PHYS 705: Classical Mechanics. Hamiltonian Formulation & Canonical Transformation

Transcription

1 1 PHYS 705: Classical Mechanics Hamiltonian Formulation & Canonical Transformation

2 Legendre Transform Let consider the simple case with ust a real value function: F x F x expresses a relationship between an independent variable x and its dependent value F This relation is encoded in the functional form of We will denote this encoding in general by: F, x F x As one has learned in math and physics, it is sometime useful to encode the information contained in a function in different ways e.g., Fourier Transform, Laplace Transform, etc.

3 3 Legendre Transform 1 F k e F x dx As in the Fourier Transform: ˆ ikx and its inverse transform: F x The original information of for a given x value is encoded in terms ˆF k of the Fourier components and a new set of variables k. In our new encoding scheme, we will say that 1 ikx Fx e Fˆ kdk Fk ˆ, encodes the SAME information as F, x

4 4 Legendre Transform The Legendre Transform is yet another (convenient) encoding scheme F x for when the following two conditions are true: 1. The function is strictly convex (concave), i.e, nd derivative always positive (negative) and smooth df dx. It is easier to measure, control, or think about than x itself df dx Since never changes sign (condition 1), there is a one-to-one df dx correspondence between and x Legendre Transform gives us a new encoding scheme with s df dx being a function of instead of x Gs, Gs

5 5 Legendre Transform Here is the definition of the Legendre Transform for F x G s G s sx s F x s Note that is a function of s and we have to express in df x terms of s by inverting the function: to get s dx F x s xs

6 6 Legendre Transform Geometric Construction of the Legendre Transform: G sxf F x slope = s F F G sx G G sx y-intercept of slope x

7 7 Legendre Transform Important Properties of the Legendre Transform: 1. The Legendre Transform is its own inverse transform: G s - Suppose we have and define s y H y ys y G s y y s dg s - Using the inverted relation, the Legendre Transform of is G syh ds G s - This can be rewritten as and comparing to our original H, y F, x G sx F transform, we can immediate see that F xsg and is F s own inverse transform.

8 8 Legendre Transform - The two independent variables are two conugate pair of variables related to each other through x s dg s ds or x, s - Note that there is only ONE independent variable (either x or s) in: F x G s x s x x s s s s F x G x s x df x dx

9 9 Legendre Transform. Properties of the Minima [note: is convex, an extrema is a minimum] F x F x At the minimum of we have the slope F min F x s df dx 0 F min So, we have Gs ( 0) x0f F min G 0 min

10 10 Legendre Transform. Properties of the Minima Similarly, at a minimum G min of We again have correspondingly Gs y-intercept coincides with tangent pt F x dg slope for G ds x 0 So, we have G min F 0 F 0 G min

11 11 Legendre Transform. Reciprocal Relation of F & G s curvature Start with the two reciprocal definitions x s dg s ds and s x df x Now take derivatives with respective to their independent variables, dx ds F x dg s ds ds dx dx d F x dx dx ds dx ds d F x d G s ds dx 1 G s So, if is with curvature, then will have curvature 1

12 1 Review of Lagrangian Formulation Recall Lagrangian Formulation of Mechanics -Pick n proper (independent) generalized coordinates to describe the state of the system this defines a n-dim Configuration Space. - Apply the Hamilton s Principle: The motion of the system from t 1 to t in config. space is such that the Action (I) has a stationary value, i.e., 1 I L q, q, t dt 0 - Using Variational Calculus, this implies the Euler-Lagrange Eq: L d L 0, 1,, n q dtq

13 13 Review of Lagrangian Formulation L d L 0, 1,, n q dtq Solving this gives the Equations of Motion Notes: -We have n independent generalized coordinates. q 1 n - The n are time derivatives of the. n q 1 -We have n nd -order ODEs for the equations of motion - We need n initial conditions to completely specify the motion. n q 1

14 14 Hamiltonian Formulation -Instead of n nd -order ODEs as EOM with n independent generalized coordinates in configuration space. - Seek n 1 nd -order ODEs as EOM with n independent generalized coordinates in phase space. (Note: In Hamiltonian Formulation, we need to start with a set of independent generalized coordinates. If they are not proper, a reduced set of m < n proper coordinates must be chosen first.) - The most natural choice for this set of n variables in phase space is: n q 1 generalized coordinates n p 1 where p L q generalized conugate momenta

15 15 Hamiltonian Formulation Notes: - As we will see this is NOT the only choice - Coordinates in this n-dim phase space are call canonical variables (In Landau & Lifshitz: they are called canonical since the EOM resulted from them has the simplicity and symmetry in form.) - Later, we will investigate invariant ways to transform the system to other n canonical variables using canonical transformations

