Recent content by Noobnoob

Okey, that's a relief I started going crazy. Since I'm not familiar with those notations I'm trying to get everything step by step and all was clear till this point. I'll take a more relaxed approach :) Thanks !

So if in that particular case (9) follows (8), by changing indices it's equivalent to \begin{equation}\frac {\partial x^{k}} {\partial x'^{h}} = R^{kh} ,\end{equation} so I don't see what's wrong :cry: Besides I understand that \begin{equation}\frac{\partial x’^i}{\partial x^j} \neq...

Yes, I surely don't derive it correctly, but I don't see how I can't get (9) from (8)

And I still don't get it...

Yes I assume this is the case at this stage of the book

So I'm just trying to figure it out, from : \begin{equation} dx’^i = \frac{\partial x’^i}{\partial x^j} dx^j = R^{ij} dx^j => \frac{\partial x’^i}{\partial x^j} = R^{ij} , \end{equation} I can't write : \begin{equation} dx^j = \frac{\partial x^j}{\partial x'^i} dx'^i = (R^T)^{ij} = R^{ji} =>...

Thanks for the tips, and so sorry about the poor notation. I'll try to be more explicit. In his paragraph he demonstrates that the derivative of a vector field \begin{equation} V'^{i}(\vec x') = R^{ij}V^{j}(\vec x) \end{equation} transforms like \begin{equation} \frac {\partial V'^{i}(\vec x')}...

Hi there, I'm just starting Zee's Einstein Gravity in a Nutshell, and I'm stuck on a seemingly very easy assumption that I can't figure out. On the Tensor Field section (p.53) he develops for vectors x' and x, and tensor R (with all indices being upper indices) : x'=Rx => x=RT x' (because R-1=RT...