 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-1/37b/i/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Assume

qE > 0

and put

α = \frac{qE}{m c} .

\overset{u}{˙}^{0} = α u^{x}, \overset{u}{˙}^{x} = α u^{0},

with

u^{0} (0) = c

and

u^{x} (0) = 0

. Hence

u^{0} = c cosh (ατ), u^{x} = c sinh (ατ) .

Integrating from the origin gives the uniformly accelerated trajectory

c t = \frac{c}{α} sinh (ατ), x = \frac{c}{α} {cosh (ατ) - 1}, y = z = 0.

The light ray has

x_{L} = - h + c t

. At interception,

x_{L} = x

, so

h = c t - x = \frac{c}{α} (1 - e^{- ατ}) .

A finite solution exists exactly when

h < h_{c} = \frac{c}{α} = \frac{m c ^{2}}{qE} .

Then

τ_{c} = - \frac{1}{α} lo g (1 - \frac{α h}{c}),

H = α h / c

t_{c} = \frac{1}{α} sinh (α τ_{c}) = \frac{H ( 2 - H )}{2 α ( 1 - H )} .

h = h_{c}

interception is approached only as

t \to \infty

, and for

h > h_{c}

it never occurs. The limiting backward light ray is the Rindler horizon of the accelerated particle.

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