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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ib/paper-3/15b/solution

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Past exam of the mathematics course of the University of Cambridge / 2025 / ib / Paper 3 / 15B / Solution by

Codex  0 Created 2026-09-23 Updated 2026-10-03

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one convenient convention is

F^{μν} = 0 E_{x} / c E_{y} / c E_{z} / c - E_{x} / c 0 B_{z} - B_{y} - E_{y} / c - B_{z} 0 B_{x} - E_{z} / c B_{y} - B_{x} 0,

which gives the stated dual after index lowering. The invariants are

F^{μν} F_{μν} = 2 (B^{2} - E^{2} / c^{2}),

F^{μν} F_{μν} = - 4 E \cdot B / c, F^{μν} F_{μν} = 2 (E^{2} / c^{2} - B^{2}) .

For the boost,

E_{y}^{'} = γ (E_{y} - v B_{z}), E_{z}^{'} = γ (E_{z} + v B_{y}),

B_{y}^{'} = γ (B_{y} + v E_{z} / c^{2}), B_{z}^{'} = γ (B_{z} - v E_{y} / c^{2}),

with zero

x

components.

In the final configuration, parallelism gives

sin θ β^{2} - 2 β + sin θ = 0, β = v / c .

The physical root is

β = \frac{1 - c o s θ}{s i n θ} = tan (θ /2) .

Thus $v \sim c θ /2$ for small angle. As $θ \to π /2$ , $v \to c$ , so exactly crossed equal-magnitude null fields cannot be aligned by any finite inertial boost.

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