Moving Frames and Conservation Laws for Euclidean Invariant Lagrangians

0.27 MB·English

by  Tania M. N. Goncalves

Moving frames and conservation laws for Euclidean
invariant Lagrangians
2
1
Tˆania M N Gonc¸alves and Elizabeth L Mansfield
0
2 SchoolofMathematics,Statistics andActuarialScience,
UniversityofKent,Canterbury,CT27NZ,UK
n
a E-mail: [e-mail], [e-mail]
J

2 Abstract. Noether’s First Theorem yields conservation laws for Lagrangians
1 with a variational symmetry group. The explicit formulae for the laws are well
known and the symmetry group is known to act on the linear space generated

] by the conservation laws. In recent work the authors showed the mathematical
G structurebehindboththeEuler-Lagrangesystemandthesetofconservationlaws,
D intermsofthedifferentialinvariantsofthegroupactionandamovingframe. In
thispaperwedemonstratethattheknowledgeofthisstructureconsiderablyeases
. finding the extremal curves for variational problems invariant under the special
h
EuclideangroupsSE(2)andSE(3).
t
a
m
PACSnumbers: 02.20.Hj,02.30.Xx,02.40.Dr
[

2
v Submitted to: J. Phys. A: Math. Gen.
4
6
1. Introduction
9
3
. In 1918Emmy Noether wrote the seminal paper “InvarianteVariationsprobleme”[1],
6
where she showed that for differential systems derived from a variational principle,
0
1 conservation laws could be obtained from Lie group actions which left the functional
1 invariant.
: Recently in [2], it was provedthat for Lagrangiansthat areinvariantunder some
v
i Liesymmetrygroup,Noether’sconservationlawscanbe writtenintermsofamoving
X
frame and vectors of invariants. Furthermore, in [2] the authors showed that for one-
r dimensional Lagrangians that are invariant under a semisimple Lie symmetry group,
a
the new format for the conservation laws could reduce considerably the calculations
needed to solve for the extremal curves. In particular, a classification is given for
variationalproblemswhichareinvariantunderthethreeinequivalentSL(2,C)actions
ontheplane(classifiedbyLie[3]). InthispaperweusethenewstructureofNoether’s
conservation laws presented in [2] to simplify variational problems that are invariant
under SE(2) and SE(3); these groups are not semisimple.
In Section 2, we will briefly give an overview on moving frames, on differential
invariants of a group action and on invariant calculus of variations. Throughout
Section 2 we will use the group action of SE(2) on the plane as our pedagogical
example.
InSection3,weshowinsomedetailhowtocomputethenewversionofNoether’s
conservation laws presented in Theorem 2.9 for one-dimensional variational problems
Moving frames and conservation laws for Euclidean invariant Lagrangians 2
that are invariant under SE(2), and then demonstrate how their invariantized
Euler-Lagrange equations and Noether’s conservation laws can be used to solve the
integration problem.
FinallyinSection4,wepresentthesimplifiedsolutiontothephysicallyimportant
one-dimensional variationalproblems that are left unchanged under the SE(3) group
action.
2. Structure of Noether’s Conservation Laws
In this section, we will give a brief overview of concepts regarding moving frames,
differentialinvariantsofagroupactionandtheinvariantcalculusofvariationsneeded
to understand the statements of our result. For more information on these subjects,
seeFelsandOlver[4,5],Mansfield[6]andKoganandOlver[7]. WewillusetheSE(2)
action on the plane as our pedagogicalexample.
2.1. Moving frames and differential invariants of a group action
Here we are using moving frames as reformulated by Fels and Olver [4,5], adapted to
the context of differential algebra.
LetX bethespaceofindependentvariableswithcoordinatesx=(x ,...,x )and
1 p
U the space of dependent variables with coordinates u = (u1,...,uq). We will use a
multiindex notation to represent the derivatives of uα, e.g.
∂|K|uα
uα = ,
K ∂xk1∂xk2 ∂xkp
1 2 ··· p
where the tuple K = (k ,...,k ), represents a multiindex of differentiation of order
1 p
K = k +k + +k . Hence, let M = Jn(X U) be the n-th jet bundle with
1 2 p
| | ··· ×
coordinates
z=(x ,...,x ,u1,...,uq,u1,...).
1 p 1
On this space, the operator ∂/∂x extends to the total differentiation operator
i
q
D ∂ ∂
D = = + uα .
i Dx ∂x Ki∂uα
i i α=1 K K
XX
In this paper we are interested in using Noether’s conservation laws to find the
solutions that extremize variational problems which are invariant under SE(2) and
SE(3). Thus, consider a Lagrangian L(z)dx that is invariant under some symmetry
