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  pdftitle={On Geodesics and Geodesic Triangles (Additional Resources)},
  pdfauthor={by jason m osborne},
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\title{On Geodesics and Geodesic Triangles (Additional Resources)}
\author{by jason m osborne}
\date{}

\begin{document}
\maketitle

{
\setcounter{tocdepth}{2}
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}
\hypertarget{on-geodesics-and-geodesic-triangles-on-a-sphere-additional-resources}{%
\section{On Geodesics and Geodesic Triangles on a Sphere (Additional Resources)}\label{on-geodesics-and-geodesic-triangles-on-a-sphere-additional-resources}}

\textbf{In these additional resources to the post \href{https://jason-m-osborne.bitbucket.io/output/posts/GeodesicTriangle.html}{Geodesic Triangles on a Sphere} we illustrate the interest that Differential Geometers (those folks who like to mix Calculus with Geometry) have with the concepts of curvature and geodesics. We illustrate in \protect\hyperlink{the-big-resulta-euclidean-limit}{\emph{The Big Result\ldots A Euclidean Limit}} the idea of curvature as that deviant quality inherent to a sphere which deforms the standard Euclidean result requiring that angles of a triangle add to \(180^{\circ}\). We also see in \protect\hyperlink{geodesicson-a-sphere}{\emph{Geodesics\ldots On a Sphere}} that defining the Differential Geometric equivalent of straight lines (geodesics), the mathematical and computational complexities increase drastically. For example, in \protect\hyperlink{the-geometry-of-a-surface-of-revolutionparticularly-the-sphere}{\emph{The Geometry of a Surface of Revolution\ldots Particularly the Sphere}} it is necessary for us to work with higher dimensional arrays (i.e.~beyond vectors and matrices) like the Christoffel Symbols \(\Gamma^{\sigma}_{\mathbf{\alpha}\beta}\) and the Riemann Curvature Tensor \(R^{\sigma}_{\mathbf{\alpha}\beta\kappa}\). To correctly handle computations involving these objects is a major reason I wrote the Maple package \emph{tensorAddOns}. This is the first post which illustrates the Maple package \emph{tensorAddOns}. In these Additional Resources we also address}

\begin{itemize}
\item
  \textbf{Two parameterizations of the Great Circle between two points \(P=r(\theta,\phi)\) and \(Q=r(\Theta,\Phi)\) on a sphere are shown in \protect\hyperlink{equation-of-a-great-circletrig}{\emph{Equation of a Great Circle\ldots Trig}} and \protect\hyperlink{equation-of-a-great-circlestretching}{\emph{Equation of a Great Circle\ldots Stretching}} }
\item
  \textbf{\protect\hyperlink{a-geodesic-triangleand-its-area-by-stokes-theorem}{\emph{A Geodesic Triangle\ldots and It's Area by Stoke's Theorem}} which gives the details behind using Stoke's Theorem and three line integrals to evaluate the area of a Geodesic Triangle.}
\item
  \textbf{\protect\hyperlink{describing-geodesics-qualitativelyclairaut-relations}{\emph{Describing Geodesics Qualitatively\ldots Clairaut Relations}} which illustrates how a conservation law on the sphere shows that Great Circles on a sphere are actually Geodesics.}
\end{itemize}

\emph{Last Updated: Wednesday, August 09, 2023 - 23:10:09.}

\begin{center}\rule{0.5\linewidth}{0.5pt}\end{center}

\newpage

\hypertarget{what-has-gone-on-before}{%
\subsection{\texorpdfstring{\emph{What Has Gone On Before\ldots{}}}{What Has Gone On Before\ldots{}}}\label{what-has-gone-on-before}}

In the post \href{https://jason-m-osborne.bitbucket.io/output/posts/CostOfCurvature.html}{\textbf{Cost of Curvature}} the area \(\mathbb{A}\) of a \(\mathbb{L} \times \mathbb{W}\) rectangle on a sphere of any radius \(\rho\) and any starting latitude \(\mathbb{M}\) was shown to be

\begin{align}
\mathbb{A}&= \int_{\phi=\mathbb{M}}^{\phi=\frac{\mathbb{L}+\mathbb{M}\rho}{\rho}}\int_{\theta=0}^{\theta=\frac{\mathbb{W}\csc(\mathbb{M})}{\rho}} \rho^{2}\sin(\phi)\; d\theta\; d\phi\\
&=\rho \mathbb{W} \csc(\mathbb{M})\left(\cos(\mathbb{M})-\cos\left(\frac{\mathbb{L}}{\rho}+\mathbb{M}\right)  \right).
\end{align}

Specifically the area of a \(\mathbb{L}=1\) by \(\mathbb{W}=1\) square on a sphere at starting latitude \(\mathbb{M}=0.616\) is

\begin{align}
\mathbb{A}&=1.731\rho\left(0.816-\cos\left(\frac{1}{\rho}+0.616 \right)  \right).
\end{align}

As seen in Figure \ref{fig:GeoTriSquareResult1a}, a result of this formula is that the area of a \(1 \times 1\) square on a sphere approaches \(A=1\) as the radius of the sphere \(\rho\) increases.

Note that what we do not see in this \emph{area--of--rectangles--on--a--sphere result} is an ideal formula that simplifies to something like

\begin{align}
\mathbb{A}&=\mathbb{L}\mathbb{W}+\mathbb{N}(\rho),
\end{align}

where \(\mathbb{N}(\rho)\) is some \(\mathbb{N}\)umber that changes with the radius \(\rho\) and diminishes to \(0\) as \(\rho\) gets larger. A formula of this type would be ideal for the following reason:

\begin{itemize}
\tightlist
\item
  The known area of a \(\mathbb{L} \times \mathbb{W}\) rectangle drawn on a flat sheet of paper (a Euclidean rectangle) is \(\mathbb{A}=\mathbb{L}\mathbb{W}\). If the area of a \(\mathbb{L} \times \mathbb{W}\) rectangle drawn on a sphere of radius \(\rho\) were in fact \(\mathbb{A}=\mathbb{L}\mathbb{W}+\mathbb{N}(\rho)\) such that \(\mathbb{N}(\rho)\rightarrow 0\) as \(\rho\rightarrow \infty\) then this would conveniently show that the \emph{Euclidean Limit} of the rectangle area on the sphere is the Euclidean rectangle area.
\end{itemize}

Unfortunately, this ideal result is not reality. Instead we have the complicated formula for rectangular area on a sphere

\begin{align}
\mathbb{A}&=\rho \mathbb{W} \csc(\mathbb{M})\left(\cos(\mathbb{M})-\cos\left(\frac{\mathbb{L}}{\rho}+\mathbb{M}\right)  \right) \\
&\neq \mathbb{L}\mathbb{W}+\mathbb{N}(\rho).
\end{align}

In this post, we illustrate that by switching to areas of (geodesic) triangles on a sphere then there is an ideal formula

\begin{align}
\pi+\mathbb{N}(\rho)=\pi+\frac{\mathbb{A}}{\rho^{2}},
\end{align}

where \(\mathbb{A}\) is the area of a (geodesic) triangle from which a natural \emph{Euclidean Limit} will follow and which illustrates curvature as that deviant quality inherent to a sphere that deforms the standard Euclidean result requiring angles of a triangle add to \(180^{\circ}\). To approach this formula requiresthe introduction of new idea called a \emph{geodesic}.



\begin{figure}

{\centering \includegraphics{./Images/CostOfCurvature-SquareResult2b} 

}

\caption{\textbf{The area of a \(1 \times 1\) square on a sphere approaches \(A=1\) as the radius of the sphere \(\rho\) increases.}}\label{fig:GeoTriSquareResult1a}
\end{figure}

\hypertarget{recall-the-definitions}{%
\subsection{\texorpdfstring{\emph{Recall the Definitions\ldots{}}}{Recall the Definitions\ldots{}}}\label{recall-the-definitions}}

Mathematically, we can describe a sphere of radius \(\rho\) parametrically as

\begin{align}
r(\theta,\phi)=\left(\rho\cos(\theta)\sin(\phi),\rho\sin(\theta)\sin(\phi),\rho\cos(\phi)\right.
\end{align}

In this parameterization, seen in Figure \ref{fig:GeoTriDefs1}, \(\theta^{R}\) is the angle (in radians) from the Prime Meridian to a Line of Longitude while the \(\phi^{R}\) is the angle (in radians) from the North Pole to a Line of Latitude. The value \(\psi^{\circ}=90^{\circ}-\phi^{\circ}\) is therefore the angle in degrees from the Equator to a Line of Latitude. The value \(\psi^{\circ}=90^{\circ}-\phi^{\circ}\) can also be called the \(\psi^{\circ}\)th Parallel.

The \emph{Meridians} \(r(\phi)=r(k,\phi)\) are the circles of constant Longitude \(\theta=k\) and the \emph{Parallels} \(r(\theta)=r(\theta,c)\) are the circles of constant Latitude \(\phi=c\).



\begin{figure}

{\centering \includegraphics{./Images/GeodesicTriangle-Def1} 

}

\caption{\textbf{Special Curves On a Sphere: Meridians and Parallels}}\label{fig:GeoTriDefs1}
\end{figure}

\hypertarget{and-the-geodesics-on-a-sphere-are}{%
\subsection{\texorpdfstring{\emph{And the Geodesics on a Sphere Are\ldots{}}}{And the Geodesics on a Sphere Are\ldots{}}}\label{and-the-geodesics-on-a-sphere-are}}

Without yet getting into technical details, let's state that the geodesics on a sphere, seen in Figure \ref{fig:GeoTriGeodesics}, are

\begin{itemize}
\tightlist
\item
  The Equator (a special Parallel)
\item
  All Lines of Longitude (including the Prime Meridian)
\item
  And any \emph{Great Circle} between two points \(P\) and \(Q\) on the sphere. A great circle can be described as the circle of intersection of the sphere and a special plane obtained from any two distinct points on the sphere \(P\) and \(Q\) and the center of the sphere \(O\).
\end{itemize}



\begin{figure}

{\centering \includegraphics{./Images/GeodesicTriangle-GeodesicExamples} 

}

\caption{\textbf{Examples of Geodesics on a Sphere: The Equator, All Lines of Longitude, and any Great Circle between two points \(P\) and \(Q\).}}\label{fig:GeoTriGeodesics}
\end{figure}

\hypertarget{equation-of-a-great-circletrig}{%
\subsection{\texorpdfstring{\emph{Equation of a Great Circle\ldots Trig}}{Equation of a Great Circle\ldots Trig}}\label{equation-of-a-great-circletrig}}

Here is one series of steps for obtaining a parametrically defined formula \(\mathbf{\alpha}(t)\) of a great circle on a sphere whose formula is

\begin{align}
r(\theta,\phi)=\left(\rho\cos(\theta)\sin(\phi),\rho\sin(\theta)\sin(\phi),\rho\cos(\phi)\right.
\end{align}

\begin{itemize}
\item
  For two latitude (\(\phi\)) and longitude (\(\theta\)) selections \(p=(\theta,\phi)\) and \(q=(\Theta,\Phi)\), find two distinct points on the sphere \(P=r(p)=r(\theta,\phi)\) and \(Q=r(q)=r(\Theta,\Phi)\).
\item
  Find the vectors \(\mathbf{v}\) and \(\mathbf{w}\) from the center of the sphere (say the origin \(O=(0,0,0)\)) defined by
\end{itemize}

\begin{align}
\mathbf{v}&=\mathbf{OP}=P-O,\\
\mathbf{w}&=\mathbf{OQ}=Q-O.
\end{align}

These vectors \(\mathbf{v}\) and \(\mathbf{w}\) both lie in the plane containing the great circle.

\begin{itemize}
\tightlist
\item
  Another, unknown vector \(\mathbf{x}\) in this plane is \(\mathbf{x}=\mathbf{\alpha} \mathbf{v}+\beta \mathbf{w}\) (a linear combination of \(\mathbf{v}\) and \(\mathbf{w}\)). To find the coefficients \(\mathbf{\alpha}\) and \(\beta\) of this linear combination, enforce the requirement that the vector \(\mathbf{x}\) be not only perpendicular to either vector \(\mathbf{v}\) or \(\mathbf{w}\) (choose \(\mathbf{v}\), and so \(\mathbf{x} \cdot \mathbf{v}=0\)), but also that \(\mathbf{x}\) is on the sphere (and so \(|\mathbf{x}|=\sqrt{\mathbf{x}\cdot \mathbf{x}}=\rho\)). This thought process gives two equations
\end{itemize}

\begin{align}
\mathbf{x} \cdot \mathbf{v}&=0,\\
\sqrt{\mathbf{x}\cdot \mathbf{x}}&=\rho,
\end{align}

which can be solved using \href{https://www.wolfram.com/mathematica/}{\textbf{Mathematica}} for the two unknowns \(\mathbf{\alpha}\) and \(\beta\).

