Find range of parameters for which a given curve is a geodesic

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,

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

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

Hence

$$=E=1$$

$$=F=0$$

$$=G=u(t)^2+1$$

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?

• 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 $\alpha$. – Ted Shifrin Aug 4 at 19:42

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

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

and

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

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

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

Both the equations become different:

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

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

Finally

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

Solving,

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

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

My calculations seems to be correct.