Hamiltonian with z as the Independent Variable Kirk T. McDonald Joseph Henry Laboratories, Princeton University, Princeton, NJ 08544 (March 19, 2011; updated June 19, 2015)
1Problem
Deduce the form of the Hamiltonian when z rather than t is considered to be the independent variable. Illustrate this for the case of a particle of charge q and mass m in an external electromagnetic field.
2Solution
This solution follows Appendix B of [1]. See also sec. 1.6 of [2]. For simplicity we consider only a single particle.
2.1 Use of t as the Independent Variable
We recall the usual Hamiltonian description of a particle of charge q and mass m in external electromagnetic fields E and B, which can be deduced from scalar and vector potentials V and A (in some gauge) according to ∂A E −∇V − 1 , B ∇ × A, = c ∂t = (1) 2 2 2 2 2 Ht(x, y, z, px,py,pz )=Emech + qV = c m c + pmech,x + pmech,y + pmech,z + qV (2) 2 2 2 2 2 = c m c +(px − qAx/c) +(py − qAy/c) +(pz − qAz/c) + qV, in Gaussian units, where c is the speed of light in vacuum, and the components of p = pmech + qA/c are the canonical momenta associated with coordinates x =(x, y, z). The subscript on Ht indicates that time t is the independent variable in this Hamiltonian. Hamilton’s equations of motion for this case are 2 dxi ∂Ht c pmech,i = = = vi, (3) dt ∂pi Emech dpi ∂Ht vj ∂Aj ∂V = − = q − q dt ∂xi j c ∂xi ∂xi
dpmech,i q dAi dpmech,i q ∂Ai vj ∂Ai = + = + + q , (4) dt c dt dt c ∂t j c ∂xj using the convective derivative dA/dt = ∂A/∂t +(v · ∇)A forthevectorpotentialatthe position of the moving particle. Hence, ⎡ ⎤ dp ∂V ∂A v ∂A ∂A v mech,i ⎣ 1 i j j i ⎦ = q − − + − = q E + × B = FLorentz,i , (5) dt ∂xi c ∂t j c ∂xi ∂xj c i
1 such that the equations of motion for the mechanical momentum pmech is gauge invariant, although the Hamiltonian (2) is not.
2.2 Use of z as the Independent Variable
In some applications, such as transport of particles in accelerators and storage rings, it is often preferable to consider a set of particles at fixed values of a spatial coordinate, say z, rather than at fixed time.1 So, we seek a Hamiltonian formalism in which z is the independent variable, and t is the third q-coordinate, along with x and y. We must identify a canonical momentum pt that is conjugate to coordinate t, and a Hamiltonian Hz (x, y, t, px,py,pt)such that the equations of motion can be deduced from this Hamiltonian in the usual way. We anticipate that the (total) energy is conjugate to the time coordinate, so we tentatively identify
? pt = Etotal = Emech + qV = Ht. (6)
We might then guess that, by analogy, the desired Hamiltonian Hz equals the canonical momentum pz, qA E2 qA ? z mech 2 2 2 2 z Hz = pz = pmech,z + = − m c − p − p + c c2 mech,x mech,y c p − qV 2 qA 2 qA 2 qA ( t ) − m2c2 − p − x − p − x z . = c2 x c x c + c (7) The test is whether the equations of motion that follow from these identifications are con- sistent with those associated with Ht.
dx ? ∂Hz pmech,x vx = = − = − . (8) dz ∂px pmech,z vz The magnitude is correct, but the sign is wrong. This suggests that there should have been a minus sign in both eqs. (6) and (7),
pt = −Etotal = −Emech − qV = −Ht, (9)