Orbits in Particle Accelerators and the Completion of Symplectic Jets

Orbits in particle accelerators and the completion of symplectic jets

Ivo Bodin 10372164

8 july 2015

1 Contents

1 Summary 3

2 Introduction 3

3 An introduction to symplectomorphisms 4 3.1 Group of linear symplectomorphisms ...... 4 3.2 Group of symplectomorphisms ...... 6 3.3 Introduction to kick maps ...... 8

4 Introduction to canonical transformations 11 4.1 Notation ...... 11 4.2 Lagrange equations of motion ...... 12 4.3 Hamilton’s equations of motion ...... 15 4.4 Relation canonical transformation and symplectomorphisms ...... 16 4.5 Liouville’s theorem ...... 17 4.6 Generating functions ...... 19

5 Completion of symplectic jets using kick maps 21

6 Completion of symplectic jets using generating functions 28

7 Comparison 33

2 1 Summary

CERN wants to know how to complete a symplectic jet. This thesis describes two diﬀerent ways to do this. One is a purely mathematical approach. This approach delivers an exact way to solve the problem. However, it turns out that it is not possible to perform this method using the computers we have today. The other approach uses theory of classical mechanics. A numerical approximation is required. This approach turns out to be useful in practice.

2 Introduction

In 1954 twelve countries in Western Europe started CERN. Today most people know CERN for their particle accelerator, the Large Hadron Collider, mostly known as LHC. A particle accelerator uses electromagnetic ﬁelds to accelerate particles to high speeds. The LHC is so-called synchrotron. This means it is cyclic. That is all a mathematician needs to know about particle accelerators to read this thesis. In this thesis I assume that the particles follow Hamiltonian mechanics. This means we threat the system in a non-relativistic way. CERN would like to know the path that particles take. To do this, they measure the particles’ place and momentum at a given point in the accelerator. This yields six coordinates. Two place coordinates, three momenta coordinates and the time. In this thesis, we look at three place coordinates and three momenta coordinates. If the length of the particle accelerator is known, we can calculate the forward momentum from the time and vice versa. These six coordinates are measured every time the particle completes a lap. CERN would like to predict what the next coordinates would be given a set of coordinates. This is an unknown function of six variables. Hamiltonian mechanics dictate that this function is a symplectomorphism. CERN can measure this function up to a certain precision. In practice they can measure all terms of the power series up to order eleven. This measured function will, in general, not be a symplectomorphism. We will call this function a symplectic jet. Experiments show that a symplectic jet will diverge if the function is iterated. Experiments also show that symplectomorphisms tend to diverge a lot slower than a similar symplectic jets. This is not a topic of this thesis. We will take it as a given. Because of this, CERN would like to ﬁnd a symplectomorphism that agrees with the measured function up to order eleven. We call this process the completion of a symplectic jet. A.J. Dragt and D.T. Abell have written an article describing this phenomenon and suggesting ways to complete a symplectic jet. In this thesis, two of their most promising ways are worked out in detail. First is a pure mathematical approach. This gives an exact completion of a symplectic jet. However, the symplectomorphism it produces is too big to work with in practice. The second way uses

3 generating functions. This way need numerical approximation but is doable in practice. Both approaches are both proven to be correct and a few advantages and disadvantages are named.

3 An introduction to symplectomorphisms

3.1 Group of linear symplectomorphisms

In this thesis, we look at function from R2n to R2n. Symplectomorphisms are not defined for odd dimensional spaces. We will first give the definition of linear symplectomorphisms.

2n 2n Deﬁnition 1. Let Ψ be a linear mapping from R to R and deﬁne J0 by   0 1   J0 =     −1 0

Then Ψ is a linear symplectomorphism if and only if

T Ψ J0Ψ = J0

Lemma 1. The set of linear symplectomorphisms S from R2n to R2n form a group.

T T Proof. Let Ψ, Φ ∈ S. Then, by deﬁnition:Ψ J0Ψ = J0 and Φ J0Φ = J0. We calculate

T T T (ΨΦ) J0(ΨΦ) =Φ (Ψ J0Ψ)Φ T =Φ J0Φ

=J0

We conclude that, by deﬁnition, ΨΦ ∈ S. Therefore we can conclude that S is closed.

We will now show that the identity matrix In ∈ S. By deﬁnition

T In J0In = J0

2n 2n This proves that In ∈ S. Since In is the identity for linear function from R to R , it also serves as the identity for the set of linear symplectomorphisms. We will now prove that Ψ ∈ S implies that Ψ−1 ∈ S. Let Ψ ∈ S. Then, by deﬁnition: T Ψ J0Ψ = J0. Doing a simple calculation

−1 T −1 −1 T T −1 (Ψ ) J0Ψ =(Ψ ) Ψ J0ΨΨ T −1 T −1 =(Ψ ) Ψ J0ΨΨ

=J0

4 Therefore, we can conclude that Ψ−1 ∈ S. Since taking compositions of functions is an associative operation, the set of linear symplectomorphisms is associative given this operation. We conclude that the set of all linear symplectomorphisms from R2n to R2n satisﬁes all axioms of a group.

Deﬁnition 2. We deﬁne the group of linear symplectomorphisms from R2n to R2n

T SymL(2n, R) = {M ∈ GL2n(R)|M J0M = J0}

From this deﬁnition, it is evident that

SymL(2n, R) ⊂ GL2n(R)

The group of linear symplectomorphisms has another less evident property, namely

SymL(2n, R) ⊂ SL(2n, R)

In other words, an element M of the group of linear symplectomorphisms has the property

det(M) = 1

This can be proven using a polynomial called the Pfaﬃan of the matrix. However, in the canonical transformation section of this thesis, I will prove that the determinant of the Ja- cobian of every canonical transformation equals one. This implies that all linear symplectomorphisms have a determinant that equals one. It is not much work to show that the absolute value of the determinant of a linear symplectomorphism equals one. Using known properties of the determinant yields

T det(J0) =det(Ψ J0Ψ) T =det(Ψ )det(J0)det(Ψ) 2 =det(Ψ) det(J0)

We can conclude that det(Ψ)2 = 1. Direct from the definition of a linear symplectomorphism, we can find a set of equations that will be useful to find other properties of SymL(2n, R). We write a linear map in a similar

way as the matrix J0. We calculate.  T         AB 0 1 AB AT CT CD                 =               CD −1 0 CD BT DT −A −B   AT C − CT AAT D − BCT   =     BT C − ADT BT D − DT B

5 We conclude that a linear map is a linear symplectomorphism if and only if:

AT C − CT A =0 BT D − DT B =0 T T A D − BC =In

This set of equation helps us to calculate the dimension of SymL(2n, R). From the ﬁrst equation, we can deduce

