I Equilibrium equation if the barrier allows particle exchange

[Kashmir]

"... two physical systems [seperated by wall], A1 and A2. A1 has $\Omega_{1}(N1,V1,E1)$ possible microstates, and the macrostate of A2 is $\Omega_{2}(N2,V2,E2)$ "

"... at any time $t$, the subsystem $A_{1}$ is equally likely to be in anyone of the $\Omega_{1}\left(E_{1}\right)$ microstates while the subsystem $A_{2}$ is equally likely to be in anyone of the $\Omega_{2}\left(E_{2}\right)$ microstates; therefore, the composite system $A^{(0)}$ is equally likely to be in anyone of the
$\Omega_{1}\left(E_{1}\right) \Omega_{2}\left(E_{2}\right)=\Omega_{1}\left(E_{1}\right) \Omega_{2}\left(E^{(0)}-E_{1}\right)=\Omega^{(0)}\left(E^{(0)}, E_{1}\right)$"
"... if A1 and A2 came into contact through a wall that allowed an exchange of particles as well, the conditions for equilibrium would [include] the equality of the parameter $\zeta_{1}$ of subsystem $A_{1}$ and the parameter $\zeta_{2}$ of subsystem $A_{2}$ where, by definition,
$\zeta \equiv\left(\frac{\partial \ln \Omega(N, V, E)}{\partial N}\right)_{V, E, N=\bar{N}}$"

• So if we've a wall that allowed an exchange of particles we have from above equation:

$\left(\frac{\partial \ln \Omega_1(N_1, V_1, E_1)}{\partial N_1}\right)_{V_1, E_1, N=\bar{N}} =\left(\frac{\partial \ln \Omega_2(N_2, V_2, E_2)}{\partial N_2}\right)_{V_2, E_2, N=\bar{N}}$

However having a wall that allows particles to be exchanged means no wall at all, then $V_1,V_2$ are not well defined, but the above equation uses them?

 

[DrClaude]

• So if we've a wall that allowed an exchange of particles we have from above equation:

$\left(\frac{\partial \ln \Omega_1(N_1, V_1, E_1)}{\partial N_1}\right)_{V_1, E_1, N=\bar{N}} =\left(\frac{\partial \ln \Omega_2(N_2, V_2, E_2)}{\partial N_2}\right)_{V_2, E_2, N=\bar{N}}$

Does that equality come from the textbook?

However having a wall that allows particles to be exchanged means no wall at all, then $V_1,V_2$ are not well defined, but the above equation uses them?

You can still conceptually consider a fixed partitioning of the full volume, even if there is no physical partition.

 

[vanhees71]

In this case of a grand-canonical ensemble the equilibrium condition is that both $T$ and $\mu$ are equal, and that's expressed above using the microcanonical description.

 

[Kashmir]

Does that equality come from the textbook?

Yes.

You can still conceptually consider a fixed partitioning of the full volume, even if there is no physical partition.

You mean $V_1= V_0$ where $V_0$ is total volume ?

 

[Kashmir]

In this case of a grand-canonical ensemble the equilibrium condition is that both $T$ and $\mu$ are equal, and that's expressed above using the microcanonical description.

Yes this an condition of equilibrium but the way the author derives it is confusing.

 

[DrClaude]

You mean $V_1= V_0$ where $V_0$ is total volume ?

No, I mean that you can split $V_0$ into $V_1 + V_2$, even if there is no actual physical partition.

 

[Kashmir]

No, I mean that you can split $V_0$ into $V_1 + V_2$, even if there is no actual physical partition.

$V_1$ was defined as the volume separated by a wall in which $N_1$ particles were present.

Now if we lift the wall then how will
$V_1$ be defined?

 

[DrClaude]

$V_1$ was defined as the volume separated by a wall in which $N_1$ particles were present.

Now if we lift the wall then how will
$V_1$ be defined?

$V_1$ is still the same: it is the volume that used to be constrained by the wall. If you want, you can imagine that someone drew a line across the container, and call the space on one side of the line $V_1$, and the space on the other side $V_2$.