With \(\mathbf{\alpha}\) and \(\beta\) so obtained, two perpendicular vectors, \(\mathbf{v}\) and \(\mathbf{x}=\mathbf{\alpha} \mathbf{v}+\beta \mathbf{w}\) (both of length \(\rho\)) are now known and the equation of the great circle containing the points \(P\) and \(Q\) is

\begin{align}
\mathbf{\alpha}(t)&=\cos(t) \mathbf{v}+ \sin(t) \mathbf{x} =\left( x(t) , y(t) , z(t) \right),
\end{align}
where

\begin{eqnarray}
&x(t) =\frac{\rho \cos (\theta) \sin (\phi)(\sqrt{\mathbb{A}(t)} \cos (t)+2 \sin (t) \cos (\Phi) \cos (\phi))}{\sqrt{\mathbb{A}(t)}}\\
&-\frac{2 \rho \sin (t) \sin (\Phi)\left(\sin (\theta) \sin ^{2}(\phi) \sin (\theta-\Theta)+\cos(\Theta)\cos ^{2}(\phi)\right)}{\sqrt{\mathbb{A}(t)}} \\ 
\nonumber \\
\nonumber \\
&y(t) = \frac{\rho \sin (t)\left(2 \sin (\Phi)\left(\cos (\theta) \sin ^{2}(\phi) \sin (\theta-\Theta)-\sin (\Theta) \cos ^{2}(\phi)\right)\right)}{\sqrt{\mathbb{A}(t)}}\\
&+\frac{\rho \sin(t)\left(\sin (\theta) \cos (\Phi) \sin (2 \phi)\right)}{\sqrt{\mathbb{A}(t)}}\\
&+\rho \sin (\theta) \cos (t) \sin (\phi) \\
\nonumber \\
\nonumber \\
&z(t)= \frac{2 \rho \sin (t) \sin (\phi)(\sin (\Phi) \cos (\phi) \cos (\theta-\Theta)-\cos (\Phi) \sin (\phi))}{\sqrt{\mathbb{A}(t)}}\\
&+\rho \cos (t) \cos (\phi)
\end{eqnarray}
where the complicated expression \(\mathbb{A}(t)\) is given by

\begin{eqnarray}
&\mathbb{A}(t)=\sin ^{2}(\Phi)\left(2 \cos (2 \phi) \cos ^{2}(\theta-\Theta)-\cos (2(\theta-\Theta))+3\right)\\
&-2 \sin (2 \Phi) \sin (2 \phi) \cos (\theta-\Theta)+4 \cos ^{2}(\Phi) \sin ^{2}(\phi).
\end{eqnarray}

This paramterization \(\mathbf{\alpha}(t)\) with ``time'' parameter \(t\) is such that \(\mathbf{\alpha}(0)=\mathbf{v}\), \(\mathbf{\alpha}(\pi/2)=\mathbf{x}\), and \(\mathbf{\alpha}(\mathbb{T})=\mathbf{w}\) where

\begin{align}
\mathbb{T}=-\sin ^{-1}\left(\frac{\sqrt{\mathbb{A}(t)}}{2}\right).
\end{align}

That is, at a starting time \(t=0\) an object moving along the curve \(\mathbf{\alpha}(t)\) is at the starting location \(P\) while a time \(t=\pi/2\) in the future places the object at location \(R\). Further, at a time \(t=\mathbb{T}\) in the past the object is at the point \(Q\). Examples of Geodesic Segments \(\mathbf{\alpha}(t)\) of the Great Circles connecting points \(P\) and \(Q\) on the surface of the Earth viewed as a sphere of radius \(1\) are shown in Figures \ref{fig:GeoTriGeodesicEx1} and \ref{fig:GeoTriGeodesicEx2}.

The parameterization \(\mathbf{\alpha}(t)\) involves a tricky expression for the time \(\mathbb{T}\) in that \(\mathbb{A}(t)\) needs to not only be positive (so as to avoid square roots of negative numbers), but also needs to be in the domain of the inverse sine function, which for \href{https://www.wolfram.com/mathematica/}{\textbf{Mathematica}} is \(\mathrm{ArcSin}[x]: -1\leq x \leq 1\). That is, the parameterization of the Geodesic Segment between points \(P=r(p)\) and \(Q=r(q)\) using the Great Circle \(\mathbf{\alpha}(t)\) above is valid only for the points \(p=(\theta,\phi)\) and \(q=(\Theta,\Phi)\) meeting the condition

\begin{align}
0 \leq \mathbb{A}(t) \leq 1.
\end{align}

Due to the complexities of this parameterization it will be a good idea to have an alternate description of a geodesic segment.



\begin{center}\includegraphics{./Images/GeodesicTriangle-GreatCircleEx1a} \end{center}
\begin{figure}

{\centering \includegraphics{./Images/GeodesicTriangle-GreatCircleEx1b} 

}

\caption{\textbf{A Geodesic Segment on the Great Circle Connecting a point \(P\) (off the coast of West Africa) to a point \(Q\) (in Iran)}}\label{fig:GeoTriGeodesicEx1}
\end{figure}



\begin{center}\includegraphics{./Images/GeodesicTriangle-GreatCircleEx1a} \end{center}
\begin{figure}

{\centering \includegraphics{./Images/GeodesicTriangle-GreatCircleEx1b} 

}

\caption{\textbf{A Geodesic Segment on the Great Circle Connecting a point \(P\) (off the coast of Greenland) to a point \(Q\) (in Brazil)}}\label{fig:GeoTriGeodesicEx2}
\end{figure}

\hypertarget{equation-of-a-great-circlestretching}{%
\subsection{\texorpdfstring{\emph{Equation of a Great Circle\ldots Stretching}}{Equation of a Great Circle\ldots Stretching}}\label{equation-of-a-great-circlestretching}}

Another method for finding the equation of a Great Circle, illustrated in Figure \ref{fig:GeoTriStretch}, is simply to stretch/scale each point on the line between the two points \(P=r(\theta,\phi)\), \(Q=r(\Theta,\Phi)\) on a sphere by an appropriate amount to reach the sphere's surface. This process can be outlined as:

\begin{itemize}
\item
  Find the equations of the straight lines between the points \(P\) and \(Q\) using the \emph{vectors} between these points. The vector between the points \(P\) and \(Q\) is \(\mathbf{PQ}=Q-P\) and therefore the line between these points is \(l(t)=P+\mathbf{PQ}\cdot t\) with values of a new parameter \(t\) ranging from \(t=0\) to \(t=1\) since \(l(0)=P\) and \(l(1)=P+(Q-P)\cdot 1=Q\).
\item
  Consider any point \(A\) on the line \(l(t)\). The vector \(\mathbf{A}=\mathbf{OA}=A-O\) from the center of the sphere \(O\) to this point \(A\) has some length \(|\mathbf{A}|\) and therefore the vector \(\mathbf{A}/|\mathbf{A}|\) has length 1. It follows that \(\rho\cdot \frac{\mathbf{A}}{|\mathbf{A}|}\) has length \(\rho\) (the radius of the desired sphere) and therefore is a vector on the sphere. Similarly, a point \(B\) on the line \(l(t)\) can be stretched to the sphere as \(\rho \mathbf{B}/|\mathbf{B}|\).
\item
  The Geodesic Segments between the points \(P\) and \(Q\) can then be found by stretching each vector on the line \(\mathbf{l}(t)\) to a new curved line \(L(t)=\rho \frac{\mathbf{l(t)}}{|\mathbf{l(t)}|}\). This process can be repeated to find two more geodesic segments, seen in Figure \ref{fig:GeoTriStretch}, between points \(P\) and \(R\) and \(Q\) and \(R\). These three geodesic segments make up the three edges of what is called a \emph{Geodesic Triangle}.
\end{itemize}

In general the equation of the Geodesic Segment between points \(P=r(\theta,\phi)\) and \(Q=r(\Theta,\Phi)\) obtained by the ``stretching-a-line-method'' is

\begin{align}
\beta(t)&= \left( x(t), y(t), z(t) \right)
\end{align}

where

\begin{align}
x(t)&=\frac{\rho^2(t \cos (\Theta) \sin (\Phi)-(t-1) \cos (\theta) \sin (\phi))}{\sqrt{\mathbb{B}(t)}} \\
\nonumber \\
y(t)&=\frac{\rho^2(t \sin (\Theta) \sin (\Phi)-(t-1) \sin (\theta) \sin (\phi))}{\sqrt{\mathbb{B}(t)}} \\
\nonumber \\
z(t)&= \frac{\rho^2(t \cos (\Phi)-t \cos (\phi)+\cos (\phi))}{\sqrt{\mathbb{B}(t)}}
\end{align}

and where the complicated function \(\mathbb{B}(t)\) is given by

\begin{eqnarray}
&\mathbb{B}(t)=\rho^2\left(((t-1) \sin (\theta) \sin (\phi)-t \sin (\Theta) \sin (\Phi))^2\right) \\
&+ \rho^{2}\left(((t-1) \cos (\theta) \sin (\phi)-t \cos (\Theta) \sin (\Phi))^2\right) \\
&+\rho^{2}\left((t \cos (\Phi)-t \cos (\phi)+\cos (\phi))^2\right).
\end{eqnarray}

In this ``stretch''-parameterization, the parameter \(t\) takes on the values \(t=0 \ldots 1\).



\begin{figure}

{\centering \includegraphics{./Images/GeodesicTriangle-StretchedRegionParam} 

}

\caption{\textbf{A Stretchy Idea: (1) The vector between the points \(P\) and \(Q\) is \(\mathbf{PQ}=Q-P\) and therefore the line between these points is \(l(t)=P+\mathbf{PQ}\cdot t\) with values of a new parameter \(t\) ranging from \(t=0\) to \(t=1\). (2) Pick a point \(A\) on the line \(l(t)\). The vector \(\mathbf{A}=\mathbf{OA}=A-O\) to this point \(A\) has some length \(|\mathbf{A}|\) and therefore the vector \(\mathbf{A}/|\mathbf{A}|\) has length 1. It follows that \(\rho\cdot \frac{\mathbf{A}}{|\mathbf{A}|}\) has length \(\rho\) (the radius of the desired sphere) and therefore is a vector on the sphere. (3) The edge of the geodesic triangle between the points \(P\) and \(Q\) can then be found by stretching each vector on the line \(l(t)\) to a new curved line \(L(t)=\rho \frac{\mathbf{l(t)}}{|\mathbf{l(t)}|}\). }}\label{fig:GeoTriStretch}
\end{figure}

Examples of Great Circles on the Earth (viewed as a sphere of radius 1) can be seen in Figures \ref{fig:GeoTriGeodesicEx3} and \ref{fig:GeoTriGeodesicEx4}.



\begin{center}\includegraphics{./Images/GeodesicTriangle-GreatCircleEx4a} \end{center}
\begin{figure}

{\centering \includegraphics{./Images/GeodesicTriangle-GreatCircleEx4b} 

}

\caption{\textbf{A Geodesic Segment on the Great Circle Connecting a point \(P\) (in India) to a point \(Q\) (in Australia)}}\label{fig:GeoTriGeodesicEx3}
\end{figure}



\begin{center}\includegraphics{./Images/GeodesicTriangle-GreatCircleEx4a} \end{center}
\begin{figure}

{\centering \includegraphics{./Images/GeodesicTriangle-GreatCircleEx4b} 

}

\caption{\textbf{A Geodesic Segment on the Great Circle Connecting a point \(P\) (in California) to a point \(Q\) (in Chile)}}\label{fig:GeoTriGeodesicEx4}
\end{figure}

\hypertarget{a-geodesic-triangleand-its-area-by-stokes-theorem}{%
\subsection{\texorpdfstring{\emph{A Geodesic Triangle\ldots and It's Area by Stoke's Theorem}}{A Geodesic Triangle\ldots and It's Area by Stoke's Theorem}}\label{a-geodesic-triangleand-its-area-by-stokes-theorem}}

A \emph{Geodesic Triangle} is created by connecting three points \(P\), \(Q\), and \(R\) on a sphere (called the vertices) with segments of Great Circles (called the edges) as shown in Figure \ref{fig:GeoTriangle}. We now have two choices for describing the edges of a geodesic triangle:

\begin{itemize}
\tightlist
\item
  either \(\mathbf{\alpha}(t)\) from Section \textbf{\protect\hyperlink{equation-of-a-great-circletrig}{\emph{Equation of a Great Circle\ldots Trig}}}
\item
  or \(\beta(t)\) from Section \textbf{\protect\hyperlink{equation-of-a-great-circlestretching}{\emph{Equation of a Great Circle\ldots Stretching}}}.
\end{itemize}



\begin{figure}

{\centering \includegraphics{./Images/GeodesicTriangle-StokesOnly1} 

}

\caption{\textbf{A Geodesic Triangle: The measure of the angles of this triangle (for example, \(\theta_{2}\)) can be found from the (tangent) vectors to the edges of the triangle (say, \(\mathbf{v}_{2}\) and \(\mathbf{w}_{2}\)) using the formula \(\theta_{2}=\mbox{arccos}\left(\frac{\mathbf{v}_{2}\cdot \mathbf{w}_{2}}{|\mathbf{v}_{2}||\mathbf{w}_{2}|}\right)\). The area of the geodesic triangle can be determined using Stokes's Theorem (computing three Line Integrals along the edges \(\mathbf{\alpha}(t),\beta(t)\) and \(\sigma(t)\)) or a Surface Area Integral.}}\label{fig:GeoTriangle}
\end{figure}

