Find The Geodesic Lines And Curves of A Surface

Question

Find the geodesic lines and curves of $x^2+y^2+z^2=r^2$

The following surface is a sphere so we can use the parametrization

$$X(\theta,\phi)=(rsin\phi cos\theta,rsin\phi sin\theta,rcos\phi)$$

1. why is it $X(\theta,\phi)$ and not $X(\phi,theta)$?

Next we have to find the metric

$$<X_{\theta},X_{\theta}>=r^2sin^2\phi sin^2\theta+r^2sin^2\phi cos^2\theta=r^2sin^2\phi(sin^2\theta+cos^2\theta)=r^2sin^2\phi$$

$$<X_{\theta},X_{\phi}>=<X_{\phi},X_{\theta}>=0$$

$$<X_{\phi},X_{\phi}>=r^2cos^2\phi cos^2\theta+r^2cos^2\phi sin^2\theta+r^2sin^2\phi=r^2cos^2\phi(co^2\theta+sin^2\theta)+r^2sin^2\phi=r^2(cos^2\phi+sin^2\phi)=r^2$$

So we have $$g_{ij}=\begin{pmatrix} r^2sin^2\phi & 0\\0 & r^2\end{pmatrix}$$

2. could we use that fact that the metric of a surface of revolution is $$g_{ij}=\begin{pmatrix} r^2 & 0\\0 & 1\end{pmatrix}$$?

Next we have to find the Christoffel coefficients:

To do so we find:

$$g^{ij}=\begin{pmatrix} \frac{1}{r^2sin^2\phi} & 0\\0 & \frac{1}{r^2}\end{pmatrix}$$

$$g_{ij;\theta}=\begin{pmatrix} 0 & 0\\0 & 0\end{pmatrix}$$

$$g_{ij;\phi}=\begin{pmatrix} 2r^2\cos\phi & 0\\0 & \frac{1}{r^2}\end{pmatrix}$$

To get

$$\Gamma_{11}^{1}=\frac{1}{2}(0-0+0)\frac{1}{r^2sin^2\phi}+\frac{1}{2}(0-2r^2cos\phi+0)0=0$$

$$\Gamma_{12}^{1}=\Gamma_{21}^{1}=\frac{1}{2}(0-0+0)\frac{1}{r^2sin^2\phi}+\frac{1}{2}(0-0+0)0=0$$

$$\Gamma_{22}^{1}=\frac{1}{2}(0-0+0)\frac{1}{r^2sin^2\phi}+\frac{1}{2}(\frac{1}{r^2}-\frac{1}{r^2}+\frac{1}{r^2})0=0$$

$$\Gamma_{11}^{2}=\frac{1}{2}(0-0+0)0+\frac{1}{2}(0-2r^2cos\phi+0)\frac{1}{r^2}=cos\phi$$

$$\Gamma_{12}^{2}=\Gamma_{21}^{2}=\frac{1}{2}(2r^2cos\phi-0+0)0+\frac{1}{2}(0-0+0)\frac{1}{r^2}=0$$

$$\Gamma_{22}^{2}=\frac{1}{2}(0-0+0)0+\frac{1}{2}(\frac{1}{r^2}-\frac{1}{r^2}+\frac{1}{r^2})r^2=1$$

3. But how do I continue from here? in the book I found 3 different formulas?

is there a written algorithm to find geodesic lines and curves of a surface?

can the process be done in matlab?

Answer 1

In the case of the sphere there are at least two easier ways to find the geodesics with calculus of variations:

Using vectors

firstly, we can avoid the metric entirely using the energy in Euclidean space, $\int \dot{x} \cdot \dot{x} \, dt$, with the constraint that $x$ lies on the sphere, i.e. $ x \cdot x = r^2$. Then the Lagrangian is $$ \int (\dot{x} \cdot \dot{x} -\lambda(x)(x \cdot x-r^2)) \, dt, $$ and the Euler–Lagrange equations are $$ \ddot{x} + \lambda(x) x = 0 $$ (if you don't believe this, substitute a variation in and integrate by parts). Also, $$ 0 = \frac{d}{dt}(x \cdot x) = 2x \cdot \dot{x}, \\ 0 = \frac{d}{dt}(x \cdot \dot{x}) = x \cdot \ddot{x} + \dot{x} \cdot \dot{x} $$ and so dotting the E–L equation with $\dot{x}$ gives $\dot{x} \cdot \ddot{x} = 0$, which gives that $\dot{x} \cdot \dot{x}=k^2$ is constant. Thus $x \cdot \ddot{x} = -k^2 = -\lambda(x) r^2 $, so $\lambda(x)=k^2/r^2$ is constant.

