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Rest fluid state, frame and geometric current definitions — Examples

1 · Prerequisites

2 · Summary

This leaf gives a direct parcel calculation from the A-page state function. Adding a constant rest-energy term per marker leaves the temperature and pressure derivatives unchanged, shifts $H$ and $μ$ by that constant and $ε$ and $h$ by number density times the constant, and gives $E_{m} = E_{0} + N m c^{2}$ for parcel totals. No material measurement is asserted.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

ExampleConstruction: AI-adaptedVerification: AI-adaptedjudge pass (gpt-6-luna)Open item page →

Parcel derivatives and rest-energy shift

Example

Let $e_{0} \in C^{2} (X, R)$ be any specified single-marker state function on a nonempty open state domain $X \subset (0, \infty) \times R$ . Fix a mass per marker $m \geq 0$ with units $k g m a r k e r^{- 1}$ and $c > 0$ with units $m s^{- 1}$ , and define $e_{m} (v, σ_{p}) : = e_{0} (v, σ_{p}) + m c^{2} .$ Then $e_{m} \in C^{2} (X, R)$ . The derived temperature and pressure are unchanged, while $H_{m} = H_{0} + m c^{2}$ , $μ_{m} = μ_{0} + m c^{2}$ , $ε_{m} = ε_{0} + n m c^{2}$ , and $h_{m} = h_{0} + n m c^{2}$ . For parcel totals, $E_{m} (S, V, N) = E_{0} (S, V, N) + N m c^{2}$ . This is a mathematical change of the state function by a constant rest-energy offset, not an empirical statement about a material.

Facts & Assumptions

Given: The state domain, state function, marker volume and entropy coordinates from Rest fluid state, frame and geometric current definitions, and the differential identities of Local Gibbs, Euler, parcel first law and response identities.

[F1]

Rest fluid state, frame and geometric current definitions: $T = e_{σ_{p}}$ , $p = - e_{v}$ , $ε = n e$ , $H = e + p v$ , $h = ε + p$ , and $μ = H - T σ_{p}$ .

[F2]

Local Gibbs, Euler, parcel first law and response identities: the parcel energy $E (S, V, N) = N e (V / N, S / N)$ has differential $d E = T d S - p d V + μ d N$ .

[A1]

$m \geq 0$ and $c > 0$ are fixed parameters, so the added energy per marker is the constant $m c^{2}$ on $X$ .

Verification

1.1F1A1algebra

Since $m c^{2}$ is constant in $(v, σ_{p})$ , $(e_{m})_{σ_{p}} = (e_{0})_{σ_{p}}$ and $- (e_{m})_{v} = - (e_{0})_{v}$ . Thus the derived temperature and pressure are unchanged.

2.1F1step 1.1A1algebra

The definitions give $H_{m} = e_{m} + p_{m} v = H_{0} + m c^{2}$ and $μ_{m} = H_{m} - T_{m} σ_{p} = μ_{0} + m c^{2}$ .

2.2F1step 1.1A1algebra

With $n = 1 / v$ unchanged, $ε_{m} = n e_{m} = ε_{0} + n m c^{2}$ and $h_{m} = ε_{m} + p_{m} = h_{0} + n m c^{2}$ .

3.1F2step 2.1A1algebra∎

At fixed $(S, V, N)$ , the parcel construction gives $E_{m} = E_{0} + N m c^{2}$ . Differentiating adds only $m c^{2} d N$ to $d E_{0}$ ; hence at fixed marker count it changes neither the entropy derivative $T$ nor the pressure-work derivative $- p$ . This agrees with the chemical-potential shift in step 2.1 and the parcel first law in [F2].