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

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

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

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 fromR2n toR2n. Symplectomorphisms are not defined for odd dimensional spaces. We will first give the definition of linear symplectomorphisms. Definition 1. Let Ψ be a linear mapping from R2n toR2n and define J0 by

J0 =     0 1 −1 0     Then Ψ is a linear symplectomorphism if and only if

ΨTJ0Ψ = J0

Lemma 1. The set of linear symplectomorphisms S from R2n to R2n form a group. Proof. Let Ψ, Φ ∈ S. Then, by definition:ΨTJ0Ψ = J0 and ΦTJ0Φ = J0. We calculate

(ΨΦ)TJ0(ΨΦ) =ΦT(ΨTJ0Ψ)Φ

=ΦTJ0Φ

=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

I_nTJ0In= J0

This proves that In ∈ S. Since In is the identity for linear function from R2n toR2n, 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: ΨTJ0Ψ = J0. Doing a simple calculation

(Ψ−1)TJ0Ψ−1=(Ψ−1)TΨTJ0ΨΨ−1

=(ΨT)−1ΨTJ0ΨΨ−1

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 fromR2ntoR2nsatisfies all axioms of a group.

Definition 2. We define the group of linear symplectomorphisms from R2n _{to R}2n

SymL(2n,R) = {M ∈ GL2n(R)|MTJ0M = 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 symplecto-morphisms have a determinant that equals one.

It is not much work to show that the absolute value of the determinant of a linear symplec-tomorphism equals one. Using known properties of the determinant yields

det(J0) =det(ΨTJ0Ψ)

=det(ΨT)det(J0)det(Ψ)

=det(Ψ)2det(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.

    A B C D     T     0 1 −1 0         A B C D     =     AT CT BT DT         C D −A −B     =     ATC − CTA ATD − BCT BT_{C − AD}T _BT_{D − D}T_B    

We conclude that a linear map is a linear symplectomorphism if and only if: ATC − CTA =0

BTD − DTB =0 ATD − BCT =In

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

ATC =CTA =(ATC)T

This implies that ATC is symmetrical. This demand constraints half of the off-diagonal elements of ATC, for a total of 1₂n(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 dim(GL2n(R)) − 2 · 1 2n(n − 1) − n 2_{= (2n)}2_{− n(n − 1) − n}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 R}2n_{. Then f is a}

symplecto-morphism if and only if the Jacobian matrix Df satisfies (Df )TJ0Df = J0

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

(Df )TJ0Df − J0 = O(zm+1)

We will now prove an equivalent statement to check if something is a symplectomorphism. Lemma 2. Let f be a total differentiable map fromR2n toR2n. Let ω(u, v) be defined as

ω(u, v) = uTJov

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.

ω(u, v) =uTJov

(Df )∗(ω(u, v)) =ω(Df (u), Df (v)) =(Df (u))TJ0Df (v))

=uT(Df )TJ0(Df )v

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

ω(u, v) = (Df )∗(ω(u, v)) ⇐⇒ (Df )TJ0(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

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)) =uT((Df )−1)TJ0(Df )−1(v)

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

uT((Df )−1)TJ0(Df )−1v =uT((Df )T)−1J0(Df )−1v

=uT((Df )T)−1(Df )TJ0(Df )(Df )−1v

=uTJ0v

=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.

Definition 5. The Poisson brackets of functions f (pi, qi) and g(pi, qi) is defined by: [f, g] =X j ∂f ∂qj ∂g ∂pj − ∂f ∂pj ∂g ∂qj

From this definition, some properties are easily deduced. [f, f ] =X j ∂f ∂qj ∂f ∂pj − ∂f ∂pj ∂f ∂qj = 0 [f, g] = −X j ∂g ∂qj ∂f ∂pj − ∂g ∂pj ∂f ∂qj = [g, f ] [af + bg, h] = a[f, h] + b[g, h] [f g, 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 j ∂f ∂qj ∂ ∂pj − ∂f ∂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. Definition 7. exp(: f :) =P

n :f :n

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 brack-ets, they reduce to a much simpler form. This will be clear from the next lemma.

