 Solution

ID: past-exam-of-the-relativity-course-of-the-university-of-cambridge/2017/1/c/solution

(+, -, -, -)

and the supplied covariant electromagnetic field tensor, so

F_{0 i} = E_{i}

and

F_{ij} = - ϵ_{ijk} B_{k}

. Raising both indices reverses the time-space components but leaves the spatial components unchanged. Its complete contraction is therefore

F_{μν} F^{μν} = 2 (B^{2} - E^{2}) .

F^{μν} = \frac{1}{2} ε^{μν ρ σ} F_{ρ σ}

, with

ε^{0123} = + 1

. Direct contraction gives

E^{2} - B^{2} = - \frac{1}{2} F_{μν} F^{μν}, E \cdot B = - \frac{1}{4} F_{μν} F^{μν} .

These electromagnetic field invariants are complete tensor contractions. The second is a pseudoscalar: it is invariant under proper Lorentz transformations and changes sign under spatial inversion.

For

T^{0 i}

the metric trace term vanishes. Since

F^{0}_{j} = E_{j}

F^{0}_{ρ} F^{i ρ} = \sum_{j} E_{j} (- ϵ_{ijk} B_{k}) = - (E \times B)_{i} .

Consequently the Poynting vector is exactly the time-space row in the stated

c = 1

units:

(T_{em}^{01}, T_{em}^{02}, T_{em}^{03}) = \frac{1}{μ _{0}} E \times B .

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