# A special case of the grand canonical ensemble

Homework Statement:
Can we consider the internal energy ##U## constant and only allow for the number of particles ##N## to vary, in the grand canonical ensemble?
Relevant Equations:
$$\mu = -T \left(\frac{\partial S}{\partial N}\right)_{U,V} \hspace{1cm} (1)$$
$$\frac{1}{T} = \left(\frac{\partial S}{\partial U} \right) \hspace{1cm} (2)$$
In addition to the homework statement and considering only the case where ##U= constant## and ##N = large## : Can we also consider the definition of chemical potential ##\mu## and temperature ##T## as in equations ##(1)## and ##(2)##, and use them in the grand partition function?

More specifically, we can take the case of an Einstein Solid and the Schroeder's definition of internal energy ##U = qhf##. Assuming that ##U## is depending only on a constant number of energy quanta ##q## and allowing only ##N## to vary. Can we use the grand partition function and ##(1), (2)##, for calculating the average value of ##N##? In this case of an Einstein Solid, we also assume that ##N \gg q##.

The grand partition function is given by: $$Q_{(\alpha, \beta)} = \sum_{N=0}^{\infty} e^{\alpha N} Z_{N}(\alpha, \beta) \hspace{1cm} (3)$$
And: $$\bar{N} =\left( \frac{\partial\ln{Q}}{\partial \alpha}\right)_\beta \hspace{1cm} (4)$$
Where: $$\beta = \frac{1}{kT} \hspace{1cm} \alpha = \frac{\mu}{kT} \hspace{1cm} (5)$$
$$Z_1 = 1/ (1 - e^{-\frac{hf}{kT}})\hspace{1cm}(6)$$
And: $$Z_N = (Z_1)^N \hspace{1cm}(7)$$
Under the above conditions and equations:
$$\mu = -kT\frac{q}{N} = \frac{kT}{1 - e^{hf/kT}}\hspace{1cm} (8)$$

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Temperature ##T## is given by equation ##(2)##, in the microcanonical ensemble and calculated in Wikipedia (https://en.wikipedia.org/wiki/Einstein_solid). The final result is: $$\frac{q}{N} = \frac{1}{e^{hf/kT} - 1}\hspace{1cm} (9)$$
The chemical potential ##\mu## is given by ##(1)##, when ##U, V## are held constant. The calculation of ##(1)## can be found in the solutions manual of D. Schroeder's book: "An Introduction to Thermal Physics". The final result is $$\mu = -kT\ln(1+ \frac{q}{N})\hspace{1cm} (10)$$ (exercise 3.36).
In the case we examine it is assumed that ##N \gg q## so ##(10)## becomes: $$\mu = -kT\frac{q}{N}\hspace{1cm} (11)$$