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The relation G Pn =const leads to Gn + kB T ln Pn eq + kB T=const. Insert this into Eq. 23 to find Gn + kB T ln Pn eq =G, then isolate Pn eq to get the desired result.
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We have ln Pm eq T= [-1 kB T] ( (Gm -G) T- (Gm -G)T) = (1 kB T) (S Pm -S+ kB) and ln Pm eq (-p) = [-1 kB T0] [ (Gm -G) (-p)] = (1 kB T0) (V Pm -V).
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If the eigenvalues of Y are denoted by λk with corresponding eigenvectors ek, there are N-1 negative eigenvalues and one zero eigenvalue.43 Equation 23 implies that for any vector x we have 〈x A (ω) x〉 = (1 kB T0) λk <0 〈 Rx ek 〉 〈 ek Rx 〉 λk (λk -iω) (where the sum is restricted to the N-1 negative eigenstates). Letting ω→0 we find that 〈x A (0) x〉 0, showing that A (0) is positive semidefinite. For the imaginary part, we find that 〈x A″ (ω) x〉 = (1 kB T0) λk <0 〈 Rx ek 〉 〈 ek Rx 〉 λk ω (λk2 + ω2) 0 whenever ω>0; thus A″ (ω) is negative semidefinite. In both cases, equality applies if and only if the vector Rx lies in the one-dimensional eigenspace of Y corresponding to the zero eigenvalue. Thus equality applies if and only if (Rx) n (Pn0) 12, or x (1,1).
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If the eigenvalues of Y are denoted by λk with corresponding eigenvectors ek, there are N-1 negative eigenvalues and one zero eigenvalue. Equation implies that for any vector x we have 〈x A (ω) x〉 = (1 kB T0) λk <0 〈 Rx ek 〉 〈 ek Rx 〉 λk (λk -iω) (where the sum is restricted to the N-1 negative eigenstates). Letting ω→0 we find that 〈x A (0) x〉 0, showing that A (0) is positive semidefinite. For the imaginary part, we find that 〈x A″ (ω) x〉 = (1 kB T0) λk <0 〈 Rx ek 〉 〈 ek Rx 〉 λk ω (λk2 + ω2) 0 whenever ω>0; thus A″ (ω) is negative semidefinite. In both cases, equality applies if and only if the vector Rx lies in the one-dimensional eigenspace of Y corresponding to the zero eigenvalue. Thus equality applies if and only if (Rx) n (Pn0) 12, or x (1,1).
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