Now, suppose that $k$ is a nonzero constant vector with $k \cdot x(0) = 0$ and $k \cdot \dot{x}(0) =0$. We have $$ \frac{d^2}{dt^2}(k \cdot x) = k \cdot \ddot{x} = -(k^2/r^2) k \cdot x, $$ which is a second-order linear homogeneous differential equation. It has solutions of the form $A\cos{(kt/r)}+B\sin{(kt/r)}$, and the initial conditions imply that the solution is exactly zero. Hence $k \cdot x(t) =0$ for all $t$, so the geodesic lies in the intersection of the sphere with $k \cdot x=0$, i.e. it's a great circle.

Alternatively,

Using the metric

The easiest way to calculate the Christoffel symbols is normally to use the Euler–Lagrange equations of the energy functional: $$ E[x,\dot{x}] = \int g_{ab}(x) \dot{x}^a \dot{x}^b \, dt, $$ and then the Euler–Lagrange equations are $$ 0 = \frac{d}{dt}(2g_{ab}(x) \dot{x}^b) - (\partial_{a} g_{bc}) \dot{x}^b \dot{x}^c \\ = 2g_{ab} \ddot{x}^b + \partial_c g_{ab} \dot{x}^{c}\dot{x}^b - (\partial_{c} g_{ab}) \dot{x}^a \dot{x}^b, $$ which some tensor algebra rearranges into $\ddot{x}^b + \Gamma^b_{ac} \dot{x}^a \dot{x}^b = 0 $, the geodesic equation. With this in mind, we have the metric $ds^2 = r^2 \, d\theta^2 + r^2\sin^2{\theta} \, d\phi^2 $, from which the energy is $$ r^2 \int (\dot{\theta}^2+\sin^2{\theta} \, \dot{\phi}^2) \, dt, $$ and so the Euler–Lagrange equations are $$ 0 = \frac{d}{dt} (\sin^2{\theta} \, \dot{\phi}) = \sin^2{\theta} \, \ddot{\phi} + 2\sin{\theta}\cos{\theta} \, \dot{\phi}\dot{\theta}, \\ 0 = \ddot{\theta} + \sin{\theta}\cos{\theta} \, \dot{\phi}^2. $$ We can then read off the Christoffel symbols $$ \Gamma^{\phi}_{\theta\phi} = \Gamma^{\phi}_{\phi\theta} = \cot{\theta}, \quad \Gamma^{\phi}_{\phi\phi}=\Gamma^{\phi}_{\theta\theta}=0 \\ \Gamma^{\theta}_{\phi\phi} = \sin{\theta}\cos{\theta}, \quad \Gamma^{\theta}_{\theta\phi} = \Gamma^{\theta}_{\phi\theta} = \Gamma^{\theta}_{\theta\theta} = 0 $$ from the quadratic part if we so desire, but the more useful statement is that $\sin^2{\theta} \, \dot{\phi}=h$ is constant, and so $\ddot{\theta} + h^2 \cot{\theta}\csc^2{\theta}=0$. We can at this point cheat and note that if $\dot{\phi}=0$ initially, $h=0$, so either $\sin{\theta}=0$ and the point doesn't move, or $\dot{\phi}=0$ and $\ddot{\theta}=0$, so $\theta$ increases linearly. We can get away with this by choosing our coordinates so that $\dot{\phi}=0$ and $\theta \neq 0$ initially. A more sophisticated method would find the equation of the plane the velocity lies in initially and show it stays there, as above.