Saturday, 3 August 2019

Geodesic equation on a sphere

I was looking for the geodesic equation on the surface of a sphere. We start with the general geodesic equation (2), the metric and the Christoffel symbol (3), (1). We arrive at the geodesic differential equations for the surface at (4), (5). I calculated these in two different ways to give me confidence. Both are fiddly but fairly simple. Since then I have found other references that confirm my equations.

At Great Circles we have also found (in three ways) an equation for a great circle (6) and checked it graphically. We know that is a geodesic. Differentiating (6) was at first difficult but when we do it and put it into (5) it disagrees. This indicates that the differentiation is wrong or the geodesic differential equations are wrong. But I have triple checked everything!

I started this at the end of March and it is now 3 August. I have had some diversions on the way:
• My niece got married in London (5 days)
• My computer died (7 days). I now have a giant 4500x3000 pixel 27.5" screen. Much better than 1920x1080 15" for formula editing and movies. I also improved the equation macros.
• Got right to reside permanently in Germany (1 day).
• Programming and managing distribution of 1064 items to eight brothers from the contents of my deceased father's house (4 days). (The programming inspired me to start programming Penrose tiles.)
• Validated the geodesic equation I have found by plotting various ones with my 3D graph plotter.
• Enhancing 3D graph plotter to show animations. (2 days)
• Writing a program to draw Penrose tiles (3 months)
So it's been a month really that I have been working on this problem. That's too long and it seems like a year.

I was in despair until Orodruin came to the rescue on Physics Forums in answer to my question. As ever he was slightly obscure and said "Your equation for the great circle is not affinely parametrised." I had not hear of this word "affine" before and poked around the internet, found a hint from Professor Govindarajan and then checked Carroll's book. There was affine parameter in the index on page 109. The geodesic equation is on page 106 and I had read to  the end of the section on page 108. Grrr! Orodruin's tip, Govindarajan's hint and Carrrol carried me over the finishing line. At last I could prove that the great circle equation did indeed satisfy the geodesic equation after a bit of indirect reparameterization. Phew!

It is interesting for many reasons including that the great circle equation could not be reparametrized directly but that only the derivative of the of its parameter with respect to an affine parameter was needed.

Here's the problem

The Christoffel symbol (torsion-free and metric compatible)
\begin{align}
\Gamma_{\mu\nu}^\sigma=\frac{1}{2}g^{\sigma\rho}\left(\partial_\mu g_{\nu\rho}+\partial_\nu g_{\rho\mu}-\partial_\rho g_{\mu\nu}\right)&\phantom {10000}(1)\nonumber
\end{align}and the geodesic equation (Carroll's 3.44)
\begin{align}
G^\sigma\equiv\frac{d^2x^\sigma}{d\lambda^2}+\Gamma_{\mu\nu}^\sigma\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}=0&\phantom {10000}(2)\nonumber
\end{align}$x^\mu\left(\lambda\right)$ is a parameterised curve. If it satisfies $G^\sigma=0$ then vectors are parallel transported on it and it is a geodesic. This is the definition of a geodesic in the manifold. Clearly $G^\sigma$ has $n$ components on an $n$-dimensional manifold and (2) is $n$ equations.

With $x^0=\phi,\ x^1=\theta$ (polar and azimuthal angle - slightly unconventional) the metric and inverse metric are
\begin{align}
g_{\mu\nu}=\left(\begin{matrix}1&0\\0&\sin^2{\phi}\\\end{matrix}\right)\ \ ,\ \ g^{\mu\nu}=\left(\begin{matrix}1&0\\0&\sin^{-2}{\phi}\\\end{matrix}\right)&\phantom {10000}(3)\nonumber
\end{align}I have checked the metric in many ways. The various ways of defining polar coordinates added to the confusion (also described in Great Circles).

The geodesic equations for a sphere and the great circle equation

\begin{align}
\frac{d^2\phi}{d\lambda^2}-\sin{\phi}\cos{\phi}\left(\frac{d\theta}{d\lambda}\right)^2=0&\phantom {10000}(4)\nonumber\\
\frac{d^2\theta}{d\lambda^2}+2\cot{\phi}\frac{d\theta}{d\lambda}\frac{d\phi}{d\lambda}=0&\phantom {10000}(5)\nonumber\\
\theta=\lambda\ \ ,\ \ \phi=\tan^{-1}{\left(\frac{C}{A\cos{\lambda}+B\sin{\lambda}}\right)}&\phantom {10000}(6)\nonumber
\end{align}It is very simple to solve the geodesic equation on a plane. I did that for an exercise when I thought I was going insane. Connecting up (4),(5),(6) was much harder. It's in

Commentary 3.3 Parallel transport and geodesics.pdf (7 pages)