Nöether s Theorem Under the Legendre Transform by Jonathan Herman

Transcription

1 Nöether s Theorem Uner the Legenre Transform by Jonathan Herman A research paper presente to the University of Waterloo in fulfilment of the research paper requirement for the egree of Master of Mathematics in Pure Mathematics Waterloo, Ontario, Canaa, 2014 c Jonathan Herman, 2014

2 AUTHOR S DECLARATION I hereby eclare that I am the sole author of this research paper. This is a true copy of the research paper, incluing any require final revisions, as accepte by my examiners. I unerstan that my research paper may be mae electronically available to the public. Jonathan Herman ii

3 Abstract In this paper we emonstrate how the Legenre transform connects the statements of Nöether s theorem in Hamiltonian an Lagrangian mechanics. We give precise efinitions of symmetries an conserve quantities in both the Hamiltonian an Lagrangian frameworks an iscuss why these notions in the Hamiltonian framework are somewhat less rigi. We explore conitions which, when put on these efinitions, allow the Legenre transform to set up a oneto-one corresponence between them. We also iscuss how to preserve this corresponence when the efinitions of symmetries an conserve quantities are less restrictive. iii

4 Acknowlegements First an foremost, a very sincere thank you goes to my supervisor Dr. Spiro Karigiannis. I am extremely grateful for his patience an the tremenous amount of time he spent teaching an helping me this summer. I woul also like to thank Dr. Shenga Hu, my secon reaer, for the very useful comments an corrections that he provie. I nee to acknowlege two of my goo friens; Janis Lazovksis an Cameron Williams. Janis is a LaTeX machine, an I am very appreciative of the time he spent helping me this summer. Cam was a great support this year, always there to help me work through any problem. I also woul like to thank my family for their constant love an interest, it means very much to me. Last, but not least (I say least goes to Cameron Williams), a thank you goes to my cousin Matt Rappoport, for without our iscussions some of the contents in this paper woul not exist. iv

5 Contents 1 Introuction 1 2 Symplectic Geometry Symplectic Vector Spaces Symplectic Manifols When is a Diffeomorphism a Symplectomorphism? Canonical Symplectic Structure of Cotangent Bunles Lifting a Diffeomorphism Constructing Symplectomorphisms Applications to Geoesic Flow Lagrangian Mechanics Conservative an Central Forces The Calculus of Variations an the Euler-Lagrange Equations Examples of Lagrangian Systems Hamiltonian Mechanics Hamiltonian Vector Fiels Hamilton s Equations The Poisson Bracket The Legenre Transform The Legenre Transform on a Vector Space The Legenre Transform on Manifols The Legenre Transform Relates Lagrangian an Hamiltonian Mechanics Nöether s Theorem Nöether s Theorem in Lagrangian Mechanics Nöether s Theorem in Hamiltonian Mechanics Nöether s Theorem Uner the Legenre Transform The Converse of Nöether s Theorem in the Lagrangian Setting Relaxing the Definitions of Symmetries an Conserve Quantities v

6 1 Introuction This paper stuies the theorem that Emmy Nöether publishe in 1918, which provies a mathematical way to see connections between symmetries an conserve quantities. As we shall see, Nöether s theorem can be state in both the Lagrangian an Hamiltonian frameworks. In Section 6 we emonstrate how the Legenre transform relates these statements an furthermore how, uner specific requirements, it gives a one-to-one corresponence between the respective notions of symmetry an conserve quantity. Section 2 is eicate to introucing the tools neee from symplectic geometry to formulate Hamiltonian mechanics. In Section 1.6 we will see how geoesic flow on a Riemannian manifol arises as a symplectomorphism generate by a specific iffeomorphism. In Section 5.3 we apply the Legenre transform to this setup an recover an equivalent way to efine geoesic flow in the Hamiltonian framework. Section 3 gives an introuction to Lagrangian mechanics. In particular, we erive the Euler- Lagrange equations using tools from the calculus of variations. A Lagrangian is just a smooth function on the tangent bunle an we will see that when this function is a natural Lagrangian, the Euler-Lagrange equations are equivalent to Newton s secon law. We also emonstrate how the Euler-Lagrange equations are a generalization of Newton s secon law; in particular, the Euler- Lagrange equations hol in non-inertial reference frames. We give many examples of Lagrangian systems an then translate these systems to the Hamiltonian framework in Section 5. In Section 4 we use the tools introuce in Section 2 to stuy some basic notions in Hamiltonian mechanics. As mentione above, the main object of stuy in Lagrangian systems is the Lagrangian, which is just a smooth function on the tangent bunle T M. In Hamiltonian mechanics the main object is the Hamiltonian, which is just a smooth function on the cotangent bunle T M. In Section 2.4 we show how the cotangent bunle always has a canonical symplectic structure an so we see that an avantage of Hamiltonian mechanics is that it incorporates the use of tools from symplectic geometry. After introucing Lagrangian an Hamiltonian mechanics, Section 5 emonstrates how the two formulations are equivalent uner the Legenre transform. Given a Lagrangian L C (T M) we get an inuce map calle the Legenre transform, which we enote by Φ L, from T M to T M. Similarily, given a Hamiltonian H C (T M) we get the inuce Legenre transform Φ H : T M T M. Uner certain conitions, which we iscuss, the Legenre transform is a iffeomorphism. We use the Legenre transform to translate examples given in Section 3 an Sections 4 into the opposing frameworks. In particular, we will see how the Legenre transform takes motions in one framework to motions in the other. In Section 6 we stuy Nöether s theorem in both the Lagrangian an Hamiltonian frameworks. We give physical examples in both settings to emonstrate the power of this theorem. We then show 1

7 how the statements of Nöether s theorem can be translate, uner the Legenre transform, from one framework to the other. We give examples of how the Legenre transform takes symmetries to symmetries an conserve quantities to conserve quantities. With the efinitions given, we use the Laplace-Runge-Lenz vector to show how the notions of symmetry an conserve quantity are not in one-to-one corresponence. However, we fix this problem by putting restrictions on the symmetries an conserve quantities. Lastly, we iscuss the problem of how to make the corresponence one-to-one when the efinitions are more general. Throughout this paper we will use the Einstein summation convention. 2

