Paper 2, Section II, E

Classical Dynamics | Part II, 2019

The Lagrangian of a particle of mass mm and charge qq moving in an electromagnetic field described by scalar and vector potentials ϕ(r,t)\phi(\mathbf{r}, t) and A(r,t)\mathbf{A}(\mathbf{r}, t) is

L=12mr˙2+q(ϕ+r˙A)L=\frac{1}{2} m|\dot{\mathbf{r}}|^{2}+q(-\phi+\dot{\mathbf{r}} \cdot \mathbf{A})

where r(t)\mathbf{r}(t) is the position vector of the particle and r˙=dr/dt\dot{\mathbf{r}}=d \mathbf{r} / d t.

(a) Show that Lagrange's equations are equivalent to the equation of motion

mr¨=q(E+v×B),m \ddot{\mathbf{r}}=q(\mathbf{E}+\mathbf{v} \times \mathbf{B}),

where

E=ϕAt,B=×A\mathbf{E}=-\nabla \phi-\frac{\partial \mathbf{A}}{\partial t}, \quad \mathbf{B}=\nabla \times \mathbf{A}

are the electric and magnetic fields.

(b) Show that the related Hamiltonian is

H=pqA22m+qϕ,H=\frac{|\mathbf{p}-q \mathbf{A}|^{2}}{2 m}+q \phi,

where p=mr˙+qA\mathbf{p}=m \dot{\mathbf{r}}+q \mathbf{A}. Obtain Hamilton's equations for this system.

(c) Verify that the electric and magnetic fields remain unchanged if the scalar and vector potentials are transformed according to

where f(r,t)f(\mathbf{r}, t) is a scalar field. Show that the transformed Lagrangian L~\tilde{L} differs from LL by the total time-derivative of a certain quantity. Why does this leave the form of Lagrange's equations invariant? Show that the transformed Hamiltonian H~\tilde{H} and phase-space variables (r,p~)(\mathbf{r}, \tilde{\mathbf{p}}) are related to HH and (r,p)(\mathbf{r}, \mathbf{p}) by a canonical transformation.

[Hint: In standard notation, the canonical transformation associated with the type-2 generating function F2(q,P,t)F_{2}(\mathbf{q}, \mathbf{P}, t) is given by

p=F2q,Q=F2P,K=H+F2t.]\left.\mathbf{p}=\frac{\partial F_{2}}{\partial \mathbf{q}}, \quad \mathbf{Q}=\frac{\partial F_{2}}{\partial \mathbf{P}}, \quad K=H+\frac{\partial F_{2}}{\partial t} .\right]

ϕϕ~=ϕft,AA~=A+f,\begin{aligned} & \phi \mapsto \tilde{\phi}=\phi-\frac{\partial f}{\partial t}, \\ & \mathbf{A} \mapsto \tilde{\mathbf{A}}=\mathbf{A}+\nabla f, \end{aligned}

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