First, the system is copied $\mathscr{N}$, and the number of systems on each microstate $j$ is $n_j$, we have:
$$
\begin{align}
\sum_{j}^{state} n_j =& \mathscr{N} \tag{1} \\
\sum_{j}^{state} n_j E_j = & \mathscr{E} \tag{2}
\end{align}
$$
And then get:
$$
n_j = e^{-\alpha - \beta E_j} \tag{3}
$$
where $\alpha$ and $\beta$ are constants from the Lagrange multiplier method.
For an arbitrary microstate $i$ with the same energy $E_j$, then it is obvious that $n_i=n_j$, that is to say, the probability of the two is equal.
So I have not used the equal probability hypothesis at the current position, but it has indeed been proved that the microstates with the same energy are equally probable, regardless of whether they belong to the same phase trajectory.
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Zhao Dazhuang
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