Tuesday 6 August 2019

Exercise 3.03 The Christoffel symbols with a diagonal metric


A diagonal metric in 4-space
Imagine we had a diagonal metric ##g_{\mu\nu}##. Show that the Christoffel symbols are given by
\Gamma_{\mu\nu}^\lambda&=0&\phantom {10000}(1)\nonumber\\
\Gamma_{\mu\mu}^\lambda&=-\frac{1}{2}\left(g_{\lambda\lambda}\right)^{-1}\partial_\lambda g_{\mu\mu}&\phantom {10000}(2)\nonumber\\
\Gamma_{\mu\lambda}^\lambda&=\partial_\mu\left(\ln{\sqrt{\left|g_{\lambda\lambda}\right|}}\right)&\phantom {10000}(3)\nonumber\\
\Gamma_{\lambda\lambda}^\lambda&=\partial_\lambda\left(\ln{\sqrt{\left|g_{\lambda\lambda}\right|}}\right)&\phantom {10000}(4)\nonumber
\end{align}In these expressions, ## \mu\neq\nu\neq\lambda## and repeated indices are not summed over.


At last I have recovered from the grind of that geodesic on a sphere and completed this question. It involved a bit of index manipulation, some knowledge about the inverse of a diagonal matrix and the chain rule (taken slowly).

All three pages in Ex 3.03 Diagonal metric.pdf.

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)
\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)
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
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

\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)

Great circles

We know that a great circle is a line between two points on a sphere which is the intersection of a plane through the sphere's centre and the two points and the surface of the sphere. We also know that the shorter distance between the two points is the shortest distance between the two points if we are confined to the surface of the sphere.

Here we find the equation for a great circle and prove it is the same as the one given (rather ambiguously) in Wolfram Mathsworld and, loosely, the one by Professor Govindarajan. I also plotted some great circles to test the equations visually. That's the video. The reason for testing the equations so rigorously is that they did not seem to agree with Carroll's geodesic equation at (3.44). Eventually I showed that they did.

Polar coordinates

First we must deal with the angular coordinates ## \phi,\theta##. Some sources have ## \theta## as the longitude (azimuthal angle) and ## \phi## as the colatitude (angle from north pole) in other sources they are swapped around. I tried to follow the majority and did the former with ## \phi,\theta## in that order. Then Wikipedia told me that "in one system frequently encountered in physics ##(r,\theta,\phi)## gives the radial distance, polar angle, and azimuthal angle, whereas in another system used in many mathematics books ##(r,\theta,\phi)## gives the radial distance, azimuthal angle, and polar angle. ... Other conventions are also used, so great care needs to be taken to check which one is being used." Not only are the meanings of ## \phi,\theta## swapped but also the order!! I have followed neither convention. Boo hoo. At Mathworld a table of seven variations is given. They use the same as me, but not in the same order. Carroll and Govindarajan follow the physicist convention. In future will follow the physicist convention but here and in the upcoming post on the parallel transport and geodesics, I stick to my own.

The great circle equations

\phi=\tan^{-1}{\left(\frac{A_z}{A_x\cos{\theta}+A_y\sin{\theta}}\right)}&\phantom {10000}\nonumber
\end{align}Wolfram Mathsworld
\cos{\theta}\sin{\phi}\sin{c_2}+\sin{\theta}\sin{\phi}\cos{c_2}+c_1\cos{\phi}=0&\phantom {10000}\nonumber
\end{align}Professor Govindarajan's
\cot{\phi}=A\cos{\left(\theta+\theta_0\right)}&\phantom {10000}\nonumber
\end{align}There are two or three constants in each equation.

Read the full details at Commentary 3.3 Great circle.pdf
The spreadsheet which generated the video is at Commentary 3.3 Parallel transport and geodesics.xlsm