# Question about geodesics on a sphere

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• George Keeling
In summary, the conversation discusses the geodesic equation and its derivation by Carroll. The equation is used to test the surface of a sphere, where great circles are known to be geodesics. However, there seems to be a discrepancy with the equation and the solution for a great circle, which is not always equal to zero. Further discussion and solutions are provided by the individuals involved in the conversation. It is suggested to use the Euler-Lagrange equations instead of the expression in terms of the metric for the Christoffel symbols.
George Keeling
Gold Member
TL;DR Summary
I wanted to test the geodesic equation on the surface of a sphere where I know that great circles are geodesics and a sphere is about the simplest non-trivial case I can think of.
I am working from Sean Carroll's Spacetime and Geometry : An Introduction to General Relativity and have got to the geodesic equation. I wanted to test it on the surface of a sphere where I know that great circles are geodesics and is about the simplest non-trivial case I can think of.

Carroll derives the geodesic equation:\begin{align}
\frac{d^2x^\sigma}{d\lambda^2}+\Gamma_{\mu\nu}^\sigma\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}=0&\phantom {10000}(1)\nonumber
\end{align}##\Gamma_{\mu\nu}^\sigma## is 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}(2)\nonumber
\end{align}In this case the indices are 0,1 and ##x^0=\phi## the colatitude (angle from north pole), ##\ x^1=\theta## , longitude. 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}Those should give me equations for ## \phi,\theta## parameterized by ## \lambda##.

I now expand the second term in (1) for ## \sigma=1## and using the fact that ## \Gamma## is torsion-free (##\Gamma_{\mu\nu}^\sigma=\Gamma_{\nu\mu}^\sigma##).\begin{align}
\Gamma_{\mu\nu}^1\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}&=\Gamma_{00}^1\frac{dx^0}{d\lambda}\frac{dx^0}{d\lambda}+2\Gamma_{10}^1\frac{dx^1}{d\lambda}\frac{dx^0}{d\lambda}+\Gamma_{11}^1\frac{dx^1}{d\lambda}\frac{dx^1}{d\lambda}&\phantom {10000}(4)\nonumber\\
&=\Gamma_{00}^1\left(\frac{d\phi}{d\lambda}\right)^2+2\Gamma_{10}^1\frac{d\phi}{d\lambda}\frac{d\theta}{d\lambda}+\Gamma_{11}^1\left(\frac{d\theta}{d\lambda}\right)^2&\phantom {10000}(5)\nonumber
\end{align}using the metric and (2) and taking each ## \Gamma## in turn, many terms vanish because the metric components are constant or 0. The last one vanishing in (13) was the subject of my previous question.\begin{align}
\Gamma_{00}^1&=\frac{1}{2}g^{1\rho}\left(\partial_0g_{0\rho}+\partial_0g_{\rho0}-\partial_\rho g_{00}\right)=g^{1\rho}\partial_0g_{0\rho}&\phantom {10000}(6)\nonumber\\
&=g^{10}\partial_0g_{00}+g^{11}\partial_0g_{01}=0&\phantom {10000}(7)\nonumber\\
2\Gamma_{10}^1&=g^{1\rho}\left(\partial_1g_{0\rho}+\partial_0g_{\rho1}-\partial_\rho g_{10}\right)=g^{1\rho}\partial_1g_{0\rho}+g^{1\rho}\partial_0g_{\rho1}&\phantom {10000}(8)\nonumber\\
&=g^{10}\partial_1g_{00}+g^{11}\partial_1g_{01}+g^{10}\partial_0g_{01}+g^{11}\partial_0g_{11}&\phantom {10000}(9)\nonumber\\
&=\frac{1}{\sin^2{\phi}}\frac{\partial}{\partial\phi}\left(\sin^2{\phi}\right)=\frac{2\sin{\phi}\cos{\phi}}{\sin^2{\phi}}=2\cot{\phi}&\phantom {10000}(10)\nonumber\\
\Gamma_{11}^1&=\frac{1}{2}g^{1\rho}\left(\partial_1g_{1\rho}+\partial_1g_{\rho1}-\partial_\rho g_{11}\right)=g^{1\rho}\partial_1g_{\rho1}-\frac{1}{2}g^{1\rho}\partial_\rho g_{11}&\phantom {10000}(11)\nonumber\\
&=g^{10}\partial_1g_{01}+g^{11}\partial_1g_{11}-\frac{1}{2}g^{10}\partial_0g_{11}-\frac{1}{2}g^{11}\partial_1g_{11}&\phantom {10000}(12)\nonumber\\
&=\frac{1}{2}\frac{1}{\sin^2{\phi}}\frac{\partial}{\partial\theta}\left(\sin^2{\phi}\right)=0&\phantom {10000}(13)\nonumber

