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

## 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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