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 Classical Mechanics - 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 Energy 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 space The state of the system at a time 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 motion 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 mass m subject to a force 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, velocity 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 Lagrangian Mechanics 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 trajectory 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 coordinate system (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 energies 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 momentum 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 Legendre transformation 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 conjugate variables 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 equations of motion 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 canonical transformation and it depends on old and new phase space 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.