- #1
George Keeling
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- 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?
I am now reading https://www.physicsforums.com/threads/geodesics-on-a-sphere-and-the-christoffel-symbols.869625/ ...
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?
I am now reading https://www.physicsforums.com/threads/geodesics-on-a-sphere-and-the-christoffel-symbols.869625/ ...
Last edited:
! 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!