6.2 Gibbs ensemble MC

Phase separations are best studies with this method. Two separate boxes are used, and in addition to particle moves and volume changes a transfer of particles between the boxes is permitted. To achieve constant and equal temperature, pressure, and chemical potential in both boxes the following procedure (due to Panagiotopoulos) is used:

Let $V_{1,2}$ be the box volumes, with a constant $V=V_{1}+V_{2}$. The particle numbers are $N_{1,2}$, again with a constant total of $N=N_{1}+N_{2}$.

• Choose a particle $i$ in one of the boxes and perform a trial move $\vec{r}_{i} \rightarrow \vec{r'}_{i}$ in the usual manner. Compute the energy difference between the trial configuration and the given one: $\Delta U= U(\vec{r'}_{i})+U(\vec{r}_{i})$, where $U(\vec{r}_{i}) = \sum_{j \neq i} U_{ij}$ etc.
  
  Accept or reject the trial configuration with probability
  

  $$P_{acc}^{move}=min\{1, \exp^{- \beta \Delta U }\}$$ (6.5)

  
  

• Repeat this basic MC step for a number of particles; usually all $N$ particles in both boxes are treated in sequence.
• Now perform a trial volume change $V_{1} \rightarrow V_{1}' = V_{1}+ \Delta V$; since the total volume is conserved the volume $V_{2}$ must change by $-\Delta V$. In each box $m=1,2$ all particle coordinates are (implicitely) scaled by $f = (V'_{m}/V_{m})^{1/3}$, which entails a change in energy of $\Delta U_{m}$.
  
  Compute the total enthalpy change
  

  $$\Delta H = \Delta U_{1}+\Delta U_{2} - kT N_{1} ln (V_{1}'/V_{1})- kT N_{2} ln (V_{2}'/V_{2})$$ (6.6)

  
  
  noting that the contributions to $P \Delta V$ from each box cancel each other (since $\Delta V_{1}=-\Delta V_{2}$).
  
  Accept/reject the volume exchange according to
  

  $$P_{acc}^{vol-exc}=min\{1, \exp^{-\beta \Delta H} \}$$ (6.7)

  
  

• Now follows the particle transfer step:
  
  Choose one of the boxes $m=1,2$ with equal probabilites. Choose any of the particles in box $m$, remove it, and place it at an arbitrary position in the other box, $m'$. The total Gibbs potential then changes according to
  

  $$\Delta G = \Delta U_{1}+\Delta U_{2} +kT \ln{\left( \frac{V_{m} (N_{m'}+1)}{N_{m}V_{m'}}\right) }$$ (6.8)

  
  
  Note that the contributions from the chemical potential cancel since $\Delta N_{1}=-\Delta N_{2}$.
  
  Accept/reject the particle transfer according to
  

  $$P_{acc}^{trans}=min\{1, \exp^{-\beta \Delta G} \}$$ (6.9)

  
  

[to be extended...]