The grand canonical ensemble is an ensemble that is not isolated from the environment, but is in thermal and chemical equilibrium with a much larger heat and particle reservoir. As such, both energy and number of particles can fluctuate. The conserved quantities are the temperature T T T and the chemical potential μ \mu μ . Its density function is
ρ ( z , V , T ) = z N Q N ( V , T ) \rho(z,V,T)=z^{N}Q_{N}(V,T) ρ ( z , V , T ) = z N Q N ( V , T )
where z = e β μ z=e^{\beta \mu} z = e β μ is the fugacity and Q N Q_{N} Q N is the canonical partition function . Its partition function is
Z ( z , V , T ) ≡ ∑ N = 0 ∞ z N Q N ( V , T ) \mathcal{Z}(z,V,T)\equiv \sum_{N=0}^{\infty} z^{N}Q_{N}(V,T) Z ( z , V , T ) ≡ N = 0 ∑ ∞ z N Q N ( V , T )
Derivation from the canonical ensemble#
Let's consider a system and a particle and heat reservoir of particle numbers N 1 N_{1} N 1 and N 2 N_{2} N 2 bound by N 2 = N − N 1 N_{2}=N-N_{1} N 2 = N − N 1 and volumes V 1 V_{1} V 1 and V 2 = V − V 1 V_{2}=V-V_{1} V 2 = V − V 1 . Let's assume V 2 ≫ V 1 V_{2}\gg V_{1} V 2 ≫ V 1 and N 2 ≫ N 1 N_{2}\gg N_{1} N 2 ≫ N 1 and their Hamiltonians are separable:
H ( q , p , N ) = H 1 ( q 1 , p 1 , N 1 ) + H 2 ( q 2 , p 2 , N 2 ) H(\mathbf{q},\mathbf{p},N)=H_{1}(\mathbf{q}_{1},\mathbf{p}_{1},N_{1})+H_{2}(\mathbf{q}_{2},\mathbf{p}_{2},N_{2}) H ( q , p , N ) = H 1 ( q 1 , p 1 , N 1 ) + H 2 ( q 2 , p 2 , N 2 )
This implicitly means that we are neglecting interactions between the systems beyond the bare minimum required to exchange particles and heat , which generally means that the interaction range is short.
Note how the grand canonical is just an extended version of the canonical ensemble , so let's start from the canonical partition function :
Q N = ∫ e β H ( q , p , N ) h 3 N N ! d q d p Q_{N}=\int \frac{e^{\beta H(\mathbf{q},\mathbf{p},N)}}{h^{3N}N!}d\mathbf{q}\,d\mathbf{p} Q N = ∫ h 3 N N ! e β H ( q , p , N ) d q d p
N N N is now a variable, so we also need to take every possible combination of N 1 N_{1} N 1 and N 2 N_{2} N 2 into account. At a fixed N 1 N_{1} N 1 , the number of states due to N N N is given by the binomial theorem as
( N N 1 ) = N ! N 1 ! ( N − N 1 ) ! = N ! N 1 ! N 2 ! \begin{pmatrix}N \\ N_{1}\end{pmatrix}=\frac{N!}{N_{1}!(N-N_{1})!}=\frac{N!}{N_{1}!N_{2}!} ( N N 1 ) = N 1 ! ( N − N 1 )! N ! = N 1 ! N 2 ! N !
but N 1 N_{1} N 1 can go anywhere between 0 0 0 and N N N , so the actual total is
∑ N 1 = 0 N ( N N 1 ) = ∑ N 1 = 0 N N ! N 1 ! N 2 ! \sum_{N_{1}=0}^{N}\begin{pmatrix}N \\ N_{1}\end{pmatrix} =\sum_{N_{1}=0}^{N} \frac{N!}{N_{1}!N_{2}!} N 1 = 0 ∑ N ( N N 1 ) = N 1 = 0 ∑ N N 1 ! N 2 ! N !
