 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2016/ib/paper-2/18d/c/solution

Codex  0 Created 2026-09-24 Updated 2026-10-06

F^{' μν} = Λ^{μ}_{α} Λ^{ν}_{β} F^{α β}

. For example,

F^{' 02} = γ F^{02} - γ (v / c) F^{12} = - \frac{γ}{c} (E_{y} - v B_{z}) .

Reading off the other components in the same way yields the transverse field transformation

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

The longitudinal components remain

E_{x}^{'} = B_{x}^{'} = 0

. These signs correspond to the stated boost in which

S^{'}

moves with velocity

v

relative to

S

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