group. Let a group G act on the space M as follows
G M M
×z → z=g z,
7→ ·
which satisfies either g (h z) = (gh) z, called a left action, or g (h z) = (hg) z,
called a right action. W· e s·ay aeLagra·ngian L(z)dx is invariant u·nde·r some gro·up
action if
L(z)dx=L(z)dx
for all g G.
∈
Consider a Lie group Geactieng smootlhy on M such that the action is free and
regular. Then for every z M there exists a neighbourhood of z, as illustrated in
∈ U
Figure 1, such that
Moving frames and conservation laws for Euclidean invariant Lagrangians 3
- the group orbits have the dimension of the group G and folliate ;
U
- thereisasurface whichintersectsthegrouporbitstransversallyatasingle
K⊂U
point. This surface is called the cross section;
- if (z)representsthegrouporbitthroughz,thenthegroupelementg Gtaking
z O to k is unique. ∈
∈U
U
K
O(z)
g
z
k
Figure 1. Alocalfoliationwithatransversecrosssection
Undertheseconditions,wecandefinearight moving frame asthemapρ: G
which sends z to the unique element ρ(z) G such that U →
∈U ∈
ρ(z) z=k, k = (z) .
· { } O ∩K
The element g G in Figure 1 corresponds to ρ(z).
To obtain∈the right moving frame, which sends z to k, we must first define the
cross section as the locus of the set of equations ψ (z) = 0, j = 1,...,r, where r is
j
K
thedimensionofG. Normally,thecrosssectionischosensoastoeasethecalculations.
Then solving the set of equations
ψ (z)=ψ (g z)=0, j =1,...,r, (1)
j j
·
knownasthe normalization equations, forthe r groupparametersdescribingG yields
the right moving fraeme in parametric form. Hence, the frame obtained satisfies
ψ (ρ(z) z)=0, j =1,...,r.
j
·
Bytheimplicitfunctiontheorem,auniquesolutionof(1)providesanequivariant
map, i.e. for a left action
ρ(z)=ρ(z)g−1
and for a right action
e
ρ(z)=g−1ρ(z).
Example 2.1. Consider the SE(2) group acting on curves in the (x,y(x))-plane as
follows, e
x x cosθ sinθ x a
= − , (2)
y 7→ y sinθ cosθ y b
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) − (cid:19)(cid:18) − (cid:19)
where θ, a and b are constantsethat parametrize the group action. Here we are using
the inverse action because it siemplifies the calculations.
Moving frames and conservation laws for Euclidean invariant Lagrangians 4
There is an induced action on the derivatives y , where K is the index of
K
differentiation with respect to x, called the prolonged action. The induced action
on y is defined to be
x
dy dy/dx
y =g y = =
x x
· dx dx/dx
by the chain rule, so the actioneof (2)eon y is
e x
e e
sinθ+y cosθ
x
y = − .
x
cosθ+y sinθ
x
Similarly,
e
d2y 1 d dy y
xx
y =g y = = = . (3)
xx · xx d(x)2 dx/dxdx dx (cosθ+y sinθ)3
(cid:18) (cid:19) x
If we consider M to be the space wieth coordinates (x,y,ey ,y ,...), then the action is
g x xx
locally free near the identity of SE(e2). Thues, taking theenormalization equations to be
x=0, y =0 and y =0, we obtain
x
a=x, b=y, and θ =arctany (4)
x
e e e
as the frame in parametric form.
Remark 2.2. In this paper we will consider all independent variables to be invariant.
If these are not invariant, then we can reparametrize and set the original independent
variables as depending on the new invariant parameters.
Theorem 2.3. Let ρ(z) be a right moving frame. Then the quantity I(z) = ρ(z) z
·
is an invariant of the group action (see [4]).
Consider z = (z ,...,z ) M and let the normalization equations z = c for
1 m i i
∈
i=1,...,r, where r is the dimension of the group G, then
ρ(z) z=(c ,...,c ,I(z ),...,I(z )), e
1 r r+1 m
·
where
I(zl)=g zg=ρ(z), l=r+1,...,m.
· |
Example 2.1 (cont.) Evaluating y given by (3) at the frame (4) yields
xx
y
xx
y = ,
xx|(a=x,b=y,θ=arctagnyx) (1+y2)3/2
x
the Euclidean curvature. So evaluating y at the frame (4) yields a differential
g K
invariant.
Definition2.4. ForanyprolongedactioninfthejetspaceJn(X U),theinvariantized
×
jet coordinates are denoted as
Ji =I(xi)=xi|g=ρ(z), IKα =I(uαK)=uαK|g=ρ(z). (5)
These are also known as the normalized differential invariants.
e f
Example 2.5. Consider the group action of SE(2) as in Example 2.1. Since x is
not invariant we reparametrize (x,y(x)) as (x(s),y(s)), where s is invariant and let
g SE(2) act on (x(s),y(s)) as in Example 2.1. Solving the normalization equations
∈
x=0, y =0 and y =0, we obtain the frame
s
y
s
a=x, b=y, θ =arctan . (6)
e e e x
(cid:18) s(cid:19)
Moving frames and conservation laws for Euclidean invariant Lagrangians 5
We have then
g zg=ρ(z) =(s,x,y,xs,ys,yss)g=ρ(z)
· | |
=(Ie(se),eIxe,Iye,I1xf,I1y,I1y1)
