 Solution

ID: past-exam-of-the-symmetry-and-relativity-course-of-the-university-of-oxford/2017/4/a/solution

For a boost along positive

x

, let

γ = (1 - v_{x}^{2} / c^{2})^{- 1/2}

. Parallel field components are unchanged, while transverse components mix. The Lorentz transformation of electromagnetic fields gives

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

These signs correspond to

S^{'}

moving with

+ v_{x}

relative to

S

, not the inverse boost. They can be obtained from

F^{'} = Λ F Λ^{T}

 New to topics? Read the docs here!