8 2 Symplectic Geometry We iscuss here the concepts in symplectic geometry which will be neee to formulate Hamiltonian mechanics in Section 4. In Riemannian geometry, manifols are equippe with a non-egenerate symmetric quaratic form, whereas in symplectic geometry the non-egenerate quaratic form is require to be skew-symmetric. Although there are some similarities between symplectic an Riemannian geometry, as we shall see there are also some vast ifferences. 2.1 Symplectic Vector Spaces Let V be an m imensional real vector space an Ω : V V R a skew-symmetric bilinear map. Let U = {u V ; Ω(u, v) = 0 for all v V }. Suppose that im U = k an that {u 1,..., u k } is a basis. Recall the stanar form theorem for skew-symmetric bilinear maps: Theorem 2.1. (Stanar Form for Skew-Symmetric Bilinear Maps) With U, V an Ω as above, we can fin n N an a basis u 1,..., u k, e 1,... e n, f 1..., f n of V such that Ω(u i, v) = 0 Ω(e i, e j ) = 0 = Ω(f i, f j ) Ω(e i, f j ) = δ j i for all i an for all v V for all i, j for all i, j Proof. This is a fairly straightforwar inuction proof. See [1], page 3 for etails. It follows that, with respect to this basis, the matrix representation of Ω is I 0 I 0 Definition 2.2. The bilinear map Ω is sai to be symplectic (or non-egenerate) if U = {0}. If this is the case then the pair (V, Ω) is calle a symplectic vector space an {e 1,..., e n, f 1,..., f n } is calle the corresponing symplectic basis. It follows from the theorem that any symplectic vector space is necessarily even imensional an the corresponing skew-symmetric bilinear map is of the form [ ] 0 I I 0 Example 2.3. (Symplectic Vector Space Prototype) The simplest example of a symplectic vector space is (R 2n, Ω 0 ) where Ω 0 is efine such that e 1 = (1, 0,..., 0),..., e n = (0,..., 0, 1, 0..., 0) together with f 1 = (0,..., 0, 1, 0..., 0),..., f n = (0,..., 0, 1) form a symplectic basis. The reason this symplectic vector space is referre to as a prototype is given by the Darboux theorem, which is state in section

9 Definition 2.4. Let (V 1, Ω 1 ) an (V 2, Ω 2 ) be symplectic vector spaces. A linear isomorphism ϕ : V 1 V 2 is calle a symplectomorphism if ϕ Ω 2 = Ω 1. Here ϕ is the pullback of ϕ meaning that (ϕ Ω 2 )(u, v) = Ω 2 (ϕ(u), ϕ(v)). In the same way that a Riemannian metric inuces the musical isomorphism between T M an T M, where M is some Riemannian manifol, so the skew-symmetric bilinear form Ω inuces a natural isomorphism between V an V. Proposition 2.5. Given a symplectic vector space (V, Ω), the non-egenerate bilinear form Ω inuces an isomorphism between V an V through the map V V v Ω(v, ) Proof. The non-egeneracy of Ω shows this map is injective, while we know that im V = im V. Hence this is inee an isomorphism. Definition 2.6. Let (V, Ω) be a finite imensional symplectic vector space an Y V a subspace. The symplectic complement of Y is efine to be the subspace Y Ω := {v V ; Ω(v, u) = 0 for all u Y }. For a subspace Y V, consier the map Φ : V Y v Ω(v, ) Y It s clear that ker Φ = Y Ω. The surjectivity of Φ follows by combining Proposition 2.5 together with the fact that any element of α Y can be extene to an element of α V such that α Y = α. It follows, by the first isomorphism theorem, V/Y Ω = Y. Since im Y = im Y, we have that im V = im Y + im Y Ω. Moreover, by efinition, Ω Y Y is non-egenerate if an only if Y Y Ω = 0. That is Ω Y Y is non-egenerate if an only if V = Y Y Ω. This leas to the following efinition. Definition 2.7. If Ω Y Y 0 then Y is calle an isotropic subspace of V. If Y is isotropic an im Y = 1 2 im V then Y is calle a Lagrangian subspace of V. The above remarks give us Proposition 2.8. A subspace Y V is Lagrangian if an only if Y = Y Ω. 2.2 Symplectic Manifols Let M be a manifol an let ω Ω 2 (M) be a 2-form. By efinition, for each p M we have that ω(p) := ω p is a skew-symmetric bilinear map ω p : T p M T p M R. 4

10 Definition 2.9. A 2-form ω Ω 2 (M) is sai to be symplectic if it is close an if ω p is symplectic (non-egenerate) for each p M. In such a case, the pair (M, ω) is calle a symplectic manifol. By the stanar form theorem, a symplectic manifol is necessarily even imensional. Definition Given a 1-form µ T M, the unique vector fiel V µ in T M such that ω(v µ, ) = µ is calle the symplectic ual of µ. That is, for each p M we set V µ (p) to be the pre-image of µ(p) uner the map efine in Proposition 2.5. In other wors, V µ is the unique vector fiel satisfying V µ ω = µ Example (Prototype of a Symplectic Manifol) Let M = R 2n with stanar coorinates x 1,..{., x n, y 1,..., y n. The form ω 0 := x i y i is symplectic, an T p M } = R 2 has symplectic basis p,..., p p p x 1 x, n y 1,..., y n so that (R 2n, ω 0 ) is a symplectic manifol. The Darboux theorem shows why the manifol above can be thought of as the prototype of symplectic manifols. The theorem locally classifies symplectic manifols up to symplecteomorphism. That is, locally every symplectic manifol is symplectomorphic to (R 2n, ω 0 ). Theorem (Darboux) Let (M, ω) be a symplectic manifol. For any p M there exists a coorinate chart (U, x 1,... x n, y 1,..., y n ) centre at p such that ω U = n x i y i i=1 The coorinates giving this local expression of ω are calle Darboux coorinates. Proof. The proof is just an application of the Frobenius theorem together with a characterization of Darboux coorinates. See [4], page 349 for the etails. In the same way we efine Lagrangian subspaces of a vector space, we can efine Lagrangian submanifols. Definition Given a symplectic manifol (M, ω), a submanifol (N, ι) of M is calle a Lagrangian submanifol if at each p N, T p N is a Lagrangian subspace of T p M. That is, N is Lagrangian if an only if ι ω = ω TpN TpN = 0 an im N = 1 2 im M. We finish this subsection with a simple yet important proposition. Proposition Let (M 1, ω 1 ) an (M 2, ω 2 ) be symplectic manifols. If (L, ι) is a Lagrangian submanifol of (M 1, ω 1 ) an f : (M, ω 1 ) (M 2, ω 2 ) is a symplectomorphism, then (f(l), f ι) is a Lagrangian submanifol of (M 2, ω 2 ). 5

