Phase coexistence implies nondifferentiability of the pressure
Consider a lattice system in finite volume with an external field coupled linearly to an extensive order parameter : . Let be the partition function and the finite-volume pressure .
Assume the thermodynamic limit pressure exists:
with the limit taken along a standard van Hove sequence.
Statement
The function is convex. Moreover:
If is differentiable at a given , then every infinite-volume Gibbs measure at parameters has the same order-parameter density
and it is fixed by the derivative:
Conversely, if there exist two Gibbs measures at the same with
then is not differentiable at ; equivalently, the left and right derivatives satisfy
This is a first-order transition / phase coexistence signature in the sense of Gibbs-phase transitions .
In the important special case (magnetization), this connects coexistence of distinct magnetized phases to a cusp in .
Key hypotheses
- Linear field coupling .
- Existence of the thermodynamic limit pressure .
- Existence of at least two infinite-volume Gibbs measures at the same with different (for the “coexistence nondifferentiability” direction).
Conclusions
- Differentiability of at rules out coexistence of Gibbs states with different densities of the conjugate order parameter.
- Coexistence forces a non-single-valued “slope” at (a nontrivial subgradient), hence nondifferentiability.
- In convex-analysis language (see convex functions ), multiple coexisting values of correspond to a nontrivial subdifferential of at .
Proof idea / significance
In finite volume, differentiate with respect to to obtain identities of the form ; this is a special case of derivatives as connected correlations (and yields fluctuation formulas such as susceptibility = variance for magnetization when differentiating again). Passing to the thermodynamic limit, convexity guarantees existence of one-sided derivatives. If distinct Gibbs measures at the same realize distinct limiting densities , then must have more than one supporting slope at , i.e. it cannot be differentiable there.