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The area of a symmetry sphere in the weak-field isotropic metric is $4\pi {\rho }^2(1+2GM/\rho )$. Equating it to $4\pi r^2$ defines the areal radius and gives the displayed change. To first order $dr=d\rho$, and replacing $1/\rho$ by $1/r$ changes the perturbation only at second order. This explains why the isotropic angular correction disappears in the linearized Schwarzschild metric without discarding spatial curvature.

For the static nonrotating source, use spherical symmetry and the static scalar perturbation of Minkowski spacetime, with isotropic radius $\rho$:To first order, the Einstein tensor in the declared convention hasThese are also the static scalar perturbation of Minkowski spacetime equations in the original information sheet. The source has $T_{00}=M{\delta }^{(3)}(\mathbf{x})$ and zero spatial stress. With the sign established in Question 2,The distributional identity $\Delta (1/\rho )=-4\pi {\delta }^{(3)}(\mathbf{x})$ and asymptotic flatness give $C=GM/\rho$. For $F=A+C$, the spatial field equation has trace $-2\Delta F=0$ and therefore ${\partial }_i{\partial }_{j}F=0$. Its decaying spherical solution is $F=0$, so $A=-GM/\rho$. This determines both potentials; the spatial correction cannot be inferred from the Newtonian gravitational potential alone.

Now use the isotropic-to-areal weak-field radius change. The angular coefficient defines $r^2=(1+2GM/\rho ){\rho }^2$, hence $r=\rho +GM$ to first order. Thus $dr=d\rho$ to the same accuracy, and replacing $1/\rho$ by $1/r$ inside a perturbation changes only second-order terms. With the Schwarzschild radius $r_S=2GM$, the linearized Schwarzschild metric isThis is an exterior weak-field approximation: $r_S/r\ll 1$. The point source is used distributionally to fix the mass, not to claim a regular or weak field at the origin.

For a null geodesic choose its orbital plane as $\theta =\pi /2$, using rotational symmetry, and let a dot denote an affine parameter derivative. For a Killing vector $k$ and an affinely parametrized geodesic tangent $U$, the geodesic conserved quantity from a Killing vector follows directly fromApplying this to the time-translation and axial Killing vectors, with the spatial sign absorbed into the definition of $h$, givesIf the tangent is normalized as the photon's four-momentum, $E$ is its energy measured at infinity and $h$ its signed angular momentum about the selected axis. Rescaling an arbitrary affine parameter rescales both; the ratio $b=|h|/E$ is the invariant impact parameter. Future-directed orbits have $E>0$. Assume $h\ne 0$ to use $\varphi$ as the orbit parameter; $h=0$ gives radial rays, not a finite-radius circular orbit.

The null vector constraint givesExpand consistently to first order in the weak-field approximation correction: $(1-u^2{)}^{-1}=1+O(u^2)$ and $(1+u{)}^{-1}=1-u+O(u^2)$. Also $du/d\varphi =-r_S\dot{r}/h$. Therefore the first-order orbital equation isThe cubic term is the first correction relative to the flat inverse-radius term; the equation is not a controlled exact equation for large $u$.

A constant solution of a first integral is not by itself sufficient to establish a circular geodesic. For a circular trajectory, the radial Euler-Lagrange equation also requires $A^{\prime }(r){\dot{t}}^2/2=r{\dot{\varphi }}^2$, where now $A(r)=1-r_S/r$ is the metric's time coefficient. Combining this with the null vector constraint gives $r{A}^{\prime }=2A$, or $u=2/3$. Equivalently, the formal potential $V(u)=u^2(1-u)$ must have a stationary point. Apart from the infinite-radius endpoint $u=0$, its only stationary point is $u_c=2/3$, with $V(u_c)=4/27$. Thus the formal circular solution requiresThe reduced radial equation is $u^{\prime \prime }=-u+\frac{3}{2}{u}^2$, with prime denoting $d/d\varphi$. It follows by differentiating the first integral on moving segments and agrees with the regular Euler-Lagrange equation at turning points. Perturbing $u=u_c+ϵ$ givesThere is an exponentially growing linear instability mode; equivalently $V^{\prime \prime }(u_c)=-2<0$. The formal circular orbit is unstable.

The weak-field photon-orbit validity limitation is essential: $u_c=2/3$ is not small. There is therefore no finite-radius circular photon orbit established within the controlled weak-field region. The formal extrapolation happens to reproduce the radius and impact parameter of the exact Schwarzschild photon sphere, but that physical result needs the exact geometry. The calculation also cannot establish a Schwarzschild event horizon from a first-order expansion at $u\sim 1$.