# Geodesic equation in 2-d: exponential metric

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 8; Problem 8.6.

This is another example of the geodesic equation in a 2-d space-time (with one time and one space dimension). The metric in this case is

$\displaystyle ds^{2}=-e^{-x/a}dt^{2}+dx^{2} \ \ \ \ \ (1)$

The metric tensor is then

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

Using the time component of the geodesic equation, we get

 $\displaystyle \frac{d}{d\tau}\left(-e^{-x/a}\frac{dt}{d\tau}\right)$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (3)$ $\displaystyle \frac{1}{a}\frac{dx}{d\tau}e^{-x/a}\frac{dt}{d\tau}-e^{-x/a}\frac{d^{2}t}{d\tau^{2}}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (4)$ $\displaystyle \frac{d^{2}t}{d\tau^{2}}$ $\displaystyle =$ $\displaystyle \frac{1}{a}\frac{dx}{d\tau}\frac{dt}{d\tau} \ \ \ \ \ (5)$

This might look impossible to solve since we don’t know ${dx/d\tau}$, but if we try (for a constant ${c}$):

 $\displaystyle \frac{dt}{d\tau}$ $\displaystyle =$ $\displaystyle ce^{x/a}\ \ \ \ \ (6)$ $\displaystyle \frac{d^{2}t}{d\tau^{2}}$ $\displaystyle =$ $\displaystyle \frac{c}{a}\frac{dx}{d\tau}e^{x/a}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{a}\frac{dx}{d\tau}\frac{dt}{d\tau} \ \ \ \ \ (8)$

Using the usual trick of requiring ${\mathbf{u}\cdot\mathbf{u}=-1}$ we get

 $\displaystyle g_{ij}u^{i}j^{j}$ $\displaystyle =$ $\displaystyle -e^{-x/a}\left(\frac{dt}{d\tau}\right)^{2}+\left(\frac{dx}{d\tau}\right)^{2}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -e^{-x/a}c^{2}e^{2x/a}+\left(\frac{dx}{d\tau}\right)^{2}=-1\ \ \ \ \ (10)$ $\displaystyle \frac{dx}{d\tau}$ $\displaystyle =$ $\displaystyle \pm\sqrt{c^{2}e^{x/a}-1} \ \ \ \ \ (11)$

Using the chain rule, we get

 $\displaystyle \frac{dx}{d\tau}$ $\displaystyle =$ $\displaystyle \frac{dx}{dt}\frac{dt}{d\tau}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle ce^{x/a}\frac{dx}{dt}\ \ \ \ \ (13)$ $\displaystyle \frac{dx}{dt}$ $\displaystyle =$ $\displaystyle \pm\frac{\sqrt{c^{2}e^{x/a}-1}}{ce^{x/a}}\ \ \ \ \ (14)$ $\displaystyle \pm\int\frac{ce^{x/a}}{\sqrt{c^{2}e^{x/a}-1}}dx$ $\displaystyle =$ $\displaystyle t+t_{0}\ \ \ \ \ (15)$ $\displaystyle \pm\frac{2a}{c}\sqrt{c^{2}e^{x/a}-1}+\alpha$ $\displaystyle =$ $\displaystyle t+t_{0} \ \ \ \ \ (16)$

where ${\alpha}$ and ${t_{0}}$ are constants of integration. If we set ${t_{0}=0}$ and consider ${x}$ to be increasing from ${x\left(0\right)}$, then we get, by inverting the above equation:

$\displaystyle x\left(t\right)=a\ln\left(\frac{1}{c^{2}}+\frac{\left(t-\alpha\right)^{2}}{4a^{2}}\right) \ \ \ \ \ (17)$

At ${t=0}$, the position is ${x_{0}}$:

$\displaystyle x_{0}=a\ln\left(\frac{1}{c^{2}}+\frac{\alpha^{2}}{4a^{2}}\right) \ \ \ \ \ (18)$

