Past exam of the general relativity course of the University of Oxford / 2021 / 1

In general relativity, gravity is the geometry of spacetime, whose curvature is related to the stress-energy tensor, and freely falling test particles follow timelike geodesics of its metric tensor.

The apple's proper time $\tau$ is the time read by a clock carried with it, with $d{\tau }^2=-d{s}^2/c^2$ for metric signature $-+++$. At any event choose a local inertial frame with coordinates ${\xi }^{\alpha }$, in which free motion satisfies $d^2{\xi }^{\alpha }/d{\tau }^2=0$ at that event. The chain rule givesMultiplying by the inverse coordinate Jacobian matrix yields the exact geodesic equationHere $x^{\mu }$ are the observer's coordinates, $u^{\mu }$ is the apple's four-velocity, and the Christoffel symbols describe the Levi-Civita connection. The last expression applies when the $\xi$ coordinates are locally inertial at the event considered; one such chart need not remain inertial throughout a finite region.

Choose $x^0=ct$, so $u^0=cdt/d\tau$. In the weak-field approximation and at small spatial speed compared with $c$, the leading spatial geodesic equation isIt is exact at the release event when all spatial $u^i$ vanish, and is the intended approximation to the displayed expression in the source. Terms $2{\Gamma }_{0j}^{\mu }{u}^0{u}^j$ and ${\Gamma }_{ij}^{\mu }{u}^i{u}^j$ cannot generally be dropped; the time component must also be retained to find changing $dt/d\tau$. For a concrete counterexample to the unrestricted expression, a straight free trajectory $x=v\gamma \tau$, $y=b$ in flat spacetime has polar radius $r=\sqrt{b^2+v^2{\gamma }^2{\tau }^2}$ and $d^{2}r/d{\tau }^2=v^2{\gamma }^2{b}^2/r^3\ne 0$, although ${\Gamma }_{00}^r=0$. The missing angular term ${\Gamma }_{\varphi \varphi }^r(d\varphi /d\tau {)}^2$ supplies that radial coordinate acceleration.

Approximate Earth by an isolated, nonrotating, spherically symmetric body. By Birkhoff's theorem, the exterior vacuum metric tensor is the Schwarzschild metric. With $x^0=ct$ it isHere $r$ is the areal radius, $\theta$ and $\varphi$ are spherical angles, $t$ is Schwarzschild time normalized to the proper time of a static observer at infinity, $M$ is Earth's mass, $G$ the gravitational constant, and $c$ the speed of light. The nonzero diagonal metric tensor components are $g_{00}=-f$, $g_{rr}=f^{-1}$, $g_{\theta \theta }=r^2$, $g_{\varphi \varphi }=r^2{\sin }^2\theta$.

The apple is a test particle; neglect its effect on the metric tensor, air resistance, Earth's rotation and nonspherical structure, and external tidal forces. Near Earth's surface, $GM/(R_E{c}^2)\simeq 7\times 10^{-10}$, $h/R_E\ll 1$, and the falling speed remains much smaller than $c$. Thus the Newtonian limit of this metric tensor is appropriate, while the exact radial geodesic is available as a check. A static observer under the tree has proper time $T=\sqrt{f(R_E)}t$, up to an additive constant. Such an observer is supported by the ground and has nonzero proper acceleration; the released apple instead has zero proper acceleration.

Write $\mu =GM/c^2$ and $r_0=R_E+h$. The Christoffel symbols follow fromThe relevant time and radial entries, using $x^0=ct$, areFor the angular geodesic equations, ${\Gamma }_{r\theta }^{\theta }={\Gamma }_{r\varphi }^{\varphi }=1/r$, ${\Gamma }_{\varphi \varphi }^{\theta }=-\sin \theta \cos \theta$, and ${\Gamma }_{\theta \varphi }^{\varphi }=\cot \theta$. Thus, with primes on coordinates here denoting proper time derivatives,Initial conditions ${\theta }^{\prime }={\varphi }^{\prime }=0$ are preserved by these ordinary differential equations. The motion is therefore radial, with constant angular coordinates. Choose the polar axis so the apple is away from the angular coordinate singularities.

The exact remaining radial geodesic equation and four-velocity normalization areEliminating $dt/d\tau$ gives the particularly simple exact result $r^{\prime \prime }=-GM/r^2$. The supported observer's proper time $T$ differs from the apple's proper time by relative corrections of order $GM/(R_E{c}^2)$ and $v^2/c^2$. Consequently, to leading order in the weak-field approximation and $h/R_E$, $d^{2}r/d{T}^2\simeq -GM/R_E^2=-g$, with $r(0)=r_0$ and zero initial radial velocity. HenceThe proper height measured along a static radial ruler is ${\int }_{R_E}^{r}dr/\sqrt{f}\simeq r-R_E$ at this accuracy. The use of $T$ means the ground clock's elapsed time, with propagation delays corrected; raw visual arrival times include the changing light travel time.

If variation of $g$ over the fall is retained but relativistic corrections are neglected, a useful parametric solution isIts small-$\eta$ expansion gives the same constant-$g$ result. For an exact check on the choice of clock, the initial coordinate acceleration measured using $T$ is $-GMf(r_0)/[r_0^{2}f(R_E)]$, which reduces to the displayed approximation.

The time-translation Killing vector field of the Schwarzschild metric gives a conserved Killing energy per unit rest mass,For unit rest mass, this is the total energy relative to infinity, including rest energy. At release $dr/d\tau =0$, so the four-velocity normalization gives $dt/d\tau =f(r_0{)}^{-1/2}$ andSubstitution into the radial four-velocity normalization yieldsChoosing the inward square root and evaluating at the ground provesThis identity is exact within the assumed Schwarzschild metric, despite its familiar Newtonian mechanics form. The quantity $dr/d\tau$ is not exactly the locally measured speed. A Schwarzschild static observer measures $v_{\mathrm{loc}}=(dr/dt)/f=(dr/d\tau )/\sqrt{f(r_0)}$. Thus the distinction disappears in the stated weak-field approximation, giving $v_{\mathrm{loc}}^2\simeq 2gh$ for $h\ll R_E$.