The Geodesics of a Sphere.

I need to find the geodesics of a sphere. Then in polar coordinates
$$x=a \sin\theta \cos\phi \\y=a \sin\theta \sin\phi\\ z=a\cos\theta$$

Then $ds^2=dx^2+dy^2+dz^2$.
Can someone please tell me how is $ds^2=a^2(d\theta^2+\sin^2\theta \, d\phi ^2)$ obtained. I don't know much about spherical coordinates.

Solution 1:

Rather than discuss via comments, here's an explicit list of steps/hints. Perhaps you can look at them only when stuck:

Step 1. Admittedly, the integral you mentioned is not easy. Here's one approach: \begin{align*} \phi &= \int \frac{k \, d\theta}{\sin \theta \sqrt{ \sin^2 \theta-k^2}} = \int\frac{k \, d \theta}{\sin^2\theta\sqrt{1-k^2\csc^2\theta}} = \int\frac{k \csc^2 \theta \, d \theta}{\sqrt{1-k^2(1+\cot^2 \theta)}}. \end{align*} Now use a $u$-sub.

Step 2. Setting $u=\cot \theta$, we have $du=d(\cot \theta)=-\csc^2 \theta \, d\theta$, hence \begin{align*} \phi &= \int \frac{-k \, du}{\sqrt{1-k^2(1+u^2)}} = -\int \frac{ \, du}{\sqrt{\left(\frac{1-k^2}{k^2}\right) -u^2}}. \end{align*} Step 3. Using the formula $\int (a^2-x^2)^{-1/2} \, dx = \sin^{-1}(x/a)$, we have \begin{align*} \phi= - \sin^{-1}\left(\frac{u}{\sqrt{(1-k^2)/k^2}}\right)+C= -\sin^{-1}\left(\frac{k \cot \theta}{\sqrt{1-k^2}}\right)+C. \end{align*}

Step 4. Compose with sine and use sum/difference formulas: $$k \cot \theta / \sqrt{1-k^2} = \sin(C-\phi)=\sin(C)\cos(\phi)-\cos(C)\sin(\phi).$$ Multiplying by $\sin \theta$, this gives $$ k \cos \theta /\sqrt{1-k^2} = \sin (C) \cos (\phi) \sin (\theta)-\cos (C)\sin(\phi) \sin(\theta).$$

Step 5. Switch back to Cartesian coordinates: $$\frac{k}{ \sqrt{1-k^2}}\cdot \frac{z}{a} = \sin(C) \cdot \frac{x}{a} - \cos(C) \cdot \frac{y}{a}.$$ Note that $k$ and $C$ are constants, so this is a linear relation of $x$, $y$, and $z$. What set of points satisfies such a relation? When you intersect the sphere with this set, you get a geodesic.

Solution 2:

Well from

$\begin{eqnarray} x &=& r \sin(\theta) \cos(\phi)\\ y &=& r \sin(\theta) \sin(\phi)\\ z &=& r \cos(\theta)\\ \end{eqnarray}$

follows

$\begin{eqnarray} dx &=& \sin(\theta) \cos(\phi) dr + r \cos(\theta) \cos(\phi) d\theta - r \sin(\theta) \sin(\phi) d\phi\\ dy &=& \sin(\theta) \sin(\phi) dr + r \cos(\theta) \sin(\phi) d\theta + r \sin(\theta) \cos(\phi) d\phi\\ dz &=& \cos(\theta) dr - r \sin(\theta) d\theta\\ \end{eqnarray}$

and for the surface we have $dr=0$, thus

$\begin{eqnarray} dx &=& r \cos(\theta) \cos(\phi) d\theta - r \sin(\theta) \sin(\phi) d\phi\\ dy &=& r \cos(\theta) \sin(\phi) d\theta + r \sin(\theta) \cos(\phi) d\phi\\ dz &=& r \sin(\theta) d\theta\\ \end{eqnarray}$

Then we find that

$\begin{eqnarray} dx^2 + dy^2 + dz^2 &=& \Big( r \cos(\theta) \cos(\phi) d\theta - r \sin(\theta) \sin(\phi) d\phi \Big)^2\\ && + \Big( r \cos(\theta) \sin(\phi) d\theta + r \sin(\theta) \cos(\phi) d\phi \Big)^2\\ && + \Big( r \sin(\theta) d\theta \Big)^2\\ &=& r^2 \Big( \cos^2(\theta) \cos^2(\phi) d\theta^2 - 2 \cos(\theta) \cos(\phi) \sin(\theta) \sin(\phi) d\theta d\phi\\ &&\hspace{2em} + \sin^2(\theta) \sin^2(\phi) d\phi^2\\ && + \cos^2(\theta) \sin^2(\phi) d\theta^2 - 2 \cos(\theta) \sin(\phi) \sin(\theta) \cos(\phi) d\theta d\phi\\ &&\hspace{2em} + \sin^2(\theta) \cos^2(\phi) d\phi^2\\ && + \sin^2(\theta) d\theta^2 \Big) \end{eqnarray}$ $\begin{eqnarray} dx^2 + dy^2 + dz^2 &=& r^2 \Big( d\theta^2 + \sin^2(\theta) d\phi^2 \Big) \end{eqnarray}$