϕ often requires ϕ to representa
r harmonicmapfromspacetimeintothetargetspace(Inthephysicsliteraturesuch
a
maps are better known under the name ’non-linear σ-model’).
Thequestionariseswhetheragivenfieldconfigurationistopologicallytrivial
(continuously deformable to the constant map). Topological defects are topolog-
ically non-trivial field configurations. If we consider a topologically trivial four
dimensionalspace-timemanifoldM,non-trivialfieldconfigurationsareingeneral
singular on a certain sub-manifold S ⊂ M. The dimension of S depends on the
first non-trivial homotopy of ϕ(M\S)≡Imϕ:
If π (Imϕ) is non-trivial, the submanifold S forms a network of ’domain walls’ of
0
space-time dimension 3. If π (Imϕ) is non-trivial, a network of ’strings’ of space-
1
time dimension 2 is formed. If π (Imϕ) is non-trivial ’monopoles’ of space-time
2
dimension 1 appear. And if π (Imϕ) is non-trivial ’textures’, singular events of
3
space-time dimension 0 appear.
1
Higherhomotopygroupsdonotleadtotopologicaldefectsinfourspace-time
dimensions.Thesimplestandmostcommonexamplesoftopologicaldefectsarethe
cases Imϕ=Sn, where Sn denotes the sphere of dimension n and S0 ={−1,1}.
But also other examples play an important role in solid state physics (helium[1],
liquid crystals[2]) and cosmology[3, 4].
InthecaseofafieldlivingonSnwithn≤2simplestaticdomainwall(n=0),
string (n = 1) and monopole (n = 2) solutions are known (see, e.g. [4] or [5]).
The question which we want to address here is whether there also exist static
texture solutions (n = 3). A static texture solution is different from monopoles,
strings and domain walls in that it is non-singular: A map from R3 to S3 which
is asymptotically constant,lim ϕ(x,t)=ϕ (t), can wind aroundS3 without
|x|→∞ 0
being singular anywhere. Derricks theorem[6] then already implies that there is
no static texture solution with finite energy. However, the simple static domain
wall, string and monopole solutions we are alluding to (which contain a singular
sheet, line and point respectively), have infinite total energy and we thus want
to allow also for infinite energy solutions. We therefore cannot apply Derricks
theorem. Nevertheless, numerical simulations[7, 8] indicate, that winding texture
configurations always shrink, leading to a singularity, the unwinding event, at
a finite time t , after which the configuration becomes topologically trivial and
c
approaches the constant solution, as predicted by Derricks theorem. The total
energy of the initial configuration is, however in general infinite so that Derricks
theorem cannot be applied.
Thisnumericalfindingpromptedustosearchforaproofforthenon-existence
ofstatictextureinflatphysicalspace.Clearly,theresultdependsonthegeometry
ofphysicalspace.Ifspaceis a 3-sphere,the identity maprepresentsa welldefined
statictexturesolution.Wewanttoinvestigatewhethersuchsolutionsareexcluded,
for example in Minkowski space.
Wedonotquitesucceedinthistask.First,weshallassumethatthesearched
for static winding solution has a spherically symmetric Lagrangian density. This
assumption does not bother us too much. It seems physically well motivated (we
can, however not use any rigorous energy arguments to justify it, since the total
energy of our solution must be infinite). Also within the class of solutions with
spherically symmetric Lagrangian densities we have a proof only for a special
ansatzfor the fieldconfigurationanditremains anopen problemhow generalour
ansatz is.
Our paper is organized as follows: In the next section we write down the
equationsofmotionforthescalarfieldandspecifyouransatz.Insection3wethen
showthatwithinthisansatznostaticsolutionwithnon-trivialwindingnumbercan
existanddiscussthenatureofthegloballyexisting(non-winding)staticsolutions.
In section 4 we present the conclusions and an outlook.
2
2 Spherically symmetric ’texture’ fields
We consider a scalar field (order parameter) ϕ : M → S3 and we ask ϕ to be
harmonic, a non-linear σ-model. A harmonic map satisfies the Euler-Lagrange
equations for the action
1
S(ϕ)= |dϕ|2 dx. (1)
2 S3
ZM
We consider the situation where M is 4-dimensional Minkowski spacetime with
the flat Lorentzian metric g and S3 is the unit three-sphere with the standard
metric which we denote by G.