11 Proof. By efinition, we have that f ω 2 = ω 1. Hence (f ι) ω 2 = ι f ω 2 = ι ω 1 = 0 since (L, ι) is a Lagrangian submanifol of (X 1, ω 1 ). Using the results from this section we can answer the question of when a iffeomorphism between two symplectic manifols is a symplectomorphism. 2.3 When is a Diffeomorphism a Symplectomorphism? Let ϕ : (M 1, ω 1 ) (M 2, ω 2 ) be a iffeomorphism of two symplectic manifols. We will see that the answer to the pose question of this subsection is if an only the graph of ϕ is a Lagrangian submanifol of the twiste symplectic manifol (M 1 M 2, ω). We first formalize the efinitions in this statement. Given the two symplectic manifols (M 1, ω 1 ) an (M 2, ω 2 ) as above, consier their Cartesian prouct M 1 M 2. Let π 1 an π 2 enote the projection maps onto the first an secon factors respectively. For any a, b R\{0}, consier the 2-form ω := a(π 1ω 1 ) + b(π 2ω 2 ) Since the exterior erivative commutes with the pull-back, it follows ω is close. Moreover, to see that ω is symplectic, let (p, q) M 1 M 2 be arbitrary an consier non-zero (V p, W q ) T p M 1 T q M 2. Without loss of generality, suppose that V p is nonzero. By the non-egeneracy of ω 1 there exists X p T p M such that (ω 1 ) p (V p, X p ) 0 so that ω((p, V p ), (q, 0 q )) = a ω 1,p (V p, X p ) 0. Definition In particular, taking a = 1 an b = 1 we obtain the twiste prouct symplectic form ω Ω 2 (M 1 M 2 ): ω = π 1ω 1 π 2ω 2 Let Γ ϕ = {(p, ϕ(p)); p M} enote the graph of ϕ. It s clear the the function f : M 1 Γ ϕ p (p, ϕ(p)) is an embeing. Since Γ ϕ is the image of M 1 uner f, it follows that Γ ϕ is a submanifol of M 1 M 2 of imension 4n 2n = 2n. Hence Γ ϕ always satisfies half of the requirements of being Lagrangian. We can now prove the statement pose at the beginning of this section. Proposition The iffeomorphism ϕ is a symplectomorphism Γ ϕ is a Lagrangian submanifol of (M 1 M 2, ω). 6

12 Proof. We alreay know that (Γ ϕ, ι) is a submanifol of (M 1 M 2, ω), where ι : Γ ϕ : M 1 M 2 is the inclusion map. Let f be as above. We have that Γ ϕ is Lagrangian ι ω = 0 f ι ω = 0 since f is a iffeomorphism (ι f) ω = 0 But (ι f) ω := ((ι f) π 1)ω 1 ((ι f) π 2)ω 2 = (π 1 ι f) ω 1 (π 2 ι f) ω 2 = ω 1 ϕ ω 2 Hence Γ ϕ is Lagrangian ϕ ω 2 = ω 1 Remark It is crucial in the above proof that the 2-form on M 1 M 2 is the twiste prouct form, otherwise this woul not work. 2.4 Canonical Symplectic Structure of Cotangent Bunles Given an arbitrary manifol M, the total space of the cotangent bunle T M can always be turne into a symplectic manifol. This subsection escribes how. Let M be an arbitrary n-imensional manifol an T M the cotangent bunle. To turn T M into a symplectic manifol we nee to fin a close symplectic 2-form ω Ω 2 (T M). Consier first the 1-form α Ω 1 (T M) efine by ( α (p,ξp)(v (p,ξp)) := ξ p π (V ) (p,ξp)) where (p, ξ p ) T p M an V (p,ξp) T (p,ξp)(t M) are arbitrary an π is the ifferential of the projection map π : T M M. Define ω := α. By efinition, both α an ω are global forms on T M. After the computation of α an ω in local coorinates, shown below, it is straightforwar to verify that ω is symplectic. It is clear that ω is close, since it is exact. Hence (T M, ω) is a symplectic manifol. Definition The 1-form α Ω 1 (T M) is calle the tautological 1-form an the 2-form ω Ω 2 (T M) is calle the canonical symplectic 2-form. For future use we compute here α an ω in local coorinates. Let (U, x 1,..., x n ) be an arbitrary coorinate chart in M an (T U, x 1,..., x n, ξ 1,..., ξ n ) the inuce chart on T M. The first thing to show is how π : T (T M) T M works. For arbitrary (p, σ p ) Tp M an W (p,σp) = W i (p,σp) x + i 7

13 W i ξ i (p,σp) T (p,σ p)(t M) there exists a i R such that π (W (p,σp)) = a i x i p. It follows a i = (π (W (p,σp)))(x i p ) = W (p,σp)(x i π) = W j x j (x i π) + W j ξ j = W i (p,σp) (p,σp) (x i π) That is, ) ( ) π W(p,σp) = (p, W i x i. p Since α is an element of Γ(T (T M)) we have functions a 1,..., a n, b 1,..., b n C (π 1 (U)) such that α = a i x i + b i ξ i. By efinition, for arbitrary (p, σ p ) Tp M ( a i (p, σ p ) = α (p,σp) x i (p,σp) ) ( = σ p (π x i (p,σp) )) ( ) = σ p x i = σ i (p) = ξ i (p, σ p ) p an ( b i (p, σ p ) = α (p,σp) ξ i (p,σp) ) ( = σ p (π ξ i (p,σp) )) = σ p (p, 0 p ) = 0 = 0(p, 0 p ). We have shown that in local coorinates α = ξ i x i an it follows ω := α = x i ξ i Given another 1-form µ Ω 1 (M) we now show that the graph of µ, consiere as a function M T M, is a Lagrangian submanifol of T M if an only if µ is close. To avoi confusion, let s µ enote the map s µ : M T M given by p (p, µ p ) an let Γ sµ enote the image of µ in T M s µ (M) := Γ sµ = {(p, µ p ) ; p M} That is, the image of µ as a map is the same thing as the graph of s µ. Let π : T M M enote the projection mapping. It s clear that π s µ = i. Proposition Let α be the tautological 1-form on T M. Then s µα = µ. 8

14 Proof. Fix arbitrary (p, µ p ) Γ sµ. By efinition, α (p,µp)(v ) = µ p (π V ). Hence for arbitrary V T p M, s µα(v ) = α((s µ ) V ) = µ p (π (s µ ) V ) = µ p ((π s µ ) V ) = µ p (i(v )) = µ p (V ). Using this we get Proposition Γ sµ is a Lagrangian submanifol of T M µ is close. Proof. Let τ : M Γ sµ is a iffeomorphism an s µ = ι τ. Hence be the same map as s µ but with range restricte to Γ sµ. It follows that τ Γ sµ is Lagrangian ι ω 0 ι α 0 τ ι α 0 because τ is a iffeomorphism (ι τ) α 0 (s µ ) α 0 (s µ ) α 0 µ 0 µ is close 2.5 Lifting a Diffeomorphism Definition Given a iffeomorphism f : M 1 M 2 between two manifols M 1 an M 2, there is an inuce symplectomorphism f : T M 1 T M 2 calle the lift of f which is constructe as follows. Since f is a iffeomorphism we have that f : T M T M is an isomorphism. For arbitrary (p 1, ξ p1 ) Tp 1 M 1 we efine f by f (p 1, ξ p1 ) := (f(p 1 ), (f ) 1 (ξ p1 )). Since f is a iffeomorphism we get that both f an f 1 are bijective an smooth. Moreover, we 9

