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

J. T. Tyler

HORIZON BOUNDARY MECHANICS16 September 2026

The 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

Their difference in is

Equilibrium FRW

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

Schwarzschild

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

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

Equilibrium FRW · energy identity

E=TS=\Pi A

Equilibrium FRW · changes

2\,dE=T\,dS=\Pi\,dA

Schwarzschild · energy identity

\frac{E}{2}=TS=\Pi A

Schwarzschild · changes

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 GR^2}=\frac{1}{k_GR^2}=\frac{\kappa^2}{8\pi G}=\frac{\rho}{3}

Here $k_G=8\pi G/c^4$ is the , and $k_GPR^2=1$ . The also has the same quadratic acceleration–stress form as the $g^2/(8\pi G)$ .

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

F=\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 $E=2\Pi A=2TS$ 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

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}

The associated is $L/c=c/\kappa$ : light travels the distance $L=c^2/\kappa$ in that characteristic time. In Rindler, the length has the concrete just described; the timescale is not a measured transit time for a signal from the horizon to the accelerated observer.

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.

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 .

Why this is a mechanical response, not just a change of units

For and a on the selected , the supplies the same coefficient of the variation. That is the canonical grounding of the in the technical manuscript. and stationary realize this sector. supplies its own calibrated screen realization and geometric work conversion.

6. The pressure and energy identity

We can now derive the as

P=\frac{F_L}{A}=\frac{U}{LA}=\frac{1}{k_GL^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 $k_GPL^2=1$ ;
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

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}

This supplies a mechanical counterpart to the thermal product in the , giving an equal thermal–mechanical decomposition suggestive of a .

We also notice that has $L=2R$ , so the same work can be written in terms of the :

F_L\,dL=M\kappa\,dR

M\kappa=2F_L=\frac{c^4}{4G}=\frac{F_{\mathrm{FRW}}}{2}

Here $M=E/c^2$ is the , while $M_U=M/2$ . The and radius-conjugate forces describe the same work with different .

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 .