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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.

Local equivalence and tides. In Newtonian mechanics, inertial mass $m_{\mathrm{i}}$ measures the response to a force, whereas passive gravitational mass $m_{\mathrm{g}}$ couples to an imposed Newtonian gravitational field. A freely falling body obeysThe observed composition independence of free fall says that this ratio is universal. With a common choice of units it is one. This is the Newtonian content of the weak equivalence principle for test particles; it does not by itself specify all nongravitational physics.

The Einstein equivalence principle further says that local nongravitational experiments in a sufficiently small freely falling laboratory have the form predicted by special relativity, independently of the laboratory's location and motion. In such a local inertial frame, the metric tensor is locally the Minkowski metric and the Christoffel symbols vanish at the selected event. The strong equivalence principle extends the universality claim to local gravitational experiments and bodies with significant gravitational self-energy. None of these statements removes measurable tidal forces across a finite apparatus.

In an ideal small freely falling lift, a released ball initially floats relative to the cabin, because both undergo the same free fall. Two separated balls, however, need not retain their separation: they sample different parts of the gravitational field. For a point source, linearizing $\mathbf{g}=-GM\mathbf{r}/r^3$ across a small separation $\mathbf{\xi }$ gives the tidal tensor predictionA radial separation stretches with coefficient $2GM/r^3$, whereas a transverse separation is compressed with coefficient $-GM/r^3$. Initially parallel velocities therefore develop relative motion even though the individual bodies experience zero proper acceleration.

The relativistic statement is geodesic deviation. With the Riemann curvature tensor conventiona connecting vector between neighboring freely falling worldlines satisfiesHere $D/d\tau$ is the covariant derivative along the reference geodesic, and $U^{\mu }$ its four-velocity. Curvature can be nonzero where the Christoffel symbols have been made zero. In the weak, static, slow-motion limit this equation reduces to ${\ddot{\xi }}^i=-{\partial }_i{\partial }_j\Phi {\xi }^j$, recovering the tidal tensor above. A finite falling lift can detect gravity through these relative accelerations; local equivalence does not imply a globally uniform field.

Weak-field connection and potential. Take $x^0=ct$ and the $+---$ metric signature. There is an index error in the printed formula for the Christoffel symbols: its first differentiated metric component must be $g_{\nu 0}$, not $g_{\mu 0}$. The correctly contracted expression isThe offending term vanishes under the stipulated time independence, so this correction does not change the intended calculation. Write $g_{\mu \nu }={\eta }_{\mu \nu }+h_{\mu \nu }$ and treat both $h$ and its derivatives as first order. To that order $g^{00}=1-h_{00}$, $g^{0i}=O(h)$, and $g^{ij}=-{\delta }^{ij}+O(h)$. A stationary metric tensor givesFor the temporal component the inverse-metric coefficient is already first order, soFor a spatial component the zeroth-order inverse metric contributes:The time component of the approximate geodesic equation says that $dt/d\tau$ is constant to the retained order. In its spatial component, convert the proper time derivatives to coordinate-time derivatives and cancel $(dt/d\tau {)}^2$:Comparison with the Newtonian equation $d^2\mathbf{x}/d{t}^2=-\nabla \Phi$ identifies $h_{00}=2\Phi /c^2+C$. For an isolated source, choose the gravitational potential to tend to zero at infinity and normalize the time coordinate there, giving $C=0$. ThusThe quantity $\Phi$ is the Newtonian gravitational potential, or gravitational potential energy per unit mass, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}={\mathrm{m}}^2{\mathrm{s}}^{-2}$. For a point source, $\Phi =-GM/r$. A static clock has $d\tau =\sqrt{g_{00}}dt\approx (1+\Phi /c^2)dt$, connecting the Newtonian limit with gravitational time dilation.

Frequency of the orbiting source. Treat the gas blob as a test particle on a timelike circular Schwarzschild geodesic, with pressure and other nongravitational forces negligible. Treat our very distant location as infinity, so the observer's own gravitational potential is negligible. These are the assumptions that make a unique answer in terms of $R_S/r$ possible.

Put $f=1-R_S/r$ and use a local static observer at the emission point as an intermediate measuring frame. The remote observer lies on the symmetry axis of the orbit. A ray arriving on that axis has zero conserved axial angular momentum, so its local direction is perpendicular to the emitter's azimuthal velocity. Bending within a meridional plane does not alter this orthogonality. The transverse relativistic Doppler effect therefore giveswhere $v$ is the orbital speed measured by that local static observer. The gravitational redshift between static Schwarzschild observers then givesA gravitational circular orbit has $v^2=GM/r+O((GM/r{)}^2/c^2)$ in the weak-field limit, so $v^2/c^2=R_S/(2r)+O((R_S/r{)}^2)$. Expanding both factors,Solving for the emitted frequency givesThe gravitational part is $R_S/(2r)$ and the transverse kinematic time dilation contributes another $R_S/(4r)$; the absence of line-of-sight motion does not remove this second effect.

As an exact check, the local speed on a timelike circular Schwarzschild geodesic satisfies $v^2/c^2=\mu /(r-2\mu )$, where $\mu =GM/c^2=R_S/2$. Therefore the face-on frequency shift of a circular Schwarzschild source iswhose expansion agrees with the boxed answer. If the gas is supported or accelerated by pressure or magnetic stresses rather than following a geodesic, its orbital speed is not fixed by the mass and radius alone; the appropriate first-order result is instead ${\nu }_{\mathrm{em}}\approx {\nu }_{\mathrm{ob}}[1+GM/(r{c}^2)+v^2/(2c^2)]$, with $v$ supplied independently.