15 have the following commutative iagram. T M 1 f T M 2 π 1 π 2 M 1 f M 2 ( ) Proposition Let α 1 an α 2 enote the tautological forms on T M 1 an T M 2 respectively. Then f (α 2) = α 1. Proof. Let (p 1, ξ p1 ) T p 1 M 1 an (p 2, ξ p2 ) T p 2 M 2 be such that p 2 = f(p 1 ) an ξ p1 = f ξ p2 It nees to be shown that (f ) (α 2 ) (p2,ξ p2 ) = (α 1 ) (p1,ξ p1 ) By efinition, (f ) (α 2 ) (p2,ξ p2 ) T (p 1,ξ p1 ) (T M 1 ) so let η T (p1,ξ p1 )(T M 1 ) be arbitrary. Then f ( ) ( ) (α 2 ) (p2,ξ p2 ) (η) := (α 2 ) (p2,ξ p2 ) ((f ) )η ) := ξ p2 (π 2 ) (f η = ξ p2 ((π 2 f ) η) = ξ p2 ((f π 1 ) η) by = ξ p2 (f (π 1 η)) = f ξ p2 (π 1 η) = ξ p1 (π 1 η) = (α 1 ) (p1,ξ p1 )η Corollary In the setup of Proposition 2.22, if we take M 1 = M 2 = M an let f : M M be a iffeomorphism, then the lift of f preserves ω. That is, f (ω) = ω. Proof. This follows immeiately from the fact that the pull back commutes with the exterior erivative. The following Lemma an Theorem will be neee in section 6 to stuy Nöether s theorem. Lemma Let M be a manifol. Fix X Γ(T M) an let θ t enote its flow. There exists a unique vector fiel X on the cotangent bunle (i.e. X Γ(T (T M))) such that the flow of X, say Ψ t, is the lift of θ t. That is, Ψ t = θ t,. Note that by Corollary 2.23, each Ψ t is a symplectomorphism. Proof. Let θ t enote the flow of X Γ(T M). We have that θ t is a iffeomorphism θ t : M M so its lift θ t, is a symplectomorphism θ t, : T M T M. Proposition 2.22 shows that θ t, preserves 10

16 α. Just let X be the infinitesimal generator of θ t,. Here the integral curves are of the form θ (p,ξ) : R T M, t θ t, (p, ξ), an so θ t, is a local flow of X. Theorem (Lifting to the Cotangent Bunle) Let M be a manifol. Let α Γ(T (T M)) enote the tautological 1-form on T M an consier the symplectic manifol (T M, ω = α). If g : T M T M is a symplectomorphism preserving α (i.e. g α = α) then there exists a iffeomorphism f : M M such that g = f. Proof. The proof of this theorem is one by combining the following claims. For what is below we let V enote the symplectic ual of α. That is, ω(v, ) = V ω = α. Claim If g α = α then g commutes with the flow of V, or equivalently g V = V. Proof. Let θ t enote the flow of V. It nees to be shown that g θ t = θ t g, or equivalently, that g θ t g 1 = θ t. By efinition, for each p M, we have that θ (p) is the unique curve satisfying θ (p) (0) = p an t t=0 θ (p) = V p. Since θ 0 is the ientity, we have that g θ 0 g 1 (p) = p. Hence, by uniqueness, it suffices to show that t t=0 g θ t g 1 (p) = V p which happens, by the non-egeneracy of ω, if an only if ( ) t g θ t g 1 (p) t=0 ω p = V p ω p = α p (2.1) By the chain rule t g θ t g 1 (p) = g,g 1 (p) t=0 = g,g 1 (p)(v g 1 (p)) ( ) t (θ t(g 1 (p))) Fix an arbitrary Y p Γ(T p M) an plug it into both sies of (2.1). The right han sie is V p ω p (Y p ) = α p (Y p ) while the left han sie becomes ( ) t g θ t g 1 ( ) (p) ω p (Y p ) = ω p g,g 1 (p)(v g 1 (p)), Y p t=0 ( = ω p g,g 1 (p)(v g 1 (p)), g,g 1 (p) g,p(y 1 p ) ) = (g ( ω) g 1 (p) Vg 1 (p), g,p(y 1 p ) ) ( = ω g 1 (p) Vg 1 (p), g,p(y 1 p ) ) = α g 1 (p)(g 1,p(Y p )) = (g α) p ((g,p ) 1 (Y p )) = α p (Y p ) 11

17 Notice that this claim ha nothing to o with the fact the symplectic manifol was a cotangent bunle. This result hols for any symplectic manifol (X, ω) for which ω = α for some 1-form α an g α = α. Claim The integral curves, γ : R T M, of V are of the form where (p, σ) T M is arbitrary. γ (p,σ) (t) = (p, σe t ) Proof. In local coorinates we know that α = ξ i x i an ω = x i ξ i. Let V = a i + b x i i ξ i a i, b i C (T U). By efinition where α = V ω = a i ξ i b i x i an so a i = 0 an b i = ξ i. Let (p, σ) T M be arbitrary an suppose γ : R T M is an integral curve of V starting at (p, σ). We can write γ(t) = (q(t), r(t)) an it follows γ (t) = V γ(t) = (q i (t)) x i + (r i (t)) γ(t) ξ i It follows that for all t R, (q i (t)) = 0 an (r i (t)) = ξ i (γ(t)) = r i (t). That is, q i (t) is a constant function while r i (t) = r i (0)e t. By assumption γ(0) = (p, σ) an so it follows γ(t) = (p, σe t ) γ(t) It immeiately follows that θ t is fibre preserving. That is, θ t (T x M) = T x M. Also, it implies that if g(p, ξ) = (q, η) then for all λ > 0, g(p, λξ) = g(q, λη). This is because the flow of V is complete an e t is surjective onto (0, ). Also, by the continuity of g an θ t we have that g(p, 0 p ) = g(p, lim t θ t (ξ)) = lim t g(θ t (p, ξ)) = lim t θ t (g(p, ξ)) = g(q, 0 q ) Hence g(p, ξ) = (q, η) = g(p, λξ) = (q, λη) for all λ 0 Supposing that g(p, ξ) = (q, η) consier another arbitrary element (p, σ p ) of T p M. Suppose that g(p, σ p ) = ( q, µ q ). Then by applying the above it follows g(p, λσ p ) = ( q, λµ p ) for λ = 0. That is, g(p, 0) = ( q, 0). But g(p, 0) = (q, 0) an so it must be that q = q. Hence g maps fibres to fibres. We are now reay to construct f : M M such that f = g. Inee, efine f by f : M M p π g(p, 0 p ). 12

