Finding the instantaneous curvature of a trajectory on a 2D surface embedded in ordinary Euclidean 3D space

Question

I have an unknown trajectory of a particle, ${\bf r}(t)$ constrained to follow a 2D surface parameterised in $\theta$ and $\varphi$: ${\bf r}(\theta(t),\varphi(t))$.

I do know the surface completely (i.e. I have explicit expressions for $r_x(\theta,\phi)$, $r_y(\theta,\phi)$, and $r_z(\theta,\phi)$), but I don't know the trajectory $\theta(t)$, $\varphi(t)$. However, at a particular point $P$ in this trajectory, say, at $t_1$, I know both (A) its instantaneous velocity $\dot{\bf r}(\theta(t_1),\varphi(t_1))$, and (B) its "parametric" accelerations $\ddot{\theta}(t_1)$ and $\ddot{\varphi}(t_1)$.

What I ultimately want to find is the instantaneous acceleration, $\ddot{\bf r}(\theta(t_1),\varphi(t_1))$, and in particular, the instantaneous curvature of the trajectory, $\kappa(t_1)$. I intuitively feel that I have enough information to compute these, but I'm new enough to differential geometry that I'm struggling to find/derive the right formulae.

For instance, for the curvature, I have found formulae for both the normal curvature and the geodesic curvature, but the formulae I found are given in terms of the arc length parameterised curve. For example, in the formula for the geodesic curvature:

$\kappa_g = \left[\Gamma^2_{11}\left(\frac{d\theta}{ds}\right)^3 + (2\Gamma^2_{12} - \Gamma^1_{11})\left(\frac{d\theta}{ds}\right)^2\frac{d\varphi}{ds} + (\Gamma^2_{22} - 2\Gamma^1_{12})\frac{d\theta}{ds}\left(\frac{d\varphi}{ds}\right)^2\right.$

$\left.\qquad -\Gamma^1_{22}\left(\frac{d\varphi}{ds}\right)^3 + \frac{d\theta}{ds}\frac{d^2\varphi}{ds^2} - \frac{d^2\theta}{ds^2}\frac{d\varphi}{ds}\right]\sqrt{EG-F^2}$,

I know how to compute everything in this formulae (at point $P$) except the second derivatives $\frac{d^2\theta}{ds^2}$ and $\frac{d^2\varphi}{ds^2}$, but I don't see how to calculate those when I only know the behaviour of the particle at the point $P$.

So my questions are:

• Do I really have enough information to calculate both the instantaneous acceleration and the instantaneous curvature at point $P$ (I don't care about other points on the trajectory)?
• If so, what are their formulae in terms of my known boundary conditions $\dot{\bf r}(\theta(t_1),\varphi(t_1)) = {\bf v}$, $\ddot{\theta}(t_1) = a_{\theta}$, and $\ddot{\varphi}(t_1) = a_{\varphi}$?

Extra information, in case it's relevant

• My surface is a surface of revolution about the $z$-axis, but my parameterisation is not orthogonal (i.e. in the first fundamental form, $F \ne 0$). The parameter $\varphi$ is, however, the "parameter of revolution".
• $a_{\varphi}$ happens to be zero in my particular case. That is, the particle is free to move arbitrarily along the iso-parametric curve corresponding to variable $\theta$, but $\varphi(t)$ is strictly of the form $\varphi(t) = At+B$, for some constants $A$ and $B$.

Answer 1

I can translate all the unknown derivatives in $s$ into derivatives in $t$ by an application of the chain rule.

$$\frac{d\theta}{ds} = \frac{d\theta}{dt}\frac{dt}{ds}$$

and

$$\begin{aligned} \frac{d^2\theta}{ds^2} &= \frac{d}{ds}\frac{d\theta}{ds} \\ &= \frac{dt}{ds} \frac{d}{dt} \left( \frac{d\theta}{dt}\frac{dt}{ds} \right) \\ &= \frac{1}{ds/dt} \left( \frac{d^2\theta}{dt^2}\frac{1}{ds/dt} + \frac{d\theta}{dt} \frac{d}{dt} \left(\frac{1}{ds/dt} \right)\right) \\ &= \frac{1}{ds/dt} \left( \frac{d^2\theta}{dt^2}\frac{1}{ds/dt} - \frac{d\theta}{dt} \left(\frac{1}{ds/dt} \right)^2 \frac{d^2s}{dt^2} \right), \end{aligned}$$ and similarly for $\varphi$. When I substitute these back into the equation for geodesic curvature, for example, I get $$ \begin{aligned} \kappa_g &= \left(\frac{1}{ds/dt}\right)^3 \left[\Gamma^2_{11}\left(\frac{d\theta}{dt}\right)^3 + (2\Gamma^2_{12} - \Gamma^1_{11})\left(\frac{d\theta}{dt}\right)^2 \frac{d\varphi}{dt} + (\Gamma^2_{22} - 2\Gamma^1_{12})\frac{d\theta}{dt}\left(\frac{d\varphi}{dt} \right)^2\right. \\ &\qquad \left. -\Gamma^1_{22}\left(\frac{d\varphi}{dt}\right)^3 + \frac{d\theta}{dt}\frac{d^2\varphi}{dt^2} - \frac{d^2\theta}{dt^2} \frac{d\varphi}{dt}\right]\sqrt{EG-F^2}, \end{aligned} $$ which resembles the original, except that all the derivatives are now taken with respect to $t$, and there is the extra factor of $1/(ds/dt)^3$ out the front.