Usage of partition function in derivation of Sackur-Tetrode

[georg gill]

Why do they introduce the partition function. I have seen it in the derivation of the Boltzmann distribution. But I don't know the physical significance of it here? And how do they get to (L.11) after that? I get everything until L.7. Including L.7.

The rest of the proof is here just in case you are interested:

 

[vanhees71]

The idea is to use the canonical ensemble. Then the classical partition sum is given by
$$Z_{\text{cl}}=\int \mathrm{d}^{3N} x \mathrm{d}^{3N} p \exp[-\beta H(x,p)]=q^N.$$
The phase-space probability distribution is the Maxwell-Boltzmann distribution
$$f(x,p)=\frac{1}{Z} \exp[-\beta H(x,p)],$$
and thus the entropy from its information-theoretical definition
$$S=-\langle \ln f(x,p) \rangle=\ln Z+\langle \beta H \rangle=\ln Z+\beta U=\ln Z-\beta \partial_{\beta} U.$$
Now $$\beta=1/T$$ and thus
$$-\beta \partial_{\beta}U = -\frac{1}{T} \frac{\partial T}{\partial \beta} \partial_T U =+T \partial_T U$$
and thus
$$S=\partial_T (T \ln Z),$$
which is (L.16).

BTW: Which book is this from? It looks like a good one, judging from the small excerpt :-)).

Now using
$$\ln Z=N \ln q$$
you get
$$S_{\text{cl}}=N \ln \left (\frac{V}{\Lambda^3} \right)+\frac{3N}{2},$$
which is not extensive.

To solve this "Gibbs paradox" one has to consider the indistinguishability of particles. Since on average for the applicability of the Boltzmann statistics you shouldn't have more than 1 particle in each one-particle state (otherwise you must take into account Bose-Einstein or Fermi-Dirac statistics, as is also argued in your text), the indistinguishability is simply accounted for by deviding the classical partition sum by ##N!## since then you count all states that just differ by an arbitrary permutation of particles as only one (as it must be due to the indistinguishability of particles), you get
$$S=S_{\text{cl}}-\ln (N!) \simeq S_{\text{cl}} - (N \ln N-N)=N \ln \left (\frac{V}{N \Lambda^3} \right) + \frac{5 N}{2},$$
which is the Sackur-Tetrode formula. Note that now ##S## is extensive since ##V/N=1/n## is intensive!

BTW: From which book is this. It looks like a good one :-).

 

[georg gill]

the book is:
A Farewell To Entropy
by Arieh Ben-Naim

https://www.amazon.com/dp/9812707077/?tag=pfamazon01-20

Thank you for the answer! I think I should look more at the theory for the canonical ensemble before doing anything else.

 

[georg gill]

Thank you for the answer. I have now read about the canonical ensemble in this link

http://faculty.uca.edu/saddison/ThermalPhysics/CanonicalEnsemble.pdf

I believe they arrive at the canonical ensemble just before they start the part that is called Partitions functions and the Boltzmann distribution.
It is as far as I can see the ratio of number of cases of one energy state divided by sum of all number of cases. In the theory that I wrote at the first post in this thread it seems that they only use the denumerator, total number of cases of energy states, and not the numerator of the canonical ensemble in (L.8). But they also multiply with ##\frac{dW}{dE}##. What physical quantity is it that they get in (L.8)?

 

[vanhees71]

I think what you want is a derivation of the canonical ensemble from the microcanonical.

The microcanonical ensemble refers to a closed system, where we know that the conserved total energy is between ##U## and ##U+\delta U##. Given all the constraints, e.g., that a gas is confined in a given container of volume ##V## and that ##U## is in the above mentioned interval, the fundamental assumption of thermodynamics is that in equilbrium all the available microstates are found with equal probability.

Now let ##W(U)## be the number of microstates with an energy ##H(x,p)<U##, which is given by
$$W(U)=\int_{\mathbb{R}^{6N}} \mathrm{d}^{3N} x \mathrm{d}^{3N} p \frac{1}{h^{3N}} \Theta[U-H(x,p)].$$
Then the number of microstates in the energy shell considered above is
$$\Omega(U)=\delta U \frac{\partial W}{\partial U} = \delta U \int_{\mathbb{R}^{6N}} \mathrm{d}^{3N} x \mathrm{d}^{3N} p \frac{1}{h^{3N}} \delta[U-H(x,p)].$$
Now consider a small partial volume separated from the gas by a heat-conducting wall but inpenetrable for the gas particles and consider that still everything is in thermal equilibrium. Let's denote this subsystem with the label 1 and the rest system ("the heat bath") with label 2.

