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For a geodesic tangent $u$ and a stationary Killing vector field $K$, the Killing energy is $E=-g(K,u)$. It is conserved along the geodesic. When $K$ is future timelike and $u$ is future causal, $E>0$ for a nonzero tangent; it need not be positive where $K$ becomes spacelike.

The Physical-process first law for a rotating black hole, in four-dimensional general relativity with $G=1$, isIt applies to a small neutral matter influx into an initially stationary, nonextremal black hole that settles to another stationary state. The background surface gravity and angular velocity are $\kappa$ and ${\Omega }_H$, and $\delta E,\delta J$ are the Killing energy and angular momentum delivered through the event horizon. Other conserved charges are held fixed. The perturbation must be small enough that its horizon generators do not develop caustics, and all equalities below are to first order.

Let $t^a$ be the stationary Killing vector field normalized at infinity, and ${\psi }^a$ the axial Killing vector field with $2\pi$-periodic orbits. The horizon generator is ${\xi }^a=t^a+{\Omega }_H{\psi }^a$. Choose an affine tangent $k^a$ and parameter $\lambda$ with zero at the past stationary limiting section, so affine horizon-generator scaling gives, on the background horizonThe background null expansion and null shear vanish, and null twist vanishes for the hypersurface-orthogonal horizon generators. Linearizing the Null Raychaudhuri equation and using the Einstein field equations givesThe quadratic null expansion and null shear terms are second order. With background cross-sectional measure $d{A}_0$, the first-order area change is $\delta A={\int }_{\mathcal{H}}\delta \theta d\lambda d{A}_0$. Multiplying the evolution equation by $\lambda$ and integrating by parts gives the first-order horizon-area response to matter fluxThe boundary term $[\lambda \delta \theta ]$ vanishes: $\lambda =0$ at the past limiting section, and the final stationary boundary condition gives zero expansion in the settled future. Equivalently, assume a sufficiently localized influx with the required late-time decay.

The energy and angular-momentum fluxes have signsThese correspond to the particle conventions $E=-p\cdot t$ and $J=p\cdot \psi$. Their combination isThis proves the requested Physical-process first law of black-hole mechanics. For charged infall the corresponding law contains an additional horizon-potential term ${\Phi }_H\delta Q$; the energy-minus-angular-momentum version assumes that work term is absent.

A spacetime is geodesically complete if every maximal geodesic has an affine parameter ranging over all of $\mathbb{R}$. For timelike geodesics this is equivalent to unbounded proper time in both directions. An extendible geodesic segment can be prolonged in the same spacetime; an inextendible geodesic cannot. A finite coordinate endpoint need not imply finite affine parameter.

For the Kruskal spacetime, use $U,V$ withIn the right exterior $U<0<V$. A truncated ray $U=U_0<0$, $V_1<V<V_2$ is an extendible geodesic of radial null type: neither artificial endpoint is a spacetime boundary. A future ray $U=U_0>0$ in the black hole reaches $UV=1$, hence $r=0$, and is inextendible geodesic and future null-geodesically incomplete. Its Killing energy $E\ne 0$ gives $dr/d\lambda =-|E|$, so the Schwarzschild singularity occurs at finite affine parameter. The maximal continuation toward the past supplies the other half of this same null geodesic.

There is no inextendible, complete radial timelike geodesic in positive-mass Kruskal spacetime. This requested example is impossible as printed. For a radial timelike geodesic, the conserved Killing energy and normalization giveIf $E^2<1$, $r$ has at most one turning point, a maximum $r_{\max }=2M/(1-E^2)$; a maximal trajectory runs from the white hole Schwarzschild singularity to the Schwarzschild singularity. If $E^2\ge 1$, there is no finite turning point; one end can lie at infinity but the other reaches $r=0$. The exceptional $E=0$ trajectory through the bifurcation surface also reaches $r=0$ in both time directions. Near $r=0$,whose integral is finite. Constant-$r$ radial timelike curves are accelerated, not geodesics.

Two plausible repairs have different meanings. Removing “radial” permits a complete circular timelike geodesic at $r=R>6M$, with nonzero angular momentum and proper time ranging over $\mathbb{R}$. Replacing “timelike” by “null” permits a complete horizon null geodesic: $U=0$, $V\in \mathbb{R}$, with $V$ an affine parameter. The Penrose diagram shows both repairs explicitly, together with the two valid requested examples; the circular trajectory is only a radial projection and is labelled as nonradial.

The horizontal boundaries are the Schwarzschild singularities, diagonal dashed lines are the Killing horizons, and outer diagonal edges are null infinity.

