Weyl tensor for the Godel metric interpretation

DrGreg
Gold Member
##x=r \, \sin\theta \, \cos\varphi##

To convert this to a transformation of differentials we differentiate
##dx= \partial_r(r \, \sin\theta \, \cos\varphi)dr##.
Unless I've completely misunderstood what you're trying to do here, there's a bit more to it than that:$$dx = \frac{\partial x}{\partial r} dr + \frac{\partial x}{\partial \theta} d\theta + \frac{\partial x}{\partial \varphi} d\varphi$$

Unless I've completely misunderstood what you're trying to do here, there's a bit more to it than that:
##dx = \frac{\partial x}{\partial r} dr + \frac{\partial x}{\partial \theta} d\theta + \frac{\partial x}{\partial \varphi} d\varphi##
Yes, we need the full diffential. I put it through Maxima and it works fine

##x=\cos\left( \phi\right) \,r\,\sin\left( \theta\right) ##
##y=\sin\left( \phi\right) \,r\,\sin\left( \theta\right) ##
##z=r\,\cos\left( \theta\right) ##
with
##dx = \frac{\partial x}{\partial r} dr + \frac{\partial x}{\partial \theta} d\theta + \frac{\partial x}{\partial \varphi} d\varphi ##
and similarly for ##dy,\ dz## gives

##dx^2 + dy^2 + dz^2 = {d\phi}^{2}\,{r}^{2}\,{\sin\left( \theta\right) }^{2}+{d\theta}^{2}\,{r}^{2}+{dr}^{2}##

I don't understand how to get basis vectors by the OPs method. It is not clear what is covariant and what is contavariant.