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Goldstein page 342,

Consider a single particle (non-relativisitic) of mass $m$ and charge $q$ moving in an electromagnetic field.

The Lagrangian is $$ L = T-V = \frac{1}{2}mv^2-q\phi +q\vec{A}\cdot \vec{v}. \tag{p.341}$$

Using Cartesian coordinates as generalized coordinates, the Lagrangian is $$ L = \frac{m}{2}\dot{x_i}\dot{x_i} + qA_i\dot{x_i}-q\phi \tag{8.32}$$

Because of the linear term in the generalized velocities, $qA_i\dot{x_i}$, the Hamiltonian is not $T+V$.

Then what really confused me is the following:

The Hamiltonian is still the total energy since the "potential" energy in an electromagnetic field is determined by $\phi$ alone.

I have known that there are two conditions must be satisfied if the Hamiltonian is the total energy:

  • the equations defining the generalized coordinates don't depend on time explicitly.
  • the forces are derivable from a conservative potential $V$.

From Eq.(8.32), I think the potential $V$ should be $q\phi-\sum_i qA_i\dot{x_i}$, and since the potential is not an explicit function of time, I think it is conservative.

I want to know how to understand "the potential energy in an EM field is determined by $\phi$ alone"? I think this potential $V$ should be the function of the position and velocity. $$ V(x_i,\dot{x_i}) = q\phi-qA_i\dot{x_i}$$

And there may be some mistakes, please point them out and correct me.

Qmechanic
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David Li
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2 Answers2

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Have you computed the Hamiltonian explicitly? It should take the form $$H(\mathbf x,\mathbf p) = \frac{1}{2m}\big(\mathbf p -q\mathbf A(\mathbf x)\big)^2 + q\phi(\mathbf x)$$

where

$$\mathbf p = \frac{\partial L}{\partial \dot {\mathbf x}} = m \dot{\mathbf x} + q\mathbf A \iff \dot{\mathbf x} = (\mathbf p - q\mathbf A)/m$$

So the first term in the Hamiltonian can be understood as the kinetic energy of the particle, and the second term as the potential energy.

J. Murray
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1

Its basically referring to the fact that the last term in the Lorentz force $F=q(\vec{E}+\vec{v}\times\vec{B})$ involves a force that is always perpendicular to the velocity and therefore doesnt contribute to the particles energy. Two uses of the word potential is being used, on the one hand the generalized potential $V$ certainly contains the term with $A\dot{x}$. On the other hand, as the term involving $A$ in the Lorentz force doesnt contribute to the energy, it isnt part of the potential energy in that sense. Note also that a velocity dependent pontential is not really considered conservative, and although the Lorentz force is conservative when using one of the "equivalent" criterias for a conservative force (via the fact that it doesnt do work), it fails to meet the others.

user330563
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