
#1
Mar2911, 07:28 AM

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?




#3
Mar2911, 08:02 AM

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)?




#4
Mar2911, 08:12 AM

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.




#5
Mar2911, 08:17 AM

P: 1,133

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




#6
Mar2911, 08:55 AM

Sci Advisor
HW Helper
PF Gold
P: 2,532





#7
Mar2911, 09:45 AM

Sci Advisor
P: 3,361





#8
Mar2911, 09:46 AM

P: 789




Register to reply 
Related Discussions  
Temperature and Entropy of an ideal gas  Introductory Physics Homework  2  
change in entropy at constant temperature  Introductory Physics Homework  5  
finding temperature from entropy  Advanced Physics Homework  1  
Same temperature and entropy  Classical Physics  2  
Black Hole Temperature and Entropy  Astrophysics  5 