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My classical mechanics is very rusty. Recently, I wondered if it is possible to replace the canonical momentum with the kinetic momentum using a canonical transformation (so this isn't homework). I tried to work this out, but somehow, the Lorenz force got lost. Maybe some of you has an idea what goes wrong:
Consider the classical hamiltonian for a particle of mass ##m## and charge ##e## in an electromagnetic field.
Its position is ##q(t)## and it's canonical momentum ##p(t)##.
$$
H=\frac{1}{2m}(p-eA(q,t))^2+e\Phi.
$$
The kinetic momentum is ##P=p-eA##.
Can we introduce it as new canonical momentum?
We need to find a generating function ##G(q,P,t)## so that
$$
p=\frac{\partial{G}}{\partial q},
$$
$$
Q=\frac{\partial{G}}{\partial P},
$$
and
$$
H'(Q,P)=H+\frac{\partial{G}}{\partial t}
$$
Integrating the first equation, we obtain
$$
G(q,P,t)=\int_0^q (P+eA)\; dq=Pq+e\int_0^q A(q,t)\;dq + f(P,t).
$$
We consider a transformation with ##f(P,t)=0##.
From the second equation, the canonical variable to ##P## is
$$
Q = q,
$$
so I will use ##q## furtheron instead of ##Q##.
With
$$
\frac{\partial A}{\partial t} =-E -\nabla \Phi.
$$
we obtain from the third equation
$$
H'= \frac{1}{2m}P^2- e\int_0^q E(q,t)dq.
$$
So the transformed hamiltonian is completely free of potentials. The integration paths seems to be arbitrary.
We get the equations of motion
$$
\dot{q}= \frac{\partial H}{\partial P}= \frac{P}{m}
$$
and
$$
\dot{P} = -\frac{\partial H}{\partial q} = eE(q)
$$
somehow, the Lorenz force got lost!
Consider the classical hamiltonian for a particle of mass ##m## and charge ##e## in an electromagnetic field.
Its position is ##q(t)## and it's canonical momentum ##p(t)##.
$$
H=\frac{1}{2m}(p-eA(q,t))^2+e\Phi.
$$
The kinetic momentum is ##P=p-eA##.
Can we introduce it as new canonical momentum?
We need to find a generating function ##G(q,P,t)## so that
$$
p=\frac{\partial{G}}{\partial q},
$$
$$
Q=\frac{\partial{G}}{\partial P},
$$
and
$$
H'(Q,P)=H+\frac{\partial{G}}{\partial t}
$$
Integrating the first equation, we obtain
$$
G(q,P,t)=\int_0^q (P+eA)\; dq=Pq+e\int_0^q A(q,t)\;dq + f(P,t).
$$
We consider a transformation with ##f(P,t)=0##.
From the second equation, the canonical variable to ##P## is
$$
Q = q,
$$
so I will use ##q## furtheron instead of ##Q##.
With
$$
\frac{\partial A}{\partial t} =-E -\nabla \Phi.
$$
we obtain from the third equation
$$
H'= \frac{1}{2m}P^2- e\int_0^q E(q,t)dq.
$$
So the transformed hamiltonian is completely free of potentials. The integration paths seems to be arbitrary.
We get the equations of motion
$$
\dot{q}= \frac{\partial H}{\partial P}= \frac{P}{m}
$$
and
$$
\dot{P} = -\frac{\partial H}{\partial q} = eE(q)
$$
somehow, the Lorenz force got lost!
Last edited: