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The Lagrangian of a charged particle of charge $e$ moving in an electromagnetic field is given by $$L=\frac{1}{2}m\dot{\textbf{r}}^2-e\phi-e\textbf{A}\cdot \textbf{v}$$ where $\phi(\textbf{r},t)$ is the scalar potential and $\textbf{A}(\textbf{r},t)$ is the vector potential. Here the "potential" $U=e(\phi-\textbf{A}\cdot \textbf{v})$ is velocity dependent. The corresponding Hamiltonian is given by $$H=\frac{(\textbf{p}-e\textbf{A})^2}{2m}+e\phi$$

  1. Is it possible to define a total energy for the the charged particle? If yes, what is the expression for total energy and is that a constant of motion?

  2. Does the expression for the Hamiltonian coincide with that of the total energy? My guess is that the Hamiltonian cannot represent the total energy because it is not gauge invariant.

SRS
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    Have a look at (for example) http://physics.stackexchange.com/q/94699/ – jim Mar 25 '16 at 19:44
  • @jim- I looked at the post but my questions are different. – SRS Mar 25 '16 at 20:09
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    InRe #2: One of the requirements to identify the Hamiltonian with the total energy is explicitly that the potential energy be independent of the generalized velocities. /been reviewing my Hamiltonian mechanics lately... – dmckee --- ex-moderator kitten Mar 25 '16 at 23:46

3 Answers3

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  1. For a particle-field system the only way to define a gauge invariant energy is to consider the energy carried by the field as well, in the form of the energy momentum tensor $T^{\mu\nu}$ in the presence of charges. $T^{\mu\nu}$ is a manifestly gauge invariant quantity. To derive this use Noether's theorem and the maxwell equations in the presence of charges. This gives a conserved energy for the system

  2. So the answer to this would be no, it does not correspond to the total energy as you must also consider the field energy to attain a conserved quantity. Intuitively, if we choose some gauge where $\phi = 0$, then the change in the energy of particle is compensated for by a change in the energy carried by the fields. Thus resulting in the same total energy for the system.

bremsstrahlung
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  • It turns out that for the free electromagnetic field $T^{00}=\frac{1}{2}(\textbf{E}^2+\textbf{B}^2)+\nabla\cdot(\phi\textbf{E})$. This is the energy carried by the field only. Why do you say that $T^{\mu\nu}$ is gauge invariant? Can you write down the expression for the total energy of the combined particle-field system? – SRS Mar 31 '16 at 16:00
  • It also turns out that you can redefine the energy momentum tensor so that it is gauge invariant by adding the term $\frac{1}{4\pi}\partial^{\mu} A^{\nu} F^{\alpha}{\mu}$. This leaves the final form of tensor as $\frac{1}{4\pi}\bigg(-F^{\nu\mu}F^{\alpha}{\mu} + \frac{1}{4}g^{\nu\alpha}F_{\mu\beta}F^{\mu\beta}\bigg)$. Which is gauge invariant. – bremsstrahlung May 05 '16 at 15:05
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This question is studied in detail in http://aapt.scitation.org/doi/abs/10.1119/1.12463. The punch line is that there is a natural way to define the total energy of the electromagnetic field and particles together, but there is no single natural way to divide it up between the two contributions.

tparker
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That Hamiltonian already represents the total energy of the particle but it is not conserved because the particle is an open system. Adding the energy of the field and of the particles associated to the Coulomb interaction gives a conserved energy for the total closed system of particles plus field

$$H_\mathrm{tot}= \sum_i \frac{(\boldsymbol{p}_i - e_i \boldsymbol{A})^2}{2m_i^\mathrm{bare}} + \frac{1}{2} \sum_i\sum_{j\neq i} \frac{e_i e_j}{4\pi\epsilon_0 | \boldsymbol{r}_i - \boldsymbol{r}_j |} + \frac{1}{8\pi} \int d^3x \> ( \boldsymbol{E}_\mathrm{T}^2 + \boldsymbol{B}^2) . $$

where $m_i^\mathrm{bare}$ is the bare mass, an infinite quantity that has to be renormalized using the "electromagnetic mass" associated to the divergent vector potential and $\boldsymbol{E}_\mathrm{T}$ is the transverse component of the electric field.

juanrga
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  • This Hamiltonian or way to resolve the inconsistency of point charges with Poynting's formulae is the standard way repeated in textbooks, but it is seriously defective - although the electric energy is made finite by the Coulomb energy substitution, the magnetic energy is still infinite. And $e_i \mathbf A$ is undefined - $\mathbf A$ diverges in the place where the particle is. One has to decide whether one wants point particles, in which case the field part is not given by Poynting's expressions, or one wants Poynting's expressions, but then particles are composed and things get complicated. – Ján Lalinský Jul 27 '18 at 17:17
  • @JánLalinský the divergence in the vector potential $\mathbf A$ is compensated by a term included in the bare mass http://www.juanrga.com/2017/07/renormalized-field-theory-from-particle.html – juanrga Jul 28 '18 at 11:40
  • This kind of endeavour with compensated infinities is not well motivated. If they compensate, why introduce them at all? We can stick, mathematically, with the model you began with, the Hamiltonian (4) on your blog page. However, it has another problem; the particle interaction is not correct. Already the Darwin Lagrangian contains also term that depends on radial components of the momenta, and we know even that is not fully consistent with Maxwell's equations. Also, you made a logical jump in 19); how did you arrive at the integral of $B^2$? For point particles, this integral is infinite. – Ján Lalinský Jul 28 '18 at 18:01
  • @JánLalinský I prefer using real mass and real potentials. Simply mentioned that the unphysical bare mass is needed when one uses the divergent field-theoretic potential. Note that the Darwin term has a 1/2 factor in the interaction, because Darwin Hamiltonian splits the whole interaction into modes. The whole integral in (19) is divergent. This divergence is compensated by $E^\mathrm{self}$ and by the kinetic term. Check (12). – juanrga Aug 01 '18 at 06:40