Here dx denotes the volume element of the metric g on M and
∂ξi ∂ξj
L(ϕ)=|dϕ|2 =Trace (ϕ∗G)=gµν(x)G (ϕ(x)) (x) (x), (2)
S3 g ij ∂x ∂x
µ ν
for some(local)coordinatesx=(x ,...,x ) onM andϕ=(ξ ,ξ ,ξ )onS3 (We
0 3 1 2 3
always assume summation over repeated indices).
We only consider regular maps ϕ, i.e. maps that have finite energy density
|dϕ|2 everywhere. In addition to being stationary points of (1) we also demand
S3
our maps to be asymptotically constant,
lim ϕ(x,t)=ϕ (t) .
0
|x|→∞
At fixed time we then canconsider them as maps ϕ from compactifiedR3, R3 =
t
R3∪{∞}≡S3 to S3, assigning ϕ (x)=ϕ(x,t) and ϕ (∞,t)=ϕ (t).
t t 0
The winding number of this extended map ϕ : S3 → S3 is a topological
t
invariant and counts how many times ϕ (S3) winds around the target S3. This
t
number cannot change under continuous time evolution. We would like to show
that there are no static solutions ϕ with non-zero winding number.
Unfortunately, we are not able to solve the problem in this generality. We
thus impose some restrictions onthe maps ϕ. One way of doing this is to demand
ϕ to obey certain symmetry properties. We want to impose spherical symmetry,
i.e., invariance under SO(3), the group of rotations of physical space.
The action of an element of the rotation group, g ∈ SO(3) on the maps
ϕ:M →S3 is given by
(g·ϕ)(x)=ϕ(g−1·x) (3)
(scalarfield).Thefixedpointsoftheaction(3)arethesphericallysymmetricfields
of the form ϕ=ϕ(r,t), r =|~x|. We might want to require spherical symmetry of
the fieldϕitself.Forourpurposeshowever,thisrestrictionistoosevere:Sincethe
imageof asmoothmapcannever haveadimensiongreaterthanthe dimensionof
the domain, this would limit us to only two-dimensional ranges (one-dimensional
in the static case) which are topologically not interesting. Instead we will only
demand that the Lagrangiandensity
L=|dϕ|2 (4)
S3
3
be SO(3) invariant, L(gϕ)=L(ϕ).
We proceedas follows:We first derive the full Euler-Lagrangeequations and
then make an ansatz for ϕ (which is inspired by our symmetry requirement).
We then insert the ansatz into the Euler-Lagrangeequations. Since the equations
remain self-consistent, we can try and solve them. The solutions we find are then
always solutions of the full Euler-Lagrange equations. It remains to investigate
whether they can be topologically nontrivial, i.e., whether there exist solutions
with nonzero winding number.
We use standardspherical coordinates (r,θ,φ) for the spatial part and write
the standard metric on flat Minkowski spacetime M as
g =−dt2+dr2+r2(dθ2+sin2(θ)dφ2). (5)
For y ∈S3 we use the standard spherical coordinates
y = (sinξ sinξ sinξ ,sinξ sinξ cosξ ,sinξ cosξ ,cosξ ).
3 2 1 3 2 1 3 2 3
In our case the ξ are functions living on spacetime M,
i
ξ :M →R, ξ =ξ (t,r,θ,φ). (6)
i i i
The standardranges for the angles ξ are ξ ∈[0,2π],ξ ∈[0,π] and ξ ∈[0,π]. It
i 1 2 3
is important to note that for a map to cover all of S3, ξ and ξ have to assume
2 3
both boundary values, 0 and π, and ξ must assume both 0 and 2π.
