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For a weak, nearly static conformal factor in Nordstrom theory of gravitation, the trace of the dust stress-energy tensor is negative in signature $(-+++)$, so its scalar field equation reduces to the Poisson equation for the Newtonian gravitational potential. The geodesic equation for a conformally flat metric gives $\ddot{\mathbf{x}}=-\nabla \Phi$ at small speeds. The constant one in $\varphi$ has no effect on the force.

Use geometrized units $G=c=1$ until converting the clock readings. By the Birkhoff theorem, the spherical vacuum exterior is a portion of the Schwarzschild spacetime with fixed mass $M$:The exterior is not flat and is not radiating merely because the surface moves. When the surface crosses $r=2M$, this exterior must be represented in regular ingoing Eddington-Finkelstein coordinates, rather than treating the Schwarzschild-coordinate divergence as a physical barrier.

The Oppenheimer-Snyder model has a homogeneous pressureless matter interior. Since its initial expansion vanishes but its density is positive, the Friedmann equation requires positive spatial curvature. Choose a comoving radial angle $\chi$ and the boundary's proper time $\tau$, with the stopwatch set to zero at release. The interior FLRW metric and density areThe density law follows from the dust stress-energy tensor and its covariant conservation. The Friedmann equation and the initial rest condition giveIts contracting branch has the parameterizationIndeed $d\tau /d\eta =a$ and $\dot{a}=-\sin \eta /(1+\cos \eta )$ satisfy the displayed equation; $\eta =0$ is the turning point and $\eta =\pi$ is the final curvature singularity.

For homogeneous dust-ball matching, the boundary has areal radius $R=a\sin {\chi }_0$. Matching its mass to the exterior givesThis $M$ is the gravitational mass parameter, not the integral of rest density over the curved proper-volume element. The matching can also be checked directly. The outward angular extrinsic curvature inside is $K_{\theta \theta }=R\cos {\chi }_0$; outside it is $R\sqrt{1-2M/R+{\dot{R}}^2}$. The boundary motion hasso these curvatures agree. The induced metric is $-d{\tau }^2+R^{2}d{\Omega }^2$ on both sides and $K_{\tau \tau }=0$ because the pressure-free boundary is geodesic. No artificial surface stress layer is required.

The observer moving with this boundary meets the event horizon when $R=2M$. From $R=R_0(1+\cos \eta )/2$, the proper-time horizon crossing in homogeneous dust collapse occurs atThe singularity is reached at ${\tau }_s=\pi a_{\max }/2$, so the remaining clock time isThese are finite proper times, even though the external static time diverges at horizon crossing.

Convert the supplied rounded units consistently: $M=10^{38}$ Planck masses and $R_0=6.25\times 10^{39}$ Planck lengths. Thus $2M/R_0=0.032$, ${\chi }_0\simeq 0.17985$, and $a_{\max }\simeq 3.4939\times 10^{40}$ in Planck units. Multiplying geometric times by $5\times 10^{-44}$ seconds givesAt the accuracy justified by the rounded constants, these are about three milliseconds to horizon crossing and about seven microseconds more to the singularity. For $R_0\gg 2M$, expanding ${\eta }_H=\pi -2{\chi }_0$ gives $\Delta \tau \simeq (2/3)a_{\max }{\chi }_0^3\simeq 4M/3$, independently confirming the microsecond scale. The exact expression above retains the finite release-radius correction.

Strictly, this boundary clock reading is when the observer crosses the horizon, not when the horizon first exists anywhere. The event horizon inside an Oppenheimer-Snyder cloud is an outgoing radial null ray. Since $d\tau =ad\eta$, the interior metric is conformal to $-d{\eta }^2+d{\chi }^2+{\sin }^2\chi d{\Omega }^2$, and that ray has $d\chi /d\eta =1$. Tracing it back from $({\eta }_H,{\chi }_0)$ givesFor this cloud it starts at the centre after release, at comoving time ${\tau }_{\mathrm{birth}}\simeq 2.722\times 10^{-3}$ seconds. This answers the literal global-formation interpretation as well; it is not an event on the outer observer's worldline. The distinction matters because an event horizon is defined by the entire future causal escape problem, not by a locally detectable signal of formation.

