Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 4 / 37E / b

Table of contents
- i b
  - Solution i
- ii b
  - Solution ii
- iii b
  - Solution iii

i

Solution

Along the radial curve at fixed

t, θ, φ

, the spatial line element is

d S = 1 + 2 M / r d r

. To first order,

S = \int_{r_{1}}^{r_{2}} (1 + \frac{M}{r}) d r,

so

S = (r_{2} - r_{1}) + M lo g \frac{r _{2}}{r _{1}} .

ii

Solution

A radial null geodesic satisfies

0 = - (1 - \frac{2 M}{r}) d t^{2} + (1 + \frac{2 M}{r}) d r^{2},

so for the outgoing ray

\frac{d t}{d r} = 1 + \frac{2 M}{r} + O (M^{2} / r^{2}) .

The coordinate-time interval is therefore

Δ t = (r_{2} - r_{1}) + 2 M lo g \frac{r _{2}}{r _{1}} .

The stationary observer at

r_{2}

measures proper time

d T = 1 - 2 M / r_{2} d t

. Hence

T = (r_{2} - r_{1}) + 2 M lo g \frac{r _{2}}{r _{1}} - \frac{M}{r _{2}} (r_{2} - r_{1}) .

iii

Solution

Put

Δ r = r_{2} - r_{1}

. Expanding the quotient

S / T

consistently to first order gives

v_{eff} = 1 + \frac{M}{r _{2}} - \frac{M}{Δ r} lo g \frac{r _{2}}{r _{1}} .

Thus

v_{eff} = 1 + \frac{M}{r _{2}} - \frac{M}{r _{2} - r _{1}} lo g \frac{r _{2}}{r _{1}} .

Because

lo g (r_{2} / r_{1}) / (r_{2} - r_{1}) \to 1/ r_{2}

r_{1} \to r_{2}

,

r_{1} \to r_{2} lim v_{eff} = 1 .

This local limit is the expected speed of a massless particle measured in a freely falling local inertial frame; the finite-interval effective velocity combines nonlocal length and clock comparisons.

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