Since state combinations are found by multiplying, we can now state what our new ensemble average over the whole system looks like for a quantity regarding system 1:
⟨ O 1 ( H 1 , N 1 ) ⟩ = 1 Q N ∑ N 1 = 0 N N ! N 1 ! N 2 ! Q N O ( H 1 , N 1 ) ( expand Q N ) = 1 Q N ∑ N 1 = 0 N N ! N 1 ! N 2 ! 1 h 3 ( N 1 + N 2 ) N ! ∫ e − β [ H 1 ( q 1 , p 1 , N 1 ) + H 2 ( q 2 , p 2 , N 2 ) ] O ( H 1 , N 1 ) d q d p = ∑ N 1 = 0 N ∫ e − β H 1 ( q 1 , p 1 , N 1 ) O ( H 1 , N 1 ) h 3 N 1 N 1 ! d q 1 d p 1 ∫ e − β H 2 ( q 2 , p 2 , N 2 ) h 3 N 2 N 2 ! d q 2 d p 2 ⏟ Q N 2 = ∑ N 1 = 0 N Q N 2 Q N ∫ e − β H 1 ( q 1 , p 1 , N 1 ) O ( H 1 , N 1 ) h 3 N 1 N 1 ! d q 1 d p 1 = … \begin{align}
\langle O_{1}(H_{1},N_{1}) \rangle &=\frac{1}{Q_{N}} \sum_{N_{1}=0}^{N} \frac{N!}{N_{1}!N_{2}!}Q_{N}O(H_{1},N_{1}) \\
(\text{expand } Q_{N})&=\frac{1}{Q_{N}}\sum_{N_{1}=0}^{N} \frac{\cancel{ N! }}{N_{1}!N_{2}!}\frac{1}{h^{3(N_{1}+N_{2})}\cancel{ N! }}\int\ e^{-\beta[H_{1}(\mathbf{q}_{1},\mathbf{p}_{1},N_{1})+H_{2}(\mathbf{q}_{2},\mathbf{p}_{2},N_{2})]}O(H_{1},N_{1}) \ d\mathbf{q}d\mathbf{p} \\
&=\sum_{N_{1}=0}^{N}\int\frac{e^{-\beta H_{1}(\mathbf{q}_{1},\mathbf{p}_{1},N_{1})}O(H_{1},N_{1})}{h^{3N_{1}}N_{1}!} d\mathbf{q}_{1}d\mathbf{p}_{1}\underbrace{ \int \frac{e^{-\beta H_{2}(\mathbf{q}_{2},\mathbf{p}_{2},N_{2})}}{h^{3N_{2}}N_{2}!} d\mathbf{q}_{2}d\mathbf{p}_{2} }_{ Q_{N_{2}} }\\
&=\sum_{N_{1}=0}^{N} \frac{Q_{N_{2}}}{Q_{N}}\int\frac{e^{-\beta H_{1}(\mathbf{q}_{1},\mathbf{p}_{1},N_{1})}O(H_{1},N_{1})}{h^{3N_{1}}N_{1}!} d\mathbf{q}_{1}d\mathbf{p}_{1}=\ldots
\end{align} ⟨ O 1 ( H 1 , N 1 )⟩ ( expand Q N ) = Q N 1 N 1 = 0 ∑ N N 1 ! N 2 ! N ! Q N O ( H 1 , N 1 ) = Q N 1 N 1 = 0 ∑ N N 1 ! N 2 ! N ! h 3 ( N 1 + N 2 ) N ! 1 ∫ e − β [ H 1 ( q 1 , p 1 , N 1 ) + H 2 ( q 2 , p 2 , N 2 )] O ( H 1 , N 1 ) d q d p = N 1 = 0 ∑ N ∫ h 3 N 1 N 1 ! e − β H 1 ( q 1 , p 1 , N 1 ) O ( H 1 , N 1 ) d q 1 d p 1 Q N 2 ∫ h 3 N 2 N 2 ! e − β H 2 ( q 2 , p 2 , N 2 ) d q 2 d p 2 = N 1 = 0 ∑ N Q N Q N 2 ∫ h 3 N 1 N 1 ! e − β H 1 ( q 1 , p 1 , N 1 ) O ( H 1 , N 1 ) d q 1 d p 1 = …
The ratio of partition functions is
Q N 2 ( V 2 , T ) Q N ( V , T ) = e − β [ A ( N − N 1 , V − V 1 , T ) − A ( N , V , T ) ] \frac{Q_{N_{2}}(V_{2},T)}{Q_{N}(V,T)}=e^{-\beta[A(N-N_{1},V-V_{1},T)-A(N,V,T)]} Q N ( V , T ) Q N 2 ( V 2 , T ) = e − β [ A ( N − N 1 , V − V 1 , T ) − A ( N , V , T )]
where A A A is the Helmholtz free energy . Using the fact that N 1 ≪ N 2 N_{1}\ll N_{2} N 1 ≪ N 2 and V 1 ≪ V 2 V_{1}\ll V_{2} V 1 ≪ V 2 , which means N 2 ≃ N N_{2}\simeq N N 2 ≃ N and V 2 ≃ V V_{2}\simeq V V 2 ≃ V , we can use a Taylor series in two dimensions to approximate A A A around N N N and V V V , so that we get a first order approximation
