Horizon Boundary Mechanics: Force, Pressure, and Work from the Einstein Boundary Potential

J. T. Tyler

HORIZON BOUNDARY MECHANICS17 September 2026

Summary

Horizon boundary mechanics — from two boundary coordinates to force and pressure

Start with the horizon boundary’s and . The established then yields , , and by one complete differential. The checks that the length response is present in the gravitational boundary action.

: $L=L_\kappa$ , $U=U_\partial$ , $F=F_L$ , $P=P_L$ , and $\Pi=\Pi_\partial$ . We use $\kappa>0$ and a .

1. Two coordinates, one boundary potential

$A$ measures transverse size. $L$ expresses the selected $\kappa$ as a length:

L=\frac{c^2}{\kappa}

Like area, $L$ is a shared across the realizations below—not a universal . Here $c$ is the speed of light and $G$ is Newton’s constant. The becomes

U=\frac{A\kappa c^2}{8\pi G}=\frac{Ac^4}{8\pi GL}

Define the potential’s , not an additional mass:

M_U\equiv\frac{U}{c^2}=\frac{A\kappa}{8\pi G}=\frac{Ac^2}{8\pi GL}

2. Let the differential reveal the mechanics

Treat $A$ and $L$ as the two . Their gives

dU=\frac{U}{A}\,dA-\frac{U}{L}\,dL=\frac{\kappa c^2}{8\pi G}\,dA-M_U\kappa\,dL

Read the coefficients as :

\Pi\equiv\left(\frac{\partial U}{\partial A}\right)_L=\frac{U}{A},\qquad F\equiv-\left(\frac{\partial U}{\partial L}\right)_A=\frac{U}{L}

The has the form $F=M_U\kappa=PA$ .

Dividing the force by gives the :

\boxed{P\equiv\frac{F}{A}=\frac{\Pi}{L}=\frac{\kappa^2}{8\pi G}=\frac{c^4}{8\pi GL^2}}

$\Pi$ is force per length; $P$ is force per area. No enclosed or is needed for these definitions.

The definitions give the mechanical products below. With the and , the same potential is also $TS$ :

Energy identity

U=FL=\Pi A=TS=PAL

The gives $\Pi\,dA=T\,dS$ . Where the geometry supplies the $dV=A\,dL$ , the takes the familiar form

First law

dU=\Pi\,dA-F\,dL=T\,dS-P\,dV

3. Check the same force in the canonical response

For on a , Einstein $(\theta=\delta\theta=0)$ , the selected is [1]

\frac{\Theta^{(0)}_\kappa}{\Delta\tau}=\frac{Ac^2}{8\pi G}\,\delta\kappa=-\frac{A\kappa^2}{8\pi G}\,\delta L=-F\,\delta L

This is the same length-work coefficient, now obtained from the . and stationary realize this . The FRW application below uses its separately selected .

REALIZATIONS AND ENTROPY17 September 2026

The same mechanics, three realizations

Each setting supplies its own rate and relation between $L$ and geometry. For the spherical cases, the at a of $R$ gives the common bridge

E_{\mathrm{MS}}=\frac{c^4R}{2G},\qquad U=E_{\mathrm{MS}}\frac{R}{L}

4. FRW: energy, pressure, and cosmic expansion

For spatially flat, expanding , with $H=\dot a/a>0$ , choose the instantaneous [2]:

R_A=\frac{c}{H}=L,\qquad \kappa=cH,\qquad U=E_{\mathrm{MS}}=\rho V

Here $A=4\pi R_A^2$ , $V=4\pi R_A^3/3$ is , and $\rho$ is total . The supplies $E_{\mathrm{MS}}=\rho V$ [3]. Because $dV=A\,dR_A=A\,dL$ , the becomes

dE_{\mathrm{MS}}=T\,dS-P\,dV,\qquad P=\frac{\rho}{3}

The then reads

H^2=\frac{8\pi G}{c^2}P=\frac{8\pi G}{3c^2}\rho

With $p_m$ and $\kappa_E=8\pi G/c^4$ , the and are

\frac{1}{R_A^2}=\kappa_E P,\qquad\frac{2\ddot a}{ac^2}=-\kappa_E(P+p_m)

becomes $\dot P+H(\rho+p_m)=0$ . The $\rho+p_m$ controls the pressure’s evolution.

Spatial curvature and the meaning of volume

: the flat FRW case is shown for clarity. For general FRW use $R_A=c/\sqrt{H^2+kc^2/a^2}$ and $\kappa=c^2/R_A$ ; $P=\rho/3$ remains unchanged. $V=4\pi R_A^3/3$ is , not generally proper spatial volume.

5. Schwarzschild: a thermal–mechanical balance

With and $R_H=2GM/c^2$ , the same mechanics gives

\kappa=\frac{c^2}{2R_H},\qquad L=2R_H,\qquad U=\frac{Mc^2}{2}

The can be written

Mc^2=TS+FL

The equal thermal and mechanical products express a . The above explains the contrast: FRW uses $L=R_A$ , while Schwarzschild uses $L=2R_H$ .

6. Rindler: the construction needs no sphere

For a uniformly accelerated observer with $a_{\mathrm{obs}}$ , select $\kappa=a_{\mathrm{obs}}$ and a finite :

L=\frac{c^2}{a_{\mathrm{obs}}},\qquad A=\Delta y\,\Delta z,\qquad P=\frac{a_{\mathrm{obs}}^2}{8\pi G},\qquad U=\frac{Aa_{\mathrm{obs}}c^2}{8\pi G}

Here $L$ is an actual , while $A$ and $L$ remain independent. No enclosed source mass or spherical projection is needed.

7. Entropy compares a thermal length with a quantum length

The positive boundary potential also defines a , denoted by $\bar\lambda_{\partial}$ :

\bar\lambda_{\partial}=\frac{\hbar c}{U}=\frac{\hbar c}{TS}

Using the $T=\hbar\kappa/(2\pi c k_B)=\hbar c/(2\pi Lk_B)$ together with $U=TS$ , we obtain

\frac{S}{k_B}=\frac{2\pi L}{\bar\lambda_{\partial}}

Horizon entropy : the macroscopic thermal circumference $2\pi L$ and the quantum inverse-energy length $\bar\lambda_{\partial}$ .