# $1/4$ coefficient in QED Lagrangian [duplicate]

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

## marked as duplicate by AccidentalFourierTransform, Thomas Fritsch, John Rennie electromagnetism StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Aug 28 at 19:17

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 $$L = K - U = \frac{1}{2} \, m v^2 - U$$.
• In the inhomogeneous case you could also drop the 1/4, but the interaction becomes $2A\cdot J$ instead of just $A\cdot J$. – AccidentalFourierTransform Aug 28 at 18:35
• If you also put $\epsilon_0$ then you find the correct energy from the Noether theorem. – my2cts Aug 28 at 19:58