differential-geometry
What is the geodesic of the helicoid?
$M=\{ (x,y,z)\in \mathbb{R}^3: x\sin z – y\cos z =0\}$
Whose tangent at the point $p = (1,0,0)$ in the line $r = \{(x, y, z) \in \mathbb {R} ^ 3: x = 1, y = z\}$.
I tried with the definition, but I don't know the parameterization, and so do not know which is the tangent plane. I can't do $\alpha ''(t) \perp T_{\alpha(t)}M$.
Any help is appreciated.
Note that for each $v$, the intersection of the helicoid with a coaxial cylinder (spiral) in cylindrical coordinates $(v,u,z)$ has the equation $z=u$, or more generally $z=\alpha u$; i.e., a "straight line through the origin".
Therefore, it makes sense to define a "circular helicoid" in toroidal coordinates $(r,u,v)$ by the equation $v=\alpha u$.
$$\begin{aligned}x= & \left(R+r\cos\alpha u\right)\cos u\\ y= & \left(R+r\cos\alpha u\right)\sin u\\ z= & r\sin\alpha u \end{aligned}$$
The problem can be solved synthetically as follows. The vertical ray, e.g. the $y$-axis, is invariant under the reflection $(x,y)\mapsto (-x,y)$. The reflection is an isometry and therefore the vertical line must be a geodesic.
To show that the semicircles are geodesics, use an isometry sending one of its endpoints to infinity. The result is a vertical ray hence a geodesic as already shown.
If the semicircle passes through the origin, just use $z\mapsto 1/z$. Otherwise use a horisontal translation first.
Best Answer
The general parametrisation is $X(u,v) = \left( u \cos(v) , u \sin(v) , kv \right)$, where $k$ is a non-zero constant (greater than zero for right, or less than for left), and looking at your definition of $M$ it appears that $k=1$. It is always important to remember that there is usually more than one way to parametrise a surface, so as long as you can argue that it is one-to-one and continuously differentiable then you're all 'g'!
Step 1: compute the first partial derivatives \begin{align*} X_u &= \left( \cos(v) , \sin(v) , 0 \right) \\ X_v &= \left( -u \sin(v) , u \cos(v) , k \right) \end{align*} Step 2: compute the coefficients of the first fundamental form: \begin{align*} E &= X_u \cdot X_u = 1; & F &= X_u \cdot X_v = 0; & G &= X_v \cdot X_v = u^2+k^2. \end{align*} Step 3: compute the Christoffel symbols by solving the following system for $\Gamma_{ij}^k$: \begin{align*} \Gamma_{11}^1 E + \Gamma_{11}^2 F &= X_{uu} \cdot X_u = \frac{E_u}{2}; \\ \Gamma_{11}^1 F + \Gamma_{11}^2 G &= X_{uu} \cdot X_v = F_u - \frac{E_v}{2}; \\ \Gamma_{12}^1 E + \Gamma_{12}^2 F &= X_{uv} \cdot X_u = \frac{E_v}{2}; \\ \Gamma_{12}^1 F + \Gamma_{12}^2 G &= X_{uv} \cdot X_v = \frac{G_u}{2}; \\ \Gamma_{22}^1 E + \Gamma_{22}^2 F &= X_{vv} \cdot X_u = F_v - \frac{G_u}{2}; \\ \Gamma_{22}^1 F + \Gamma_{22}^2 G &= X_{vv} \cdot X_v = \frac{G_v}{2}. \end{align*} Therefore, \begin{align*} \Gamma_{11}^1 &= X_{uu} \cdot X_u = \frac{E_u}{2} = 0; \\ \Gamma_{11}^2 \left( u^2+k^2 \right) &= X_{uu} \cdot X_v = F_u - \frac{E_v}{2}=0; \\ \Gamma_{12}^1 &= X_{uv} \cdot X_u = \frac{E_v}{2}=0; \\ \Gamma_{12}^2 \left( u^2+k^2 \right) &= X_{uv} \cdot X_v = \frac{G_u}{2}=u; \\ \Gamma_{22}^1 &= X_{vv} \cdot X_u = F_v - \frac{G_u}{2}=-u; \\ \Gamma_{22}^2 \left( u^2+k^2 \right) &= X_{vv} \cdot X_v = \frac{G_v}{2}=0. \end{align*} This yields $\Gamma_{22}^1 = -u$, $\Gamma_{12}^2 = u/\left( u^2+k^2 \right)$, and the rest are equal to zero.
Step 4: compute the geodesic curves $\alpha(t) = (u(t),v(t))$ by solving \begin{align} \frac{\mathrm{d}^2 u}{\mathrm{d}t^2} + \Gamma_{11}^1 \left( \frac{\mathrm{d}u}{\mathrm{d}t} \right)^2 + 2 \Gamma_{12}^1 \frac{\mathrm{d}u}{\mathrm{d}t} \frac{\mathrm{d}v}{\mathrm{d}t} + \Gamma_{22}^1 \left( \frac{\mathrm{d}v}{\mathrm{d}t} \right)^2 &= 0; \\ \frac{\mathrm{d}^2 v}{\mathrm{d}t^2} + \Gamma_{11}^2 \left( \frac{\mathrm{d}u}{\mathrm{d}t} \right)^2 + 2 \Gamma_{12}^2 \frac{\mathrm{d}u}{\mathrm{d}t} \frac{\mathrm{d}v}{\mathrm{d}t} + \Gamma_{22}^2 \left( \frac{\mathrm{d}v}{\mathrm{d}t} \right)^2 &= 0. \end{align} Therefore, \begin{align} \frac{\mathrm{d}^2 u}{\mathrm{d}t^2} - u \left( \frac{\mathrm{d}v}{\mathrm{d}t} \right)^2 &= 0; \\ \frac{\mathrm{d}^2 v}{\mathrm{d}t^2} + 2 \frac{u}{u^2+k^2} \frac{\mathrm{d}u}{\mathrm{d}t} \frac{\mathrm{d}v}{\mathrm{d}t} &= 0. \end{align} The last equation above gives (in shorthand notation) \begin{align} \frac{v''}{v'} &= -\, \frac{2uu'}{u^2+k^2} \\ \implies \int \; \frac{v''}{v'} \, \mathrm{d}v' &= \int \; - \, \frac{2uu'}{u^2+k^2} \, \mathrm{d}u \\ \implies \ln \left| v' \right| &= - \ln \left| u^2+k^2 \right| + \text{constant} \\ \implies v' &= \frac{\Theta}{u^2+k^2}, \end{align} where $\Theta$ is just some constant which is leftover from the integration.
Note: \begin{align} \left( v' \right)' &= \left( \frac{\Theta}{u^2+k^2} \right)' \\ \implies v'' &= \frac{- 2u'u \Theta}{\left( u^2+k^2 \right)^2} = \frac{-2u'uv'}{u^2+k^2} \\ \implies v'' + 2 \frac{u}{u^2+k^2} u' v' &= 0, \end{align} which is exactly what we needed.
Given the above, are you now able to compute the geodesic curves?