I am trying to understand a part of I.E.P.'s answer here.

I.E.P. states that one can see from the following Hamiltonian, $$ H = \frac{1}{2m}|{\bf p}+q{\bf A}|^2 +q \phi \tag{8.35} $$ that the magnetic field does not work on the charged particle, and thus does not contribute to the energy.

  1. How can one see that directly from this Hamiltonian?

  2. The corresponding Lagrangian for this system is

\begin{equation} L = \frac{1}{2} m\dot{\bf r}^2 - q \phi + q \dot{\bf r}\cdot {\bf A}. \tag{8.32} \end{equation}

by Goldstein's matrix formulation to the Hamiltonian formalism, since $L$ is not a homogeneous function of degree 2, $H$ is not equal to kinetic energy + potential energy. HOWEVER, Goldstein does state that

There is now a linear term in the generalized velocities such that the matrix $\mathbf{a}$ has the elements $q A_i$. Because of this linear term in $V$, the Hamiltonian is not $T + V$. However, it is still in this case the total energy since the “potential” energy in an electromagnetic field is determined by $\phi$ alone.

Can I have some help also in how to reconcile the above quote of Goldstein's with his comment about $H\neq E$ unless $L$ is a homogeneous function of the velocities squared?


1 Answer 1

  1. Consider the (Lagrangian) energy function $$ h(q,\dot{q},t)~=~\left(\sum_j\dot{q}^j\frac{\partial }{\partial \dot{q}^j}-1 \right)L(q,\dot{q},t), \tag{2.53} $$ which should not be confused with the Hamiltonian function $H(q,p,t)$. They are different functions, although their values agree.

  2. OP's quote from Ref. 1 is closely related to the following fact. If $L_n$ denotes the part of the Lagrangian $L$ that is a homogeneous polynomial of $n$'th degree in the generalized velocities $\dot{q}^j$, and if the Lagrangian is of the form $L=L_2+L_1+L_0$, then the energy is $$h~=~L_2-L_0.\tag{2.57}$$

  3. In particular, for a non-relativistic charge in an E&M background, Ref. 1 denotes $L_2=T$ and $L_1+L_0=-V$. The energy function is $$ h({\bf r},\dot{\bf r},t)~=~ \frac{m}{2}\dot{\bf r}^2 +q \phi({\bf r}) $$ is different from $T+V$. Note that the energy $h$ is independent of the magnetic potential ${\bf A}$, i.e. the magnetic force produces no work, cf. the work-energy theorem.

  4. Concerning the relationship between Hamiltonian and energy, see also e.g. this Phys.SE posts and links therein.


  1. H. Goldstein, Classical Mechanics, 3rd edition; Chapter 2 + 8.
  • $\begingroup$ Thank you for the answer! :D One follow-up: so it is indeed the energy function h(q,$\dot{q}$,t) which we can more directly compare to the energy (E=T+U)? I.e. it is much more easy note the absence of the q($\dot{r}\cdot A$) term from h, whereas it is in E. $\endgroup$
    – Lopey Tall
    Commented Aug 2, 2020 at 0:02

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