18 Claim 2.27 shows that f is well efine, while it reaily follows that f π = π g. (2.2) To prove f = g, we will show H := g f 1 is the ientity map. By efinition, H is a map from T M to T M. Let (p, σ p ) T M be arbitrary an suppose that H(p, σ p ) = (q, η q ). Let V (p,σp) Γ(T (p,σp)(t M)) be arbitrary. By hypothesis, g preserves α while Proposition 2.22 shows that f also preserves α. Hence H preserves α. That is, (H α (q,ηq))(v (p,σp)) = (H α) (p,σp)(v (p,σp)) = α (p,σp)(v (p,σp)) := σ p (π (V (p,σp))) On the other han, (H α (q,ηq))(v (p,σp)) = α (q,ηq)(h (V (p,σp))) = η q (π (H (V (p,σp)))) = η q ((π g f 1 ) (V (p,σp))) = η q ((f π f 1 ) (V (p,σp))) by (2.2) = η q (π (V (p,σp))) by But if σ p (π (V (p,σp))) = η q (π (V (p,σp))) for all V (p,σp) T (p,σp)(t M) then it must be that (p, σ p ) = (q, η q ) = H(p, σ p ). That is, H is the ientity map. 2.6 Constructing Symplectomorphisms Given two manifols M 1 an M 2, we emonstrate in Section 2.4 that their cotangent bunles have a canonical symplectic structure. Let α 1 an α 2 enote the tautological 1-forms on T M 1 an T M 2 respectively. Let ω 1 = α an ω 2 = α 2 enote the canonical 2-forms on T M 1 an T M 2 respectively. A straightforwar calculation shows that the tautological 1-form on T M 1 T M 2 is α = π 1α 1 + π 2α 2. implying that the canonical symplectic form on (T M 1 T M 2 ) = T (M 1 M 2 ) is ω = π 1ω 1 + π 2ω 2. By Proposition 2.16 we know that if the graph of a iffeomorphism ϕ : (T M 1, ω 1 ) (T M 2, ω 2 ) is a Lagrangian submanifol of the twiste prouct (T M 1 T M 2, ω), then ϕ is a symplectomorphism. While by Proposition 2.20 we know that the graph of a 1-form µ Ω 1 (M 1 M 2 ) is a Lagrangian submanifol of T (M 1 M 2 ) if the 1-form is close. Hence, in particular, we have that 13

19 Γ f is a Lagrangian submanifol of (T M 1, ω 1 ) (T M 2, ω 2 ) = (T (M 1 M 2 ), ω) for any smooth function f C (M 1 M 2 ). Also, by Proposition 2.14 we know that symplectomorphisms take Lagrangian submanifols to Lagrangian submanifols. With this in min, we first fin a symplectomorphism from (T (M 1 M 2 ), ω) to (T (M 1 M 2 ), ω) so that we can fin Lagrangian submanifols of the later. After oing this, we construct a symplectomorphism between T M 1 an T M 2 by fining a iffeomorphism τ : T M 1 T M 2 whose graph, which is a subset of T (M 1 M 2 ), equals the graph of f. We will see that the existence of such a iffeomorphism is governe by the implicit function theorem. Consier the functions σ 2 : T M 2 T M 2, (p, ξ) (p, ξ) an σ := i σ 2 : T M 1 T M 2 T M 1 T M 2 Proposition The map σ : (T M 1 T M 2, ω) (T M 1 T M 2, ω) is a symplectomorphism. That is, σ ω = ω Proof. First note that σ is involutive an so bijective. Moreover in local coorinates x 1,..., x n, ξ 1,..., ξ n on T M 2 we have σ 2α 2 = σ 2(ξ i x i ) = (ξ i σ 2 )((x i σ 2 )) = ξ i x i = α 2 an so σ ω = σ (π 1ω 1 ) + σ (π 2ω 2 ) = (π 1 σ) ω 1 + (π 2 σ) ω 2 = π 1ω 1 π 2ω 2 = ω Definition If Y is a Lagrangian submanifol of (T M 1 T M 2, ω) we efine the twist of Y, enote Y σ to be the image of Y uner σ. That is, Y σ := σ(y ). Proposition If Y is a Lagrangian submanifol of (T M 1 T M 2, ω) then the twist of Y is a Lagrangian submanifol of (T M 1 T M 2, ω) Proof. Since σ : (T M 1 T M 2, ω) (T M 1 T M 2, ω) is a symplectomorphism, this result is a corollary of Proposition As mentione at the beginning of this section, we now want to fin a iffeomorphism whose graph equals the graph of the close 1-form f, where f C (M 1 M 2 ). We call the graph of f the Lagrangian submanifol generate by f. Before stating this formally, we introuce some notation so that we can write this submanifol in a way that will allow us to fin conitions on when it is the graph of a iffeomorphism ϕ. By efinition, we have that (f) (x,y) = π1 (( 1f) x ) + π2 (( 2f) y ) 14

20 where x 1,..., x n an y 1,..., y n are local coorinates on M 1 an M 2 respectively, π 1 an π 2 are the natural projections on T M 1 T M 2 an 1 f = f x i an x i 2 f = f y i. y i Definition The Lagrangian submanifol generate by f is the Lagrangian submanifol of M 1 M 2 efine by Y f := { ((x, y), (f) (x,y) ) ; (x, y) M 1 M 2 } = {((x, y), ((1 f) x, ( 2 f) y )) ; (x, y) M 1 M 2 } Definition If there exists a iffeomorphism ϕ : T M 1 T M 2 such that Yf σ = Γ ϕ then, by Proposition 2.16, ϕ is a symplectomorphism. If such a symplectomorphism exists we call it the symplectomorphism generate by f. Recall that here Y σ f is the twist of Y f. We are trying to fin a iffeomorphism ϕ : T M 1 T M 2 such that Γ ϕ = Yf σ. But notice that Γ ϕ = Y σ f {((x, ξ), (y, η)) : ϕ(x, ξ) = (y, η)} = {((x, ( 1 f) x ), (y, ( 2 f) y ))} ξ = ( 1 f) x an η = ( 2 f) y ξ i = f x i (x, y) ( ) an η i = f y i (x, y) ( ) That is, writing ϕ(x, ξ) = (ϕ 1 (x, ξ), ϕ 2 (x, ξ)), for Γ ϕ to equal Yf σ all we nee is that ϕ 2(x, ξ) = ( 2 f) y, for then it automatically follows ϕ 1 (x, ξ) = y. Given any (x, ξ) the implicit function theorem says (locally) when a solution to ( ) exists. That is, it tells us when one can write y as a function of both x an ξ. If (x 1,..., x n, y 1,..., y n ) are local coorinates on M 1 M 2 the implicit function theorem says that we can write y as a function of x an ξ locally if an only if [ ( )] f et y j x i 0. Also, note that if we have a solution to ( ) say, y = ϕ 1 (x, ξ) then we can plug this solution into ( ) an thus completely etermine the map ϕ satisfying Γ ϕ = Yf σ. We give some examples of this process in the next section. 2.7 Applications to Geoesic Flow Recall the efinition of geoesic flow. Definition Let (M, g) be a Riemannian manifol. The geoesic flow of M is the local R-action on T M efine by Θ : R T M T M, (t 0, V p ) t γ Vp (t) t=t0 where γ Vp is the unique geoesic starting at p M with initial velocity V p. Recall that a Riemannian manifol is calle geoesically complete if every geoesic is efine for all t R an is calle geoesically convex if for any two points in the manifol there exists 15

