Hamiltonian for Charged Particles + EM-Field

[Cryo]

Summary:

I have found the Hamiltonian for the free particles and the electromagnetic field ($\mathbf{E}$ - electric, $\mathbf{B}$ - magnetic) to be (non-relativistic !):

$H=\sum_i \frac{m \dot{r}_i^2}{2} + \int d^3 r \left(\frac{\epsilon_0}{2} E^2 + \frac{1}{2\mu_0} B^2\right)$ (1)

where $\sum_i$ is over all the particles, $m$ is the mass of the particles (identical), and $\dot{\mathbf{r}}_i$ is the velocity of $i$-th particle. Integral $\int d^3 r$ runs over volume large enough for fields to vanish outside it (all particles are containted within this volume).

Is this a correct result? Normally one has things like $\mathbf{J}.\mathbf{A}$ (for current density $\mathbf{J}$ and vector potential $\mathbf{A}$), but here I am considering both charges and fields to be completely free, thus it seems reasonable that one either stores energy as kinetic energy of charges, or in the electro-magnetic field, which includes the field produced by the charges.

Details. Lagrangian and equations of motion

I start with the Lagrangian for the full system:

$L=\sum_i \frac{m\dot{r}_i^2}{2} + \int d^3 r \left(\frac{\epsilon_0}{2} E^2 - \frac{1}{2\mu_0} B^2 + \mathbf{A}.\mathbf{J} - \phi \rho\right)$

Where scalar and vector potentials are $\phi$ and $\mathbf{A}$, whilst charge and current densities are $\rho$ and $\mathbf{J}$. All of these are to be varied in order to bring action to the extremum.

Varying the scalar and vector potential we can readily obtain the Maxwell's equations for vacuum + charge and current density, assuming $\mathbf{E}=-\boldsymbol{\nabla}\phi-\dot{\mathbf{A}}$ and $\mathbf{B}=\boldsymbol{\nabla}\times\mathbf{A}$. Substituting

$\mathbf{J}=\sum_i q\dot{\mathbf{r}} \delta\left(\mathbf{r}-\mathbf{r_i}\right)$
$\rho=\sum_i q\delta\left(\mathbf{r}-\mathbf{r_i}\right)$

With $q$ being the electrical charge of each particle, we can reduce the Lagrangian to:

$L=\sum_i \left[\frac{m\dot{r}_i^2}{2} + q\dot{\mathbf{r_i}}.\mathbf{A}_i-q\phi_i\right] + \int d^3 r \left(\frac{\epsilon_0}{2} E^2 - \frac{1}{2\mu_0} B^2 \right)$ (2)

Where $\phi_i=\phi\left(t,\,\mathbf{r}_i \right)$ and same for vector potential.

This, in turn permits deriving the forces on each charge by variation with respect to $\mathbf{r_i}^k$

$m\ddot{\mathbf{r}}_i=q\mathbf{E}_i + q \mathbf{\dot{r}}_i\times\mathbf{B}_i$ (3)

Details. Hamiltonian

I have obtained Hamiltonian by finding what quantity was conserved as a result of invariance with respect to time-translations, i.e. I started with

$\frac{dL}{dt}=\dots$

and simplified it until I had $\frac{d}{dt}\left(L-\sum_i \left(q\dot{\mathbf{r}}_i.\mathbf{A}_i - q\phi_i\right)+\int d^3 r\frac{1}{\mu_0}B^2 \right)=0$

It is also easy to verify the reverse:

$\frac{dH}{dt}=\sum_i \mathbf{\dot{r}}_i.m\mathbf{\ddot{r}}_i + \int d^3 r \left(\epsilon_0 \mathbf{E}.\mathbf{\dot{E}}+\frac{1}{\mu_0} \mathbf{B}.\mathbf{\dot{B}}\right)$

From Maxwell's equations we find (ignoring surface effects):

$\int d^3 r \left(\mathbf{E}.\epsilon_0 \mathbf{\dot{E}}\right)=\int d^3 r \left(\frac{-1}{\mu_0}\mathbf{B}.\mathbf{\dot{B}}-\mathbf{E}.\mathbf{J}\right)$

Thus, using Eq. (3):

$\frac{dH}{dt}=\sum_i \mathbf{\dot{r}}_i.\left(q\mathbf{E}_i+q\mathbf{\dot{r}}_i\times\mathbf{B}_i\right)+ \int d^3 r \left(-\mathbf{E}.\mathbf{J}\right)=0$

So, indeed this quantity is conserved. Is this the Hamiltonian?

 

[vanhees71]

Looks good!