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).

    Reply

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