Past exam of the relativity course of the University of Cambridge / 2018 / Section A / 1 / b / Solution

The required invariant is obtained by tensor contraction, rather than by an ordinary sum of squares of the entries of the electromagnetic field tensor. With ${\eta }_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$, raising both indices givesThe electromagnetic field tensor is an antisymmetric second-rank tensor, so every independent off-diagonal entry occurs twice in its contraction. The three electric contributions each acquire a minus sign, whereas the magnetic contributions retain their sign. ConsequentlyAll indices are contracted, so this quantity is a Lorentz scalar. Since the speed of light is invariant,is Lorentz invariant. Both terms have the units of squared electric field, ${\mathrm{V}}^2{\mathrm{m}}^{-2}$ in SI units.