What does Schwarzschild refer to as Einstein's approximation regarding his exact solution?

Question

In 1916, Schwarzschild published his $R$-metric solution that differs from the $r$-metric solution we are all familiar with. The relation between $R$ and $r$ is $R^3=r^3 + α^3$ with $r$ been the distance marker and $\alpha$ being the well-known $α=2GM$.

I quote from his paper: "Actually Mr. Einstein’s approximation for the orbit goes into the exact solution when one substitutes for $r$ the quantity $R$." http://old.phys.huji.ac.il/~barak_kol/Courses/Black-holes/reading-papers/SchwarzschildTranslated.pdf (4)

The approximation Schwarzschild refers to is the one presented by Einstein in 1915, available in https://einsteinpapers.press.princeton.edu/vol6-trans/125 . This contains the term $g_{tt}= 1-(α/r)$ which is the $g_{tt}$ from the $r$-metric we are all familiar with.

Is Schwarzschild suggesting that both metrics are different? If not, then what differs exactly between the Schwarzschild exact solution and Einstein's approximation?

Answer 1

Einstein used a linearized version of the Schwarzschild metric to calculate the precession of the perihelion of Mercury. This was a measured quantity that was not adequately explained by the quadrupole moment of the sun, and it was an important early check of his general theory of relativity. (Einstein probably could have found Schwarzschild's exact solution if he had made an effort, but he was apparently more concerned with checking that the first nontrivial experimental prediction of his new theory agreed with the real-world data.) The linearized theory has the correct $g_{tt}$, but it effectively approximates the radial metric component as $$g_{rr}=\left(1-\frac{\alpha}{r}\right)^{-1}\approx1+\frac{\alpha}{r}.$$

Answer 2

In his 1915 paper "Perihelion motion of Mercury", Einstein solves his new found field equations for a point mass in a large distance approximation. More specifically, he finds the metric to first order in $1/r$ and (some of) the Christoffel symbols to the next order.

He then proceeds to find the equation of motion for any orbit around this mass (further invoking a slow motion approximation. He finds (equation 11 of his paper)

$$ \left( \frac{dx}{d\phi}\right)^2 \approx \frac{2A}{B^2}+\frac{\alpha}{B^2}x-x^2+\alpha x^3,$$

where $x=1/r$, $\alpha$ is a constant related to the mass being orbited ($2GM/c^2$), and $A$ and $B$ are constants of motion. In Einstein's derivation is an approximation, and would receive further corrections if higher order terms were included.

In Schwarzschild's 1916 paper, he finds the exact metric for a point mass. His paper features two radial quantities $R$ and $r$ related by $R^3 = r^3+\alpha^3$. Here is $R$ is the familiar radial coordinate. However, for reason irrelevant to this answer Schwarzschild preferred to view $r$ as the "real" radius (although both are just coordinates). Note that there is no a priori relationship between Schwarzschild's $r$ and the $r$ in Einstein's approximation, other then them agreeing in the first order approximation.

With the exact metric in hand Schwarzschild proceeds to find the exact equation of motion for a geodesic. He finds

$$ \left( \frac{dX}{d\phi}\right)^2 = \frac{1-h}{c^2}+\frac{h\alpha}{c^2}X-X^2+\alpha X^3,$$

where now $X=1/R$, and $h$ and $c$ are different constants of motion. After identifying $B=c^2/h$ and $2A = (1-h)/h$, this exact solution is identical to the approximation found by Einstein, accept that it feature $X=1/R$ instead of $x=1/r$. In different words, if in Einstein's approximation for the equation of motion you replace $r$ by $R$ the approximate result becomes exact.

Answer 3

This is the Schwarzschild line element

$$ds^2=- \left( 1-{\frac {\alpha}{R}} \right) {{\dot t}}^{2}+{{\dot r}}^{2} \left( 1-{\frac {\alpha}{R}} \right) ^{-1}+{R}^{2} \left( {\dot \vartheta }^{2}+ \left( \sin \left( \vartheta \right) \right) ^{2}{\dot \varphi }^{2} \right) $$

where

$$R=r\left(1+{\frac {{\alpha}^{3}}{{r}^{3}}}\right)^{1/3}$$

in his paper Schwarzschild wrote that because $\frac{\alpha}{r}\approx 10^{-12}$ is very small one can use the Einstein metric simplification , $\frac{\alpha}{R}\mapsto 0~,R=r$

Schwarzschild quotation [1]:

"Da $\frac{\alpha}{r}$ nahe gleich dem doppelten Quadrat der Planetengeschwindigkeit (Einheit die Lichtgeschwindigkeit) ist, so ist die Klammer selbst für Merkur nur um Größen der Ordnung $ 10^{-12}$ von 1 verschieden. Es ist also praktisch R mit r identisch und Hrn. EINSTEINS Annäherung für die entferntesten Bedürfnisse der Praxis ausreichend."

[1] Über das Gravitationsfeld eines Massenpunktes nach der EINSTEINschen Theorie

Edit

Schwarzschild Metric

coordinates are $~t\,,r\,,\vartheta\,,\varphi$

$$\left[ \begin {array}{cccc} {\frac {\sqrt [3]{-{\alpha}^{3}+{r}^{3}}- \alpha}{\sqrt [3]{-{\alpha}^{3}+{r}^{3}}}}&0&0&0\\ 0 &-{\frac {{r}^{4}}{ \left( -{\alpha}^{3}+{r}^{3} \right) \left( \sqrt [3]{-{\alpha}^{3}+{r}^{3}}-\alpha \right) }}&0&0 \\ 0&0&- \left( -{\alpha}^{3}+{r}^{3} \right) ^{2/3} &0\\ 0&0&0&- \left( -{\alpha}^{3}+{r}^{3} \right) ^{ 2/3} \left( \sin \left( \vartheta \right) \right) ^{2}\end {array} \right] $$

Einstein Metric

$$\left[ \begin {array}{cccc} 1-{\frac {\alpha}{r}}&0&0&0 \\ 0&- \left( 1-{\frac {\alpha}{r}} \right) ^{-1}&0&0 \\ 0&0&-{r}^{2}&0\\ 0&0&0&-{r}^{2} \left( \sin \left( \vartheta \right) \right) ^{2}\end {array} \right]$$