21 a minimizing geoesic connecting them. For the rest of this section, unless state otherwise, all manifols are assume to be compact. Since compact metric spaces are complete, it follows from the Hopf-Rinow theorem that all of our Riemannian manifols are geoesically complete an geoesically convex. Example (Free Translational Motion) Let M 1 = R n = M 2 with coorinate charts (R n, x 1,..., x n ) an (R n, y 1,..., y n ) respectively. Enow M 1 an M 2 with the stanar metric. We have the respective inuce coorinate charts (T R n, x 1,..., x n, ξ 1,..., ξ n ) an (T R n, y 1,..., y n, η 1,..., η n ). Let f C (M 1 M 2 ) be given by f(x, y) = 1 2 (x, y)2 = 1 2 n (x i y i ) 2. i=1 Since the metric is assume to be the stanar one it follows that (x, y) is the usual Eucliean istance. By efinition, {( Yf σ = a, b, f x i (a, b)xi, f ) (a, b)yi yi a, b R n }. We woul like to fin the symplectomorphism genereate by f. That is, we woul like to fin a map ϕ : R n R n such that Yf σ equals Γ ϕ. In this case ( ) is ξ i = f (a, b) = b i a i an ( ) is x i η i = f = b i a i. Since f = y i x i we have that y i x i [ ( )] f y j x i = δ ij ij so that the implicit function theorem guarantees a solution to ( ). We have shown that Y σ f = {(a, b, b a, b a) a, b Rn }. For fixe a R n, the implicit function theorem has shown the existence of a function ϕ 1 : R n R n where for every b R n there exists ξ R n such that b = ϕ 1 (a, ξ). In this case, it is obvious that every element ξ T a R n = R n is of the form b a for some b R n. We set an ϕ 1 (a, ξ) = ϕ 1 (a, b a) = b η = ϕ 2 (a, ξ) := f (a, b) = ξ yi so that ϕ(a, ξ) = (b, η) = (ξ + a, ξ) an Γ ϕ = Yf σ. That is, ϕ is the symplectomorphism generate by f. Ientifying Tp R n with T p R n = {vectors emanating from p} we see that ϕ is free translational motion. The above example is a special case of the following. 16

22 Example (Geoesic Flow) Let (M, g) be an arbitrary compact Riemannian manifol. As in the previous example, let : M M R enote the Riemann istance function an f C (M M) be given by f = 1 2 (a, b)2. To fin the sympelctomorphism generate by f we nee to solve ( ) ξ i = a f an ( ) η i = b f. Using the musical isomorphism we can ientify T M with T M. That is, for a M, we have : T a M Ta M V V := g(v, ). Let V an W in T M be the unique vector fiels such that V = ξ an W = η. Then the above equations become ( ) g(v, ) = a f( ) an ( ) g(w, ) = b f( ). We now show that uner this ientification the symplectomorphism generate by f is the geoesic flow. First we nee the following Lemma. Lemma Let U, V T a M. Then t 1 t=0 2 ((exp) a(u), (exp) a (tv )) 2 =< V, U >. Proof. First notice that for s R small enough so that (exp) a (su) is containe in a geoesic ball centre at (exp) a (U), we have that ((exp) a (U), (exp) a (su)) = 1 s U. Inee, let γ : R M be the unique geoesic starting at a with initial velocity U. Since geoesics have constant spee it follows that the length of γ over [s, 1], enote L ( ) γ [s,1], is L ( ) 1 γ [s,1] = γ (t) t s = U = U 1 s Since the raial geoesic from (exp) a (U) to (exp) a (su) is the unique minimizing curve from (exp) a (U) to (exp) a (su) (for a proof of this see [10], Proposition 6.10) it follows that ((exp) a (U), (exp) a (su)) = 1 s U. Next, fix (exp) a (U) M an consier the function, also enote f, efine by 1 s t f : M R b f((exp) a (U), b). Notice that given V T a M with V U the Gauss Lemma shows that there exists a curve 17

23 β : R M starting at a with initial velocity V such that (β(t), b) is constant. That is, f(β(t)) = c implying that f (β (0)) = 0. Observe that we can write V = V + λu for some λ R. Letting G enote the function G : R M given by t (exp) a (tv ) it follows that t 1 t=0 2 ((exp) a(u), (exp) a (tv )) 2 = t f(g(t)) t=0 = (f ) G(0) G (0) = (f ) a (V ) = λf (U) = λ s 1 s=0 2 (((exp) a(su), (exp) a U)) 2 = λ s 1 s=0 2 1 s 2 U 2 = λ U 2 =< V, U > With this lemma we now reconsier ( ). Evaluating the left han sie at V we get V 2. By geoesic convexity, there exists some U T M such that (exp) a (U) = b. Using Lemma 2.36, the right han sie is a f(v ) = (f ) a (V ) = t ((exp) a (tv ), y) t=0 = t ((exp) a (tv ), (exp) a (U)) t=0 =< V, U > Now take any vector V T M such that V V. Plugging V into ( ) the left han sie becomes 0 an the right han sie is < V, U >. Hence, < V, U >=< V, V > an < V, U >= 0 =< V, V > for any V V. It follows that U = V an so b = (exp) a (V ). We now nee to solve ( ). We will see the solution is given by W = γ V (1) = t t=1 (exp) a (tv ). Inee, let W = t t=1 (exp) a (tv ) an fix any W W. Again, by the Gauss Lemma, we have that b f(w ) = 0 an so W = k W for some k R. But since geoesics have constant spee it follows that V 2 = W 2. Therefore the left han sie of ( ) is < W, W > = k < W, W > 18

24 = k W 2 = k V 2 while the right han sie is ( b f)( W ) = (f ) b ( W ) = 1 s s=0 2 (a, (exp) a((1 + s)(v ))) 2 = 1 s s=0 2 ((exp) a(0v ), (exp) a ((1 + s)v )) 2 = 1 s s=0 2 (1 + s2 ) V 2 = V 2 Hence k = 1 showing that W = t t=1 (exp) a (V ). In summary, for Γ ϕ to equal Yf σ it nees to be that V is the unique vector fiel such that b = (exp) a (V ) (so that b is a function of a an V ) an further that W = t t=1 (exp) a (tv ). That is, the map ϕ is given by (a, V ) ( (exp) a (V ), ) t (exp) a (tv ) = ( γ V (1), γ V (1) ) t=1 In the section on Hamiltonian mechanics, we will return to the concept of geoesic flow an give some insight as to why it arose as the symplectomorphism generate by the Riemann istance function. 19

25 3 Lagrangian Mechanics Recall that Newton s secon law states that in an inertial reference frame the motion of a particle, with position γ(t), is given by the solution to the ODE F (γ(t)) = p (t) where F is the net force acting on the particle an p(t) = m(t)v(t) is the particle s momentum. Lagrangian mechanics is a reformulation of Newtonian mechanics in which motions are given by solutions to the Euler-Lagrange equations. In any situation where Newton s secon law can be applie, so can the Euler-Lagrange equations. We will see that, in any such system, the Euler- Lagrange equations are equivalent to Newton s secon law. However, the equations also hol in settings in which Newton s secon law oes not hol. For example, Newton s secon law only hols in an inertial reference frame, while the Euler-Lagrange equations are vali in any coorinate system. Another avantage in using the Euler-Lagrange equations comes with the way in which they allow constraint forces on a mechanical system to be ignore. For example, if stuying the motion of a bea on a wire, in the Lagrangian setting we o not nee to worry about the forces keeping the bea constraine to the wire. For the rest of this paper all forces are assume to be conservative, an so we first recall this efinition. 3.1 Conservative an Central Forces Consier R n equippe with stanar coorinates x 1,..., x n. Let g be a metric on R n an let x 1,..., x n, v 1..., v n enote the inuce coorinates on T R n = R 2n. Recall the kinetic energy is efine by K : R 2n R (x, v) 1 2 mg(v, v) = 1 2 mg ijv i v j = 1 2 m v 2 where m is some positive constant, calle the mass of the particle. Fix two points r 1, r 2 R n an consier a curve γ : [a, b] R n such that γ(a) = r 1 an γ(b) = r 2. Then the work one by a force F on a particle moving along γ is efine to be the integral of F over the curve γ W (r 1 r 2, γ) := γ F s = b a F (γ(t)) γ (t)t Theorem 3.1. (Work-Kinetic Energy Theorem) Given a system of k particles of masses m i with position r i, the change in kinetic energy of the system is equal to the sum of the work one on each particle. Proof. Let F i enote the force acting on the i-th particle an W i the work one by F i on the i-th 20

