I am a mathematician and I am trying to understand the intension behind the inverse temperature and the activity (or intensity).

Consider e.g. the Hamiltonian [with (12,6)-Lennard-Jones pair potential] with some intensity parameter $z > 0$ and interaction parameters $\sigma > 0,$ $\varepsilon > 0,$ that is,

$$ H_{\Lambda}(\omega) := \sum_{x\in\omega_{\Lambda}} z + 4 \sigma^{12} \varepsilon \sum_{\substack{\{x,y\} \subset \omega\\ \{x,y\} \cap \Lambda \neq \emptyset }} \frac{1}{\|x-y\|^{12}} - 4 \sigma^{6} \varepsilon \sum_{\substack{\{x,y\} \subset \omega\\ \{x,y\} \cap \Lambda \neq \emptyset }} \frac{1}{\|x-y\|^{6}}, $$

where $\Lambda \subset \mathbb{R}^k$ is some compact set. Then, to study uniqueness (or phase transitions) of a corresponding (grand canonical) Gibbs measure one consider additionally the inverse temperature $\beta$ and tries to understand the low temperature behaviour.

My question: Where exactly must the inverse temperature be placed and why?

Some authors study particle models with $\beta$ acting on the whole Hamiltonian, that is, $\beta H_{\Lambda}.$

Some authors consider the intensity part $z |\omega_{\Lambda}| = \sum_{x\in\omega_{\Lambda}} z$ separately. Then, the inverse temperature $\beta$ is considered to act on the terms of pair potentials only, that is,

$$ \beta \cdot \left(4 \sigma^{12} \varepsilon \sum_{\substack{\{x,y\} \subset \omega\\ \{x,y\} \cap \Lambda \neq \emptyset }} \frac{1}{\|x-y\|^{12}} - 4 \sigma^{6} \varepsilon \sum_{\substack{\{x,y\} \subset \omega\\ \{x,y\} \cap \Lambda \neq \emptyset }} \frac{1}{\|x-y\|^{6}}\right). $$

Is there a specific reason to avoid considering an inverse temperature before the whole Hamiltonian (for example in the Lennard-Jones model)?


1 Answer 1


I've never heard of "intensity" but it looks something like the chemical potential $\mu$. The grand canonical partition function is usually defined as $$ Z= \sum_{{\rm states},N}\exp\{\beta(\mu N -H)\}= \exp\{-\beta(E-TS-\mu N)\}. $$ For a gas the Grand Potential (or the Landau free energy) is $E-TS-\mu N=-PV$.


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