Ch. 8: Hamilton Equations of Motion Sect. 8.1: Legendre Transformations

2. Ch. 8: Hamilton Equations of MotionSect. 8.1: Legendre Transformations • Lagrange Eqtns of motion: n degrees of freedom (d/dt)[(∂L/∂qi)] - (∂L/∂qi) = 0(i = 1,2,3, … n) • n 2nd order, time dependent, differential equations.  The system motion is determined for all time when 2n initial values are specified: n qi’s & n qi’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 qi). • PHYSICS:In the Lagrangian Formulation of Mechanics,a system with n degrees of freedom = a problem inn independent variables qi(t). The generalized velocities, qi(t) are simply determined by taking the time derivatives of the qi(t). The velocities are not independent variables.

3. Hamiltonian Formulation of Mechanics • Hamiltonian Mechanics: A fundamentally different picture! • Describes the system motion in terms of 1st order, time dependent equations of motion. The number of initial conditions is, of course, still 2n. • We must describe the system motion with2n independent1st order, time dependent, differential equations expressed in terms of2n independent variables. • We choose n of these =n generalized coordinates qi. • We choose the other n =n generalized (conjugate) momenta pi.

4. Hamiltonian Mechanics: • Describes the system motion in terms of n generalized coordinates qj & n generalized momenta pi. It gets 2n1st order, time dependent equations of motion. • Recall that byDEFINITION:The generalized Momentumassociated with the generalized coordinate qj: pi (∂L/∂qi) • (q,p)  “conjugate” or “canonical” variables. • See footnote, p 338, which discusses the historical origin of the word “canonical”.

5. Legendre Transformations • Physically,the Lagrange formulation assumes the coordinates qiare independent variables & the velocities qi are dependent variables & only obtained by taking time derivatives of the qi once the problem is solved! • Mathematically, the Lagrange formalism treats qi& qias independent variables. e.g., in Lagrange’s equations, (∂L/∂qi) means take the partial derivative of L with respect to qikeeping all other q’s &ALSOall q’s constant. Similarly (∂L/∂qi) means take the partial derivative of L with respect to qikeeping all other q’s &ALSOall 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)

6. 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 functionf(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

7. 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.

8. 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 aLegendre 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/qi)dqi + (L/qi)dqi + (L/t)dt (1) • Canonical Momentumis defined:(d/dt)[(∂L/∂qi)] - (∂L/∂qi) = 0 pi (L/qi)  pi (L/qi) (2) • Put (2) into (1): dL= pidqi + pidqi + (L/t)dt (3)

9. dL= pidqi + pidqi + (L/t)dt (3) • DefinetheHamiltonian Hby theLegendre Transformation:(sum on i) H(q,p,t)  qipi– L(q,q,t) (4)  dH = qidpi + pidqi – dL (5) • Combining (3) & (5):  dH = qidpi - p dqi - (L/t)dt (6) • Since H = H(q,p,t) we can also write: dH  (H/qi)dqi + (H/pi)dpi + (H/t)dt (7) • Directly comparing (6) & (7)  qi(H/pi), - pi(H/qi), - (L/t)  (H/t)

10. Hamiltonian H:(sum on i) H(q,p,t)  qipi– L(q,q,t) (a)  qi(H/pi) (b) - pi(H/qi) (c) - (L/t)  (H/t) (d) • (b) & (c) together  Hamilton’s Equations of Motion or theCanonical Equations of Hamilton • 2n 1st order, time dependent equations of motionreplacing then 2nd orderLagrange Equations of motion

11. Discussion of Hamilton’s Eqtns • Hamiltonian: H(q,p,t) = qipi– L(q,q,t) (a) Hamilton’s Equations of Motion: qi=(H/pi) (b), - pi =(H/qi) (c), -(L/t) = (H/t) (d) • 2n 1st order, time dependent equations of motion replacing the n 2nd 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): qi=(H/pi): Gives qi’s as functions of (q,p,t). Given initial values, integrate to getqi= qi(q,p,t) They form the “inverse” relations of the equations pi= (L/qi) which give pi= pi(q,q,t). “No new information”.

12. Hamiltonian: H(q,p,t) = qipi– L(q,q,t) (a) Hamilton’s Equations of Motion: qi=(H/pi) (b), -pi =(H/qi) (c), - (L/t) = (H/t) (d) • (b): qi=(H/pi): qi= qi(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): pi= - (H/qi):  Given the initial values, integrate to getpi= pi(q,p,t) • (d): -(L/t) = (H/t): This is obviously only important in time dependent problems!

13. Recall the “energy function” h from Ch. 2 (Eq. (2.53):Define the Energy Functionh: h  qi(L/qi) - L = h(q1,..qn,q1,..qn,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 properHamiltonian(for use in Hamiltonian dynamics) isALWAYS (!!!!)written as a function of the generalized coordinates & momenta: H  H(q,p,t). Similarly, a properLagrangian(for use in Lagrangian dynamics) isALWAYS (!)written as a function of the generalized coordinates & velocities:L L(q,q,t)

14. TO EMPHASIZE THIS: Consider a single free particle (p = mv): Energy = KE = T = (½)mv2only. So, h = T and H = T But, if it is a PROPER HAMILTONIAN (!!!), can it be written H =(½)mv2? NO!!!!!!HMUST be expressed in terms of the momentum p, NOTthe velocity v! So the PROPER HAMILTONIAN(!!!)is H = p2/(2m) !!!!!

15. Recipe for Hamiltonian Mechanics • Hamiltonian: H(q,p,t) = qipi– L(q,q,t) (a) Hamilton’s Equations of Motion: qi=(∂H/∂pi) (b), - pi =(∂H/∂qi) (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:pi (∂L/∂qi) 3.Form the Hamiltonian H from (a). This is of the “mixed” form H = H(q,q,p,t) 4.Invert the npi (∂L/∂qi) to get qi qi(q,p,t). 5. Apply the results of 4to eliminate the qi from H to get a proper HamiltonianH = H(q,p,t). Then & only then can you properly & correctly use (b) & (c) to get the equations of motion!

16. 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 1st order differential equations of motion qi=(H/pi) (b) & - pi =(H/qi) (c) gives the SAME n 2nd 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!

17. Hamiltonian Mechanics (In Most Cases of Interest!) • 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 = L0(q,t) + L1(q,t)qk+ L2(q,t)qkqm • Use this form to construct the Hamiltonian: H = qipi– L(q,q,t)  H = qipi– [L0(q,t) + L1(q,t)qk+ L2(q,t)qkqm] • We’ve also seen (in Ch. 2) thatin many cases:The equations defining the generalized coordinates don’t depend on the time explicitly:  L2(q)qkqm = T(the kinetic energy) & L1= 0 • We’ve also seen (in Ch. 2)that in many cases: The forces are conservative & a potential V exists:  L0 = - V

18. We’ve seen (in Ch. 2) thatin 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 theMOMENTApi(not the velocities qi !)! Often it is easy to see how the pidepend on the qi& thus its easy to do this. Once this is done, we can go ahead & do Hamiltonian Dynamics without ever having written the Lagrangian down!!

19. Often, we can go further! For large classes of problems, the 1st & 2nd degree Lagrangian terms can be written (sum on i): L1(q,t)qk + L2(q,t)qkqm = qiai(q,t) + (qi)2Ti(q,t) So: L = L0(q,t) + qiai(q,t) + (qi)2Ti(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!