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

In summary, the Lagrangian for charged particles in an electromagnetic field is written like: -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} and the Lorentz force equation is: f_\nu = q F_{\nu\mu} \dot{y}^\mu + \frac{1}{4} \frac{\partial}{\partial y^\nu} \int d^3 x F_{\
  • #1
DuckAmuck
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How would you unify the two Lagrangians you see in electrodynamics?
Namely the field Lagrangian:
Lem = -1/4 Fμν Fμν - Aμ Jμ
and the particle Lagrangian:
Lp = -m/γ - q Aμ vμ
The latter here gives you the Lorentz force equation.
fμ = q Fμν vν

It seems the terms - q Aμ vμ and - Aμ Jμ account for the same thing. So if you were to add in the kinetic term from Lem to Lp, you get:
L+ = -m/γ - q Aμ vμ -1/4 Fμν Fμν
The subsequent Lorentz force equation is:
fμ = q Fμν vν + 1/4 ∂μ(Fαβ Fαβ)

This looks weird and makes me think I am missing something. What do you all think? Thanks in advance.
 
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  • #2
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").
 
  • #3
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.
 
  • #4
But the last term is 0, because nothing depends on ##y##!
 
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  • #5
vanhees71 said:
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:
$$ \partial_\nu \rho \sqrt{\dot{y}_\mu \dot{y}^\mu} + \partial_\nu (A_\mu J^\mu) + \frac{1}{4} \partial_\nu (F_{\rho\sigma} F^{\rho\sigma}) = \frac{d}{d\tau} \left( \rho \frac{\dot{y}_\nu}{\sqrt{\dot{y}_\mu \dot{y}^\mu}} \right)$$
 

1. What is a Lagrangian in electrodynamics?

A Lagrangian is a mathematical function that describes the dynamics of a physical system. In electrodynamics, it is used to describe the behavior of electric and magnetic fields and their interactions with charged particles.

2. How is the Lagrangian used to unify Fμν, Aμ Jμ, and the Lorentz force?

The Lagrangian for electrodynamics combines the equations for the electromagnetic field tensor (Fμν), the four-potential (Aμ), and the four-current (Jμ) to derive the equations of motion for charged particles under the influence of electromagnetic fields, known as the Lorentz force.

3. What is the significance of unifying these equations?

Unifying these equations allows us to describe the behavior of charged particles in electromagnetic fields using a single, comprehensive framework. This makes it easier to understand and analyze complex systems and phenomena involving electricity and magnetism.

4. How does the Lagrangian approach differ from the traditional Maxwell's equations?

The traditional Maxwell's equations describe the behavior of electric and magnetic fields in terms of their sources (charges and currents). In contrast, the Lagrangian approach describes the behavior of these fields in terms of their dynamics, making it more intuitive and generalizable.

5. What are some practical applications of unifying Lagrangians in electrodynamics?

The unified Lagrangian approach has been used in various fields, including particle physics, plasma physics, and astrophysics, to study the behavior of charged particles in complex electromagnetic environments. It has also been used to develop advanced technologies such as particle accelerators and magnetic confinement fusion reactors.

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