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The Lagrangian of a charge $q$ in a constant (in time) external magnetic field $\boldsymbol{B}=\boldsymbol{\nabla}\times\boldsymbol{A}$ is according to Goldstein $$ L=T-U = \tfrac12 m\dot{\boldsymbol{x}}^2 + q\dot{\boldsymbol{x}}\cdot\boldsymbol{A} \qquad\text{with}\qquad U= -q\dot{\boldsymbol{x}}\cdot\boldsymbol{A}. $$ The corresponding conjugate momentum and Hamiltonian are \begin{align} \boldsymbol{p} &= \frac{\partial L}{\partial\dot{\boldsymbol{x}}} &&= m\dot{\boldsymbol{x}} + q\boldsymbol{A}, & \\ H &=\boldsymbol{p}\cdot\dot{\boldsymbol{x}} - L &&= \tfrac12m\dot{\boldsymbol{x}}^2 &&= \frac{(\boldsymbol{p}-q\boldsymbol{A})^2}{2m}. \end{align} The Hamiltonian is just the kinetic energy and is conserved since $$ \frac{\partial H}{\partial t}=0. $$ So what is the correct interpretation of $U$? Is it the potential energy (of the charge in the external field)? But then shouldn't it be part of the conserved total energy (according to Noether, if $\partial L/\partial t=0$ the total energy is conserved)?

Note, a similar question was asked before, but only one answer appears relevant to my question. Unfortunately, this answer is not much more than a link to an academic article from 1979. What I'm essentially asking for is a concise summary of the argument from that article.

EDIT: The mentioned paper is "Gauge‐invariant classical Hamiltonian formulation of the electrodynamics of nonrelativistic particles" by Donald H. Kobe, American Journal of Physics 49, 581 (1981).

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  • $\begingroup$ It is "part of the conserved energy". Expand your rightmost expression for the Hamiltonian, the cross-term $-\mathbf{p}\cdot q\mathbf{A}/m$ will give you back $U$. $\endgroup$ – Demosthene Dec 1 '17 at 14:10
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The author of the paper you mention does the following:

  • They first show that both the Lagrangian and Hamiltonian for a system of particles and electromagnetic fields are invariant under gauge transformations (sections II and III)
  • They then note that since the system is isolated, the Hamiltonian gives the energy for the system of particles and electromagnetic fields. However, since particles and fields can exchange energy, the energy of the fields and the energy of the particles are not individually conserved. (Section IV)
  • They then discuss whether it is nonetheless possible to distinguish between the energy of the particles and the energy of the fields. The answer is negative. However, the author shows that it is possible to separate the power gained or lost by particles and the power gained or lost by the fields, and that these two cancel out (also Section IV)
  • They finally consider the Hamiltonian for particles in a background electromagnetic field. (in the previous sections, the fields were treated dynamically) This is probably the most relevant part for your question. In the time dependent case, the Hamiltonian should not be interpreted as the energy of the particles (even though this is not really well-defined). However in the time-indepedent case the two coincide. (Section V)

Note that your Hamiltonian is conserved because you're assuming that the field $\textbf{A}$ is time-independent. Moreover, as mentioned in the paper these potentials cannot really be interpreted as potentials, because a potential is something whose gradient gives you a force, and this is not the case here.

In formulas, eq. $(4.1)$ is the energy of the system of fields and particles: $$\varepsilon = H = \frac{1}{2} \int d^3 x (E^2+B^2)+\sum_\alpha \frac{1}{2}m_\alpha \dot{\textbf{r}}_\alpha^2$$ The first term is the energy of the fields and the second term is the kinetic energy of the particles The paper also involves an external static potential energy $U$ (for instance due to gravity), but we'll ignore it.

Above we've given an obvious separation for the energies of the particles and of the fields. However, it is sometimes more natural to think of the usual (Coulombian) potential energy between particles as being energy of the particles, rather than of the fields. In fact one can take the Coulombian interaction out of the $E^2$ term, to get eq.s $(4.18)$ and $(4.20)$: $$\varepsilon = H = \frac{1}{2} \int d^3 x (E_T^2+B^2)+\sum_\alpha \frac{1}{2}m_\alpha \dot{\textbf{r}}_\alpha^2+V$$ where $V$ is the sum of Coulombian interaction between particles, and $E_T$ is the field component which is not radial (i.e. it is transverse). This illustrates the fact that there is no unique separation between the energy of the particles and the energy of the fields.

Finally, if one includes the scalar potential in the Lagrangian (which for example you do not), you'd get a Hamiltonian which looks like the energy of the particles in the distinction that was just obtained, plus the scalar potential times the charge. Therefore they argue that the actual energy should in this case be the hamiltonian minus the scalar potential terms. This illustrates the last point.

So to answer your question, is your $U$ a potential? The answer is no, at least not in the definition of the paper, because its gradient does not give you a force. The physical reason is that the magnetic field does no work. So it's not surprising to not see it in the Hamiltonian.

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