AT C =CT A =(AT C)T

This implies that AT C is symmetrical. This demand constraints half of the oﬀ-diagonal T 1 elements of A C, for a total of 2 n(n − 1). The second demand constraints an equal amount of elements. The last demand constraints n2 elements. We conclude that the dimension of SymL(2n, R) is equal to 1 dim(GL (R)) − 2 · n(n − 1) − n2 = (2n)2 − n(n − 1) − n2 2n 2 = 2n2 + n

For n = 1, the ﬁrst two restrictions are empty. The third restriction reduces to det(Ψ) = 1. Therefore, SymL(2, R) =SL(2, R). This is not true for n > 1. This can easily be seen in the following calculation

dim(SL(2n, R)) =3n2 dim(SymL(2, R)) =2n2 + n

For n 6= 1, the dimensions are diﬀerent. Therefore, the groups cannot be equal. It will however stay true that

SymL(2, R) ⊂ SL(2n, R)

3.2 Group of symplectomorphisms

We will now extend the definition to the definition of a linear symplectomorphism to the definition of a symplectomorphism

Definition 3. Let f be a total differentiable map from R2n to R2n. Then f is a symplectomorphism if and only if the Jacobian matrix Df satisfies

T (Df) J0Df = J0

6 If f is a linear symplectomorphism, then Df is a symplectic matrix. Therefore this definition is equivalent to the first definition for linear mappings. We will now define a symplectic jet.

Definition 4. Let f be a total differentiable map from R2n to R2n. Then f is a symplectic jet of order m if and only if the Jacobian matrix Df satisfies

T m+1 (Df) J0Df − J0 = O(z )

We will now prove an equivalent statement to check if something is a symplectomorphism.

Lemma 2. Let f be a total diﬀerentiable map from R2n to R2n. Let ω(u, v) be deﬁned as

T ω(u, v) = u Jov

Then f is a symplectomorphism if and only if for all u, v ∈ R2n

ω(u, v) = Df ∗(ω(u, v)) := ω(Df(u), Df(v))

We call f ∗ω the pullback of f.

Proof.

T ω(u, v) =u Jov (Df)∗(ω(u, v)) =ω(Df(u), Df(v)) T =(Df(u)) J0Df(v)) T T =u (Df) J0(Df)v

Since u, v are arbitrary, we can easily see that

∗ T ω(u, v) = (Df) (ω(u, v)) ⇐⇒ (Df) J0(Df) = Jo

The collection of symplectomorphisms forms a group. We deﬁne

Sym(R2n) = {f : R2n → R2n|ω(u, v) = (Df)∗(ω(u, v))}

We will need a few properties of the Jacobian matrix Df.

D(Id) = 1 D(f −1) = (Df)−1 D(f ◦ g) = Df ◦ Dg

7 We can now check all the group axioms hold: We will ﬁrst prove that the group is closed. Let f, g ∈ Sym(R2n). Then

(D(f ◦ g))∗(ω(u, v)) =(Df ◦ Dg))∗ω(u, v) =Df ∗ω(Dg(u), Dg(v)) =Df ∗ω(u, v) =ω(u, v)

Therefore f ◦ g ∈ Sym(R2n). Let f be the identity, then

(Df)∗(ω(u, v)) =ω(Df(u), Df(v)) =ω(1(u), 1(v)) =ω(u, v)

Thus Id ∈ Sym(R2n) We will now prove that the inverse of a symplectomorphism is again a symplectomorphism. Let f ∈ Sym(R2n), then

(Df −1)∗(ω(u, v)) =ω(Df −1(u), Df −1(v)) =ω((Df)−1(u), (Df)−1(v)) T −1 T −1 =u ((Df) ) J0(Df) (v)

Switching the inverse and the transpose and using the fact that f is a symplectomorphism yields

T −1 T −1 T T −1 −1 u ((Df) ) J0(Df) v =u ((Df) ) J0(Df) v T T −1 T −1 =u ((Df) ) (Df) J0(Df)(Df) v T =u J0v =ω(u, v)

Proving that f −1 ∈ Sym(R2n). Associativity holds for all maps f , therefore also for all symplectomorphisms f. We may conclude that Sym(R2n) is a group.

3.3 Introduction to kick maps

This section, we will introduce kick maps. However, before we can do this, Poisson brackets have to be introduced.

8 Deﬁnition 5. The Poisson brackets of functions f(pi, qi) and g(pi, qi) is deﬁned by:

X ∂f ∂g ∂f ∂g [f, g] = − ∂q ∂p ∂p ∂q j j j j j

From this deﬁnition, some properties are easily deduced.

X ∂f ∂f ∂f ∂f [f, f] = − = 0 ∂q ∂p ∂p ∂q j j j j j X ∂g ∂f ∂g ∂f [f, g] = − − = [g, f] ∂q ∂p ∂p ∂q j j j j j [af + bg, h] = a[f, h] + b[g, h] [fg, h] = [f, h]g + f[g, h]

The last two expressions are easily proven by writing down the deﬁnition. We can inter- pret them as the sum rule and product rule of Poisson brackets. There are three easy, but important expressions to be made about the coordinates:

[qi, qj] = 0

[pi, pj] = 0

[qi, pj] = δij

Using the notation η = (q, p) leads to a more common notation:

[ηi, ηj] = Jij

Associated with the Poisson bracket is a diﬀerential operator called a Lie operator, denoted by :f:

Deﬁnition 6. : f := P ∂f ∂ − ∂f ∂ j ∂qj ∂pj ∂pj ∂qj

It is worth noting that: : f : g = [f, g]. Powers of : f : are repeated Poisson brackets. For example:

: f :3 g = [f, [f.[f.g]]]

There is a particular power series consisting of Lie operators called a Lie transformation.

P :f:n Deﬁnition 7. exp(: f :) = n n!

9 Deﬁnition 8. Let g = h(q) be a function dependent of only the ﬁrst n coordinates. Then exp(: g :)η is called a kick map.

Kick maps may seem like diﬃcult functions. However, by properties of the Poisson brackets, they reduce to a much simpler form. This will be clear from the next lemma.

Lemma 3. exp(:g:)=1+:g:

Proof.

∂h(q) ∂qj ∂h(q) ∂qj : g : qj =[h(q), qj] = − = 0 ∂qj ∂pj ∂pj ∂qj ∂h(q) ∂pj ∂h(q) ∂pj ∂h(q) : g : pj =[h(q).pj] = − = ∂qj ∂pj ∂pj ∂qj ∂qj

2 ∂h(q) : g : pj =[h(q), ] = 0 ∂qj

The last expression follows from the fact that both h(q) and ∂h(q) are independent of p . From ∂qj j 2 2 the ﬁrst expression follows that : g : qj = 0. We conclude that : g : ηi = 0 ∀i. Therefore: ∞ X : g :n X : g :n exp(: g :) = = (1+ : g :) + = 1+ : g : n! n! n n=2

These kick maps are very useful because exp(: g :)η turns out to be a symplectomorphism.

