# Question about Spherical Metric and Approximations

## Homework Statement

This is Problem 2 from Chapter 1, Section V of A. Zee's Einstein Gravity in a Nutshell. Zee asks us to imagine a colony of "eskimo mites" that live at the north pole. The geometers of the colony have measured the following metric of their world to second order (with the radius of the sphere equal to 1):

$$ds^{2}=\left (1-\frac{y^{2}}{3} \right )dx^{2}+\left (1-\frac{x^{2}}{3} \right )dy^{2}+\frac{2}{3}xy\, dx\, dy+\cdots$$

Zee states: "For ## x,y \ll 1 ##, the space is flat and as Euclidean as it could be. But note that in the second order the metric is not diagonal."

The problem asks us to derive the above metric, given that we know (1) that the mite coordinates ## (x,y) ## are related to the usual spherical coordinates ## x = \theta \cos \phi ## and ## y=\theta \sin \phi ## and (2) that the metric for spherical coordinates is:

$$ds^{2}=d\theta^{2}+\sin^{2}\theta\, d\phi^{2}.$$

## Homework Equations

I believe that the most important equation allows us to calculate the metric in the primed coordinates ## (x,y) ## given the metric in the unprimed coordinates ## (\theta,\phi) ##:

$${g}'_{\rho\sigma}({x}')=g_{\mu\nu}(x)\frac{\partial x^{\mu} }{\partial {x}'^{\rho}}\frac{\partial x^{\nu} }{\partial {x}'^{\sigma}}.$$

I believe that I also need to find expressions for the unprimed coordinates as a function of the primed coordinates. In these equations, I assume that the radius of the sphere is 1.

$$\sin \theta=\frac{\sqrt{x^{2}+y^{2}}}{r}=\sqrt{x^{2}+y^{2}} .$$

$$\tan \phi=\frac{x}{y}.$$

## The Attempt at a Solution

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So far, I have been unable to calculate ## {g}'_{xx} ##, which according to the problem statement should be (to second order) ## 1-\frac{y^{2}}{3} ##. According to our general relation for calculating the primed metric, I have:

$${g}'_{xx}={g}_{\theta\theta}\left (\frac{\partial\theta}{\partial x} \right )^{2}+g_{\phi\phi}\left (\frac{\partial\phi}{\partial x} \right )^{2} .$$

Then taking the (I hope) correct partial derivatives from above and substitute in some trigonometric identities.

$$\frac{\partial\theta}{\partial x}=\frac{x}{\sqrt{x^{2}+y^{2}}}\cdot \frac{1}{\cos\theta}=\frac{x}{\sqrt{x^{2}+y^{2}}\sqrt{1-x^{2}-y^{2}}} .$$

$$\frac{\partial\phi}{\partial x}=-\frac{y}{x^{2}}\cos^{2}\phi=-\frac{y}{x^{2}}\cdot \frac{x^{2}}{x^{2}+y^{2}}=-\frac{y}{x^{2}+y^{2}}$$

I can now plug these results back into my expression for ## {g}'_{xx} ##:

$${g}'_{xx}=\frac{x^{2}}{(x^{2}+y^{2})(1-x^{2}-y^{2})}+\sin^2\theta\frac{y^{2}}{(x^{2}+y^{2})^{2}}.$$

Simple trigonometry tells us that ## \sin^2 \theta = x^{2}+y^{2} ##. Thus I end up with the result, before simplification:

$${g}'_{xx}=\frac{x^{2}}{(x^{2}+y^{2})(1-x^{2}-y^{2})}+\frac{y^{2}}{(x^{2}+y^{2})}.$$

I have tried to simplify this to the correct second-order approximation, but cannot get there. It is very well possible that I am approaching this problem from the completely wrong direction. In the following chapter, which provides a brief introduction to Riemann Normal Coordinates, Zee makes use of Taylor Series Powers expansions. However, I was unsure of how to apply those to this problem.

Nonetheless, I take it that this problem emphasizes the fact that we can select coordinates such that the local metric -- for infinitesimal displacements -- is Euclidean to the zeroth order. Moreover, we can remove any first-order terms by careful selection of new coordinates. However, we cannot get rid of the second-order deviations.

I hope this provides enough information for some exterior guidance.

TSny
Homework Helper
Gold Member
Welcome to PF!

I believe the mistake is with the equation ## \sin \theta=\sqrt{x^{2}+y^{2}} ##. This would be applicable if the "mite coordinates" ##x## and ##y## obeyed

##x = \sin\theta \cos \phi##
##y = \sin\theta \sin \phi##

But the mite coordinates are given to obey

##x = \theta \cos \phi##
##y = \theta \sin \phi##

If you use this second set of relations, then it seems to work out.

Thank you for the suggestion. In addition to that change, I also used the Taylor series ## \sin ^2 \theta = \theta^2 + \frac{1}{3}\theta^4 + ... ## and it worked out.