Equilibrium equation if the barrier allows particle exchange

  • #1
Kashmir
468
74
IMG_20220321_235505.JPG


"... 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?
 
Science news on Phys.org
  • #2
Kashmir said:
• 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?

Kashmir said:
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.
 
  • #3
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.
 
  • #4
DrClaude said:
Does that equality come from the textbook?
Yes.
DrClaude said:
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 ?
 
  • #5
vanhees71 said:
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.
 
  • #6
Kashmir said:
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.
 
  • #7
DrClaude said:
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?
 
  • #8
Kashmir said:
##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##.
 
Back
Top