Proof.

T ∂h(q) ∂h(q) T (exp(: g :)η) = (q1, ..., qn, p1 + , ..., pn + ) ∂q1 ∂qn This expression allows us to calculate the Jacobian of d(exp(: g :)).   1 0   d(exp(: g :) =     dH −1 where   ∂h(q) ∂h(q) ··· ∂h(q)  ∂q1∂q1 ∂q1∂q2 ∂q1∂qn       ∂h(q) ∂h(q) ··· ∂h(q)   ∂q2∂q1 ∂q2∂q2 ∂q2∂qn  T dH =   = (dH)  . . . .   . . .. .        ∂h(q) ∂h(q) ··· ∂h(q) ∂qn∂q1 ∂qn∂q2 ∂qn∂qn

10 Partial derivatives commute, so dH is symmetric. Using this, we can verify the symplectic condition:       1 (dH)T 0 1 1 0 T       d(exp(: g :)) · J · d(exp(: g :)) =             0 1 −1 0 dH 1     (dH)T − dH 1 0 1     =   =   = J     −1 0 −1 0

Therefore, exp(: g :) is a symplectomorphism

We will use kick maps to complete a symplectic jet.

4 Introduction to canonical transformations

4.1 Notation

It will be very useful that all physics that we will be discussing in this thesis is in the same notation as before. We have discussed function from R2n to R2n. To make the connection with physics, we can think of these functions as functions from Rn × Rn to Rn × Rn. The ﬁrst n coordinates represent n space coordinates while the next n coordinates represent the n momentum coordinates. In physics space and momentum coordinates are parameterized by the time t. The functions go from coordinates (q, p) to (Q, P ). Physicists call these functions co¨ordinatetransformations. In this thesis, we will use the following notation.

T T (η) =(q1(t), ..., qn(t), p1(t), ..., pn(t)) T q =(q1(t), ..., qn(t)) T p =(p1(t), ..., pn(t)) (q, p) =(Q(0),P (0)) (Q, P ) =(Q(t),P (t))

For a given t, there exists a function that gives the new coordinates (Q(t),P (t)) as a function of the initial coordinates (q, p). These functions are continuous and diﬀerentiable in t, so we can talk aboutq, ˙ p,˙ Q˙ and P˙ . They are deﬁned as

Deﬁnition 9. Let f be a function from Rn to Rn. Then df f˙ = dt Deﬁnition 10. For every t, we can calculate (q(t), p(t)). The curve parameterized by t is called a path.

11 4.2 Lagrange equations of motion

We will start with the three physical laws that laid the foundation for classical mechanics, namely Newton’s laws of motion. We will use these laws to deﬁne forces.

Deﬁnition 11. Let q be the place coordinates, v is given by

v =q ˙

Here v is representing the velocity of the particle.

Deﬁnition 12. The force F is deﬁned by

F = mq¨

Here m represents the mass of the system.

This deﬁnition can be written as

F − mq¨ = 0

For a system of particles, D’Alembert’s principle states that

X (Fi − mqï)dqi = 0 i I will explain this notation. It states that if we change the place coordinates infinitesimal, then the sum of the differences between the forces acting on a system of mass particles and the time derivatives of the momenta of the system is zero. In other words, the path (q(t)) P is a solution to the minimalisation problem of | i(Fi − mqï)|. In practice, this means that the system follows a path such that Newton’s laws are hold. Here Fi is the force acting on particle i, −mqï is called the reversed effective force of particle i. We will now transform this principle into an expression of independent generalized coordinates. The transformation equations are given by

qi = qi(q1, q2, ..., qn, t)

Using the transformation equations and the chain rule, we can express dri in terms of dqi

X ∂qi dq = dq i ∂q j j j

We can now rewrite the first part of D’Alembert’s principle and define a generalized force Kj. Generalized forces are defined as

12 Deﬁnition 13.

X ∂qi K = F · j i ∂q i j

Normally Qj is used to represent generalized forces. However, I used Q(t) already as the result of a canonical transformation. Using this deﬁnition, we can write

X X X ∂qi X F · dq = F · dq = K dq i i i ∂q j j j i j i j j We will now rewrite the second part of D’Alembert’s principle by the chain rule and substituting dqi

X X X ∂qi m q¨ · dq = m q¨ · dq i i i i i ∂q j i j i j Before we continue, let us ﬁrst consider the following relation by use of the product rule

d ∂qi d ∂qi d ∂qi miq˙i · = miq˙i · + miq˙i · dt ∂qj dt ∂qj dt ∂qj Shuﬄing the terms

∂qi d ∂qi d ∂qi miq¨i · = miq˙i · − miq˙i · ∂qj dt ∂qj dt ∂qj We can substitute this equality to obtain

X X ∂qi X X h d ∂qi d ∂qi i m q¨ · dq = m q˙ · − m q˙ · dq i i ∂q j dt i i ∂q i i dt ∂q j j i j j i j j We will now try to introduce the velocity into these equations. By deﬁnition and use of the chain rule and the product rule we ﬁnd expressions for the velocity and its derivatives

dqi X ∂qi ∂qi vi = = q˙k + dt ∂qk ∂t k P ∂qi ∂q ∂ q˙k i 2 2 ∂vi k ∂qk ∂t X ∂ qi ∂ qi = + = q˙k + ∂qj ∂qj ∂qj ∂qj∂qk ∂qj∂t k ∂q˙ d ∂q = i = i ∂qj dt ∂qj P ∂qi ∂qi ∂v ∂ k ∂q q˙k + ∂t ∂q˙ ∂q i = k = i = i ∂q˙j ∂q˙j ∂q˙j ∂qj Substituting these expressions and deﬁne

Deﬁnition 14. X 1 T = m v2 2 i i i Function T represents the kinetic energy of the set of particles.

13 Leads to the following calculation

X X ∂qi X X h d ∂vi ∂vi i m q¨ · dq = m v · − m v · dq i i ∂q j dt i i ∂q˙ i i ∂q j j i j j i j j X h d ∂ X 1 ∂ X 1 i = m v2 − m v2 dq dt ∂q˙ 2 i i ∂q 2 i i j j j i j i X h d ∂T ∂T i = − dq dt ∂q˙ ∂q j j j j

All qj’s are independent, so the only way that D’Alembert’s principle holds, if for all j d ∂T ∂T − = Kj dt ∂q˙j ∂qj These equations are referred to as Lagrange’s equations. However, if we assume that all forces are derivable from a potential V , we can use this to deﬁne V and rewrite the Lagrange equations in a more recognizable form.

