I'm working on the geodesics of a helicoid $S$ parametrised by $X(u,v) = \left( u \cos(v) , u \sin(v) , v \right)$, with $\alpha(t) = \left( u(t) , v(t) \right) : I \subset \mathbb{R} \to S$ a unit speed curve on $S$, i.e. parametrised by arc length. Thus far I have \begin{align} X_u &= \frac{\mathrm{d}X(u,v)}{\mathrm{d}u} = \left( \cos(v) , \sin(v) , 0 \right) \\ X_v &= \frac{\mathrm{d}X(u,v)}{\mathrm{d}v} = \left( -u\sin(v) , u\cos(v) , 1 \right) \end{align} Giving the coefficients of the first fundamental form $E = X_u \cdot X_u = 1$, $F = X_u \cdot X_v = 0$, and $G = X_v \cdot X_v = u^2 +1$, such that \begin{align} \mathrm{d}s^2 &= E \mathrm{d}u^2 + 2F \mathrm{d}u \mathrm{d}v + G \mathrm{d}v^2 \\ &= \mathrm{d}u^2 + \left(u^2 +1 \right)\mathrm{d}v^2 . \end{align} Here, as $\alpha$ has unit speed, I have reasoned that $\mathrm{d}s^2/\mathrm{d}t^2 = 1$ implying that $1 = \left( u'(t) \right)^2 + \left( u^2+1 \right) \left( v'(t) \right)^2$; is this correct reasoning?
By computing the derivatives $E_u = E_v = F_u = F_v = G_v = 0$ and $G_u = 2u$, we can compute the Christoffel symbols \begin{align} \Gamma_{11}^1 = \Gamma_{11}^2 = \Gamma_{12}^1 = \Gamma_{22}^2 = 0; \\ \Gamma_{12}^2 = \frac{u}{u^2+1}; \qquad \Gamma_{22}^1 = -u. \end{align} It is given that $\alpha$ is a geodesic on $S$, which is equivalent to the satisfaction of the following system: \begin{align} u'' + \Gamma_{11}^1 \left( u'(t) \right)^2 + 2 \Gamma_{12}^1 u'(t) v'(t) + \Gamma_{22}^1 \left( v'(t) \right)^2 &= 0 \\ v'' + \Gamma_{11}^2 \left( u'(t) \right)^2 + 2 \Gamma_{12}^2 u'(t) v'(t) + \Gamma_{22}^2 \left( v'(t) \right)^2 &= 0 \\ \implies u'' - u \left( v'(t) \right)^2 &= 0 \\ v'' + 2 \frac{u}{u^2+1} u'(t) v'(t) &= 0 \end{align} The last equation yields by integration $v' = \Theta/ \left( u^2 +1 \right)$, where $\Theta$ is an integration constant. If $\Theta = 0$ then $v'\equiv 0$ and subsequently $v = $ constant, i.e. it is a necessary condition that $u' \ne 0$ for $v = $ constant and $u = u(s)$ to be a geodesic. Is this correct?
Furthermore, if $\Theta =1$ then $v' = 1/\left(u^2 +1 \right)$, such that \begin{align} \mathrm{d}v^2 \left( u^2 +1 \right)^2 &= \mathrm{d}s^2 \\ &= \mathrm{d}u^2 + \left( u^2 + 1 \right) \mathrm{d}v^2 \\ \implies \mathrm{d}v^2 \left( u^4 + u^2 \right) &= \mathrm{d}u^2 \\ \implies \mathrm{d}v &= \frac{\mathrm{d}u}{u \sqrt{u^2 +1 }} \\ \implies v(s) &= \text{constant} \pm \int \; \frac{\mathrm{d}u}{u \sqrt{u^2 +1 }} \\ &= \text{constant} \pm \ln\left| \csc(s) + \cot(s) \right| \end{align} The above equation is the geodesic for $\Theta = 1$, in terms of $u$ or explicitly in terms of arc length $s$; is this correct?
I am feeling like a noob with geodesics so any and all help is appreciated. Also, if you would like to extend to explain something that you find interesting about this then I would really like to hear from you - the more interested I am the easier it is to learn!