Lemma 3. exp(:g:)=1+:g: Proof. : g : qj =[h(q), qj] = ∂h(q) ∂qj ∂qj ∂pj − ∂h(q) ∂pj ∂qj ∂qj = 0 : g : pj =[h(q).pj] = ∂h(q) ∂qj ∂pj ∂pj −∂h(q) ∂pj ∂pj ∂qj = ∂h(q) ∂qj : g :2pj =[h(q), ∂h(q) ∂qj ] = 0

The last expression follows from the fact that both h(q) and ∂h(q)_∂q

j are independent of pj. From

the first expression follows that : g :2 qj = 0. We conclude that : g :2 ηi= 0 ∀i. Therefore:

exp(: g :) =X n : g :n n! = (1+ : g :) + ∞ X n=2 : g :n n! = 1+ : g :

These kick maps are very useful because exp(: g :)η turns out to be a symplectomorphism. Proof. (exp(: g :)η)T = (q1, ..., qn, p1+ ∂h(q) ∂q1 , ..., pn+ ∂h(q) ∂qn )T This expression allows us to calculate the Jacobian of d(exp(: g :)).

d(exp(: g :) =     1 0 dH −1     where dH =             ∂h(q) ∂q1∂q1 ∂h(q) ∂q1∂q2 · · · ∂h(q) ∂q1∂qn ∂h(q) ∂q2∂q1 ∂h(q) ∂q2∂q2 · · · ∂h(q) ∂q2∂qn .. . ... . .. ... ∂h(q) ∂qn∂q1 ∂h(q) ∂qn∂q2 · · · ∂h(q) ∂qn∂qn             = (dH)T

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

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 R}n_×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¨ordinate transformations. In this thesis, we will use the following notation.

(η)T =(q1(t), ..., qn(t), p1(t), ..., pn(t))T

qT =(q1(t), ..., qn(t))

pT =(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 about ˙q, ˙p, ˙Q and ˙P . They are defined as

Definition 9. Let f be a function from Rn to Rn. Then ˙

f = df dt

Definition 10. For every t, we can calculate (q(t), p(t)). The curve parameterized by t is called a path.

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 = m¨q Here m represents the mass of the system.

This definition can be written as

F − m¨q = 0

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

i

(Fi− m ¨qi)dqi = 0

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)) is a solution to the minimalisation problem of |P

i(Fi− m ¨qi)|. 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, −m ¨qi 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

dqi= X j ∂qi ∂qj dqj

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

Definition 13. Kj = X i Fi· ∂qi ∂qj

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 i Fi· dqi = X j X i Fi· ∂qi ∂qj dqj = X j Kjdqj

We will now rewrite the second part of D’Alembert’s principle by the chain rule and substi-tuting dqi X i miq¨i· dqi= X j X i miq¨i· ∂qi ∂qj dqj

Before we continue, let us first consider the following relation by use of the product rule d dt  miq˙i· ∂qi ∂qj  = d dt  miq˙i  · ∂qi ∂qj + miq˙i· d dt ∂qi ∂qj

Shuffling the terms

miq¨i· ∂qi ∂qj = d dt  miq˙i· ∂qi ∂qj  − m_iq˙i· d dt ∂qi ∂qj

We can substitute this equality to obtain X j X i miq¨i· ∂qi ∂qj dqj = X j X i hd dt  miq˙i· ∂qi ∂qj  − miq˙i· d dt ∂q_i ∂qj i dqj

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

vi= dqi dt = X k ∂qi ∂qk ˙ qk+ ∂qi ∂t ∂vi ∂qj = ∂ P k ∂qi ∂qkq˙k ∂qj + ∂qi ∂t ∂qj =X k ∂2qi ∂qj∂qk ˙ qk+ ∂2qi ∂qj∂t = ∂ ˙qi ∂qj = d dt ∂q_i ∂qj  ∂vi ∂ ˙qj = ∂P k ∂qi ∂qkq˙k+ ∂qi ∂t  ∂ ˙qj = ∂ ˙qi ∂ ˙qj = ∂qi ∂qj

Substituting these expressions and define Definition 14. T =X i 1 2miv 2 i

Leads to the following calculation X j X i miq¨i· ∂qi ∂qj dqj = X j X i hd dt  mivi· ∂vi ∂ ˙qj
− mivi· ∂vi ∂qj i dqj =X j hd dt  ∂ ∂ ˙qj  X i 1 2miv 2 i  − ∂ ∂qj  X i 1 2miv 2 i i dqj =X j hd dt ∂T ∂ ˙qj  − ∂T ∂qj i dqj