\end{align}Putting the expressions for ## \Gamma## back into (5) we get\begin{align}

\Gamma_{\mu\nu}^1\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}&=2\cot{\phi}\frac{d\phi}{d\lambda}\frac{d\theta}{d\lambda}&\phantom {10000}(14)\nonumber

\end{align}Putting that back into (1) we have\begin{align}

\frac{d^2\theta}{d\lambda^2}+2\cot{\phi}\frac{d\phi}{d\lambda}\frac{d\theta}{d\lambda}=0&\phantom {10000}(15)\nonumber

\end{align}I can use the same method to get a solution for ## \sigma=0##, but that's not relevant yet. Put together they show that lines of longitude are always geodesics as is the equator. So far so good… but

A great circle equation (which we know is a geodesic) can be written as\begin{align}

\theta=\lambda\ \ ,\ \ \phi=\tan^{-1}{\left(\frac{C}{A\cos{\lambda}+B\sin{\lambda}}\right)}&\phantom {10000}(16)\nonumber

\end{align}where ## A,B,C## are constants depending on the start and end points. I calculated this and checked it against the slightly ambiguous Wolfram maths.

From (16) clearly\begin{align}

\frac{d^2\theta}{d\lambda^2}=0&\phantom {10000}(17)\nonumber

\end{align}so the other term in (15) should always be zero. It is\begin{align}

2\cot{\phi}\frac{d\phi}{d\lambda}\frac{d\theta}{d\lambda}=-2\frac{\left(B\cos{\lambda}-A\sin{\lambda}\right)\left(A\cos{\lambda}+B\sin{\lambda}\right)}{C^2+\left(A\cos{\lambda}+B\sin{\lambda}\right)^2}&\phantom {10000}(18)\nonumber
\end{align}which is not always zero! So something is wrong.

It's either
(15) which I have calculated in another way to check it or
(16) which I have also plotted and gives good results or
my differentiation of ##\frac{d\phi}{d\lambda}## which I checked with Symbolab. (Is there a better tool than Symbolab?)

Help! What has gone wrong?

Last edited:
George Keeling said:
So something is wrong
The geodesic equations are for an affinely parametrised geodesic. Your equation for the great circle is not affinely parametrised.

Edit: I also strongly suggest against computing the geodesic equations from using the expression in terms of the metric for the Christoffel symbols. It is much more straightforward to write down the Euler-Lagrange equations for
$$\int g_{ab} \dot x^a \dot x^b ds.$$

DEvens and George Keeling
Orodruin said:
Your equation for the great circle is not affinely parametrised.
Is that the same as
Orodruin said:
a parametrisation which fixes the length of the tangent vector.
which you wrote on the other thread.

George Keeling said:
Is that the same as
which you wrote on the other thread.
Yes.

George Keeling
Orodruin said:
Your equation for the great circle is not affinely parametrised.
I rummaged the internet to find out what an affine parameter is and got a good hint from Prof Govindarajan. That also enlightens me on
Orodruin said:
It is much more straightforward to write down the Euler-Lagrange
Then I decided to check Carroll's book. There is one entry for affine parameter in the index and it took me to one page further than I have read ! I turned the handles and if ##s## is the affine parameter which replaces ##\lambda## in my geodesic equations (one of which is 15 above) I get$$s=\int{\frac{D}{C^2+A^2\cos^2{\lambda}+2AB\sin{\lambda}\cos{\lambda}+B^2\sin^2{\lambda}}d\lambda}$$which looks nasty. But luckily(?) I only needed ##d\lambda / ds## which is much simpler and was able to reach my objective. Everything runs like clockwork. Thanks very much for the help!

## 1. What is a geodesic on a sphere?

A geodesic on a sphere is the shortest path between two points on the surface of a sphere. It is the equivalent of a straight line on a flat surface.

## 2. How are geodesics on a sphere calculated?

Geodesics on a sphere can be calculated using various mathematical equations, such as the Haversine formula or the Vincenty formula. These equations take into account the curvature of the sphere and the coordinates of the two points to determine the shortest path between them.

## 3. Why are geodesics important in studying spheres?

Geodesics are important in studying spheres because they represent the shortest distance between two points on the surface. This has applications in navigation, cartography, and other fields that involve measuring distances on a curved surface.

## 4. Can geodesics on a sphere be curved?

Yes, geodesics on a sphere can be curved. This is due to the fact that a sphere is a curved surface, and the shortest path between two points on a curved surface is not always a straight line. Geodesics can also be great circles, which are curved paths that follow the circumference of the sphere.

## 5. How do geodesics on a sphere differ from geodesics on other curved surfaces?

Geodesics on a sphere differ from geodesics on other curved surfaces in that they are always closed curves. This means that they form a complete loop and return to their starting point. On other curved surfaces, geodesics may be open curves that do not form a complete loop.

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