Fi := − ∇iV

X ∂qi X ∂qi ∂V Q = F · = − ∇ V · = − j i ∂q i ∂q ∂q i j i j i In this case, D’Alembert’s principle can be put in the form d ∂T ∂T − V − = 0 dt ∂q˙j ∂qj The potential V does not depend on the generalized velocities. So the partial derivative of V with respect toq ˙j is zero, therefore d ∂T − V ∂T − V − = 0 dt ∂q˙j ∂qj We now deﬁne a new function, the Lagrangian

Deﬁnition 15. If such potential V exists, the Lagrangian L of a system is given by

L = T − V

Then D’Alembert’s principle reduces to the ﬁnal result, the Lagrange equations of motion. d ∂L ∂L − = 0 dt ∂q˙j ∂qj

14 4.3 Hamilton’s equations of motion

The Lagrangian is a function of q,q ˙ and t. We have not spoken yet of the momenta p. This will be done in terms of the Lagrangian. A similar function called the Hamiltonian will be deﬁned in terms of q, p and t.

Deﬁnition 16. We deﬁne pi by

∂L(qj, q˙j, t) pi = ∂q˙i As mentioned before, p represents the momentum of a particle.

Deﬁnition 17. The Hamiltonian of a system is given by

X H(q, p, t) = q˙ipi − L(q, q˙i, t) i

In this setting, it seems that H still a function ofq ˙i. However, we can eliminateq ˙i by inverting the definition of the generalized momenta to obtainq ˙i as functions of q, p, and t. From the definition of the Hamiltonian, we can derive a set of equations called Hamilton’s equation. This will be done by calculating the differential of H in two different ways. A direct calculation of the differential of H gives as result

X ∂H ∂H ∂H dH = dq + dp + dt ∂q i ∂p i ∂t i i i We can also calculate the diﬀerential of H by the deﬁnition and using the Lagrange equation

X ∂L ∂L ∂L dH = q˙ dp + p dq˙ − dq˙ − dq − dt i i i i ∂q˙ i ∂q i ∂t i i i X ∂L = q˙ dp + p dq˙ − p dq˙ − p˙ dq − dt i i i i i i i i ∂t i X ∂L = q˙ dp − p˙ dq − dt i i i i ∂t i Comparing these two expressions for the diﬀerential of H terms by terms gives us Hamilton’s equations and a diﬀerential relation between the Hamiltonian and the Lagrangian. Poisson brackets can be used to shorten the notation ∂H q˙i = = {q, H} ∂pi ∂H p˙i = − = {p, H} ∂qi ∂L ∂H − = ∂t ∂t

15 4.4 Relation canonical transformation and symplectomorphisms

We can write the Hamilton equations of motion in a more compact way. Using earlier deﬁned matrix J0 and vector η, we can reduce Hamilton equations to ∂H η˙ = J 0 ∂η These equations hold for every tuple of classical coordinates η. We will now look at a coordinate transformation given by

Q1 =f1(q1, ..., qn, p1, ..., pn) . .

Qn =fn(q1, ..., qn, p1, ..., pn)

P1 =fn+1(q1, ..., qn, p1, ..., pn) . .

Pn =f2n(q1, ..., qn, p1, ..., pn)

2n 2n We deﬁne f as the vector of functions fi. This is a map from R to R . It will be interesting to look at the time derivative of f. We ﬁnd by use of the chain rule:

dfi(η) ∂fi(η) ∂fi(η) ∂fi(η) ∂fi(η) = q˙1 + ... + q˙n + p˙1 + ... + p˙n dt ∂q1 ∂qn ∂p1 ∂pn It will be useful to write the time derivative in matrix notation. Let M be the Jacobi matrix of f

∂fi Mij = ∂ηj Then the time derivative of f is given by: df ∂H = Mη˙ = MJ dt 0 ∂η For the last equality, we substituted the Hamilton equations. We can also verify that, again by the chain rule ∂H ∂H ∂f = j ∂ηi ∂fj ∂ηi Which we can also put in matrix notation ∂H ∂H = M T ∂η ∂f

16 Substituting this in the expression for the time derivative yields df ∂H ∂H f˙ = = Mη˙ = MJ = MJ M T dt 0 ∂η 0 ∂f We have already seen that every set of classical coordinates η satisﬁes the Hamilton equations. Our function f describes a new set of classical coordinates (Q, P ), so (Q, P ) satisﬁes the Hamilton equations. ∂H f˙ = J 0 ∂f

T Comparing these two expressions for f˙, we ﬁnd that J0 = M J0M. We conclude that f is a symplectomorphism. This coordinate transformation is called a canonical transformation. Every canonical transformation can be expressed by a symplectomorphism.

4.5 Liouville’s theorem

Theorem 1. Volume in phase space in conserved under canonical transformations, meaning ZZ ZZ dqdp = dQdP

Proof. We can calculate the right hand side of the equation treating it as a normal coordinate transformation. ZZ ZZ dQdP = Jdqdp

Here J equals the Jacobian, the determinant of the Jacobian matrix M. We do the following calculation

J = det(M) = eT rlnM

We can use this expression to calculate the time derivative of J. 1 dM J˙ =T r eT rlnM M dt 1 dM =T r J M dt n n X X dMji =J M −1 ij dt i=1 j=1 Recall the notation

η =(q, p) ζ =(Q, P )

17 −1 dM We can express matrices M,M and dt in terms of these coordinates.

∂ζi Mij = ∂ηj −1 ∂ηi (Mij ) = ∂ζj dM ∂ζ˙ ji = j dt ηi Substituting this and using the chain rule yields ˙ X X ∂ηi ∂ζj J˙ =J ∂ζ ∂η j i j i ˙ X ∂ηi ∂ζj ∂ζk =J ∂ζj ∂ζk ∂ηi i,j,k The sum over i can now be performed n X ∂ηi ∂ζk = ∂ζ ∂η i=1 k i n X −1 = Mij Mki i=1 n X −1 = MkiMij i=1

=δkj

This reduces the equation to ˙ X ∂ζj J˙ =J δkj ∂ζk j,k ˙ X ∂ζj = ∂ζ j k

=J∇x · x˙

We now need to do two observations. The ﬁrst one we look what happens at t = 0. Then, M = 1, concluding that the Jacobian equals one.

The second observation is that for a Hamiltonian system ∇x ·x˙ = 0. The Jacobian is constant in time. We conclude that the Jacobian equals one proving Liouville’s theorem. ZZ ZZ dqdp = dQdP

This also proves that every linear symplectomorphism has determinant one, as promised before.

18 4.6 Generating functions

We will introduce Hamilton’s principle. We will prove that this principle is equivalent to D’Alemberts principle, meaning that they both lead to the Lagrange and Hamilton equations of motion. We will ﬁrst deﬁne the action of the path.

