diff options
Diffstat (limited to 'content/know/concept/microcanonical-ensemble')
| -rw-r--r-- | content/know/concept/microcanonical-ensemble/index.pdc | 41 |
1 files changed, 22 insertions, 19 deletions
diff --git a/content/know/concept/microcanonical-ensemble/index.pdc b/content/know/concept/microcanonical-ensemble/index.pdc index 89d114b..16628b8 100644 --- a/content/know/concept/microcanonical-ensemble/index.pdc +++ b/content/know/concept/microcanonical-ensemble/index.pdc @@ -34,8 +34,8 @@ $$\begin{aligned} = \sum_{U_A \le U} c_A(U_A) \: c_B(U - U_A) \end{aligned}$$ -Where $c_A$ and $c_B$ are the number of microstates of -the subsystems at the given energy levels. +Where $c_A$ and $c_B$ are the numbers of subsystem microstates +at the given energy levels. The core assumption of the microcanonical ensemble is that each of these microstates has the same probability $1 / c$. @@ -43,18 +43,20 @@ Consequently, the probability of finding an energy $U_A$ in $A$ is: $$\begin{aligned} p_A(U_A) - = \frac{c_A(U_A) \:c_B(U - U_A)}{c(U)} + = \frac{c_A(U_A) \: c_B(U - U_A)}{c(U)} \end{aligned}$$ If a certain $U_A$ has a higher probability, then there are more $A$-microstates with that energy, -meaning that $U_A$ is "easier to reach" or "more comfortable" for the system. -Note that $c(U)$ is a constant, because $U$ is given. +so, statistically, for an *ensemble* of many boxes, +we expect that $U_A$ is more common. -After some time, the system will reach equilibrium, -where both $A$ and $B$ have settled into a "comfortable" position. +The maximum of $p_A$ will be the most common in the ensemble. +Assuming that we have given the boxes enough time to settle, +we go one step further, +and refer to this maximum as "equilibrium". In other words, the subsystem microstates at equilibrium -must be maxima of their probability distributions $p_A$ and $p_B$. +are maxima of $p_A$ and $p_B$. We only need to look at $p_A$. Clearly, a maximum of $p_A$ is also a maximum of $\ln p_A$: @@ -66,13 +68,13 @@ $$\begin{aligned} Here, in the quantity $\ln{c_A}$, we recognize the definition of -the entropy $S_A \equiv k_B \ln{c_A}$, -where $k_B$ is Boltzmann's constant. -We thus multiply by $k_B$: +the entropy $S_A \equiv k \ln{c_A}$, +where $k$ is Boltzmann's constant. +We thus multiply by $k$: $$\begin{aligned} - k_B \ln p_A(U_A) - = S_A(U_A) + S_B(U - U_A) - k_B \ln{c(U)} + k \ln p_A(U_A) + = S_A(U_A) + S_B(U - U_A) - k \ln{c(U)} \end{aligned}$$ Since entropy is additive over subsystems, @@ -87,7 +89,7 @@ more concrete, equilibrium condition: $$\begin{aligned} 0 - = k_B \dv{(\ln{p_A})}{U_A} + = k \dv{(\ln{p_A})}{U_A} = \pdv{S_A}{U_A} + \pdv{S_B}{U_A} = \pdv{S_A}{U_A} - \pdv{S_B}{U_B} \end{aligned}$$ @@ -107,13 +109,14 @@ $$\begin{aligned} Recall that our partitioning into $A$ and $B$ was arbitrary, meaning that, in fact, the temperature $T$ must be uniform in the whole box. - We get this specific result because heat was the only thing that $A$ and $B$ could exchange. -The key point, however, -is that the total entropy $S$ must be maximized. -We also would have reached that conclusion if our imaginary wall -allowed changes in volume $V_A$ and particle count $N_A$. + +The point is that the most likely state of the box +maximizes the total entropy $S$. +We also would have reached that conclusion +if our imaginary wall was permeable and flexible, +i.e if it allowed changes in volume $V_A$ and particle count $N_A$. |