A ( N − N 1 , V − V 1 , T ) ≃ A ( N , V , T ) + ∂ A ∂ N ∣ V , T ( N − N 1 − N ) + ∂ A ∂ V ∣ N , T ( V − V 1 − V ) = A ( N , V , T ) − μ N 1 + P V 1 \begin{align}
A(N-N_{1},V-V_{1},T)&\simeq A(N,V,T)+\left.\frac{ \partial A }{ \partial N }\right|_{V,T} (N-N_{1}-N)+ \left.{\frac{ \partial A }{ \partial V } }\right|_{N,T}(V-V_{1}-V) \\
&=A(N,V,T)-\mu N_{1}+PV_{1}
\end{align} A ( N − N 1 , V − V 1 , T ) ≃ A ( N , V , T ) + ∂ N ∂ A V , T ( N − N 1 − N ) + ∂ V ∂ A N , T ( V − V 1 − V ) = A ( N , V , T ) − μ N 1 + P V 1
where we got the chemical potential μ \mu μ from:
μ = ∂ A ( N − N 1 , V − V 1 , T ) ∂ N ∣ N 2 = N \mu=\left.{ \frac{ \partial A(N-N_{1},V-V_{1},T) }{ \partial N } }\right|_{N_{2}=N} μ = ∂ N ∂ A ( N − N 1 , V − V 1 , T ) N 2 = N
which is the rate of change of free energy with respect to number of particles, and the pressure P P P from the Maxwell relations :
P = − ∂ A ( N − N 2 , V − V 2 , T ) ∂ V ∣ V 2 = V P=- \left. \frac{ \partial A(N-N_{2},V-V_{2},T) }{ \partial V } \right|_{V_{2}=V} P = − ∂ V ∂ A ( N − N 2 , V − V 2 , T ) V 2 = V
Through this, our ratio looks like
Q N 2 ( V 2 , T ) Q N ( V , T ) = e β ( μ N 1 − P V 1 ) \frac{Q_{N_{2}}(V_{2},T)}{Q_{N}(V,T)}=e^{\beta(\mu N_{1}-PV_{1})} Q N ( V , T ) Q N 2 ( V 2 , T ) = e β ( μ N 1 − P V 1 )
Back to the ensemble average
… = ∑ N 1 = 0 N e β ( μ N 1 − P V 1 ) ∫ e − β H 1 ( q 1 , p 1 , N 1 ) O ( H 1 , N 1 ) h 3 N 1 N 1 ! d q 1 d p 1 = e − β P V 1 ∑ N 1 = 0 N e β μ N 1 Q N 1 1 Q N 1 ∫ … d q 1 d p 1 = … \begin{align}
\ldots&=\sum_{N_{1}=0}^{N} e^{\beta(\mu N_{1}-PV_{1})} \int\frac{e^{-\beta H_{1}(\mathbf{q}_{1},\mathbf{p}_{1},N_{1})}O(H_{1},N_{1})}{h^{3N_{1}}N_{1}!} d\mathbf{q}_{1}d\mathbf{p}_{1} \\
&=e^{-\beta PV_{1}}\sum_{N_{1}=0}^{N} e^{\beta\mu N_{1}} Q_{N_{1}} \frac{1}{Q_{N_{1}}} \int\ldots d\mathbf{q}_{1}d\mathbf{p}_{1} \\
&=\ldots
\end{align} … = N 1 = 0 ∑ N e β ( μ N 1 − P V 1 ) ∫ h 3 N 1 N 1 ! e − β H 1 ( q 1 , p 1 , N 1 ) O ( H 1 , N 1 ) d q 1 d p 1 = e − βP V 1 N 1 = 0 ∑ N e β μ N 1 Q N 1 Q N 1 1 ∫ … d q 1 d p 1 = …
The integral and the denominator Q N 1 Q_{N_{1}} Q N 1 now make a canonical ensemble average:
… = e − β P V 1 ∑ N 1 = 0 N e β μ N 1 Q N 1 ⟨ O 1 ⟩ canon \begin{align}
\ldots&=e^{-\beta PV_{1}}\sum_{N_{1}=0}^{N} e^{\beta \mu N_{1}}Q_{N_{1}}\langle O_{1} \rangle_\text{canon}
\end{align} … = e − βP V 1 N 1 = 0 ∑ N e β μ N 1 Q N 1 ⟨ O 1 ⟩ canon
Note how there no longer is any mention of the second system. It is either system 1 or the total N N N . This justifies a thermodynamic limit , for which N → ∞ N\to \infty N → ∞ :
