Motivation: I am trying to understand how one in practice computes geodesics on a simple space by means of the orthogonality of the acceleration vector to the tangent space at that point. I consider a cylinder in $\Bbb R^3$ and compute what I believe is the 2nd order non-linear ode that describes geodesics:

The cylinder $$C=\{(x,y,z)\in\Bbb R^3\mid x^2+y^2=1\}$$ is parametrised in $\Bbb R^3$ by: $$\Phi(\theta,z)=(\cos\theta,\sin\theta,z).$$

A geodesic $c(t)$ of the surface has $c''(t)$ orthogonal to $T_{c(t)}C$ at each time $t$. Take a curve $$c(t) = (\cos(\theta(t)),\sin(\theta(t)),z(t))$$ then we require that $c''(t)$ is orthogonal to the plane $$\{(-\sin\theta(t)d,\cos\theta(t)d,z)\mid d,z\in\Bbb R\}\subset\Bbb R^3.$$ which means $c''(t)$ will have to be orthogonal to $$(0,0,1)\times (-\sin(\theta(t)),\cos(\theta(t)),0)= (-\cos(\theta(t)),\sin(\theta(t)),0)=\mathbf{Q}$$

We compute $c''(t)$: $$c'(t) = (-\sin(\theta(t))\dot\theta(t),\cos(\theta(t))\dot\theta(t),\dot z(t))$$ $$c''(t) = (-\cos(\theta(t))\dot\theta^2(t) -\sin(\theta(t))\ddot\theta(t), \sin(\theta(t)\dot\theta^2(t)+\cos(\theta(t))\ddot\theta(t),\ddot z(t))$$ Because this isn't in the tangent space, I would have to orthogonally project (compute covariant derivative) or otherwise check orthogonality in the ambient $\Bbb E^3$?

I think orthogonality just comes from $c''(t)\cdot \mathbf{Q}=0$: $$\cos^2(\theta(t))\dot{\theta}^2(t)+\sin(\theta(t))\cos(\theta(t))\ddot\theta(t)+\sin^2(\theta(t))\dot\theta^2(t) +\sin\theta(t)\cos(\theta(t))\ddot{\theta}(t)=0$$ $$=\dot\theta^2(t)+2\cos(\theta(t))\sin(\theta(t))\ddot\theta(t)$$

and then I just have to solve the ode: $$\frac{d^2\theta(t)}{dt^2}(2\sin(\theta(t))\cos(\theta(t))) + \frac{d\theta(t)}{dt} = 0$$

Am I doing this correctly? (I'll try approaching this using connections later, so I'd rather not see that form in an answer)

  • $\begingroup$ Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. $\endgroup$ – dantopa Jun 11 '19 at 20:07
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    $\begingroup$ A cylinder is locally perfectly flat. So a geodesic is just a helix. $\endgroup$ – TonyK Jun 11 '19 at 20:14
  • $\begingroup$ @TonyK My desire to compute the geodesics here isn't to know what they are, but to practice computing geodesics in general. (Although that's good to know, and I thank you for sharing your intuition!) $\endgroup$ – user681391 Jun 11 '19 at 20:17

You have a sign mistake when computing $\ddot{c}(t)$. However, since $N:\Bbb S^1\times \Bbb R\to \Bbb S^2$ given by $N(x,y,z)=(x,y,0)$ is a unit normal vector to the cylinder, we know that if $c$ is a geodesic then the tangent projection $D\dot{c}/{\rm d}t$ of $\ddot{c}(t)$ vanishes, which is to say that the latter is a multiple of $N(c(t))$ for all $t$. So we write $\ddot{c}(t)=\lambda(t)N(c(t))$ for some (automatically) smooth function $\lambda$ -- that we don't really need to solve for. So this relation becomes $$(-\sin(\theta(t))\ddot{\theta}(t)-\cos(\theta(t))\dot{\theta}(t)^2, \cos(\theta(t))\ddot{\theta}(t)-\sin(\theta(t))\dot{\theta}(t)^2,\ddot{z}(t))=\lambda(t)(\cos(\theta(t)),\sin(\theta(t)),0)$$Equating components and solving the system gives $\theta(t)=at+b$ and $z(t)=ct+d$ for some constants $a,b,c,d\in \Bbb R$. If $a=c=0$ we get a point; if $a=0$ but $c\neq 0$ we get a straight vertical line. If $a\neq 0$ and $c=0$, a horizontal circle. Else, helices.

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    $\begingroup$ Thank you, this is massively helpful. $\endgroup$ – user681391 Jun 12 '19 at 7:30

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