sign in sign up
<<
Home , Canonical coordinates

Hamiltonian Dynamics Lecture 1

David Kelliher

RAL

November 12, 2019

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 1 / 59 Bibliography

The Variational Principles of Mechanics - Lanczos - Goldstein, Poole and Safko A Student’s Guide to Lagrangians and Hamiltonians - Hamill Classical Mechanics, The Theoretical Minimum - Susskind and Hrabovsky Theory and Design of Charged Particle Beams - Reiser Accelerator Physics - Lee Particle Accelerator Physics II - Wiedemann Mathematical Methods in the Physical Sciences - Boas Beam Dynamics in High Particle Accelerators - Wolski

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 2 / 59 Content

Lecture 1 Comparison of Newtonian, Lagrangian and Hamiltonian approaches. Hamilton’s equations, symplecticity, integrability, chaos. Canonical transformations, the Hamilton-Jacobi equation, Poisson brackets. Lecture 2 The “accelerator” Hamiltonian. Dynamic maps, symplectic integrators. Integrable Hamiltonian.

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 3 / 59 Configuration

The state of the system at a q 3 t can be given by the value of the t2 n generalised coordinates qi . This can be represented by a point in an n dimensional space which is called “configuration space” (the system t1 is said to have n degrees of free- q2 dom). The of the system as a whole is then characterised by the line this system point maps out in q1 configuration space.

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 4 / 59 Newtonian Mechanics

The equation of motion of a particle of m subject to a F is d (mr˙) = F(r, r˙, t) (1) dt In Newtonian mechanics, the dynamics of the system are defined by the force F, which in general is a function of position r, r˙ and time t. The dynamics are determined by solving N second order differential equations as a function of time. Note: coordinates can be the vector spatial coordinates ri(t) or generalised coordinates qi (t).

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 5 / 59

In Lagrangian mechanics the key function is the Lagrangian

L = L(q, q˙, t) (2)

The solution to a given mechanical problem is obtained by solving a set of N second-order differential equations known as the Euler-Lagrange equations, d ∂L ∂L − = 0 (3) dt ∂q˙ ∂q

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 6 / 59 Action y

t2

t1

x

The action S is the integral of L along the Z t2 S = L(q, q˙, t)t (4) t1

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 7 / 59 Principle of least action

The principle of least action or Hamilton’s principle holds that the system evolves such that the action S is stationary. It can be shown that the Euler-Lagrange equation defines a path for which.

Z t2  δS = δ L(q, q˙, t)t = 0 (5) t1

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 8 / 59 Derivation of Euler-Lagrange equation Adding a perturbation , ˙ to the path one obtains

Z t2 δS = [L(q + , q˙ + ˙) − L(q, q˙)] dt (6) t1 Z t2  ∂L ∂L =  + ˙ dt (7) t1 ∂q ∂q˙ Integration by parts leads to

 ∂Lt2 Z t2  ∂L d ∂L δS =  +  −  dt (8) ∂q˙ ∂q dt ∂q˙ t1 t1

Since we are varying the path but not the end points, (t1) = (t2) = 0

Z t2 ∂L d ∂L δS =  − dt (9) t1 ∂q dt ∂q˙

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 9 / 59 Advantages of Lagrangian approach

The Euler-Lagrange is true regardless of the choice of (including non-inertial coordinate systems). We can transform to convenient variables that best describe the symmetry of the system. It is easy to incorporate constraints. We formulate the Lagrangian in a configuration space where ignorable coordinates are removed (e.g. a mass constrained to a surface), thereby incorporating the constraint from the outset.

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 10 / 59 Conservative force

In the case of a convervative force field the Lagrangian is the difference of the kinetic and potential

L(q, q˙) = T (q, q˙) − V (q) (10)

where ∂V (q) F = (11) ∂q

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 11 / 59 Example: mass on a spring

The Lagrangian 1 1 L = T − V = mx˙ 2 − kx2. (12) 2 2 Plugging this into Lagrange’s equations

d ∂L ∂L − = 0 (13) dt ∂x˙ ∂x d ∂ 1 1  ∂ 1 1  mx˙ 2 − kx2 − mx˙ 2 − kx2 = 0 (14) dt ∂x˙ 2 2 ∂x 2 2 d (mx˙) + kx = 0 (15) dt mx¨ + kx = 0. (16)

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 12 / 59 Relativistic free particle - Lagrangian

