 Solution

ID: past-exam-of-the-general-relativity-and-cosmology-course-of-the-university-of-oxford/2018/1/d/solution

Put

F (r) = 1 - r_{s} / r

and

r_{s} = 2 GM / c^{2}

. For radial motion

u^{θ} = u^{ϕ} = 0

a^{0} = \frac{d u ^{0}}{d τ} + \frac{2 GM}{c ^{2} r ^{2} F} u^{0} u^{r},

and

a^{r} = \frac{d u ^{r}}{d τ} + \frac{GM}{c ^{2} r ^{2}} [F (u^{0})^{2} - F^{- 1} (u^{r})^{2}] .

- F (u^{0})^{2} + F^{- 1} (u^{r})^{2} = - c^{2} .

Its bracketed combination is therefore exactly

c^{2}

, independently of the instantaneous velocity or the thrust history. Consequently

a^{r} = \frac{d u ^{r}}{d τ} + \frac{GM}{r ^{2}} .

The derivative here is with respect to proper time. This expression does not say that the ship measures a gravitational acceleration

GM / r^{2}

during free fall; setting

a^{r} = 0

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