The Clausius coefficient and the directions that determine it

4. Extract the leading Clausius coefficient

Use the curvature convention in which positive $R_{kk}=R_{ab}k^ak^b$ focuses the congruence. For affine parameter on the twist-free pencil, the Raychaudhuri and area equations are

$$\dot\theta=-R_{kk}-\frac12\theta^2-\sigma_{ab}\sigma^{ab}, \qquad \frac{d}{d\lambda}dA_\lambda=\theta\,dA_\lambda.$$

With $\theta(0,p)=0$, the central generator has $\theta(\lambda,p)=-\lambda R_{kk}(p)+O(\lambda^2)$. Let $P_\ell\subset P$ be a centered transverse domain of size $\ell$ and area $A_0$. Use a fixed smooth congruence with bounded local derivatives in screen coordinates. Centering means that those coordinates have zero first moment with respect to $dA_0$. Taylor expansion then gives

$$\frac{1}{A_0}\int_{P_\ell}\theta(0,y)\,dA_0(y)=O(\ell^2).$$

This estimate allows a nonzero transverse derivative of the initial expansion. Without centering, that derivative can contribute at leading order. If $\ell=O(h)$, integrating the area equation gives

$$\frac{\delta A}{A_0}=\frac{h^2}{2}R_{kk}(p)+o(h^2).$$

The initial-expansion term contributes $O(h\ell^2)$, the transverse variation of the linear focusing coefficient vanishes as $\ell$ shrinks, and the quadratic optical terms vanish at the center of the initial section. The heat flux on the same domain satisfies

$$\frac{\delta Q}{A_0}=\frac{\kappa h^2}{2}T_{kk}(p)+o(h^2).$$

The normalized residual therefore has the limit

$$C(p,k) =\lim_{h\to0}\frac{2(\pi_Q-\pi_T\pi_S)}{\kappa A_0h^2} =T_{kk}(p)-\frac{\hbar\eta}{2\pi}R_{kk}(p).\tag{1}$$

A relative $o(1)$ error in the thermal readout leaves this limit intact. The local scaling-limit temperature is needed at this order. The area-entropy law and $C=0$ are constitutive inputs. The theorem gives their consequence.

Theorem 4.1. Leading equivalence

Suppose $g$ is $C^3$, $T$ is $C^1$ and conserved, and the thermal and entropy readouts hold at leading order with constant $\eta>0$. If $C(p,k)=0$ at every $p$ for all future null directions, or on a determining set, then

$$R_{ab}-\tfrac12 Rg_{ab}+\Lambda g_{ab}=\frac{2\pi}{\hbar\eta}T_{ab}$$

for a constant $\Lambda$ on $U$. Conversely, this tensor equation implies $C(p,k)=0$ at every aperture.

Proof. Set $E_{ab}=R_{ab}-2\pi T_{ab}/(\hbar\eta)$. In an orthonormal frame a normalized future null vector is $k=(1,n)$ with $|n|=1$. Subtracting $E(k,k)=0$ for $n$ and $-n$ gives $E_{0i}=0$. The remaining equality $E_{ij}n^in^j=-E_{00}$ for every unit $n$ gives $E_{ij}=-E_{00}\delta_{ij}$. Thus $E=\phi g$. A determining set gives the same implication. Take the divergence of $R_{ab}-\phi g_{ab}=2\pi T_{ab}/(\hbar\eta)$. Conservation and the contracted Bianchi identity give $d(\phi-R/2)=0$. Connectedness yields $\phi=R/2-\Lambda$. For the converse, a metric-proportional $E$ has zero contraction with every null vector.

Without conservation the readouts recover the trace-free equation and leave an undetermined scalar function. A constant vacuum-energy term proportional to $g$ has zero null flux and can be absorbed into $\Lambda$.

Restore units using $T_\chi=\hbar c\kappa/(2\pi k_B)$ for geometric $\kappa$ and physical entropy density $\eta$. The curvature coupling is then $2\pi k_B/(\hbar c\eta)=8\pi G/c^4$, hence

$$G=\frac{k_Bc^3}{4\hbar\eta}, \qquad \eta=\frac{k_B}{4\ell_P^2}.$$

The supplied entropy density fixes the coupling through this equality.

5. Determine the tensor from null directions

Aperture completeness is a finite-dimensional question at each point. In the orthonormal frame above write

$$E(k,k)=F(n)=A+b_i n^i+Q_{ij}n^in^j, \qquad Q_{ij}=Q_{ji},\quad \operatorname{tr}Q=0.$$

Here $A=E_{00}+\frac13\sum_i E_{ii}$, $b_i=2E_{0i}$, and $Q$ is the Euclidean trace-free spatial block. The kernel of $E\mapsto F$ is precisely $\mathbb Rg$. Thus $F$ determines the spacetime trace-free tensor. This function space has dimension $1+3+5=9$, and a set of directions is determining exactly when evaluation on that set is injective. Any open set suffices by real analyticity. A change of orthonormal observer relabels the sphere and rescales the normalized null contraction by a nonvanishing weight, preserving the criterion.

Reconstruct from nine evaluations

Use the six directions $\pm e_i$ and the three directions $(e_i+e_j)/\sqrt{2}$ for $i<j$, where $e_i$ form an orthonormal spatial basis. Set $s_i=[F(e_i)+F(-e_i)]/2$. Then

$$b_i=\tfrac12[F(e_i)-F(-e_i)], \qquad A=\tfrac13\sum_i s_i, \qquad Q_{ii}=s_i-A,$$

and

$$Q_{ij}=F\!\left(\frac{e_i+e_j}{\sqrt 2}\right) -\tfrac12(s_i+s_j)-\frac{b_i+b_j}{\sqrt 2}.$$

This is a constructive full-rank inversion. The evaluation determinant is analytic and nonzero at this tuple, so a generic tuple of nine directions has full rank. Eight evaluations leave at least one dimension undetermined.

What a circle misses

For a nondegenerate circle, rotate the axes so $n_3=c$ with $|c|<1$. The restriction is a trigonometric polynomial of order at most two. Restricting $1,n_1,n_2,n_1^2-n_2^2,n_1n_2$ supplies all five Fourier modes. The rank is five, and the four-dimensional kernel is spanned by

$$(n_3-c)\{1,n_1,n_2,n_3\}.$$

These are pointwise statements. With noisy data, reconstruction depends on the conditioning of the chosen evaluation matrix.