Derivation of Lagrangian for Classical Electrodynamics

AI Thread Summary
The discussion centers on the derivation of the classical electrodynamic Lagrangian, particularly in relation to current and charge sources. The original poster seeks a deeper understanding beyond the standard presentation in textbooks, expressing interest in incorporating fictitious magnetic charges and currents into the Lagrangian density. They reference the non-relativistic Lagrangian density and contemplate the implications of adding terms from duality principles. A suggestion is made to consult "Quantum Field Theory" by Itzykson and Zuber for a derivation. The conversation highlights the complexities of formulating a robust Lagrangian that accurately reflects the physical principles involved.
Born2bwire
Science Advisor
Gold Member
Messages
1,780
Reaction score
24
Is there a derivation for the classical electrodynamic Lagrangian? I have taken a look at a few textbooks that I have on hand but all of them just state the Lagrangian (in the voodoo four-vector talk, \glares) without explaining the reasoning behind it. I know that the Lagrangian for a charged particle can be found by working it out but I am interesed in the Lagrangian from current and charge sources. What I want to do is apply the non-relativistic Lagrangian density,

\mathcal{L} = \frac{1}{2}\left(\epsilon E^2-\frac{1}{\mu}B^2\right) - \phi\rho + \mathbf{J}\cdot\mathbf{A}

and add in the contribution due to fictious magnetic charges and currents. We often use magnetic currents in our work to simplify the solution process and increase robustness and though I am tempted to just add in the analogue terms from the dual I do not want to just haphazardly cram in terms that look like they are correct without knowing that the principles are sound.
 
Physics news on Phys.org
If you know how to work out the Lagrangian for a charged particle, you can just use that! Treat each infinitesimal piece of your extended charge/current distribution as a point charge dq=\rho dV with current density \textbf{J}=\rho\textbf{v}.
 
gabbagabbahey said:
If you know how to work out the Lagrangian for a charged particle, you can just use that! Treat each infinitesimal piece of your extended charge/current distribution as a point charge dq=\rho dV with current density \textbf{J}=\rho\textbf{v}.

Ok, I was figuring that was going to be it, I got as far as that in Jackson before I saw something shiny and then before I knew it's bedtime. I'll give this a go when I reboot in the morning. Unfortunately this isn't looking like it's going to set itself up the way I would like it to unless I can play around with the gauges... Eh screw it, I'm going to bed.
 
Born2bwire said:
...What I want to do is apply the non-relativistic Lagrangian density,

\mathcal{L} = \frac{1}{2}\left(\epsilon E^2-\frac{1}{\mu}B^2\right) - \phi\rho + \mathbf{J}\cdot\mathbf{A}

and add in the contribution due to fictious magnetic charges and currents.

This looks like the relativistic Lagrangian density to me, just not in four vector and tensor form?
 
A derivation is given in the book "Quantum Field Theory" by Claude Itzykson and Jean-Bernard Zuber, page 7 and further.
 
Back
Top