The for a free particle is 1 pi = γmx˙i , i = 1, 2, 3; γ = (17) p1 − β2

where m = m0 the rest mass and β is the velocity relative to c. To ensure

∂L mx˙i pi = = p . (18) ∂x˙i 1 − β2 the Lagrangian should be of the form r p 1 L(x, x˙, t) = −mc2 1 − β2 = −mc2 1 − x˙ 2 +x ˙ 2 +x ˙ 2. (19) c2 1 2 3

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 13 / 59 General electromagnetic fields

Now include general EM fields U(x, x˙, t) = e(φ − v · A) p L(x, x˙, t) = −mc2 1 − β2 − eφ + ev · A. (20)

The conjugate momentum is

∂L mx˙i Pi = = p + eAi (21) ∂x˙i 1 − β2

i.e. the field contributes to the conjugate momentum.

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 14 / 59

1 The Legendre transform takes us from a convex function F (ui ) to another function G(vi ) as follows. Start with a function

F = F (u1, u2,..., un). (22)

Introduce a new set of through the following transformation ∂F vi = . (23) ∂ui We now define a new function G as follows

n X G = ui vi − F (24) i=1

1 ∂2F F is convex in u if ∂u2 > 0 David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 15 / 59 Apply Legendre’s transformation to the Lagrangian

Start with the Lagrangian

L = L(q1,..., qn, q˙1,..., q˙n, t), (25)

and introduce some new variables we are going to call the pi s ∂L pi = . (26) ∂q˙i We can then introduce a new function H defined as

n X H = pi q˙i − L (27) i=1

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 16 / 59 We now have a function which is dependent on q, p and time.

H = H(q1,..., qn, p1,..., pn, t) (28)

L and H have a dual nature: X X H = pi q˙i − L, L = pi q˙i − H, ∂L ∂H pi = , q˙i = . ∂q˙i ∂pi

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 17 / 59 We can also use the passive variables q and t to demonstrate that ∂L ∂H = − , (29) ∂qi ∂qi ∂L ∂H = − . (30) ∂t ∂t

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 18 / 59 Hamilton’s canonical equations

Starting from Lagrange’s equation

d  ∂L  ∂L = dt ∂q˙i ∂q and combining with ∂L pi = ∂q˙i leads to ∂L ∂H p˙i = = − (31) ∂qi ∂qi

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 19 / 59 So we have ∂H ∂H q˙i = (32)p ˙i = − (33) ∂pi ∂qi which are called Hamilton’s canonical equations. They are the of the system expressed as 2n first order differential equations.

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 20 / 59 Symmetry and Conservation Laws

A cyclic coordinate in the Langrangian is also cyclic in the Hamiltonian. Since H(q, p, t) =q ˙i pi − L(q, q˙, t), a coordinate qj absent in L is also absent in H. A symmetry in the system implies a cyclic coordinate which in turn leads to a conservation law (Noether’s theorem).

∂L ∂H = 0 =⇒ = 0 (34) ∂qj ∂qj Hence p˙j = 0 (35)

so the momentum pj is conserved. Often we wish to simplify our problem by applying a transformation that exploits any symmetry in the system.

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 21 / 59 Canonical transformations

Transform from one set of canonical coordinates (pi , qi ) to another (Pi , Qi ). The transformation should preserve the form of Hamilton’s equations.

Old coordinates New coordinates Hamiltonian: H(q, p, t) Hamiltonian: K(Q, P, t)

∂H ∂K q˙i = (36) Q˙i = (38) ∂pi ∂Pi ∂H ∂K p˙i = − (37) P˙i = − (39) ∂qi ∂Qi

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 22 / 59 Preservation of Hamiltonian form

For the old Hamiltonian H it was true that ! Z t2 X δ pi q˙i − H(qi , pi , t) dt = 0 (40) t1 i Likewise, for the new Hamiltonian K ! Z t2 X δ Pi Q˙i − K(Qi , Pi , t) dt = 0 (41) t1 i

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 23 / 59 In order for this to be true we require dF λ(pq˙ − H) = PQ˙ − K + (42) dt where, in order to ensure no variation at the endpoints it is required

Z t2 dF δ dt = δ(F (t2) − F (t1)) = 0 (43) t1 dt

If λ = 1 the transformation is said to be canonical2. We assume this condition in the following.