According to the fundamental principle the probability that part 1 has precise energy ##E_1## is proportional to the number of available microstates the heat bath can be in, i.e.,
$$P(E_1)=\frac{\Omega_2(U-E_1)}{\Omega(U)}. \qquad (1)$$
##\Omega(U)## is the total number of available states of the total system in equilibrium. Now the number of states for the total system, given that system 1 has energy ##E_1## is
$$\Omega(U|H_1=E_1)=\Omega_1(E_1) \Omega_2(U-E_1),$$
and both ##\Omega_1## and ##\Omega_2## are steeply raising functions of their argument. Thus the function above has a very sharp maximum, and thus the equilibrium state is charactrized by ##E_1=U_1##, where ##U_1## is the value, where the above function takes a maximum. For simplicity we define the entropy as
$$S(U|H_1=E_1)=\ln \Omega(U|H_1=E_1)=S_1(E_1) + S_2(U-E_1).$$
The condition for a maximum at ##E_1=U_1## is
$$\partial_{E_1} S(U|H_1=E_1)=\partial_{U'} S_1(U')|_{U'=U_1}-\partial_{U'} S_1(U')|_{U'=U-U_1}=0.$$
From phenomenological thermodynamics the condition for equilibrium for two systems in heat contact is the equality of temperature, and thus one defines
$$\partial_{U'} S_1(U')|_{U'=U_1}=\beta_1=\frac{1}{T_1}$$
and anlogously for system 2 (the heatbath). The equilibrium condition then is indeed
$$T_1=T_2$$
So we have with very good accuracy
$$\Omega(U)=\Omega_1(U_1) \Omega_2(U-U_1).$$
Taking the logarithm of (1) then gives
$$\ln P(E_1)=S_2(U-E_1) -S_2(U-U_1) -S_1(U_1)=S_2(U-U_1+U_1-E_1)-S_2(U-U_1) -S_1(U_1) = S_2(U-U_1) + \beta_2 (U_1-E_1)- S_2(U-U_1) -S_1(U_1) =(\beta_1 U_1-S_1) - \beta_1 E_1.$$
Now we have
$$(\beta_1 U_1 - S_1)=\beta_1(U_1-T_1 S_1)=\beta_1 F_1,$$
where ##F_1## is the free energy of part 1 of the gas. So finally we have
$$P(E_1)=\exp(\beta_1 F_1) \exp(-\beta_1 E_1).$$
Since
$$\int \mathrm{d}^{3N_1} x \mathrm{d}^{3N_1} p \frac{1}{h^{3N}} P(E_1)=1$$
we finally get
$$Z_1=\int \mathrm{d}^{3N_1} x \mathrm{d}^{3N_1} p \frac{1}{h^{3N_1}} \exp(-\beta_1 E_1)=\exp(-\beta_1 F_1)$$
or
$$F_1=-T_1 \ln Z_1.$$
From
$$\mathrm{d} F_1=\mathrm{d} (U_1-T_1 S_1)=T_1 \mathrm{d} S_1 - P_1 \mathrm{d} V_1 -T_1 \mathrm{d} S_1-S_1 \mathrm{d} T_1=-S_1 \mathrm{d} T_1 -P_1 \mathrm{d} V_1$$
we get
$$S_1=-\left (\frac{\partial F_1}{\partial T_1} \right )_{V_1,N_1}, \quad P_1=-\left (\frac{\partial F_1}{\partial T_1} \right )_{T_1,N_1}.$$
This we have used above in the derivation of ##S_1##.

Again we note that the correction due to indistinguishability of the particles in the sense of QT, one should put the ##1/N_1!## correction in front of ##Z_1## to get rid of the Gibbs paradox for the entropy, as discussed in my previous posting.

 

[georg gill]

I have a problem with the last rewriting in the proof. Here is the problem:

I get: ##Q=\frac{V}{\Lambda ^3}## Here we have used the integral ##\int_\infty^0 \varpi e^{-\beta \varepsilon} d \varepsilon##

But then: From this page

http://faculty.uca.edu/saddison/ThermalPhysics/CanonicalEnsemble.pdf

it is obtained that

##\Omega_2(E_0-E_r)=e^{\frac{S_2}{k}-\beta E_r}=e^{\beta(TS_2-E_r)}=e^{\beta(-A)}##

hence ##lne^{\beta(-A)}=\beta(-A)##

##-\frac{1}{\beta}ln\Omega_2(E_0-E_r)=A##
Where A is Helmholtz's energy

They use later in (L.16) in the first post that

that ##-\frac{\partial}{\partial T}(\frac{1}{\beta}lnQ)=\frac{\partial A}{\partial T}##
But here they use Q and not ##\Omega_2(E_0-E_r)## that I have derived this relation for Helmholtz's energy. Can someone derive why ##-\frac{\partial}{\partial T}(\frac{1}{\beta}lnQ)=A## as Q is the integral over all the different ##\Omega_2(E_0-E_r)##.?

 

[vanhees71]

The only difference between what's written in my previous posting and this writeup is that they call the Helmholtz free energy ##A## instead of ##F##. I've also derived that ##A=-T \ln Z## (I call it ##F## instead of ##A##; sorry if that causes the confusion).