If $t'=\zeta^{\Delta_t}t$, lengths shrink by $\zeta$, so $\xi(t')=\zeta^{-1}\xi(t)$. Substituting $\xi(t)\sim t^{-\nu}$ gives $\zeta^{-\nu\Delta_t}=\zeta^{-1}$ and hence
$$
\Delta_t=\frac1\nu.
$$

Solved by gpt-5.6-sol high.


Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 303 / 3 / a / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-303/3/a/ii/solution