All qj’s are independent, so the only way that D’Alembert’s principle holds, if for all j

d dt ∂T ∂ ˙qj  − ∂T ∂qj = Kj

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 Qj = X i Fi· ∂qi ∂qj = −X i ∇iV · ∂qi ∂qj = −∂V ∂qi

In this case, D’Alembert’s principle can be put in the form d dt ∂T ∂ ˙qj  −∂T − V ∂qj = 0

The potential V does not depend on the generalized velocities. So the partial derivative of V with respect to ˙qj is zero, therefore

d dt ∂T − V ∂ ˙qj  −∂T − V ∂qj = 0 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 dt ∂L ∂ ˙qj  − ∂L ∂qj = 0

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

pi=

∂L(qj, ˙qj, t)

∂ ˙qi

As mentioned before, p represents the momentum of a particle. Definition 17. The Hamiltonian of a system is given by

H(q, p, t) =X

i

˙

qipi− L(q, ˙qi, t)

In this setting, it seems that H still a function of ˙qi. However, we can eliminate ˙qiby inverting

the definition of the generalized momenta to obtain ˙qi 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

dH =X i ∂H ∂qi dqi+ ∂H ∂pi dpi  +∂H ∂t dt

We can also calculate the differential of H by the definition and using the Lagrange equation dH =X i  ˙ qidpi+ pid ˙qi− ∂L ∂ ˙qi d ˙qi− ∂L ∂qi dqi  −∂L ∂tdt =X i  ˙ qidpi+ pid ˙qi− pid ˙qi− ˙pidqi  −∂L ∂tdt =X i  ˙ qidpi− ˙pidqi  −∂L ∂tdt

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

˙ qi = ∂H ∂pi = {q, H} ˙ pi = − ∂H ∂qi = {p, H} −∂L ∂t = ∂H ∂t

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

˙ η = J0

∂H ∂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)

We define f as the vector of functions fi. This is a map fromR2n toR2n. It will be interesting

to look at the time derivative of f. We find by use of the chain rule: dfi(η) dt = ∂fi(η) ∂q1 ˙ q1+ ... + ∂fi(η) ∂qn ˙ qn+ ∂fi(η) ∂p1 ˙ p1+ ... + ∂fi(η) ∂pn ˙ pn

It will be useful to write the time derivative in matrix notation. Let M be the Jacobi matrix of f

Mij =

∂fi

∂ηj

Then the time derivative of f is given by: df

dt = M ˙η = M J0 ∂H

∂η

For the last equality, we substituted the Hamilton equations. We can also verify that, again by the chain rule

∂H ∂ηi = ∂H ∂fj ∂fj ∂ηi

Which we can also put in matrix notation ∂H

∂η = M

T∂H

Substituting this in the expression for the time derivative yields ˙ f = df dt = M ˙η = M J0 ∂H ∂η = M J0M T∂H ∂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.

˙ f = J0

∂H ∂f

Comparing these two expressions for ˙f , we find that J0 = MTJ0M . 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 Z Z

dqdp = Z Z

dQdP

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

Z Z

dQdP = Z Z

J dqdp

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 . ˙ J =T r 1 M dM dt  eT rlnM =T r 1 M dM dt  J =J n X i=1 n X j=1 M_ij−1dM ji dt Recall the notation

η =(q, p) ζ =(Q, P )

We can express matrices M, M−1 and dM_dt in terms of these coordinates. Mij = ∂ζi ∂ηj (M_ij−1) = ∂ηi ∂ζj dMji dt = ∂ ˙ζj ηi

Substituting this and using the chain rule yields ˙ J =JX j X i ∂ηi ∂ζj ∂ ˙ζj ∂ηi =JX i,j,k ∂ηi ∂ζj ∂ ˙ζj ∂ζk ∂ζk ∂ηi

The sum over i can now be performed

n X i=1 =∂ηi ∂ζk ∂ζk ∂ηi = n X i=1 M_ij−1Mki = n X i=1 MkiMij−1 =δkj

This reduces the equation to

˙ J =JX j,k δkj ∂ ˙ζj ∂ζk =X j ∂ ˙ζj ∂ζk =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. Z Z

dqdp = Z Z

dQdP

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

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

S[q] = Z t2

t1

L((q(t), ˙q(t), t)dt

Hamilton’s principle states that the evolution of a system is the solution to the equation dS

dq(t) = 0

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

dS = Z t2 t1 ∂L ∂qdt + h ∂L ∂ ˙q it2 t1 − Z t2 t1 d dt ∂L ∂ ˙qdt By definition of , the second terms equals 0. The equation reduces to

dS = Z t2 t1 ∂L ∂q −  d dt ∂L ∂ ˙qdt = 0 Hamilton’s principle can only hold for every  if