2If λ 6= 1 the transformation is said to be extended canonical. David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 24 / 59 The function F is called the generating function of the and it depends on old and new coordinates. It can take 4 forms corresponding to combinations of (qi , pi ) and (Qi , Pi ):

F = F1(qi , Qi , t) (44)

F = F2(qi , Pi , t) (45)

F = F3(pi , Qi , t) (46)

F = F4(pi , Pi , t) (47)

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 25 / 59 Generating function F1(q, Q, t)

dF p q˙ − H = P Q˙ − K + 1 (48) i i i i dt ∂F1 ∂F1 ∂F1 = Pi Q˙i − K + q˙i + Q˙i + (49) ∂qi ∂Qi ∂t       ∂F1 ∂F1 ∂F1 pi − q˙i − Pi + Q˙i + K − H + = 0 (50) ∂qi ∂qi ∂t The old and new coordinates are separately independent so the coefficients ofq ˙i and Q˙i must each vanish leading to

∂F1 pi = (51) ∂qi ∂F1 Pi = − (52) ∂Qi ∂F K = H + 1 (53) ∂t

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 26 / 59 F1 example

F1(q, Q, t) = qQ (54) This does not depend on time, so by equation 53 the new and original Hamiltonians are equal. ∂F p = 1 = Q (55) ∂q ∂F P = − 1 = −q (56) ∂Q This generating function essentially swaps the coordinates and momenta.

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 27 / 59 Generating function F2(q, P, t)

Look for a function of the form

F = F2(q, P, t) − Qi Pi (57)

so that d p q˙ − H = P Q˙ − K + (F − Q P ) (58) i i i i dt 2 i i dF = P Q˙ − K + 2 − Q˙ P − Q P˙ (59) i i dt i i i i ∂F2 ∂F2 ∂F2 = −K + q˙i + P˙i + − Qi P˙i (60) ∂qi ∂Pi ∂t

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 28 / 59 Equating terms we find

∂F2 pi = (61) ∂qi ∂F2 Qi = (62) ∂Pi ∂F K = H + 2 . (63) ∂t

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 29 / 59 F2 example

X F2(q, P, t) = qi Pi (64) i This example generating function also does not depend on time so the new and original Hamiltonians are again equal. So

∂F2 pi = = Pi (65) ∂qi ∂F2 Qi = = qi (66) ∂Pi This generating function is just the identity transformation, the coordinates and Hamiltonian are swapped into themselves.

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 30 / 59 Generating function F3(p, Q, t)

We define F = qi pi + F3(p, Q, t) (67) so can then say d p q˙ − H = P Q˙ − K + (q p + F (p, Q, t)) (68) i i i i dt i i 3 dF (p, Q, t) p q˙ − H = P Q˙ − K +q ˙ p + q p˙ + 3 (69) i i i i i i i i dt ∂F3 ∂F3 ∂F3 −H = Pi Q˙i − K + qi p˙i + p˙i + Q˙i + . (70) ∂pi ∂Qi ∂t

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 31 / 59 Equating the terms

∂F3 qi = − (71) ∂pi ∂F3 Pi = − (72) ∂Qi ∂F K = H + 3 . (73) ∂t

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 32 / 59 F3 example

F3(p, Q, t) = pi Qi (74) We can express

∂F3 qi = − = −Qi (75) ∂pi ∂F3 Pi = − = −pi . (76) ∂Qi This generating function inverts the coordinates and momenta and keeps the Hamiltonians the same. Is anyone’s head spinning yet?

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 33 / 59 Generating function F4(p, P, t)

Let F = qi pi − Qi Pi + F4(p, P, t) (77) so we can express d p q˙ − H = P Q˙ − K + (q p − Q P + F (p, P, t)) (78) i i i i dt i i i i 4 pi q˙i − H = Pi Q˙i − K +q ˙i pi + qi p˙i − Q˙i Pi − Qi P˙i (79) ∂F4 ∂F4 ∂F4 + p˙i + P˙i + ∂pi ∂Pi ∂t

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 34 / 59 Equating terms

∂F4 qi = − (80) ∂pi ∂F4 Qi = (81) ∂Pi ∂F K = H + 4 . (82) ∂t

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 35 / 59 F4 example

F4(pi , Pi , t) = pP (83)

K = H (84) q = −P (85) Q = p (86)

This transformation flips the coordinates and momenta and inverts the transformed momenta.

