PHY.4:5.3 - Exchange two participants

Two identical chambers at the same temperature exchange one kind of particle through a symmetric passage. Let a and b be their current concentrations and let J(a,b) be the instantaneous mean transfer rate from the first chamber to the second. Assume the proposed regime needs no additional passage state, and the surroundings introduce no directional bias.

Keep the chamber labels and the positive counting direction fixed, and exchange the concentrations in the two preparations. Because the chambers and passage are symmetric and the surroundings supply no bias, this physical exchange predicts a reversed measured transfer rate:

J(b,a)=-J(a,b).

At equal concentrations, J(a,a)=0. A proposed law J(a,b)=k*(a-b)^2 with k>0 fails the interchange condition: it gives a positive rate in the same counted direction after the preparations are exchanged.

Both J(a,b)=k1*(a-b) and J(a,b)=k3*(a-b)^3 satisfy the interchange condition when their constants have the corresponding units. Symmetry has not chosen between them or determined their coefficients. Directional thermodynamic claims would additionally use the physical driving potentials and the applicable dissipation law.

If the passage stores no particles, the chamber particle numbers satisfy dN1/dt=-J and dN2/dt=J, preserving their sum. If appreciable particles accumulate in the passage, include its particle number and separate inlet and outlet rates. The earlier two-chamber balance then omits a relevant participant.

This comparison uses exchange and balance rather than rotation. It opens the same kind of result: an admissible law family and the physical condition that would require revising it.