# Show that the total weight w of an assembly in thermal equilibrium is

1. Apr 8, 2014

### VHAHAHA

1. The problem statement, all variables and given/known data
Show that the total weight w of an assembly in thermal equilibrium is a maximum.

2. Relevant equations
S= k ln W?
dS = dQ/ T ?
S is maximize when thermal equilibrium is reached.

3. The attempt at a solution
First of all, i don't know what does "total weight w" means.
Dose it means the number of ways W that a total of N particles can be classified into energy levels?
I think that I have to establish a equation W = f(S) and use differentiation to prove that it is a maximum.
However, i have no clue to start my work.
Can someone give me some clues so that i can start my work? Thank you.

2. Apr 9, 2014

### tman12321

I would assume that the "weight" is a statistical weight, i.e. the relative probability. Here is a related example to get you started. Suppose you have two systems A and A' with energies E and E', respectively. These systems can exchange energy, and form a total system A0 with energy E0. The probability, say Pr(E0) that the total system A0 is in a particular state (configuration of A and A') depends on how the energy is distributed between A and A'. That is Pr(E0) = Pr(E,E'). The maximum probability occurs when the derivative vanishes, i.e. dPr(E0)/dE0 = 0, or equivalently dln(Pr(E0))/dE0 = 0 because the natural logarithm is a monotonically increasing function. In general, ln(Pr(E,E')) is proportional to ln(W(E)) + ln(W'(E')) + C, where W(E) and W'(E') are the number of microstates accessible to A and A' respectively, and C is a normalization constant. We recall that dln(W(E))/dE is proportional to 1/T, and likewise for the other system. This means that maximal probability implies equality of temperatures, which is the condition for thermal equilibrium. I have glossed over some details, leaving them to you. One last point is that you must be able to generalize this to any thermodynamic system. I hope this helps.

Last edited: Apr 9, 2014