For the observer at infinity, use the boundary's conserved Killing energy $E$ in the exterior:As $\tau \uparrow {\tau }_H$, $\dot{R}\to -E$ and $R-2M\simeq E({\tau }_H-\tau )$. Consequentlywhere ${\tau }_{\ast }>0$ fixes a harmless dimensionless logarithm. For actual outgoing light signals the relevant retarded time is $u=t-r_{\ast }(R)$. Since $r_{\ast }\simeq 2M\log (R-2M)+O(1)$,Therefore she never receives a signal showing the surface cross the horizon at any finite asymptotic time. Radiation emitted ever closer to crossing arrives ever later and is increasingly redshifted and diluted; photons emitted at or inside the event horizon cannot reach her. The formal crossing time at infinity is infinite, not the finite boundary stopwatch reading. Nor can the earlier global birth of the horizon be observed directly as a local flash from its central birth event.

Use geometric units $G=c=1$, metric signature $(-+++)$, and a four-velocity normalized by the physical metric. Write $\varphi =1+\Phi$, where $|\Phi |\ll 1$. For a pressureless perfect fluid, the dust stress-energy tensor is $T_{ij}=\rho U_i{U}_j$ and $g^{ij}{U}_i{U}_j=-1$. Since ${\eta }^{ij}={\varphi }^2{g}^{ij}$, its background trace isConsequently the field equation is $(-{\partial }_t^2+\Delta )\varphi =4\pi \rho {\varphi }^3$. Neglecting time derivatives and higher weak-field orders gives the Newtonian limit of Nordstrom gravity:The constant shift between $\varphi$ and $\Phi$ does not affect the Poisson equation or the force. Here $\rho$ is the physical rest density; if fluid indices are instead defined using the background metric, the intermediate powers of $\varphi$ change but the same leading-order equation results.

For the motion, the Levi-Civita connection of the conformally flat metric isIn particular ${\Gamma }^{\alpha }{}_{00}={\partial }_{\alpha }\log \varphi$. Replacing the affine parameter in the geodesic equation by coordinate time gives, with $w^i=(1,\mathbf{v})$ and $\mathbf{v}=d\mathbf{x}/dt$,At small speeds and for a nearly static field, this becomesThus the Newtonian gravitational potential and freely falling trajectories agree to the stated weak-field, slow-motion order. The comparison is an approximation, not an exact equivalence of relativistic trajectories.

In the local rest frame the four-velocity is $(c,0,0,0)$, the current is $(q{n}_{0}c,0,0,0)$, and the pressureless particle fluid has only $T^{00}=m{n}_0{c}^2$ nonzero. Covariance therefore givesThe absence of random rest-frame velocities is what makes this a dust stress-energy tensor rather than a fluid with pressure. The Lorentz four-force on a particle and its equation of motion are $f^{\nu }=q{F}^{\nu }{}_{\mu }{u}^{\mu }$ and $m{u}^{\mu }{\partial }_{\mu }{u}^{\nu }=q{F}^{\nu }{}_{\mu }{u}^{\mu }$.

Particle-number conservation gives ${\partial }_{\mu }(n_0{u}^{\mu })=0$. It also follows from charge conservation ${\partial }_{\mu }{J}^{\mu }=0$ when $q\ne 0$, obtained by antisymmetric double differentiation of Maxwell's equations. ConsequentlyAdding part (i) proves ${\partial }_{\mu }(T_{\mathrm{em}}^{\mu \nu }+T_{\mathrm{fluid}}^{\mu \nu })=0$. Energy and momentum exchanged with the field cancel in the total tensor.

For static scalar perturbations with no mixed time-space or spatial shear terms, the displayed metric tensor uses signature $(+---)$. With the reversed convention in curvature-sign convention in Killing derivative identities, its first-order Einstein tensor is $\delta G_{00}=2\Delta C$, $\delta G_{0i}=0$, and $\delta G_{ij}={\partial }_i{\partial }_j(A+C)-{\delta }_{ij}\Delta (A+C)$. These follow by substituting the metric into the first-order Christoffel symbols and Ricci tensor, discarding products of perturbations. A static source with the dust stress-energy tensor obeys $\Delta C=-4\pi G\rho$ and the decaying spatial equation makes $A=-C$. For a point mass the Newtonian potential of a point mass gives $A=-GM/r$, $C=GM/r$. The spatial potential is necessary to determine light propagation; a time potential alone is insufficient.