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

The metric tensor is then

Using the time component of the geodesic equation, we get

This might look impossible to solve since we don’t know , but if we try (for a constant ):

Using the usual trick of requiring we get

Using the chain rule, we get

where and are constants of integration. If we set and consider to be increasing from , then we get, by inverting the above equation:

At , the position is :

From above, the initial velocity is then

We can replace by in the expression for to get

If we require then from above we must have , which gives (since appears in denominators, it can’t be zero). This in turn requires , so the geodesic curve is given by

If we plot versus we get the following:

We can also work out and using the above equations. First, look at :

where is a constant of integration. If we require at with the initial conditions above ) then using the expression for above, we find that . Therefore (taking )

From this we have from 6 (again, taking ):

Integrating we get

where we’ve chosen the constant of integration so that . As , so

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Chris KranenbergThe 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).

growescienceIt seems I did the squaring at the wrong point. Fixed now.