⟨ O 1 ( H 1 , N 1 ) ⟩ = e − β P V 1 ∑ N 1 = 0 ∞ e β μ N 1 Q N 1 ⟨ O 1 ⟩ canon \langle O_{1}(H_{1},N_{1}) \rangle =e^{-\beta PV_{1}}\sum_{N_{1}=0}^{\infty} e^{\beta \mu N_{1}}Q_{N_{1}}\langle O_{1} \rangle_\text{canon} ⟨ O 1 ( H 1 , N 1 )⟩ = e − βP V 1 N 1 = 0 ∑ ∞ e β μ N 1 Q N 1 ⟨ O 1 ⟩ canon
Now that we only have mentions N 1 N_{1} N 1 and V 1 V_{1} V 1 , we can safely drop the index 1 1 1 , since it is longer relevant, and define the fugacity as z ≡ e β μ z\equiv e^{\beta \mu} z ≡ e β μ :
⟨ O ( H , N ) ⟩ grand = e − β P V ∑ N = 0 ∞ z N Q N ⟨ O ⟩ canon \boxed{\langle O(H,N) \rangle_\text{grand} =e^{-\beta PV}\sum_{N=0}^{\infty} z^{N}Q_{N}\langle O \rangle_\text{canon}} ⟨ O ( H , N ) ⟩ grand = e − βP V N = 0 ∑ ∞ z N Q N ⟨ O ⟩ canon
This is the ensemble average of the grand canonical ensemble. By comparing this to the discrete definition of ensemble average, we see
⟨ O ⟩ = 1 Z ∑ ρ O = 1 e β P V ∑ N = 0 ∞ z N Q N O \langle O \rangle =\frac{1}{Z}\sum \rho\ O =\frac{1}{e^{\beta PV}}\sum_{N=0}^{\infty} z^{N}Q_{N}\ O ⟨ O ⟩ = Z 1 ∑ ρ O = e βP V 1 N = 0 ∑ ∞ z N Q N O
We intuit that the grand canonical density function must be
ρ ( q , p , N ) = z N Q N \boxed{\rho(\mathbf{q},\mathbf{p},N)=z^{N}Q_{N}} ρ ( q , p , N ) = z N Q N
and that the grand canonical partition function must be
Z = e β P V \mathcal{Z}=e^{\beta PV} Z = e βP V
There is a better statement for this. It can be found by taking O O O to be the constant 1 1 1 , which reduces the ensemble average to
⟨ 1 ⟩ grand = 1 = e − β P V ∑ N = 0 ∞ z N Q N ⇒ ∑ N = 0 ∞ z N Q N = e β P V \langle 1 \rangle _\text{grand}=1=e^{-\beta PV}\sum_{N=0}^{\infty} z^{N}Q_{N}\qquad\Rightarrow \qquad \sum_{N=0}^{\infty} z^{N}Q_{N}=e^{\beta PV} ⟨ 1 ⟩ grand = 1 = e − βP V N = 0 ∑ ∞ z N Q N ⇒ N = 0 ∑ ∞ z N Q N = e βP V
So evidently, the partition function must be
Z ( z , V , T ) = ∑ N = 0 ∞ z N Q N \boxed{\mathcal{Z}(z,V,T)=\sum_{N=0}^{\infty} z^{N}Q_{N}} Z ( z , V , T ) = N = 0 ∑ ∞ z N Q N
Energy#
Just like in the canonical ensemble, the energy is
U = ⟨ H ⟩ = − ∂ ∂ β ln Z ( Z , V , T ) U=\langle H \rangle =-\frac{ \partial }{ \partial \beta } \ln \mathcal{Z}(Z,V,T) U = ⟨ H ⟩ = − ∂ β ∂ ln Z ( Z , V , T )
Particle number fluctuations#
Since N N N varies, we can express the mean number of particles as the ensemble average
⟨ N ⟩ = 1 Z ∑ N = 0 ∞ N z N Q N ( V , T ) \langle N \rangle =\frac{1}{\mathcal{Z}}\sum_{N=0}^{\infty}Nz^{N}Q_{N}(V,T) ⟨ N ⟩ = Z 1 N = 0 ∑ ∞ N z N Q N ( V , T )
We employ the common method using a derivative to make a quantity (here N N N ) disappear:
N z N = z ∂ ∂ z z N Nz^{N}=z\frac{ \partial }{ \partial z } z^{N} N z N = z ∂ z ∂ z N
so
