Classical first-law product leaves entropy-temperature calibration free

Let $X$ be a nonempty connected smooth manifold, let $E,A,\kappa \in C^{\mathrm{\infty }}(X,\mathbb{R})$, and let $C>0$ be a constant with $\kappa (x)>0$ for every $x\in X$. Here $dE$ and $dA$ are the differential one-forms of the indicated functions, and equality of one-forms means pointwise equality on $X$. Assume the classical first-law product identity $dE=C\kappa dA\mathrm{.}$ For every constant $b>0$ and offset $s_{\ast }\in \mathbb{R}$, set $T_b:=b\kappa$ and $S_{b,s_{\ast }}:=(C\mathrm{/}b)A+s_{\ast }$. Then $T_{b}d{S}_{b,s_{\ast }}=dE$. Distinct positive $b$ give distinct functions $T_b$, and changing $s_{\ast }$ changes $S_{b,s_{\ast }}$ without changing its differential. Thus this product identity alone does not determine the temperature scale or the absolute entropy offset.

The names $T_b$ and $S_{b,s_{\ast }}$ denote real-valued functions here; no operational temperature or physical entropy is assumed. If the symbols carry dimensions, consistency requires $[T]=[b][\kappa ]$, $[S]=[C][A]\mathrm{/}[b]$, and $[E]=[T][S]$. No particular unit system is used.

This is only a mathematical factorization result. It does not assert that every pair $(T_b,S_{b,s_{\ast }})$ describes measured thermodynamic quantities or derive a quantum calibration. The cited treatments distinguish the classical black-hole-mechanics identity from the additional quantum calculation that fixes a Hawking-temperature/entropy normalization. Any application with charge, angular momentum, or other work terms must retain those terms and its stated restrictions.