Calculate the average number of oscillators of an Einstein solid, in the grand canonical ensemble, when q>>N

  • #1
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Homework Statement:

Calculate the average number of oscillators of an Einstein solid, in the grand canonical ensemble, when q>>N. N is the number of oscillators and q the total number of energy quanta.

Relevant Equations:

I am quoting from "Equilibrium and non-Equilibrium Statistical Thermodynamics", by M. Bellac.
$$Q_{(\alpha, \beta)} = \sum_{N=0}^{\infty} e^{\alpha N} Z_{N}(\alpha, \beta) \hspace{1cm} (3.127)$$ $$\beta = \frac{1}{kT} \hspace{1cm} \alpha = \frac{\mu}{kT} \hspace{1cm} (3.128)$$ $$\bar{N} = \frac{\partial{\ln Q}}{\partial \alpha} \hspace{1cm} (3.129)$$
$$q >> N \rightarrow q \approx 10^z N, z \geq 2$$
In this case: $$\mu = -kT\ln(\frac{q}{N}) \hspace{1cm} (1)$$ Reference: Daniel V. Schroeder, An Introduction to Thermal Physics,
(Addison-Wesley, 2000) - Problems 3.35 - 3.36.
$$Z_{N} = \frac{N}{2 \sinh(\frac{hf}{2kT})} = N\cdot C \hspace{1cm} (2)$$
$$Q_{(\alpha, \beta)} = \sum_{N=0}^{\infty} e^{\alpha N} Z_{N}(\alpha, \beta) \hspace{1cm} (3.127)$$
Where ##Q## is the grand partition function, ##Z_N## is the canonical partition function and:
$$\beta = \frac{1}{kT} \hspace{1cm} \alpha = \frac{\mu}{kT} \hspace{1cm} (3.128)$$
In the case of an Einstein solid, ##N## is the number of oscillators and ##q## the total quanta of energy. We are interested in the case: $$q >> N \rightarrow q \approx 10^z N, z \geq 2$$ In this case: $$\mu = -kT\ln(\frac{q}{N}) \hspace{1cm} (1)$$ Reference: Daniel V. Schroeder, An Introduction to Thermal Physics,
(Addison-Wesley, 2000) - Problems 3.35 - 3.36.
From https://en.wikipedia.org/wiki/Einstein_solid, we get: $$Z_{N} = \frac{N}{2 \sinh(\frac{hf}{2kT})} = N\cdot C \hspace{1cm} (2)$$
Where $$C = \frac{1}{2\sinh(\frac{hf}{2kT})} \hspace{1cm} (3)$$
##C## depends only on ##\beta##. Substituting in ##(3.127)##, we get:
$$Q = C \cdot \sum_{N=0}^{\infty} e^{\alpha N} \cdot N = C \cdot \sum_{N=0}^{\infty} e^{\ln((\frac{N}{q})^N)} \cdot N = $$ $$ = C \cdot \sum_{N=0}^{\infty} N (\frac{N}{q})^N \hspace{1cm} (4)$$
Since ##q >> N##, (4) becomes:
$$Q \cong \left(C\cdot \frac{N}{q}\right) \Rightarrow$$ $$ \Rightarrow \ln{Q} \cong \ln{\frac{C}{ q}} + \ln{N} \hspace{1cm} (5)$$

Quoting ##(3.129)##: $$\bar{N} = \frac{\partial{\ln{Q}}}{\partial \alpha} = \frac{\partial{\ln{Q}}}{\partial N}\cdot \frac{1}{\frac{\partial \alpha}{\partial N}} = \frac{\frac{1}{N}}{\frac{1}{N}} \Rightarrow$$ $$\Rightarrow \bar{N} = 1 \hspace{1cm} (6)$$
I would like you to let me know of my errors in calculations, comments, etc. If I am correct in my calculations, is it equilibrium or non-equilibrium statistics? How about ##\Delta N##, when we start from ##N \approx 10^{10}##?
 
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Answers and Replies

  • #2
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The main question is whether ##q## should be considered a constant, or else, dependent on every value ##N## takes, in the sum.
 
  • #3
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Should ##N## have an upper bound other than infinity? Should we consider an interaction between systems (instead of the isolated Einstein solid), governed by the grand canonical ensemble? Is the derivation of the chemical potential ##(\mu)## in the microcanonical ensemble, not applicable in the case of the grand canonical ensemble?
 
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