Solution
= Solution
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
$$
\boxed{\Delta_t=\frac1\nu.}
$$
= Solution
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
$$
\boxed{\Delta_t=\frac1\nu.}
$$