Intuition about the lack of a quadratic term in geometric expansions

Question

Let $(\Sigma,g)$ be a Riemannian 2-manifold and let $p\in\Sigma$. It turns out that the circumference $C(r)$ of a geodesic circle $S_r(p)$ of radius $r$ around $p$ satisfies $$ C(r)=2\pi r-\frac{\pi}{3}Kr^3+O(r^4), $$ where $K$ denotes the Gaussian curvature of $\Sigma$ in $p$. On the other hand, if $\gamma$ is a smooth arc length parametrized curve in $\Sigma$ with $\gamma(0)=p$, then the geodesic distance from $p$ to $\gamma(t)$ satisfies $$ d(p,\gamma(t))=t-\frac{k_g^2}{24}t^3+O(t^4), $$ where $k_g$ denotes the geodesic curvature of $\gamma$ in $p$. Now here's the question. Can somebody give me an intuitive reason why there are no quadratic terms in these expansions? At least for the second expansion I have a vague idea why this is plausible but maybe there is a neat explanation for this. Of course, the proofs of the statements "explain" somehow, that/why there are no quadratic terms but I'm really looking for some heuristics/an intuitive explanation.

Answer 1

(1) Consider $$ f(0)=f'(0)=0,\ f(t)=f(-t),\ f'(t)>0\ (t>0)$$

With $t=0$, we rotate $f$. Then $$ r(s):={\rm length}\ \{ (t,f(t))| 0\leq t\leq s \} =\int_0^s \sqrt{ 1+(f')^2 },\ r'(0)=1 $$

If $f'(t)=c_1t+c_2t^2+\cdots$ (cf. Taylor series), $$r(s)=\int_0^s (1+c_3t^2 + c_4t^3 +\cdots )\ dt =s+c_5s^3 +\cdots $$

(2) That is in the revolution surface (i.e., fixed point is corresponded to $t=0$), length of geodesic sphere of radius $r$ is $$ l(r)/(2\pi )=s= r - c_6 r^3 + \cdots $$

(3) Consider two spheres $S^2(r),\ S^2(R),\ r< R $ If $M$ is a surface and if curvature $K$ at $p\in M$ has $K(p)>0$, then for given $\epsilon >0$, we have $\delta$ s.t. on $B_\delta (p)$, $$ |K-K(p) | <\epsilon $$

So we have $$ \frac{1}{R^2} < K < \frac{1}{r^2} $$

That is three surfaces are tangent at fixed point and on $\delta$-balls except fixed point do not intersect. So intuitively $$ l_{S^2(r)} (t) < l_M (t) < l_{S^2(R)} (t) $$

That is $M$ does not have quadratic term