# $q\mathbf{A}\cdot\mathbf{v}$ term in potential energy

In the famous Goldstein mechanics book, there is an example about a single (non-relativistic) particle of mass m and charge q moving in an E&M field.

It says the force on the charge can be derived from the following velocity-dependent potential energy $$U=q\phi-q\mathbf{A}\cdot\mathbf{v} .\tag{1.62}$$

(eq 1.62 of 3rd ed.) I can see where the expression came from my E&M knowledge. So far it's OK. Next

$$L=T-V=\frac{1}{2}mv^2-q\phi+q\mathbf{A}\cdot\mathbf{v}.$$

(p.341) (It changed notation from $U$ to $V$ without mention.) It says that because of the $q\mathbf{A}\cdot\mathbf{v}$ term in $V$, the Hamiltonian is not $T+V$. However, it says it's still total energy since the "potential" energy in an E&M filed is determined by $\phi$ alone.

I'm confused by the sentence. Is it insisting that potential energy is only $V=\phi$? Then why it introduced velocity-dependent potential earlier? What's the role of $q\mathbf{A}\cdot\mathbf{v}$ term?

1. Firstly, Goldstein uses the letters $V$ and $U$ for velocity-independent and velocity-dependent potentials, respectively, as explained in the beginning of section 1.5,

2. Both the 2nd edition (p. 346) & the 3rd edition (p. 341) wrongly state that the Lagrangian for a point charge in an E&M field is $$L~=~T-V$$ rather than $$L~=~T-U.$$ It seems that Goldstein forgets his own notation convention from Section 1.5!

3. The 2nd edition states (p. 346)

Because of this linear term in $U$, the Hamiltonian is not $T+U$.

While the 3rd edition states (p. 342)

Because of this linear term in $V$, the Hamiltonian is not $T+V$.

The 2nd edition is here correct, while the 3rd edition is wrong, as the Hamiltonian $H$ is indeed the sum of the kinetic energy $T$ and the electric potential energy $V=q\phi$. It seems that the initial error in the 2nd edition caused a new error in the 3rd edition!

References:

1. H. Goldstein, Classical Mechanics, 2nd edition, p. 346.

2. H. Goldstein, Classical Mechanics, 3rd edition, p. 341-342.

• OK that clears the matter, I think. So Is U not actual potential energy and V is the true potential energy? – Septacle Aug 8 '18 at 22:05
• Well, be prepared that the Devil's advocate may disagree about what true potential energy is :-) – Qmechanic Aug 9 '18 at 8:09

I think Goldstein made a mistake (or is at least being misleading).

The Hamiltonian for a charged particle in an electromagnetic field is

$$H=\frac{1}{2m}(\vec{p}-q\vec{A})^2+q\phi(\vec x)$$ We also know, from the canonical transformation, that the canonical momentum is given by $\vec{p}=m\dot{\vec{x}}+q\vec{A}$. So in fact, the Hamiltonian is nothing more than $$H=\frac{1}{2}m\dot x^2+q\phi(\vec{x})$$ written in terms of the canonical momentum $\vec{p}$. Thus, not only is the Hamiltonian the total energy of the particle, but it is in fact exactly $T+V$.

What I think Goldstein is referring to is that earlier, in chapter 1, he described the Lagrangian for a charged particle as arising from a "velocity dependent potential energy" $U$, from which he could write $L=T-U$. This $U$ is NOT a real potential energy, but it makes the Lagrangian work out. In terms of this $U$, he is saying that we cannot write $H=T+U$, where $U$ here is an artificial "velocity dependent potential energy." But we emphatically CAN write $H=T+V$, where $V$ is the boring regular potential energy of a particle in an electric field.