Sect. 8.3: Routh’s Procedure These examples were simple, but they illustrate the procedure which is followed in many more complex problems! In most cases: 1) The 1st Hamilton Equation qi = (H/pi), usually merely gives the relation between pi & qi. This is usually already known  The 1st Equation usually gives no new information. 2) The 2nd Hamilton Equation pi = - (H/qi), usually gives Newton’s 2nd Law. To solve the problem, take the time derivative of the result of the 1st Equation & equate it to the result of the 2nd Equation. This gives a 2nd order time dependent differential equation of motion which is IDENTICAL to what is obtained by applying the Lagrange formalism (& also to the result obtained from the Newtonian formalism)!

All generalized velocities qi still occur in L. Most often the Hamiltonian procedure gives identically the same equations of motion as the Lagrangian procedure & provides no new information. It is also more lengthy & tedious than the Lagrangian procedure! Exception: The Hamiltonian method is very well adapted to the treatment of problems involving cyclic coordinates. Consider the Lagrangian formalism with one cyclic coordinate, say, qn. The Lagrangian (with qn missing) is: L = L(q1,q2, …., qn-1;q1,q2, ….qn;t) All generalized velocities qi still occur in L.  We still have to solve a problem with n degrees of freedom  n 2nd order differential equations of motion!

H = H(q1,q2, …., qn-1;p1,p2, ….pn-1,α;t) On the other hand, consider the Hamiltonian formalism with cyclic coordinate qn. The conjugate momentum pn to the coordinate qn is conserved (constant). Define: pn  α = constant  The Hamiltonian: H = H(q1,q2, …., qn-1;p1,p2, ….pn-1,α;t) A function of n-1 qi & n-1 pi!  The problem is reduced from one with n degrees of freedom to one with n-1 degrees of freedom!  2(n-1) 1st order differential equations of motion. We can solve the problem by completely IGNORING qn except for its time dependence.  qn truly deserves the description “IGNORABLE” because it can be ignored in getting the equations of motion. The time dependence of qn is obtained by integrating the Hamilton Equation: qn = (H/pn)  (H/α)

Routh’s Procedure: Devised to take advantage of the Hamiltonian method’s simpler handling of cyclic coordinates PLUS to take advantage of the Lagrangian method’s simpler treatment of non-cyclic coordinates. A “hybrid” Hamiltonian-Lagrangian procedure! Consider a problem with s non-ignorable coordinates q1,q2, …., qs & n-s ignorable coordinates, qs+1,qs+2, …., qn. Use the Hamiltonian procedure for the n-s ignorable (cyclic) coordinates & the Lagrangian procedure for the s non-ignorable coordinates. Carry out the Legendre transformation from coordinates & velocities q,q to coordinates & momenta q,p only for the n-s cyclic coordinates qs+1,qs+2, …., qn. Get their equations of motion using the Hamiltonian method & Hamilton’s equations of motion. For the s non-ignorable coords q1,q2, …., qs use Lagrange’s method & Lagrange’s equations of motion.

Do this formally as follows: Define a new function (the Routhian) as: R  R(q1, …., qn;q1, ..,qs;ps+1, ….,pn;t)  ∑(i = s+1, n)piqi – L This is equivalent to writing: R  R(q1, …., qn;q1, ..,qs;ps+1, ….,pn;t)  Hcyc(qs+1, …,qn;ps+1,…,pn;t) – Lnoncyc(q1, ..,qs;q1, .,qs;t) The equations of motion for the s non-cyclic coordinates have the Lagrangian form: (d/dt)[(∂R/∂qi)] - (∂R/∂qi) = 0, (i = 1,…s) The equations of motion for the n-s cyclic coordinates have the Hamiltonian form: qi = (R/pi), pi = - (R/qi), (i = s+1,…n)

R(r,r,pθ) = (2m)-1(pθ)2r-2 – (½)mr2 – kr-n Example: The Kepler (Central Force) problem (Ch. 3): A single particle moving in a plane under a Central Force f(r) obtained from a potential V(r) = -kr-n. Lagrangian: L = (½)m(r2 + r2θ2) + kr-n Clearly (as in Ch. 3) the ignorable coordinate is θ The conjugate momentum is pθ  (L/θ). Hamiltonian: H = (2m)-1[(pr)2 + (pθ)2r-2] – kr-n The Routhian is thus (from the general formalism): R(r,r,pθ) = (2m)-1(pθ)2r-2 – (½)mr2 – kr-n

R(r,r,pθ) = (2m)-1(pθ)2r-2 – (½)mr2 – kr-n Routhian: R(r,r,pθ) = (2m)-1(pθ)2r-2 – (½)mr2 – kr-n Lagrangian Equation of motion for the non-cyclic coord r: (d/dt)[(∂R/∂r)] - (∂R/∂r) = 0  (d/dt)[-mr] + [d(-kr-n)/dr] = 0 Simplifying: r – (pθ)2(mr3)-1 + (nk)r-(n+1) = 0 Hamiltonian Equations of motion for the cyclic coordinate θ: pθ = - (R/θ) = 0  pθ = constant θ = (R/pθ) = pθmr-2  pθ = mr2θ Taken together, this gives: pθ = mr2θ   = constant. As we already knew! This procedure “makes the analysis more automatic” than that in Ch. 3?

