Ch. 8: Hamilton Equations of Motion Sect. 8.1: Legendre Transformations Lagrange Eqtns of motion: n degrees of freedom (d/dt)[(∂L/∂q i )] - (∂L/∂q i ) = 0 (i = 1,2,3, … n) n 2 nd order, time dependent, differential equations.  The system motion is determined for all time when 2n initial values are specified: n q i ’s & n q i ’s We can represent the state of the system motion by the time dependent motion of a point in an abstract n-dimensional configuration space (coords = n generalized coords q i ). PHYSICS: In the Lagrangian Formulation of Mechanics, a system with n degrees of freedom = a problem in n independent variables q i (t). The generalized velocities, q i (t) are simply determined by taking the time derivatives of the q i (t). The velocities are not independent variables.

Hamiltonian Formulation of Mechanics Hamiltonian Mechanics: A fundamentally different picture! Describes the system motion in terms of 1 st order, time dependent equations of motion. The number of initial conditions is, of course, still 2n.  We must describe the system motion with 2n independent 1 st order, time dependent, differential equations expressed in terms of 2n independent variables. We choose n of these = n generalized coordinates q i. We choose the other n = n generalized (conjugate) momenta p i.

Hamiltonian Mechanics: Describes the system motion in terms of n generalized coordinates q j & n generalized momenta p i. It gets 2n 1 st order, time dependent equations of motion. Recall that by DEFINITION: The generalized Momentum associated with the generalized coordinate q j : p i  (∂L/∂q i ) (q,p)  “conjugate” or “canonical” variables. –See footnote, p 338, which discusses the historical origin of the word “canonical”.

Legendre Transformations Physically, the Lagrange formulation assumes the coordinates q i are independent variables & the velocities q i are dependent variables & only obtained by taking time derivatives of the q i once the problem is solved! Mathematically, the Lagrange formalism treats q i & q i as independent variables. e.g., in Lagrange’s equations, (∂L/∂q i ) means take the partial derivative of L with respect to q i keeping all other q’s & ALSO all q’s constant. Similarly (∂L/∂q i ) means take the partial derivative of L with respect to q i keeping all other q’s & ALSO all q’s constant. Treated as a pure math problem, changing from the Lagrange formulation to the Hamilton formulation corresponds to changing variables from (q,q,t) (q,q, independent) to (q,p,t) (q,p independent)

To change from the Lagrange to the Hamilton formulation  Change (or transform) variables from (q,q,t) (q,q, independent) to (q,p,t) (q,p independent). Mathematicians call such a procedure a Legendre Transformation. Pure math for a while: Consider a function f(x,y) of 2 independent variables (x,y) The exact differential of f: df  u dx + v dy Obviously: u  (  f/  x) v  (  f/  y) Now, change variables to u & y, so that the differential quantities are expressed in terms of du & dy. Let g = g(u,y) be a function defined by g  f - ux

Change from f(x,y)  df  u dx + v dy u  (  f/  x) v  (  f/  y) To g(u,y)  f - ux. The exact differential of g: dg  df - u dx - x du = v dy – x du Obviously: v  (  g/  x) x  - (  g/  u) This is a Legendre Transformation. Such transformations are used often in thermodynamics. See examples in Goldstein, pp 336 & 337.

Change from the Lagrange to the Hamilton formulation.  Changing variables from (q,q,t) (q,q, independent) to (q,p,t) (q,p independent) is a Legendre Transformation. However, it’s one where many variables are involved instead of just 2. Consider the Lagrangian L = L(q,q,t) (n q’s, n q’s) The exact differential of L (sum on i): dL  (  L/  q i )dq i + (  L/  q i )dq i + (  L/  t)dt (1) Canonical Momentum is defined: (d/dt)[(∂L/∂q i )] - (∂L/∂q i ) = 0 p i  (  L/  q i )  p i  (  L/  q i ) (2) Put (2) into (1):  dL = p i dq i + p i dq i + (  L/  t)dt (3)

dL = p i dq i + p i dq i + (  L/  t)dt (3) Define the Hamiltonian H by the Legendre Transformation: (sum on i) H(q,p,t)  q i p i – L(q,q,t) (4)  dH = q i dp i + p i dq i – dL (5) Combining (3) & (5):  dH = q i dp i - p dq i - (  L/  t)dt (6) Since H = H(q,p,t) we can also write: dH  (  H/  q i )dq i + (  H/  p i )dp i + (  H/  t)dt (7) Directly comparing (6) & (7)  q i  (  H/  p i ), - p i  (  H/  q i ), - (  L/  t)  (  H/  t)

Hamiltonian H: (sum on i) H(q,p,t)  q i p i – L(q,q,t) (a)  q i  (  H/  p i ) (b) - p i  (  H/  q i ) (c) - (  L/  t)  (  H/  t) (d) (b) & (c) together  Hamilton’s Equations of Motion or the Canonical Equations of Hamilton 2n 1 st order, time dependent equations of motion replacing the n 2 nd order Lagrange Equations of motion

Hamiltonian: H(q,p,t) = q i p i – L(q,q,t) (a) Hamilton’s Equations of Motion: q i = (  H/  p i ) (b), - p i = (  H/  q i ) (c), -(  L/  t) = (  H/  t) (d) 2n 1 st order, time dependent equations of motion replacing the n 2 nd order Lagrange Eqtns of motion. (a): A formal definition of the Hamiltonian H in terms of the Lagrangian L. However, as we’ll see, in practice, we needn’t know L first to be able to construct H. (b): q i = (  H/  p i ): Gives q i ’s as functions of (q,p,t).  Given initial values, integrate to get q i = q i (q,p,t)  They form the “inverse” relations of the equations p i = (  L/  q i ) which give p i = p i (q,q,t).  “No new information”. Discussion of Hamilton’s Eqtns

