Temperature and Entropy

Gear300

The fundamental temperature is defined so that 1/τ = ∂σ/∂U. This relation occurs as an equilibrium state, so wouldn't that imply that ∂σ/∂U = 0, leaving the temperature undefined?

Mapes

Why?

Gear300

∂σ/∂U = 1/g(∂g/∂U), in which g is the multiplicity of states. Let us say that the system is uncoupled, then wouldn't the equilibrium condition imply ∂g/∂U = 0 (or is it that temperature is defined specifically for a coupling between systems)?

Mapes

I think I see what you mean. I suppose one would have a problem defining temperature for a system that one absolutely could not add energy too, even in theory. Then ∂σ/∂U would have no meaning, U being constant. But every system dealt with in practice can conceivably be heated and/or have work performed on it. If heated, the entropy of the system would increase; if reversible work were to be done, the entropy would not increase.

Gear300

I see...so does that make it unreasonable to speak of the equilibrium states of an isolated system?

Mapes

No, but it seems unreasonable to speak of the temperature of a system whose energy could not be altered, even in theory.

DrDu

Temperature is defined by the zeroth law of thermodynamics, i.e. transitivity of equilibrium: If two systems A and B are in equilibriumwhen brought into contact and B and C are also in equilibrium when brought into contact, then A and C will also be in equilibrium when brought into contact. This allows the introduction of temperature specifically for coupled systems.

Rap

Yes, temperature is defined in terms of systems being coupled. If two systems are thermally connected and allowed to equilibrate, their temperatures are, by the zeroth law of thermodynamics, the same. An isolated system that can never be connected to another system, can never be connected to a thermometer, and so its temperature will remain unknown, but thats a technological problem. It still has a temperature, it just cannot be practically measured.