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Hamiltonian Mechanics (For Most Cases of Interest)

Hamiltonian Mechanics (For Most Cases of Interest). We just saw that, for large classes of problems, the Lagrangian terms can be written (sum on i ): L = L 0 (q,t) + q i a i (q,t) + (q i ) 2 T i (q,t) (A)

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Hamiltonian Mechanics (For Most Cases of Interest)

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  1. Hamiltonian Mechanics (For Most Cases of Interest) • We just saw that, for large classes of problems, the Lagrangian terms can be written (sum on i): 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 Hamiltonian mechanics recipe in general, once & for all! Do this by matrix manipulation, as follows: • Form all qiinto an n dimensional column vector q. Its transpose is a row vector q. Define also a column vector a = a(t), made up of the ai(q,t) from (A) & a n  n square matrix T = T(t), made up of the Ti(q,t).

  2. The Lagrangian can then be written: L = L(q,q,t) = L0(q,t) + qa(t) + (½)qT(t)q (B) • For example, in the special caseqi = (x,y,z) & T is diagonal: m 0 0 x (½)qT(t)q = (½)(x,y,z) 0 m 0 y = (½)m(x2+y2+z2) 0 0 m z and ax qa(t) = (x,y,z) ay = ax x + ay y + az z = ar az • The Hamiltonian in this notation is (using (B)): H = qp – L = q(p – a) – (½)qT(t)q – L0 (C)

  3. Lagrangian: L(q,q,t) = L0(q,t) + qa(t) + (½)qT(t)q (B) • Hamiltonian: H = q(p – a) – (½)qT(t)q – L0 (C) • Conjugate momentum:p = Tq + a • Generalized Velocity: q = T-1(p –a)(assuming T-1 exists!) Transpose of Generalized Velocity:q = (p - a)T-1 • Combine all this into Hamiltonian H(C) to eliminate the dependence of H on q & express it as a function of p only!  H = (½)(p – a)T-1(p –a) -L0(q,t) (D)  If the Lagrangian L can be written in the form (C), we can immediately skip all intermediate steps & write the Hamiltonian H in the form (D). That is: Steps 2-5 in the Hamiltonian mechanics recipe have been done in general, once & for all & can be skipped & (D) can be used immediately!

  4. Consider: H = (½)(p – a)T-1(p – a) -L0(q,t) (D) • Formally obtain the inverse KE matrixT-1as: T-1  (Tc)|T|-1 where |T| determinant of the KE matrix T Tc  cofactor matrix with elements = (Tc)jk = (-1)j+k|Mjk| where |Mjk| determinant of Mjk= matrix obtained from T by leaving out the jth row & the kthcolumn  H = (½)(p – a)(Tc)|T|-1(p – a) -L0(q,t) (E) • Summary: If the Lagrangian is L(q,q,t) = L0(q,t) + qa(t) + (½)qT(t)q the Hamiltonian has the form (E). This happens for almost every case of practical interest! So, in MANY cases we can skip 4 of the 5 steps in the recipe & go directly to (E)!

  5. In the simple example from before: m 0 0 m-1 0 0 m2 0 0 T = 0 m 0  T-1 = 0 m-1 0 Tc = 0 m2 0 0 0 m 0 0 m-1 0 0 m2 |T| = m3 • EXAMPLE 1:A particle in a conservative Central Force Field, using the spherical coordinates(r,θ,). Potential V = V(r). KE: T = (½)mv2 = (½)m(r2 + r2θ2 + r22sin2θ) = T(r,r,θ,θ,) Clearly, since force is conservative, Hamiltonian H = T + V. However, is writing H directly in terms of the above T correct? NO!!!!!HMUST be expressed in terms of the momenta p, not the velocities(r,θ,)!