Recall that Stokes's Theorem relates a ``Flux'' integral to a ``Work'' integral. Specifically, the flux of the curl of a vector field \(\mathbf{F}\) through a part of a surface \(S\) parameterized by \(r(u,v)\) is equal to the work done by the \(\mathbf{F}\) along the boundary curve of \(\partial S\) (parameterized by \(\mathbf{r}(t)\) (counter-clockwise) ) of the surface according to the formula

\begin{align}
\int\int_{S} \left((\nabla \times \mathbf{F})|_{\mathbf{r}(u,v)} \cdot (\mathbf{r}_{u} \times \mathbf{r}_{v})\right)\; du\; dv &= \int_{\partial S}\mathbf{F}|_{\mathbf{r}(t)} \cdot \mathbf{r}_{t} \; dt. 
\end{align}

Recalling also that the definition \(\mathbf{N}\) of the unit normal vector to a surface \(\mathbf{r}(u,v)\) is

\begin{align}
\mathbf{N}&=\frac{\mathbf{r}_{u} \times \mathbf{r}_{v}}{|\mathbf{r}_{u} \times \mathbf{r}_{v}|}\cdot \mathbf{r}_{u} \times \mathbf{r}_{v}
\end{align}

then Stoke's Theorem can be conveniently written as

\begin{align}
\int\int_{S} \left((\nabla \times \mathbf{F})|_{\mathbf{r}(u,v)} \cdot \mathbf{N}\right)\; |\mathbf{r}_{u} \times \mathbf{r}_{v}| \; du\; dv &= \int_{\partial S}\mathbf{F}|_{\mathbf{r}(t)} \cdot \mathbf{r}_{t} \; dt. 
\end{align}

What is convenient here is the recognition that if we can find a vector field on the surface \(\mathbf{F}|_{\mathbf{r}(u,v)}\) for which \((\nabla \times \mathbf{F})|_{\mathbf{r}(u,v)} \cdot \mathbf{N}=1\) then the left-hand-side of Stoke's Theorem reduces to a Surface Area Integral

\begin{align}
\int \int_{S} \left(1 \right)\;|\mathbf{r}_{u} \times \mathbf{r}_{v}| \; du\; dv &= \int_{\partial S}\mathbf{F}|_{\mathbf{r}(t)} \cdot \mathbf{r}_{t} \; dt, 
\end{align}

That is, for a special choice of a vector field \(\mathbf{F}\) then the surface area of \(S\) can be obtained by a Line Integral along the boundary curve \(\partial S\). On a sphere described in general as a level surface \(f(x,y,z)=c\) and specifically by \(x^{2}+y^{2}+z^{2}=\rho^{2}\) then the (unit) normal vector to the sphere is

\begin{align}
\mathbf{N}&=\frac{\nabla f}{|\nabla f|}=\frac{\mathrm{grad}(f)}{|\mathrm{grad}(f)|}\\
&=\left(\begin{array}{l}
\frac{2 x}{\sqrt{4 x^2+4 y^2+4 z^2}} \\
\frac{2 y}{\sqrt{4 x^2+4 y^2+4 z^2}} \\
\frac{2 z}{\sqrt{4 x^2+4 y^2+4 z^2}}
\end{array}\right).
\end{align}

Note also that by selecting the vector field \(\mathbf{F}\)

\begin{align}
\mathbf{F}(x,y,z) &=\left(\begin{array}{c}
0 \\
\frac{\rho(x+\pi) z}{y^2+z^2} \\
-\frac{\rho(x+\pi) y}{y^2+z^2}
\end{array}\right)
\end{align}

then \(\nabla \times \mathbf{F}=\mathrm{curl}(\mathbf{F})\) is equal

\begin{align}
\nabla \times \mathbf{F}&=\left(\begin{array}{c}
\frac{2 \rho(x+\pi) y^2}{\left(y^2+z^2\right)^2}-\frac{2 \rho(x+\pi)}{y^2+z^2}+\frac{2 \rho(x+\pi) z^2}{\left(y^2+z^2\right)^2} \\
\frac{\rho y}{y^2+z^2} \\
\frac{\rho z}{y^2+z^2}
\end{array}\right),
\end{align}

and therefore the dot product of the \(\mathrm{curl}(\mathbf{F})\) with the (unit) normal to the sphere is

\begin{align}
(\nabla \times \mathbf{F})\cdot \mathbf{N}&=\frac{\rho}{\sqrt{x^2+y^2+z^2}},
\end{align}

which restricts to the surface of the sphere as the desired result,

\begin{align}
(\nabla \times \mathbf{F})|_{x^{2}+y^{2}+z^{2}=\rho^{2}} \cdot \mathbf{N}&=\frac{\rho}{\sqrt{x^2+y^2+z^2}}\biggr \rvert _{x^{2}+y^{2}+z^{2}=\rho^{2}}\\
&=\frac{\rho}{\sqrt{\rho^{2}}}=1.
\end{align}

It is very exciting then that the area enclosed by the geodesic triangle can be computed by the three Line Integrals along the edges of the triangle,

\begin{align}
&\int \int_{S}\;|\mathbf{r}_{u} \times \mathbf{r}_{v}| \; du\; dv = \\
&\int_{\mathbf{\alpha}} \mathbf{F}|_{\mathbf{\alpha}} \cdot \mathbf{\alpha}_{t}\; dt +\int_{\sigma} \mathbf{F}|_{\sigma} \cdot \sigma_{t}\; dt +\int_{\beta} \mathbf{F}|_{\beta} \cdot \beta_{t}\; dt
\end{align}

where, for example, \(\mathbf{\alpha}(t)\) and \(\beta(t)\) are the parameterizations we have discussed so far and \(\sigma(t)\) is any parameterization you wish so long as that parameterization proceeds from \(Q\) to \(R\) ( which fits into the grand scheme of progressing around the boundary \(\partial S\) by starting at \(P\) proceeding to \(R\) to \(Q\) and back to \(P\) in a counter-clockwise fashion). To illustrate Stoke's Theorem using the relatively simple region (each edge is a longitude or the equator) shown in Figure \ref{fig:GeoTriangle} consider the curves

\begin{align}
&\mathbf{\alpha}(t)=\mathbf{r}(a(t))=\mathbf{r}(1,t),\;\; t=0 \ldots \pi/2,\\
&\mathbf{\sigma}(t)=\mathbf{r}(s(t))=\mathbf{r}(1+t,\pi/2),\;\; t=0 \ldots 1, \\
&\mathbf{\beta}(t) =\mathbf{r}(b(t))=\mathbf{r}(2,t), \;\; t=\pi/2 \ldots 0,
\end{align}

where, per usual for a sphere

\begin{align}
\mathbf{r}(\theta,\phi)=\left(\rho\cos(\theta)\sin(\phi),\rho\sin(\theta)\sin(\phi),\rho\cos(\phi)\right.
\end{align}

The First Line Integral in complete detail is

\begin{align}
\mathbf{\alpha}(t)=\mathbf{r}(1,t)=\left(\begin{array}{c}
\rho \cos (1) \sin (t) \\
\rho \sin (1) \sin (t) \\
\rho \cos (t)
\end{array}\right),
\end{align}

\begin{align}
\mathbf{F}(\mathbf{\alpha}(t))=\mathbf{F}|_{\mathbf{\alpha}(t)}=\left(\begin{array}{c}
0 \\
\frac{\cos (t)(\rho \cos (1) \sin (t)+\pi)}{\sin ^2(1) \sin ^2(t)+\cos ^2(t)} \\
-\frac{\sin (1) \sin (t)(\rho \cos (1) \sin (t)+\pi)}{\sin ^2(1) \sin ^2(t)+\cos ^2(t)}
\end{array}\right),
\end{align}

\begin{align}
\mathbf{F}(\mathbf{\alpha}(t))\cdot \frac{d\mathbf{\alpha}}{dt}=\mathbf{F}|_{\mathbf{\alpha}(t)}\cdot \mathbf{\alpha}_{t}=\frac{\rho \sin (1)(\rho \cos (1) \sin (t)+\pi)}{\sin ^2(1) \sin ^2(t)+\cos ^2(t)},
\end{align}

\begin{align}
\int_{t=0}^{t=\pi/2}\mathbf{F}|_{\mathbf{\alpha}(t)}\cdot \mathbf{\alpha}_{t}\; dt=\frac{1}{2} \rho\left(\pi^2+(-2+\pi) \rho\right).
\end{align}

Similarly, the two other Line Integrals are

\begin{align}
&\int_{t=0}^{t=1}\mathbf{F}|_{\mathbf{\sigma}(t)}\cdot \mathbf{\sigma}_{t}\; dt=0\\
&\int_{t=\pi/2}^{t=0}\mathbf{F}|_{\mathbf{\beta}(t)}\cdot \mathbf{\beta}_{t}\; dt= -\frac{1}{2} \rho\left(\pi^2-4 \rho+\pi \rho\right).
\end{align}

It follows that the surface area of the geodesic triangle shown on the sphere of radius \(\rho=1\) is

\begin{align}
\mathbb{A}&= \int \int_{S}\;|\mathbf{r}_{u} \times \mathbf{r}_{v}| \; du\; dv = \\
&\frac{1}{2} \rho\left(\pi^2+(-2+\pi) \rho\right)+ 0 -\frac{1}{2} \rho\left(\pi^2-4 \rho+\pi \rho\right)=\rho^{2}=1.
\end{align}

\hypertarget{technical-difficulties-in-stokes-theorem}{%
\subsection{\texorpdfstring{\emph{Technical Difficulties in Stoke's Theorem\ldots{}}}{Technical Difficulties in Stoke's Theorem\ldots{}}}\label{technical-difficulties-in-stokes-theorem}}

When trying something new (like Stoke's Theorem to compute Area as seen in Figure \ref{fig:GeoTriangle}), it is a good idea to start with an example that you already know the answer too. When writing this post I already knew the result

\begin{align}
\Sigma_{i=1}^{3}\theta_{i}=\pi+\frac{A}{\rho^{2}},
\end{align}
which is a consequence of the Gauss-Bonnet Theorem (see \emph{Barrett Oneill, Elementary Differential Geometry, revised Second Edition, pg. 379}). Understanding this result is the main focus of this post.

This result in the case \(\rho=1\) for a geodesic triangle whose edges are two lines of longitude and an arc of the equator (subtended by an angle of \(\theta=1\)) and therefore the region \(NPQ_{1}\) shown in Figure \ref{fig:GeoTriStokesOnly} yields a sum of angles \(4.14\), and consequently,
\begin{align}
\Sigma_{i=1}^{3}\theta_{i}&=\pi+\frac{A}{1^{2}}=\pi+A=4.14 \\
&\implies A=4.14-\pi=1.
\end{align}

This area is easily corrobarated using a Surface Area Integral (in Spherical Coordinates)

\begin{align}
A=\int_{\phi=0}^{\phi=\pi/2} \int_{\theta=0}^{\theta=1} \rho^{2}\sin(\phi)\; d\theta\;d\phi=1.
\end{align}

Both results, corroborate the Stoke's Theorem computation from earlier

\begin{align}
\mathbb{A}&= \int \int_{S}\;|\mathbf{r}_{u} \times \mathbf{r}_{v}| \; du\; dv = \\
&\frac{1}{2} \rho\left(\pi^2+(-2+\pi) \rho\right)+ 0 - \frac{1}{2} \rho\left(\pi^2-4 \rho+\pi \rho\right)=\rho^{2}=1.
\end{align}

Anticipating a time when I would need to compute the area of a region \(NPQ_{2}\) as shown in Figure \ref{fig:GeoTriStokesOnly}, I continued using Stoke's Theorem. While this approach did correctly compute the area in this case, it turns out that the line integrals of Stoke's Theorem along the edges of the geodesic triangle are problematic computationally (from a time expense point of view) if the edge between the two points \(P\) and \(Q_{2}\) is parameterized by \(\mathbf{\alpha}(t)=\cos(t)\mathbf{v}+\sin(t)\mathbf{Q_{2}}\). Recall from a previous Section \textbf{\protect\hyperlink{equation-of-a-great-circletrig}{\emph{Equation of a Great Circle\ldots Trig}}} that in this parameterization \(\mathbf{v}\) is a vector in the plane defined by \(\mathbf{P}\) and \(\mathbf{Q_{2}}\) and that the values of the parameter are \(t=0 ... -\sin^{-1}(\sqrt{\mathbb{A}(t)}/2)\) where \(\mathbb{A}(t)\) is evidently computationally not-nice/fast as the upper limit of the Stokes' Theorem Line Integral. The Line Integral computational speed was improved using the parameterization \(\mathbf{\beta}(t)\) between points \(P\) and \(Q_{2}\) as developed in Section \textbf{\protect\hyperlink{equation-of-a-great-circlestretching}{\emph{Equation of a Great Circle\ldots Stretching}}}, but still somewhat slow. For these parametrization reasons, we go back to the drawing board to come up with a better method for describing and computing the area of a geodesic triangle.



