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

J. T. Tyler

HORIZON BOUNDARY MECHANICS20 October 2026

The intuitive path to the mechanics

From two horizon questions to force, pressure, and work

and share the same for , , , , and . describes a universe that is , with or .

Here $\Pi$ denotes the : the mechanical coefficient multiplying a change in .

E=\frac{c^4}{2G}R,\qquad R=\frac{2GM}{c^2}

\Pi=\frac{\kappa c^2}{8\pi G}

T=\frac{\hbar\kappa}{2\pi k_Bc}

S=\frac{Ak_B}{4\ell_P^2}

use $M=E/c^2$ , a , and the positive value.

1. The factor of two that raises the questions

We can see a factor of two difference in their .

Equilibrium FRW

\kappa=\frac{c^4}{2GM}=\frac{c^2}{R}

Schwarzschild

\kappa=\frac{c^4}{4GM}=\frac{c^2}{2R}

And this results in a difference in how their mechanical and thermal products $\Pi A$ and $TS$ relate to $E$ :

Energy identities and changes for equilibrium FRW and Schwarzschild horizons
Horizon	Energy identity	Changes
Equilibrium FRW	$E=TS=\Pi A$	$2\,dE=T\,dS=\Pi\,dA$
Schwarzschild	$\frac{E}{2}=TS=\Pi A$	$dE=T\,dS=\Pi\,dA$

Where is the in $dE=T\,dS-P\,dV$ for , given that the contribution appears as $2\,dE$ ?
What accounts for the of the energy when $TS=E/2$ ?

2. Treating the changing FRW horizon as a whole

relates the change to , using an equilibrium-form . Other formulations include matter work and the changing enclosed volume. The question we want to ask is more specific: since the , , , , and of the entire changing sphere can all be written in terms of the or , can a in the familiar described by the or also be found?

Looking at the through this , we see that given $E=(c^4/2G)R$ and $\kappa=c^4/(2GM)$ ,

M\kappa=\frac{c^4}{2G}=\frac{E}{R}

We can ask how this may play into the mechanics and thermodynamics, and how it may handle the relation $2\,dE=T\,dS$ . Since every quantity except is already specified, we can first ask which makes the familiar hold for a changing sphere:

dE=T\,dS-P\,dV

Since $2\,dE=T\,dS$ , we immediately find

dE=P\,dV

Solving for $P$ , where $\rho=E/V$ is , gives

P=\frac{c^4}{8\pi G R^2}=\frac{\kappa^2}{8\pi G}=\frac{\rho}{3}

The same pressure follows directly by differentiating the and with respect to radius. This does not depend on a stationary null horizon.

Given $P=F/A$ , we can solve for the and find

F=PA=\frac{c^4}{8\pi G R^2}\times A=\frac{c^4}{2G}=\frac{E}{R}=M\kappa

This is the we found above. The and now have a direct place in the changing ’s .

3. The FRW results in pressure variables

This relation between and also simplifies the presentation of the . In the spatially flat case, the is $R=c/H$ , giving

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

Using the Einstein coupling $k_G=8\pi G/c^4$ , the and relations become

\frac{1}{R^2}=k_GP,\qquad\frac{2\ddot a}{ac^2}=-k_G(P+p_m)

Here $a(t)$ is the , $H=\dot a/a$ , and $p_m$ is the , distinct from the $P$ . sets the inverse-square of the , while its combination with determines . Expansion accelerates when $p_m<-P$ .

,

\dot\rho+3H(\rho+p_m)=0

becomes

\dot P+H(\rho+p_m)=0

The combination $\rho+p_m$ , the , therefore controls how evolves. These are the familiar cosmological relations expressed in horizon-pressure variables. Only the identification $R=c/H$ assumes ; the relations written in terms of the also apply with .

4. What is shared by the different horizons?

So this and clearly play a role in , but how do they relate to , or the even simpler , given their different energy descriptions? ’s $dE=\Pi\,dA$ does not need an additional work term in its , though $\frac{E}{2}=\Pi A=TS$ raises the question of what mechanical product accompanies the thermal half in the .

The contribution $\Pi\,dA$ describes the response at a given . The gives a common mechanical description through the shared and structure:

U=\frac{A\kappa c^2}{8\pi G}=\Pi A

For , this value is exactly the . The has the same area–rate structure with its own rate and work allocation. Matching values on a horizon family does not automatically identify their independent source variations.

With the , this gives

U=\Pi A=TS

As $U$ is the common mechanical , we can take its to see how it changes:

dU=\frac{U}{A}\,dA+\frac{U}{\kappa}\,d\kappa=T\,dS+S\,dT

The coefficient $U/A=\Pi$ is the familiar , and $T\,dS$ is the contribution familiar from thermodynamics. The other contribution, $S\,dT$ , describes a change in the scale. But $d\kappa$ does not yet display the length or volume displacement we expect in .

