# What are the exact physical parameters used to calculate Mercury precession with Einstein theory?

NASA measured 43,13 arc seconds per century. General relativity predicts 42,98 arc seconds per century.

I try to find out what the parameters' values such as $G$, $M_{sun}$, $\omega_{min}$ at aphelion radius, etc. used to get this exact 42,98.

With a new model developed by R. Plamondon, we get exactly 43,13 and also it matched perfectly with Mercury's period of revolution around Sun.

My problem is the following. When I put my parameters into Einstein metric, I got something unexpected.

• Who is Plamondon/what's the model? – Stan Liou Feb 2 '15 at 22:41

In very a general post-Newtonian metric for a two-body system with the first body oblate, where $M\equiv m_1+m_2$ is the total mass, $\mu\equiv m_1m_2/M$ is the reduced mass, and $p\equiv a(1-e^2)$ is the semi-latus rectum of the orbit, the perihelion advance per orbit is $$\small\delta\varpi = 6\pi\frac{GM}{pc^2}\left[\underbrace{\frac{2-\beta+2\gamma}{3}}_{\text{GTR}=1} + \underbrace{\frac{2\alpha_1-\alpha_2+\alpha_3+2\zeta_2}{6}}_{\text{GTR}=0}\frac{\mu}{M} + \frac{J_2R^2c^2}{2GMp}\right]\text{.}$$ According to GTR, $\beta=\gamma=1$ exactly. The second term contains various parameters relating to preferred-frame effects and energy-momentum non-conservation. In GTR, they are all identically zero, while experimentally, $\alpha_1\lesssim10^{-4}$ and the rest are many orders of magnitude less than that. Finally, in the third term, $J_2$ is the quadrupole moment of the first body; $J_2R^2 = (C-A)/m_1$ where $C$ and $A$ are moments of inertia about rotational and equatorial axes, respectively. A derivation can be found Will §7.3.

I try to find out what the parameters' values such as $G$, $M_{sun}$, $\omega_{min}$ at aphelion radius, etc. used to get this exact 42,98.

This is just the front coefficient in the above: if $T$ is the orbital period of Mercury, then $$6\pi\frac{GM/c^2}{a(1-e^2)}\frac{1}{T} = \frac{42.98''}{\mathrm{century}}\text{,}$$ and we can restate the GTR prediction of Mercury as $$\dot\varpi_{\text{☿}}^{\scriptsize\text{GTR}} = \left.\frac{42.98''}{\mathrm{century}}\right.\left[1+2958J_2\right]\text{,}$$ where $J_2$ is the gravitational quadrupole moment of the Sun.

NASA measured 43,13 arc seconds per century. General relativity predicts 42,98 arc seconds per century.

If you're referring to one of Anderson et al.'s analyses, the value is $43.13\pm0.14$. Under GTR, this value is compatible with any $J_2$ up to about $$\begin{eqnarray*}\frac{J_2R^2c^2}{2GMp}\lesssim 0.0067 &\Longleftrightarrow & J_2\lesssim 2.3\times10^{-6}\text{.}\end{eqnarray*}$$ But what is the Sun's quadrupole moment? This is a somewhat sticky question, but the results of lunar laser ranging experiments are incompatible with $J_2$ of more than about $3\times10^{-6}$, while using helioseismology, Pijpers (1998) estimates $J_2 = (2.18\pm0.06)\times 10^{-7}$. Using Pijper's value yields a GTR prediction of $43.01''$ per century, which is compatible with Anderson et al.'s analysis.

With a new model developed by R. Plamondon, we get exactly 43,13 and also it matched perfectly with Mercury's period of revolution around Sun.

I have no idea what Plamondon's model is, but I would be rather suspicious of any model that gives "exactly" $43.13$ when that's just the middle of an interval of statistical uncertainty according to one of the many prior analyses of Mercury's orbit.

When I put my parameters into Einstein metric, I got something unexpected.

If we ignore the post-Newtonian parameters other than $\beta$ and $\gamma$, the static post-Newtonian metric is $$\mathrm{d}s^2 = -\left(1+2\Phi+2\beta\Phi^2\right)\mathrm{d}t^2 + \left(1-2\gamma\Phi\right)\mathrm{d}S_\text{Euclid}^2\text{,}$$ where the Newtonian potential of the Sun is modeled up to the quadrupole term $$\Phi = -\frac{M_\odot}{r}\left[1-J_2\frac{R_\odot^2}{r^2}\frac{3\cos^2-1}{2}\right]\text{.}$$ Deriving the perihelion advance of Mercury with this metric is Exercise 40.5 in MTW, and it matches the general result referenced above with $M\equiv M_\odot+M_{☿} \approx M_\odot$ and sans the middle term. For the Sun-Mercury system, $\mu/M\sim 10^{-7}$, so the middle term is even more irrelevant than bounds on the PPN parameters would suggest.

References:

1. Will, C. M., Theory and experiment in gravitational physics, (Cambridge University Press, Cambridge, 1993), 2nd edition.
2. Anderson, J. D., Campbell, J. K., Jurgens, R. F., Lau, E. L., Newhall, X. X., Slade, M. A., Standish Jr, E. M., in Proceedings of the Sixth Marcel Grossmann Meeting on General Relativity, ed. by Sato, H. and Nakamura, T., (World Scientific, Singapore, 1992).
3. Pijpers, F. P., Mon. Not. R. Astron. Soc. 297, L76 (1998). [arXiv:astro-ph/9804258]
4. Misner, C. W., Thorne, K. S., and Wheeler, J. A., Gravitation, (Freeman, San Francisco, 1973).
• Good answear. But by the way: Why is this β = γ = 1 in general relativity? From special relativity we are used to β = v/c and γ = √(1-β²) which is a bit confusing. Why aren't we using 1 instead of double defining this variables, are there also situations where β or γ in a general relativistic sense were not 1? – Yukterez Feb 6 '15 at 2:28
• @СимонТыран The post-Newtonian parameters $\beta$ and $\gamma$ are the coefficients in the metric above, not the STR meaning. In GTR, can't be anything but $1$, but other theories may differ. I referenced them because (a) most tests of GTR actually work in PPN, so that the results can be readily compared to other experiments and perhaps other theories, and (b) if Plamondon's is a metric theory, the OP will be aware of a systematic framework to compare their experimental predictions. – Stan Liou Feb 6 '15 at 5:32
• Does the PPN equation include the effects of retarded gravity (the changes in the field only propagate with c instead of instantaneous)? – Yukterez Feb 9 '15 at 0:47
• @СимонТыран The simplifier metric at the end of the answer (neglecting Mercury's mass) is static and there are no changes to be propagated. The more general result quoted at the front doesn't explicitly include them because it only makes a difference at 2.5PN order and above, and so is unobservably low in our solar system. See also this question on why gravity in the solar system can be accurately modeled as essentially instantaneous, as if at speeds $\gtrsim 10^{10}c$. – Stan Liou Feb 9 '15 at 1:18