Why is the electrodynamic Lagrangian E^2 - B^2?

[heinz]

Is there a simple way to understand why the Lagrangian of the classical electrodynamic field is (in SI units)

E^2/2 e0 - B^2/2 mu0 ?

Why is there a minus in it? Is there some simple, intuitive explanation for it?

Heinz

 

[robphy]

One reason: this Lagrangian density has to be a Lorentz-invariant scalar. A plus-sign there would not be Lorentz-invariant.

It has a resemblance to T-V...
"E involves time-derivatives of A" is similar to "v involves time-derivatives of x".

Why this minus sign? A common answer is that it is what gives the correct equations of motion.

 

[GiftOfPlasma]

http://www.hep.caltech.edu/~peck/lecture_EMRelativisticSymmetry.pdf"

This link has a little more detail about what robphy was saying.