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Set $f(r)=1-2M/r$ and use proper time along a timelike geodesic. In the equatorial plane, the radial Christoffel symbols needed for a circular orbit areThe radial geodesic equation with $\dot{r}=0$ gives $(fM/r^2){\dot{t}}^2-fr{\dot{\phi }}^2=0$. Outside the horizon this implies

For stability use the conserved energy $E=f\dot{t}$ and angular momentum $L=r^2\dot{\phi }$. The normalization of the timelike tangent in the Schwarzschild metric gives ${\dot{r}}^2+V(r)=E^2$, with the effective potential for timelike Schwarzschild geodesicsAt a circular orbit $V^{\prime }=0$, so $L^2=M{r}^2/(r-3M)$, requiring $r>3M$. Differentiate $V$ holding this conserved $L$ fixed and then substitute the circular value:The linear radial perturbation equation is $\delta \ddot{r}=-V^{\prime \prime }(r)\delta r/2$. Thus circular timelike orbits are stable for $r>6M$, unstable for $3M<r<6M$, with the marginal case at $6M$.

For the orbiting astronaut at radius $R$, use $d\phi /dt=\sqrt{M/R^3}$ to obtainThe static Earth twin has $d{\tau }_0/dt=\sqrt{1-2M/R_0}$. During equal coordinate-time intervals of the long circular-orbit phase, the astronaut ages less precisely whenThe outward and return transfers have finite additional proper times; the stated long-orbit comparison is the limit in which the circular phase dominates them. At equality the circular phase contributes no age difference, so transfer details can decide the finite-trip result.

The Killing vectors ${\partial }_t$ and ${\partial }_{\varphi }$ give conserved specific energy and angular momentum,The normalization $g_{\mu \nu }{\dot{x}}^{\mu }{\dot{x}}^{\nu }=-1$ then rearranges toso the displayed radial energy is also constant. This is the effective potential for timelike Schwarzschild geodesics.

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.