Geodesic Equation 풀어보기

Thu 01 March 2018

Introduction

In this manuscript, we are going to derive a geodesic curve on the Euclidean manifold \(\mathbb{R}^{2}\) using polar coordinate chart. The curve is obviously a straight line but it is a good practice to check it by hand. Before going into the problem, let us derive the geodesic equation first.

If we are given a Riemannian metric \(g_{mn}\), the length of a curve \(\gamma(t):[a,b]\to M\) can be computed as

$$L(\gamma)=\int_{a}^{b}\Vert\frac{d\gamma}{dt}\Vert dt=\int_{a}^{b}\sqrt{g_{mn}(\gamma)\dot{x}^{m}\dot{x}^{n}}dt.$$

The geodesic curve \(\gamma^{*}\) from \(p\) to \(q\) (\(p,q\in M\)) is the curve that minimizes the length among all the curves \(\gamma\) from \(p\) to \(q\) (i.e. \(\gamma(a)=p\), \(\gamma(b)=q\)),

$$\gamma^{*}=\arg\min_{\gamma}L(\gamma).$$

Direct minimization of the length is hard because of the square root. However, the problem can be simplified using Cauchy-Schwarz inequality for \(L^{2}\) inner product.

$$(\int_{a}^{b}\Vert1\cdot\frac{d\gamma}{dt}\Vert dt)^{2}\le(\int_{a}^{b}\Vert1\Vert^{2}dt)(\int_{a}^{b}\Vert\frac{d\gamma}{dt}\Vert^{2}dt)$$
$$L(\gamma)^{2}\le2(b-a)E(\gamma),$$

where \(E(\gamma)=\frac{1}{2}\int_{a}^{b}\Vert\frac{d\gamma}{dt}\Vert^{2}dt\) is the energy function. The equality holds when the two vectors are parallel to each other, that is \(d\gamma/dt=const.\) For any curve, we can make it satisfy the constant condition without changing the shape or length of the curve. In fact, we can just put \(d\gamma/dt=1\), and find \(\gamma^{*}\) by minimizing the energy \(E(\gamma)\)

$$\gamma^{*}=\arg\min_{\gamma}E(\gamma)=\arg\min_{\gamma}\frac{1}{2}\int_{a}^{b}g_{mn}(\gamma)\dot{x}^{m}\dot{x}^{n}dt.$$

Using Euler-Lagrange equation,

$$\frac{d}{dt}(\frac{\partial E}{\partial\dot{x}^{i}})-\frac{\partial E}{\partial x^{i}}=0$$
$$\begin{aligned} 0 & =\frac{d}{dt}(g_{in}\dot{x}^{n}+g_{mi}\dot{x}^{m})-\frac{\partial g_{mn}}{\partial x^{i}}\dot{x}^{m}\dot{x}^{n}\\ & =g_{in}\frac{d}{dt}\dot{x}^{n}+g_{mi}\frac{d}{dt}\dot{x}^{m}+\dot{x}^{n}\frac{d}{dx^{m}}g_{in}\frac{dx^{m}}{dt}+\dot{x}^{m}\frac{d}{dx^{n}}g_{mi}\frac{dx^{n}}{dt}-g_{mn,i}\dot{x}^{m}\dot{x}^{n}\\ & =2g_{il}\ddot{x}^{l}+g_{in,m}\dot{x}^{m}\dot{x}^{n}+g_{mi,n}\dot{x}^{m}\dot{x}^{n}-g_{mn,i}\dot{x}^{m}\dot{x}^{n}.\end{aligned}$$

Now multiplying both sides by \(\frac{1}{2}g^{il}\) ,

$$\ddot{x}^{l}+\frac{1}{2}g^{il}(g_{in,m}+g_{mi,n}-g_{mn,i})\dot{x}^{m}\dot{x}^{n}=0$$
$$\ddot{x}^{l}+\Gamma_{nm}^{l}\dot{x}^{m}\dot{x}^{n}=0,$$

where \(\Gamma_{nm}^{l}=\frac{1}{2}g^{il}(g_{in,m}+g_{mi,n}-g_{mn,i})\) is the Christoffel Symbol. The last differential equation is called geodesic equation.