⟨ N ⟩ = 1 Z ∑ N = 0 ∞ z ∂ ∂ z z N Q N ( V , T ) = z Z ∂ ∂ z ∑ N = 0 ∞ z N Q N ( V , T ) = z Z ∂ ∂ z Z = z ∂ ∂ z ln Z \langle N \rangle =\frac{1}{\mathcal{Z}}\sum_{N=0}^{\infty} z\frac{ \partial }{ \partial z } z^{N}Q_{N}(V,T)=\frac{z}{\mathcal{Z}}\frac{ \partial }{ \partial z } \sum_{N=0}^{\infty} z^{N}Q_{N}(V,T)=\frac{z}{\mathcal{Z}}\frac{ \partial }{ \partial z } \mathcal{Z}=z\frac{ \partial }{ \partial z } \ln \mathcal{Z} ⟨ N ⟩ = Z 1 N = 0 ∑ ∞ z ∂ z ∂ z N Q N ( V , T ) = Z z ∂ z ∂ N = 0 ∑ ∞ z N Q N ( V , T ) = Z z ∂ z ∂ Z = z ∂ z ∂ ln Z
And so
⟨ N ⟩ = z ∂ ∂ z ln Z ( z , V , T ) \boxed{\langle N \rangle =z\frac{ \partial }{ \partial z } \ln \mathcal{Z}(z,V,T)} ⟨ N ⟩ = z ∂ z ∂ ln Z ( z , V , T )
Now that we have the mean, we can also find the Variance . The trick is the same as before. If doing z ∂ ∂ z z\frac{ \partial }{ \partial z } z ∂ z ∂ gave us ⟨ N ⟩ \langle N \rangle ⟨ N ⟩ , doing it twice should give us ⟨ N 2 ⟩ \langle N^{2} \rangle ⟨ N 2 ⟩ :
z ∂ ∂ z z ∂ ∂ z ln Z = z ∂ ∂ z 1 Z ∑ N = 0 ∞ N z N Q N = z [ 1 Z ∑ N = 0 ∞ N ( ∂ ∂ z z N ) Q N + ( ∂ ∂ z 1 Z ) ∑ N = 0 ∞ N z N Q N ] = z [ 1 Z 1 z ∑ N = 0 ∞ N 2 z N Q N − 1 Z 2 1 z ( ∑ N = 0 ∞ N z N Q N ) ( ∑ N = 0 ∞ N z N Q N ) ] = = ∑ N = 0 ∞ N 2 z N Q N Z − ( ∑ N = 0 ∞ N z N Q N Z ) 2 = ⟨ N 2 ⟩ − ⟨ N ⟩ 2 \begin{align}
z\frac{ \partial }{ \partial z } z\frac{ \partial }{ \partial z } \ln \mathcal{Z}&=z\frac{ \partial }{ \partial z } \frac{1}{\mathcal{Z}} \sum_{N=0}^{\infty}Nz^{N}Q_{N} \\
&=z\left[ \frac{1}{\mathcal{Z}}\sum_{N=0}^{\infty} N\left( \frac{ \partial }{ \partial z } z^{N} \right)Q_{N}+ \left( \frac{ \partial }{ \partial z } \frac{1}{\mathcal{Z}} \right)\sum_{N=0}^{\infty} Nz^{N}Q_{N} \right] \\
&=z\left[ \frac{1}{\mathcal{Z}} \frac{1}{z}\sum_{N=0}^{\infty} N^{2}z^{N}Q_{N}- \frac{1}{\mathcal{Z}^{2}} \frac{1}{z}\left( \sum_{N=0}^{\infty} Nz^{N}Q_{N} \right)\left( \sum_{N=0}^{\infty} Nz^{N}Q_{N} \right) \right]= \\
&=\frac{\sum_{N=0}^{\infty}N^{2}z^{N}Q_{N}}{\mathcal{Z}}- \left( \frac{\sum_{N=0}^{\infty}Nz^{N}Q_{N}}{\mathcal{Z}} \right)^{2} \\
&=\langle N^{2} \rangle -\langle N \rangle ^{2}
\end{align} z ∂ z ∂ z ∂ z ∂ ln Z = z ∂ z ∂ Z 1 N = 0 ∑ ∞ N z N Q N = z [ Z 1 N = 0 ∑ ∞ N ( ∂ z ∂ z N ) Q N + ( ∂ z ∂ Z 1 ) N = 0 ∑ ∞ N z N Q N ] = z [ Z 1 z 1 N = 0 ∑ ∞ N 2 z N Q N − Z 2 1 z 1 ( N = 0 ∑ ∞ N z N Q N ) ( N = 0 ∑ ∞ N z N Q N ) ] = = Z ∑ N = 0 ∞ N 2 z N Q N − ( Z ∑ N = 0 ∞ N z N Q N ) 2 = ⟨ N 2 ⟩ − ⟨ N ⟩ 2
which is precisely the variance. However, though we understand what the variance is, the left hand side of the equation is still cryptic. To find the physics in it, we can express the Z Z Z derivative in terms of the chemical potential as