x y y x
= s,0,0, x2+y2,0, s ss− s ss . (7)
s s x2s+ys2 !
p
The second, third and fifth components of (7) coprrespond to the normalization
equations x= 0, y =0 and y =0 respectively. The fourth and sixth components, Ix
s 1
and Iy respectively, are the lowest order differential invariants and all higher order
11
invariants can be obtained in terms of them and their derivatives.
e e e
Theorem 2.6. (Replacement Theorem [5]) If f(z) is an invariant, then
f(z)=f(I(z)).
This theorem allows one to find the Iα in terms of historically well-known
K
invariants without having to solve for the frame.
Example 2.5 (cont.) We know that SE(2) preserves x , thus applying the
s
| |
Replacement Theorem we obtain
x = x2+y2 = (Ix)2+(Iy)2 = Ix ,
| s| s s 1 1 | 1|
q
which yields that x2p+y2 = Ix up to a sign. Next we know that the Euclidean
s s 1
curvature κ is also invariant under SE(2), and we obtain
p
x y y x IxIy IyIx Iy
κ= s ss− s ss = 1 11− 1 11 = 11 ,
(x2+y2)3/2 ((Ix)2+(Iy)2)3/2 (Ix)2
s s 1 1 1
which gives us Iy in terms of κ and x .
11 | s|
Since we are considering all independent variables to be invariant, all total
differential operators will also be invariant. Hence, the invariantized differential
operators
=D z =D .
Di i|g=ρ( ) i
We know that
f
∂
uα =uα ,
∂x K Ki
i
although the same cannot be said about its invariantized version and in general
Iα =Iα ; indeed we have that
Di K 6 Ki
Iα =Iα +Mα , (8)
Di K Ki Ki
whereMα isknownasthecorrection term. Wewillnotgointoitscalculationsinceit
Ki
wouldtakeustofarafield,butformoreinformationoncorrectiontermssee 4.5[6]. In
§
anycase,softwareexiststocalculatethesecorrectionterms[8]. Equation8showsthat
theprocessesofinvariantizationanddifferentiationdonotcommute. Consideringtwo
generatingdifferentialinvariantsIα andIα andlettingJK =LM sothatIα =Iα ,
J L JK LM
then this implies that
Iα Mα = Iα Mα . (9)
DK J − JK DM L − LM
Moving frames and conservation laws for Euclidean invariant Lagrangians 6
These equations are called syzygies or differential identities. These will play a crucial
role in the obtention of the invariantized Euler-Lagrange equations and Noether’s
conservation laws.
Example 2.5 (cont.) Ifwe setx=x(s,t) and y =y(s,t) andtake thenormalization
equations as before, we obtain
x x +y y x y y x
xt|g=ρ(z) =I2x = s xt2+ys2t, yt|g=ρ(z) =I2y = s xt2−+sy2t.
s s s s
Furthermore, sinece both s and t arpe invariant, s aned t commute. pFrom Figure 2,
D D
we can see that there are two ways in which we can obtain Ix
12
b b b b
Ix Ix
22 122 b b
Ix DxIx Ix
2 12 112 b
t
D
0 Ix Ix
1 11 b
Figure 2. Paths toI1x2
and since both ways must beequal, we get a syzygy between Ix and η =Ix. The syzygy
2 1
is
η = Ix κηIy. (10)
Dt Ds 2 − 2
Similarly, we have a syzygy between Iy and Iy and the syzygy is
2 11
η
Iy = 2Iy s Iy κ2η2Iy+2κη Ix+κ ηIx. (11)
Dt 11 Ds 2 − η Ds 2 − 2 Ds 2 s 2
2.2. Invariant calculus of variations
KoganandOlverin[7]studiedinvariantcalculusofvariationsfromageometricpoint
of view, we instead do it from a differential algebra point of view.
AssumeLagrangianstobesmoothfunctionsofx, uandfinitelymanyderivatives
of the uα and denote these as L[u] = L[u]dx. Furthermore, suppose these are
invariant under some group action and let the κ , j =1,...,N, denote the generating
j
R
differential invariants of that group action. Also, assume that the action leaves the
independentvariablesinvariantsothatthe Lagrangianscanberewrittenas L[κ]dx.
ToobtaintheinvariantizedEuler-Lagrangeequationsweproceedinasimilarway
R
as for finding the Euler-Lagrangeequations in the original variables (x,u).
Recall that if x (x,u) extremizes the functional L[u], then for a small
7→
perturbation of u,
d
0= L[u+εv]
dε
(cid:12)ε=0
(cid:12)q
(cid:12) D ∂L
= (cid:12) Eα(L)vα+ vα+ dx
Z α=1" i Dxi (cid:18)∂uαi ···(cid:19)#
X X
Moving frames and conservation laws for Euclidean invariant Lagrangians 7
after differentiation under the integral sign and integration by parts, where
D|K| ∂
Eα = ( 1)|K|
XK − Dxk11···Dxkpp ∂uαK
istheEuleroperatorwithrespecttothedependentvariablesuα. Theboundaryterms
correspond to Noether’s conservation laws and the variation v to the infinitesimals.
ToobtaintheinvariantizedEuler-Lagrangeequations,wefirstintroduceadummy
invariant independent variable t and set the uα = uα(x,t). The introduction of this
nsyezwyginiedsepenκd=ent vIa(uria)btlehartesiuslts in q new invariants Itα = g·uαt|g=ρ(z) and a set of
t t
D H
κ I1
1 t
. .
. = . , (12)
t .   . 
D H
κ Iq
 N   t 
   
