Geodesics on the two-sphere

The two-sphere (of radius $R$) can be defined by the following equation in ${\mathbb R}^3$:

(1)
\begin{align} x^2 + y^2 + z^2 = R^2\ . \end{align}

We can solve the equation by working in spherical polar coordinates:

(2)
\begin{align} x=R \sin\theta \cos \varphi\quad,\quad y=R \sin\theta \sin\varphi\quad,\quad z=R \cos\theta\ , \end{align}

where $\theta\in[0,\pi]$ and $\varphi\in [0,2\pi)$.The induced metric on the two-sphere, $S^2$, can be easily shown to be

(3)
\begin{align} ds^2 = R^2(d\theta^2 + \sin^2\theta\ d\varphi^2)\ , \end{align}

in spherical polar coordinates. We set $R=1$ for the rest of the discussion.

Geodesics on the two-sphere

Geodesics are the analog of straight lines in ${\mathbb R}^n$ — they are curves corresponding to the shortest length between any two points. It will be shown that these curves necessarily lie on great circles1. Let us first write out the equation of a great circle in spherical polar coordinates. A great circle can be obtained by intersecting a plane passing through the origin in ${\mathbb R}^3$ with the two-sphere. The equation of such a plane is

(4)
\begin{align} a x + b y + c z =0\ , \end{align}

for some constants $a,\ b,\ c$. A small amount of algebra converts this into the following equation in spherical polar coordinates:

(5)
\begin{align} \cot \theta = A\ \cos (\varphi + \varphi_0)\ , \end{align}

where $A$ and $\varphi_0$ are constants.

The action for the geodesic

Consider the following action for a particle:

(6)
\begin{align} S=\int dt \ \sqrt{\dot{\theta}^2 + \sin^2\theta\ \dot{\varphi}^2}\ . \end{align}

This action has an unusual symmetry i.e., the reparametrisation of ‘time’. Let us write a change of variable given by $t=t(\xi)$. Then, it is easy to see that

(7)
\begin{align} S=\int d\xi \ \sqrt{{\theta'}^2 + \sin^2\theta\ {\varphi'}^2}\ , \end{align}

where the prime indicates differentiation w.r.t. to the variable $\xi$. In other words, one has the freedom to redefine the time variable. Let us, for instance, choose to parametrise time by the azimuthal angle $\varphi$. Then, one has

(8)
\begin{align} S=\int d\varphi \ \sqrt{{\theta'}^2 + \sin^2\theta}\ , \end{align}

where now we consider $\theta(\varphi)$ and the prime indicates differentiation w.r.t. $\varphi$. Thus, the system now effectively has one degree of freedom. The Euler-Lagrange equation of motion is given by

(9)
\begin{align} \sin\theta\ \theta'' - 2 \cos\theta (\theta')^2 = (\sin\theta)^2\ \cos\theta\ . \end{align}

This can be rewritten as

(10)
\begin{align} \frac{d^2\cot \theta}{d\varphi^2}=- \cot\theta\ . \end{align}

It is now easy to see that $\cot \theta = A\ \cos (\varphi + \varphi_0)$ is indeed the most general solution to the Euler-Lagrange equation. $\varphi$ plays the role of the affine parameter.

A second and more elegant parametrization is to chose the proper length $ds$ as the time coordinate. One then has

(11)
\begin{align} \frac{ds}{dt} = \sqrt{\dot{\theta}^2 + \sin^2\theta\ \dot{\varphi}^2}\ . \end{align}

Using

(12)
\begin{eqnarray} \frac{\partial L}{\partial \dot\theta} = \frac{\dot{\theta}}{ \sqrt{\dot{\theta}^2 + \sin^2\theta\ \dot{\varphi}^2}} = \frac{dt}{ds}\ \dot{\theta}=\frac{d\theta}{ds}\ . \\ \frac{\partial L}{\partial \dot\varphi} =\frac{\sin^2 \theta \ \dot{\varphi}}{\sqrt{\dot{\theta}^2 + \sin^2\theta\ \dot{\varphi}^2}}=\sin^2\theta \frac{d\varphi}{ds} \end{eqnarray}

The Euler-Lagrange equation for $\theta$ is

(13)
\begin{eqnarray} \frac{d}{dt} \frac{\partial L}{\partial \dot\theta}=\frac{\partial L}{\partial \theta} \\ \frac{ds}{dt} \frac{d^2 \theta}{ds^2} = \frac{\sin\theta \cos\theta\ \dot{\theta}^2}{ \sqrt{\dot{\theta}^2 + \sin^2\theta\ \dot{\varphi}^2}} \end{eqnarray}

can be rewritten as

(14)
\begin{align} \boxed{ \frac{d^2 \theta}{ds^2} - \sin\theta\cos\theta \left(\frac{d\varphi}{ds}\right)^2 =0\ . } \end{align}

Similarly, the Euler-Lagrange equation for $\varphi$ is

(15)
\begin{align} \frac{d}{ds} \left(\sin^2\theta \frac{d\varphi}{ds}\right)=0 \end{align}

which can be written as

(16)
\begin{align} \boxed{ \frac{d^2 \varphi}{ds^2} +2 \cot\theta\ \frac{d\theta}{ds}\frac{d\varphi}{ds}=0\ . } \end{align}

Remarks

The form of the Euler-Lagrange equations that appear when we choose the affine parameter as "time" leads to what is known as the geodesic equation on a space with coordinates $x^\mu$:

(17)
\begin{align} \frac{d^2x^\mu}{ds^2} + \Gamma^{\mu}_{\nu\rho} \frac{dx^\nu}{ds} \frac{dx^\rho}{ds} = 0\ , \end{align}

where $\Gamma^{\mu}_{\nu\rho} =\Gamma^{\mu}_{\rho\nu}$ is the Christoffel symbol/connection. We see that for a two sphere, with $x^\mu=(\theta,\varphi)$, $\Gamma^\theta_{\varphi\varphi}=- \sin\theta\cos\theta$ and $\Gamma^\varphi_{\theta\varphi}=\cot\theta$ are the only two non-vanishing Christoffel symbols.

We can also understand the use of the term affine parameter. An affine transformation of a parameter $s$ is the change of variable $s\rightarrow s'=as+b$. It is a simple exercise to show that the (form of the) geodesic equation, Eq.(17), is unchanged under affine transformations. Any other transformation will generate extra terms.

(18)
\begin{align} \frac{d^2{x'}^{\mu}}{ds^{'~2}} + \Gamma^{\mu}_{\nu\rho} \frac{d{x'}^{\nu}}{ds'} \frac{d{x'}^{\rho}}{ds'} = 0\ , \end{align}

where ${x'}^{\mu}(s'):=x^\mu(s)$. Thus the affine parameter for a geodesic is unique up to an affine change of variables.

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