 Solution

ID: 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/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.

 New to topics? Read the docs here!