Riemann Normal Coordinates and the metric

[kalphey]

Homework Statement

Consider a 2D spacetime where space is a circle of radius R and time has the usual description as a line. Thus spacetime can be pictured as a cylinder of radius R with time running vertically. Take the metric of this spacetime to be [itex]ds^{2}=-dt^{2}+R^{2}d\phi^{2}[/itex] in the coordinates [itex](t,\phi)[/itex], where phi is an angular coordinate along the circle which runs over [-pi, pi].



a. Derive Riemann normal coordinates valid near the origin of this space time (t=0, phi=0)

b. Check that the first derivative of the metric vanishes in these coordinates. Do higher derivatives vanish? If not, why not? If so, why do they vanish? What is the Gaussian curvature of this spacetime? Is this Gaussian curvature consistent with what you are finding about the derivatives of the metric, and if so, why?

Homework Equations

The Attempt at a Solution

Since the Riemann normal coordinates can be expressed as [itex]x^{\alpha}=sn^{\alpha}[/itex], I'm guessing that [itex]n^{A}=(t,\phi)[/itex] and [itex]s=\sqrt{t^{2}+R^{2}\phi^{2}}[/itex], so [itex]x^{A}=(t\sqrt{t^{2}+R^{2}\phi^{2}}, \phi\sqrt{t^{2}+R^{2}\phi^{2}}))[/itex] but I'm not sure if it's right. For part B I'm guessing that the Gaussian curvature is just the radius of the circle. I'm not sure how do find the metric in the Riemann normal coordinates. Can anyone give me a hint?

 

[clamtrox]

Why are you guessing like that?

The point with Riemann normal coordinates is that the metric components are locally flat, that is [itex] g_{A B}= \eta_{A B} + \mathcal{O}(x^2) [/itex].

Now your line element is [itex] ds^2 = -dt^2 + R^2 d\phi^2 [/itex]. Can you find a transformation which would make it into [itex]ds^2 = -d\tilde{t}^2 + d\tilde{\phi}^2[/itex]?

Also, think carefully about the curvature - don't guess. What shape is this universe, and is it actually curved at all?