# GR Tetrads & ZAMO example

I am self-learning GR.

Intro: Tetrads are a way of representing general relativity in a coordinate-independent fashion.

I am having trouble understanding tetrad notations. Basically, I know that I can transform e.g. 4-velocities between tetrad frames by: $e^m_{\ \mu} x^\mu=x^m$.

Problem: Most sources, however, give tetrads in some funny form by simply introducing vectors (example: equation 12 for Zero Angular Momentum Observer tetrad):

e.g. $\gamma^{(t)}=|g_{tt}-\omega^2 g_{\phi phi}|^{1/2} dt$

Which is somehow related to the tetrad basis vectors.

Question: I) Is there any simple way to understand tetrad basis vectors, II) how can I relate the tetrad basis to vierbeins $e^m_{\ \mu}$ and III) are there any inherent symmetries in vierbeins e.g. $e^m_{\ \mu}$?

Note: I have mainly read physics books which did not deal with differential geometry

• I guess my biggest confusion comes from the fact that $e^m_{\ \ \mu}$ is not the same as $e_{\mu}^{\ \ m}$? – Otto Feb 22 '16 at 2:12
• One's the "inverse" of the other (if we pretend they're both square matrices), recall ${e^{\mu}}_{m}{e^{\nu}}_{n}g_{\mu\nu}=\eta_{mn}$, then ${e^{\mu}}_{m}{e^{m}}_{\nu}=\delta^{\mu}_{\nu}$. – Alex Nelson Feb 24 '16 at 0:57

The tetrad and vielbein are exactly the same thing. The tetrad is just a set of four vectors $e_0,e_1,e_2,e_3$ which is orthonormal at each point: $$g(e_i,e_j)=\eta_{ij}=\operatorname{diag}(-1,1,1,1)$$ Here $g$ is the metric tensor of spacetime. The vectors $e_i$ have components with respect to some coordinates $x^\mu$, we denote these coordinate components by $e_i{}^\mu$. Here $i$ denotes which one of the four vectors we are talking about and $\mu$ which coordinate component. Since $\mu$ and $i$ go over $4$ values, this is actually a matrix, and can be shown to be invertible. We denote its matrix inverse by $e^i{}_\mu$.
The vectors $e_i$ have a "symmetry" in the following sense. If $\Lambda_j{}^i$ is a Lorentz transformation, then the new set of vectors $\Lambda_i{}^je_j$ is a tetrad: $$g(\Lambda_i{}^ke_k,\Lambda_j{}^le_l)=\Lambda_i{}^k\Lambda_j{}^lg(e_k,e_l)=\Lambda_i{}^k\Lambda_j{}^l\eta_{kl}=\eta_{ij}$$