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Ma443 Statistical Physics

Lecturer: Dr. Stefan Sint

Website: Link

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Derivation of canonical and grand canonical ensembles from maximising information entropy

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Canonical ensemble

This is the ensemble for a system in contact with a heat bath, the constraints are

$\int d^{3N}\!pd^{3N}\!q\, \rho = 1 \qquad \int d^{3N}\!pd^{3N}\!q\, \mathcal{H} \rho = E$

In what follows we will put $dpdq \equiv d^{3N}\!pd^{3N}\!q$ for brevity. The information entropy in this continuous case can be defined by

$S = - k \int dpdq\, \rho \ln \rho$

and we maximise it using Lagrange multipliers $\lambda$ and $\beta$ , that is we seek to maximise

$- k \int dpdq\, \left( \rho \ln \rho + \lambda \rho + \beta \mathcal{H} \rho \right)$

Differentiating with respect to $\rho$ and setting the result equal to zero we find

$1 + \ln \rho + \lambda + \beta \mathcal{H} = 0 \Rightarrow \rho = \frac{\exp(-\beta \mathcal{H} )}{z}$

where $z = \exp(1+\lambda)$ , or, using the first constraint, we see that

$z = \int dpdq\,\exp(-\beta \mathcal{H} )$

This is known as the partition function.

Substituting for our probability into $S$ we have

$S = -k \int dpdq\, \rho ( - \beta \mathcal{H} - \ln z) = \beta k \int dp dq \, \rho \mathcal{H} + k \ln z \int dp dq \rho$

giving

$S = k \beta E + k \ln z \Rightarrow E = \frac{S}{k \beta} - \frac{1}{\beta} \ln z$

Identifying the energy $E$ with the internal energy $U$ of thermodynamics, and comparing with the expression $U = TS + A$ , we see that we must have

$\beta = \frac{1}{kT} \qquad A = -kT \ln z$

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Grand canonical ensemble - not quite correct

This is the ensemble for a system in contact with a heat and particle bath, the constraints are

$\int dpdq\, \rho = 1 \qquad \int dpdq\, \mathcal{H} \rho = E \qquad \int dpdq\, n \rho = N$

Notice that this claims that particle number is continuous, which is wrong. I should fix this sometime.

We now wish to maximise

$- k \int dpdq\, ( \rho \ln \rho + \lambda \rho + \beta \mathcal{H} \rho + \gamma n \rho)$

Differentiating with respect to $\rho$ and setting the result equal to zero we find

$1 + \ln \rho + \lambda + \beta \mathcal{H} + \gamma n = 0 \Rightarrow \rho = \frac{\exp(-\beta \mathcal{H} - n \gamma)}{z}$

where $z = \exp(1+\lambda)$ , or, using the first constraint as before,

$z = \int dpdq\,\exp(-\beta \mathcal{H} - n \gamma )$

Subbing into the entropy for $\ln \rho$ gives

$S = k \beta \int dpdq \, \rho \mathcal{H} + k \int dpdq \, \rho n + k \ln z \int dpdq\, \rho$

$S = k \beta E + k \gamma N + k \ln z$

which can be rewritten, identifying $E \equiv U$ ,

$U - TS + k T \ln z = - kT \gamma N$

Recalling that $G = U - TS + PV = \mu N$ where $\mu$ is the chemical potential, we see that we must have

$PV = kT \ln z \qquad \mu = - \gamma k T$

and also

$\rho = \frac{\exp( -\beta [ \mathcal{H} - \mu n ])}{z}$

with

$z = \int dpdq\,\exp( -\beta [ \mathcal{H} - \mu n ])$

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Possibly useful references

Thermodynamics and Statistical Mechanics, Greiner (very good, clear book)
James Binney's lecture notes
Statistical Mechanics, Huang
Elements of Statistical Mechanics, Sachs, Sexton, Sen
SklogWiki

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443

From Mathsoc wiki

Derivation of canonical and grand canonical ensembles from maximising information entropy

Canonical ensemble

Grand canonical ensemble - not quite correct

Possibly useful references

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