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.
Asked
Active
Viewed 1,496 times
1 Answers
9
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 $$\Omega\approx 532\, \mathrm{arcseconds\ per\ century}.$$
Diracology
- 18,168