The isotropic coordinate form of the Schwarzschild exterior metric and the isothermal radius shift

(The coordinate change and its inverse.) Expanding $R(\rho )=\rho (1+\frac{s}{4\rho }{)}^2=\rho +\frac{s}{2}+\frac{s^2}{16\rho }$ gives the algebraic identity of clause (a), and differentiating termwise gives $R^{\mathrm{\prime }}(\rho )=1-\frac{s^2}{16{\rho }^2}=(16{\rho }^2-s^2)\mathrm{/}(16{\rho }^2)>0$ for $\rho >s\mathrm{/}4$ by [L3]. Hence $R$ is strictly increasing and continuous on the interval $(s\mathrm{/}4,\mathrm{\infty })$ and $R(\rho )\to s$ as $\rho \downarrow s\mathrm{/}4$, while $R(\rho )\to +\mathrm{\infty }$ as $\rho \to \mathrm{\infty }$; by [L4] the image of $R$ is the whole interval $(s,\mathrm{\infty })$ and $R$ is a bijection onto it. For the displayed inverse, substitution gives $2{\rho }_0(r)-r+\frac{s}{2}=\sqrt{r(r-s)}$, so ${\rho }_0(r)$ satisfies the quadratic identity ${\rho }_0(r{)}^2-{\rho }_0(r)(r-\frac{s}{2})+\frac{s^2}{16}=0$; dividing this identity by ${\rho }_0(r)>0$ gives ${\rho }_0(r)+\frac{s^2}{16{\rho }_0(r)}=r-\frac{s}{2}$, hence $R({\rho }_0(r))={\rho }_0(r)+\frac{s}{2}+\frac{s^2}{16{\rho }_0(r)}=r$. Thus ${\rho }_0$ is a left inverse of $R$, hence the inverse of the bijection $R$; it is smooth on $(s,\mathrm{\infty })$ by [L2], and ${\rho }_0(r)>s\mathrm{/}4$ because $2{\rho }_0(r)=r-\frac{s}{2}+\sqrt{r(r-s)}>\frac{s}{2}$ for $r>s$.

(Metric coefficients in the new chart.) In the chart with coordinates $(x^0,\rho ,\vartheta ,\phi )$, where $r=R(\rho )$, the components of $g$ are determined by the transformation $r\mapsto \rho$: the coefficient of $d{\rho }^2$ is $g_{11}(R^{\mathrm{\prime }}(\rho ){)}^2$ and the angular coefficient ${\rho }^2$ acquires the factor $(R\mathrm{/}\rho {)}^2$ in addition to $g_{22}$ and $g_{33}$, because $r^2=R(\rho {)}^2={\rho }^2(1+\frac{s}{4\rho }{)}^4$; the time component $g_{00}$ is unchanged since $x^0$ is not transformed. Writing $a=s\mathrm{/}(4\rho )$, i.e. $R=\rho (1+a{)}^2$ and $R^{\mathrm{\prime }}=1-a^2$ by [L3], we obtain $1-f(R)=s\mathrm{/}R=4a\mathrm{/}(1+a{)}^2$, hence $f(R)=1-\frac{4a}{(1+a{)}^2}=\frac{(1-a{)}^2}{(1+a{)}^2},\frac{R^{\mathrm{\prime }2}}{f(R)}=\frac{(1-a^2{)}^2(1+a{)}^2}{(1-a{)}^2}=(1+a{)}^4\mathrm{.}$ Thus $g_{00}=-(1-a{)}^2\mathrm{/}(1+a{)}^2$, $g_{\rho \rho }=(1+a{)}^4$ and the angular components are $R^2{\gamma }_{ij}=(1+a{)}^4{\rho }^2{\gamma }_{ij}$ with ${\gamma }_{22}=1$, ${\gamma }_{33}={\sin }^2\vartheta$. All off-diagonal components remain zero. This is clause (b).

(Conformal flatness and the expansion.) Setting $\mathrm{\Phi }:=(1+a{)}^4$, the spatial metric is $\mathrm{\Phi }\delta$, whose Cartesian components are $\mathrm{\Phi }{\delta }_{ij}$; in $(\rho ,\vartheta ,\phi )$ it is conformal to $\delta =d{\rho }^2+{\rho }^2(d{\vartheta }^2+{\sin }^2\vartheta d{\phi }^2)$ in spherical coordinates on ${\mathbb{R}}^3$; this is the conformally flat form of clause (c). By the binomial identity, $(1+\frac{s}{4\rho }{)}^4=1+\frac{s}{\rho }+\frac{3s^2}{8{\rho }^2}+\frac{s^3}{16{\rho }^3}+\frac{s^4}{256{\rho }^4},$ so the stated inequality holds with $C({\rho }_{\ast })=\frac{3}{8}+\frac{s}{16{\rho }_{\ast }}+\frac{s^2}{256{\rho }_{\ast }^2}$. For $r\ge 2s$ set $t=s\mathrm{/}r\in [0,1\mathrm{/}2]$. The explicit inverse in clause (a) and the estimate in [L2] give ${\rho }_0(r)=\frac{r-s\mathrm{/}2+rh(t)}{2}=r-\frac{s}{2}-\frac{s^2}{16r}+\frac{r}{2}{R}_3(t)\mathrm{.}$ The displayed bound on $R_3$ implies $\mathrm{\mid }r{R}_3(t)\mathrm{/}2\mathrm{\mid }\le C{s}^3\mathrm{/}{r}^2$ for a constant $C$ independent of $r$; this proves the expansion in clause (c) as $r\mathrm{/}s\to \mathrm{\infty }$ with fixed $s$.

(The comparison warning.) The areal coefficient is $f(r{)}^{-1}=(1-s\mathrm{/}r{)}^{-1}=1+\frac{s}{r}+\frac{(s\mathrm{/}r{)}^2}{1-s\mathrm{/}r}$, an exact identity whose last term is $O(r^{-2})$, while the isotropic coefficient has the exact expansion $\mathrm{\Phi }=1+\frac{s}{\rho }+O({\rho }^{-2})$ of step 2.1; the two expressions are both $1+s\mathrm{/}(\text{radius})$ at leading order but involve different radial coordinates related by $\rho =r-\frac{s}{2}+O(1\mathrm{/}r)$. Substituting $\rho =r$ in the first-order terms would replace $s\mathrm{/}r$ by $s\mathrm{/}\rho =s\mathrm{/}(r-\frac{s}{2}+\dots )=s\mathrm{/}r+O(r^{-2})$, so the mismatch appears at relative order $(s\mathrm{/}r{)}^2$; the comparison of coefficients must therefore be performed after the conversion $r=R(\rho )$ (or $\rho ={\rho }_0(r)$) with the error terms retained. This is clause (d); no statement is made here about the numerical size of $s$ or about any observed system.