# What happens to the form basis after making the metric time orthogonal

1. Sep 16, 2013

### center o bass

Given a basis for spacetime $\{e_0, \vec{e}_i\}$ for which $\vec{e}_0$ is a timelike vector. Of these vectors one can make a new basis for which all vectors are orthogonal to $\vec{e}_0$. I.e. the vectors $$\hat{\vec{e}}_i = \vec{e}_i - \frac{\vec{e}_i \cdot \vec{e}_0}{\vec{e}_0 \cdot \vec{e}_0} \vec{e}_0 =\vec{e}_i - \frac{g_{i 0}}{g_{00}} \vec{e}_0$$

Now at page 35 in the following lecture notes: http://www.uio.no/studier/emner/matnat/fys/FYS4160/v06/undervisningsmateriale/kompendium.pdf

one use the basis $\{\vec{e}_\hat{0}, \hat{e}_i\}$ basis to define a "spatial metric" $\gamma_{ij} = \hat{\vec{e}}_i \cdot \hat{\vec{e}}_j$ for which it is trivial to show that

$$\gamma_{ij} = g_{ij} - \frac{g_{i0}g_{j0}}{g_{00}}$$ such that in this basis the spacetime metric becomes

$$ds^2 = - d\hat{t}^2 + dl^2$$

where $dl^2 = \gamma_{ij} dx^i dx^j$ and $\hat{\vec{e}}_0$ is normalized. Note that $\gamma_{i0} = \gamma_{00} = 0$. Now what the author does next is to express $d \hat{t}^2$ in terms of the arbitrary basis by

$$d \hat{t}^2 = dl^2 - ds^2 = (\gamma_{\mu \nu} - g_{\mu \nu})dx^\mu dx^\nu$$

and by plugging in the result above for $\gamma_{ij}$ he arrives at

$$d\hat{t}^2 = - g_{00}(dx^0 + \frac{g_{i0}}{g_{00}} dx^i)^2.$$

Here is my problem with this derivation: when writing
$$dl^2 - ds^2 = (\gamma_{\mu \nu} - g_{\mu \nu})dx^\mu dx^\nu$$
one is using the same forms $dx^\mu$ to express $dl^2$ for the spatial metric as one uses to express $ds^2$ (the not time-orthogonal metric). Is there a rationale behind doing this? I.e. can one use the same basis of forms for expressing the spatial metric as one does with the spacetime metric when using components defined by $\vec{e}_i \cdot \vec{e}_j$ and $\hat{\vec{e}}_i \cdot \hat{\vec{e}}_j$ respectively?