Geodesic Curve of \(\mathbb{R}^{2}\)

Computation of Christoffel Symbols

Now let us solve the original problem. For the Euclidean Manifold \(\mathbb{R}^{2}\) presented in polar coordinate, the Riemannian metric is given as follows,

$$g=\left(\begin{array}{cc} 1 & 0\\ 0 & r^{2} \end{array}\right)$$

such that infinitesimal length \(ds\) is

$$ds=dr^{2}+r^{2}d\theta^{2}.$$

Now that we have the metric tensor, we can compute the Christoffel symbols. Though there are many different combinations of indices of Christoffel symbols, it is not hard to calculate them because most of them are zero. First of all there is only diagonal entries in \(g\) and \(g^{-1}\), so \(g_{r\theta}=0\) and \(g^{r\theta}=0\). Secondly, the derivatives of \(g\) are all zero except

$$g_{\theta\theta,r}=2r.$$

Now, to compute \(\Gamma_{mn}^{l}\), let us first inspect where \(l=r\). The \(g^{il}\) will be non-zero only when \(i=r\) because \(g^{-1}\) is diagonal,

$$\Gamma_{mn}^{r}=\frac{1}{2}g^{rr}(g_{rn,m}+g_{mr,n}-g_{mn,r}).$$

Here, only the last term \(g_{mn,r}\) can survive when \(m=n=\theta\), as we have seen earlier. So,

$$\Gamma_{\theta\theta}^{r}=\frac{1}{2}g^{rr}(-g_{\theta\theta,r})=-r.\label{eq:chris1}$$

When \(l=\theta\), \(g^{il}\) will be non-zero only when \(i=\theta\) in similar sense, so,

$$\Gamma_{mn}^{\theta}=\frac{1}{2}g^{\theta\theta}(g_{\theta n,m}+g_{m\theta,n}-g_{mn,\theta}).$$

Here, the last term \(g_{mn,\theta}\) goes away and the first and the second terms remain when \(n=\theta\), \(m=r\) or \(n=r\), \(m=\theta\). So,

$$\Gamma_{\theta r}^{\theta}=\Gamma_{r\theta}^{\theta}=\frac{1}{2}g^{\theta\theta}g_{\theta\theta,r}=\frac{1}{2}(\frac{1}{r^{2}})(2r)=\frac{1}{r}.\label{eq:chris2}$$

Solving Geodesic Equation

Now, by substituting Christoffel symbols [eq:chris1]{reference-type="ref" reference="eq:chris1"},[eq:chris2]{reference-type="ref" reference="eq:chris2"}, the geodesic equation is reduced down to

$$\ddot{r}-r\dot{\theta}^{2}=0\label{eq:rdiff}$$
$$\ddot{\theta}+\frac{2}{r}\dot{r}\dot{\theta}=0.\label{eq:thetadiff}$$

Additionally, we have constant (one in this case) velocity condition,

$$\dot{s}=\dot{r}^{2}+r^{2}\dot{\theta}^{2}=1.\label{eq:const_vel}$$

Let us first multiply \(r^{2}\) both sides of [eq:thetadiff]{reference-type="ref" reference="eq:thetadiff"}. Then,

$$r^{2}\ddot{\theta}+2r\dot{r}\dot{\theta}=\frac{d(r^{2}\dot{\theta})}{dt}=0.$$

Which means that \(r^{2}\dot{\theta}\) is constant throughout time. So, let us define a constant \(p=r^{2}\dot{\theta}\). Substituting this into [eq:const_vel]{reference-type="ref" reference="eq:const_vel"}, we get

$$1=\dot{r}^{2}+\frac{p^{2}}{r^{2}}$$

Rearranging for \(\dot{r}\),

$$\dot{r}=\pm\sqrt{1-\frac{p^{2}}{r^{2}}}$$

Separating the variables,

$$\int_{r_{0}}^{r}\frac{rdr}{\sqrt{r^{2}-p^{2}}}=\pm\int_{0}^{t}dt$$
$$\sqrt{r^{2}-p^{2}}-\sqrt{r_{0}^{2}-p^{2}}=\pm t$$

Rearranging for \(r\),

$$r=\sqrt{p^{2}+(\sqrt{r_{0}^{2}-p^{2}}\pm t)^{2}}.$$

Now we can substitute \(r\) into \(p=r^{2}\dot{\theta}\) to solve for \(\theta\). However, as it gets too complicated, here we only calculate it for some special initial conditions. Firstly, let us consider an initial condition: \(\dot{\theta}(0)=0\). Then, \(p=0\) so \(r=r_{0}\pm t\), \(\theta=const.\) which is a straight line of radial direction. Secondly, consider a initial condition: \(\dot{r}(0)=0\). Then, from the constant velocity condition, \(p=r_{0}\) and \(r=\sqrt{r_{0}^{2}+t^{2}}\), \(\dot{r}=\frac{t}{\sqrt{r_{0}^{2}+t^{2}}}\) therefore, \(\dot{\theta}=\pm\frac{r_{0}}{(r_{o}^{2}+t^{2})}\), \(\theta=\theta_{0}+\arctan(t/r_{0})\). The solution is also a straight line tangent to the circle of radius \(r_{0}\).

Category: Riemannian Geometry Tagged: Geodesic

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