z ∂ ∂ z = z ∂ μ ∂ z ∂ ∂ μ = z ( ∂ ∂ z k B T ln z ) ∂ ∂ μ = k B T ∂ ∂ μ z\frac{ \partial }{ \partial z } =z\frac{ \partial \mu }{ \partial z } \frac{ \partial }{ \partial \mu } =z\left( \frac{ \partial }{ \partial z } k_{B}T\ln z \right)\frac{ \partial }{ \partial \mu } =k_{B}T\frac{ \partial }{ \partial \mu } z ∂ z ∂ = z ∂ z ∂ μ ∂ μ ∂ = z ( ∂ z ∂ k B T ln z ) ∂ μ ∂ = k B T ∂ μ ∂
Plugging this back into the previous equation gives us a more interesting look on particle number fluctuations:
var ( N ) = ⟨ N 2 ⟩ − ⟨ N ⟩ 2 = k B 2 T 2 ∂ 2 ∂ μ 2 ln Z \boxed{\text{var}(N)=\langle N^{2} \rangle -\langle N \rangle ^{2}=k_{B}^{2}T^{2}\frac{ \partial ^{2} }{ \partial \mu ^{2} } \ln \mathcal{Z}} var ( N ) = ⟨ N 2 ⟩ − ⟨ N ⟩ 2 = k B 2 T 2 ∂ μ 2 ∂ 2 ln Z
Unsurprisingly, higher temperatures mean higher variances.
Canonical ensemble equivalence#
We can use the previous result to our advantage to prove that the grand canonical and canonical ensembles are equivalent in the Thermodynamic limit . In fact, if we divide the particle variance by the square of the volume V V V occupied the ensemble and we define the particle density n = N / V n=N/V n = N / V , we get
⟨ n 2 ⟩ − ⟨ n ⟩ 2 = k B 2 T 2 V 2 ∂ 2 ∂ μ 2 ln Z \langle n^{2} \rangle -\langle n \rangle ^{2}=\frac{k_{B}^{2}T^{2}}{V^{2}}\frac{ \partial ^{2} }{ \partial \mu ^{2} } \ln \mathcal{Z} ⟨ n 2 ⟩ − ⟨ n ⟩ 2 = V 2 k B 2 T 2 ∂ μ 2 ∂ 2 ln Z
When V → ∞ V\to \infty V → ∞ in the limit, this equation goes to zero, which means that particle density fluctuations nullify, which in turn also means that particle number fluctuations are also zero.
Thermodynamics#
To derive thermodynamics relations in the grand canonical, we can do something similar to keeping only the highest entropy state in the microcanonical ensemble . The grand canonical partition function is a sum:
Z = ∑ N = 0 ∞ z N Q N \mathcal{Z}=\sum_{N=0}^{\infty} z^{N}Q_{N} Z = N = 0 ∑ ∞ z N Q N
If one term (call it ⟨ N ⟩ \langle N \rangle ⟨ N ⟩ ) is considerably larger than all the others, we can keep only that one and approximate the sum away. This happens if the number variance is vanishingly small, which is to say in the thermodynamic limit. In this case, we can state
Z = ∑ N = 0 ∞ z N Q N ≃ z ⟨ N ⟩ Q ⟨ N ⟩ \mathcal{Z}=\sum_{N=0}^{\infty} z^{N}Q_{N}\simeq z^{\langle N \rangle}Q_{\langle N \rangle} Z = N = 0 ∑ ∞ z N Q N ≃ z ⟨ N ⟩ Q ⟨ N ⟩
The logarithm simplifies considerably:
ln Z ≃ ln ( z ⟨ N ⟩ Q ⟨ N ⟩ ) = ⟨ N ⟩ ln z + ln Q ⟨ N ⟩ = ⟨ N ⟩ β μ − β A ⟨ N ⟩ \ln \mathcal{Z}\simeq \ln(z^{\langle N \rangle}Q_{\langle N \rangle})=\langle N \rangle\ln z+\ln Q_{\langle N \rangle}=\langle N \rangle\beta\mu-\beta A_{\langle N \rangle} ln Z ≃ ln ( z ⟨ N ⟩ Q ⟨ N ⟩ ) = ⟨ N ⟩ ln z + ln Q ⟨ N ⟩ = ⟨ N ⟩ β μ − β A ⟨ N ⟩