16 16 Configuration Space vs. Phase Space 1 q q A given point in configuration space prescribes fully the 1,, n configuration of the system at a given time t. However, the specification of a point in this space does NOT specify the time evolution of the system completely! (a unique soln for a n-dim nd order ODE needs n ICs) Many different paths can go thru a given point in config space q P Different paths crossing P will have n n q 1 q 1 the same set of but diff q 1

17 17 Configuration Space vs. Phase Space To specify the state AND time evolution of a system uniquely at a given time, one needs to specify BOTH AND q, p or equivalently, The n-dim space where both and are independent variables is called phase space. q q q p Thru any given point in phase space, there can only be ONE unique path! p GOAL: to find the EOM that applies to points in phase space. q

18 18 Hamiltonian Formulation - Instead of using the Lagrangian,, we will introduce a new function that depends on,, and t: L Lq, q, t q p H H q, p, t - This new function is call the Hamiltonian and it is defined by: L H q L q - Plugging in the definition for the generalized momenta: H p q L One can think of this as a coordinate transformation from to To be more specific, is the Legendre Transform of. H is defined similarly to the Jacobi (energy) function h BUT h is a q, q q, p (Einstein s Convention: function of and H is a function of. Repeated indices are summed) ( sum) p L q Hq, p Lqq, q, q q, p

19 19 L and H are a Legendre Transform Pair Consider a simple Lagrangian for a single particle m under the influence of a conservative potential Then, we have mq V q L p mq dq, L qq V q Inverting the above relation and arriving at Then, the Legendre Transform of is: L qq,,, H q p pq p L q q p q p p m L, q H, p p m p p p V q V q m m m xq sl q p F L G H G sxf H pq L (q is an irrelevant variable in this L-Trans)

20 0 Hamiltonian Formulation - Taking the differential of our definition for, we have - Now, we require that so that we should have - To be consistent, let try to resolve this by expanding dl: p - Plug in : dh p dq q dp dl H H q, p, t H H H dh dq dp dt q p t L L L dl dq dq dt ( sum) q q t L q L L dl dq p dq dt ( sum) q t H pq L ( sum) ( sum)

21 1 Hamiltonian Formulation - Now, since the system satisfies the Euler-Lagrange Equation, we have L d L q dt q - Substituting this into our previous equation, we have: L dl p dq p dq dt t - Now, plug this into our equation for dh: p (** this is an important step if we want our formalism to describe the same dynamic.) ( sum) dh q q dq p dq dp pd p L dh qdp p dq dt ( sum) t L dt t

22 Hamiltonian Formulation -If H describes the same dynamics as is given by the EL equation, the two expressions must equal to each other: L H H H dh p dq qdp dt dh dq dp dt t q p t - Comparing gives the condition: H q H p p q and H t L t These are called the Hamilton s Equations of Motion and they are the desired set of equations describing the EOM in phase space.

23 3 Hamiltonian Formulation Summary of Steps: 1. Pick a proper set of and form the Lagrangian L. Obtain the conugate momenta by calculating 3. Form 4. Eliminate from H using the inverse of so as to have 5. Apply the Hamilton s Equation of Motion: q H pq L q H H( q, p, t) H q H p and q p p L q As you will see, this formulation does not necessary simplify practical calculations but it forms a theoretical bridge to QM and SM. p L q

24 4 Hamilton Equations in Matrix (Symplectic) Notation The pair of Hamilton equations look almost symmetric (except the - sign). The following is an elegant way to write the Hamilton equations into a single matrix equation: For a system with n dofs, we group all of our q i s and p i s into a n-dim vector η: q, p ; 1,, n n Similarly, we will define another n-dimensional column vector : H H H H, ; 1,, n η q η p n H η

25 5 Hamilton Equations in Matrix (Symplectic) Notation n n Now, we define a anti-symmetric matrix, 0 I J I 0 Note that the transpose of J is its own inverse (orthogonal): J T 0 I I 0 and In summary, we have the following properties for J: T 1, J J J J n n where I is a identity matrix 0 is a n n zero matrix T T I 0 JJJJ 0 I J I, J det 1

26 6 Hamilton Equations in Matrix (Symplectic) Notation The Hamilton equation can then be written in a compact form: H η J η As an example, with only dofs, this matrix equation expands into: q H q p1 q H q p p H p q 1 p H p q This is typically referred to as the matrix or symplectic notation for the Hamilton equations.

