When scientists measured the position of Mercury in the 18th century, they interpreted the results assuming a Euclidean background, because they did not know general relativity. So they measured $r$ and $\phi$ in function of time attributing to these coordinates an Euclidean meaning, that is, assuming that the relations of Euclidean geometry hold for these coordinates.

In Schwarzschild solution

$$ ds^{2} = \left(1 - \frac{2 m}{r} \right) dt^{2} - \frac{1}{1 - \frac{2 m}{r}} dr^{2} - r^{2} d\Omega^{2} \tag{1} $$

$r$ and $\phi$ don't have the same meaning as the Euclidean $r$ and $\phi$, in mind of scientists of the 18th century; so my doubt is:

If general relativity predicts a precession of Mercury perihelion with $\Delta \phi = 43^{\prime \prime}$, why we should compare this quantity with the $\Delta \phi$ measured in the 18th century, surely measured with reasoning that assumes a Euclidean space?

  • 1
    $\begingroup$ Please use Mathjax to write equations math.meta.stackexchange.com/q/5020 $\endgroup$
    – alanf
    Commented Apr 25 at 9:08
  • $\begingroup$ 43" per century that is. $\endgroup$
    – my2cts
    Commented May 4 at 14:15

4 Answers 4


This is a good question.

The 3D space part of a 3+1D flat Minkowski spacetime is a Euclidean space. We can make an identification when it is so.

You are correct that the coördinates that we are using for Schwarzchild spacetime is not the same as that of Euclidean space. However, you are wrong in the specifics. The angular coördinates are the same. Only $r$ has a new definition.

And it is even nicer than that. Schwarzchild spacetime, at long distance limits, i.e. large $r$, reduce to that of flat Minkowski spacetime. This means that the space part will also tend towards the Euclidean limit. We can thus estimate the difference and see that it is miniscule.

And it is even nicer than even that. The radial part of the problem is integrated out of the problem, and thus the precession angle thus computed does not depend upon this difference.

For the above reasons, there is no problem for the pioneers to have worked in their ways.

  • $\begingroup$ Thanks so much for the reply, you were very clear. BUT, what if $\phi$ (the angle relative to the position of mercury relative to the sun) was the result of an indirect measurement? that is, what if direct measurements were made and then through quantitative (probably trigonometric, and so Euclidean) relationships the $\phi$ was measured? What if $r$ also appeared in these quantitative relationships? $\endgroup$
    – Giovanni
    Commented Apr 27 at 14:53

It's still perfectly fine to use a Euclidean background for weak field gravity.

It's quite easy to get the necessary precession for Mercury if you take gravitational and kinematic time dilation / length contraction into account.

The Newtonian acceleration vector is: \begin{equation} \vec{g}_n = \frac{\hat{d} G M}{{\lvert\lvert \vec{d} \rvert\rvert}^2}. \end{equation}

One important value is closely related to the kinematic time dilation: \begin{equation} \label{eq_kinematic} \alpha = 2 - \sqrt{1 - \frac{\lvert\lvert \vec{v}_{o}\rvert\rvert^2}{c^2}}. \end{equation} Another important value is the gravitational time dilation: \begin{equation} \beta = \sqrt{1 - \frac{R_{s}}{\lvert \lvert \vec{d} \rvert \rvert}}. \end{equation}

Finally, the semi-implicit Euler integration is: \begin{align} \vec{v}_{o}(t + \delta_t) &= \vec{v}_{o}(t) + \delta_{t} \alpha \vec{g}_n, \\ \ell_{o}(t + \delta_t) &= \ell_{o}(t) + \delta_{t} \beta \vec{v}_{o}(t + \delta_t). \end{align}


Indeed, the space-time in the Solar system is very slightly non-Minkowski (non-Euclidean spatial part). At the position of Mercury, this deviation is on the level of a few parts in tens of millions. The spatial curvature is even a part of the computation of the precession and if you convert the perihelion precession of Mercury to radians per orbits, you get an angle of $\sim 5$ radians per ten million orbits. If we wanted to describe relativistic corrections to the period of the orbit, or the length of the orbit traced out by Mercury, we would have to be careful to describe our measurement protocol in the non-flat space-time. Would we be talking about proper period in the frame of Mercury or some other time period? How would we establish measures of length by pointing our telescopes, which catch light travelling to us through curved space-time?