1
The standard metric on S3 expressed in terms of (ξ ) is
i
G=dξ2+dξ2sin2(ξ )+dξ2sin2(ξ )sin2(ξ ). (7)
3 2 3 1 3 2
The equations of motion corresponding to the Lagrangian(1) are
∂L ∂L
∇ = , 1≤i≤3. (8)
µ
∂(∇ ξ ) ∂ξ
(cid:18) µ i (cid:19) i
With the standard metric (7) on S3 they become:
0 = ∇µ∇ ξ −sin(ξ )cos(ξ ) (∇ ξ )(∇µξ )+sin2(ξ )(∇ ξ )(∇µξ )
µ 3 3 3 µ 2 2 2 µ 1 1
0 = ∇µ∇ ξ +2cot(ξ )(∇ ξ )(∇µξ )−sin(ξ )cos(ξ )(∇ ξ )(∇µξ ) (9)
µ 2 3 µ 3(cid:2) 2 2 2 µ 1 1 (cid:3)
0 = ∇µ∇ ξ +2cot(ξ )(∇ ξ )(∇µξ )+2cot(ξ )(∇ ξ )(∇µξ )
µ 1 2 µ 2 1 3 µ 3 1
To obtain a spherically symmetric Lagrangian density we use a generalized
hedgehog Ansatz:
ξ =ξ (φ,θ), i=1,2 and ξ =ξ (r,t). (10)
i i 3 3
The idea behind this Ansatz is that we want to make use on the vaste knowledge
onharmonicmaps on2-dimensionalspaces,which willhelp us to firstsolvethe 2-
dimensionalproblemforξ andξ inthe coordinatesθ andφandthen afterwards
1 2
4
solve the equation for the remaining function ξ . The crucial point in order for
3
this to work is that the equations of motion respects this ‘decomposition’ of ϕ in
a(r,t)-dependent anda(θ,φ)-dependent part.Thisis the subjectofthe theoryof
(ρ,σ)-equivariant maps which is explained in great detail in [11].
With the Ansatz (10) we may introduce a map
Ω:S2 →S2 , Ω(φ,θ)=(ξ ,ξ ) (11)
1 2
with Lagrangiandensity
|dΩ|2 =(∇ ξ )(∇µξ )+sin2(ξ )(∇ ξ )(∇µξ ). (12)
µ 2 2 2 µ 1 1
The total Lagrangiandensity of the map ϕ is then
|dϕ|2 =(∇ ξ )(∇µξ )+sin2(ξ )|dΩ|2. (13)
µ 3 3 3
Our Ansatz (10) yields
λ(φ,θ)
|dΩ|2 = , (14)
r2
Spherical symmetry of the Lagrangian density then requires λ = constant(≥ 0).
Inserting our Ansatz (10) into the Euler-Lagrange equations (10), we find after
multiplyingwithr2 thatforthecomponentsξ andξ theseequationsarejustthe
1 2
Euler-Lagrange equations of the map Ω : S2 → S2. Thus Ω has to be harmonic
(on S2!) with constant Lagrangian density |dΩ|2 = λ, where |.|2 now denotes
S2 S2
the Lagrangian density on S2. But this means that Ω has to be an eigenmap of
the Laplacian △ on S2 with eigenvalue λ (in the sense of [11]). Therefore the
components Ω have to be given by linear combinations of spherical harmonics of
i
fixed degree k, Y ,
km
Ω = a Y .
i im km
m
X
If we apply this to our map Ω : S2 → S2, we obtain for |dΩ|2 (with respect
to the metric g on spacetime)
k(k+1)
|dΩ|2 = , k ∈N. (15)
r2
We willalwaysassumeλ=k(k+1)>0sinceweareonlyinterestedinspherically
non-trivial maps. Note that λ = 2 just corresponds to Ω = id, the identity map,
used for example in the ‘hedgehog’ monopole.
The remaining Euler-Lagrangeequation for the last component ξ is now
3
k(k+1)
0=∇µ∇ ξ − sin(2ξ ), k ∈N. (16)
µ 3 2r2 3
Itisthisequationthatwewouldliketoanalyzeinthispaper.Inthenextsec-
tionwewillprovethenon-existenceofstatic solutionswithnonzerowindingnum-
ber, whereas an infinite family of time-dependent solutions of (16) in Minkowski
space for arbitrary winding number n∈N can be found [13].
5
In[10]solutionsfromageometricalS3 toS3 arestudiedwiththeansatz(16)
for k = 0,1. Here we consider compactified R3, which is topologically equivalent
toS3,butwithflatgeometry.Usingouransatz(10)onthegeometricalS3,wecan
alsogeneralizethe resultsof[10]byshowingthatthereareactuallytwocountable
families of such maps for every k >0, k ∈N, where k(k+1) is the eigenvalue of
the map Ω defined in (11). This will be detailed in a subsequent paper [13].
All of these results can also be found in [12].
3 Non-existence of static winding solutions
We consider maps to the standard three sphere where spacetime is parametrized
bystandardCartesiancoordinatesrandt.Inwhatfollowswewillusethenotation
∂ ∂
˙= , and ′ = . (17)
∂t ∂r
Then (16) becomes
2 k(k+1)
ξ′′−ξ¨ + ξ′ − sin(2ξ )=0, k∈N (18)
3 3 r 3 2r2 3
An exact solution to (18) which describes a winding time dependent texture for
the case k = 1 has been found in [14]. (Time-dependent solutions to (18) can in
fact be found for any k∈N and for any winding number n∈N [13]).
