I am working on the following problem;

Given parameterised surface $X(u,v)=(u \cdot \cos v,u \cdot \sin v,v)$ determine for which values $\alpha$ the curve $\gamma_{\alpha}=(t \cdot \cos (\alpha t),t \cdot \sin(\alpha t),\alpha t)$ is a geodesic.

According to a theorem a curve $\gamma = X \circ \beta $ is a geodesic if and only if $\beta(t)=(u(t),v(t))$ satisfy

$\frac{d}{dt}(E\dot{u}+F\dot{v})=\frac{1}{2}(E_{u}(\dot{u})^{2}+2F_{u}\dot{u}\dot{v} +G_{u}(\dot{v})^{2})$

$\frac{d}{dt}(F\dot{u}+G\dot{v})=\frac{1}{2}(E_{v}(\dot{u})^{2}+2F_{v}\dot{u}\dot{v} +G_{v} (\dot{v})^{2})$.

Given the data in this problem we have,

$X_{u}=(\cos v,\sin v,0)$

$X_{v}=(-u \cdot \sin v, u \cdot \cos v, 1 )$





Therefore we get the system

$0=\alpha^2\cdot t$

$\alpha\cdot 2t=0$

Which means that $\alpha$ must be zero. This however dosn't seem quite right. Can anyone see where I go wrong and what the right answer should be?

  • 1
    $\begingroup$ There's also a typo in the definition of the curve $\gamma_\alpha$. Note that as you typed it, the curve does not lie on the surface $X$ for any value of $\alpha$. $\endgroup$ – Ted Shifrin Aug 4 at 19:42

There are some mistakes on the LHS of your equations. The corrected equations are as follows:

$\displaystyle \frac{d(E \dot{u} + F \dot{v})}{dt} = \frac{1}{2}\left(E_u \dot{u}^2 + 2F_u \dot{u}\dot{v} + G_u \dot{v}^2 \right)$


$\displaystyle \frac{d(F \dot{u} + G \dot{v})}{dt} = \frac{1}{2}\left(E_v \dot{u}^2 + 2F_v \dot{u}\dot{v} + G_v \dot{v}^2 \right)$

If $\alpha$ is a constant, it must be zero.

What if $\alpha$ is a function of $t$?

Both the equations become different:

$\frac{1}{2} \left( 2t(\alpha + \dot{\alpha}t)^2\right) = 0$

$(t^2+1) (\alpha + \dot{\alpha}t) = 0$


$\alpha + \dot{\alpha}t = 0$


$\displaystyle \alpha = \frac{C}{t}$ where $C$ is a constant

  • $\begingroup$ No, I dont see how that would work out :/ $\endgroup$ – user1 Aug 5 at 6:29
  • $\begingroup$ I edited the answer a bit. Please let me know if it is clear now. $\endgroup$ – PTDS Aug 5 at 17:49
  • $\begingroup$ $\alpha$ is a constant $\endgroup$ – user1 Aug 6 at 10:44
  • $\begingroup$ @DownVoter: What went wrong? $\endgroup$ – PTDS Aug 9 at 2:09

My calculations seems to be correct.


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