26 particle. By Newton s secon law we have that t K = r(t) n i=1 (m i r i (t)) r i(t) = n F i (r i (t)) r (t). i=1 It follows that K(t 2 ) K(t 1 ) = t2 t 1 K t t = n i=1 t2 t 1 F i r it = n i=1 W i As the next example shows, given two paths α, β : [a, b] R n with α(a) = r 1 = β(a) an α(b) = r 2 = β(b), it may be that W (r 1 r 2, α) W (r 1 r 2, β). Example 3.2. (A Non-Conservative Force) Consier a particle moving aroun the unit circle uner the following force fiel. This vector fiel (force) has integral curve γ(t) = (cos t, sin t). It follows that F (x, y) = ( y, x). Fix the points (1, 0) an ( 1, 0). Consier the curve α : [0, π] R 2 given by t (cos t, sin t) an β : [0, π] R 2 given by t (cos t, sin t). Then α an β are both curves starting at (1, 0) an ening at ( 1, 0); however, W ((1, 0) ( 1, 0), α) = while W ((1, 0) ( 1, 0), β) = α β F s = F s = π 0 π 0 ( sin t, cos t) ( sin t, cos t)t = π (sin t, cos t) ( sin t, cos t) = π Contrary to this example, for certain forces (such as the gravitational an electrostatic forces) it is the case that W oes not epen on the path traverse by the particle. This leas to the following efinition. Definition 3.3. A force is calle conservative if the work one is path inepenent. Letting r 1, r 2 R n be arbitrary, this means that for any two curves α, β : [a, b] R n with α(a) = r 1 = β(a) 21

27 an α(b) = r 2 = β(b), W (r 1 r 2, α) = α F s = β F s = W (r 1 r 2, β) In this case the work one by F is justifiably enote W (r 1 r 2 ). A mechanical system is calle conservative if the net force is conservative. Recall that the graient of a function U : R n R n is efine to be U := (U) = ( ) ji U g x i x j where : T R n T R n is the musical isomorphism an g ji = [g 1 ] ji. With the stanar metric, the above efinition reuces to the stanar notion of the graient. Theorem 3.4. A force F is conservative if an only if there exists a continuously ifferentiable function U : R n R such that F = U. Proof. First suppose that the work one by F is conservative. Fix a point r 0 R n an efine U : R n R r F s γ where γ : [a, b] R n is an arbitrary curve with γ(a) = r 0 an γ(b) = r. By hypothesis this function is well efine. By the funamental theorem of line integrals it follows γ U s = U(r) U(r 0 ) = F s U(r 0 ). γ In particular, taking γ to be the curve γ : [0, t] R n given by t (1 t)r 0 + tr, it follows that t 0 (γ(t))(r r 0 )t = t 0 F (γ(t))(r r 0 )t U(r 0 ). But since U(r 0 ) is a constant, it follows from the funamental theorem of calculus that F = U. Conversely, suppose that there exists U : R n R such that F = U. Then for arbitrary γ as above, F s = U s = U(r) U(r 0 ) γ γ That is, W (r 0 r, γ) = U(r) U(r 0 ) an so only epens on the en points r 0 an r. Hence, the work one is path inepenent showing F is conservative. Definition 3.5. For a conservative force F, the scalar function U : R n R such that F = U is calle the potential energy. The reason that a force with this property is calle conservative comes from Theorem 3.7 below. 22

28 Definition 3.6. The total energy of a conservative system is efine to be E = T + U, the kinetic energy plus the potential energy. Theorem 3.7. (Conservation of Total Energy) In a conservative mechanical system, the total energy is conserve. That is, t E = 0. Proof. Suppose that F = U. For arbitrary t 1, t 2 R, by the Work-Kinetic Energy Theorem K(γ(t 2 )) K(γ(t 1 )) = W (γ(t 1 ) γ(t 2 )) = γ F s = γ U s = U(γ(t 1 )) U(γ(t 2 )) Hence (K + U)(γ(t 2 )) = (K + U)(γ(t 1 )) an so E := K + U is inepenent of t. In section 6 we will o some interesting computations with the Laplace-Runge-Lenze vector an so we recall here the two-boy central force problem. Definition 3.8. A central force on a particle with position vector r is a conservative force for which the corresponing potential energy is only a function of r. The classical example of a central force is one given by a potential of the form U( r) = k r where k is some constant. Given a system of two particles, (say r 1 an r 2 ) in a close system, by fixing one of the particles an consiering the relative position vector r = r 1 r 2 we can give explicit formulas for the potential energy corresponing to the gravitational an Coulomb force. The gravitational potential is given by U( r) = Gm 1m 2 r an the Coulomb by U( r) = 1 q 1 q 2 4πε 0 r where q 1 an q 2 are the charges of the two particles. Here G an ε 0 are two constants whose explicit values epen on the units being use. Proposition 3.9. Consier a close system of two particles moving in R 3 with the stanar metric. If the particles are subject to a central force fiel then the force is always parallel to the relative position of the two particles. Proof. By efinition, U is a function of only r. That is U = U( r ). In spherical coorinates we have that U = U r r where r is associate with the vector ˆr = r r. 23

29 Definition If r is the position vector of a particle, then its angular momentum is efine to be L := r p = m( r r) an the torque is τ := r F Using Newton s secon law, it immeiately follows that torque is the time erivative of angular momentum. Proposition If two particles are subject to a central force fiel then the angular momentum of their relative position vector is constant. Proof. By proposition 3.9 the force is parallel to r. Hence tl = m( r r) + ( r F ) = 0. Remark By efinition, the angular momentum is always orthogonal to the momentum an position vector. In the two-boy central force problem, Proposition 3.11 showe that the angular momentum vector is constant. Hence it must be that the plane etermine by the momentum an position of the relative position vector is constant. That is, the movement of the two particles is always restricte to a plane. Uner translation an rotation, it is no loss of generality to assume that the particles motion is restricte to the xy-plane. 3.2 The Calculus of Variations an the Euler-Lagrange Equations The calculus of variations stuies functionals on a given space X. In this section X will be the set of smooth curves γ : [a, b] M, where M is some manifol. The Euler-Lagrange equations will arise as the extreme points of a specific function, calle the action. After eriving the Euler-Lagrange equations we will see how they are a generalization of Newton s secon law. Definition Let M be an n-imensional manifol with tangent bunle T M. Given L C (T M) the pair (M, L) is calle a Lagrangian system an L is calle the Lagrangian. Definition A Riemannian manifol (M, g) with Lagrangian L = K U is calle a natural Lagrangian system. At first, it may seem that the Lagrangian efine in a natural system is ranom; however, we will see that in these systems the Euler-Lagrange equations are equivalent to Newton s secon law. The Euler-Lagrange equations have proven to be more effective than Newton s secon law in many ifferent natural Lagrangian systems. Definition Given a smooth curve γ : [a, b] M we get an inuce curve γ, calle the lift of γ, into the tangent bunle efine by γ : [a, b] T M t (γ(t), γ (t)) 24

