Recover the perihelion dependency

The orbit calculation uses the declared higher-order static completion and the cited orbital dependency. Its controlled expansion makes the perihelion result part of the complete recovered chain.

17. Perihelion advance and the cited dependency

The 1916 article states Einstein (75) and refers the calculation to Einstein’s 1915 perihelion paper and Schwarzschild’s 1916 solution. The dependency used here is the cubic orbit polynomial and turning-point operation, not a source lift of every line of those separate papers [4], [12]. The radial completion in Section 14 supplies the higher-order field information that Einstein (70) alone lacks.

On an equatorial timelike geodesic of (14.7), the time and angular source equations yield constants

$$\mathcal E=A\frac{\mathrm{d}t}{\mathrm{d}s},\qquad \ell=r^2\frac{\mathrm{d}\varphi}{\mathrm{d}s}. \tag{17.1}$$

The equatorial plane is preserved. Combining the constants with the calibrated norm one gives

$$\left(\frac{\mathrm{d}r}{\mathrm{d}s}\right)^2 =\mathcal E^2-A\left(1+\frac{\ell^2}{r^2}\right). \tag{17.2}$$

For $w=1/r$ and $\ell\ne0$, $\mathrm{d}r/\mathrm{d}s=-\ell\,\mathrm{d}w/\mathrm{d}\varphi$, so

$$\boxed{(w')^2=\frac{\mathcal E^2-1}{\ell^2} +\frac\alpha{\ell^2}w-w^2+\alpha w^3.} \tag{17.3}$$

This is Einstein 1915 equation (11), with the polynomial dictionary $B_{1915}^2=\ell^2$ and $2A_{1915}=\mathcal E^2-1$. It does not identify every earlier approximation-dependent parameter in that paper. Schwarzschild’s equation (18) gives the same cubic after its stated radial-variable transformation. Differentiating away from a turning point and continuing smoothly gives

$$w''+w=\frac\alpha{2\ell^2}+\frac{3\alpha}{2}w^2. \tag{17.4}$$

No division by $w'=0$ at the turning point is required to assert the continued equation.

Turning roots before expansion

Let the two positive turning values be $w_-<w_+$, and put $S=w_-+w_+$, $P=w_-w_+$. Requiring these roots in (17.3) gives

$$\ell^2=\frac\alpha{S-\alpha(S^2-P)},\qquad \mathcal E^2=1-\ell^2P(1-\alpha S). \tag{17.5}$$

Expansion verifies the exact factorization

$$(w-w_-)(w_+-w)[1-\alpha(S+w)]. \tag{17.6}$$

Define turning-radius elements

$$a_o=\frac{r_{\max}+r_{\min}}2,\quad e=\frac{r_{\max}-r_{\min}}{r_{\max}+r_{\min}},\quad p=a_o(1-e^2),\quad \delta=\frac\alpha p. \tag{17.7}$$

Then $w_\pm=(1\pm e)/p$. Substitution $w=(1+e\cos\chi)/p$ regularizes the integral before any expansion:

$$\boxed{\Delta\varphi_{\rm radial} =2\int_0^\pi[1-\delta(3+e\cos\chi)]^{-1/2}\mathrm{d}\chi.} \tag{17.8}$$

For $0<e<1$, $\delta>0$ and $q_*=\delta(3+e)<1$, the final factor in (17.6) is positive across the interval. The denominator of $\ell^2$ is positive because $2-\delta(3+e^2)>0$ on this domain. Moreover

$$\mathcal E^2=\frac{2[1-\delta(1+e)][1-\delta(1-e)]} {2-\delta(3+e^2)}>0,$$

and $\mathcal E^2<1$ follows from (17.5) since $1-2\delta>0$. The turning radii lie in the static exterior. This specifies a regular outer bound branch; radial orbits and a weak expansion at the separatrix are not included. An exactly circular curve has no distinguished perihelion direction, although a circular limit exists.

Theorem 17.1 (Perihelion coefficient and remainder)

For the branch just specified, let $\Delta\varpi=\Delta\varphi_{\rm radial}-2\pi$. Then

$$\boxed{\Delta\varpi=3\pi\delta+O(\delta^2),} \tag{17.9}$$

with the explicit bound

$$\boxed{0\le\Delta\varpi-3\pi\delta \le\frac{3\pi\delta^2(9+e^2/2)}{4(1-q_*)^{5/2}}.} \tag{17.10}$$

Proof.

Expand the regular integrand in (17.8), not the original singular turning-point expression. Its linear term is $\delta(3+e\cos\chi)/2$. Multiplication by two and integration gives $3\pi\delta$. Convexity makes the remainder nonnegative. The Taylor bound is $3\delta^2(3+e\cos\chi)^2/[8(1-q_*)^{5/2}]$. Use the exact integral $\int_0^\pi(3+e\cos\chi)^2\mathrm{d}\chi=\pi(9+e^2/2)$ and the factor two in (17.8) to obtain (17.10).

$\square$

The same-mass normalization (14.2) and the leading Kepler relation $T^2=4\pi^2a_o^3/(KM)$ yield

$$\boxed{\Delta\varpi_{\rm leading} =\frac{3\pi\alpha}{a_o(1-e^2)} =\frac{6\pi KM}{a_o(1-e^2)c^2} =\frac{24\pi^3a_o^2}{T^2c^2(1-e^2)}.} \tag{17.11}$$

This recovers the expression in Einstein (75). The Kepler substitution is a leading-order conversion, not an exact relativistic period identity. The bound (17.10) applies to the turning-radius elements defined here. Applying it to a differently fitted observational element set requires the corresponding conversion analysis.

Scope of the numerical statements

The original last page reports approximately $1.7$ arcseconds for a solar-grazing ray, $0.02$ arcseconds for Jupiter and $43$ arcseconds per century for Mercury [5] (p. 822). They are retained as historical source-reported numbers. The article does not give a complete numerical input table sufficient for an independent audit of those values. No current astronomical data have been substituted, no mass or radius fitted to reproduce them, and no present-day observational claim is made. The verification below uses declared dimensionless diagnostic parameters to check the regularized integrals and bounds, not astronomical observations.