Deﬁnition 18. Let q(t) be a path from t1 to t2. Then the action functional of that path is given by

Z t2 S[q] = L((q(t), q(˙t), t)dt t1 Hamilton’s principle states that the evolution of a system is the solution to the equation dS = 0 dq(t) It states that the followed path q(t) is a stationary point of the action functional. We show that the Lagrange equations of motion can also be deduced from Hamilton’s principle. Let be a small variation of q(t) such that

(t1) = (t2) = 0

Since is small, we can use ﬁrst order Taylor expansion

Z t2 dS = L((q(t) + (t), q˙(t) + ˙(t), t) − L((q(t), q(˙t), t)dt t1 Z t2 ∂L ∂L Z t2 ∂L Z t2 ∂L = + ˙ dt = dt + ˙ dt t1 ∂q ∂q˙ t1 ∂q t1 ∂q˙ Using integration by parts on the second term yields

t t Z 2 ∂L h ∂Lit2 Z 2 d ∂L dS = dt + − dt t t1 ∂q ∂q˙ 1 t1 dt ∂q˙ By deﬁnition of , the second terms equals 0. The equation reduces to

Z t2 ∂L d ∂L dS = − dt = 0 t1 ∂q dt ∂q˙ Hamilton’s principle can only hold for every if ∂L d ∂L − = 0 ∂q dt ∂q˙ Therefore we see that D’Alembert’s principle and Hamilton’s principle are equivalent. We will now introduce generating functions. To do this we have to rewrite Hamilton’s principle

Z t2 Z t2 ˙ X dS = d L((q(t), q(t), t)dt = d q˙ipi − H(q, q˙i, t)dt t1 t1 i

19 We now deﬁne a new set of coordinates

Qi =Qi(q, p, t)

Pi =Pi(q, p, t)

Corresponding to these new coordinates is a new Hamiltonian K. The Hamilton equations of motion now state ∂K Q˙ i = ∂Pi ∂K P˙i = − ∂Qi They also satisfy Hamilton’s principle, so

Z t2 X d Q˙ iPi − K(Q, Q˙ i, t)dt = 0 t1 i Both the diﬀerential of integrals are zero, however this does not imply that the integrands are equal. We can multiply with a constant λ and add a derivative of any two times continuous diﬀerentiable function F. This function F adds only a constant to the integral, so the solutions to Hamilton’s principle do not change by adding this function. This means that: X X dF λ( q˙ p − H(q, q˙ , t)) = Q˙ P − K(Q, Q˙ , t) + i i i i i i dt i i

−1 This λ is easily set to one using Qi = λ qi, Pi = pi. If above equation holds for λ = 1, we speak of a canonical transformation. In theory, F can depend of q, p, Q and P . However, there are four interesting options, in all of which F depends on halve the old and half on the new coordinates. Assume that F = F (q, Q, t). Then

X X ∂F ∂F ∂F q˙ p − H(q, q˙ , t)) = Q˙ P − K(Q, Q˙ , t) + q˙ + Q˙ + t i i i i i i ∂q˙ ˙ ∂t i i ∂Q Since the coordinates are independent, this can only hold if ∂F p = ∂q ∂F P = − ∂Q ∂F K = H + ∂t To complete a symplectic jet, we will use a generating function of type 3. A generating function of type 3 is of the form:

F = q · p + G(Q, p, t)

20 Doing a similar calculation as above, we ﬁnd that

X X dF q˙ p − H(q, p, t)) = Q˙ P − K(Q, P, t) + i i i i dt i i X d(q · p) ∂G ∂G ∂G = Q˙ P − K(Q, P, t) + + p˙ + Q˙ + i i dt ∂p ∂Q ∂t i X ∂G ∂G ∂G = Q˙ P − K(Q, P, t) +q ˙ p + q p˙ + p˙ + Q˙ + i i i i i i ∂p ∂Q ∂t i X X ∂G ∂G ∂G −q p˙ − H(q, p, t)) = Q˙ P − K(Q, P, t) + p˙ + Q˙ + i i i i ∂p ∂Q ∂t i i Since all the coordinates are independent, this can only hold if ∂G q = − ∂p ∂G P = − ∂Q ∂G K = H + ∂t Similar equalities hold for the two other interesting types of F . However, we will only use generating functions of type 3 to complete a symplectic jet.

5 Completion of symplectic jets using kick maps

Theorem 2. Let T be a homogeneous symplectic jet of order m. With a combination of homogeneous kick maps and conjugates of linear symplectomorphisms, we can construct a symplectic function Φ such that Φ−1 ◦T = Id+O(zm+1) which implies that T −Φ = O(zm+1).

This implication follows from the following calculation Φ ◦ Φ−1 ◦ T = T Φ ◦ Id + O(zm+1) = Φ + O(zm+1)

Combining these results yields: T − Φ = O(zm+1). We have to be careful in the second equation. This is not obvious because Φ has not to be linear. The equation holds because Φ is polynomial. In general, O(zm+1) will be different at both sides of the equation. We can prove that the equation holds by writing down the functions as a power series. However, this is not very insightful so I decided not to present it here To prove the main theorem, we need a few lemmas. The first lemma describes a different

basis for Vn,d.

21 Lemma 4. Let d ∈ N, then the elements in

E = {(a · z)d|a ∈ Cn} span the vector space of d-homogeneous polynomials in n variables if and only if there exists no d-homogenous polynomial P that vanishes on E. We will denote this vector space from now on as Vn,d. Since the dimension of this space is ﬁnite, it is possible to choose a ﬁnite amount of element of E such that the collection of these element form a basis for Vn,d.

Proof. We will prove the contrapositive of the lemma. α P A basis of Vn,d is given by z where kαk = d. With kαk = d, we mean that ai = d. This means we can write every element, therefore in particular (a · z)d, as a linear combination of zα. We write b(α) for the corresponding multi-binomial coeﬃcients. Then

X (a · z)d = b(α)aαzα α

Now assume that the elements of E do not span Vn,d, then there exists a form Λ that anni- hilates all elements (a · z)m. We note that

m X α Λ(a · z) = λα · b(α)a ≡ 0 α

α Since a is a vector just as z, a as a basis for Vn,d. We observe that we have found a d- homogenous polynomial that vanishes on C. The other way around, assume we have been given a polynomial that vanishes on C. Deﬁne

X α X α Λ( cαa ) = λαcαa ≡ 0 α α By the linearity of the summation, we easily check that Λ is a form. Since we have found a non-zero linear form that is identically 0, the elements of E do not form a basis of C. We conclude that the elements of E do not form a basis if and only if the exists a polynomial that vanishes on C. The lemma follows by taking the contrapositive of this statement.

We will now prove a handy property of a special kind of polynomial maps.

Lemma 5. Let P be a polynomial map from C2n to C2n of the form

d+1 P (z) = z + Pd(z) + O(kzk )

∗ d then (P ω −ω)(z) = O(kzk ) if and only if Pd is Hamiltonian, so per deﬁnition, if J(DPd(z)) is symmetrical.