\begin{figure}

{\centering \includegraphics{./Images/GeodesicTriangle-StokesOnly2} 

}

\caption{\textbf{Computational Complexity: Computing area of a geodesic triangle using Stoke's Theorem with one edge being the great circle between the two points \(P\) and \(Q_{2}\) parameterized by \(r(t)=\cos(t)\mathbf{v}+\sin(t)\mathbf{Q_{2}}\) is computationally intensive for a couple of reasons, not least of which being that the upper limit of the Line Integral \(t=-\sin^{-1}(\sqrt{\mathbb{A}(t)}/2)\) where \(\mathbb{A}(t)\). Computational speed was improved using a parameterization \(\mathbf{\beta}(t)\), but still somewhat slow. For these parametrization reasons, we go back to the drawing board to come up with a better method for describing and computing the area of a geodesic triangle.}}\label{fig:GeoTriStokesOnly}
\end{figure}

\hypertarget{region-enclosed-by-a-geodesic-trianglemore-stretching}{%
\subsection{\texorpdfstring{\emph{Region Enclosed by a Geodesic Triangle\ldots More Stretching}}{Region Enclosed by a Geodesic Triangle\ldots More Stretching}}\label{region-enclosed-by-a-geodesic-trianglemore-stretching}}

It is also possible to describe the region enclosed by a geodesic triangle using the stretching idea as seen in Figure \ref{fig:GeoTriRegion}.



\begin{figure}

{\centering \includegraphics{./Images/GeodesicTriangle-Seq2} 

}

\caption{\textbf{A Stretchy Idea: (4) The equation of the flat plane is \(l(t,s)=(1-t-s)P+Qt+Rs\) and therefore the region enclosed by the geodesic triangle can be described parametrically as \(r(t,s)=\rho \cdot \frac{l(t,s)}{|l(t,s)|}\) where \(t=0\ldots 1\) and \(s=0 \ldots 1-t\). The area of a the geodesic triangle can be obtained from a Surface Area Integral \(\int_{t=0}^{t=1} \int_{s=0}^{s=1-t} \left|\frac{\partial r}{\partial t} \times \frac{\partial r}{\partial s}\right| \; ds \;dt\).}}\label{fig:GeoTriRegion}
\end{figure}

\begin{itemize}
\tightlist
\item
  The equation of the flat plane between the points \(P\), \(Q\), and \(R\) (given by the black triangle in Figure \ref{fig:GeoTriRegion}). The equation of this flat plane is, for example, given parameterically as
\end{itemize}

\begin{align}
\mathbf{l}(t,s)&=P+\mathbf{PQ}\cdot t+\mathbf{PR}\cdot s\\
&= P+(Q-P)\cdot t + (R-P)\cdot s\\
&= P-Pt-Ps+Qt+Rs \\
&=(1-t-s)P+Qt+Rs.
\end{align}

\begin{itemize}
\tightlist
\item
  Further, this flat plane \(\mathbf{l}(t,s)\) can be stretched to the surface of the sphere (the geodesic triangle region) as
\end{itemize}

\begin{align}
\mathbf{r}(t,s)=\rho\cdot \frac{\mathbf{l}(t,s)}{|\mathbf{l}(t,s)|}
\end{align}

where the parameters \(t\) and \(s\) take on the values \(t=0 \ldots 1-s\) and \(s=0 \ldots 1\). The region enclosed by a geodesic triangle is illustrated in Figure \ref{fig:GeoTriRegion}. The surface area of this parameterized regioin \(\mathbf{r}(t,s)\) can now be obtained from a Surface Area Integral

\begin{align}
\mathbb{A}&=\int_{t=0}^{t=1} \int_{s=0}^{s=1-t} \left|\frac{\partial \mathbf{r}}{\partial t} \times \frac{\partial \mathbf{r}}{\partial s}\right| \; ds \;dt.
\end{align}

For example, for the geodesic triangle shown in Figure \ref{fig:GeoTriRegion}, with vertices

\begin{eqnarray}
P &=\mathbf{r}(p)=\begin{pmatrix} 0 \\ 0 \\ 5  \end{pmatrix}\;\; p=\begin{pmatrix} \theta \\ \phi \end{pmatrix}=\begin{pmatrix} \theta \\ 0 \end{pmatrix}, \\ \\
Q &=\mathbf{r}(q)=\begin{pmatrix} 2.4 \\ 3.74 \\ 2.29 \end{pmatrix} \;\; q=\begin{pmatrix} \theta \\ \phi \end{pmatrix}=\begin{pmatrix} 1 \\ 1.09 \end{pmatrix}, \\
R &=\mathbf{r}(r)=\begin{pmatrix} -1.73 \\3.79 \\2.77\end{pmatrix} \;\; r=\begin{pmatrix} \Theta \\ \Phi \end{pmatrix}=\begin{pmatrix} 2 \\ 0.984 \end{pmatrix},
\end{eqnarray}

then the integrand of the Surface Area Integral is obtained in \href{https://www.wolfram.com/mathematica/}{\textbf{Mathematica}} as

\begin{align}
&\left|\frac{\partial \mathbf{r}}{\partial t} \times \frac{\partial \mathbf{r}}{\partial s}\right|= \mathrm{Sqrt[Cross[D[r[t,s],t],D[r[t,s],s]].Cross[D[r[t,s],t],D[r[t,s],s]]]}\\
&=15.6 \sqrt{\frac{1}{\left(0.892 s^2+1.69 t s-0.892 s+2 . t^2-2 . t-0.916 t(0.446 s+t-1 .)+1 .\right)^3}}
\end{align}

and the value of the Surface Area Integral itself is obtained from \href{https://www.wolfram.com/mathematica/}{\textbf{Mathematica}} as

\begin{align}
&\mathbb{A}=\int_{s=0}^{s=1} \int_{t=0}^{t=1-s} \left|\frac{\partial \mathbf{r}}{\partial t} \times \frac{\partial \mathbf{r}}{\partial s}\right| \; dt \;ds.\\
&=\mathrm{NIntegrate[} \\
&\mathrm{Integrate[N[\left|\frac{\partial \mathbf{r}}{\partial t} \times \frac{\partial \mathbf{r}}{\partial s}\right| ],\{t,0,1-s\}, Assumptions \rightarrow s > 0],} \\
&\mathrm{\{s,0,1\}]} \\
&=11.5
\end{align}

\hypertarget{geodesic-triangle-with-areaon-the-earth}{%
\subsection{\texorpdfstring{\emph{Geodesic Triangle With Area\ldots On the Earth}}{Geodesic Triangle With Area\ldots On the Earth}}\label{geodesic-triangle-with-areaon-the-earth}}

Moving forward, we assume then that the area of any Geodesic Triangle can be obtained and be placed into context as shown on the Earth in Figure \ref{fig:GeoTriOnEarth}.



\begin{figure}

{\centering \includegraphics{./Images/GeodesicTriangle-OnEarthWithData} 

}

\caption{\textbf{If the Earth was a Sphere of Radius \(\rho=5\)\ldots then this geodesic triangle of area \(11.5\) covers approximately \(3.6\%\) of the Earths surface area \(4\pi \rho^{2}=100\pi \approx 314.15\). Throughout the rest of this post, when Stoke's Theorem is used it is with edges parameterized by \(\mathbf{\beta}(t)\) (the stretching method).}}\label{fig:GeoTriOnEarth}
\end{figure}

\hypertarget{the-big-resulta-euclidean-limit}{%
\subsection{\texorpdfstring{\emph{The Big Result\ldots A Euclidean Limit}}{The Big Result\ldots A Euclidean Limit}}\label{the-big-resulta-euclidean-limit}}

In Figures \ref{fig:GeoTriLimitOne} and \ref{fig:GeoTriLimitTwo} the relationship between the sum of angles of a geodesic triangle on a sphere of radius \(\rho\) with its area is shown to be,

\begin{align}
\Sigma_{i=1}^{3}\theta_{i}=\pi+\frac{A}{\rho^{2}}.
\end{align}

From the Apple Book \href{https://books.apple.com/us/book/on-maps-and-maths-and-making-friends/id1524370100?ign-itscg=30200\&ign-itsct=books_toolbox}{\textbf{A Curvature Story}} we know that the curvature \(\kappa\) of a sphere of radius \(\rho\) is the constant

\begin{align}
\kappa=\frac{1}{\rho^{2}},
\end{align}

from which it follows that

\begin{align}
\Sigma_{i=1}^{3}\theta_{i}=\pi+\kappa A.
\end{align}

As the radius of the sphere increases (\(\rho \rightarrow \infty\)) then the curvature of the sphere decreases (\(\kappa \rightarrow 0\)) which means that, similar to the prior post \href{https://jason-m-osborne.bitbucket.io/output/posts/CostOfCurvature.html}{\textbf{The Cost of Curvature}}, the effects of curvature become negligable on the sum of the angles of a geodesic triangle.



\begin{figure}
\centering
\includegraphics{./Images/GeodesicTriangle-SeqMovieReel1.png}
\caption{\label{fig:GeoTriLimitOne}\textbf{The Geodesic Triangles Star in a Film\ldots As the radius \(\rho\) increases (progressing to the right in the film), the areas increase while the sum of the interior angles \(\theta_{1},\theta_{2}\) and \(\theta_{3}\) decrease. A natural question is what happens to the sum of angles as the radius continues to grow? Does the sum keep decreasing? Or does the sum approach some limit? Since the sum of the angles are \(\Sigma_{i=1}^{3}\theta_{i}=4, 3.86, 3.72, 3.6\) and \(3.49\), it's hard to tell in this film sequence what is going to happen.}}
\end{figure}



\begin{figure}
\centering
\includegraphics{./Images/GeodesicTriangle-SeqMovieReel2.png}
\caption{\label{fig:GeoTriLimitTwo}\textbf{A Limit Appears\ldots The sum of the angles for this segment of film are \(\Sigma_{i=1}^{3}\theta_{i}=3.39,3.31,3.25,3.19\), and \(3.16\). It turns out that each sum \(\Sigma_{i=1}^{3}\theta_{i}=\theta_{1}+\theta_{2}+\theta_{3}\) can also be obtained (without knowing each individual angle \(\theta_{i}\)) if one simply knows the geodesic triangle area and the radius of the sphere on which the triangle is drawn. That is, \(\Sigma_{i=1}^{3}\theta_{i}=\pi+\frac{A}{\rho^{2}}\). Note that the number \(\pi\) is about \(3.14\) which is the value the sum of angles approaches as the radius of the sphere continutes to grow.}}
\end{figure}

This is the Euclidean Limit:

\begin{align}
\Sigma_{i=1}^{3}\theta_{i}&=\pi+\frac{A}{\rho^{2}}=\pi+\kappa A \rightarrow \pi \\ &\mbox{as}\;\; \kappa \rightarrow 0 \;\; \mbox{(or}\;\; \rho \rightarrow \infty ).
\end{align}

That is, as the radius of the sphere increases then any geodesic triangle drawn on this sphere will look more and more like a normal (Euclidean) triangle whose angles are known to sum to \(\pi^{R}\) or \(180^{\circ}\) (see Figure \ref{fig:GeoTriEucLimit} ).



\begin{figure}

{\centering \includegraphics{./Images/GeodesicTriangle-EuclideanLimit} 

}

\caption{\textbf{The Euclidean Limit: On a sphere of large radius, this triangle looks a lot like a normal (Euclidean) triangle. In a normal triangle, the angles add to \(\pi^{R}\) or \(180^{\circ}\) and the side lengths satisfy the Phythagorean Theorem \(a^{2}+b^{2}=c^{2}\).}}\label{fig:GeoTriEucLimit}
\end{figure}

\hypertarget{a-hint-at-some-technical-details-following-from-the-sphere-being-a-surface-of-revolution}{%
\subsection{\texorpdfstring{\emph{A Hint at Some Technical Details Following From the Sphere Being a Surface of Revolution\ldots{}}}{A Hint at Some Technical Details Following From the Sphere Being a Surface of Revolution\ldots{}}}\label{a-hint-at-some-technical-details-following-from-the-sphere-being-a-surface-of-revolution}}

In Figure \ref{fig:GeoTriGeodesics}, examples of geodesics on a sphere were shown without first giving a definition of a geodesic. The reason for this omission is simply that a formal definition requires a lot mathematical and symbological machinery. Rather than continuing to omit these details, I will try to summarize some of the technical issues as these tools and techniques will appear in future posts.

