 Solution

ID: 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

η_{μν} = diag (1, - 1, - 1, - 1)

, raising both indices gives

F^{0 i} = - F_{0 i}, F^{ij} = F_{ij} (i, j = 1, 2, 3) .

The 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. Consequently

F_{μν} F^{μν} = 2 [(B^{1})^{2} + (B^{2})^{2} + (B^{3})^{2} - \frac{( E ^{1} ) ^{2} + ( E ^{2} ) ^{2} + ( E ^{3} ) ^{2}}{c ^{2}}] .

All indices are contracted, so this quantity is a Lorentz scalar. Since the speed of light is invariant,

c^{2} ∣ B ∣^{2} - ∣ E ∣^{2} = \frac{c ^{2}}{2} F_{μν} F^{μν}

is Lorentz invariant. Both terms have the units of squared electric field,

V^{2} m^{- 2}

in SI units.

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