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 36 / 59 Summary of generating functions

Generating function Transformation equations

∂F1 ∂F1 F = F1(q, Q, t) pi = Pi = − ∂qi ∂Qi

∂F2 ∂F2 F = F2(q, P, t) − Qi Pi pi = Qi = ∂qi ∂Pi

∂F3 ∂F3 F = F3(p, Q, t) + qi pi qi = − Pi = − ∂pi ∂Qi

∂F4 ∂F4 F = F4(p, P, t) + qi pi − Qi Pi qi = − Qi = ∂pi ∂Pi

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 37 / 59 Example:

The Hamiltonian for a harmonic oscillator is given ω H = q2 + p2 . (87) 2 This Hamiltonian is the sum of two squares, which suggest that one of the new coordinates is cyclic. Try a transformation of the form

r 2 q = f (P) sin Q (88) ω r 2 p = f (P) cos Q. (89) ω Then the new Hamiltonian

K = H = f 2(P)(sin2 Q + cos2 Q) = f 2(P). (90)

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 38 / 59 Take the ratio of the transformation equations

p = q cot Q. (91)

This is independent of f(P), and has the form of the F1(q, Q, t) type of generating function ∂F p = 1 . (92) ∂q

The simplest form for F1 agreeing with the above is 1 F (q, Q) = q2 cot Q. (93) 1 2

We can then find P using the other transformation equation for F1

∂F 1 1 q2 P = − 1 = q2 csc2 Q = . (94) ∂Q 2 2 sin2 Q

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 39 / 59 Rearrange for q p √ q = 2P sin2 Q = 2P sin Q. (95) Comparing this with equation 88 gives the function f (P) √ f (P) = ωP. (96)

The new Hamiltonian is therefore

K = ωP. (97)

This is cyclic in Q, so P is constant. The energy is constant and equal to K so E P = . (98) ω ∂K Q˙ = = ω (99) ∂P

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 40 / 59 The solution for Q is Q = ωt + α (100) for some constant α. Finally the solution is r 2E q = sin ωt + α. (101) ω

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 41 / 59 The Hamilton-Jacobi equation

The Hamilton-Jacobi equation (HJE) is written

∂S ∂S ∂S H(q1,..., qn, ,..., , t) + = 0. (102) ∂q1 ∂qn ∂t where S is called Hamilton’s principle function. S is equivalent to the action. In order to obtain the HJE we seek a transformation to coordinates and momenta from the known initial condition, keeping the end points fixed.

q = q(q0, p0, t) (103)

p = p(q0, p0, t). (104)

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 42 / 59 We can achieve this by choosing a generator function of type F2 that satisfies ∂F H + 2 = 0. (105) ∂t

For historical reasons F2 is denoted by S (in fact it is equivalent to the action) S = S(q, P, t). (106) In Hamilton-Jacobi theory, the function S is called Hamilton’s principle function.

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 43 / 59 We can then use the transformation equations to state ∂S pi = (107) ∂qi ∂S Qi = (108) ∂Pi ∂S K = H + . (109) ∂t From Equation 105 we can write ∂S H(q ,..., q , p ,..., p , t) + = 0. (110) 1 n 1 n ∂t

Replacing the pi s we obtain the Hamilton-Jacobi equation. ∂S ∂S ∂S H(q1,..., qn, ,..., , t) + = 0. (111) ∂q1 ∂qn ∂t

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 44 / 59 Phase space

In , the canonical momenta pi = δL are promoted to coordinates on equal footing with the qi . The coordinates (q, p) are canonical variables, and the space of canonical variables is known as phase space.

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 45 / 59 Symplecticity

A symplectic transformation M satisfies

MT ΩM = Ω (112)

where  0 I  Ω = (113) −I 0 Hamilton’s equations in matrix form are

    ∂H ! q˙i 0 1 ∂qi = ∂H (114) p˙i −1 0 ∂pi

or in vector form ζ˙ = Ω∇H(ζ) = ΩJζ (115) where ζ is the vector of phase space coordinates.

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 46 / 59 This has solution tΩJ ζ(t) = Mz(t0) = e (116) From here its easy to show

MT (t)ΩM(t) = e−tΩJ ΩetΩJ = e−tΩJ etΩJ Ω = Ω (117)

In Hamiltonian systems the equations of motion generate symplectic maps of coordinates and momenta and as a consequence preserve volume in phase space. This is equivalent to Liouville theorem which asserts that the phase space distribution function is constant along the of the system.

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 47 / 59 Action-angle variables

A Hamiltonian system can be written in action-angle form if there is a set of canonical variables (θ, I ) such that H only depends on the action

H = H(I ) (118)

Then

θ˙ = ∇H(I ) = Ω(I ), I˙ = 0 (119)

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 48 / 59 Liouville Integrability

The Liouville-Arnold theorem states that existence of n invariants of motion is enough to fully characterize a for an n degree-of-freedom system. In that case a canonical transformation exists to action angle coordinates in which the Hamiltonian depends only on the action.

