Ideal gas in the microcanonical ensemble: I'm puzzled

Scientist XIn summary, the conversation discussed problem 8.2 in Huang's Statistical Mechanics, which required the use of the microcanonical formalism to derive the equations of state for an ideal quantum gas. The solution to the problem was provided and a discussion was raised about the Maxwell-Boltzmann distribution. The conversation clarified the concept of internal energy and explained the discrepancy in calculations for the Maxwell-Boltzmann statistics. It was also noted that in the microcanonical ensemble, PV=0 for an ideal gas.
  • #1
FranzDiCoccio
342
41
Hi all,

this is about problem 8.2 in Huang's Statistical Mechanics.
I think I've been able to solve it, but the solution raised a question about
the Maxwell-Boltzmann distribution. So first I provide my solution to the
problem, then discuss the apparently weird point.

Homework Statement



The problem requires to use the microcanonical formalism to derive the
equations of state of the ideal quantum gas, i.e.

[tex]N = \sum_j \frac{1}{z^{-1} e^{\beta \epsilon_p}\mp 1},\qquad
\frac{PV}{kT} = \mp \sum_p \log(1\mp z e^{-\beta \epsilon_p})[/tex]

where the upper and lower signs refer to Bose-Einstein and Fermi-Dirac
statistics, respectively.

Homework Equations



I think one should use the constraints inherent in the microcanonical ensemble

[tex] N = \sum_p n_p, \qquad E = \sum_p \ve_p n_p [/tex]

along with the formula for the set of occupation numbers maximizing the entropy

[tex]n_p = \frac{1}{z^{-1} e^{\beta \epsilon_p}\mp 1} \qquad (**) [/tex]

and the the formula for the entropy thereby

[tex]S = k \sum_p \left[\frac{\beta \epsilon_p - \log z }{z^{-1} e^{\beta \epsilon_p}\mp 1}\mp \log(1\mp z e^{-\beta \epsilon_p}) \right] [/tex]

All these results are derived in section 8.5.

The Attempt at a Solution



The first equation of state is trivially obtained by plugging (**) in the constraint on [tex]N[/tex].
The second equation of state was a bit harder, but at some point I recalled that E should be identified with the total internal energy, and [tex]z = e^{\beta \mu}[/tex]. Using this information and the constraints in the formula for the entropy one gets

[tex] S = \frac{1}{T}(U-\mu N) \mp k \sum_p \log(1\mp z e^{-\beta \epsilon_p})[/tex]

The second equation of state is obtained after recalling that the general form of the
internal energy is [tex]U = TS - PV + \mu N[/tex].

4. The weird point <===================

So far, so good. However, it seems to me that adopting the same approach with the Maxwell-Boltzmann statistics produces a weird result. The MB entropy is

[tex]S = k \sum_p z e^{-\beta \epsilon_p} (\beta \epsilon_p - \log z) [/tex]

so that, if I identify the same quantities as above ( total energy and number) I get [tex]TS = E - \mu N[/tex]. But, assuming that [tex]U = TS - PV + \mu N[/tex] is true, wouldn't this mean [tex]PV=0[/tex] instead of the expected [tex]PV = NkT[/tex] ?

Is there something I'm overlooking?

Thanks a lot for any insight

F
 
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  • #2
ellow scientist,

Thank you for sharing your solution to the problem! It seems like you have a good understanding of the microcanonical formalism and have correctly derived the equations of state for the ideal quantum gas. However, I believe the issue you are encountering with the Maxwell-Boltzmann distribution is due to a misunderstanding of the concept of internal energy.

In the microcanonical ensemble, the total energy E is the sum of the kinetic and potential energies of all the particles in the system. This includes the energy associated with the motion of the particles (kinetic energy) as well as any interactions between the particles (potential energy). In the case of an ideal gas, there are no interactions between the particles, so the internal energy U is equal to the kinetic energy.

In the case of the Bose-Einstein and Fermi-Dirac statistics, the expression for the internal energy is derived from the constraints of the microcanonical ensemble, as you have correctly noted. However, for the Maxwell-Boltzmann statistics, the expression for the internal energy is simply the average kinetic energy of the particles, which is given by U = 3/2 NkT. This is where the discrepancy arises in your calculations.

To address your question about PV=0, this is indeed the case for an ideal gas in the microcanonical ensemble. In this ensemble, the volume V is constant and therefore does not have any effect on the internal energy or entropy. It is only when we consider other ensembles, such as the canonical or grand canonical ensembles, that the volume V becomes a variable and affects the equations of state.

I hope this helps clarify your understanding. Keep up the good work in your studies of statistical mechanics!


 

Related to Ideal gas in the microcanonical ensemble: I'm puzzled

1. What is the microcanonical ensemble?

The microcanonical ensemble is a statistical mechanics model used to study systems that are isolated and have a fixed number of particles, volume, and energy. It assumes that the system is in a state of thermal equilibrium, meaning that all microstates are equally likely.

2. What is an ideal gas in the microcanonical ensemble?

An ideal gas in the microcanonical ensemble refers to a system of non-interacting particles that follow the ideal gas law, PV = NkT, where P is pressure, V is volume, N is the number of particles, k is the Boltzmann constant, and T is the temperature. This model allows for the study of the statistical behavior of an ideal gas in a fixed volume and energy.

3. How is the microcanonical ensemble different from other ensembles?

The microcanonical ensemble differs from other ensembles, such as the canonical and grand canonical ensembles, in that it assumes a fixed energy instead of a fixed temperature or chemical potential. This allows for the study of systems that are not in contact with a heat bath or reservoir.

4. What is the significance of the microcanonical ensemble in statistical mechanics?

The microcanonical ensemble is significant in statistical mechanics because it allows for the calculation of thermodynamic properties, such as entropy and temperature, from the fundamental principles of statistical mechanics. It also provides a basis for understanding the behavior of systems that are isolated and have a fixed energy.

5. How is the microcanonical ensemble related to the second law of thermodynamics?

The microcanonical ensemble is related to the second law of thermodynamics in that it assumes that all microstates of a system are equally likely. This means that the system has a higher entropy in a macrostate with more microstates, and the second law states that the entropy of an isolated system will tend to increase over time.

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