where is a N q matrix of operatorsdepending only on the , for i=1,...,p, the
i
H × D
κ ,forj =1,...,N,andtheirinvariantderivatives. Sinceallindependentvariablesare
j
invariant, we have that all differential operators commute, specifically [ , ] = 0,
i t
D D
for all i=1,...,p.
Remark 2.7. Up to this moment we have represented the independent variables as
x and the dependent variables as u. Since the examples in this paper only involve
two independent variables, s and t, we will represent the coordinates of the space
of dependent variables as x, where the dependent variables will be the usual space
coordinates.
For simplicity, consider a one-dimensional Lagrangian L(x,y,y ,y ,...)dx with
x xx
a finite number of arguments which is invariant under the SE(2) group action (2).
Suchvariationalproblemscanbe rewrittenintermsofthegeneratinginvariantsofits
group action, in this case the Euclidean curvature, κ, and its derivatives with respect
to s, the Euclidean arc length. We reparametrize (x,y(x)) as (x(s),y(s)), and to fix
parametrizationas arc length, we introduce η = x2+y2 =1 as a constraint. Thus,
s s
we consider the invariantized variational problem
p
[L(κ,κ ,κ ,...) λ(s)(η 1)]ds, (13)
s ss
− −
Z
where λ(s) is a Lagrange multiplier. This constraint does not reduce the solution set
and it will simplify the calculations. Symbolically, we know that
d D
L[x+εv]= L[x].
dε Dt
(cid:12)ε=0 (cid:12)xt=v
(cid:12) (cid:12)
Hence, after differen(cid:12)tiating (13) under th(cid:12)e integral sign and integrating by parts we
(cid:12) (cid:12)
obtain
[L(κ,κ ,κ ,...) λ(s)(η 1)]ds
t s ss
D − −
Z
∂L ∂L ∂L
= κ+ κ+ 2 κ+ λ(s) η ds
∂κDt ∂κ DsDt ∂κ DsDt ···− Dt
Z (cid:20) s ss (cid:21)
∂L ∂L ∂L ∂L
= + 2 3 + κ λ(s) η
∂κ −Ds ∂κ Ds ∂κ −Ds ∂κ ··· Dt − Dt
Z h(cid:18) (cid:18) s(cid:19) (cid:18) ss(cid:19) (cid:18) sss(cid:19) (cid:19)
Moving frames and conservation laws for Euclidean invariant Lagrangians 8
m−1
∂L
+ ( 1)n n m−1−n κ ds
Ds − Ds ∂κ Ds Dt
(cid:16)mX=1nX=0 (cid:18) m(cid:19) (cid:17)i
m−1
∂L
= Eκ(L) κ λ(s) η + ( 1)n n m−1−n κ ds,
Dt − Dt Ds − Ds ∂κ Ds Dt
Z h (cid:16)mX=1nX=0 (cid:18) m(cid:19) (cid:17)i
where
∂mκ
κ = .
m ∂sm
Next we substitute the circled η and κ by their respective syzygies
t t
D D
κ Ix
η Ds − 2 H1
= = I(x ), (14)
Dt(cid:18) κ (cid:19)  κs Ds2+κ2  I2y   H2  t
    
