By J. Yvon (translated from French by H.S.H. Massey)

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**Additional resources for Correlations and Entropy in Classical Statistical Mechanics (International series of monographs in natural philosophy 21)**

**Example text**

13) Lemma. Let U ∈ R3×3 be symmetric positive definite and let V = RURt = U with R = −I + 2e ⊗ e, |e| = 1. 27) admits two solutions given by U −1 e ) , m = e, |U −1 e|2 1 2ρ U e , m = (U 2 e − |U e|2 e), a= 2 |U e| ρ a = 2(U e − where ρ > 0 is a constant chosen as to have |m| = 1. 27). 13. 27) is equivalent to finding a and m, |m| = 1, such that U −1 a, m = 0 and U 2 − V 2 = (U a − |a|2 |a|2 m) ⊗ m + m ⊗ (U a − m) . 28) Moreover, a simple calculation gives U 2 − V 2 = 2[(U 2 e − |U e|2 e) ⊗ e + e ⊗ (U 2 e − |U e|2 e)].

We mention that Friesecke has a formula for the polyconvex hull of three non-trivially rank-one connected wells of equal determinant [77]. However, the quasiconvex hull remains unknown. We conclude this section by noting that in the examples above, all the wells were assumed to have positive determinant. This is of paramount importance: if the sign of the determinants is not constant, then even in two dimensions two incompatible wells can support a non-trivial gradient Young measure, see [47]. , metals, alloys, ceramics and even biological systems).

Without loss of generality we can assume that R1 = I. We therefore base our study on the following problem: find a rotation Q ∈ SO(n) and two vectors a, m ∈ Rn×n , |m| = 1, such that U − QV = a ⊗ m. 5) where U −t = (U −1 )t = (U t )−1 . Moreover, since U and V have positive determinant, 1 − a, U −t m = det (QV U −1 ) > 0 , where ·, · denotes the Euclidean scalar product in Rn . 6) where C = U −t V t V U −1 . 6). 5). The next result is due to Ball and James [27, 28] and Khachaturyan [92]. Throughout this chapter, λ1 (ξ) ≤ .