Past exam of the relativity course of the University of Cambridge / 2018 / Section B / B3

Use the $+---$ metric signature and abbreviate $f(r)=1-2\mu /r$, with $\mu =GM/c^2$ a length. Dots initially mean differentiation with respect to an affine parameter $\lambda$. The geodesic Lagrangian in the equatorial plane isThe planarity of geodesics in spherical symmetry permits this plane to be chosen. More explicitly, the full angular Euler-Lagrange equation is $\ddot{\theta }+2\dot{r}\dot{\theta }/r-\sin \theta \cos \theta {\dot{\varphi }}^2=0$, so initial data $\theta =\pi /2$, $\dot{\theta }=0$ preserve it.

Both $t$ and $\varphi$ are cyclic coordinates. Their Euler-Lagrange equations give conserved conjugate momenta $p_t=c^{2}f\dot{t}$ and $p_{\varphi }=-r^2\dot{\varphi }$. Defining constants $k,h$ therefore givesThe corresponding second-order geodesic equations areand variation of $r$ givesThe geodesic Lagrangian has no explicit $\lambda$ dependence. Its conserved Lagrangian energy function is ${\dot{x}}^{\alpha }\partial \mathcal{L}/\partial {\dot{x}}^{\alpha }-\mathcal{L}=\mathcal{L}$, since it is quadratic in the tangent components. Thus twice this function is a constant:This is the constant squared tangent norm of an affinely parametrized geodesic.

For a massive particle choose $\lambda =\tau$, its proper time, and obtain $\kappa =c^2$. For a photon, the tangent is null and $\kappa =0$. A general rescaling of a massive particle's affine parameter rescales a positive $\kappa$; the numerical value $c^2$ requires the proper time choice. With that choice, $h=L_z/m$ is signed specific angular momentum, of units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$, and $k=E/(m{c}^2)$ is dimensionless conserved specific Killing energy. The time coordinate is normalized to proper time at infinity, so this $E$ is the particle's conserved energy referred to infinity. A photon's constants scale with the normalization of its null tangent; it has no rest frame or proper time parametrization.

Substitution of $\dot{t}=k/f$ and $\dot{\varphi }=h/r^2$ into the norm gives the general radial first integralFor a massive particle this becomesAdopting the usual half-normalization for the effective potential for timelike Schwarzschild geodesics, defineAll terms have units ${\mathrm{m}}^2{\mathrm{s}}^{-2}$, or energy per unit mass. The first two terms are the Newtonian gravitational and centrifugal contributions; the last is the relativistic correction.

A circular orbit requires both $\dot{r}=0$ and $V_{\mathrm{eff}}^{\prime }(R)=0$. Differentiating at fixed $h$ givesThe radial first integral also givesFor the future-directed circular orbit at $R=4\mu$, these constants are $|h|=4\mu c$, $k=1$, and $f=1/2$. ThereforeSince Schwarzschild time agrees with the clock time of a static observer at infinity, the intended infinity-referred tangential orbital speed isThis uses the circumferential areal radius $R$ with the clock at infinity; it is not a local measurement made at the orbit. A local static observer's clock instead has $d{\tau }_{\mathrm{stat}}=\sqrt{f}dt$ and measures $v_{\mathrm{local}}=v_{\infty }/\sqrt{f}=\boxed{c/\sqrt{2}}$. These two operationally different speeds explain the significance of the phrase “at infinity”.

The effective-potential stability of a circular orbit follows fromConsequently the $R=4\mu$ orbit exists but is unstable to radial perturbations. The original plot below shows its position at a maximum, for $h=4\mu c$.

For a weak-field elliptic orbit, the relativistic correction makes the direction of closest approach advance after each radial cycle. The derivations under i and ii below giveper orbit to first order, where $a_{\mathrm{orb}}$ is the semi-major axis and $e$ the orbital eccentricity. For Mercury, the additional Schwarzschild perihelion precession is approximately 43 arcseconds per century, over and above the much larger Newtonian precession produced primarily by other planets. It is the unexplained residual after those effects have been included that provides the classic quantitative test of general relativity; one should not identify the whole observed precession with this relativistic correction.^[ref]

The labels i and ii in the original paper introduce optional orbit-equation aids, rather than separate mandatory subquestions. They are retained here for derivations of both aids and of the perihelion result.

The relativistic orbit-shape equation can be derived without dividing by $\dot{r}$ at a turning point. Substitute the norm constraint into the radial geodesic equation to eliminate $\dot{t}$ and ${\dot{r}}^2$. Using $f^{\prime }=2\mu /r^2$ givesFor a massive particle with proper time as parameter, $\kappa =c^2$. Put $u=1/r$, and use primes in this paragraph for derivatives with respect to $\varphi$. The constant specific angular momentum gives $\dot{\varphi }=h{u}^2$, so the chain rule yieldsSubstituting these expressions into the radial equation and dividing by $h^2{u}^2$ for a nonradial orbit givesThe extra $3\mu u^2$ term is responsible for Schwarzschild perihelion precession. It vanishes in the Newtonian weak-field limit $\mu /r\to 0$. The derivation uses a massive particle; for a null geodesic, $\kappa =0$ and the constant $GM/h^2$ on the right is absent.

In Newtonian gravity, the radial equation with conserved specific angular momentum is $\ddot{r}=-GM/r^2+h^2/r^3$. The substitution $u=1/r$ therefore gives the linear differential equationChoosing $\varphi =0$ along perihelion, its elliptic orbit solution isHere $p$ is the semilatus rectum. One can check the last equality directly from $r_{\mathrm{peri}}=p/(1+e)$, $r_{\mathrm{apo}}=p/(1-e)$, and $a_{\mathrm{orb}}=(r_{\mathrm{peri}}+r_{\mathrm{apo}})/2$.

To find Schwarzschild perihelion precession, write $u=u_0+\delta u$ in the relativistic equation and retain first order in $\mu /p$. The linear ordinary differential equation for the correction isThe constant and double-frequency forcing change the orbit's shape without accumulating an angular shift. The resonant $\cos \varphi$ term producessince $(d^2/d{\varphi }^2+1)(\varphi \sin \varphi )=2\cos \varphi$. The oscillating part of the perturbed orbit is consequently the first-order expansion ofSuccessive perihelia are separated by $2\pi /(1-3\mu /p)$, slightly more than a full revolution. Their forward angular advance is thusin radians per orbit. This leading expression applies when $\mu /p\ll 1$. For a perfectly circular orbit there is no uniquely located perihelion; the same expression describes the limiting mismatch of radial and azimuthal frequencies for a small nonzero orbital eccentricity.