  6. V = V(r), T = (½)m(r2 + r2θ2 + r22sin2θ) = T(r,r,θ,θ,) H = T + V. We can either follow the matrix formalism or the recipe & find that the proper way to write H is: H(r,θ,pr, pθ,p) = (2m)-1[(pr)2 + (pθ)2r-2 + (p)2r-2sin-2 θ] + V(r) • The same problem in Cartesian Coordinates. (Sum on i = 1,2,3): The KE is then: T = (½)mv2 = (½)mxixi The PE is: V(r) = V([xixi]½) Is writing H = T + V directly in terms of the above T correct? NO!!!!!HMUST be expressed in terms of the momenta p, not the velocities(xi)! Can either follow the matrix formalism or the recipe & find that the proper way to write H is: H = (2m)-1(pipi) +V([xixi]½) = (2m)-1(pp) + V([xixi]½)

  7. We can take components of the vector prelative to any coordinate system we wish: spherical coordinates, etc. In spherical coordinates,(r,θ,) the components of p are denoted:(p)r, (p)θ, (p) • Note some notational confusion might arise here!Don’t confuse the canonical momenta with these components of the vector p!For example, the canonical momentum corresponding to the spherical coordinate θis(from the Lagrange or matrix formalisms): pθ (∂L/∂θ)  (Tq + a)θ Has units of & is angular momentum! The θ component of the vector p is (θ = unit vector): (p)θ pθHas units of & is linear momentum! Clearly, pθ (p)θ ! It’s similar, of course for the components!

  8. EXAMPLE 2:A nonrelativistic particle, mass m & charge q moving in an electromagnetic field. From Ch. 1, the Lagrangian (with velocity dependent potential) is: L = (½)mv2 - q + qAv (I) • The general Lagrangian we had was of the form (q here is a coordinate, NOT a charge!): L = L0(q,t) + qiai(q,t) + (qi)2Ti(q,t) (A) • Comparing gives: L0 = - qand qiai(q,t) = qAv • Using Cartesian coords, the Lagrangian (I) can be written (Sum on i = 1,2,3): L = (½)mxixi + qAixi- q (II)

  9. L = (½)mxixi + qAixi- q (II) • The 2nd term in L is obviously linear in xi.  The vector a in the formalism we just developed has elements qAi.  Hamiltonian H T + V. However, H = total energy, since for the EM field, the energy is determined by the scalar potential alone. • Canonical momenta (from the Lagrange or matrix formalisms): pi (∂L/∂xi)  (Tq + a)i = mxi + qAi(III) (Note: pi mxi !) • We had the Hamiltonian: H = (½)(p – a)T-1(p – a) -L0(q,t) (D) • Using (III) in (D) gives (Sum on i = 1,2,3): H = (pi – qAi)(pi – qAi)(2m)-1 + q With vectors, the proper Hamiltonian (for a particle in an EM field) is: H = (p – qA)2(2m)-1 + q (IV)

  10. Obviously, Hamilton’s equations aren’t symmetric in the coordinates q & the momenta p: qi(H/pi) pi - (H/qi) • Physicists like symmetry!!! • Various schemes have been devised to come up with more symmetric equations of motion. Goldstein discusses one scheme: • n degrees of freedom. Construct a column matrix η with 2n elements, the 1stn = the q’s & the 2ndn = the p’s: ηi qiηi+n pi i  n Also: define the column vector (H/η): (H/η)i = (H/qi), (H/η)i+n = (H/pi) i  n

  11. Define a 2n  2n square matrix made of 4 n  n square matrices 1, -1, & 0: 0 1 0 -1 J  -1 0 transpose: J = 1 0 Clearly: JJ = JJ = 1 = 1 0 Also: J = -J = J-1; J2 = -1 0 1 |J| = 1 In this notation, Hamilton’s eqtns (matrix form) become: η= J(H/η) (1) • Example:n = 2: Coords q1, q2, momenta p1, p2, (1) becomes (using Hamilton’s eqtns): • What advantage does this have ? I don’t see any!

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