Perihelion Precession

Relativistic perihelion advance is the secular mismatch between the radial and azimuthal periods of a weakly relativistic bound orbit. For a test body outside a spherical, nonrotating source, the Schwarzschild effective potential contains a term absent from Newtonian gravity. Perturbing about a Kepler ellipse gives the invariant accumulated advance $6\pi GM/[a(1-e^2)]$ per radial period, even though intermediate radial coordinates and orbital elements require a stated approximation.

Required background. We use The Einstein Equation for the vacuum exterior geometry, Events, Intervals, and Minkowski Geometry for asymptotic energy and angular momentum, and The Newtonian Gravitational Field and Potential for the Kepler ellipse and its semi-latus rectum.

Helpful background. Tensor Fields and Natural Operations supports the symmetry argument; Affine and Variational Geodesics supplies the timelike geodesic equation; Killing Fields and Isometry Groups supplies the conserved energy and angular momentum; Differential Equations for Relativity prepares the perturbative orbit equation and resonant secular term.

Bound motion in the Schwarzschild exterior

Let $M$ be the physical source mass and set $\mu=GM$. The calculation assumes a structureless test body, a spherical nonrotating source, a bound orbit outside the source, and

$$\frac{\mu}{p}\ll1, \qquad v^2\sim\frac{\mu}{p}\ll1, \qquad p=a(1-e^2).$$

Source multipoles, spin, finite mass ratio, radiation reaction, and higher post-Newtonian corrections are omitted. In Schwarzschild coordinates the exterior line element is

$$ds^2=-\left(1-\frac{2\mu}{r}\right)dt^2 +\left(1-\frac{2\mu}{r}\right)^{-1}dr^2 +r^2\left(d\theta^2+\sin^2\theta\,d\phi^2\right).$$

Here $r$ is the areal radius. This page uses the exterior geometry as input; Deriving the Schwarzschild Metric establishes its exact-solution origin and domain. Spherical symmetry lets the orbit be placed in the equatorial plane $\theta=\pi/2$. The timelike and axial Killing fields then give the energy $E$ and angular momentum $L$ per unit test mass,

$$E=\left(1-\frac{2\mu}{r}\right)\frac{dt}{d\tau}, \qquad L=r^2\frac{d\phi}{d\tau}.$$

The four-velocity normalization then becomes

$$\left(\frac{dr}{d\tau}\right)^2 =E^2-\left(1-\frac{2\mu}{r}\right) \left(1+\frac{L^2}{r^2}\right).$$

Expanding the right-hand side gives $E^2-1+2\mu/r-L^2/r^2+2\mu L^2/r^3$. Equivalently, in the energy form $\tfrac12\dot r^2+V_{\rm eff}=(E^2-1)/2$, the potential is

$$V_{\rm eff}=-\frac{\mu}{r}+\frac{L^2}{2r^2}-\frac{\mu L^2}{r^3}.$$

The relativistic $r^{-3}$ term is therefore negative in the potential and positive in $\dot r^2$. Reall derives this potential in Reall 2013, §§ 12.2 and 12.4, printed pp. 134–140, Eqs. (12.12)–(12.17), PDF.

The relativistic orbit equation

Set $u(\phi)=1/r$. Using $dr/d\tau=-L\,du/d\phi$ and differentiating the first integral removes $E$ and gives

$$\frac{d^2u}{d\phi^2}+u =\frac{\mu}{L^2}+3\mu u^2.$$

The Newtonian equation is obtained by deleting $3\mu u^2$. Its bound solution is

$$u_0(\phi)=\frac{1}{p}(1+e\cos\phi), \qquad p=\frac{L^2}{\mu}=a(1-e^2)$$

at the retained order. Write $u=u_0+u_1$ and keep only first order in $\mu/p$. The correction satisfies

$$u_1''+u_1 =\frac{3\mu}{p^2} \left(1+2e\cos\phi+e^2\cos^2\phi\right).$$

Constant and $\cos2\phi$ terms merely shift the nonsecular shape at this order. The resonant source $6\mu e\cos\phi/p^2$ produces

$$u_{1,\rm sec}=\frac{3\mu e}{p^2}\,\phi\sin\phi,$$

because $(d^2/d\phi^2+1)(\phi\sin\phi)=2\cos\phi$. Combining the resonant terms,

$$u(\phi) \simeq\frac1p \left[1+e\cos\!\left[\left(1-\frac{3\mu}{p}\right)\phi\right]\right] +\text{bounded corrections}.$$

