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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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Two entropy-temperature calibrations preserve the same energy differential

Example

Use the abstract state family $X = (0, \infty)$ with coordinate $r$ , and define $E (r) = r$ , $A (r) = r$ , $κ (r) = 1$ , and $C = 1$ . These are formal real-valued functions with no physical units assigned. Then $d E = C κ d A = d r$ . The choices $b = 1$ and $b = 2$ , with offset $s_{*} = 0$ , give $(T_{1}, S_{1, 0}) = (1, r)$ and $(T_{2}, S_{2, 0}) = (2, r / 2)$ .

Facts & Assumptions

Given: The stated smooth state family and the calibration-freedom lemma.

[A1]

On $X = (0, \infty)$ , $E (r) = A (r) = r$ , $κ (r) = 1$ , and $C = 1$ .

[F1]

Classical first-law product leaves entropy-temperature calibration free: For every constant $b > 0$ and offset $s_{*} \in R$ , set $T_{b} : = b κ$ and $S_{b, s_{*}} : = (C / b) A + s_{*}$ . Then $T_{b} d S_{b, s_{*}} = d E$ . Distinct positive $b$ give distinct functions $T_{b}$ , and changing $s_{*}$ changes $S_{b, s_{*}}$ without changing its differential.

Verification

Given: [A1] and [F1].

1.1givenA1algebra

The family $X = (0, \infty)$ is a nonempty connected smooth manifold, the displayed functions are smooth, $C = 1 > 0$ , and $κ (r) = 1 > 0$ . Direct differentiation gives $d E = d r = C κ d A$ , so the lemma's hypotheses hold.

2.1F1A1step 1.1algebra∎

Apply [F1] with $(b, s_{*}) = (1, 0)$ and $(2, 0)$ . The resulting functions are $(T_{1}, S_{1, 0}) = (1, r)$ and $(T_{2}, S_{2, 0}) = (2, r / 2)$ ; hence $T_{1} d S_{1, 0} = 1 \cdot d r = d r$ and $T_{2} d S_{2, 0} = 2 \cdot (1 / 2) d r = d r = d E$ , although the temperatures and entropy gradients differ.

Depends on

Classical first-law product leaves entropy-temperature calibration free

Two entropy-temperature calibrations preserve the same energy differential

Example

Facts & Assumptions

Verification

Depends on

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