The following equation (which I shall term the Planetary Precession Formula, PPF for short) famously appeared in a 1915 publication by Einstein where he indicated how it could be derived from his General Theory of Relativity (GTR).

$$\epsilon = \frac{24 \, \pi^3a^2}{c^2 T^2(1-e^2)}$$

where $\epsilon$ is the (anomalous, non-Newtonian) angular precession per orbit, $a$ is the orbit semi-major axis, $c$ is the speed of light, $T$ is the orbital period, $e$ is the orbit ellipticity.

The PPF formula accurately predicts the (anomalous, non-Newtonian) precession of Mercury and other Solar planets.

The formula was known in scientific circles well before 1915. For example Gerber (1898) derived it from his own (widely-derided) model of gravity. In the internet article Gerber's Gravity it is written that

It became a fairly popular activity in the 1890s for physicists to propose various gravitational potentials based on finite propagation speed in order to account for some or all of Mercury's orbital precession. Oppenheim published a review of these proposals in 1895. The typical result of such proposals is a predicted non-Newtonian advance of orbital perihelia per revolution of... >

$$k\,\frac {\pi\,m}{L \,c^2} = k \frac{4 \, \pi^3a^2}{c^2 T^2(1-e^2)}.$$

where $L = a(1 - e^2)$ is the semi-latus rectum of an ellipse, $m$ is a function of the angular speed $\omega$ of an orbiting planet: $m = a^3 \omega^2$ with $\omega = 2\pi/T$ and $k$ is a constant which might be derived from theory.

Clearly with $k = 6$ we get the PPF formula given above.

I wish to know where the $k\pi m/Lc^2$ expression comes from. From the article it would appear to come from the 28-page review paper by Oppenheim, 1895 which is scanned here. I have been through the scans of this paper but without finding that equation explicitly (the paper is in German which I know very poorly, Google Translate helps a bit but leaves a lot of ambiguity). It might be that the anonymous author of the article extracted the expression from a review of Oppenheim's paper or even the original (French & German) papers themselves, but he is not contactable. Maybe someone here is familiar with this era of astrophysical history and can point me in the right direction?

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    $\begingroup$ Interesting question. Minor pedantic note: if you place the period inside the large math environments, as in $$formula\text{.}$$, then you won't get a trailing period all alone on a single line. $\endgroup$
    – Stan Liou
    Commented Nov 12, 2014 at 7:28
  • $\begingroup$ @Stan Liou. Good style aids communication so I'm happy to receive any such tips :). $\endgroup$
    – steveOw
    Commented Nov 12, 2014 at 11:53

1 Answer 1


I do not know where that formula was first published in full, but Oppenheim at least does something very close to it. First, keep in mind some of the relevant symbols in Oppenheim, though they're rather standard: $$\def\anode{☊}\begin{eqnarray*} k &=& \small\sqrt{G} = \small\text{Gaussian gravitational constant}\\ \anode &=& \small\text{longitude of the ascending node}\\ \omega &=& \small\text{argument of perihelion}\\ \varpi &=& \omega+\anode = \small\text{longitude of perihelion}\\ n &=& \small{k\sqrt{(m_0+m_1)/a^3}} = \small\text{mean motion} \end{eqnarray*}$$ We can see that if we massage the notation, the proportionality we're looking for is equivalently stated as: $$\frac{\pi m}{Lc^2} \leadsto \frac{\pi GM}{a(1-e^2)c^2} \leadsto \frac{\pi n^2a^2}{(1-e^2)c^2}\text{.}$$ Since the mean motion is $n\equiv 2\pi/T$, where $T$ is the orbital period, $$\frac{n^3a^2}{c^2} = 2\pi\frac{n^2a^2}{c^2}\frac{1}{T}\text{,}$$ which means that if we want to talk about a perihelion advance of $\delta\varpi \propto \frac{\pi n^2a^2}{(1-e^2)c^2}$ per orbit, it is equivalent to talk about a term in the form $$\frac{\mathrm{d}\varpi}{\mathrm{d}t} \propto \frac{n^3a^2}{(1-e^2)c^2} = \frac{n^3a^2}{c^2}\left(1+e^2+\mathcal{O}(e^4)\right)\text{.}$$ I cannot find where, if indeed anywhere, Oppenheim considers the missing factor of $(1-e^2)$ as belonging to the anomalous precession, but otherwise the formula is definitely there. I suspect that he only uses a circular orbit approximation, $e^2\approx 0$, because it's nowhere to be found in Sec. IV (Weber's 1846 theory) and performing his calculation (sans $1-e^2$) gives me $\delta\varpi = 13.72''$ per century for Mercury, in good agreement with his stated result of $\delta\varpi = 13.65''$, whereas putting in a factor of $(1-e^2)$ by hand gives $\delta\varpi = 14.32''$ instead.

Perhaps Oppenheim did not consider it explicitly and the author of the MathPages took it as obvious that the eccentricity factor should be there. Or perhaps there's a side comment in the text that I don't see; unfortunately, I'm nowhere fluent enough in German to understand much of what's going on.

  • $\begingroup$ Thanks for pointing out the meaning of $k$ and the omission of $(1-e^2)$. I see $e$ appearing in $dL_o/dt = (1/2)e^2.n^3a^2/c^2$ near bottom of Oppenheim page 22. $e$ also appears in $d\varpi/dt$ page 27 (von Clausius). But these are not in the expected form $1/(1-e^2)$, as you point out. $\endgroup$
    – steveOw
    Commented Nov 12, 2014 at 11:48

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