From above, the initial velocity ${\frac{dx}{d\tau}\left(0\right)\equiv u_{0}}$ is then

 $\displaystyle u_{0}$ $\displaystyle =$ $\displaystyle \sqrt{c^{2}e^{x_{0}/a}-1}\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \pm\frac{c\alpha}{2a} \ \ \ \ \ (20)$

We can replace ${\alpha}$ by ${u_{0}}$ in the expression for ${x\left(t\right)}$ to get

 $\displaystyle x\left(t\right)$ $\displaystyle =$ $\displaystyle a\ln\left(\frac{1}{c^{2}}+\frac{\left(t\pm\frac{2au_{0}}{c}\right)^{2}}{4a^{2}}\right)\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle a\ln\left(\frac{1}{c^{2}}+\left(\frac{t}{2a}\pm\frac{u_{0}}{c}\right)^{2}\right) \ \ \ \ \ (22)$

If we require ${x_{0}=u_{0}=0}$ then from above we must have ${a\ln\left(1/c^{2}\right)=0}$, which gives ${c=\pm1}$ (since ${a}$ appears in denominators, it can’t be zero). This in turn requires ${\alpha=0}$, so the geodesic curve is given by

$\displaystyle x\left(t\right)=a\ln\left(1+\frac{t^{2}}{4a^{2}}\right) \ \ \ \ \ (23)$

If we plot ${\frac{x}{a}}$ versus ${\frac{t}{a}}$ we get the following:

We can also work out ${x\left(\tau\right)}$ and ${t\left(\tau\right)}$ using the above equations. First, look at ${x}$:

 $\displaystyle \frac{dx}{d\tau}$ $\displaystyle =$ $\displaystyle \pm\sqrt{c^{2}e^{x/a}-1}\ \ \ \ \ (24)$ $\displaystyle \pm\int\frac{dx}{\sqrt{c^{2}e^{x/a}-1}}$ $\displaystyle =$ $\displaystyle \tau\ \ \ \ \ (25)$ $\displaystyle \pm2a\arctan\left(\sqrt{c^{2}e^{x/a}-1}\right)+\beta$ $\displaystyle =$ $\displaystyle \tau \ \ \ \ \ (26)$

where ${\beta}$ is a constant of integration. If we require ${\tau_{0}=0}$ at ${t=0}$ with the initial conditions above ${x_{0}=u_{0}=0}$) then using the expression for ${x_{0}}$ above, we find that ${\beta=0}$. Therefore (taking ${c=1}$)

$\displaystyle x\left(\tau\right)=a\ln\left(1+\tan^{2}\frac{\tau}{2a}\right) \ \ \ \ \ (27)$

From this we have from 6 (again, taking ${c=1}$):

 $\displaystyle \frac{dt}{d\tau}$ $\displaystyle =$ $\displaystyle e^{x/a}\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1+\tan^{2}\frac{\tau}{2a}\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sec^{2}\frac{\tau}{2a} \ \ \ \ \ (30)$

Integrating we get

 $\displaystyle t\left(\tau\right)$ $\displaystyle =$ $\displaystyle \int\sec^{2}\frac{\tau}{2a}d\tau\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2a\tan\frac{\tau}{2a} \ \ \ \ \ (32)$

where we’ve chosen the constant of integration so that ${t\left(0\right)=0}$. As ${\tau\rightarrow\pi a}$, ${t\rightarrow\infty}$ so

$\displaystyle \tau_{max}=\pi a \ \ \ \ \ (33)$

## 2 thoughts on “Geodesic equation in 2-d: exponential metric”

1. Chris Kranenberg

The expression for x(tau) should be found by the following steps after integrating and taking c squared equal to 1 (the integral for tau(x) is correct): divide both sides by +/- 2a, take the tangent of both sides, square both sides, add 1 to both sides, take the natural log of both sides, then multiply both sides by a. Then an expression for t of tau can be found by integration (it’s the integral of a secant squared function which is a closed form).