5. The natural displacement coordinate

We can look for a natural associated with this complementary contribution. have no spherical like and . Instead, a in flat spacetime has a horizon a $c^2/a_{\mathrm{obs}}$ away, measured on the observer’s . With $\kappa=a_{\mathrm{obs}}$ , this gives the :

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

defines a $t_\kappa=c/\kappa$ . Its corresponding is $L=ct_\kappa=c^2/\kappa$ . In , this length is also the observer–horizon on the observer’s instantaneous rest slice.

This makes $L$ a natural . It expresses the normal acceleration scale as a length, so the potential’s response to a change in it can take the familiar . Using only $c$ and $\kappa$ , it is the available length apart from a numerical factor, which the fixes. We can use that same coordinate in the other examples— $L=R$ for and $L=2R$ for —without requiring a universal proper-distance interpretation. The differential below identifies its force coefficient; the supplies the canonical check in its stated setting.

Rewriting the using this length gives

U=\frac{Ac^4}{8\pi GL}

Differentiating $U$ , we find

dU=\frac{U}{A}\,dA-\frac{U}{L}\,dL=T\,dS+S\,dT

As $L$ is now the mechanical , we recognize the contribution $-(U/L)\,dL$ as the , with

F_L=\frac{U}{L}=M_U\kappa,\qquad M_U=\frac{U}{c^2}

This generalizes the definition $F=E/R=M\kappa$ to the three descriptions using the . Our common boundary-potential becomes

dU=\Pi\,dA-F_L\,dL=T\,dS-F_L\,dL

-F_L\,dL=S\,dT

The negative length-work contribution is therefore exactly $S\,dT$ : increasing the appears thermally as a .

Two GR foundations: stationary action and cosmological screen

The can be written directly in acceleration length. For and a , it gives the same coefficient after dividing by the chosen duration. and stationary realize this . The instead follow directly from the Einstein constraint and Misner–Sharp horizon energy. Both have a GR foundation; the of a moving FRW boundary is a further question.

6. The pressure and energy identity

We can now derive the as

P=\frac{F_L}{A}=\frac{U}{LA}=\frac{c^4}{8\pi G L^2}=\frac{\kappa^2}{8\pi G}

This reveals the as

the familiar macroscopic , $P=F_L/A$ ;
the , $P=\Pi/L$ , or $\Pi=PL$ ;
part of the $PL^2=\frac{c^4}{8\pi G}$ ;
the same quadratic acceleration–stress form as the , $g^2/(8\pi G)$ ;
one third of the in , $P=\rho/3$ .

The common is

U=F_LL=\Pi A=TS=PAL

Where the geometry supplies $dV=A\,dL$ , as it does for , the boundary-potential becomes

dU=T\,dS-P\,dV

In this , $L=R_A$ , so $F_R=F_L$ and $P_R=P_L=P$ . The length work is already the work associated with the radius of the changing horizon sphere.

7. Returning to the black-hole question

As $U=E/2$ for , we have

E=F_LL+TS,\qquad F_LL=TS=\frac{E}{2}

The new force product supplies a mechanical representation of the second equal product in the Schwarzschild mass identity. Both $TS$ and $F_LL$ equal $U$ ; they are not two independent energy reservoirs. This answers the original “other half” question at the level of the , while retaining the standard mass first law.

The Schwarzschild work can now refer directly to the same horizon sphere. Since $L=2R$ , the : $dL/L=dR/R$ . The same work is

F_L\,dL=F_R\,dR=M\kappa\,dR=P_R\,dV

\begin{gathered}F_R=M\kappa=2F_L=\frac{c^4}{4G}=\frac{F_{\mathrm{FRW}}}{2}\\[1.1em]P_R=\frac{F_R}{A}=2P_L=\frac{\kappa^2}{4\pi G}\end{gathered}

Here $M=E/c^2$ is the full black-hole mass, while $M_U=M/2$ . The common pressure is still $P=P_L$ ; the is $P_R=2P_L$ . Since $V=4\pi R^3/3$ and $dV=A\,dR$ , both area and work now describe the same changing horizon sphere.

dU=\Pi\,dA-F_R\,dR=\Pi\,dA-P_R\,dV

The potential remains $U=E/2$ , so the remains $dE=T\,dS$ . Area and volume are linked by the same . Changing $L$ in fixed proportion to $R$ describes this horizon-size change; it does not move another surface at $2R$ .

8. Entropy in mechanical and quantum lengths

Further, we can define the associated with the ,

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

Given

T=\frac{\hbar\kappa}{2\pi k_Bc}=\frac{\hbar c}{2\pi Lk_B}

we have

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

is therefore $2\pi$ times the ratio of the ’s to the of its . Equivalently, it measures the in units of the ’s , $k_BT$ . This expresses the usual through the mechanical length and the associated with the same .

Questions and objections: what is new, why L, and the scope of the result →