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.

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