The r-dependence of the last term in (18) shows that nontrivial solutions of
the form ξ =ξ (t) are impossible for k 6=0. However,for static maps the Ansatz
3 3
(10) becomes ξ =ξ (r) and equation (18) reduces to (ξ ≡ξ )
3 3 3
2 k(k+1)
ξ′′+ ξ′− sin(2ξ)=0 , (19)
r 2r2
the equation we will discuss in the following.
3.1 Local properties
Equation (19) has a singular point at r = 0. Since we require solutions to be
regularforallr andt,weassumethatξ(r)isdescribedinaneighborhoodofr =0
by some power series
∞
ξ(0+ǫ)= a ǫj. (20)
j
j=0
X
If we multiply equation (19) by r2 and insert (20) at r = 0 + ǫ we obtain an
equation in powers of ǫ
∞
0 = (j+2)(j+1)a ǫj+2+2(j+1)a ǫj+1
j+2 j+1
j=0
X(cid:0) (cid:1)
6
a = mπ (2m+1)π/2
0
a = (0<j <k) 0 0
j
a = free 0
k
a = 0 0
k+1
a = − k(k+1) a3 0
k+2 3(2k+3) k
Table 1: Expansion coefficients for ξ at r =0 where k(k+1) is the eigenvalue of
3
the map Ω:S2 →S2 and m∈Z.
∞ ∞
k(k+1)
− cos(2a )sin 2 a ǫj +sin(2a )cos 2 a ǫj .(21)
0 j 0 j
2
j=1 j=1
X X
From comparison of the lowest order terms ǫ0 we obtain immediately
sin(2a )=0 ⇒a =mπ/2, m∈Z. (22)
0 0
Let l ≥ 1 be the smallest integer for which a 6= 0. Then the lowest order
l
terms ǫl yield
k(k+1)
l(l−1)a +2la − cos(2a )·2a =0 (23)
l l 0 l
2
Therefore
a (l(l+1)−k(k+1)cos(2a ))=0. (24)
l 0
Combining this with (22) leads to
a (l(l+1)−(−1)mk(k+1))=0, (25)
l
thus for m odd, a = 0 for all l > 0 and thus ξ is constant, but for m even, ξ(r)
l
has a nontrivial power series expansion at r = 0 for every eigenmap Ω : S2 →S2
witheigenvaluek(k+1),wherethefirstnon-vanishingexpansioncoefficientisjust
a . For the next higher orders ǫk+1 and ǫk+2 we get in a similar way
k
k(k+1)
a =0 and a =− a3. (26)
k+1 k+2 3(2k+3) k
Table 1 summarizes this information.
3.2 Global Properties
For a global analysis of the behaviour of solutions of (19) it is convenient to
transform the equation into an autonomous one by the transformation
1 2
s= lnr, where β = . (27)
β sk(k+1)
7
The new variable s runs from −∞ to +∞. Remember that we always require
k > 0, k ∈ N, so that 0 < β ≤ 1. Denoting the derivative w.r.t. s again by a
prime, Eq. (19) transforms into
ξ′′+βξ′−sin(2ξ)=0 (28)
This differential equation describes the motion of a particle with constant
damping β and potential sin(2ξ). If we switch off the damping, we obtain the
conservative system
ξ′′ =−gradU(ξ), (29)
where the potential U(ξ) is given by
ξ
U(ξ)= −sin(2ξ)dξ =−sin2ξ−U . (30)
0
Zξ0
The ‘energy’ of this system
1
E(ξ,ξ′)= ξ′2−sin2ξ−U (31)
0
2
is conserved, and all solutions are periodic, lying on the sets E=const. When we
switch on the damping, the solutions no longer remain on levels with E=const.,
but ‘fall down into the potential wells’ at ((2m+1)π/2,0) (see Fig. 1).
1
¢x
0
-1
0 1 2 3
x
Figure 1: Phase diagram for the solutions ξ(s) and ξ′ = dξ of the damped system
ds
(k =1,β =1).Becauseoftheenergylossthesolutions‘falldown’intothepotential
wells at ((2m+1)π/2,0)
Ifthedampingisweak(β ≪1)theyspiralseveraltimesaroundthosepoints,
gradually loosing energy and moving towards the center (under-damped motion),
8
whereas if the damping is strong (β ≫ 1) they move towards the center quickly
(over-damped motion). In any case, since the energy is no longer conserved, non-
trivial periodic solutions are not possible.