where A ⟨ N ⟩ A_{\langle N \rangle} A ⟨ N ⟩ is the Helmholtz free energy in state ⟨ N ⟩ \langle N \rangle ⟨ N ⟩ . We can state A N = N a ( v , T ) A_{N}=Na(v,T) A N = N a ( v , T ) by defining a a a as the free energy per particle, using v = V / N v=V/N v = V / N as the specific volume of the ensemble. Doing this, we get
ln Z ≃ ⟨ N ⟩ β μ − β ⟨ N ⟩ a ( ⟨ v ⟩ , T ) = ⟨ N ⟩ β ( μ − a ( ⟨ v ⟩ , T ) ) \ln \mathcal{Z}\simeq \langle N \rangle\beta\mu-\beta\langle N \rangle a(\langle v \rangle,T)=\langle N \rangle\beta(\mu-a(\langle v \rangle,T)) ln Z ≃ ⟨ N ⟩ β μ − β ⟨ N ⟩ a (⟨ v ⟩ , T ) = ⟨ N ⟩ β ( μ − a (⟨ v ⟩ , T ))
where ⟨ v ⟩ = V / ⟨ N ⟩ \langle v \rangle=V/\langle N \rangle ⟨ v ⟩ = V / ⟨ N ⟩ . Unfortunately, a ( ⟨ v ⟩ , T ) a(\langle v \rangle,T) a (⟨ v ⟩ , T ) is a somewhat cryptic quantity and it would be nice to express it in terms of better understood ones. Fortunately, this is possible by calculating a couple of thermodynamic quantities. Pressure is
P = − ( ∂ A ∂ V ) N , T = − N ∂ a ( V T , T ) ∂ V = − ∂ a ∂ v P=-\left( \frac{ \partial A }{ \partial V } \right)_{N,T}=-N\frac{ \partial a\left( \frac{V}{T},T \right) }{ \partial V } =-\frac{ \partial a }{ \partial v } P = − ( ∂ V ∂ A ) N , T = − N ∂ V ∂ a ( T V , T ) = − ∂ v ∂ a
and the chemical potential is
μ = ( ∂ A ∂ N ) V , T = ∂ N ∂ N a + N ∂ a ( V N , T ) ∂ N = a − N V N 2 ∂ a ∂ v = a − v ∂ a ∂ v = a + v P (1) \mu=\left( \frac{ \partial A }{ \partial N } \right)_{V,T}=\frac{ \partial N }{ \partial N } a+N\frac{ \partial a\left( \frac{V}{N},T \right) }{ \partial N } =a-N \frac{V}{N^{2}}\frac{ \partial a }{ \partial v }=a-v\frac{ \partial a }{ \partial v } =a+vP\tag{1} μ = ( ∂ N ∂ A ) V , T = ∂ N ∂ N a + N ∂ N ∂ a ( N V , T ) = a − N N 2 V ∂ v ∂ a = a − v ∂ v ∂ a = a + v P ( 1 )
Inverting this relation gives us a = μ − v P a=\mu-vP a = μ − v P . If we put this back in the partition function, we find
ln Z ≃ ⟨ N ⟩ β ( μ − μ + ⟨ v ⟩ P ) = ⟨ N ⟩ β ⟨ v ⟩ P = ⟨ N ⟩ β V ⟨ N ⟩ P = β P V = P V k B T \ln \mathcal{Z}\simeq \langle N \rangle\beta(\mu-\mu+\langle v \rangle P)=\langle N \rangle\beta \langle v \rangle P=\langle N \rangle\beta \frac{V}{\langle N \rangle}P=\beta PV=\frac{PV}{k_{B}T} ln Z ≃ ⟨ N ⟩ β ( μ − μ + ⟨ v ⟩ P ) = ⟨ N ⟩ β ⟨ v ⟩ P = ⟨ N ⟩ β ⟨ N ⟩ V P = βP V = k B T P V
Thus, in thermodynamic limit, we find an equation of state :
ln Z = P V k B T (2) \boxed{\ln \mathcal{Z}=\frac{PV}{k_{B}T}}\tag{2} ln Z = k B T P V ( 2 )
Note that it is independent from the number of particles ⟨ N ⟩ \langle N \rangle ⟨ N ⟩ . Since ⟨ N ⟩ \langle N \rangle ⟨ N ⟩ is also the only state that matters for the number of particles (as long as we are in the thermodynamic limit), we can write ⟨ N ⟩ = N \langle N \rangle=N ⟨ N ⟩ = N without much loss.