where we have already set η = 1; it makes no difference to the final result to do this
at this point. Hence,
m−1
∂L
(Eκ(L) λ(s) )I(x )+ ( 1)n n m−1−n κ ds
H2− H1 t Ds − Ds ∂κ Ds Dt
Z h (cid:16)mX=1nX=0 (cid:18) m(cid:19) (cid:17)i
= ( ∗(Eκ(L)) ∗(λ(s)))I(x )+ λ(s)Ix+Eκ(L) Iy Eκ(L)Iy
H2 −H1 t Ds − 2 Ds 2 −Ds 2
Z h (cid:16)
m−1
∂L
+ ( 1)n n m−1−n κ ds, (15)
− Ds ∂κ Ds Dt
mX=1nX=0 (cid:18) m(cid:19) (cid:17)i
after a second set of integration by parts, and where ∗ and ∗ are the respective
H1 H2
adjoint operators of and . The vector I(x ) correspondsto the variation v and
1 2 t
thus the coefficientsHof Ix andHIy representrespectively the Euler-Lagrangeequations
2 2
Ex(L)=κ Eκ(L)+λ , (16)
s s
Ey(L)= 2Eκ(L)+κ2Eκ(L)+λ(s)κ. (17)
Ds
We will use Ex(L) = 0 to eliminate λ(s) instead of Ey(L) = 0, as it contains
derivatives of κ of lower order. Since L does not depend on s explicitly, by the result
in page 220 of [6], κ Eκ(L) is a total derivative, specifically
s
m−1
∂L
κ Eκ(L)= L ( 1)j j κ ,
s Ds − − Ds ∂κ m−j
m=1 j=0 (cid:18) m(cid:19)
X X
 
