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

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SUMMARY

The Lagrangian density for the classical electrodynamic field is expressed as \( \frac{E^2}{2\epsilon_0} - \frac{B^2}{2\mu_0} \). The minus sign is essential for maintaining Lorentz invariance, as a positive sign would violate this principle. Additionally, this formulation ensures that the resulting equations of motion are correct. For a deeper understanding, refer to the provided link detailing electromagnetic relativistic symmetry.

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

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