27 7 Example: Spherical Pendulum Thus, ds U T mv mb mb E-L equations gives two nd order ODEs: Pick Generalized Coordinates:, b dsbd θˆ bsin dˆ m ds v b θˆ bsin ˆ dt sin 1 1 cos L T U mb mb sin mgb g sincos sin 0 b sin sincos 0 U mgbcos

28 8 Example: Spherical Pendulum U 0 Practically, we can do one integration immediately b through the two constants of motion: ds m Conservation of E: 1 1 E mb mb sin mgbcos d L 0 dt L p mb sin const is cyclic, i.e : These two 1 st order ODEs can be solved easily by substitute from the equation into the E equation. p

29 9 Example: Spherical Pendulum U cos L mb mb sin mgb Now, let consider the Hamiltonian Formalism. b ds m We first need to calculate the conugate momenta: L p p mb mb L p mb sin mb p sin The boxed equations are the desired transformation from the config. space to the canonical variables in phase space.

30 30 Example: Spherical Pendulum Construct the Hamiltonian: H q p L sum H p p L p p 1 1 mb mb sin mb mb sin mgb cos p p p p mgb cos mb mb sin mb mb sin p p mb mb sin mgbcos U 0 b ds m (H is properly expressed in terms of the canonical variables.),, p, p, t

31 31 Example: Spherical Pendulum H p p mb mb sin mgbcos U 0 Apply Hamilton s Equations to get four 1 st order ODEs: H p 3 p sin cos mgbsin p mb p cos p mgbsin 3 mb sin H p p 0 H p H p p mb mb p sin Just repeated what we had b ds Since is cyclic, P is a constant and can be integrated to get (t). We will exploit this later. m

32 3 Symmetry and Conservation Theorem Again - Consider the full-time derivative of H, dh H H H q p dt q p t Using the Hamilton s Equations: dh dt pq q p H t dh dt H t (so if time does not explicitly appears in H (time is cyclic), H is conserved!) Note: H = E if 1. U does not depend on the generalized velocities. Transformation defining does not depend on t explicitly. q

33 33 Subtle Difference between h and H Recall that the Jacobi integral h is a function of instead of q, q, t q, p, t L hq, q, t q L q Also, from the E-L Eq, we know that Thus, we can rewrite the above equation as, h L L q t tq t L dh t dt h L q t t q dh dt Thus, unless the red term is explicitly zero, h t 0 dh dt 0

34 34 Subtle Difference between h and H Example: 1 L mq t q L hq L q L mq qt V q 1 1 mq qt So, hqmqt mq qt V q mq 1 tq h mq V q V q

35 35 Subtle Difference between h and H 1 L mq qt V q 1 h mq V q So, L q t 0 h t but 0 From our previous calculations, we have the following relationship, h L dh q t tq dt 0 So, the LHS is zero and we can also explicitly check that the RHS=0 too dh 0 dt but with.

36 36 Subtle Difference between h and H h t q t L q dh dt L dh d 1 q q tmq t q mq t dt dt q V mqq q L dh V q qmq 0 tq dt q Thus, we have, h t 0 V q q This has to be zero since and indeed, it does. From EL Eq, we have d L L dt q q 0 d dt V q tmq 0 V mq 0 q

37 37 Subtle Difference between h and H 1 L mq qt V q So, for this specific example, we have 1 h mq V q h t 0 but L t 0 q One can also calculate using H H q, p, t 0 dh and 0 dt V mqq q q L p tmq q pt H dh V q 0 m t dt 0

38 38 Symmetry and Conservation Theorem Again - Other cyclic coordinates: q n missing Lagrangian Formalism q n If is cyclic, the Lagrangian is L q n const L L( q,, q,, q,, q, t) 1 n1 1 The n-th EOM is then given by: q n is still here (The LHS is usually a complicated function of q & q and much efforts still needed to solve for q () n t.) n Hamiltonian q n In contrast, if is cyclic in the Hamiltonian Formalism, H q n 0 pn p const (conserved) (This equation is immediately solved from IC n p n is gone from the prob.) (The Hamiltonian Formalism has a much nicer structure for cyclic coords)

39 39 Routh s Procedure Goal: Transform only the cyclic coordinates to take advantage of simplicity in phase space for the cyclic variables (missing) L L( q,, q,, q,, q, t) Let say we have: 1 s 1 n s1 n Define the Routhian, R q p R pq L s1 where R depends on where q,, q are cyclic q, 1, qs d R R 0 for 1,, s dt q q q,, s1 q R p 0 and q for s 1,, n p const similar to Hamiltonian q,, q, q,, q, p,, p, t 1 s 1 s s1 n Then, apply the EL equation to non-cyclic variables, And, apply the Hamilton s equation to cyclic variables, n n

40 40 Routh s Procedure: example A single particle in a central potential: V() r k r n From previous chapter, we have: is cyclic and r is not, so we define m m k L r r r p L mr n p mr The Routhian is: R pq L n p m m p k p m k R r r r mr mr r mr r n Depends on rrp,, only as advertised