then we obtain
m−1
∂L
λ(s)= L+ ( 1)j j κ , (18)
− − Ds ∂κ m−j
m=1 j=0 (cid:18) m(cid:19)
X X
where the constant of integration has been absorbed into λ(s) (see Remark 7.1.9.
of [6]). Hence, we are left with one invariantized Euler-Lagrange equation in one
unknown
m−1
∂L
Ey(L)= 2Eκ(L)+κ2Eκ(L) κ L ( 1)m m κ . (19)
Ds −  − − Ds ∂κ m−j
m=1 j=0 (cid:18) m(cid:19)
X X
 
This particular equation agrees with the one appearing in Kogan and Olver [7].
Moving frames and conservation laws for Euclidean invariant Lagrangians 9
2.3. New version of Noether’s conservation laws
If one calculates the conservation laws for one-dimensional variational problems that
are invariant under SE(2) from Noether’s First Theorem (for the formulae of these,
see Theorem4.29 of [9]) and then rewrite these in terms of the Euclideancurvature κ
and its derivatives with respect to the Euclidean arc length s, one obtains
cosθ sinθ 0 λ(s) κEκ(L)
sinθ −cosθ 0 − −Eκ(L) =c, (20)
s
 asinθ bcosθ acosθ+bsinθ 1 (cid:12)(cid:12)g=ρ(z)−1 −DEκ(L) 
− (cid:12)
 R(g) (cid:12)(cid:12)  υ(I) 
w|here ρ(z)−1 is the{rzight moving frame (}6), λ(s) is|the Lagra{nzge multip}lier obtained
in (18) and c the constant vector.
WenotethatR(g)istheAdjointrepresentationofSE(2)onitsLiealgebrase(2).
To see how calculations in Section 2.2 can yield the result in (20), we first show how
the Adjoint representation is calculated in the context we will need.
Consider the SE(2) group action
x cosθ sinθ x a
= − + (21)
y sinθ cosθ y b
(cid:18) (cid:19) (cid:18) (cid:19)(cid:18) (cid:19) (cid:18) (cid:19)
with generating infineitesimal vector fields
e
∂ , ∂ , y∂ +x∂ .
x y x y
−
Let g SE(2) act on
∈
v=α∂ +β∂ +γ( y∂ +x∂ )
x y x y
−
as in (21), where α, β and γ are constants. Hence,
g v=α∂xe+β∂ye+γ( y∂xe+x∂ye)
· −
cosθ sinθ 0 ∂
e e − x
= α β γ sinθ cosθ 0 ∂ .
y
  