22 Proof. Let u, v, z ∈ C2n. Here we see z as a point, while u, v are direction vector in the point z. We calculate

P ∗ω(z)(u, v) =ω(DP (z)u, DP (z)v) d+1 d+1 =ω(D(z + Pd(z) + O(kzk )u, D(z + Pd(z) + O(kzk )v) d d =ω(Id + DPd(z) + O(kzk )u, Id + DPd(z) + O(kzk )v) d d =ω(u + DPd(z)u + O(kzk )u, v + Pd(z)v + O(kzk )v) d =ω(u, v) + ω(u, DPd(z)v) + ω(DPd(z)u, v) + O(kzk )

This means that

∗ d P ω(z)(u, v) =ω(u, v) + ω(u, DPd(z)v) + ω(DPd(z)u, v) + O(kzk ) T d =ω(u, v) + u J · DPd(z)v + J(DPd(z))u · v) + O(kzk ) d =ω(u, v) + Ju · DPd(z)v + J(DPd(z))u · v) + O(kzk )

We need another identity. This one is easily proved by just writing down the deﬁnition of standard inner product and matrix multiplication.

X T x · (Ay) = xiAijyj = (A x) · y i,j Using this identity, we ﬁnd that

∗ d P ω(z)(u, v) =ω(u, v) + Ju · DPd(z)v + J(DPd(z))u · v) + O(kzk ) T d =ω(u, v) + (DPd(z)) Ju · v + J(DPd(z))u · v) + O(kzk )

Because u,v are arbitrary, (P ∗ω − ω)(z) = O(kzkd) if and only if

T −(DPd(z)) J = J(DPd(z))

This is equivalent of J(DPd(z)) being symmetric. This is true since       A ··· A A ··· A  1,1 1,2n   n,1 n,2n  0, 1        . . .   . . .  JA =    . .. .  =  . .. .        −1, 0         A2n,1 ··· A2n,2n −An−1,1 · · · −An−1,2n     A ··· A   A · · · −A  1,1 2n,1   n,1 n−1,1    0, 1   T  . . .     . . .  −A J = −  . .. .    =  . .. .          −1, 0       A1,2n ··· A2n,2n An,2n · · · −An−1,2n

Therefore, they are only equal if JA is symmetric. This proves the lemma.

23 We used the mathematical deﬁnition of Hamiltonian. To prove the following lemma’s, the physical deﬁnition of Hamiltonian will be more useful.

Deﬁnition 19. A function F is called Hamiltonian if there exists a function H such that:

F = J∇H

Of course we have to prove that the mathematical deﬁnition and the physical deﬁnition are equivalent. This will be done in the following lemma.

2n 2n Lemma 6. Let Pd be a homogeneous function from R to R . Then

J(DPd(z)) is symmetrical ⇐⇒ ∃ a function H such that: Pd = J∇H

2n Proof. Write Pd = (Q, R). Here Q and R are vectors of n functions from R to R. In this notation   ( ∂Q )( ∂R )  ∂q ∂q  DPd =    ∂Q ∂R  ( ∂p )( ∂p ) the matrix elements are total derivatives. Therefore, the matrix elements are n by n matrices. In the same notation

  ( ∂Q )( ∂R )  ∂p ∂p  JDPd =    ∂Q ∂R  −( ∂q )(− ∂q )

This means that JDPd is symmetrical if

∂Q ∂QT = ∂p ∂p ∂R ∂RT = ∂q ∂q ∂R ∂QT = − ∂p ∂q Now assume that there exists a function H of degree d + 1 such that

Pd = J∇Hd+1

In our notation, this implies that ∂H ∂H Q = ∧ R = − ∂p ∂q

24 We can now deduce ∂Q ∂H ∂H ∂Q = = = ∂p ij ∂pi∂pj ∂pj∂pi ∂p ji We can switch the partial derivatives since H is polynomial and therefore C∞. In a same way, we can show that another matrix element is symmetrical

∂R ∂H ∂H ∂R = = = ∂q ij ∂qi∂qj ∂qj∂qi ∂q ji The last thing we need to show is that ∂R ∂H ∂H ∂Q = = = − ∂p ij ∂pi∂qj ∂qj∂pi ∂q ji

We can conclude that J(DPd(z)) is symmetrical.

We will now prove the implication the other way around. Assume that J(DPd(z) is symmetrical. We want to prove that there exists a function H such that

Pd(z) = J∇H(z)

This means that H has to satisfy the following set of equations. Note that H has to meet more requirements, this set of equations is only a part of the demands: ∂H Z = Q1 ⇒ H = Q1dp1 + f1(q) ∂p1 ∂H Z = Qi ⇒ H = Qidpi + fi(q) ∂pi ∂H Z = Qn ⇒ H = Qndpn + fn(q) ∂pn We will now prove that, if H satisﬁes any of the equations above, all equations above hold. Using the fundamental theorem of calculus, we ﬁnd: Z Z Z ∂ ∂Qi ∂Qj Qidpi + fi(q) = dpi = dpi = Qj ∂pj ∂pj ∂pi ∂ Z = Qjdpj + fj(q) ∂pj

We used that J(DPd(z)) is symmetrical to switch the indices in the partial derivative. Since R the partial derivative to pj is equal for every Qidpi, we may conclude that a function H exists and that this set of equations determines H up to a function that depends only on q. We will now look at another set of equations that has to hold:

25 ∂H˜ Z − = R1 ⇒ H˜ = − R1dq1 + f1(p) ∂q1 ∂H˜ Z − = Ri ⇒ H˜ = − Ridqi + fi(p) ∂qi ∂H˜ Z − = Rn ⇒ H˜ = − Rndqn + fn(p) ∂qn For now, we will not threat H and H˜ as the same function. We will ﬁrst prove that a function H˜ exists and then we will prove that it is possible that H = H˜ . With a similar calculation as before: Z Z Z ∂ ∂Ri ∂Rj Ridqi + fi(p) = dqi = dqi = Rj ∂qj ∂qj ∂qi ∂ Z = Rjdqj + fj(p) ∂qj This implies that H˜ exists and the equations determine H˜ up to a function that only depends on p. We will now prove that H = H˜ Z ∂H ∂Qj ∂fj(q) = dpj + ∂qi ∂qi ∂qi Z ∂Ri ∂fj(q) = − dpj + ∂pj ∂qi Z ∂Ri ∂fj(q) = − dqi + ∂qi ∂qi ∂H˜ ∂f (q) = + j ∂qi ∂qi Apparently ∂fj (q) is equal for all j. This implies that f (q) is equal for all j. From now on ∂qi j this will be noted as f(q). Therefore, we ﬁnd that:

H = H˜ + f(q) + f(p)

This means that H and H˜ can only differ in pure terms of p and q. We now define a new function H¯ to have the pure terms in q of H˜ and the pure terms in p of H, while every other terms agrees with both H and H˜ . Then H¯ satisfies ∂H ∂H Q = ∧ R = − ∂p ∂q And thus

Pd(z) = J∇H(z)

We conclude that the mathematical deﬁnition and the physical deﬁnition of Hamiltonian are equivalent.