41 41 Routh s Procedure: example p m k R r n mr r Applying the EL equation to the non-cyclic variable r : d R R d p 1 k 0 mr 3 1 n 0 n dt r r dt m r r p nk mr mr r 0 3 n1 And, the Hamilton s Equation for : R p 0 p const l R p p mr l mr

42 4 Cyclic Variables: Action-Angle Coordinates - Note that a given system can be described by several different sets of generalized coordinates Generalized coordinates are not unique! - Recall also that the # of cyclic variables can depend on the choice of the generalized coordinates e.g., in the previous central force problem: Rect coord (x, y): both x, y are not cyclic polar coord (r, ): is cyclic

43 43 Cyclic Variables: Action-Angle Coordinates - Although one might not be able to pick a set of generalized coordinates (from a given physical system) to have ALL q being cyclic, one can imagine transforming them to an ideal set such that they are all cyclic. Canonical Transformation (next Chapter) - If possible, then, all the conugate momenta are constant: additionally, if H is a constant of motion, then p const H H,, 1 n cannot depend on t and q explicitly! constant in time cyclic

44 44 Cyclic Variables: Action-Angle Coordinates consequently, the EOM for the are simple: q q H H func of consts, 1 n p q () t t are integration constants depending on IC - Recall that the natural choice of the n-dim phase space variables is with q being the regular generalized coordinates and being their conugate momenta. BUT, this is NOT the only choice! The Hamiltonian Formalism can be extended to other possibilities:,,,, Q Q q p t and P P q p t p (indices for q, p are suppressed here)

45 45 Cyclic Variables: Action-Angle Coordinates Q Q, P, P,,,, Q Q q p t and P P q p t are the canonically transformed variables from the original ones needs to satisfy the corresponding Hamilton s Equations in the transformed coordinates. There are no preference between the transformation above for & P The canonical transformation treats both equally. Q Q & P Q & P are on the same theoretical footing in the Hamiltonian Formalism! The Hamiltonian Formalism is the starting point in analyzing QM systems.

46 46 Connection to Statistical Mechanics - The Hamilton s Equations describe motion in phase space q, - a point in phase space uniquely determines the state of the system p AND its future evolution. - nearby points represent system states with similar but slightly different initial conditions. - One can imagine a cloud of points bounded by a closed surface S with nearly identical initial conditions moving in time. q0, p0 q', p' S t q, p q', p' t t

47 47 Connection to Statistical Mechanics -Let say at time t, a small subset of these points crosses a differential area da on S in a given outward normal direction ˆn such that da daˆ n S da ˆn t t - Let v q, p be the mean velocity for points in da - Then, after a given time dt, this small subset of points will trace out a differential volume in phase space given by vdt da v vdt da

48 48 Connection to Statistical Mechanics - Summing up all points in the cloud bounded by S, we have, or, dv v d a dt S dv dt v d a S surface integral of the velocity field v for the points in the cloud rate of change of phase-space volume due to the motion of the points - Then, by Gauss s Law, we have dv dt v d a v S V dv V enclosed by S

49 49 Liouville s Theorem dv dt v d a v - In phase space, we have -So that, V S V dv v qqˆ q qˆ p pˆ p pˆ q q 1 1 n n 1 1 n n k k v dv dv (*) V p p k k k - Using the Hamilton s Equations, we have: q k H H q q p q p k k k k k p k H H pk pk q p q k k k

50 50 Liouville s Theorem - Substituting into Eq (*), we have, dv H dt k qkp V k p k H q k dv (H is a smooth function) dv dt 0! This is the Liouville s Theorem: collection of phase-space points move as an incompressible fluid. Phase space volume occupied by a set of points in phase space is constant in time.

51 51 Liouville s Theorem This is the starting point for statistical mechanics - Imagine many (N) identical mechanical systems but with different initial conditions (ensemble of systems). - Each is a different point in phase space with q, p - Statistical properties can be specified by a density of states function per unit volume in phase space q, p, t dv # system points in phase space volume dv q, located at at time t. p

52 5 Liouville s Theorem At statistical equilibrium, The # of points in the ensemble (N) does not change Then, since N is fixed, V stays constant, = N/V = constant as well! Thus, Liouville s Thm implies: density in a neighborhood of any system state = const as the system evolutes in phase space. Thus, equilibrium is uniform along the flow lines of the system points. (SM s Master Eq. or Fokker-Planck Eq.) This condition is typically used to solve for the equilibrium distribution eq for a statistical mechanical problem in phase space with which various statistical averages, P, T, U, S, can be calculated.