2. 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) 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 q i into 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 a i (q,t) from (A) & a n  n square matrix T = T(t), made up of the T i (q,t).

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

4. Lagrangian: L(q,q,t) = L 0 (q,t) + qa(t) + (½)qT(t)q (B) Hamiltonian: H = q(p – a) – (½)qT(t)q – L 0 (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) - L 0 (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!

5. Consider: H = (½)(p – a)T -1 (p – a) - L 0 (q,t) (D) Formally obtain the inverse KE matrix T -1 as: T -1  (T c )|T| -1 where |T|  determinant of the KE matrix T T c  cofactor matrix with elements = (T c ) jk = (-1) j+k |M jk | where |M jk |  determinant of M jk = matrix obtained from T by leaving out the j th row & the k th column  H = (½)(p – a)(T c )|T| -1 (p – a) - L 0 (q,t) (E) Summary: If the Lagrangian is L(q,q,t) = L 0 (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)!

6. In the simple example from before: m 0 0 m -1 0 0 m 2 0 0 T = 0 m 0  T -1 = 0 m -1 0 T c = 0 m 2 0 0 0 m 0 0 m -1 0 0 m 2 |T| = m 3 EXAMPLE 1: A particle in a conservative Central Force Field, using the spherical coordinates (r,θ,  ). Potential V = V(r). KE: T = (½)mv 2 = (½)m(r 2 + r 2 θ 2 + r 2  2 sin 2 θ) = 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!!!!! H MUST be expressed in terms of the momenta p, not the velocities (r,θ,  )!

7. V = V(r), T = (½)m(r 2 + r 2 θ 2 + r 2  2 sin 2 θ) = 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,p r, p θ,p  ) = (2m) -1 [(p r ) 2 + (p θ ) 2 r -2 + (p  ) 2 r -2 sin -2 θ] + V(r) The same problem in Cartesian Coordinates. (Sum on i = 1,2,3): The KE is then: T = (½)mv 2 = (½)mx i x i The PE is: V(r) = V([x i x i ] ½ ) Is writing H = T + V directly in terms of the above T correct? NO!!!!! H MUST be expressed in terms of the momenta p, not the velocities (x i )! Can either follow the matrix formalism or the recipe & find that the proper way to write H is: H = (2m) -1 (p i p i ) +V([x i x i ] ½ ) = (2m) -1 (p  p) + V([x i x i ] ½ )

8. We can take components of the vector p relative 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 θ is (from the Lagrange or matrix formalisms): p θ  (∂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  θ Has units of & is linear momentum! Clearly, p θ  (p) θ ! It’s similar, of course for the  components!

9. 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)mv 2 - q  + qA  v (I) The general Lagrangian we had was of the form (q here is a coordinate, NOT a charge!): L = L 0 (q,t) + q i a i (q,t) + (q i ) 2 T i (q,t) (A) Comparing gives: L 0 = - q  and q i a i (q,t) = qA  v Using Cartesian coords, the Lagrangian (I) can be written (Sum on i = 1,2,3): L = (½)mx i x i + qA i x i - q  (II)

10. The 2 nd term in L is obviously linear in x i.  The vector a in the formalism we just developed has elements qA i.  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): p i  (∂L/∂x i )  (Tq + a) i = mx i + qA i (III) (Note: p i  mx i !) We had the Hamiltonian: H = (½)(p – a)T -1 (p – a) - L 0 (q,t) (D) Using (III) in (D) gives (Sum on i = 1,2,3): H = (p i – qA i )(p i – qA i )(2m) -1 + q  With vectors, the proper Hamiltonian (for a particle in an EM field) is: H = (p – qA) 2 (2m ) -1 + q  (IV)

11. Obviously, Hamilton’s equations aren’t symmetric in the coordinates q & the momenta p: q i  (  H/  p i ) p i  - (  H/  q i ) 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 1 st n = the q’s & the 2 nd n = the p’s: η i  q i η i+n  p i i  n Also: define the column vector (  H/  η): (  H/  η) i = (  H/  q i ), (  H/  η) i+n = (  H/  p i ) i  n

12. 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 ; J 2 = -1 0 1 |J| = 1 In this notation, Hamilton’s eqtns (matrix form) become: η = J(  H/  η) (1) Example: n = 2: Coords q 1, q 2, momenta p 1, p 2, (1) becomes (using Hamilton’s eqtns): What advantage does this have ? I don’t see any!