 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ib/paper-3/15d/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

E_{x} = E_{x}^{'}, E_{y} = γ E_{y}^{'}, E_{z} = γ E_{z}^{'}

and

B_{x} = 0, B_{y} = - \frac{γ v}{c ^{2}} E_{z}^{'}, B_{z} = \frac{γ v}{c ^{2}} E_{y}^{'} .

Equivalently,

B = γ v \times E^{'} / c^{2}

for

v = v e_{x}

Using

1/ μ_{0} = ϵ_{0} c^{2}

and

γ^{2} = (1 - v^{2} / c^{2})^{- 1}

w = \frac{ϵ _{0}}{2} [E_{x}^{'2} + γ^{2} (E_{y}^{'2} + E_{z}^{'2}) + γ^{2} \frac{v ^{2}}{c ^{2}} (E_{y}^{'2} + E_{z}^{'2})] = \frac{ϵ _{0}}{2} [E_{x}^{'2} + \frac{c ^{2} + v ^{2}}{c ^{2} - v ^{2}} (E_{y}^{'2} + E_{z}^{'2})] .

The primed field is static, so

w

depends on

t, x

through

x^{'} = γ (x - v t)

and has no explicit

t^{'}

dependence. The chain rule gives

\partial_{t} w = - γ v \partial_{x^{'}} w, \partial_{x} w = γ \partial_{x^{'}} w .

Consequently

\frac{\partial w}{\partial t} + \nabla \cdot (w v e_{x}) = \partial_{t} w + v \partial_{x} w = 0 .

Thus the field-energy profile is transported rigidly with velocity $v e_{x}$ .

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