Fortuitously, the observation of the perihelion precession is largely impervious to the non-flatness of space-time. The key reasons are the following:

  1. The perihelion is identified from a qualitative feature of the orbit (pericenter is when you are the closest to the Sun).
  2. The precession angle is defined essentially as an average over a century (over 400 orbital periods of Mercury).
  3. The Newtonian two-body problem has no precession in itself, the effect is relativistic at leading order.
  4. The precession rates of Mercury caused by other planets are small.

Point 1. means that you are not asking "how far" Mercury is precisely, and getting an answer which is off by $10^{-7}$. When it is closest it is closest.

Of course, subtle effects with the propagation of light might lead you to pinpointing the position when it is closest incorrectly on the level of $10^{-7}$ when looking at a single pericentre passage. This means that subtle interpretation effects will be of the same order as the precession itself when observed over a single orbit. However, you are not looking at a single passage, you are looking at the effect adding up over hundreds of orbits (point 2). This means that only the long-term "secular" effect of the precession angle adding up prevails (which is $\sim 3\cdot 10^{-4}$ of a full angle after a century) and subtle interpretation biases in single measurements at the level of $10^{-7}$ are not that important.

Why is point 3 important? This is because the precession rate depends on the orbital radius and eccentricity. If the Newtonian two-body problem had its own precession rate depending on orbital radius and eccentricity, the observation of these orbital parameters entering the Newtonian precession formula would be biased at $10^{-7}$ because of incorrect Euclidean interpretation of timing and length measurements. As a result, to obtain a correct leading-order prediction, we would have to understand these subtleties. However, the Newtonian precession vanishes and the relativistic perihelion precession per orbit is roughly $$\Delta \phi_{\rm prec.}\sim 2\pi \frac{3 G M_{\rm sun}}{c^2 a_{\rm mercury}} + \mathcal{O}(e)$$ Where $a_{\rm Mercury}$ is Mercury's semi-major axis. If we measure $a_{\rm mercury}$ (or $M_{\rm sun}$ for the matter) incorrectly with an error $10^{-7}$, it adds a $\sim 10^{-14}$ error to the leading $10^{-7}$ precession rate. That is acceptable.

As to point 4. Mercury orbit precesses ten times more because of other planets in the solar system than because of relativity. Its real rate of precession is $\sim 500$ arcseconds per century, only roughly $40$ of those are due to GR, the rest can be accounted by Newtonian perturbation theory. Relativistic effects do indeed add a relative correction of order $10^{-7}$ to those $\sim $460$ arcseconds, but again, this is negligible.


if general relativity predicts a precession of Mercury perihelion with Δφ=43'', why we should compare this quantity with the Δφ measured in 18th century, surely measured with reasoning that assumes a Euclidean space?

Physicists were interested in predicting and explaining the angle by which they would see Mercury's perihelion precess when they looked at it through a telescope. So they would have models of Mercury and the observations made on it in each theory and those models make different predictions about what you will see down the telescope. A theory that didn't match what physicists saw down the telescope would have a problem. Newtonian gravity had a problem with observations of Mercury and general relativity didn't. What is relevant for a test is whether each theory predicts what astronomers see when they look down a telescope and you can work that out in each theory independently of the other theory. If there is some physical effect that changes how the observations work between the theories, then that will be used as part of your model of the measurement itself in each theory.

  • $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review $\endgroup$ Commented Apr 26 at 8:07
  • $\begingroup$ @GiorgioP-DoomsdayClockIsAt-90 There are times when a question includes a fundamental misconception or lack of clarity about some topic. In this case, the misconception is about experimental testing. In such cases, pointing out the misconception or lack of clarity in an answer can be the right way to go since a comment may be too short to address the issue. And as it happens my answer does explain how to compare the two theories. Come up with a model of the experiment and the system in each theory, use them to work out what you will see and then compare that with what you actually see. $\endgroup$
    – alanf
    Commented Apr 26 at 8:29
  • $\begingroup$ It should be noted that the other answer states that Euclidean space is a good approximation with no references and no actual calculation but the author hasn't been dinged for that. My answer doesn't depend on whether Euclidean space is a good approximation. $\endgroup$
    – alanf
    Commented Apr 26 at 8:35
  • $\begingroup$ Your answer is a too-generic comment. It could equally be applied to every comparison between theory and experiments. The question is much more specific. $\endgroup$ Commented Apr 26 at 8:55
  • $\begingroup$ @alanf Thanks so much for the reply $\endgroup$
    – Giovanni
    Commented Apr 27 at 14:57

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