Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It only takes a minute to sign up.
Sign up to join this community
In case of non-isolates system (NVT or NPT ensemble), I learned I can calculate the entropy,
$$S=-k_B\sum_jp_j\ln(p_j)$$ where $p_j$=probability at $j$ state.
but I saw that the entropy is also can be calculated by
$$S=-k_B\ln(Z).$$
I think this equation applicable for both of isolated system and non-isolated system.
These two equations are same?
I learned I can calculate the entropy $$S=-k_B\sum_jp_j\ln(p_j)$$ where $p_j$ is the probability at state $j$.
but I saw that the entropy is also can be calculated by $$S=-k_B\ln(Z)$$ I think this equation applicable for both of isolated system and non isolated system
these two equations are same ?
Given the number of states $\Omega(E,\Delta E)$ within $\Delta E$ of the thermodynamic energy $E$, the entropy is: $$ S=k\ln(\Omega)\;. $$
If the quantity $Z$ is defined such that $$ \Omega =\frac{1}{Z(E)}\;, $$ I.e., if $Z(E)$ is the statistical distribution function, i.e. $Z(E_n)$ is your probability $p_n$, then $$ S=-k\ln(Z) $$
But, we know that the log of the statistical distribution function is a linear function of the constants of motion (actually just $E$, ignoring total momentum and total angular momentum as usual) $$ \ln Z(E)=a + bE\;, $$ so that the averaging that was performed to obtain the average energy $E$ $$ E=\sum_n Z(E_n)E_n $$ can be "pulled outside" the log $$ \ln Z(E)=\sum_n p_n\ln(p_n)\;, $$ where we have written $p_n=Z(E_n)$.
Thus, so too, $$ S=-k\sum_n p_n\ln(p_n) $$