26 We will now prove the last lemma before we will prove the theorem.

Lemma 7. A basis for the vector space of d-homogenous Hamilton vector ﬁelds on C2n is given by vector ﬁelds of the form (Ja · z)da.

Proof. Let Pd be a Hamiltonian vector ﬁeld. By deﬁnition, we can write

Pd(z) = J∇Q(z)

For a certain (d+1)-homogenous polynomial Q(z). By the lemma above, we can write

N X j d+1 Q(z) = cj(a · z) j=1

This means that, by the chain rule

N N X j d+1 X j d j Pd(z) = J∇ cj(a · z) = J cj(d + 1)(a · z) a j=1 j=1

Now let a = Jb, then b = −Ja, so

J(a · z)da = (a · z)dJa = (−Jb · z)db = −(Jb · z)db

Note that since −(Jb · z)db spans the same space as (Jb · z)db, this proves the lemma.

We will now prove the theorem.

Proof. Let P(z) be a symplectic d-jet. It is not hard to match the ﬁrst term of P(z). We can simply take the ﬁrst order terms of P(z) since DP (0) ∈ SymL(2n, R). k Suppose we have found Fk ∈ Sym(2n, R) such that (Fk − P )(z) = O(kzk) for all 2 ≤ k ≤ d and note that

−1 k+2 Gk(z) := P ◦ Fk (z) = z + Pk+1 + O(kzk)

By previous lemma’s , it follows that Pk+1 is Hamiltonian and therefore we can write

N X j m j Pk+1(z) = cj · (Ja · z) · a j=1

j m j We will now deﬁne Sj = z + cj · (Ja · z) · a . Then Sj ∈ Sym(2n, R) and Hk+1 = SN ◦ ... ◦ S1

matches Gk up to order k+1. This implies that Fk+1 := Hk+1 ◦ Fk matches P to order k+1. By induction, the theorem follows.

We conclude that it is possible to ﬁnd an exact completion of a symplectic jet.

27 6 Completion of symplectic jets using generating functions

In the generating function section, we have seen the equations of a type 3 generating function. ∂G q = − ∂p ∂G P = − ∂Q ∂G K = H + ∂t We claim that these equations define the canonical transformation and therefore, they define the symplectic function corresponding to the canonical transformation. The procedure is as follows. We begin with the first set of equations ∂G q = − ∂p This are n equations in variables (q, Q, p). These can, under some light conditions, be inverted and written as

Qi = f(q, p)

We got formulae for each Qi as a function of the old coordinates. We can substitute these formulae into the second n equations. ∂G P = − ∂Q

This yields formulae for each Pi as functions of the old coordinates. Combining all the formulae yields:

(Q, P ) = f(q, p)

This function f describes the canonical transformations and is therefore a symplectomorphism. We conclude that the sets of equations deﬁne a symplectic function. It will be insightful to explain why the equations can be inverted. It makes use of the implicit function theorem:

Theorem 3. Implicit Function Theorem If the equation R(x, y) satisﬁes that det(DR(x, y)) 6= 0. Then there exist an open set V containing (x,y) where we can ﬁnd a function f such that y = f(x)

In our case:

y =Q x =(q, p) ∂G R(x, y) =R(q, Q, p) = q + = 0 ∂p

28 This means we can invert the equations if det(DR(x, y)) 6= 0. Since there is no reason it should be zero, in the general case, inverting the functions will be no problem. We have seen that a generating function corresponds with a symplectomorphism. How- ever, we do not know this generating function yet. The generating function depends on the symplectic jet. For now, we assume the generating function is of the following form.

X m+1 F = qjpj − Qjpj − G (Q, p)

Note that we have not proven this yet. We will prove that for every canonical transformation, we can ﬁnd a function Gm+1(Q, p) such that we can ﬁnd a generating function of this form. Until then, we simply assume F has this form. However, we can already give some information about the generating function. We got a type 3 generating function, since it has the form

X F = qjpj + G(Q, p)

Here G can be any function of (Q, p). In our case, G consist of a part that yields the identity plus a part of terms of degree three till m + 1. I will explain the part that gives the identity. We know that ∂G qj = − ∂pj ∂G Pj = − ∂Qj Substituting our generating function and shuﬄing some terms yields

∂Gm+1 Qj =qj − ∂pj ∂Gm+1 Pj =pj + ∂Qj

We can now clearly see that if Gm+1 = 0, then the equations reduce to

Qj = qj

Pj = pj

This is the identity. It is useful to think why this should be the case. If the first order terms are not the identity, repeating the function cannot possibly be stable. From a mathematical point of view, we can always compose the symplectic jet with a linear symplectomorphism such that the first order term is the identity. Then, after the symplectic completion, we can compose the symplectomorphism with the inverse of the first linear symplectomorphism to get our final result.

29 We now want to ﬁnd the function Gm+1. To do this, we need to keep in mind that we only

know a symplectic jet of order m, namely Tm. We will use the following notation:

1 2 (Q, P ) =Tm = (Tm,Tm) 1 m+1 Q =Tm + O(z ) 2 ˜ m+1 P =Tm + O(z )

The tilde lets us now that the higher order terms might be diﬀerent. Substituting this into the diﬀerential equations yields,

m+1 1 m+1 1 m+1 ∂G (Tm + O(z ), p) (Tm + O(z ))j =qj − ∂pj

The function Gm+1 is has a degree of m+1. This gives us an expression of Gm+1 in terms of (q, p), namely Z m+1 1 1 G (Tm, p) = qj − (Tm)jdpj

If we substitute this back into the equation, we ﬁnd:

R 1 1 m+1 ∂ qj − (Tm)jdpj (Tm + O(z ))j =qj − ∂pj 1 =qj − qj + (Tm)j 1 =(Tm)j

Function Gm+1 holds up to order m + 1. Since we only got a symplectic jet of order m, we cannot ask to hold for higher order. This generating function satisfies the first differential equation. However, it seems to depend on j which cannot be the case since it has to hold for all j. It also looks like Gm+1

has a term of order 2, namely qj integrated. However, we have already seen that the ﬁrst

term of (Tm)j is equal to qj. These two terms will cancel each other. We will now show that Gm+1 does not depend on j. The new coordinates (Q,P) are canonical and therefore obey the Hamilton equations