Number fluctuations, again#
Now that we have a sensible expression for ln Z \ln \mathcal{Z} ln Z , we can give a more satisfactory description of particle number fluctuations. In fact, we get
⟨ n 2 ⟩ − ⟨ n ⟩ 2 = ( k B T V ) 2 ∂ 2 ∂ μ 2 ( P V k B T ) = k B T V ∂ 2 P ∂ μ 2 \langle n^{2} \rangle -\langle n \rangle ^{2}=\left( \frac{k_{B}T}{V} \right)^{2}\frac{ \partial ^{2} }{ \partial \mu ^{2} } \left( \frac{PV}{k_{B}T} \right)=\frac{k_{B}T}{V}\frac{ \partial ^{2}P }{ \partial \mu ^{2} } ⟨ n 2 ⟩ − ⟨ n ⟩ 2 = ( V k B T ) 2 ∂ μ 2 ∂ 2 ( k B T P V ) = V k B T ∂ μ 2 ∂ 2 P
We can use the following relation
∂ μ ∂ v = − v ∂ 2 a ( v ) ∂ v 2 = v ∂ P ∂ v → ∂ P ∂ v ∂ v ∂ μ = 1 v \frac{ \partial \mu }{ \partial v } =-v\frac{ \partial ^{2}a(v) }{ \partial v^{2} } =v\frac{ \partial P }{ \partial v } \quad\to \quad \frac{ \partial P }{ \partial v } \frac{ \partial v }{ \partial \mu } =\frac{1}{v} ∂ v ∂ μ = − v ∂ v 2 ∂ 2 a ( v ) = v ∂ v ∂ P → ∂ v ∂ P ∂ μ ∂ v = v 1
and so
∂ P ∂ μ = ∂ P ∂ v ∂ v ∂ μ = 1 v \frac{ \partial P }{ \partial \mu }=\frac{ \partial P }{ \partial v } \frac{ \partial v }{ \partial \mu } =\frac{1}{v} ∂ μ ∂ P = ∂ v ∂ P ∂ μ ∂ v = v 1
However, v v v is itself dependent on μ \mu μ , which gives us
∂ 2 P ∂ μ 2 = − 1 v 2 ∂ v ∂ μ = κ T v 2 \frac{ \partial ^{2}P }{ \partial \mu ^{2} }= - \frac{1}{v^{2}}\frac{ \partial v }{ \partial \mu } =\frac{\kappa_{T}}{v^{2}} ∂ μ 2 ∂ 2 P = − v 2 1 ∂ μ ∂ v = v 2 κ T
where we defined the isothermal compressibility κ T \kappa_{T} κ T . As such, our fluctuations become
⟨ n 2 ⟩ − ⟨ n ⟩ 2 ⟨ n ⟩ 2 = k B T κ T V \frac{\langle n^{2} \rangle -\langle n \rangle ^{2}}{\langle n \rangle ^{2}}=\frac{k_{B}T\kappa_{T}}{V} ⟨ n ⟩ 2 ⟨ n 2 ⟩ − ⟨ n ⟩ 2 = V k B T κ T
In the thermodynamic limit, these again vanish.
Virial expansion#
The equation of state ( 2 ) (2) ( 2 ) can be rewritten using ( 1 ) (1) ( 1 ) as
n = 1 V z ∂ ∂ z ln Z ( z , V , T ) n=\frac{1}{V}z\frac{ \partial }{ \partial z } \ln \mathcal{Z}(z,V,T) n = V 1 z ∂ z ∂ ln Z ( z , V , T )
If the temperature is sufficiently high and the particle density is sufficiently low, the equation in both forms can be expanded as a power series in z z z :
P k B T = 1 λ 3 ∑ l = 1 ∞ b l z l , n = 1 λ 3 ∑ l = 1 ∞ l b l z l \frac{P}{k_{B}T}=\frac{1}{\lambda ^{3}}\sum_{l=1}^{\infty} b_{l}z^{l},\qquad n=\frac{1}{\lambda ^{3}}\sum_{l=1}^{\infty} lb_{l}z^{l} k B T P = λ 3 1 l = 1 ∑ ∞ b l z l , n = λ 3 1 l = 1 ∑ ∞ l b l z l
using the de Broglie thermal wavelength λ \lambda λ . The coefficients b l b_{l} b l are known as cluster integrals and represents the internal correlation of a group of l l l particles due to interactions. By definition, b 1 ≡ 1 b_{1}\equiv1 b 1 ≡ 1 .