The radial coordinate returns to perihelion when the cosine phase advances by $2\pi$. The corresponding azimuthal angle is $2\pi(1+3\mu/p)$ to first order, so the excess over one full turn is

$$\boxed{ \Delta\varpi =\frac{6\pi GM}{p} =\frac{6\pi GM}{a(1-e^2)} }.$$

MIT’s explicit perturbative solution displays the same resonant term in Guth–Turner 2017, §§ 7.2–8.2, printed pp. 33–36, Eqs. (7.42)–(8.9), PDF. Will gives the broader post-Newtonian expression and its reduction to the general-relativistic test-body term in Will 2014, § 4.2, printed pp. 45–46, Eqs. (64)–(65), PDF.

Sign, limits, and observable meaning

The correction is positive: successive perihelia advance in the direction of orbital motion. That sign agrees with the effective potential. The negative $-\mu L^2/r^3$ term strengthens the inward relativistic effect at small radius and makes the azimuthal phase accumulate slightly faster than the radial oscillation.

As $G\to0$ or $p\to\infty$, the advance vanishes and the Kepler ellipse closes. The formal $e\to0$ limit is finite, $6\pi GM/a$, but a perfectly circular orbit has no distinguished perihelion. Its invariant interpretation is the mismatch between the azimuthal frequency and the frequency of an infinitesimal radial epicycle. The formula becomes unreliable when $GM/p$ is not small, and it is not a modern model of a spinning, multipolar, comparable-mass, or radiating system.

The secular angle is more robust than the coordinate form of $r(\phi)$: asymptotic directions and successive radial turning points define the comparison. At higher order, one must state precisely how orbital elements are defined. No current residual or parameter bound is needed to establish this evergreen leading result.

An independent check comes from perturbing a circular orbit of areal radius $r_0$. Put $u=1/r_0+\delta u$ in the exact orbit equation and subtract the circular-orbit relation. To first order in the radial displacement,

$$\delta u''+\left(1-\frac{6\mu}{r_0}\right)\delta u=0.$$

For $r_0\gt6\mu$, one radial oscillation takes azimuthal angle $2\pi/\sqrt{1-6\mu/r_0}$. Its excess over $2\pi$ is $6\pi\mu/r_0+O((\mu/r_0)^2)$, agreeing with the small-eccentricity limit above. The oscillation ceases to be stable at $r_0=6\mu$; extrapolating the weak-field advance into that regime would miss this change of behavior.

Worked check

Two weak-field test-body orbits have $(a_1,e_1)$ and $(a_2,e_2)$. Under what condition do they have the same leading perihelion advance, and what happens if both semi-major axes are scaled by a factor $\lambda$ at fixed eccentricity?

Solution

The leading result depends only on the semi-latus rectum $p=a(1-e^2)$. The advances are equal when

$$a_1(1-e_1^2)=a_2(1-e_2^2).$$

At fixed eccentricity, $a\mapsto\lambda a$ gives $p\mapsto\lambda p$ and therefore $\Delta\varpi\mapsto\Delta\varpi/\lambda$. Increasing the orbit’s size weakens the advance, and the $\lambda\to\infty$ limit recovers a closed Newtonian orbit.

References

• Guth, Alan H., lectures; Andrew P. Turner, notes. 8.962 General Relativity, Spring 2017. Massachusetts Institute of Technology, 2017, §§ 7.2–8.2, printed pp. 33–36. PDF.
• Reall, Harvey S. General Relativity. Cambridge Part III lecture notes (2013), §§ 12.2–12.4, printed pp. 134–141. PDF.
• Will, Clifford M. “The Confrontation between General Relativity and Experiment.” Living Reviews in Relativity 17, article 4 (2014), §§ 4.1–4.4, printed pp. 40–51. DOI; arXiv.