\begin{itemize}
\tightlist
\item
  \textbf{Kernel-Index Notation, Numbering, and Naming} Rather than write the sphere as a parameterized surface of revolution in the notation
\end{itemize}

\begin{align}
r(\theta,\phi)=(f(\phi) \cos (\theta) , f(\phi) \sin (\theta), h(\phi))
\end{align}

where

\begin{align}
f(\phi)=\rho\sin(\phi)\;\; \mbox{and}\;\; h(\phi)=\rho\cos(\phi).
\end{align}

we will write the sphere equivalently as a vector-valued function in kernel-index notation

\begin{align}
F^{i}(q^{\alpha})=\begin{pmatrix} f(\phi) \cos (\theta) \\ f(\phi) \sin (\theta) \\ h(\phi) \end{pmatrix},
\end{align}

with coordinate vector

\begin{align}
q^{\alpha}=\begin{pmatrix} \theta \\ \phi \end{pmatrix}
\end{align}

and

\begin{align}
f(\phi)=\rho\sin(\phi)\;\; \mbox{and}\;\; h(\phi)=\rho\cos(\phi).
\end{align}

This is the kernel-index notation:

\begin{align*}
F^{i}:& \;\;\mbox{the symbol}\;\; F \;\; \mbox{is the kernel and} \\
&\;\;\mbox{and the symbol} \;\; i \;\; \mbox{is the row index}\\
&\;\;\mbox{which runs i=1,2,3}\\
&\\
q^{\alpha}:& \;\; q \;\; \mbox{the kernel} \;\; \alpha \;\; \mbox{the row index}\\ \; &\;\;\mbox{which runs $\alpha$=1,2}\\
\end{align*}

The \emph{upper} index \(i\) refers to the ith-\emph{row} of the vector \(F^{i}\) and the \emph{upper} \(\alpha\) index refers to the \(\alpha\)th-\emph{row} of the vector \(q^{\alpha}\). For example, the 1st (row) entry of the vector \(q^{\alpha}\) is \(q^{1}=\theta\) with 2nd (row) entry \(q^{2}=\phi\) while the 1st (row) entry of the vector \(F^{i}\) (also called the x-component of the vector \(\Psi^{i}\)) is \(x=F^{1}=f(\phi)\cos(\theta)\). The vector \(q^{\alpha}\) is called the coordinate vector with first coordinate \(q^{1}=\theta\) and second coordinate \(q^{2}=\phi\).

\begin{itemize}
\tightlist
\item
  \textbf{Coordinate Curves and Their Mappings to a Surface.} If the coordinates change with a parameter (say time, t) then a curve in the coordinate space (a coordinate curve) can be written as \(q^{\alpha}(t)\) and two vectors can be obtained by differentiation \(q_{t}^{\alpha}(t)\) (the coordinate velocity vector) and \(q_{tt}^{\alpha}(t)\) (the coordinate acceleration vector)
\end{itemize}

\begin{align}
q^{\alpha}(t)=\left(\begin{array}{l}
\theta(t) \\ \phi(t)
\end{array}\right),q_{t}^{\alpha}=\left(\begin{array}{l}
\theta_{t} \\ \phi_{t}
\end{array}\right), q_{tt}^{\alpha}=\left(\begin{array}{l}
\theta_{tt} \\ \phi_{tt}
\end{array}\right)
\end{align}

where, for example \(\phi_{t}=d\phi/dt\) is the first derivative of \(\phi\) with respect to \(t\) and \(\theta_{tt}=d^{2}\theta/dt^{2}\) is the second time derivative of \(\theta\). For example, in coordinates, the Equator is a line

\begin{align}
q^{\alpha}(t)&=\left(\begin{array}{l}
\theta(t) \\ \phi(t)
\end{array}\right)=\left(\begin{array}{l}
t \\ \pi/2
\end{array}\right) \;\; \mbox{for}\;\; t=0\ldots 2\pi \\
q_{t}^{\alpha}&=\left(\begin{array}{l}
\theta_{t} \\ \phi_{t}
\end{array}\right)=\left(\begin{array}{l}
1 \\ 0
\end{array}\right), 
\end{align}

and the Prime Meridian, in coordinates, is also a line

\begin{align}
q^{\alpha}(t)&=\left(\begin{array}{l}
\theta(t) \\ \phi(t)
\end{array}\right)=\left(\begin{array}{l}
0 \\ t
\end{array}\right) \;\; \mbox{for} \;\; t=0\ldots 2\pi\\
q_{t}^{\alpha}&=\left(\begin{array}{l}
\theta_{t} \\ \phi_{t}
\end{array}\right)=\left(\begin{array}{l}
0 \\ 1
\end{array}\right). 
\end{align}

In both cases (the Equator and Prime Meridian) the coordinate acceleration vector is

\begin{align}
q_{tt}^{\alpha}=\left(\begin{array}{l}
\theta_{tt} \\ \phi_{tt}
\end{array}\right)=\left(\begin{array}{l}
0 \\ 0
\end{array}\right).
\end{align}

The coordinate curves \(q^{\alpha}(t)\) are mapped (or drawn) onto the surface by applying the vector function \(F^{i}\) to \(q^{\alpha}(t)\) via composition \(F^{i}(q^{\alpha}(t))\). For example, the Equator maps to the surface according to

\begin{align}
F^{i}(q^{\alpha}(t))&=\begin{pmatrix} \cos (\theta)\sin(\phi) \\ \sin (\theta)\sin(\phi) \\ \cos(\phi) \end{pmatrix}\biggr \rvert_{\theta=t,\phi=\pi/2}\\
&=\begin{pmatrix} \cos (t) \\ \sin (t) \\ 0 \end{pmatrix},
\end{align}

while the Prime Meridian maps as

\begin{align}
F^{i}(q^{\alpha}(t))&=\begin{pmatrix} \cos (\theta)\sin(\phi) \\ \sin (\theta)\sin(\phi) \\ \cos(\phi) \end{pmatrix}\biggr \rvert_{\theta=t,\phi=\pi/2}\\
&=\begin{pmatrix} \sin(t) \\ 0 \\ \cos(t) \end{pmatrix}.
\end{align}

These mappings can be seen in Figures \ref{fig:GeoTriCoordCurveEquator} and Figures \ref{fig:GeoTriCoordCurveMeridian}



\begin{figure}

{\centering \includegraphics{./Images/GeodesicTriangel-CoordinateCurveEquator} 

}

\caption{\textbf{Mapping the Equator in Coordinate Space to Real 3D Space: A point \(P=(0,\pi/2)\) in 2D coordinate space maps to the point \(F^{i}(P)=(1,0,0)\) in 3D space. The black line \(q^{\alpha}(t)=(t,\pi/2)\) in coordinate space goes through the point \(P\) at time \(t=0\) and maps to the black curve in 3D space \(F^{i}(q^{\alpha}(t))=(\cos(t),\sin(t),0)\) (the Equator). We are also seeing the two pink coordinate lines through \(P\) mapped onto the surface as the two pink curves.}}\label{fig:GeoTriCoordCurveEquator}
\end{figure}



\begin{figure}

{\centering \includegraphics{./Images/GeodesicTriangle-CoordinateCurveMeridian} 

}

\caption{\textbf{Mapping the Prime Meridian in Coordinate Space to Real 3D Space: A point \(P=(0,0)\) in 2D coordinate space maps to the point \(F^{i}(P)=(0,0,1)\) in 3D space. The black line \(q^{\alpha}(t)=(t,\pi/2)\) in coordinate space goes through the point \(P\) at time \(t=0\) and maps to the black curve in 3D space \(F^{i}(q^{\alpha}(t))=(\cos(t),\sin(t),0)\) (the Prime Meridian). We are also seeing the two pink coordinate lines through \(P\) mapped onto the surface as the two pink curves (one of the pink curves is a single point at the North Pole.}}\label{fig:GeoTriCoordCurveMeridian}
\end{figure}

\hypertarget{the-geometry-of-a-surface-of-revolutionparticularly-the-sphere}{%
\subsection{\texorpdfstring{\emph{The Geometry of a Surface of Revolution\ldots Particularly the Sphere}}{The Geometry of a Surface of Revolution\ldots Particularly the Sphere}}\label{the-geometry-of-a-surface-of-revolutionparticularly-the-sphere}}

In future posts we will explain in detail the geometric computations of and the meaning of each of the geometric objects shown below. These computations were done using a Maple package I wrote called tensorAddOns. Our goal here to record this data and use what we need to define the geodesic equations on a surface of revolution with specfic application to a sphere.

\begin{itemize}
\tightlist
\item
  Surface of Revolution with Jacobian and Hessian:
\end{itemize}

\begin{align}
&F^{i}(q^{\alpha})=\begin{pmatrix} f(\phi) \cos (\theta) \\ f(\phi) \sin (\theta) \\ h(\phi) \end{pmatrix} \;\; \mbox{with}\;\; q^{\alpha}=\begin{pmatrix} \theta \\ \phi \end{pmatrix}, \\
&F^{i}_{,\alpha}=\frac{\partial F^{i}}{\partial q^{\alpha}}=\left(\begin{array}{cc}
-f(\phi) \sin (\theta) & \left(f_{\phi}\right) \cos (\theta) \\
f(\phi) \cos (\theta) & \left(f_{\phi}\right) \sin (\theta) \\
0 & h_{\phi}
\end{array}\right), \\
&F^{i}_{, \alpha \beta}=\frac{\partial^{2} F^{i}}{\partial q^{\alpha} \partial q^{\beta}}=\left(\begin{array}{c}
{\left(\begin{array}{cc}
-f(\phi) \cos (\theta) & -\left(f_{\phi}\right) \sin (\theta) \\
-\left(f_{\phi}\right) \sin (\theta) & \left(f_{\phi, \phi}\right) \cos (\theta)
\end{array}\right)} \\
{\left(\begin{array}{cc}
-f(\phi) \sin (\theta) & \left(f_{\phi}\right) \cos (\theta) \\
\left(f_{\phi}\right) \cos (\theta) & \left(f_{\phi, \phi}\right) \sin (\theta)
\end{array}\right)} \\
{\left(\begin{array}{cc}
0 & 0 \\
0 & h_{\phi, \phi}
\end{array}\right)}
\end{array}\right)
\end{align}

\begin{itemize}
\tightlist
\item
  Metric Tensors (First Fundamental Form) of a Surface of Revolution:
\end{itemize}

\begin{align}
g_{\alpha\beta}&=\left(\begin{array}{cc}
f(\phi)^{2} & 0 \\
0 & \left(f_{\phi}\right)^{2}+\left(h_{\phi}\right)^{2}
\end{array}\right) \\
g^{\alpha\beta}&=\left(\begin{array}{cc}
\frac{1}{f(\phi)^{2}} & 0 \\
0 & \frac{1}{\left(f_{\phi}\right)^{2}+\left(h_{\phi}\right)^{2}}
\end{array}\right)
\end{align}

\begin{itemize}
\tightlist
\item
  Normal Vector to a Surface of Revolution:
\end{itemize}

\begin{align}
u_{i}=\left(\frac{\cos (\theta)\left(h_{\phi}\right)}{d(\phi)} \frac{\sin (\theta)\left(h_{\phi}\right)}{d(\phi)}-\frac{f_{\phi}}{d(\phi)}\right)
\end{align}

where the denominator function is

\begin{align}
d(\phi)=\sqrt{\left(f_{\phi}\right)^{2}+\left(h_{\phi}\right)^{2}}.
\end{align}

\begin{itemize}
\tightlist
\item
  Second Fundamental Form of a Surface of Revolution:
\end{itemize}

\begin{align}
b_{\alpha\beta}&=u_{i}F_{,\alpha\beta}^{i}\\
&=\left(\begin{array}{cc}
-\frac{f(\phi)\left(h_{\phi}\right)}{\sqrt{\left(f_{\phi}\right)^{2}+\left(h_{\phi}\right)^{2}}} & 0 \\
0 & \frac{-\left(f_{\phi}\right)\left(h_{\phi, \phi}\right)+\left(h_{\phi}\right)\left(f_{\phi, \phi}\right)}{\sqrt{\left(f_{\phi}\right)^{2}+\left(h_{\phi}\right)^{2}}}
\end{array}\right)
\end{align}

\begin{itemize}
\tightlist
\item
  Shape Operator of a Surface of Revolution:
\end{itemize}

\begin{align}
b_{\alpha}^{\beta}&=g^{\sigma \beta} b_{\alpha \sigma}\\
&=\left(\begin{array}{cc}
-\frac{h_{\phi}}{f(\phi) \sqrt{\left(f_{\phi}\right)^{2}+\left(h_{\phi}\right)^{2}}} & 0 \\
0 & \frac{-\left(f_{\phi}\right)\left(h_{\phi, \phi}\right)+\left(h_{\phi}\right)\left(f_{\phi, \phi}\right)}{\left(\left(f_{\phi}\right)^{2}+\left(h_{\phi}\right)^{2}\right)^{3 / 2}}
\end{array}\right)
\end{align}