∂K P˙ = − ∂Q ∂K Q˙ = ∂P This means that there exists a function H such that

(Q, P ) = J∇H

30 We have seen before that this implies that J(DTm(z)) is symmetrical. Meaning:

∂T 1 ∂T 1 T m = m ∂p ∂p ∂T 2 ∂T 2 T m = m ∂q ∂q ∂T 2 ∂T 1 T m = − m ∂p ∂q Let’s go back to the generating function. Doing a quick calculation Z m+1 1 1 G (Tm, p) = qj − (Tm)jdpj Z Z d 1 = qj − (Tm)jdpjdpi dpi ZZ 1 dqj d(Tm)j = − dpidpj dpi dpi ZZ 1 dqi d(Tm)i = − dpjdpi dpj dpj Z 1 = qi − (Tm)idpi

m+1 1 We see that it does not matter which j we choose since G (Tm, p) is equal for all j. This also implies that Gm+1 is unique up to a constant. The only thing that is left is to show that the second part of the diﬀerential equations is satisﬁed ∂Gm+1 Pj =pj + ∂Qj R 1 2 ∂ qj − (Tm)jdpj Tm =pj + ∂Qj Before we continue, we need an identity, doing some calculations ∂Q ∂Q P˙ = q˙ + p˙ ∂q ∂p ∂P ∂H ∂P ∂H = − ∂q ∂p ∂p ∂q ∂H ∂H ∂q ∂H ∂p − = − − ∂Q ∂q ∂Q ∂p ∂Q

˙ ∂H Since P = − ∂Q , we ﬁnd that ∂q ∂P = ∂Q ∂p ∂P ∂p = − ∂q ∂Q

31 We will now check that Gm+1 satisﬁes the second part of the equations.

R 1 2 ∂ qj − (Tm)jdpj m+1 Tm =pj + + O(z ) ∂Qj R R 1 ∂ qjdpj ∂ (Tm)jdpj m+1 =pj + − + O(z ) ∂Qj ∂Qj

1 m+1 We now use that Tm + O(z ) = Q. We can add these higher order terms and subtract them to ﬁnd R R 1 2 ∂ qjdpj ∂ (Tm)jdpj m+1 Tm =pj + − + O(z ) ∂Qj ∂Qj R R 1 m+1 ∂ qjdpj ∂ (Tm)j + O(z )dpj m+1 =pj + − + O(z ) ∂Qj ∂Qj R R ∂ qjdpj ∂ Qjdpj m+1 =pj + − + O(z ) ∂Qj ∂Qj R ∂ qjdpj m+1 =pj + − pj + O(z ) ∂Qj Z ∂qj m+1 = dpj + O(z ) ∂Qj We can now use one of the identities above to rewrite this to Z 2 ∂qj m+1 Tm = dpj + O(z ) ∂Qj Z ∂Pj m+1 = dpj + O(z ) ∂pj m+1 =Pj + O(z )

We conclude that Gm+1 satisﬁes all necessary equations. However, to produce a symplectic map, we need Gm+1 as a function of (Q, p). We will now describe how this function is acquired. The symplectic jet gives us Q = Q(q, p) up to order m. We need to invert this into q = q(Q, p) up to order m. Deﬁne

F (q, Q, p) = Q − Q(q, p)

We need to solve F (q, Q, p) = 0 for q(Q, p). We use Newton’s method to obtain derivatives of q = (Q, p) to ﬁnd the Taylor expansion. We calculate the ﬁrst order Taylor expansion:

∂F (q, q(Q, p)) ∂F (q, q(Q, p)) F (q, q(Q, p)) = · q0(Q, p) + ∂q ∂p This has to be equal to zero. This can only be true if ∂F (q, q(Q, p)) ∂F (q, q(Q, p)) q0(Q, p) = −( )−1 ∂q ∂p

32 We can do this eleven more times to obtain the ﬁrst eleven terms of the Taylor expansion of q = q(Q, p). This expansion can then be substitutes for q in Gm+1(q, p) to obtain Gm+1(Q, p). To make sure that Gm+1(Q, p) consists of terms of order 3 to m + 1, all terms with higher order are neglected.

It is also clear that for every symplectic jet of order m we can ﬁnd Gm+1 such that the generating function is in the assumed form. We have already seen how a generating function leads to a canonical transformation and therefore to a symplectomorphism. This symplectomorphism completes the symplectic jet Tm.

7 Comparison

The kick map method has the clear advantage that it produces an exact symplectomorphism while the generating function method needs a numerical approximation. Before we continue, it will be useful to look at the method used for the numerical approximation. It is called Newton’s method. Let F be a function from Rn to Rn. We want to ﬁnd the roots of F . We

write down the ﬁrst order Taylor polynomial around x0.

2 F (x) = F (x0) + DF (x0)(x − x0) + O(||(x − x0)|| )

We will calculate when the ﬁrst two terms together equal zero. This yields

−1 x1 = x0 − (DF (x0)) F (x0)

We will now use the ﬁrst order Taylor polynomial of F around point x1. In general the equation is

−1 xi+1 = xi − (DF (xi)) F (xi)

It can be proven that this method has quadratic convergence if x0 is close enough to the root of F . Experiments show this is true in practice. The numerical approximation is fast enough to calculate up to the machine precision of 10−16. Since we cannot know the exact initial coordinate due to Heisenberg’s principle. It states that the uncertainty of the place and position vector always satisﬁes

h¯ σ σ ≥ ≈ 8 · 10−32 q p 2 This means the initial conditions cannot be known a lot more precise than machine precision. Therefore, the numerical approximation is no real disadvantage to the generating functions method.

33 To match a symplectic jet of order 11 using the kick map method is in practice not possible. A homogeneous polynomial of order 11 has

2n + d 6 + 11 17 = = = 4368 d + 1 12 12 terms. This means that Hk+1 = S4368 ◦ ... ◦ S1. Every Si is of degree 11, so the total degree of Hk+1 equals

4368 4548 deg(Hk+1) = 11 ≈ 6, 357 · 10

This number is too large to work with in practice. However, it is worth noting that if d 6= 4, it is possible to reduce the power of the degree of the symplectic jet to

2n + d n + d 17 14 / = / = 68 d + 1 d + 1 12 12

This means we end up in our particular problem with a polynomial of degree

1168 ≈ 6.63 · 1070

This does not solve the problem. This is still too large for the computational power we got today. Only the generating function method gives a practical solution to the problem of CERN. This is the way they use. However, if an exact completion is needed for another problem or the computational power of computers increases enough, the kick map method is the one to choose.

References

Arnold, V. 1997, Mathematical Methods of Classical Mechanics (Springer)

Dragt, A. & Abell, D. 1996, Symplectic maps and Particle accelerators

Goldstein, H. 1950, Classical mechanics (Addison-Wesley)

Løw, E., Pereira, J., Peters, H., & Wold, E. 2013, Polynomial Completion of Symplectic Jets and Surfaces Containing Involutive Lines