Past exam of the chemistry course of the year 4 of the natural sciences course of the University of Cambridge / 2022 / Paper 3 / Section D / 43 / b / ii

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- Solution ii

Solution

Expanding the scalar product gives

j_{α} = [n Q^{2} / (2 k_{B} T)] \sum_{j} f r_{ij, α} \sum_{β} r_{ij, β} E_{β}

. Comparing coefficients of the independently variable field components yields

σ_{α β} = \frac{n Q ^{2}}{2 k _{B} T} j = nn (i) \sum f (ω_{0}, Δ E_{ij}, T) r_{ij, α} r_{ij, β} .

The tensor is symmetric and positive semidefinite because its contraction with any vector twice is a nonnegative sum of squares.

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