We can formulate this more precisely: If we differentiate the ‘energy’ (31)
with respect to s and insert (28) we obtain
dE
=ξ′ξ′′−sin(2ξ)ξ′ =−βξ′2. (32)
ds
Thus the energy is always decreasing with growing s (for non-constant ξ). With
this we can show the following Lemma:
Lemma 3.1 If ξ(s )=mπ, m∈Z, for some s >−∞, and if ξ is not constant,
0 0
it tends monotonically to ±∞ for s→−∞.
Proof: (Theideaoftheproofistakenfrom[10]).Letξ(s )=mπwithξ′(s )=a.
0 0
For s<s we have
0
1 1 1 1
0<E(s)−E(s )= ξ′2−sin2ξ− a2 ≤ ξ′2− a2 (33)
0
2 2 2 2
and thus ξ′2 >a2, ∀s<s . Therefore E is monotonically increasing for s→−∞
0
and ξ′2 → ∞ with ξ′2 > 0, ∀s < s . Correspondingly, ξ tends monotonically to
0
±∞, the sign depending on the sign of ξ′.
Lemma 3.2 If ξ(s ) = mπ, m ∈ Z for some s , and ξ′(s ) = 0, then if ξ is not
0 0 0
constant either mπ <ξ(s)<(m+1)π or (m−1)π <ξ(s)<mπ ∀s>s .
0
Proof: Let ξ(s ) = mπ with ξ′(s ) = 0. If ξ is not constant then for s > s we
0 0 0
have
1
0>E(s)−E(s )= ξ′2−sin2ξ (34)
0
2
and thus sin2ξ > 1ξ′2 ≥0, ∀s>s . Especially, sin2ξ(s)6=0 ∀s>s which
2 0 0
implies our statement.
Corollary 3.1 There is no static texture solution which satisfies the ansatz (10)
( a texturesolution being a solution with homotopy degree 6=0, i.e., one that really
winds).
Proof: From Lemma 3.1: The only non-constant regular solutions through a
point mπ are the ones with
ξ(−∞)= lim ξ(s)=mπ. (35)
s→−∞
Furthermore, for a regular solution
dξ
ξ′(−∞)= lim ξ′(s)= limβr =0. (36)
s→−∞ r→0 dr
9
To fully wind around S3, ξ =ξ with ξ =mπ at r = 0 (s =−∞) would have to
3
assume either the value (m−1)π or (m+1)π which, according to Lemma 3.2 is
not possible.
3.3 Existence and stability of static solutions
Finally,wewouldliketobrieflyverifythatnon-constantsolutionsto(28)doindeed
exist globally, and we want to review their properties. With the substitution
x = ξ (37)
y = ξ′ (38)
equation (28) is equivalent to the autonomous system of first order differential
equations
x′ = y (39)
y′ = sin(2x)−βy (40)
Together with some initial conditions x(0)=u , y(0)=v , this is an initial value
0 0
problem (IVP) of the form
x′
= f(x,y)
y′
(cid:18) (cid:19)
(x(0),y(0)) = (u ,v ). (41)
0 0
The local existence and uniqueness of a solution to the IVP (41) follow from
standard theorems on ordinary differential equations.
All regular solutions (x(s),y(s)) with
lim x(s)=mπ, m∈Z(and thus lim y(s)=0) (42)
s→−∞ s→−∞
are bounded for every s ∈ R as follows directly from Lemma 3.2 and its proof.
Thereforethesesolutionsexistglobally(cf.[15],Corollary3.2).(Thelocalexistence
of solutions fulfilling (42) follows from our series expansion in Section 3.1.)
Now let us discuss the stability of the critical points (x ,y ) for this system
0 0
which are given by
(x ,y )=(mπ/2,0), m∈Z. (43)
0 0
Insomeneighborhoodofacriticalpointwecanapproximate(41)bythelinearized
system
x′ x
=(Df)(x ,y ) (44)
y′ 0 0 y
(cid:18) (cid:19) (cid:18) (cid:19)
where Df is the first derivative of f (with respect to x and y). If we calculate the
eigenvalues of Df(x ,y ) and use the principle of linearized stability [9] we find
0 0
that the critical points (x ,y ) = (mπ,0), m ∈ Z are unstable, while the critical
0 0
10
