# Geodesic equation: 2-d space-time

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 8; Problem 8.5.

This is an example of the geodesic equation in a 2-d space-time (with one time and one space dimension). The metric is given in a general way as

$\displaystyle ds^{2}=-dt^{2}+f^{2}\left(q\right)dq^{2} \ \ \ \ \ (1)$

where ${q}$ is the generalized spatial coordinate, and ${f}$ is an arbitrary function. The metric tensor is then

$\displaystyle g_{ij}=\left[\begin{array}{cc} -1 & 0\\ 0 & f^{2}\left(q\right) \end{array}\right] \ \ \ \ \ (2)$

Using the time component of the geodesic equation, we set ${a=t}$ in:

$\displaystyle \frac{d}{d\tau}\left(g_{aj}\frac{dx^{j}}{d\tau}\right)-\frac{1}{2}\partial_{a}g_{ij}\frac{dx^{i}}{d\tau}\frac{dx^{j}}{d\tau}=0 \ \ \ \ \ (3)$

We get

$\displaystyle -\frac{d^{2}t}{d\tau^{2}}=0 \ \ \ \ \ (4)$

From this we conclude that

$\displaystyle \frac{dt}{d\tau}=k \ \ \ \ \ (5)$

for some constant ${k}$.

Using the condition ${\mathbf{u}\cdot\mathbf{u}=-1}$ we get

 $\displaystyle g_{ij}\frac{dx^{i}}{d\tau}\frac{dx^{j}}{d\tau}$ $\displaystyle =$ $\displaystyle -\left(\frac{dt}{d\tau}\right)^{2}+f^{2}\left(\frac{dq}{d\tau}\right)^{2}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -k^{2}+f^{2}\left(\frac{dq}{d\tau}\right)^{2}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -1\ \ \ \ \ (8)$ $\displaystyle \frac{dq}{d\tau}$ $\displaystyle =$ $\displaystyle \pm\frac{1}{f}\sqrt{\left(k^{2}-1\right)} \ \ \ \ \ (9)$

We can write this in terms of ${dq/dt}$:

 $\displaystyle \frac{dq}{d\tau}$ $\displaystyle =$ $\displaystyle \frac{dq}{dt}\frac{dt}{d\tau}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle k\frac{dq}{dt}\ \ \ \ \ (11)$ $\displaystyle \frac{dq}{dt}$ $\displaystyle =$ $\displaystyle \pm\frac{1}{f\left(q\right)}\sqrt{\left(1-\frac{1}{k^{2}}\right)} \ \ \ \ \ (12)$

That is, the geodesic is the solution of this differential equation.

If we define a new coordinate system in which ${t^{\prime}=t}$ and ${q^{\prime}\equiv F\left(q\right)}$ is the antiderivative (integral) of ${f\left(q\right)}$ then we can transform the metric tensor to this new coordinate system using the standard transformation formula

$\displaystyle g_{ij}^{\prime}=\frac{\partial x^{k}}{\partial x^{\prime i}}\frac{\partial x^{l}}{\partial x^{\prime j}}g_{kl} \ \ \ \ \ (13)$

By implicit differentiation:

 $\displaystyle \frac{\partial q^{\prime}}{\partial q^{\prime}}$ $\displaystyle =$ $\displaystyle \frac{\partial F}{\partial q}\frac{\partial q}{\partial q^{\prime}}+\frac{\partial F}{\partial t}\frac{\partial t}{\partial q^{\prime}}\ \ \ \ \ (14)$ $\displaystyle 1$ $\displaystyle =$ $\displaystyle f\left(q\right)\frac{\partial q}{\partial q^{\prime}}+0\ \ \ \ \ (15)$ $\displaystyle \frac{\partial q}{\partial q^{\prime}}$ $\displaystyle =$ $\displaystyle \frac{1}{f\left(q\right)} \ \ \ \ \ (16)$

The only other non-zero derivative is

$\displaystyle \frac{\partial t}{\partial t^{\prime}}=1 \ \ \ \ \ (17)$

The new metric is therefore

$\displaystyle g_{ij}^{\prime}=\left[\begin{array}{cc} -1\left(\frac{\partial t}{\partial t^{\prime}}\right)^{2} & 0\\ 0 & f^{2}\left(q\right)\left(\frac{\partial q}{\partial q^{\prime}}\right)^{2} \end{array}\right]=\left[\begin{array}{cc} -1 & 0\\ 0 & 1 \end{array}\right] \ \ \ \ \ (18)$

This is the metric of flat space-time in rectangular coordinates. Thus any metric with ${g_{tt}=-1}$ represents flat space-time, since ${f\left(q\right)}$ is just a transformation of the flat metric using different coordinates.