Computing the geodesics on a cylinder in $\Bbb R^3$ directly using orthogonality of the acceleration vector

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)

• 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. – dantopa Jun 11 '19 at 20:07
• A cylinder is locally perfectly flat. So a geodesic is just a helix. – TonyK Jun 11 '19 at 20:14
• @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!) – 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.