- #1
ergospherical
- 1,019
- 1,298
Consider the ellipsoid:$$\mathcal{Q} := \{ \mathbf{x} \in \mathbb{R}^3, \ x^2 + a^2(y^2 + z^2) = 1 \}$$We have local coordinates ##\chi^A = (\rho, \phi)## on the ellipsoid surface defined by ##y = \rho \cos{\phi}## and ##z = \rho \sin{\phi}##. First we look for the metric ##\gamma := \phi^{*} g## induced from the Euclidean metric, specifically:$$\gamma_{AB} = \frac{\partial x^{i}}{\partial \chi^A} \frac{\partial x^j}{\partial \chi^B} \delta_{ij}$$Using also that ##x(\rho, \phi) = \sqrt{1-a^2 \rho^2}##, I obtain that the pull back of the metric is:$$\gamma = \left( 1 + \frac{a^4 \rho^2}{1-a^2 \rho^2} \right) d\rho^2 + \rho^2 d\phi^2$$We want to find a non-zero function ##\Omega## such that the conformal metric ##\Omega^2 \gamma_{ij}## is flat, i.e. that there is some transformation that brings ##\Omega^2 \gamma_{ij}## into a form resembling ##\delta_{ij}##. I've had no luck with my guesses, but there must be an intuitive way of seeing this?