small τ$_{0}$ (= large a/b ) we have [16] so that because α$_{m}$ = β$_{m}$ for large m . Here n is odd or even for the Ce or Se modes, respectively. We conclude therefore that has to be large and justify large q ∼ m $^{2}$. Each mode contributes Since a is large, the sum over m can be replaced by an integral over ω = m/a like in Ref. [18] and for large q 1 / 2 ∼ m ∼ a/b , we have where so that For large r the integral on the right-hand side is proportional to r and we can get the coefficient of proportionality by differentiating with respect to r . This gives 18 ∑ m ln ( λ − 1 / 2 n,m ) = − ar ∫ ∞ 0 d x [1 − xf ′ ( x ) /f ( x )] (C19) ∑ m ln ( λ − 1 / 2 n,m ) = − a ∫ ∞ 0 d ω ln [ rf ( ω r )] . (C18) r = πn 2 b , (C17) λ 1 / 2 n,m = rf ( ω r ) , (C16) det ( Δ − 1 / $^{2}$) = ∏ n,m λ − 1 / 2 n,m $_{= e}$∑ $_{n,m}$ln ( λ $_{−}$1 / 2 n,m ) . (C15) 2. Evaluating the determinant 2 q − α$_{m}$ = ( πna 2 b ) 2 (C14) √ 2 q − α$_{m}$τ$_{0}$ = πn 2 (C13) Ce$_{m}$ ( τ ) ≈ const . × cos ( √ 2 q − α$_{m}$τ ) (C12a) Se$_{m}$ ( τ ) ≈ const . × sin 2 q − β$_{m}$τ (C12b)