I Reading course in statistical physics

[Philip Koeck]

From this
$$Y' \partial_x \sigma = X' \partial_y \sigma,$$
and this is a function of $(x,y)$, which I can write in the Form $X' Y'/\tau$. This shows that for functions with 2 independent variables for any inexact differential there's always an integrating factor, $\tau$, such that
$$\mathrm{d} G/\tau=\mathrm{d} \sigma$$
is an exact differential.

Got it. The two expressions are the same so I can replace both with a 3rd expression.
The integration factor τ, which depends on x and y in general, is used to make the replacement correct. Something like that.

It will be interesting to read in the next chapter how this is applied to dS = d^-Q / T.
(Couldn't find a crossed-out d.)
Somehow one also has to realize that this equation is only valid for a d^-Q in a reversible process.

 

[Philip Koeck]

I have a rather general question about the definition of entropy used in chapter 5:
S = k ln Ω, where Ω is the number of available microstates.
Boltzmann wrote W rather than Ω, and I believe this stood for probability (Wahrscheinlichkeit).
Obviously this is not a number between 0 and 1, so it's more like something proportional to probability.

Probability would be number of available microstates divided by total number of microstates (including those that are not available).
Now for distinguishable particles both these numbers are bigger than for indistinguishable particles, by a factor N!, where N is the number of particles, in the case of low occupancy.

Would it make sense therefore to use the following definition of entropy for distinguishable particles to make sure that this "probability" W is calculated correctly?
S = k ln (Ω / N!) for distinguishable particles at low occupancy.

 

[Philip Koeck]

I have a rather general question about the definition of entropy used in chapter 5:
S = k ln Ω, where Ω is the number of available microstates.
Boltzmann wrote W rather than Ω, and I believe this stood for probability (Wahrscheinlichkeit).
Obviously this is not a number between 0 and 1, so it's more like something proportional to probability.

Probability would be number of available microstates divided by total number of microstates (including those that are not available).
Now for distinguishable particles both these numbers are bigger than for indistinguishable particles, by a factor N!, where N is the number of particles, in the case of low occupancy.

Would it make sense therefore to use the following definition of entropy for distinguishable particles to make sure that this "probability" W is calculated correctly?
S = k ln (Ω / N!) for distinguishable particles at low occupancy.

Maybe I should move this to a new post.