Hamiltonian: H(q,p,t) = q i p i – L(q,q,t) (a) Hamilton’s Equations of Motion: q i = (  H/  p i ) (b), -p i = (  H/  q i ) (c), - (  L/  t) = (  H/  t) (d) (b): q i = (  H/  p i ):  q i = q i (q,p,t).  “No new info”. –This is true in terms of SOLVING mechanics problems. However, within the Hamiltonian picture of mechanics, where H = H(q,p,t) is obtained NO MATTER HOW (not necessarily by (a)), this has equal footing (& contains equally important information as (c)). (c): p i = - (  H/  q i ):  Given the initial values, integrate to get p i = p i (q,p,t) (d): -(  L/  t) = (  H/  t): This is obviously only important in time dependent problems!

Recall the “energy function” h from Ch. 2 (Eq. (2.53): Define the Energy Function h: h  q i (  L/  q i ) - L = h(q 1,..q n,q 1,..q n,t) The Hamiltonian H & the energy function h have identical (numerical) values. However, they are functions of different variables! h  h(q,q,t) while H = H(q,p,t) NOTE!!!!! A proper Hamiltonian (for use in Hamiltonian dynamics) is ALWAYS (!!!!) written as a function of the generalized coordinates & momenta: H  H(q,p,t). Similarly, a proper Lagrangian (for use in Lagrangian dynamics) is ALWAYS (!) written as a function of the generalized coordinates & velocities: L  L(q,q,t)

TO EMPHASIZE THIS: Consider a single free particle (p = mv): Energy = KE = T = (½)mv 2 only. So, h = T and H = T But, if it is a PROPER HAMILTONIAN (!!!), can it be written H = (½)mv 2 ? NO!!!!!! H MUST be expressed in terms of the momentum p, NOT the velocity v! So the PROPER HAMILTONIAN(!!!) is H = p 2 /(2m) !!!!!

Hamiltonian: H(q,p,t) = q i p i – L(q,q,t) (a) Hamilton’s Equations of Motion: q i = (∂H/∂p i ) (b), - p i = (∂H/∂q i ) (c), - (∂L/∂t) = (∂H/∂t) (d) Recipe: (CONSERVATIVE FORCES!) 1. Set up the Lagrangian, L = T – V = L(q,q,t) 2. Compute n conjugate momenta using: p i  (∂L/∂q i ) 3. Form the Hamiltonian H from (a). This is of the “mixed” form H = H(q,q,p,t) 4. Invert the n p i  (∂L/∂q i ) to get q i  q i (q,p,t). 5. Apply the results of 4 to eliminate the q i from H to get a proper Hamiltonian H = H(q,p,t). Then & only then can you properly & correctly use (b) & (c) to get the equations of motion! Recipe for Hamiltonian Mechanics

If you think that this is a long, tedious process, you aren’t alone! Personally, this is why I prefer the Lagrange method! –This requires that you set up the Lagrangian first! –If you already have the Lagrangian, why not go ahead & do Lagrangian dynamics instead of going through all of this to do Hamiltonian dynamics? –Further, combining the 2n 1 st order differential equations of motion q i = (  H/  p i ) (b) & - p i = (  H/  q i ) (c) gives the SAME n 2 nd order differential equations of motion that Lagrangian dynamics gives! However, for many physical systems of interest, it is fortunately possible to considerably shorten this procedure, even eliminating many steps completely!

We’ve seen (in Ch. 2) that in many cases: The Lagrangian = a sum of functions which are homogeneous in the generalized velocities of degree 0, 1, & 2. That is (schematically): L = L 0 (q,t) + L 1 (q,t)q k + L 2 (q,t)q k q m Use this form to construct the Hamiltonian: H = q i p i – L(q,q,t)  H = q i p i – [L 0 (q,t) + L 1 (q,t)q k + L 2 (q,t)q k q m ] We’ve also seen (in Ch. 2) that in many cases: The equations defining the generalized coordinates don’t depend on the time explicitly:  L 2 (q)q k q m = T (the kinetic energy) & L 1 = 0 We’ve also seen (in Ch. 2) that in many cases: The forces are conservative & a potential V exists:  L 0 = - V Hamiltonian Mechanics (In Most Cases of Interest!)

We’ve seen (in Ch. 2) that in many cases: All of the conditions on the previous slide hold simultaneously.  H = T + V That is, in this case, the Hamiltonian is automatically the total mechanical energy E If that is the case, we can skip many steps of the recipe and write H = T + V immediately. Express T in terms of the MOMENTA p i ( not the velocities q i !)! Often it is easy to see how the p i depend on the q i & thus its easy to do this. Once this is done, we can go ahead & do Hamiltonian Dynamics without ever having written the Lagrangian down!!

Often, we can go further! For large classes of problems, the 1 st & 2 nd degree Lagrangian terms can be written (sum on i): L 1 (q,t)q k + L 2 (q,t)q k q m = q i a i (q,t) + (q i ) 2 T i (q,t) So: L = L 0 (q,t) + q i a i (q,t) + (q i ) 2 T i (q,t) (A) If the Lagrangian can be written in the form of (A), we can do the algebraic manipulations in steps 2-5 in the recipe in general, once & for all. Do this by matrix manipulation!