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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.

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  • $\begingroup$ I need a bit of help understanding the question. Suppose $\alpha(t) = (t, 0, 1)$ for example. Then $\alpha(t_0)$ is lightlike if $t_0 = 1$, but the curvature is zero everywhere. Is this case excluded somehow? $\endgroup$
    – apt1002
    Commented Mar 14, 2017 at 17:43
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    $\begingroup$ @apt1002 your $\alpha$ is a space curve. Also, the causal type of a curve is defined as the causal type of the velocity vector (you look at the pull-back of the metric via $\alpha$) $\endgroup$
    – Ivo Terek
    Commented Mar 15, 2017 at 0:06
  • $\begingroup$ Question on convention: do you mean $\mathbb{R}^2_1$ to be the 2 space dimension + 1 time dimension Minkowski space or the 1 space + 1 time dimension Minkowski space? In either case, what do you mean by $\det$ of two vectors? Their wedge product? $\endgroup$
    – Willie Wong
    Commented Mar 22, 2017 at 3:04
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    $\begingroup$ @Willie I mean $(\Bbb R^2, dx^2-dy^2)$, and the determinant is obtained putting the vectors in the columns of a $2\times 2$ matrix. $\endgroup$
    – Ivo Terek
    Commented Mar 22, 2017 at 3:45

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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.

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