Expand knowledge base

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 $N_A$.