Geodesics on a generalized cylinder

Question

I want to prove that given a generalized cylinder $C(s,t)=\alpha(s)+t\hat{z}$ , where $\alpha$ is a curve on the $xy$ plane and $\hat{z}$ is the $z$-axis vector, then a geodesic curve $\gamma$ has the following property: $k/\tau=constant$.
I tried to brute force the problem (calculate explicitly curvature and torsion) but I end up with a monstruous espression which is impossible to interpret.. I'm sure there is a much better way to tackle this, I was thinking it could be useful to notice that the geodesic makes constant angle with $\hat{z}$.. but so far it's not really getting me anywhere.

Any idea?

Answer 1

As far as I see you are already able to show that any geodesic curve $\gamma(s)$ makes a constant angle with the $z$ axis. So I will use this to show that $\frac{k}{\tau}$ is constant. Let $T(s),N(s),B(s)$ be respectively the unit tangent to $\gamma(s)$, the normal and the binormal. Here we assume the parameter $s$ to be the arclength of $\gamma$ i.e. $T(s) = \gamma'(s)$. As we said above the angle between $T(s)$ and $\hat{z}$ is constant. Namely, the scalar product: $$ T(s) \cdot \hat{z} = cte $$ by taking derivative $\frac{d}{ds}$ we get $$k(s) N(s) \cdot \hat{z} = 0 \, .$$ If $k(s) \equiv 0$ then you get $\frac{k}{\tau} \equiv 0$ and we are done. So we can assume $k(s) \neq 0 $ and so $$N(s) \cdot \hat{z} = 0 \, \hspace{1cm} (*).$$ Taking another derivative $\frac{d}{ds}$ we get (by using Frenet-Serret formulae) $$-k(s)T(s) \cdot \hat{z} + \tau(s)B(s)\cdot \hat{z} = 0 \, .$$ We know that $T(s) \cdot \hat{z} = cte$ . If $cte=0$ then $\gamma(s)$ is plane curve hence $\tau=0$. I guess you discarded this case, so I discard it too. So I assume $cte \neq 0$ which implies $$\frac{k(s)}{\tau(s)} = \frac{B(s) \cdot \hat{z}}{T(s) \cdot \hat{z}}$$ Finally, notice that $B(s) \cdot \hat{z}$ is constant because $$\frac{d}{ds} B(s) \cdot \hat{z} = -\tau(s) N(s) \cdot \hat{z} = 0 $$ were the last $=$ follows from the above equation (*).

Answer 2

$\newcommand{tg}{{\bf\hat t}_{\gamma}} \newcommand{ta}{{\bf\hat t}_{\alpha}} \newcommand{ng}{{\bf\hat n}_{\gamma}} \newcommand{na}{{\bf\hat n}_{\alpha}} \newcommand{bg}{{\bf\hat b}_{\gamma}} \newcommand{ba}{{\bf\hat b}_{\alpha}} \newcommand{zz}{{\bf\hat z}} \newcommand{ka}{\kappa_\alpha} \newcommand{tora}{\tau_\alpha} \newcommand{kg}{\kappa_\gamma} \newcommand{torg}{\tau_\gamma}$

Here is a fairly direct approach. The first thing to do would be to look at our intrinsic geometry, to be able to write the geodesic equations. This is very straightforward and we get that the first fundamental form is $$I =\begin{pmatrix} \|\alpha'(s)\| & 0 \\ 0 & 1\end{pmatrix}$$

From this I take that you meant $\alpha$ to be parametrized by the arc length. In fact I don't think the result follows if this isn't the case. Anyhow, let's suppose it is, so the first fundamental form is the identity. From here we easily get that all Christoffel symbols are identically $0$, and a geodesic curve must satisfy $$\left\{\begin{align}s''(r) = 0 \\ t''(r) = 0\end{align}\right.$$ where the curve is $\gamma(r) = C(s(r),t(r))$. Even though it is trivial, it won't be necessary to integrate these equations! As is good practice, we can argue with the equation itself and not worry about the explicit expression for $s,t$.

We now proceed as usual to calculate curvature and torsion. Taking derivatives, $$\begin{align}\gamma'(r) & = \alpha'(s(r))s'(r) + t'(r){\bf\hat z}\end{align}$$

But from our equations (yes we do have to integrate once, I suppose) we know that $s'(r) = c_1$ and $t'(r) = c_2$ where $c_1,c_2$ are constants. I'll drop the explicitation of where the functions are evaluated to make things less clumsy. Observe that $$\begin{align}\|\gamma\|^2 & = \langle c_1\alpha' + c_2{\bf\hat z},c_1\alpha'+ c_2{\bf\hat z}\rangle \\ &=(c_1)^2 + (c_2)^2 + c_1c_2\langle \alpha',{\bf\hat z}\rangle \end{align}$$

Since $\alpha$ is in the $xy$-plane, the last scalar product is zero. Therefore $\|\gamma\|$ is constant. We could have also proven this in general for any geodesic, and maybe you have, but It's nice to verify things every once in a while too. From here on you should notice that the final result will not change if we suppose $\|\gamma\|= 1$, so I will do that.

We differentiate once more: $$\gamma' = c_1\alpha''$$

From these two equations (the first and second derivatives) we see that there is the following relationship between the Frenet trihedra of $\gamma$ and $\alpha$: $$\begin{align}\tg &= \ta + c_2\zz \\ \ka\na &= c_1\kg\ng\end{align}$$

In particular, $\kg = c_1\ka$ and $\ng = \na$. Our goal now is to calculate the torsion, given by $-\torg = \langle \bg',\ng\rangle$. The binormal will be $$\bg = \tg\times\ng = (\ta + c_2\zz)\times\na = \ba + c_2\zz\times\na$$ Again, $\alpha$ is in the $xy$-plane, so its binormal is just $\zz$. This means that $$\bg' = c_2\zz\times\na'$$ Hence, the torsion is $$\begin{align}-\torg &= c_2\langle \zz\times\na',\na\rangle \\ &= c_2\langle\zz\times(-\ka\ta),\na\rangle \\ &= -c_2\ka\langle\zz,\ta\times\na\rangle\end{align}$$

by cyclically permuting the triple product. Observe that the last expression is just $-c_2\ka\langle\zz,\zz\rangle$, so $$\torg = c_2\ka$$

Recalling the expression we derived for $\kg$, we get $$\frac \kg\torg = \frac{c_1\ka}{c_2\ka} = \frac {c_1}{c_2}$$ as was required. Also, if $c_2$ is zero this corresponds to a planar geodesic on the cylinder i.e. a "parallel". The expression $\kg/\torg$ isn't defined in this case so we needn't worry about it.