The same-ball detection theorem

4. Detection at every radius

Theorem 4.1 (All-radii same-ball detection). In AdS$_4$ with $G=\ell=1$, take $h_{xy}=h_{yx}=z(t^2+z^2/5+1/3)$ and set the other components to zero. For every rest-frame ball at $t=0$, both first-order readouts vanish. Its wedge canonical energy is

$$\mathcal E_B(h)=\frac{\pi R^6}{138600}(29R^4+110R^2+165)>0,\qquad R>0. \tag{14}$$

The result is wedge-local. If an admissible vacuum family has this tangent on the wedge, (14) gives its $F_B''(0)$.

Proof: the first-order cancellation. Write $h_{xy}=zH(t,z)$. The linearized vacuum equation reduces to $H_{tt}-H_{zz}-4H_z/z=0$. For the polynomial, both $H_{tt}$ and $H_{zz}+4H_z/z$ equal $2$. Its boundary normalizable coefficient has no $tt$ component, so the energy variation is zero. In the area variation the hemisphere projector contracts with the shear to give $(x-x_0)(y-y_0)$ times an axially symmetric factor. Reflect either coordinate about its central diameter and the integral changes sign on the same hemisphere. It vanishes. The companion edition derives the more general fixed-frame linear kernel.

Proof: the extremal-surface response. The tangent is translation invariant, so compute at the centred ball. In spherical coordinates $(\varrho,\theta,\varphi)$ on the slice, $z=\varrho\cos\theta$. Write the radial graph as $\varrho=R+\lambda w+O(\lambda^2)$ and set $s=\sin\theta$, $c=\cos\theta$. The minimizing first-order displacement is

$$w_*=f(\theta)\sin(2\varphi),\qquad f=-\frac{R^4}{120}s^2c^3(2R^2c^2+R^2+5). \tag{15}$$

After integrating out the azimuthal dependence, the forced sector satisfies

$$R^4(21R^2c^2+25)s^4c^4-30(s^3+s)f'+120cf-30s^2cf''=0. \tag{16}$$

Substituting (15) gives zero. Its $O(s^2)$ behaviour at the pole is regular; its $O(c^3)$ behaviour at the anchor keeps the anchors fixed. The positive shape Hessian gives uniqueness in the admissible space.

Proof: evaluate the bulk and surface terms. The bulk term in the second-order charge–area identity uses

$$\left.D^2\mathcal F_{tt}[h,h]\right|_{t=0}=\frac{z^4}{30}(69z^4+130z^2+45). \tag{17}$$

To see the bulk coefficient, use time symmetry: the extrinsic curvature of the slice vanishes along the affine metric path, so the varying part of $\mathcal F_{tt}$ is $z^{-2}\mathcal R^{(3)}/2$, with $\mathcal R^{(3)}$ the scalar curvature of the spatial metric. Rotate $x,y$ by $45^\circ$ to diagonalize the shear. In proper radial coordinate $\sigma=-\log z$, the two transverse scale factors are $a_\pm=z^{-1}\sqrt{1\pm\lambda z^3(z^2/5+1/3)}$. Differentiate the spatial scalar curvature $-2(a_+''/a_++a_-''/a_-+a_+'a_-'/(a_+a_-))$ twice in $\lambda$, with primes here denoting $\sigma$ derivatives, to obtain (17).

At fixed $z$ the transverse disk has radius $\sqrt{R^2-z^2}$. On it $\xi_B^t=\pi(R^2-z^2-r^2)/R$; integrating over the disk gives $\pi^2(R^2-z^2)^2/(2R)$. The fixed-surface term comes from differentiating the induced area density $\sqrt{q_{\theta\theta}q_{\varphi\varphi}-q_{\theta\varphi}^2}$ of the radial graph. With $X=q_0^{-1}\delta q$ on the fixed hemisphere, its second derivative is the reference density times $(\operatorname{tr}X)^2/4-\operatorname{tr}(X^2)/2$. For the shape term, insert the proper normal displacement $w_* /(R c)$ in the shape Hessian. Denote the bulk integral including its $1/(8\pi)$ prefactor by $I$. The three terms are

$$A_{gg}=-\frac{4\pi R^6}{2027025}(3717R^4+15210R^2+16445), \tag{18}$$ $$Q[w_*]=\frac{\pi R^6}{4054050}(567R^4+2730R^2+3575), \tag{19}$$ $$I=\frac{\pi R^6}{41580}(69R^4+286R^2+297). \tag{20}$$

Then $-A_{gg}/4+Q[w_*]/4-I$ simplifies to (14). $\square$

Positivity without solving for the surface

Set the trial displacement to zero in the shape minimization. This gives a lower bound directly:

$$\mathcal E_B(h)\ge -\frac{A_{gg}}4-I =\frac{\pi R^6}{8108100}(1413R^4+5070R^2+7865)>0. \tag{21}$$

The bound holds at every $R>0$. The fixed-surface value is already positive, and allowing the surface to relax adds $Q[w_*]/4$.

Independent symplectic evaluations

The direct symplectic calculation uses (15) as the normal gauge data at $\widetilde B$. The mixed and normal-block Hollands–Wald conditions give the required first normal derivatives of the other components. Integrate the current and the surface correction. The totals are $38\pi/17325$ at $R=1$ and $8552\pi/17325$ at $R=2$, agreeing with (14). At $R=1$ the bulk and surface split is $\pi/675+37\pi/51975$. Varying the free gauge data leaves the sum unchanged.

For the AdS$_3$ black-brane benchmark from [1], the split in units $G=\ell=1$ is $-2R^4/15+7R^4/45=R^4/45$. This is also the second derivative of the leading relative entropy $\lambda^2R^4/90$ for the exact family. Dropping the surface correction would reverse the benchmark's sign. The vacuum and a boundary-trivial pure-gauge tangent both give zero. The controls check the orientation, the Hessian normalization and cancellation of the gauge-dependent terms.