asinθ bcosθ acosθ+bsinθ 1 y∂ +x∂
x y
(cid:0) (cid:1) − −
Thus, R(g) is the Adjoint representation of SE(2), denoted as d(g). 
A
Remark 2.8. In Example 2.1 we used the right action of SE(2) on the plane to
calculate the right moving frame, as it simplified its calculation. However, to compute
d(ρ)−1 we considered the left action of SE(2) on the plane, avoiding in this way the
A
need to calculate the inverse of d(ρ).
A
Next recall the boundary terms (15) obtained in the calculation of the
invariantized Euler-Lagrangeequations,
m−1
∂L
λ(s)Ix+Eκ(L) Iy Eκ(L)Iy+ ( 1)n n m−1−n κ=c,
− 2 Ds 2 −Ds 2 − Ds ∂κ Ds Dt
m=1n=0 (cid:18) m(cid:19)
X X
where c is a constant. Substituting Iy and m−1−n κ for all m in the above
Ds 2 Ds Dt
expression by the differential formulae
Iy =Iy κIx,
Ds 2 12− 2
κ =Iy 2κIx ,
Dt 112− 12
κ=Iy 2κ2Iy 3κIx 2κ Ix ,
DsDt 1112− 12− 112− s 12
.
.
.
Moving frames and conservation laws for Euclidean invariant Lagrangians 10
where these were obtained from (8), and rewriting it as
Eκ(L)
λ(s) κEκ(L) −Ds
− − Eκ(L) 2κ2 ∂L + 
2κ∂L 2κ ∂L + − ∂κss ···
(I2xI1x2···)− ∂−κs3κ−∂∂κLs...ss+∂κ·ss·· ··· +(I2yI1y2···) ∂∂∂∂κκLLsss+...+······ =k (22)
   
Cx  
Cy
| {z }
yields the boundary terms in a form that is linear in the Iα .
| 2K {z }
We now let t be a group parameter. If the parameters are (a ,...,a ) and
1 r
t = a , then from Theorem 3 of [2], it is shown that the vectors ( Iα Iα )
j 2 12 ···
can be written as the product of row j of d(ρ)−1 and the matrix of invariantized
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infinitesimals
∂z
Ωα(I)= ζi(I) , ζi = i ,
j j ∂a
j(cid:12)g=e
(cid:16) (cid:17) e (cid:12)
where α represents a depenedent variable, a a gr(cid:12)oup parameter, and e the identity
j (cid:12)
element. Thevectorofinvariantsin(20)equalsthesumoftheproductsofthematrices
of invariantized infinitesimals Ωα(I) with the vectors α,
C
λ(s) κEκ(L)
Ωx(I) x+Ωy(I) y = − −Eκ(L) ,
s
C C  −DEκ(L) 
 
where
x x x y y y y
s ss s ss sss
··· ···
a 1 0 0 a 0 0 0 0
··· ···
Ωx(I)= b 0 0 0 , Ωy(I)= b 1 0 0 0 .
 ···  ···
θ 0 0 κ θ 0 1 0 κ2
− ··· − ···
   
Noether’s conservation laws for one-dimensional Lagrangians invariant under
SE(2) can be written as
x y 0 λ(s) κEκ(L)
s s
y −x 0 − −Eκ(L) =c, (23)
s s s
 xy yx xx +yy 1  −DEκ(L) 
s s s s
−
  
where x2+y2 =1 was used to simplify the conservation laws and
s s
m−1
∂L
λ(s)= L+ ( 1)j j κ .
− − Ds ∂κ m−j
m=1 j=0 (cid:18) m(cid:19)
X X
The following theorem generalizes what we have just seen for one-dimensional
variational problems that are invariant under SE(2); it states that the conservation
lawsfromNoether’s FirstTheoremcanbe writtenasthe divergenceofthe productof
a moving frame with vectors of invariants.
Theorem 2.9. Let L(κ ,κ ,...)dx be invariant under G M M, where M =
1 2
× →
Jn(X U), with generating invariants κ , for j = 1,...,N, and let x = g x = x ,
j i i i
× R ·
for i = 1,...,p. Let (a ,...,a ) be coordinates of G near the identity e, and v , for
1 r i
i = 1,...,r, the associated infinitesimal vector fields. Furthermore, let d(g) be the
e
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