The Geodesics of a Sphere.

Question

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.

Do you know how to write, for example, $dx$ in terms of $\theta,\phi,d\theta,$ and $d\phi$? — Kyle, Jun 16 '14 at 13:16
This is sometimes called the "total derivative" of $x$: $dx=\frac{\partial x}{\partial \theta} , d\theta + \frac{\partial x}{\partial\phi} , d\phi$. Try computing that, as well as the corresponding terms for $y$ and $z$. — Kyle, Jun 16 '14 at 13:29
@user153841 -- No problem. When you have computed $dy$ and $dz$ and add up $dx^2+dy^2+dz^2$, there should be some sign cancellation and trig identities taking you to $ds^2=a^2(d\theta^2+\sin^2\theta d\phi^2)$. Do you know how to tackle the geodesic part of the question? — Kyle, Jun 16 '14 at 13:44
@user153841 Haven't gone to the rest of the problem.Was stuck here.Have to try to do the geodesic part — sam_rox, Jun 16 '14 at 13:46
@user153841 In calculating the geodesic part I came upto $\phi prime= {k\over sin\theta\sqrt{sin^2\theta-k^2}}$.Can you please tell me so far whether I have done correctly.And if so how to compute this integral — sam_rox, Jun 16 '14 at 14:05
That's correct so far. The $\sin \theta$ in the denominator is probably irking you, right? Try moving it to the numerator then play around with recasting in terms of $\csc \theta$. — Kyle, Jun 16 '14 at 14:35

score 6 · Accepted Answer · answered Jun 16 '14 at 15:34

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.

Thanks a lot.That integration part is a lot difficult.I wouldn't be able to think of it myself — sam_rox, Jun 16 '14 at 16:24
Suppose I need to find the geodesic on a sphere but cartesian coordinates using euler lagrange eqyation then is the following correct - in the expression for ds^2 I will evaluate d(theta) and d(phi) in terms of dx dy dz and then use the equation of sphere to eliminate 1 of dx dy dz, say dz... And then use that ds in terms of dx dy and I should get the cartesian equation of a circle... Is that right... I just want to know this.... The only problem is what will be the limits of the integral — Shashaank, Mar 23 '20 at 08:52

johannesvalks · Answer 2 · 2014-06-16T16:10:21.613

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

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

Suppose I need to find the geodesic on a sphere but cartesian coordinates using euler lagrange eqyation then is the following correct - in the expression for ds^2 I will evaluate d(theta) and d(phi) in terms of dx dy dz and then use the equation of sphere to eliminate 1 of dx dy dz, say dz... And then use that ds in terms of dx dy and I should get the cartesian equation of a circle... Is that right... I just want to know this.... The only problem is what will be the limits of the integral — Shashaank, Jan 01 '20 at 12:33

The Geodesics of a Sphere.

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