Thermal joins force a common mathematical energy orientation

Suppose two component signs are opposed. Choose a join state $Z$ and a reversible split $(x,y)$ by T2 in [A2]. Write $x=(u_x,v_x)$ and $y=(u_y,v_y)$. Openness in [A1] permits a sufficiently small nonzero $\delta$ with $x^{\mathrm{\prime }}=(u_x+\delta ,v_x)$ and $y^{\mathrm{\prime }}=(u_y-\delta ,v_y)$ in their respective charts. Choose the sign of $\delta$ so ${\epsilon }_{{\mathrm{\Gamma }}_x}\delta >0$; opposed signs imply ${\epsilon }_{{\mathrm{\Gamma }}_y}(-\delta )>0$. Thus [F1] gives $x\prec \prec x^{\mathrm{\prime }}$ and $y\prec \prec y^{\mathrm{\prime }}$. These pairs have the same summed energy and work coordinates and the same represented aggregate composition and mass.

If $(x^{\mathrm{\prime }},y^{\mathrm{\prime }})\prec (x,y)$, consistency with $y\prec y^{\mathrm{\prime }}$ gives $(x^{\mathrm{\prime }},y)\prec (x^{\mathrm{\prime }},y^{\mathrm{\prime }})\prec (x,y)$. Cancel the common represented spectator $y$ by [F2] to get $x^{\mathrm{\prime }}\prec x$, contrary to step 1.1. But T1 gives $(x^{\mathrm{\prime }},y^{\mathrm{\prime }})\prec Z$, and the chosen T2 split gives $Z\prec (x,y)$, so transitivity produces exactly this forbidden reverse comparison. Opposed signs are impossible. Every pair has an admitted join, hence one common sign $\epsilon$ works for the whole family. The affine change $u^{\mathrm{\prime }}=\epsilon u$ preserves openness and convexity and sends $u=u_x+u_y$ to $u^{\mathrm{\prime }}=\epsilon u_x+\epsilon u_y$; its positive orientation describes the same accessibility relation. No physical energy, unit or calibration premise is redefined.