∂L ∂q − d dt ∂L ∂ ˙q = 0

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

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

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

˙ Qi = ∂K ∂Pi ˙ Pi = − ∂K ∂Qi

They also satisfy Hamilton’s principle, so d Z t2 t1 X i ˙ QiPi− K(Q, ˙Qi, t)dt = 0

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:

λ(X i ˙ qipi− H(q, ˙qi, t)) = X i ˙ QiPi− K(Q, ˙Qi, t) + dF dt

This λ is easily set to one using Qi = λ−1qi, 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 i ˙ qipi− H(q, ˙qi, t)) = X i ˙ QiPi− K(Q, ˙Qi, t) + ∂F ∂ ˙qq +˙ ∂F ∂ ˙Q ˙ Q +∂F ∂tt Since the coordinates are independent, this can only hold if

p = ∂F ∂q P = −∂F ∂Q K = H + ∂F ∂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:

Doing a similar calculation as above, we find that X i ˙ qipi− H(q, p, t)) = X i ˙ QiPi− K(Q, P, t) + dF dt =X i ˙ QiPi− K(Q, P, t) + d(q · p) dt + ∂G ∂pp +˙ ∂G ∂QQ +˙ ∂G ∂t =X i ˙ QiPi− K(Q, P, t) + ˙qipi+ qip˙i+ ∂G ∂pp +˙ ∂G ∂Q ˙ Q + ∂G ∂t X i −q_ip˙i− H(q, p, t)) = X i ˙ QiPi− K(Q, P, t) + ∂G ∂pp +˙ ∂G ∂QQ +˙ ∂G ∂t Since all the coordinates are independent, this can only hold if

q = −∂G ∂p P = −∂G ∂Q K = H + ∂G ∂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(z}m+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.

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.

A basis of Vn,d is given by zα where kαk = d. With kαk = d, we mean thatP 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}

(a · z)d=X

α

b(α)aαzα

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 Λ(a · z)m=X

α

λα· b(α)aα≡ 0

Since a is a vector just as z, aα as a basis for Vn,d. We observe that we have found a

d-homogenous polynomial that vanishes onC.

The other way around, assume we have been given a polynomial that vanishes onC. Define Λ(X α cαaα) = X α λαcαaα ≡ 0

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 ofC.

We conclude that the elements of E do not form a basis if and only if the exists a polynomial that vanishes onC. 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 P (z) = z + Pd(z) + O(kzkd+1)

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

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)

=ω(D(z + Pd(z) + O(kzkd+1)u, D(z + Pd(z) + O(kzkd+1)v)

=ω(Id + DPd(z) + O(kzkd)u, Id + DPd(z) + O(kzkd)v)

=ω(u + DPd(z)u + O(kzkd)u, v + Pd(z)v + O(kzkd)v)

=ω(u, v) + ω(u, DPd(z)v) + ω(DPd(z)u, v) + O(kzkd)

This means that

P∗ω(z)(u, v) =ω(u, v) + ω(u, DPd(z)v) + ω(DPd(z)u, v) + O(kzkd)

=ω(u, v) + uTJ · DPd(z)v + J (DPd(z))u · v) + O(kzkd)

=ω(u, v) + J u · DPd(z)v + J (DPd(z))u · v) + O(kzkd)

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

x · (Ay) =X

i,j

xiAijyj = (ATx) · y

Using this identity, we find that

P∗ω(z)(u, v) =ω(u, v) + J u · DPd(z)v + J (DPd(z))u · v) + O(kzkd)

=ω(u, v) + (DPd(z))TJ u · v + J (DPd(z))u · v) + O(kzkd)

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

This is equivalent of J (DPd(z)) being symmetric. This is true since

J A =     0,1 −1, 0             A1,1 · · · A1,2n .. . . .. ... A2n,1 · · · A2n,2n         =         An,1 · · · An,2n .. . . .. ... −An−1,1 · · · −An−1,2n         −AT_{J = −}         A1,1 · · · A2n,1 .. . . .. ... A1,2n · · · A2n,2n             0,1 −1, 0     =         An,1 · · · −An−1,1 .. . . .. ... An,2n · · · −An−1,2n         Therefore, they are only equal if J A is symmetric. This proves the lemma.

