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Half-capacity bath attracts the selected exchange state

Physical scope

A single common-temperature point of the adopted finite-bath scalar exchange model with C_b=|C_G|/2; no causal transport or generic star-stability claim.

Example

At a selected subsystem state with $T_{*} > 0$ and $C_{G} < 0$ , choose the finite bath capacity $C_{b} = ∣ C_{G} ∣ / 2$ and bath energy $E_{b} = E_{0} + C_{b} T_{*}$ . In the adopted scalar exchange model, the common-temperature state is a strict local entropy maximum and is locally exponentially attracting; its entropy curvature is $S_{t o t}^{''} (x_{*}) = - \frac{1}{T_{*}^{2} ∣ C_{G} ∣} < 0.$

Facts & Assumptions

Given: The finite-bath exchange theorem Finite bath stabilizes the selected negative-capacity exchange model and its adopted matched-temperature point.

[F1]

Finite bath stabilizes the selected negative-capacity exchange model gives $S_{t o t}^{''} (x_{*}) = - T_{*}^{- 2} (1 / C_{G} + 1 / C_{b})$ and the strict local maximum, forward-invariant local basin, exponential attraction, and entropy production $L (S_{t o t}^{'})^{2} \geq 0$ whenever $C_{G} < 0$ and $0 < C_{b} < ∣ C_{G} ∣$ .

Verification

1.1givenF1algebra

With $C_{b} = ∣ C_{G} ∣ / 2$ , the bath capacity is positive and satisfies $C_{b} < ∣ C_{G} ∣$ . Setting $E_{b} = E_{0} + C_{b} T_{*}$ gives $T_{b} = (E_{b} - E_{0}) / C_{b} = T_{*}$ , so the selected state is common-temperature as required in [F1].

2.1F1step 1.1algebra∎

Since $C_{G} < 0$ , $1 / C_{G} = - 1 / ∣ C_{G} ∣$ and $1 / C_{b} = 2 / ∣ C_{G} ∣$ . Hence [F1] gives $S_{t o t}^{''} (x_{*}) = - 1 / (T_{*}^{2} ∣ C_{G} ∣) < 0$ and its stated local exponential attraction and nonnegative entropy production. This is the specified exchange ODE, not a stellar evolution law.

Depends on

Finite bath stabilizes the selected negative-capacity exchange model

Dependency tree · 2 levels

Sources

P.-H. Chavanis, Statistical mechanics of self-gravitating systems in general relativity: I. The quantum Fermi gas, arXiv:1908.10806v1 (2019), Sec. III.E