In statistical mechanics, entropy is defined in terms of the probability distribution of the microstates of the system, by the Gibbs formula, $S = -k_B\sum_i P_i \ln P_i$. How does that reduce to $dS = \frac{\delta Q}{T}$ in case of heat engines or other classical thermodynamical systems? Also, is $dE = TdS$? If yes, how?


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Let's assume we have a Boltzmann distribution,

$$ p_i = \frac{1}{Z}e^{-\beta E_i} $$

with $\beta = (k_B T)^{-1}$ With this, it follows from the Gibbs formula that

$$ dS = -k_B \sum_i dp_i \ln p_i = -k_B \sum_i dp_i (-\beta E_i -\ln Z) $$

Using $\sum_i dp_i = 0$ we find

$$ dS = \frac{1}{T}\sum_i E_i dp_i = \frac{1}{T}\sum_i d(E_ip_i) - d(E_i)p_i $$

The first term in the Equation is the change in total Energy $dE$, and the second term is work done on the system for small changes $\delta w$. Using the first law of thermodynamics $dE = \delta w + \delta Q$ you find

$$ dS = \frac{\delta Q}{T} $$

If you assume no work is done on or by the system, the third equation will give you

$$ dS = \frac{dE}{T} $$

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