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.

Lemma 6. Let Pd be a homogeneous function from R2n to R2n. Then

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

Proof. Write Pd = (Q, R). Here Q and R are vectors of n functions from R2n toR. In this

notation DPd=     (∂Q_∂q) (∂R_∂q) (∂Q_∂p) (∂R_∂p)    

the matrix elements are total derivatives. Therefore, the matrix elements are n by n matrices. In the same notation

J DPd=     (∂Q_∂p) (∂R_∂p) −(∂Q_∂q) (−∂R_∂q)     This means that J DPdis symmetrical if

∂Q ∂p  =∂Q ∂p T ∂R ∂q  =∂R ∂q T ∂R ∂p  = − ∂Q ∂q T

Now assume that there exists a function H of degree d + 1 such that Pd= J ∇Hd+1

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

We can now deduce ∂Q ∂p  ij =  ∂H ∂pi∂pj  = ∂H ∂pj∂pi  =∂Q ∂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 ∂q  ij =  ∂H ∂qi∂qj  = ∂H ∂qj∂qi  =∂R ∂q  ji

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

symmet-rical. 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 ∂p1 = Q1⇒ H = Z Q1dp1+ f1(q) ∂H ∂pi = Qi⇒ H = Z Qidpi+ fi(q) ∂H ∂pn = Qn⇒ H = Z Qndpn+ fn(q)

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:

∂ ∂pj Z Qidpi+ fi(q)  = Z ∂Qi ∂pj dpi = Z _∂Q j ∂pi dpi= Qj = ∂ ∂pj Z Qjdpj+ fj(q) 

We used that J (DPd(z)) is symmetrical to switch the indices in the partial derivative. Since

the partial derivative to pj is equal for every R 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:

−∂ ˜H ∂q1 = R1 ⇒ ˜H = − Z R1dq1+ f1(p) −∂ ˜H ∂qi = Ri ⇒ ˜H = − Z Ridqi+ fi(p) −∂ ˜H ∂qn = Rn⇒ ˜H = − Z Rndqn+ fn(p)

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: ∂ ∂qj Z Ridqi+ fi(p)  = Z _∂R i ∂qj dqi = Z _∂R j ∂qi dqi = Rj = ∂ ∂qj Z Rjdqj + fj(p) 

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 ∂H ∂qi = Z _∂Q j ∂qi dpj+ ∂fj(q) ∂qi = − Z _∂R i ∂pj dpj+ ∂fj(q) ∂qi = − Z ∂Ri ∂qi dqi+ ∂fj(q) ∂qi = ∂ ˜H ∂qi +∂fj(q) ∂qi Apparently ∂fj(q)

∂qi is equal for all j. This implies that fj(q) is equal for all j. From now on

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

Q =∂H ∂p  ∧ R = −∂H ∂q  And thus Pd(z) = J ∇H(z)

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

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 (J a · z)da.

Proof. Let Pdbe 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

Q(z) =

N

X

j=1

cj(aj · z)d+1

This means that, by the chain rule

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

Now let a = J b, then b = −J a, so

J (a · z)da = (a · z)dJ a = (−J b · z)db = −(J b · z)db

Note that since −(J b · z)d_{b spans the same space as (J b · z)}d_{b, 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).

Suppose we have found Fk∈ Sym(2n,R) such that (Fk− P )(z) = O(kzk)k for all 2 ≤ k ≤ d

and note that

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

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

Pk+1(z) = N

X

j=1

cj· (J aj· z)m· aj

We will now define Sj = z + cj· (J aj· z)m· aj. 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.

6

Completion of symplectic jets using generating functions

In the generating function section, we have seen the equations of a type 3 generating function. q = −∂G ∂p P = −∂G ∂Q K = H + ∂G ∂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

q = −∂G ∂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.

P = −∂G ∂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 symplectomor-phism. 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)

R(x, y) =R(q, Q, p) = q +∂G ∂p = 0

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.

