# Details of Newtonian Prediction for Mercury's Precession

Could anyone point me to a book or outline the methods used to actually calculate the 532 arcseconds per century that Newtonian theory apparently predicts for Mercury's precession. I am completely comfortable with getting GR's 43 but don't even know where to start for Newton. I feel rather ridiculous given that state of affairs.

• "I feel rather ridiculous given that state of affairs." ::chuckles:: It's very easy to overlook how developed physics was at the beginning of the 20th century in the excitement of studying the radical rise of Einstein's relativities and quantum stuff, isn't it? Marion and Thorton give only a sketch, so you might try a graduate mechanics book in search of more detail. – dmckee --- ex-moderator kitten Jun 6 '16 at 21:01
• See physics.stackexchange.com/q/814 and links therein; this overview was recommended, although I don't think it's quite what you're looking for. This focuses mainly on the Newtonian precession. – HDE 226868 Jun 6 '16 at 21:04
• @HDE 226868 Thank you for the reference. The idea of Taylor expanding around a circular orbit is great. But surely there exists a detailed exposition of the effects of all the other planets somewhere. – Tomas Smith Jun 6 '16 at 21:27
• It's not clear where this figure of 5557 arcseconds per century comes from. After many years of observations of the orbit of Mercury, the average disagreement in precession was calculated to be 43 arcseconds per century, a figure which is explained by including the effects of relativity on calculating the orbit of Mercury. That relativity could account for such a vanishingly small discrepancy in observational data went a long way in convincing physicists that the theory was valid. – user16622 Jun 7 '16 at 0:35
• @user16622 That number comes from Einstein Gravity in a Nutshell, page 368. I'm already quite familiar with the watershed moment Einstein had with Mecrury. But thank you. – Tomas Smith Jun 7 '16 at 18:55

The Newtonian Mechanics prediction for the Mercury's Precession is actually $532''$ per century.

## The general result

If the central force is attractive, there is a circular orbit of radius $r_0$. This circular orbit is stabble if it correspond to a minimum of the effective potential, i.e. $$U_{ef}''(r_0)>0.$$ Using that $U_{ef}=L^2/2mr^2+U$ and $F(r_0)=-L^2/mr_0^3$, where $L$ is the angular momentum and $F=-U'$, we get $$U_{ef}''(r_0)=-\frac{3F(r_0)+r_0F'(r_0)}{r_0}.$$ If $r_0$ gives a minimum of $U_{ef}$ than after expanding up to second order $U_{ef}$ around $r_0$ we can compute the period of the radial oscillations, $$T_r=2\pi\sqrt{\frac{m}{U_{ef}''(r_0)}}=2\pi\sqrt{\frac{-mr_0}{3F(r_0)+r_0F'(r_0)}}.$$ For small perturbations around the circular orbit we can approximate the angle swept in the interval $T_r$ by $$\Delta\phi=\dot\phi T_r,$$ where $$\dot\phi=\frac{L}{mr^2}=\sqrt{-\frac{F(r_0)}{mr_0}}.$$ Hence $$\Delta\phi=2\pi\sqrt{\frac{m}{U_{ef}''(r_0)}}=2\pi\sqrt{\frac{F(r_0)}{3F(r_0)+r_0F'(r_0)}}.$$ If $\Delta\phi=2\pi$ it means the particle revolves exactly once during a radial oscillation. There is no precession. It is convenient to define the precessed angle by $\Phi=\Delta\phi-2\pi$ and the rate of this precession, $$\Omega=\frac{\Phi}{T_r}.$$

## The particular result

We are left to calculate the total force on Mercury which I decompose as $$F(r)=F_0(r)+F_p(r),$$ where $F_0$ is the force due to the Sum and $F_p$ is the small (perturbative) force due to the other planets. The nicest calculation I have seen of $F_p$ is presented in

Price, Rush - Nonrelativistic contribution to Mercury's perihelion precession - AJP 47, 531 (1979);

The idea is that since the precession is too slow (295000 years for a complete revolution) compared to period of revolution of the solar system bodies, the other planets effectively act as a uniform ring of mass. Notice this is only regarded to the Mecrury's precession. Following the paper is quite easy to calculate the force a ring $i$ of density $\lambda_i$ and radius $R_i$ does on Mercury which is located at $r$. The total force due to all other planets reads $$F_p(r)=\sum_i\frac{Gm\pi\lambda r}{R_i^2-r^2}.$$ Noticing that $r_0F_0'(r_0)=-2F_0(r_0)$ and using these expressions in the general form of $\Delta\phi$ above we get $$\Delta\phi=2\pi\left[\frac{F_0(r_0)+F_p(r_0)}{F_0(r_0)+3F_p(r_0)+r_0F_p'(r_0)}\right]^{1/2}.$$ Since $|F_p(r_0)|\approx |r_0F_p'(r_0)|\ll |F_0(r_0)|$ we can Taylor expand up to first order in $F_p(r_0)/F(r_0)$ and $F_p'(r_0)/F(r_0)$, i.e. $$\Delta\phi=2\pi\left[1-\frac{F_p(r_0)}{F(r_0)}-\frac{r_0F_p'(r_0)}{F(r_0)}\right].$$ Therefore, the precessed angle is $$\Phi=-\frac{F_p(r_0)}{F(r_0)}-\frac{r_0F_p'(r_0)}{F(r_0)}.$$ Plugging in the astronomical data and dividing by the sideral year of Mercury we get the rate of precession $$\Omega=7.060\cdot 10^{-8}\, \mathrm{rad\ per\ day},$$ or $$\Omega\approx 532\, \mathrm{arcseconds\ per\ century}.$$

• +1, I really learnt something from the second section - how elegant. Like the OP I probably wouldn't know where to begin, and imagined it would be a detailed tallying up of forces from planets separately over long time periods. That fast-compared-with-precession insight to replace the planets with rings is wonderful. – WetSavannaAnimal Jul 9 '16 at 15:17
• @WetSavannaAnimalakaRodVance IMHO the beauty of the general result is that it does not specify the nature of the perturbation $F_p$. If one uses $F_p=\beta/r^ 3$ one gets the 43 arcsec per century of GTR. – Diracology Jul 9 '16 at 15:36
• Very nice indeed! – NickD Feb 27 '17 at 20:47
• BTW, I think there is a factor of $\dot{\phi}$ after the first equals sign in the last equation for $\Delta\phi$ in the "General Result" section. But everything is fine after the second equals sign. – NickD Apr 11 '17 at 3:18
• ... a factor of $\dot{\phi}$ missing... There's some kind of irony here :-) – NickD Apr 11 '17 at 5:01