# Temperature and Entropy

by Gear300
Tags: entropy, temperature
 P: 1,133 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?
HW Helper
PF Gold
P: 2,532
 Quote by Gear300 This relation occurs as an equilibrium state, so wouldn't that imply that ∂σ/∂U = 0
Why?
 P: 1,133 ∂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)?
 Sci Advisor HW Helper PF Gold P: 2,532 Temperature and Entropy 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.
 P: 1,133 I see...so does that make it unreasonable to speak of the equilibrium states of an isolated system?
HW Helper
PF Gold
P: 2,532
 Quote by Gear300 I see...so does that make it unreasonable to speak of the equilibrium states of an isolated system?
No, but it seems unreasonable to speak of the temperature of a system whose energy could not be altered, even in theory.