For radial free fall, let $E=f{\dot{t}}_A>0$ be Alice's conserved Killing energy. The inward branch of the timelike geodesic has $\dot{r}=-\sqrt{E^2-f}$, henceWith $\delta =r-2M$, the near-horizon geodesic equations giveThe constants in the logarithms are understood to make their arguments dimensionless. Therefore $u_A=t_A-r_{\ast }=2t_A+O(1)$. For an outgoing signal reaching the fixed-radius Bob, $t_B=u_A+r_{\ast }(r_B)$, so $t_B=2t_A+O(1)$. The redshift consequently behaves asThis is the answer when $t$ is Bob's Schwarzschild time, as appropriate to the stated observation. It is the Schwarzschild surface gravity $a=\kappa =1/(4M)$. If $t$ instead means the emission Schwarzschild time $t_A$, the same redshift is proportional to $e^{t_A/(2M)}$ and gives $M=1/(2a)$. The two answers use different clocks, not different dynamics. If the measured exponential uses Bob's proper time ${\tau }_B$, its rate is $1/(4M\sqrt{f_B})$. In SI units the reception-time result is $M=c^3/(4Ga)$.

The reception-time exponent also holds for smooth radial infall crossing the future Schwarzschild event horizon with finite nonzero $dr/d\tau$: regular Ingoing Eddington-Finkelstein coordinates give finite $v$ there, $u=v-2r_{\ast }\sim -4M\log ({\tau }_H-\tau )$. Thus no special value of $E$ is needed.

Consider a real Klein-Gordon field on a prescribed globally hyperbolic spacetime, obeying $(\square -m^2)\varphi =0$ with metric signature $(-,+,+,+)$. A curvature coupling can be included as $m^2\mapsto m^2+\xi R$. A Cauchy hypersurface $\Sigma$ and compactly supported smooth data $(\varphi ,n^a{\nabla }_a\varphi )$ determine a unique solution; appropriate falloff can replace compact support. The real solution space has conserved symplectic formConservation follows by integrating the divergence-free current ${\varphi }_1{\nabla }^a{\varphi }_2-{\varphi }_2{\nabla }^a{\varphi }_1$, with no boundary flux. On complex solutions the conserved Klein-Gordon inner product isIt is indefinite on the full complex solution space.

Choose a complete positive-norm mode subspace, with modes $u_j$ satisfyingEquivalently choose a compatible complex structure on the Klein-Gordon solution space. The mode labels may be continuous, in which case sums and Kronecker symbols become integrals and Dirac delta functions. Construct the one-particle Hilbert space from these modes and its bosonic Fock space. Promote the field to the operator-valued distributionFor a foliation with spatial metric determinant $h$, the conjugate momentum density is $\hat{\pi }=\sqrt{h}{n}^a{\nabla }_a\hat{\varphi }$. Mode completeness gives the equal-time canonical commutation relationsThe Fock vacuum obeys $a_j|0⟩=0$, and $a_j^{†}{a}_j$ counts particles in the chosen mode. For local products and a renormalized stress-energy tensor, physically admissible states are further restricted by the Hadamard condition. The field algebra exists without a preferred Fock vacuum.

A different admissible mode splitting can mix positive and negative norms:Orthonormality imposes the canonical identities for a bosonic Bogoliubov transformationThe same field then hasWithout a preferred notion of positive frequency, a nonstationary spacetime supplies no distinguished mode splitting: particle number and vacuum depend on the choice of modes, although the field equation and field algebra do not. With infinitely many modes the Bogoliubov transformation need not be unitarily implementable; finite total mixing requires a Hilbert-Schmidt operator $\beta$.

For a stable strictly stationary spacetime, a chosen future globally timelike Killing vector field $K$ gives a preferred time translation. Choose modes withWhen the corresponding conserved Killing energy is positive and the spectral problem has suitable boundary conditions and no problematic zero modes, this gives the preferred vacuum state in a stationary spacetime and particles relative to $K$. Positive-frequency mode mixing within that same subspace leaves the Fock vacuum unchanged. Rescaling $K$ by a positive constant changes the frequency units but not their sign.

Stationarity alone, if it only means a Killing field timelike near infinity, is insufficient for a global unique particle interpretation. In the Kerr ergoregion, ${\partial }_t$ is spacelike, so positive frequency relative to $t$ does not automatically select a positive-norm subspace throughout the geometry; superradiance illustrates the difficulty. Additional vacuum and boundary choices remain necessary. The customary stationary answer therefore assumes a suitable timelike stationary flow and a stable positive-energy quantization; it does not assert that every stationary black-hole extension has one globally preferred vacuum.