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What is the reason 1/4 coefficient in the tensor multiplication of the electromagnetic field strength? $$\mathscr{L} = -\, \frac{1}{4} \, F_{\mu \nu} \, F^{\mu \nu}. \tag{1}$$

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marked as duplicate by AccidentalFourierTransform, Thomas Fritsch, John Rennie electromagnetism Aug 28 at 19:17

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It's a convention, to simplify some calculations. You could add an arbitrary factor. In SI units for example: $$\mathscr{L}_{\text{EM}} = -\: \frac{1}{4 \mu_0} \, F_{\mu \nu} \, F^{\mu \nu}. \tag{1}$$ However, to recover the non-homogeneous Maxwell equation: $\partial_{\mu} \, F^{\mu \nu} = \mu_0 \, J^{\nu}$, you need the $-\, \frac{1}{4}$ factor.

Also, you could write the following: $$-\: \frac{1}{4} \, F_{\mu \nu} \, F^{\mu \nu} = \frac{1}{2} (\, E^2 - B^2), \tag{2}$$ which is similar to the classical mechanical expression $L = K - U = \frac{1}{2} \, m v^2 - U$.

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  • $\begingroup$ In the inhomogeneous case you could also drop the 1/4, but the interaction becomes $2A\cdot J$ instead of just $A\cdot J$. $\endgroup$ – AccidentalFourierTransform Aug 28 at 18:35
  • $\begingroup$ If you also put $\epsilon_0 $ then you find the correct energy from the Noether theorem. $\endgroup$ – my2cts Aug 28 at 19:58

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