F =Xqjpj− Qjpj− Gm+1(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

F =Xqjpj + 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 qj = − ∂G ∂pj Pj = − ∂G ∂Qj

Substituting our generating function and shuffling some terms yields Qj =qj − ∂Gm+1 ∂pj Pj =pj + ∂Gm+1 ∂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.

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:

(Q, P ) =Tm = (Tm1, Tm2)

Q =T_m1 + O(zm+1) P =T_m2 + ˜O(zm+1)

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

(T_m1 + O(zm+1))j =qj−

∂Gm+1(T_m1 + O(zm+1), p) ∂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

Gm+1(T_m1, p) = Z

qj − (Tm1)jdpj

If we substitute this back into the equation, we find: (T_m1 + O(zm+1))j =qj−

∂R qj− (Tm1)jdpj

∂pj

=qj− qj+ (Tm1)j

=(T_m1)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 ˙ P = −∂K ∂Q ˙ Q = ∂K ∂P This means that there exists a function H such that

We have seen before that this implies that J (DTm(z)) is symmetrical. Meaning: ∂T1 m ∂p  =∂T 1 m ∂p T ∂T2 m ∂q  = ∂T2 m ∂q T ∂T2 m ∂p  = − ∂T1 m ∂q T

Let’s go back to the generating function. Doing a quick calculation Gm+1(T_m1, p) = Z qj− (Tm1)jdpj = Z _d dpi Z qj − (Tm1)jdpjdpi = Z Z _dq j dpi − d(T 1 m)j dpi  dpidpj = Z Z _ dqi dpj −d(T 1 m)i dpj  dpjdpi = Z qi− (Tm1)idpi

We see that it does not matter which j we choose since Gm+1(T_m1, 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 Pj =pj+ ∂Gm+1 ∂Qj T_m2 =pj+ ∂R qj − (Tm1)jdpj ∂Qj

Before we continue, we need an identity, doing some calculations ˙ P =∂Q ∂qq +˙ ∂Q ∂pp˙ =∂P ∂q ∂H ∂p − ∂P ∂p ∂H ∂q −∂H ∂Q = − ∂H ∂q ∂q ∂Q− ∂H ∂p ∂p ∂Q Since ˙P = −∂H_∂Q, we find that

∂q ∂Q = ∂P ∂p ∂P ∂q = − ∂p ∂Q

We will now check that Gm+1 satisfies the second part of the equations. T_m2 =pj+ ∂R qj− (Tm1)jdpj ∂Qj + O(zm+1) =pj+ ∂R qjdpj ∂Qj − ∂R (T 1 m)jdpj ∂Qj + O(zm+1)

We now use that T_m1 + O(zm+1) = Q. We can add these higher order terms and subtract them to find T_m2 =pj + ∂R qjdpj ∂Qj − ∂R (T 1 m)jdpj ∂Qj + O(zm+1) =pj + ∂R qjdpj ∂Qj − ∂R (T 1 m)j+ O(zm+1)dpj ∂Qj + O(zm+1) =pj + ∂R qjdpj ∂Qj − ∂R Qjdpj ∂Qj + O(zm+1) =pj + ∂R qjdpj ∂Qj − pj+ O(zm+1) = Z _∂q j ∂Qj dpj + O(zm+1)

We can now use one of the identities above to rewrite this to T_m2 = Z ∂qj ∂Qj dpj+ O(zm+1) = Z _∂P j ∂pj dpj + O(zm+1) =Pj+ O(zm+1)

We conclude that Gm+1 satisfies all necessary equations. However, to produce a symplectic map, we need Gm+1as 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)) ∂q · q

0_{(Q, p) +}∂F (q, q(Q, p))

∂p This has to be equal to zero. This can only be true if

q0(Q, p) = −(∂F (q, q(Q, p))

∂q )

−1∂F (q, q(Q, p))

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 toRn. We want to find the roots of F . We write down the first order Taylor polynomial around x0.

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

We will calculate when the first two terms together equal zero. This yields x1 = x0− (DF (x0))−1F (x0)

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

equation is

xi+1= xi− (DF (xi))−1F (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

σqσp ≥

¯ h

2 ≈ 8 · 10

−32

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.

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 d + 1  =6 + 11 12  =17 12  = 4368

terms. This means that Hk+1 = S4368◦ ... ◦ S1. Every Si is of degree 11, so the total degree

of Hk+1 equals

deg(Hk+1) = 114368≈ 6, 357 · 104548

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

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