Graduate Unifying Lagrangians in Electrodynamics: Fμν, Aμ Jμ, & Lorentz Force

[DuckAmuck]

How would you unify the two Lagrangians you see in electrodynamics?
Namely the field Lagrangian:
L_em = -1/4 F_μν F^μν - A_μ J^μ
and the particle Lagrangian:
L_p = -m/γ - q A_μ v^μ
The latter here gives you the Lorentz force equation.
f_μ = q F_μν v^ν

It seems the terms - q A_μ v^μ and - A_μ J^μ account for the same thing. So if you were to add in the kinetic term from L_em to L_p, you get:
L₊ = -m/γ - q A_μ v^μ -1/4 F_μν F^μν
The subsequent Lorentz force equation is:
f_μ = q F_μν v^ν + 1/4 ∂_μ(F_αβ F^αβ)

This looks weird and makes me think I am missing something. What do you all think? Thanks in advance.

 

[vanhees71]

I don't understand your last paragraph. Obviously you consider point-particle mechanics of charged particles and the electromagnetic field as a dynamical system, which is more complicated (and in a sense still ill-defined after more than 100 years of struggle since Lorentz) than you might think. Formally it's as follows:

Your Lagrangian consists of 3 parts: the free matter Lagrangian (where "matter" here means a single particle)
$$L_0^{(\text{matter})}=-m \sqrt{\dot{y}_{\mu} \dot{y}^{\mu}},$$
where the dot means the derivative wrt. an arbitrary world-line parameter (which can in this form of the Lagrangian NOT be proper time); $y^{\mu}(\lambda)$ is the parametrization of the point-particle worldline.

Then the Lagrangian of the free electromagnetic field,
$$L_0^{(\text{field})}=\int_{\mathbb{R}^3} \mathrm{d}^3 x [-\frac{1}{4} F_{\mu \nu} F^{\mu \nu}],$$
where
$$F_{\mu \nu} =\partial_{\mu} A_{\nu} -\partial_{\nu} A_{\mu}.$$

Finally you have the interaction Lagrangian
$$L_{\text{int}}=-q A_{\mu}(y) \dot{y}^{\mu}.$$
This you can formally write also in the other form you have in mind
$$L_{\text{int}}=-\int \mathrm{d}^3 \vec{x} A_{\mu}(x) J^{\mu},$$
where for a point particle
$$J^{\mu}=q \int_{-\infty}^{\infty} \mathrm{d} \lambda \dot{y}^{\mu} \delta^{(4)}[x-y(\lambda)].$$
The equations of motion are very problematic since for the equation of motion of the particle you evaluate the field due to this particle at the space-time point of this same particle, which is ill-defined.

For a discussion of this, see Sects. 4.7 and 4.8 (the former is about the Lagrangian in the continuum-theoretical case, which is well-defined, the latter about point particles and the radiation-reaction problem) in

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

For a discussion of the Landau-Lifshitz approximation of the LAD equation, see the paper by Nakleh quoted in my manuscript (which is still far from being in final form, particularly concerning this "poin-particle enigma").

 

[DuckAmuck]

The last paragraph is basically asking, how do I write the full Lagrangian of a massive charged particle in an electromagnetic field?

From what you've said, I gathered that it would be written like:
$$ L = -m\sqrt{\dot{y}_\mu \dot{y}^\mu} - q A_\mu (y) \dot{y}^\mu - \frac{1}{4} \int d^3 x F_{\mu\nu} F^{\mu\nu} $$
You can derive the Lorentz force from this, but the term containing the field tensor F seems like it would add an extra term to the force equation due to its dependence on position.
Evaluate the Euler-Lagrange equation:
$$ \frac{\partial L}{\partial \dot{y}^\nu} = -\frac{m}{\sqrt{\dot{y}_\mu \dot{y}^\mu}}\dot{y}_\nu - q A_\nu (y) $$
$$\frac{d}{d\tau} \frac{\partial L}{\partial \dot{y}^\nu} = - f_\nu - q A_{\nu,\mu} \dot{y}^\mu $$
$$ \frac{\partial L}{\partial y^\nu} = -q A_{\mu,\nu} \dot{y}^\mu - \frac{1}{4} \frac{\partial}{\partial y^\nu} \int d^3 x F_{\rho\sigma} F^{\rho\sigma} $$
Put together to get the Lorentz force:
$$ f_\nu = q F_{\nu\mu} \dot{y}^\mu + \frac{1}{4} \frac{\partial}{\partial y^\nu} \int d^3 x F_{\rho\sigma} F^{\rho\sigma} $$
I want to say the last term vanishes, but it seems like there could be a condition where it doesn't. For example, if the limits on the integral depend on y.

 

[vanhees71]

But the last term is 0, because nothing depends on $y$!

 

[DuckAmuck]

But the last term is 0, because nothing depends on $y$!

Okay so the Lagrangian behavior is straightforward then. What about the Lagrangian density? Where rho is the mass density of a particle cloud.
$$ \mathcal{L} = -\rho(y) \sqrt{\dot{y}_\mu \dot{y}^\mu} - A_\mu J^\mu -\frac{1}{4} F_{\rho\sigma} F^{\rho\sigma}$$
$$ \frac{\partial \mathcal{L}}{\partial y^\nu} = -\partial_\nu \rho \sqrt{\dot{y}_\mu \dot{y}^\mu} -\partial_\nu (A_\mu J^\mu)- \frac{1}{4} \partial_\nu (F_{\rho\sigma} F^{\rho\sigma}) $$
$$ \frac{\partial \mathcal{L}}{\partial \dot{y}^\nu} = -\rho \frac{\dot{y}_\nu}{\sqrt{\dot{y}_\mu \dot{y}^\mu}} $$
Put together to get:
$$ \partial_\nu \rho \sqrt{\dot{y}_\mu \dot{y}^\mu} + \partial_\nu (A_\mu J^\mu) + \frac{1}{4} \partial_\nu (F_{\rho\sigma} F^{\rho\sigma}) = \frac{d}{d\tau} \left( \rho \frac{\dot{y}_\nu}{\sqrt{\dot{y}_\mu \dot{y}^\mu}} \right)$$