\begin{itemize}
\tightlist
\item
  Christoffel Symbols on a Surface of Revolution, \(\Gamma_{\alpha \beta}^{\sigma}=\frac{1}{2} g^{\sigma \kappa}\left(g_{\{\alpha \beta, \kappa\}}\right)=\frac{1}{2} g^{\sigma \kappa}\left(-g_{\alpha \beta, \kappa}+g_{\beta \kappa, \alpha}+g_{\kappa \alpha, \beta}\right)\) displayed here as a vector of symmetric matrices (with the upper index \(\sigma\) the vector index)
\end{itemize}

\begin{align}
\Gamma^{\sigma}_{\alpha \beta}=\left(
\begin{gathered}
\left(\begin{array}{cc}
0 & \frac{f_{\phi}}{f(\phi)} \\
\frac{f_{\phi}}{f(\phi)} & 0
\end{array}\right)\\
\left(\begin{array}{cc}
-\frac{\left(f_{\phi}\right) f(\phi)}{\left(f_{\phi}\right)^{2}+\left(h_{\phi}\right)^{2}} & 0 \\
0 & \frac{\left(f_{\phi}\right)\left(f_{\phi, \phi}\right)+\left(h_{\phi}\right)\left(h_{\phi, \phi}\right)}{\left(f_{\phi}\right)^{2}+\left(h_{\phi}\right)^{2}}
\end{array}\right)
\end{gathered}\right).
\end{align}

\begin{itemize}
\tightlist
\item
  The (lowered) Riemann Curvature Tensor on a Surface of Revolution, \(R_{\alpha \beta \rho \eta}=2g_{\sigma \eta}\left(\Gamma_{[\alpha|\rho|, \beta]}^{\sigma}+\Gamma_{[\alpha|\rho|}^{\kappa} \cdot \Gamma_{\beta] \kappa}^{\sigma}\right)\) displayed here as a matrix (indexed by \(\rho\) and \(\eta\)) of anti-symmetric matrices (indexed by \(\alpha\) and \(\beta\))
\end{itemize}

\begin{align}
R_{\alpha \beta \rho \eta}&=
\begin{pmatrix}
\begin{pmatrix}
0 & 0 \\
0 & 0
\end{pmatrix}
&
\begin{pmatrix}
0 & R_{1212} \\
-R_{1212} & 0
\end{pmatrix}\\
\begin{pmatrix}
0 & -R_{1212} \\
R_{1212} & 0
\end{pmatrix}
& 
\begin{pmatrix}
0 & 0 \\
0 & 0
\end{pmatrix}
\end{pmatrix},
\end{align}

where

\begin{align}
R_{1212}&=\operatorname{det}\left(b_{\alpha \beta}\right)\\
&=-\frac{f(\phi)\left(h_{\phi}\right)\left(-\left(f_{\phi}\right)\left(h_{\phi, \phi}\right)+\left(h_{\phi}\right)\left(f_{\phi, \phi}\right)\right)}{\left(f_{\phi}\right)^{2}+\left(h_{\phi}\right)^{2}}.
\end{align}

\begin{itemize}
\tightlist
\item
  Gauss Curvature on a Surface of Revolution, \(K=R_{1212}/det(g_{\alpha\beta})\):
\end{itemize}

\begin{align}
K=-\frac{\left(h_{\phi}\right)\left(-\left(f_{\phi}\right)\left(h_{\phi, \phi}\right)+\left(h_{\phi}\right)\left(f_{\phi, \phi}\right)\right)}{f(\phi)\left(\left(f_{\phi}\right)^{2}+\left(h_{\phi}\right)^{2}\right)^{2}}
\end{align}

\begin{itemize}
\tightlist
\item
  The Geodesic Equations on a Surface of Revolution: \(q_{tt}^{\sigma}+\Gamma_{\alpha \beta}^{\sigma}q_{t}^{\alpha}q_{t}^{\beta}=0\)
\end{itemize}

\begin{align}
&\theta_{tt}+\left(\frac{2f_{\phi}}{f(\phi)}\right)\phi_{t}\theta_{t}=0,\\
&\phi_{tt}+\mathbb{A}(\phi)\cdot \left(\phi_{t}\right)^{2}-\mathbb{B}(\phi)\cdot \left(\theta_{t}\right)^{2}=0,
\end{align}

where

\begin{align}
\mathbb{A}(\phi)&=
\left(\frac{\left(h_{\phi}\right)\left(h_{\phi, \phi}\right)}{\left(f_{\phi}\right)^{2}+\left(h_{\phi}\right)^{2}}+\frac{\left(f_{\phi}\right)\left(f_{\phi, \phi}\right)}{\left(f_{\phi}\right)^{2}+\left(h_{\phi}\right)^{2}}\right),
\end{align}

and

\begin{align}
\mathbb{B}(\phi)&=\left(\frac{\left(f_{\phi}\right) f(\phi)}{\left(f_{\phi}\right)^{2}+\left(h_{\phi}\right)^{2}}\right).
\end{align}

As a demonstration of the tensorAddOns package I wrote in Maple (and a more specific instance of the geometric formulas above), we show screenshots of the calling sequence and output of the code in the case of a sphere. A sphere is surface of revolution

\begin{align}
F^{i}(q^{\alpha})=\begin{pmatrix} f(\phi) \cos (\theta) \\ f(\phi) \sin (\theta) \\ h(\phi) \end{pmatrix},
\end{align}

with functions \(f(\phi)=\rho\sin(\phi)\) and \(h(\phi)=\rho\cos(\phi)\).



\begin{center}\includegraphics{./Images/GeodesicTriangle-tensorAddOns1} \end{center}

\begin{center}\includegraphics{./Images/GeodesicTriangle-tensorAddOns2} \end{center}

\begin{center}\includegraphics{./Images/GeodesicTriangle-tensorAddOns3} \end{center}
\begin{figure}

{\centering \includegraphics{./Images/GeodesicTriangle-tensorAddOns4} 

}

\caption{\textbf{Calling sequence and output of Maple package \emph{tensorAddOns}: Surface Data and Geodesic Equations on a Sphere.}}\label{fig:GeoTriSphereData}
\end{figure}

\hypertarget{geodesicson-a-sphere}{%
\subsection{\texorpdfstring{\emph{Geodesics\ldots On a Sphere}}{Geodesics\ldots On a Sphere}}\label{geodesicson-a-sphere}}

From the computations in Section \textbf{\protect\hyperlink{the-geometry-of-a-surface-of-revolutionparticularly-the-sphere}{\emph{The Geometry of a Surface of Revolution\ldots Particularly the Sphere}}} we see that in the parameterization of the sphere

\begin{align}
F^{i}(q^{\alpha})=\begin{pmatrix} \rho\cos (\theta)\sin(\phi) \\ \rho\sin (\theta)\sin(\phi) \\ \rho\cos(\phi) \end{pmatrix}  \;\; \mbox{where} \;\; q^{\alpha}=\begin{pmatrix} \theta \\ \phi \end{pmatrix},
\end{align}

with coordinate velocity vector and coordinate acceleration vector

\begin{align}
q^{\alpha}(t)=\left(\begin{array}{l}
\theta(t) \\ \phi(t)
\end{array}\right),q_{t}^{\alpha}=\left(\begin{array}{l}
\theta_{t} \\ \phi_{t}
\end{array}\right), q_{tt}^{\alpha}=\left(\begin{array}{l}
\theta_{t, t} \\ \phi_{t, t}
\end{array}\right),
\end{align}

then the Christoffel (Levi-Civita) Symbols are

\begin{align}
\Gamma^{\sigma}_{\alpha \beta}=\left(\begin{array}{cc}
{\left(\begin{array}{cc}
0 & \frac{\cos (\phi)}{\sin (\phi)} \\
\frac{\cos (\phi)}{\sin (\phi)} & 0
\end{array}\right)} \\
{\left(\begin{array}{cc}
-\sin (\phi) \cos (\phi) & 0 \\
0 & 0
\end{array}\right)}
\end{array}\right),
\end{align}

from which it follows that the geodesic equations \(q_{tt}^{\sigma}+\Gamma_{\alpha \beta}^{\sigma}q_{t}^{\alpha}q_{t}^{\beta}=0\) are

\begin{align}
\theta_{tt}+\frac{2 \cos (\phi)}{\sin (\phi)}\left(\phi_{t}\right)\left(\theta_{t}\right)=0 \\
\phi_{tt}-\sin (\phi) \cos (\phi)\left(\theta_{t}\right)^{2}=0.
\end{align}

A geodesic (in coordinates) is a special curve

\begin{align}
c^{\alpha}(t)=\left(\begin{array}{l}
\Theta(t) \\ \Phi(t)
\end{array}\right)
\end{align}

for which \(c^{1}(t)=\Theta(t)\) and \(c^{2}(t)=\Phi(t)\) solve the geodesic equations as a system of ordinary differential equations. The mapping of the curve \(c^{\alpha}(t)\) (geodesic in coordinates) onto the surface via \(F^{i}\) is the curve \(C^{i}(t)=F^{i}(c^{\alpha}(t))\)) which is called the geodesic on the surface.

Recall from Section \textbf{\protect\hyperlink{and-the-geodesics-on-a-sphere-are}{\emph{And the Geodesics on a Sphere Are\ldots{}}}} that the equator can be written as

\begin{align}
c^{\alpha}(t)=\left(\begin{array}{l}
\Theta(t) \\ \Phi(t)
\end{array}\right)=\left(\begin{array}{l}
t \\ \pi/2
\end{array}\right).
\end{align}

Since

\begin{align}
&\Theta_{tt}=0, \;\; \mbox{and} \;\; \Phi_{t}=0 \;\; \mbox{then} \\
&\implies \Theta_{tt}+\frac{2 \cos (\phi)}{\sin (\phi)}\left(\Phi_{t}\right)\left(\Theta_{t}\right)=0 
\end{align}

and

\begin{align}
& \Phi_{tt}=0 \;\; \mbox{and} \;\; \cos(\pi/2)=0 \;\; \mbox{then} \\
&\implies \Phi_{tt}-\sin (\Phi) \cos (\Phi)\left(\Theta_{t}\right)^{2}=0
\end{align}

then it follows that the curve (the Equator)

\begin{align}
C^{i}(t)&=F^{i}(c^{\alpha}(t))=\begin{pmatrix} \cos (\theta)\sin(\phi) \\ \sin (\theta)\sin(\phi) \\ \cos(\phi) \end{pmatrix}\biggr \rvert_{\Theta=t,\Phi=\pi/2}\\
&=\begin{pmatrix} \cos (t) \\ \sin (t) \\ 0 \end{pmatrix},
\end{align}

is a formulaic solution to the geodesic equations on the sphere. Geodesics are the straightest possible (zero-acceleration) curves on a surface. It is often hard to find formulaic solutions like this to the geodesic equations as seen in Figure \ref{fig:GeoTriGeodesicSolve1}. But don't worry numerical solutions can be found as seen in Figures \ref{fig:GeoTriGeodesicSolve2} and \ref{fig:GeoTriGeodesicSolve3}.



\begin{center}\includegraphics{./Images/GeodesicTriangle-GeodesicSolve1a} \end{center}

\begin{center}\includegraphics{./Images/GeodesicTriangle-GeodesicSolve1b} \end{center}

\begin{figure}

{\centering \includegraphics{./Images/GeodesicTriangle-GeodesicSolve1c} 

}

\caption{\textbf{Even Computers Can Have Troubles: Despite having found and demonstrated (by hand) that the equator \((\theta(t),\phi(t))=(t,\pi/2)\) is a solution to the geodesic equations, Mathematica cannot find this solution exactly via DSolve. It's OK\ldots these equations are evidently hard to solve. However, solutions can be found using numerical methods NDSolve as seen in the plots.}}\label{fig:GeoTriGeodesicSolve1}
\end{figure}



\begin{center}\includegraphics{./Images/GeodesicTriangle-GeodesicSolve2a} \end{center}

\begin{figure}

{\centering \includegraphics{./Images/GeodesicTriangle-GeodesicSolve2b} 

}

\caption{\textbf{Guess and Check 1: Solution to the Geodesic Equations Seems to be a Great Circle.}}\label{fig:GeoTriGeodesicSolve2}
\end{figure}



\begin{center}\includegraphics{./Images/GeodesicTriangle-GeodesicSolve3a} \end{center}

\begin{figure}

{\centering \includegraphics{./Images/GeodesicTriangle-GeodesicSolve3b} 

}

\caption{\textbf{Guess and Check 2: Solution to the Geodesic Equations is Confirmed to be Great Circle.}}\label{fig:GeoTriGeodesicSolve3}
\end{figure}

