Lagrangian of a moving charge distribution under action of an eletromagnetic field

Question

We all know that for a single charged particle, we can derive the Lagrangian starting from Lorentz law of force:

$$ \mathbf{F}=q(\mathbf{E}+\mathbf{v}\times\mathbf{B}). $$

and by using the definition of generalized potential functions, you know that you need a function $U$ that satisfies:

$$ F_{k}=-\frac{\partial U}{\partial q_{k}}+\frac{d}{dt}\left(\frac{\partial U}{\partial\dot{q}_{k}}\right) $$

And the function $U$ below satisfies the condition: $$ U=q \phi-\frac{q}{c}\left(\mathbf{v}\cdot\mathbf{A}\right). $$

If I try to generalize that for a moving distribution of charge we will have the following equation for Lorentz force ($\rho$ is the charge density):

$$ \mathbf{F}=\int dV\rho(x,y,z,t)(\mathbf{E}+\mathbf{v}\times\mathbf{B}) $$

and doing some algebra I can get to

$$ \mathbf{F}=\int dV\rho(x,y,z,t)(\nabla(\phi-\mathbf{v}\cdot\mathbf{A})-\frac{d\mathbf{A}}{dt}). $$

On the single particle case we can reduce that to the second equation and then we can find the potential $U$, but in this case we cannot do that because charge density varies with time (and position!). I feel that I could find some answer on fluid dynamics, but I'm thinking on that problem for quite a few days and I can't find a answer.

In molecular dynamics we use what is called the "length Gauge", invented by physicist Maria Goppert-Mayer in 1930. In her article, she shows how to derive the equations for bounded electrons interacting with a laser beam. However, a molecule (or a pair of scattered heteronuclear atoms that will collide) cannot be described by a mere one-particle Lagrangian, and for sake of generalization I thought at consider the two atoms classically as a continuous charge distribution, and this is why I got to this problem. Maybe it's the wrong way to deal with that... lol

How would you guys proceed? Can we derive a general potential $U$ for a continuous charge distribution under a electromagnetic field? Could be easier to try to describe the atoms colliding by using discrete coordinates (for the nuclei and the electrons)?

Answer 1

Hint: The velocity-dependent potential term $$-U~=~\sum_{i=1}^Nq_i\dot{x}_i^{\mu}A_{\mu}(x_i), \qquad x_i^0~=~t, \qquad c=1,$$ for $N$ point charges turns into the interaction/source term $$\int d^3x ~J^{\mu}(x) A_{\mu}(x)$$ in the continuum limit.

Answer 2

I think I got it by applying the continuum limit AFTER deriving length gauge for discrete charged points.

I will call electron's coordinates "u", and nuclei coordinates "r". The interaction Hamiltonian (I'm not writing here the Coulombian terms nor the kinect energy terms) that I got is:

$$ \mathit{H}_{int}=-((\sum_{i=1}^{n}(-e\mathbf{u_{i}\cdot}))+q_{1}\mathbf{r}_{1}\cdot+q_{2}\mathbf{r}_{2}\cdot)\mathbf{E} $$

Calling the bracket term a permanent dipole moment of the pair of atoms, we can write the usual interaction Hamiltonian on length gauge:

$$ H_{int}=-\mathbf{\mathbb{\overrightarrow{\mu}}}\cdot\mathbf{E} $$

The first equation is equivalent to that one:

$$ \mathit{H}_{int}=-\{\intop\mathbf{r'}(\sum_{i=1}^{n}-\delta^{3}(\mathbf{r'}-\mathbf{u}_{i})e)+q\delta^{3}(\mathbf{r'}-\mathbf{r_{1}})+q\delta^{3}(\mathbf{r'}-\mathbf{r_{2}}))dV'\}\cdot\mathbf{E} $$

And then you can exchange the delta charge densities for a arbitrary one, $\rho(x,y,z,t)$, and then you will get:

$$ H_{int}=-\{\intop\rho(\mathbf{r'},t)\mathbf{r}'dV'\}\cdot\mathbf{E} $$

I also got the Lagrangian for the continuous case on Velocity or Coulomb Gauge, but it's not of great use. On discrete case, you only need to sum a total diferential $\frac{d}{dt}(q\frac{d\mathbf{r}}{dt}\cdot\mathbf{A})$ and then you would get to the length gauge. On continous case, I will have to work with a Lagrangian density and in this case I don't know how can I find a equivalent Lagrangian (because in this case space and time are put on equal footing).