Sect. 8.4: Hamiltonian Formulation of Relativistic Mechanics Similar to the question we asked earlier (in Ch. 7) about Lagrangian mechanics: A natural question is: How to generalize Hamiltonian mechanics to Relativity? Or, what is an appropriate Relativistic Hamiltonian? There are two ways people have done this: 1. Try to obtain a covariant Hamilton’s Principle & Hamilton’s Equations. Space & time treated on an equal footing in 4d Minkowski space. 2. Don’t worry about 1. Try to find a Hamiltonian which will (in some inertial frame) reproduce the Newton’s 2nd Law Equations of motion obtained by using the spatial part of the Minkowski 4-force: K = (dp/dτ) = γu(dp/dt)

L  T* - V = -mc2[1 - β2]½ - V β = (u/c) Here, we focus on method 2. Goldstein, Sect. 8.4, p. 351-353 has the details of method 1. Both methods lead to the same PHYSICS! Sect. 7.9, p. 313, shows (for conservative forces) that a suitable Relativistic Lagrangian (for a single particle with velocity u) is (watch out for Goldstein’s notation here; it’s confusing, at least to me!): L  T* - V = -mc2[1 - β2]½ - V β = (u/c) Recall that T* = -mc2[1 - β2]½  T (the KE). & that the KE is T = ([1 - β2]-½ -1) mc2  (γ - 1)mc2 = E – E0 E  γmc2 = “total energy” & E0  mc2 = “rest energy”. Also another form for E in terms of the relativistic (3-vector) momentum p = γmu, E2 = p2c2 + (E0)2 = p2c2 + (mc2)2 (1) Here the notation may be confusing! In this section, Goldstein uses the symbol T for (1) instead of E. In this section T  KE, but T = total energy (exclusive of the potential V). To try to avoid this confusion, in what follows I replace E in the above with the symbol ET (“KE + rest energy”)

So, start with the Relativistic Lagrangian: L  T* - V = -mc2[1 - β2]½ - V β = (u/c) Starting here, in Sect. 7.9, p 314, Goldstein shows that the relativistic “energy function” h is (my notation for the 1st term): h = ET + V = E (total energy) where ET  γmc2 = mc2[1 - β2]-½ As we’ve said all along in Ch. 8: the Hamiltonian H is the SAME as the energy function h with the dependence on the velocities removed so that H is a function of the momenta. So, carrying that over to Relativity, we can write for the Hamiltonian: H = ET + V & use (ET)2 = p2c2 + m2c4 for ET  The Relativistic Hamiltonian is: H = [p2c2 + m2c4]½ + V

Starting with the Relativistic Hamiltonian, H = [p2c2 + m2c4]½ + V Apply Hamilton’s Equations of motion as usual. Working in Cartesian coordinates (x1,x2,x3) with conjugate momentum components (p1,p2,p3) the Hamiltonian is: H = [{(p1)2+(p2)2+(p3)2}c2 + m2c4]½ + V(x1,x2,x3) Hamilton’s Equations of motion: xi = (H/pi), pi = - (H/xi), (i = 1,2,3) These give: xi = pic2[p2c2 + m2c4]-½ (i = 1,2,3) (a) and: pi = - (dV/dxi) (i = 1,2,3) (b) Solve by taking the time derivative of (a) & substituting (b) into the result. A non-linear differential equation, even for fairly simple potentials V!

For a charged particle in an EM field, the potential is velocity dependent, as we’ve seen. The Relativistic Lagrangian is (from Ch. 7): L = -mc2[1 - β2]½ - q + qAv Goldstein shows that the corresponding Relativistic Hamiltonian is: H = ET + V. However, now, ET has the form: (ET)2 = P2c2 + m2c4 , where P  (p – qA) That is, (as we’ve seen before) the conjugate momentum in the presence of an EM field is not p = γmu. Instead it is P = (p – qA) . So, the Relativistic Hamiltonian has the form: H = [(p - qA)2c2 + m2c4]½ + V

Read Sect. 8.6: The Principle of Least Action. Read the rest of Sect. 8.4 (pp 351 – 353) on the covariant formulation of Hamiltonian mechanics. In order to be covariant, a theory must treat the time & space coordinates on an equal footing. So, there must be a “conjugate momentum” to the time “coordinate”. That section shows that the “conjugate momentum” to the time is the negative Hamiltonian – H! This is not surprising if you recall (Ch. 7) that the timelike part of the 4-vector momentum is E/c. Further, the “time” versions of Hamilton’s Equations have the same math form as the space versions. Read Sect. 8.5: The derivation of Hamilton’s Equations from a Variational Principle. Read Sect. 8.6: The Principle of Least Action.