Limit of curvature near lightlike points

Question

Let $\alpha \colon I \to \Bbb R^2_1$ be a regular curve and $t_0 \in I$ be such that $\alpha$ is lightlike at $t_0$, and not lightlike at $]t_0-r,t_0[$ for some $r>0$. Then, in that interval the curvature is given by $$k(t) = \frac{\det(\alpha'(t),\alpha''(t))}{\|\alpha'(t)\|^3}.$$I think that $\lim_{t \to t_0^-}|k(t)| = +\infty$, but I do not know how to prove it, nor I'm 100% sure if this is true. For every concrete example I tried until now, it was true. If suffices to check that $\det(\alpha'(t),\alpha''(t)) \not\to 0$, but I don't know how to estimate that determinant.

Help?

Answer 1

I think your conjecture is correct (assuming that "regular" means "smooth"), with one possible caveat that I'll mention below.

Since curvature is independent of parametrization, without loss of generality you can assume that $$ \alpha(t) = (t, F(t)) $$ for some function $F$. Then $$ \alpha'(t) = (1, f(t)) $$ where $f(t) = F'(t)$, and your hypothesis says that $f(t_0) = 1$ and $f(t) \neq 1$ for $t \in ]t_0 - r, t_0[$.

Here's the caveat: Suppose that the function $g(t) = f(t) - 1$ vanishes to finite order at $t_0$. Then we can write $$ f(t) = 1 + (t-t_0)^n h(t) $$ for some integer $n \geq 1$ and some smooth function $h$ with $h(t_0) \neq 0$. Then $$ \frac{\det(\alpha'(t), \alpha''(t))}{\|\alpha'(t)||^3} = \frac{f'(t)}{(1 - f(t)^2)^{3/2}} = \frac{O((t-t_0)^{n-1})}{O((t-t_0)^{3n/2})} = O((t-t_0)^{-n/2-1}) $$ as $t \to t_0$, so the limit is indeed infinite.

I suspect that the result is still true even if $g(t)$ vanishes to infinite order at $t_0$; e.g., it works if $g(t) = e^{-1/(t-t_0)^2}$. But I'm not sure sure how to give a rigorous proof in that case.