Is the Entropy Correction Term Calculated Correctly?

[Gerenuk]

I assumed that entropy derives from combinatorics so I calculated
$$S=\ln \frac{\prod_{i=1}^S n_k!}{n!}$$
for n particles in S states. The result is
$$S=\ln \frac{\prod_{i=1}^S n_k!}{n!}=n\sum_{i=1}^S p_i\ln p_i+\frac12 s\ln n$$
if $p_i=n_i/n$ and $\ln n/n\ll S$
What about the second correction term? Can it not play a role?

For one particle in each state for example
$$S=-s\ln n+\frac12 s\ln n$$

(oh weird; my browser doesn't show the equations. I hope they are displayed correctly)

 

[Aaron Barnard]

No, the second correction term cannot play a role in this case. The reason is that the entropy of a system with one particle in each state will always be zero, regardless of the correction term involved. This is because the entropy of a system with one particle in each state is determined by the logarithm of the number of available states, which is always equal to 1 in this case. Thus, the entropy is always zero regardless of the correction term.

 

[astrokreng]

There is a correction term that is missing in this calculation, known as the "entropy correction term". This term takes into account the number of ways in which the particles can be distributed among the states, and is crucial in accurately calculating the entropy of a system. It is calculated by taking the logarithm of the number of possible microstates, which is given by the formula S = k ln W, where k is the Boltzmann constant and W is the number of microstates. In your calculation, it seems that this term has been omitted, which could lead to an inaccurate result. Therefore, it is important to include this correction term in your calculation to ensure that the entropy is calculated correctly.