# Building the Lagrangian of electromagnetism from the Lorentz invariant?

The definition of the relativistic Action is

$$S=\int_a^b ds$$

The Lorentz invariant of electromagnetism is

$$s^2=\frac{1}{c^2}||\mathbf{E}||^2-||\mathbf{B}||^2-2i\frac{1}{c}(\mathbf{B}\cdot \mathbf{E})$$

Therefore, could we write the action as:

$$S=\int_a^b d\sqrt{\frac{1}{c^2}||\mathbf{E}||^2-||\mathbf{B}||^2-2i\frac{1}{c}(\mathbf{B}\cdot \mathbf{E})}$$

This may not be the prettiest way to write it I concede, but is it possible? Clearly, this is not the field description, but perhaps the movement of a test particle/photon in an electromagnetic field?

• You should probably look up the covariant notation for electromagnetism, which is written in terms of the scalar and vector potentials. It is exactly what you are looking for. – Feel My Black Hole Aug 26 '19 at 13:37
• @FeelMyBlackHole I am familiar with the covariant notation, but what I am looking for it to know if the Lagrangian can be built from the Lorentz invariant in the manner stated in the question. – Alexandre H. Tremblay Aug 26 '19 at 13:40
• Possible duplicates: physics.stackexchange.com/q/87817/2451 and links therein. – Qmechanic Aug 26 '19 at 15:44

Your first equation is the relativistic action for a free point particle, where $$ds$$ is the infinitesimal invariant arc length along its world line.
This invariant geometric spacetime interval has nothing to do with the electromagnetic field invariant $$s^2$$ you write in the next equation.
The final equation makes no sense. The variable of integration in a normal integral cannot itself be a function. And what are $$a$$ and $$b$$ now? Not only does it not describe the motion of either a test particle or a photons in an EM field (and, by the way, photons just go straight and don’t feel the field), there is no way that it could describe these things. No world line appears in this integral, nor any interaction between the field and a charged test particle.