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 49 / 59 Henon-Heiles system

The H´enon-Heilespotential can be written 1 1 V (x, y) = x2 + y 2 + x2y − y 3 (120) 2 3 with Hamiltonian 1 1 H = p2 + p2 + x2 + y 2 + x2y − y 3 = E (121) 2 x y 3 The Hamiltonian is integrable only for limited number of initial conditions.

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 50 / 59 The motion is bounded for energy E ≤ 1/6. As E increases, the dynamics look increasingly chaotic3.

3S. A. Antipov and S. Nagaitsev, Proc. of IPAC2017, Cophenhagen, Denmark, WEOAB1. David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 51 / 59 Poisson brackets

Let p and q be canonical variables and let u and v be functions of p and q. The of u and v is defined as ∂u ∂v ∂u ∂v [u, v] = − . (122) p,q ∂q ∂p ∂p ∂q Generalising to a system of n variables this becomes

X  ∂u ∂v ∂u ∂v  [u, v] = − . (123) ∂qi ∂pi ∂pi ∂qi i Using the Einstein summation convention this is just ∂u ∂v ∂u ∂v [u, v] = − . (124) ∂qi ∂pi ∂pi ∂qi

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 52 / 59 From the definition of the Poisson bracket

[qi , qj ] = [pi , pj ] = 0 (125)

[qi , pj ] = −[pi , qj ] = δi,j . (126)

A Poisson bracket is invariant under a change in canonical variables

[u, v]p,q = [u, v]P,Q . (127)

In other words, Poisson brackets are canonical invariants, which gives us an easy way to determine whether a set of variables is canonical.

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 53 / 59 Equations of motion with brackets

Hamilton’s equations may be written in terms of Poisson brackets For a function u = u(qi , pi , t) the total differential is du ∂u ∂u ∂u = q˙i + p˙i + . (128) dt ∂qi ∂pi ∂t

We can replaceq ˙i andp ˙i with their Hamiltonian solutions to obtain du ∂u ∂H ∂u ∂H ∂u = − + (129) dt ∂qi ∂pi ∂pi ∂qi ∂t which is just du ∂u = [u, H] + . (130) dt ∂t

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 54 / 59 du ∂u If u is constant, then dt = 0 and [u, H] = − ∂t . If u does not depend explicitly on t [u, H] = 0. If u = q q˙ = [q, H]. (131) If u = p p˙ = [p, H]. (132) Which are just the equations of motion in terms of Poisson brackets.

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 55 / 59 Lie Transformations Suppose we have some function of the phase space variables

f = f (xi , pi ) (133)

which has no explicit dependence on the independent variable, s. However if we evaluate f for a particle moving along a beamline, the value of f will evolve with s as the dynamical variables evolve. The rate of change of f with s is

n df X dxi ∂f dpi ∂f = + . (134) ds ds ∂xi ds ∂pi i=1 Using Hamilton’s equations

n df X ∂H ∂f ∂H ∂f = − . (135) ds ∂pi ∂xi ∂xi ∂pi i=1

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 56 / 59 We now define the Lie : g : for any function g(xi , pi )

n X ∂g ∂ ∂g ∂ : g := − . (136) ∂xi ∂pi ∂pi ∂xi i=1 Compare with the definition of a Poisson bracket ∂u ∂v ∂u ∂v [u, v] = − . (137) p,q ∂q ∂p ∂p ∂q If the Hamiltonian H has no explicit dependence on s we can write df = − : H : f . (138) ds

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 57 / 59 We can express f at s = s0 + ∆s in terms of f at s = s0 in terms of a Taylor series

2 2 df ∆s d f f |s=s +∆s = f |s=s + ∆s + + ... (139) 0 0 ds 2 ds2 s=s0 s=s0 ∞ m m X ∆s d f = (140) m! dsm m=0 s=s0 ∆s d = e ds f . (141) s=s0 This suggests the solution for equation 138 can be written as

−∆s:H: f |s=s +∆s = e f . (142) 0 s=s0

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 58 / 59 The operator e−∆s:g: is known as a Lie transformation, with generator g. In the context of accelerator beam dynamics, applying a Lie transformation with the Hamiltonian as the generator to a function f produces a transfer map for f . f can be any function of the dynamical variables Any Lie transformation represents the evolution of a conservative dynamical system, with Hamiltonian corresponding to the generator of the Lie transformation The map represented by a Lie transformation must be symplectic

David Kelliher (RAL) Hamiltonian Dynamics November 12, 2019 59 / 59