30 Definition Let C = {γ : [a, b] M ; γ is smooth}. Consier the function A : C R efine by A(γ) := A γ := b a γ L(t)t = b a L(γ(t), γ (t))t The function A : C R is calle the action of the Lagrangian system. The goal of this section is to fin curves γ : [a, b] M which are critical points for A. Just as minimum an maximum points are extreme points in elementary calculus, we seek to fin curves for which the erivative of the action vanishes. Intuitively, for a curve γ : [a, b] M to be a minimum we nee the value of A to be no greater on γ than on curves close to γ, say within ε, as picture below. γ ε γ More precisely, fix a coorinate chart (U, x 1,..., x n ) in M an consier (T U, x 1,..., x n, v 1,..., v n ), the inuce chart on T M. Let γ : [a, b] U be a curve. Given ε R, pick arbitrary c 1,..., c n C ([a, b]) with c i (a) = 0 = c i (b). Define γ ε (t) := (γ 1 (t) + εc 1 (t),..., γ n (t) + εc n (t)). Note that we can choose ε small enough so that γ ε is containe in U. We have that γ 0 = γ an so if γ is a minimum of A then ε A γε = 0 ε=0 On the other han, the chain rule, prouct rule an funamental theorem of calculus give that ε A γε = ε=0 ε = = = = b a b a b a b a b L(γ ε (t), γ ε(t))t a (( ) ( ) ) L x i c i L (t) + γ(t) v i γ(t) t ci (t) t (( ) (( L x i c i (t) + ) ) ( L γ(t) t v i c i (t) L γ(t) t v i (( ) ( L x i c i (t) ) ) [( L γ(t) t v i c i L (t) t + γ(t) v i (( ) ( L x i c i (t) ) ) L t v i c i (t) t γ(t) 25 γ(t) γ(t) γ(t) ) ) ) c i (t) t c i (t) ] b a

31 = b a [( L x i γ(t) ) ( L t v i γ(t) )] c i (t) But since this expression is equal to zero for all c 1 (t),..., c n (t) with c i (a) = c i (b) = 0, the Funamental Lemma of Calculus of Variations (see [2] page 57) implies ( L x i L γ(t) t v i γ(t) ) = 0 for all i = 1,, n These n secon orer ODE s are calle the Euler-Lagrange equations. Hence we have shown that a necessary conition for a curve to minimize the action is that it nees to satisfy the Euler-Lagrange equations. Remark If the Lagrangian is strictly convex, meaning for fixe x M, arbitrary v, w T x M an 0 < t < 1 we have L(x, tw + (1 t)v) < tl(x, w) + (1 t)l(x, v), then the converse is locally true. That is, if a curve γ : [a, b] M satisfies the Euler-Lagrange equations, then there exists a subinterval [a 1, b 1 ] [a, b] such that γ [a1,b 1 ] minimizes the action. See [1] page 117 for a proof of this. Remark In the above erivation of the Euler-Lagrange equations, it was assume that the Lagrangian was time inepenent. However, there are many situations in which the Lagrangian oes epen on time, some of which we will see in subsequent sections. But notice that even if the Lagrangian were of the form L(x 1,..., x n, v 1,..., v n, t) the above calculation woul be exactly the same an the Euler-Lagrange equations erive above woul not change. 3.3 Examples of Lagrangian Systems Example The Action as the Length Functional In the case that we are working in a natural system for which the net force is zero, our Lagrangian reuces to L = K : T M R (p, V p ) 1 2 mg p(v p, V p ). In this case A(γ) is just a constant times the length of γ. That is, A(γ) = 1 2 m γ g γ(t) (γ (t), γ (t))t. A stanar result from Riemannian geometry is that the critical points of the length functional are geoesics. Hence any geoesic is a critical point of A. In particular, if the Riemannian manifol is R n with the stanar metric, an the net force is zero, then we get that the solutions of the Euler-Lagrange equations are straight lines. That is, the shortest path between two points is a straight line. Using this mechanical system we will give, in the section on Hamiltonian mechanics, another interpretation of geoesic flow. 26

32 Example (Natural System with Stanar Coorinates) Consier a natural Lagrangian system in R 2 with the stanar metric. Let x 1, x 2 enote the stanar coorinates an let x 1, x 2, v 1, v 2 be the inuce coorinates on T R 2 = R 4. In this coorinate system we have that g = (x 1 ) 2 + (x 2 ) 2. Suppose that a particle of mass m is moving in R 2 uner a conservative force fiel F = U. By efinition the Lagrangian L : R 4 R is efine by L(x 1, x 2, v 1, v 2 ) := K(x 1, x 2, v 1, v 2 ) U(x 1, x 2, v 1, v 2 ) = 1 2 m(v1 ) m(v2 ) 2 U(x 1, x 2 ). The elements of the Euler-Lagrange equations are an t L U ( γ(t)) = x1 x 1 (γ(t)), ( ) L v 1 ( γ(t)) = ( mv 1 ( γ(t)) ) = m γ 1 (t), t L U ( γ(t)) = x2 x 2 (γ(t)) ( ) L t v 2 ( γ(t)) = ( mv 2 ( γ(t)) ) = m γ 2 (t). t Hence, in this setting, the Euler-Lagrange equations are equivalent to Newton s secon law. This equivalence easily extens to conservative force fiels on R n. The same result also hols for k particles moving in R n. To see this, just take the manifol to be R kn so that the motion of the k-particles can be escribe by the motion of one particle. Example (Natural System with Polar Coorinates) Consier the setup of the previous example, but with polar coorinates (r, θ). Let r, θ, r, θ enote the inuce coorinates on T R 2 = R 4. By the chain rule we have r = cos θ + sin θ x 1 x 2 an θ = r sin θ + r cos θ. It follows that in polar coorinates x 1 x 2 g = [ r 2 ] an g 1 = [ r 2 In these coorinates, the velocity of γ(t) = (r(t), θ(t)) is given by ṙ r + θ θ. Therefore the kinetic energy of the particle at time t is K( γ(t)) = K(r, θ, ṙ, θ) = 1 ( 2 mg ṙ r, θ ) θ = 1 2 m(ṙ) mr2 ( θ) 2 ] 27