Hamiltonian with z as the Independent Variable

1 Problem 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. 2 Solution 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 E = −∇V − 1 c ∂A ∂t , B = ∇ × A, (1) H t (x, y, z, p x , p y , p z) = E mech + qV = c m 2 c 2 + p 2 mech,x + p 2 mech,y + p 2 mech,z + qV (2) = c m 2 c 2 + (p x − qA x /c) 2 + (p y − qA y /c) 2 + (p z − qA z /c) 2 + qV, in Gaussian units, where c is the speed of light in vacuum, and the components of p = p mech + qA/c are the canonical momenta associated with coordinates x = (x, y, z). The subscript on H t indicates that time t is the independent variable in this Hamiltonian. Hamilton's equations of motion for this case are dx i dt = ∂H t ∂p i = c 2 p mech,i E mech = v i , (3) dp i dt = − ∂H t ∂x i = q j v j c ∂A j ∂x i − q ∂V ∂x i = dp mech,i dt + q c dA i dt = dp mech,i dt + q c ∂A i ∂t + q j v j c ∂A i ∂x j , (4) using the convective derivative dA/dt = ∂A/∂t + (v · ∇)A for the vector potential at the position of the moving particle. Hence,