\hypertarget{describing-geodesics-qualitativelyclairaut-relations}{%
\subsection{\texorpdfstring{\emph{Describing Geodesics Qualitatively\ldots Clairaut Relations}}{Describing Geodesics Qualitatively\ldots Clairaut Relations}}\label{describing-geodesics-qualitativelyclairaut-relations}}

Recall that for a surface of revolution (particularly the sphere)

\begin{align}
\mathbf{r}(\theta,\phi)=(f(\phi) \cos (\theta) , f(\phi) \sin (\theta), h(\phi))
\end{align}

with

\begin{align}
f(\phi)=\rho\sin(\phi)\;\; \mbox{and}\;\; h(\phi)=\rho\cos(\phi),
\end{align}

then \(g_{11}=\mathbf{r}_{\theta}\cdot \mathbf{r}_{\theta}=f^{2}(\phi)\) and \(g_{22}=\mathbf{r}_{\phi}\cdot \mathbf{r}_{\phi}=f_{\phi}^{2}+g_{\phi}^{2}\) with \(g_{12}=g_{21}=\mathbf{r}_{\theta}\cdot \mathbf{r}_{\phi}=0\) which builds the metric tensor on a surface of revolution (sphere) as

\begin{align}
g_{\alpha\beta}&=\left(\begin{array}{cc}
f(\phi)^{2} & 0 \\
0 & \left(f_{\phi}\right)^{2}+\left(h_{\phi}\right)^{2}
\end{array}\right) = \left(\begin{array}{cc}
\rho^{2}\sin^{2}(\phi) & 0 \\
0 & \rho^{2}
\end{array}\right).
\end{align}

Borrowing the language from the sphere in particular and applying it to surfaces of revolution in general, the (rotation) angle \(u=\theta\) parameterizes the \emph{Parallels}, the angle \(v=\phi\) parameterizes the \emph{Meridians}, and the vector \(\mathbf{r}_{\theta}\) points tangent to the \emph{Parallels} as seen in Figures \ref{fig:GeoTriSurfRev1} and \ref{fig:GeoTriSurfRev2}.



\begin{figure}

{\centering \includegraphics{./Images/GeodesicTriangle-SurfRev1} 

}

\caption{\textbf{Think of this Surface as a General Surface of Revolution: Rotating a reference curve by an angle \(\theta\) (the Parallel angle) creates a surface. The vectors \(\mathbf{r}_{\theta}\) point tangent to the Parallels but also perpendicular to the Meridians.}}\label{fig:GeoTriSurfRev1}
\end{figure}



\begin{figure}

{\centering \includegraphics{./Images/GeodesicTriangle-SurfRev2} 

}

\caption{\textbf{The Sphere as a Specific Surface of Revolution: Rotating a reference curve (a circle) by an angle \(\theta\) (the Parallel angle) creates a surface. The vectors \(\mathbf{r}_{\theta}\) point tangent to the Parallels but also perpendicular to the Meridians.}}\label{fig:GeoTriSurfRev2}
\end{figure}

The geodesic equations on a surface of revolution (sphere) are

\begin{align}
&\theta_{tt}+\left(\frac{2f_{\phi}}{f(\phi)}\right)\phi_{t}\theta_{t}=0,\\
&\left( \theta_{tt}+\frac{2 \cos (\phi)}{\sin (\phi)}\left(\phi_{t}\right)\left(\theta_{t}\right)=0\right) \\
\nonumber \\
&\phi_{tt}+\mathbb{A}(\phi)\cdot \left(\phi_{t}\right)^{2}-\mathbb{B}(\phi)\cdot \left(\theta_{t}\right)^{2}=0.\\
&\left(\phi_{tt}-\sin (\phi) \cos (\phi)\left(\theta_{t}\right)^{2}=0\right)
\end{align}

Finding a solution to these equations can be attempted by choosing four initial (position and velocity) conditions \(\theta(0)=\theta_{0}\), \(\phi(0)=\phi_{0}\), \(\theta_{t}(0)=\theta'_{0}\), and \(\phi_{t}(0)=\phi'_{0}\).

The geodesic equations are often hard to find non-numerical solutions to! However, if one looks at the sphere as a surface of revolution then the qualitative methods of Clairaut can be used to describe enough of the behavior of geodesics to effectively know a solution.

The key to the whole Clairaut process is to recognize that the \(\theta_{tt}\) geodesic equation can be written as a conservation equation (constant, c)

\begin{align}
&\frac{d}{dt}\left(f^{2}\theta_{t} =c \right),
\end{align}

\begin{align}
\frac{d}{dt}\left(f^{2}\theta_{t}\right)&=2ff_{t}\theta_{t}+\theta_{tt}f^{2}\\
&=2f(f_{\theta}\theta_{t}+f_{\phi}\phi_{t})\theta_{t}+\theta_{tt}f^{2},  \;\;\; (\mbox{but}\; f_{\theta}=0)\\
&=2ff_{\phi}\theta_{t}\phi_{t}+\theta_{tt}f^{2}\\
\frac{d}{dt}\left(  c\right)&=0,\\
&\implies \theta_{tt}+\frac{2f_{\phi}}{f}\theta_{t}\phi_{t}=0.
\end{align}

That is, a coordinate curve \(Q(t)=(\Theta(t),\Phi(t))\) is a geodesic if the function \(f^{2}(\phi)\theta_{t}\) is a constant along this curve, meaning \(f^{2}(\Phi(t))\theta_{t}(t)=c\).

We can gather further information about the constant \(c\) if we look into the angle between a curve on the surface and a parallel. For example, let \(q(t)=(\theta(t),\phi(t))\) be \emph{some} coordinate curve (not necessarily a geodesic) and let the image of this curve on the surface be \(\mathbf{\alpha}(t)=\mathbf{r}(q(t))=\mathbf{r}(\theta(t),\phi(t))\). It follows again from the Chain Rule and Properties of the Dot Product that

\begin{align}
\mathbf{\alpha}_{t}(t)\cdot \mathbf{r}_{\theta}&=(\mathbf{r}_{\theta}\theta_{t}+\mathbf{r}_{\phi}\phi_{t})\cdot \mathbf{r}_{\theta}\\
&=\mathbf{r}_{\theta}\cdot \mathbf{r}_{\theta}\theta_{t}\;\; (\mbox{since} \; g_{12}=\mathbf{r}_{\theta}\cdot \mathbf{r}_{\phi}=0)\\
&=g_{11}\theta_{t} \\
&=f^{2}\theta_{t}.
\end{align}

If the curve \(\mathbf{\alpha}(t)\) is not just some curve on the surface but actually a geodesic \(\mathbf{\alpha}(t)=\mathbf{r}(\Theta(t),\Phi(t))\)
then along this geodesic there is a conserved quantity,

\begin{align}
\mathbf{\alpha}_{t}(t)\cdot \mathbf{r}_{\theta}&=f^{2}(\Phi(t))\theta_{t}=c.
\end{align}

What is interesting is that there is another method of computing the \emph{Dot Product} \(\mathbf{v} \cdot \mathbf{w}\) as \(|\mathbf{v}||\mathbf{w}|\cos(\Psi)\) where \(\Psi\) is the angle between \(\mathbf{v}\) and \(\mathbf{w}\) which will give more information about the constant \(c\). That is,

\begin{align}
\mathbf{\alpha}_{t}(t)\cdot \mathbf{r}_{\theta}&=|\mathbf{\alpha}_{t}(t)||\mathbf{r}_{\theta}|\cos(\Psi)\\
&=|\mathbf{\alpha}_{t}(t)|\sqrt{\mathbf{r}_{\theta}\cdot \mathbf{r}_{\theta}}\cos(\Psi)\\
&=|\mathbf{\alpha}_{t}(t)|\sqrt{g_{11}}\cos(\Psi)\\
&=|\mathbf{\alpha}_{t}(t)|f\cos(\Psi),
\end{align}

where \(\Psi\) is the angle between the path of the curve on the surface and the parallel crossed by this curve. This expression can be simplified further assuming that the curve on the surface \(\mathbf{\alpha}(t)=\mathbf{r}(q(t))=\mathbf{r}(\theta(t),\phi(t))\) has unit speed, meaning \(|\mathbf{\alpha}_{t}(t)|=1\). The unit speed requirement can be expressed as

\begin{align}
|\mathbf{\alpha}_{t}(t)|&=\sqrt{(\mathbf{r}_{\theta}\theta_{t}+\mathbf{r}_{\phi}\phi_{t})\cdot (\mathbf{r}_{\theta}\theta_{t}+\mathbf{r}_{\phi}\phi_{t})}\\
&=\sqrt{\mathbf{r}_{\theta}\cdot \mathbf{r}_{\theta} \theta_{t}^{2}+\mathbf{r}_{\phi}\cdot \mathbf{r}_{\phi}\phi_{t}^{2} } \\
&=\sqrt{g_{11}\theta_{t}^{2} +g_{22}\phi_{t}^{2}}\\
&=1\\
&\implies g_{11}\theta_{t}^{2} +g_{22}\phi_{t}^{2}=1.
\end{align}

Putting all the requirements together leads to a qualitative conclusion about a geodesic \(\mathbf{\alpha}(t)\) with unit speed \(|\mathbf{\alpha}_{t}(t)|=1\) on a surface of revolution

\begin{align}
& \mathbf{r}(\theta,\phi)=(f(\phi) \cos (\theta) , f(\phi) \sin (\theta), h(\phi))\\
\nonumber \\
& (\mbox{sphere,} \; f(\phi)=\rho\sin(\phi) \; \mbox{and} \;  h(\phi)=\rho\cos(\phi))
\end{align}

that is simply stated as a conservation law:

\textbf{A Clairaut Relation:} Along the path \(\mathbf{\alpha}(t)\) of a unit-speed geodesic on the surface of revolution (sphere) \(r(\theta,\phi)\) then

\begin{align}
& f^{2}\theta_{t}=f\cos(\Psi)=c \;\;\mbox{and} \;\; f^{2}\theta_{t}^{2} +\left(\left(f_{\phi}\right)^{2}+\left(h_{\phi}\right)^{2}\right)\phi_{t}^{2}=1, \\
\nonumber \\
& (\; \rho^{2}\sin^{2}(\phi)\theta_{t}=\rho\sin(\phi)\cos(\Psi)=c \;\; \mbox{and}\;\;\rho^{2}\sin^{2}(\phi)\theta_{t}^{2} +\rho^{2}\phi_{t}^{2}=1 )
\end{align}

where \(\Psi\) is the angle between the between the path of the geodesic and the parallel crossed by this geodescic (see Figure \ref{fig:PsiDef}).



\begin{figure}

{\centering \includegraphics{./Images/GeodesicTriangle-PsiDef} 

}

\caption{\textbf{\(\Psi\) is the angle between the path of the geodesic (it's direction of motion) and the parallel crossed by this geodesic. This angle changes with latitude, \(\phi\).}}\label{fig:PsiDef}
\end{figure}

\hypertarget{clairaut-relation-analysison-parallels-and-meridians}{%
\subsection{Clairaut Relation Analysis\ldots On Parallels and Meridians}\label{clairaut-relation-analysison-parallels-and-meridians}}

To begin to get a feel for these Clairaut Relations it is helpful to begin asking which of the coordinate curves (meridians and parallels) are geodesics, since along these curves \(\cos(\Psi)\) will itself be a constant.

\begin{itemize}
\item
  \textbf{All Meridians \(\mathbf{\alpha}(t)\) are geodesics} since the angle between \(\mathbf{\alpha}_{t}(t)\) (the direction the geodesic is pointing) and \(\mathbf{r}_{\theta}\) (the direction of the parallel begin crossed) is \(\Psi=\pi/2 \implies \cos(\Psi)=0\). Along a meridian, we know also that \(\theta_{t}=0\). The conservation law is then \(f^{2}\theta_{t}=f\cos(\Psi)=c\) where \(c=0\).
\item
  \textbf{All Parallels for which \(f_{\phi}=0\) are geodesics} since the angle between \(\mathbf{\alpha}_{t}(t)\) (the direction the geodesic is pointing) and \(\mathbf{r}_{\theta}\) (the direction of the parallel being crossed) is \(\Psi=0 \implies \cos(\Psi)=1\). The conservation law is then \(f(\phi)\cos(\Psi)=f(\phi)=c\), which will hold if the parallel satisifies \(f_{\phi}=0\). The specific value of the constant \(c\) can be obtained from \(f^{2}(\phi)\theta_{t}=c\) once the specific geodesic parallel(s) are known.
\end{itemize}

As an example and in the specific case of the sphere, when seeking the parallels \(0 \leq \phi \leq \pi\) that are geodesics we only need to solve the simple equation,

\begin{align}
&f(\phi)=\rho\sin(\phi) \implies f_{\phi}=\rho\cos(\phi)=0.
\end{align}

The only solution to the equation \(f_{\phi}=0\) on the sphere is \(\phi=\pi/2\) and consequently the only Parallel which is a geodesic on the sphere is the Equator!

