Physics Asked by Ted Ali on April 18, 2021
I am quoting from "Equilibrium and non-Equilibrium Statistical Thermodynamics", by M. Bellac.
$$Q_{(alpha, beta)} = sum_{N=0}^{infty} e^{alpha N} Z_{N}(alpha, beta) hspace{1cm} (3.127)$$
Where $Q$ is the grand partition function, $Z_N$ is the canonical partition function and:
$$beta = frac{1}{kT} hspace{1cm} alpha = frac{mu}{kT} hspace{1cm} (3.128)$$
$$bar{N} =left( frac{partialln{Q}}{partial alpha}right)_beta hspace{1cm} (3.129)$$
In the case of an Einstein solid, $N$ is the number of oscillators and $q$ the number of quanta of energy. We are interested in the case: $$q >> N rightarrow q approx 10^z N, z geq 2 hspace{1cm} (1)$$ In this case: $$mu = -kTln(frac{q}{N}) hspace{1cm} (2)$$ (Reference for $(2)$: 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})} = Ncdot C hspace{1cm} (3)$$
Where $$C = frac{1}{2sinh(frac{hf}{2kT})} hspace{1cm} (4)$$
$C$ depends only on $beta$. Substituting in $(3.127)$, we get:
$$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} (5)$$
Since $q >> N$, $(5)$ becomes:
$$Q cong left(Ccdot frac{N}{q}right) Rightarrow $$ $$ Rightarrow ln{Q} cong ln{C} + ln{frac{N}{q}} hspace{1cm} (6)$$
From $(3.128)$ and $(2)$: $$ alpha = ln{frac{N}{q}}hspace{1cm} (7)$$
Quoting $(3.129), (6), (7)$: $$bar{N} = left( frac{partial ln{Q}}{partial alpha}right)_{beta} = 1 hspace{1cm} (8)$$
I would be grateful if you could provide the correct solution and let me know of my errors in calculations, comments, etc. The main question is whether $q$ should be considered a constant, or else, if $q$ is dependent on each value $N$ takes in the sum. If I am correct in my calculations, is it equilibrium or non-equilibrium statistics?
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