0
$\begingroup$

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)?

$\endgroup$
  • $\begingroup$ Could you simply take an infinitesimal portion of charge in space, write out the Lagrangian for that portion as if it was a particle, then sum over all such portions (integrating over space) to obtain the Lagrangian of the system as a function of time and the charge density function? $\endgroup$ – Cicero Jul 31 '18 at 16:35
  • $\begingroup$ It's probably worth noting that $\rho$ and $\mathbf{v}$ are not arbitrary, but must satisfy $\partial \rho/\partial t = - \mathbf{\nabla} \cdot (\rho \mathbf{v})$ via the continuity equation. This might cause your expression to simplify somewhat. $\endgroup$ – Michael Seifert Jul 31 '18 at 16:39
  • $\begingroup$ You will find therein : Deriving Lagrangian density for electromagnetic field a derivation of the Lagrangian for charge distribution based on Maxwell equations only, that is without any use of the Lorentz force equation, the later going together with the former, being independent of them and could not be derived from them. $\endgroup$ – Frobenius Aug 2 '18 at 12:37
  • $\begingroup$ So, if you could reach this Lagrangian based on the Lorentz force equation only then we must suspect that there exists some mysterious still undiscovered connection between the Lorentz force equation and Maxwell equations. $\endgroup$ – Frobenius Aug 2 '18 at 12:38
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.