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Let $f(r)=1-2m/r$ and work in the exterior $r>2m$, with $G=c=1$. The affinely parametrized geodesic Lagrangian isIts polar-angle equation isThe initial data $\theta =\pi /2$, $\dot{\theta }=0$ therefore remain in that plane. Rotational symmetry allows the axes to be chosen so that any nonradial initial trajectory has those data: its conserved angular-momentum direction fixes an orbital plane through the centre. Radial trajectories can also be placed in an equatorial plane.

The coordinates $t,\varphi$ are cyclic in $L$, so their canonical momenta are constant:The corresponding Killing vector fields are ${\partial }_t$ and ${\partial }_{\varphi }$. With the displayed signature, $g({\partial }_t,\dot{x})=E$ and $g({\partial }_{\varphi },\dot{x})=-J$, explaining the sign in the second canonical momentum. Thus $E$ is conserved energy per unit rest mass for a massive particle parametrized by proper time, and $J$ is its axial angular momentum per unit rest mass. For a photon, affine rescaling rescales both constants; after the tangent is normalized to the physical four-momentum they represent its energy at infinity and axial angular momentum. Their ratio $b=J/E$ is the signed impact parameter.

In the equatorial plane, the photon satisfies the null geodesic constraintwhich givesFor $J\ne 0$, put $u=m/r$ and $\alpha =Em/J$. Thenand consequentlyThe $J=0$ radial case is not represented by $\varphi$ as an orbit parameter.

There is a circular photon orbit at $r=3m$, but it is unstable. The radial effective potential isA circular orbit requires $V^{\prime }(r)=0$ and $E^2=V(r)$, givingIts second derivative isIt is a maximum, rather than a minimum, so radial perturbations depart from the orbit. Equivalently the second-order orbit equation is $u^{\prime \prime }=3u^2-u$. Linearizing about $u_c=1/3$ gives $\delta u^{\prime \prime }=\delta u$, with modes $e^{\varphi }$ and $e^{-\varphi }$. The existence of a growing mode directly contradicts the printed stability claim. Exact initial data can remain circular, but that does not make the orbit stable.

The supplied implicit expression describes the critical-energy family with ${\alpha }^2=1/27$, as fixed by the circular orbit. DefineRationalizing once gives $F=(\sqrt{3}-s)/(2(\sqrt{3}+s))$ andDifferentiating $F(u)=A{e}^{\varphi }$ gives $F^{\prime }{u}^{\prime }=F$, soThis verifies the first-order equation for every allowed integration constant $A$ on its real branch. For $A=0$ the derivative argument is replaced by direct substitution: $u=1/3$ is constant and solves the critical equation. The displayed family does not solve the equation at an independently arbitrary value of $\alpha$.

For $A>0$, write $q=A{e}^{\varphi }$. On the physical exterior branch,As $\varphi \to -\infty$, $q\to 0^{+}$ andThus for $A=0$ the photon remains exactly on the circular orbit. For $A>0$ it approaches $r=3m$ from outside through indefinitely many revolutions in the negative-$\varphi$ limit. With increasing $\varphi$, this same branch peels away outward and reaches infinity when $q=1-\sqrt{3}/2$. Reversing the direction of motion gives an incoming critical ray asymptotically approaching the photon circle. These are separatrix trajectories, not evidence of stability.

The potential maximum and the outward-growing radial separation display the same instability in two complementary descriptions.

For this question use the Levi-Civita connection, its metric compatibility and the usual curvature symmetries. In the index ordering of Question 1, put $R_{\mu \nu \rho \sigma }=g_{\sigma \lambda }{R}_{\mu \nu \rho }{}^{\lambda }$ and $R_{\nu \rho }=g^{\mu \sigma }{R}_{\mu \nu \rho \sigma }$. Contract the second Bianchi identity in its derivative index $\mu$ and curvature index $\rho$. The curvature symmetries giveNow contract $\lambda ,\sigma$. The left-hand side becomes $-{\nabla }^{\rho }{R}_{\nu \rho }$, while the right-hand side becomes ${\nabla }^{\sigma }{R}_{\nu \sigma }-{\nabla }_{\nu }R$. ThusUsing $\nabla g=0$ yields the contracted Bianchi identityTaking the covariant divergence of the Einstein field equations, with constant $G$, then gives $8\pi G{\nabla }^{\mu }{T}_{\mu \nu }=0$. The matter equation is therefore consistent with the geometric identity rather than imposing an extra incompatible condition on the metric.

Stress-energy conservation expresses the local balance of energy and momentum, including their transport and the effects of the gravitational Levi-Civita connection. In a freely falling frame at a point it reduces there to the familiar ordinary divergence law. A global conserved energy requires additional geometric structure, such as a timelike Killing vector field; it does not follow in an arbitrary spacetime just by replacing a covariant derivative with an ordinary derivative.

For relativistic dust,With signature $(+,-,-,-)$, $u^{\nu }{u}_{\nu }=1$ implies $u_{\nu }{u}^{\mu }{\nabla }_{\mu }{u}^{\nu }=0$. Contracting gives ${\nabla }_{\mu }(\rho u^{\mu })=0$; where $\rho >0$, substitution then gives $u^{\mu }{\nabla }_{\mu }{u}^{\nu }=0$. Thus the continuity law combines rest-mass conservation and free geodesic motion for this matter model.

For the Newtonian limit, take $c=1$ and $\eta =\mathrm{diag}(1,-1,-1,-1)$, consistently with Question 3. Use the static weak field sourced by the time-independent, slowly moving matter, retaining first order in $ϵ$ and leading order in the spatial speed. Write spatial gradients with ${\partial }^i={\delta }^{ij}{\partial }_j$ here; these are Euclidean spatial indices in the Newtonian equation. The Levi-Civita connection formula givesThe exact change from proper time to $t$ in the spatial geodesic equation isTerms containing spatial velocities are negligible at leading order, leavingAn additive constant in $\Phi$ is immaterial to this acceleration and is normally fixed by the asymptotic time normalization. With units restored, $g_{00}=1+2{\Phi }_{\mathrm{physical}}/c^2$.

At the same order, staticity and the curvature definition giveThe trace of the four-dimensional Einstein equations is $R=-8\pi GT$, so their trace-reversed form isFor nonrelativistic dust, $u^0\simeq 1$, $u^i\simeq 0$, so $T_{00}\simeq \rho$ and $T\simeq \rho$. Keeping leading matter terms therefore gives $R_{00}=4\pi G\rho$. Combining this with the geometric expression yieldsThis trace reversal is essential: equating $R_{00}$ directly to $8\pi G\rho$ would miss the factor of two. The approximation is the quasistatic Newtonian sector; arbitrary independently imposed gravitational radiation is not part of that sector.