# General Geodesic Curvature Formula

Suppose $$(M, g)$$ is a Riemannian manifold, and $$\gamma: I \rightarrow M$$ is a regular (but not necessarily unit-speed) curve in $$M$$. Show that the geodesic curvature of $$\gamma$$ at $$t\in I$$ is $$\kappa(t)=\frac{\sqrt{|\gamma'(t)|^2|D_t\gamma'(t)|^2 - \langle \gamma'(t), D_t \gamma'(t) \rangle^2}}{|\gamma'(t)|^3}.$$

I've been searching for an error in my solution for hours but I can't find it. Here's my solution:

Let $$\tilde\gamma = \gamma \circ\varphi$$ be a unit-speed parametrization of $$\gamma$$. We start by listing identities.

By the product rule, $$D_t \tilde\gamma'(\varphi^{-1}(t))= D_t \frac{\gamma'(t)}{|\gamma'(t)|}\implies\\ D_t \tilde\gamma'(\varphi^{-1}(t))= -\frac{|\gamma'(t)|'}{|\gamma'(t)|^2}\gamma'(t) + \frac{1}{|\gamma'(t)|}D_t \gamma'(t)\tag{i}$$

Since $$|\tilde\gamma'|$$ is constant, $$\langle \tilde\gamma'(\varphi^{-1}(t))\, , \, D_t \tilde\gamma'(\varphi^{-1}(t))\rangle = 0 \implies \\ \langle \frac{1}{|\gamma'(t)|}\gamma'(t)\, , \, D_t \tilde\gamma'(\varphi^{-1}(t))\rangle =0\tag{ii}$$

By combining (i) and (ii), we have $$\langle \frac{1}{|\gamma'(t)|}\gamma'(t)\, , \, -\frac{|\gamma'(t)|'}{|\gamma'(t)|^2}\gamma'(t) + \frac{1}{|\gamma'(t)|}D_t \gamma'(t)\rangle =0 \implies \\ -\frac{|\gamma'(t)|'}{|\gamma'(t)|^3}\langle\gamma'(t), \gamma'(t) \rangle + \frac{1}{|\gamma'(t)|^2} \langle\gamma'(t), D_t\gamma'(t) \rangle=0 \implies\\ \frac{\langle\gamma'(t), D_t\gamma'(t) \rangle}{|\gamma'(t)|}=|\gamma'(t)|' \tag{iii}$$

Finally we compute: $$(\kappa(t))^2=\langle D_t \tilde\gamma'(\varphi^{-1}(t))\, , \, D_t \tilde\gamma'(\varphi^{-1}(t)) \rangle \\ =\langle -\frac{|\gamma'(t)|'}{|\gamma'(t)|^2}\gamma'(t) + \frac{1}{|\gamma'(t)|}D_t \gamma'(t)\, , \, D_t \tilde\gamma'(\varphi^{-1}(t))\rangle \quad\quad \text{by (i)}\\ =\langle \frac{1}{|\gamma'(t)|}D_t \gamma'(t), \, -\frac{|\gamma'(t)|'}{|\gamma'(t)|^2}\gamma'(t) + \frac{1}{|\gamma'(t)|}D_t \gamma'(t)\rangle \quad\quad \text{by (i) and (ii)}\\ = -\frac{|\gamma'(t)|'}{|\gamma'(t)|^3}\langle \gamma'(t), D_t\gamma'(t)\rangle + \frac{|D_t\gamma'(t)|^2}{|\gamma'(t)|^2}\\ = -\frac{\langle \gamma'(t), D_t\gamma'(t)\rangle^2}{|\gamma'(t)|^4}+\frac{|D_t\gamma'(t)|^2}{|\gamma'(t)|^2} \quad\quad \text{by (iii)}\\ =\frac{|\gamma'(t)|^2|D_t\gamma'(t)|^2 - \langle \gamma'(t), D_t \gamma'(t) \rangle^2}{|\gamma'(t)|^4}$$

But that final denominator is clearly supposed to be $$|\gamma'(t)|^6$$. Any help would be greatly appreciated.

• The error is at the beginning. Write $\tilde\gamma(s)$, and differentiate with respect to $s$. You’re missing a crucial chain rule step. Commented Sep 4, 2023 at 22:26

You probably did the same mistake as me, since we both arrived at the same conclusion: $$D_t$$ does not only depend on a given curve, it also depends on its parametrization. So to get the right formula you need to take into account that there is a different operator $${D_t}'$$ for the reparametrized curve $$\tilde{\gamma}$$.