Looking a bit deeper, consider the following steps:

\begin{itemize}
\tightlist
\item
  Take into account the parallel one is on (in this case, \(\phi=\pi/2\)) and an initial velocity (\(\phi_{t}=0\), to stay on the Equator) from which the unit speed condition \(\rho^{2}\sin^{2}(\phi)\theta_{t}^{2} +\rho^{2}\phi_{t}^{2}=1\) can be used to solve for \(\theta_{t}\):
\end{itemize}

\begin{align}
&\rho^{2}\sin^{2}(\pi/2)\theta_{t}^{2}+\rho^{2}(0)^{2}=1\\
&\rho^{2}\theta_{t}^{2}=1\\
&\implies \theta_{t}=1/\rho.
\end{align}

\begin{itemize}
\tightlist
\item
  Using the initial conditions \(\phi=\pi/2\), \(\theta_{t}=1/\rho\) one can find the constant of this curve using the equation \(f^{2}(\phi)\theta_{t}=c\)
\end{itemize}

\begin{align}
\rho^{2}\sin^{2}(\pi/2)(1/\rho)=\rho=c
\end{align}

\begin{itemize}
\tightlist
\item
  Now the other conservation equation \(f(\phi)\cos(\Psi)=c\) can be used to gather information about the angle \(\Psi\).
\end{itemize}

\begin{align}
&\rho \sin(\phi)\cos(\Psi)=\rho \\
&\cos(\Psi)=1/\sin(\phi).
\end{align}

We have already show that the coordinate curve \(q(t)=(\theta(t),\phi(t))=(t,\pi/2)\) is a solution to the geodesic equations. It follows that the conservation law \(f(\phi)\cos(\Psi)=c\) along this solution simplifies to

\begin{align}
&\cos(\Psi)=1/\sin(\pi/2)\\
&\implies \cos(\Psi)=1 \implies \Psi=0.
\end{align}

That is, along the Equator, we have shown the unit speed condition and the conservation equations are all consistent with each other.

\hypertarget{clairaut-relation-analysisgreat-circle}{%
\subsection{Clairaut Relation Analysis\ldots Great Circle}\label{clairaut-relation-analysisgreat-circle}}

Follow the same steps above but start on a parallel that is \emph{not} a geodesic like \(\phi_{0}=1\).

\begin{itemize}
\tightlist
\item
  Take into account the parallel you want to start on (in this case, \(\phi_{0}=1\)) and an initial velocity (\(\phi'_{0}=0\), to try to stay on this Parallel) from which the unit speed condition \(\rho^{2}\sin^{2}(\phi)\theta_{t}^{2} +\rho^{2}\phi_{t}^{2}=1\) can be used to solve for \(\theta'_{0}\):
\end{itemize}

\begin{align}
&\rho^{2}\sin^{2}(\phi_{0})\theta_{t}^{2}+\rho^{2}(0)^{2}=1\\
&\rho^{2}\sin^{2}(\phi_{0})\theta_{t}^{2}=1\\
&\implies \theta_{t}=1/(\rho\sin(\phi_{0})).
\end{align}

\begin{itemize}
\tightlist
\item
  Using the initial conditions \(\phi_{0}=1\), \(\theta'_{0}=1/(\rho\sin(1))\) one can find the constant of this curve using the equation \(f^{2}\theta_{t}=c\)
\end{itemize}

\begin{align}
\rho^{2}\sin^{2}(\phi_{0})(1/(\rho\sin(\phi_{0}))=\rho\sin(\phi_{0})=c
\end{align}

\begin{itemize}
\tightlist
\item
  Now the other conservation equation \(f(\phi)\cos(\Psi)=c\) can be used to gather information about the angle \(\Psi\).
\end{itemize}

\begin{align}
&\rho\sin(\phi)\cos(\Psi)=\rho\sin(\phi_{0}),
\end{align}

from which we can conclude that, since the Parallel \(\phi_{0}=1\) is not a geodesic, the solution to the geodesic equations (with initial conditions \(\phi_{0}=1\), \(\theta_{0}=0\), \(\theta'_{0}=1/(\rho\sin(\phi_{0}))\), and \(\phi'_{0}=0\)) will cross other Parallels \(\phi\) and the angle \(\Psi\) will vary according to the equation

\begin{align}
\Psi(\phi)=\cos^{-1}\left( \frac{\sin(\phi_{0})}{\sin(\phi)}  \right).
\end{align}

While this equation is good to know it requires a knowledge \(\phi(t)\) which is possible only if we can solve (formulaically or numerically) the geodesic equations. What is interesting is that by backing up to the Clairaut Relation

\begin{align}
f(\phi)\cos(\Psi)=c \;\; (\;c=\rho\sin(\phi_{0} \;\; \mbox{and}\;\; f(\phi)=\rho\sin(\phi))
\end{align}

and recognizing that, in the case of the sphere \(f(\phi)\) is the radius of the Parallel/Circle of Latitude crossed by the geodesic, information can be gleaned from this conservation law about this geodesic. For example, as seen in Figure \ref{fig:Clairaut1}, observe qualitatively that in \textbf{Case 1 (Northern Hemisphere)}

\begin{itemize}
\item
  By starting on an initial parallel in the Northern Hemisphere \(\phi_{0}=1\) (at any longitude, say \(\theta_{0}=0\)) then by also selecting an initial velocity \(\phi'_{0}=0\) we are guaranteed from the unit speed condition that \(\theta'_{0}=1/(\rho\sin(\phi_{0}))\). This means that, initially, the geodesic is trying to point in the same direction as the Parallel, which implies \(\Psi=0 \implies \sin(\Psi)=1\). Since this is the largest \(\sin(\Psi)\) can be it follows, by conservation, that this is also the smallest that the radius \(f(\phi)\) can be.
\item
  Since the initial (\(t_{0}=0\)) parallel corresponds to the largest value of \(\sin(\Psi)\), in the next moment of time (\(t_{1}>t_{0}\)) along the geodesic, \(\sin(\Psi)\) will have to decrease and by conservation the radius \(f(\phi)\) will have to increase to maintain constancy of \(f(\phi)\sin(\Psi)\). The geometric implication being that, in this next moment, the geodesic will pass to a Parallel \(\phi_{1} > \phi_{0}\) with larger radius (meaning, South towards the Equator).
\item
  At some time (\(t_{2}\)) the radius \(f(\phi)\) will reach it's maximum value of \(\rho\), the radius of the sphere, at \(\phi=\pi/2\) (the Equator) and the angle between the geodesic and parallel \(\Psi\) is a maximum (with \(\cos(\Psi)\) a minimum).
\item
  At some moment of time (\(t_{3}>t_{2}\)) the radius is decreasing and continuing to the Southern Hemisphere since \(\phi_{t}\) is still positive at this time.
\item
  As we continue heading into the Southern Hemisphere, the radius \(f(\phi)\) will keep decreasing until some time \(t_{4}\) when it's smallest value is reached and when the largest value of \(\cos(\Psi)\) is obtained, namely when the angle \(\Psi\) returns to \(0\) and the geodesic is again tangent to the largest parallel.
\item
  This process, summarized in Figure \ref{fig:ClairautSummary}, continues as the geodesic heads back towards the Northern Hemisphere from it's lowest southern point.
\end{itemize}



\begin{center}\includegraphics{./Images/GeodesicTriangle-Clairaut1a} \end{center}

\begin{figure}

{\centering \includegraphics{./Images/GeodesicTriangle-Clairaut1b} 

}

\caption{\textbf{The Conservation Law \(f(\phi)\cos(\Psi)=c\) Explains The Great Circles as Geodesics on a Sphere.}}\label{fig:Clairaut1}
\end{figure}



\begin{center}\includegraphics{./Images/GeodesicTriangle-Clairaut2a} \end{center}

\begin{figure}

{\centering \includegraphics{./Images/GeodesicTriangle-Clairaut2b} 

}

\caption{\textbf{The Conservation Law \(f(\phi)\cos(\Psi)=c\) Explains the Behavior of any Geodesic Starting in the Northern Hemisphere.}}\label{fig:Clairaut2}
\end{figure}



\begin{center}\includegraphics{./Images/GeodesicTriangle-Clairaut3a} \end{center}

\begin{figure}

{\centering \includegraphics{./Images/GeodesicTriangle-Clairaut3b} 

}

\caption{\textbf{The Conservation Law \(f(\phi)\cos(\Psi)=c\) Also Explains the Behavior of any Geodesic Starting in the Southern Hemisphere.}}\label{fig:Clairaut3}
\end{figure}



\begin{figure}

{\centering \includegraphics{./Images/GeodesicTriangle-ClairautSummary} 

}

\caption{\textbf{Qualitative Description of Geodesics on Sphere via the Conservation Law \(f(\phi)\cos(\Psi)=c\). At time the initial time \(t_{0}\), the geodesic starts out tangent to the \(\phi_{0}\) parallel which implies that the angle \(\Psi=0\) and further that \(\cos(\Psi)\) is the largest it can be. By the conservation law, \(f(\phi)\) is the smallest it can be. At some small moment of time \(t_{1} \rangle t_{0}\) later, it follows that \(\cos(\Psi)\) is smaller and therefore, by the conservation law, \(f(\phi)\) is larger to compensate. This implies that the new Parallel \(\phi_{1}\) crossed by the geodesic is further south than Parallel \(\phi_{0}\). The geodesic proceeds south until it passes the Equator (\(\phi=\pi/2\)) at which point \(f(\phi)\) is maximized while \(\cos(\Psi)\) is minimized at the maximum angle \(\Psi\). The geodesic continues to the lowest southern point where again the angle \(\Psi=0\) in which case \(\cos(\Psi)\) is maximized and \(f(\phi)\) is minimized. This qualitative description of the geodesics on a sphere holds for any starting value \(\phi_{0}\) in the Northern Hemisphere as seen in Figure \ref{fig:Clairaut2}. Figure \ref{fig:Clairaut3} shows how this same conservation law applies to geodesics starting in the Southern Hemisphere.}}\label{fig:ClairautSummary}
\end{figure}

\hypertarget{conclusion}{%
\subsection{\texorpdfstring{\emph{Conclusion\ldots{}}}{Conclusion\ldots{}}}\label{conclusion}}

In this post we have tried to illustrate the interest Differential Geometers (those folks that like to mix Calculus with Geometry) have with the concept of curvature and geodesics. The technical and computational challenges are numerous in Differential Geometry. In this post we have illustrated a really interesting formula in \textbf{\protect\hyperlink{the-big-resulta-euclidean-limit}{\emph{The Big Result\ldots A Euclidean Limit}}},

\begin{align}
\Sigma_{i=1}^{3}\theta_{i}=\pi+\frac{A}{\rho^{2}},
\end{align}

that not only defines the curvature of a sphere but also in the same formula shows the implications of this definition (a two for one so to speak). Namely, curvature is that deviant quality inherent to a sphere which deforms the standard Euclidean result that angles of a triangle are required to add to \(180^{\circ}\). We have also seen that in \textbf{\protect\hyperlink{geodesicson-a-sphere}{\emph{Geodesics\ldots On a Sphere}}} that defining the Differential Geometric equivalent of straight lines (the geodesics), the mathematical and computational complexity go up drastically. For example, in \textbf{\protect\hyperlink{the-geometry-of-a-surface-of-revolutionparticularly-the-sphere}{\emph{The Geometry of a Surface of Revolution\ldots Particularly the Sphere}}} it was necessary for us to work with higher dimensional arrays (i.e.~beyond vectors and matrices) like the Christoffel Symbols \(\Gamma^{\sigma}_{\mathbf{\alpha}\beta}\) and the Riemann Curvature Tensor \(R^{\sigma}_{\mathbf{\alpha}\beta\kappa}\). To correctly handle computations involving these objects was a major reason I wrote the Maple package \emph{tensorAddOns}. We will see more on tensors and usage of tensorAddOns package in future posts (see Figure \ref{fig:GeoTriMatrixView}). This post also featured additional technical details about \textbf{\protect\hyperlink{a-geodesic-triangleand-its-area-by-stokes-theorem}{\emph{A Geodesic Triangle\ldots and It's Area by Stoke's Theorem}}} and about \textbf{\protect\hyperlink{describing-geodesics-qualitativelyclairaut-relations}{\emph{Describing Geodesics Qualitatively\ldots Clairaut Relations}}} (particularly on a sphere).



\begin{center}\includegraphics{./Images/GeodesicTriangle-tensorAddOns1} \end{center}

\begin{figure}

{\centering \includegraphics{./Images/GeodesicTriangle-tensorAddOns5} 

}

\caption{More tensors and the Maple \emph{tensorAddOns} package to come in future posts.}\label{fig:GeoTriMatrixView}
\end{figure}

\begin{tcolorbox}
Easter Egg
\end{tcolorbox}

\end{document}
