Geodesic Equation

Prerequisites: Levi-Civita Connection. Full path (8 pages)

• Intuitive
• Concrete
• Abstract
• Why is it interesting?

Intuitive

Solutions of the geodesic equations are called geodesics.

Geodesics are the "shortest" paths between two points in a flat spacetime and the straightest path between two points in a curved spacetime.

On a sphere the geodesics are "great circles".

Concrete

Derivation

The Lagrangian for a free point particle in a spacetime $Q$ is $$\begin{align*} L(q,\dot{q}) &= m\sqrt{g(q)(\dot{q},\dot{q})} \\ &= m\sqrt{g_{ij}\dot{q}^i\dot{q}^j} \end{align*}$$ just like in special relativity but instead of the Minkowski metric $\eta_{ij}$, we now have a general metric $g_{ij}$. Alternatively we can use $$\begin{align*} L(q,\dot{q}) &= \tfrac{1}{2}m g(q)(\dot{q},\dot{q}) \\ &= \tfrac{1}{2}m g_{ij}\dot{q}^i\dot{q}^j \end{align*}$$

We now want, as usual, to find the equations of motion. Using the Euler-Lagrange equations we get

$$\begin{align*} p_i = \frac{\partial L}{\partial\dot{q}^i} &= mg_{ij}\dot{q}^j \\ F_i = \frac{\partial L}{\partial q^i} &= \frac{\partial}{\partial q^i}\Bigl(\tfrac{1}{2}mg_{jk}(q)\dot{q}^j\dot{q}^k\Bigr)\\ &=\tfrac{1}{2}m\partial{i}g_{jk}\dot{q}^j\dot{q}^k, \quad(\text{where } \partial_i=\frac{\partial}{\partial q^i}). \end{align*}$$ So the Euler–Lagrange equations say $$\frac{d}{dt}mg_{ij}\dot{q}^j = \tfrac{1}{2}m\partial_{i}g_{jk}\dot{q}^j\dot{q}^k.$$ An important observation is that the mass factors away. Therefore, the motion is independent of the mass!

We can rewrite the geodesic equation as follows $$\begin{align*} \frac{d}{dt}g_{ij}\dot{q}^j &= \tfrac{1}{2}\partial_{i}g_{jk}\dot{q}^j\dot{q}^k \\ \hspace{-3ex}\rightarrow\quad \partial_{k}g_{ij}\dot{q}^k\dot{q}^j + g_{ij}\ddot{q}^j &= \tfrac{1}{2}\partial_{i}g_{jk}\dot{q}^j\dot{q}^k \\ \hspace{-3ex}\rightarrow\quad g_{ij}\ddot{q}^j &= \bigl(\tfrac{1}{2}\partial_{i}g_{jk}-\partial_{k}g_{ij}\bigr)\dot{q}^j\dot{q}^k \\ &= \tfrac{1}{2}\bigl(\partial_{i}g_{jk}-\partial_{k}g_{ij}-\partial_{j}g_{ki}\bigr)\dot{q}^j\dot{q}^k \end{align*}$$ where the last line follows since $g_{ik}=g_{ki}$. Now we define the so-called Christoffel symbols $$\Gamma_{ijk} \equiv -\bigl(\partial_{i}g_{jk}-\partial_{k}g_{ij}-\partial_{j}g_{ki}\bigr)$$

Using this definition, we can write the geodesic equation as $$\begin{align*} \ddot{q}_i = g_{ij}\ddot{q}^j &= -\Gamma_{ijk}\dot{q}^j\dot{q}^k \\ \hspace{-3ex}\rightarrow\quad \ddot{q}^i &= -\Gamma^i_{jk}\dot{q}^j\dot{q}^k. \end{align*}$$

Abstract

Geodesics are paths $q:[t_0,t_1]\rightarrow Q$ that are critical points of the action $$S(q) = \int_{t_0}^{t_1}\sqrt{g_{ij} \dot{q}^i\dot{q}^j}\,dt$$ This action is exactly the proper time when $(Q,g)$ is a Lorentzian manifold, or arclength when $(Q,g)$ is a Riemannian manifold.

Why is it interesting?

Almost every problem in classical mechanics can be regarded as geodesic motion.

In the geometric optics approximation light acts like particles tracing out geodesics, i.e. the shortest paths.

Especially in general relativity particles always follow geodesics. To be precise, a free particle in general relativity traces out a geodesic on the Lorentzian manifold, i.e. spacetime.

Path to this page

1. Calculus
2. Linear Algebra
3. Group Theory
4. Manifold
5. Fiber Bundles
6. Parallel Transport
7. Connections
8. Levi-Civita Connection