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How to get null tetrads ${l^a,n^a,m^a,\overline{m}^a}$ for this metric? This on is from Ryder's book (Introduction to general relativity) page 268

$g^{\mu\nu}=\begin{pmatrix} 0 & \frac{1}{c} & 0 & 0\\ \frac{1}{c} & 1-\frac{2m}{r} & 0 & 0\\ 0& 0 & \frac{1}{r^2} &0 \\ 0& 0 &0 & \frac{1}{r^2sin^2\theta} \end{pmatrix}$

null tetrads are already found in that book, but how to find that?

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If $t^\mu, x^\mu, y^\mu, z^\mu$ are an orthonormal tetrad with $t^\mu$ lightlike and the others spacelike, then $$\begin{align} l^\mu & = t^\mu + z^\mu \\ n^\mu & = t^\mu - z^\mu \\ m^\mu & = x^\mu + i y^\mu\\ \overline{m}^\mu & = x^\mu - i y^\mu \end{align}$$ are a standard null tetrad.

You can find an orthonormal tetrad starting from any basis vectors by modifying the Gram-Schmidt process to account for that the metric isn't positive definite.

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The solution can be found on this wiki page in section "Example: Null tetrad for Schwarzschild metric in Eddington-Finkestein coordinates". Your metric is exactly in the same form (to see that just calculate the inverse of the given $g^{\mu\nu}$). The value of $F$ is slightly different and there is $c$ in several places, but because the metric is exactly in the same form, you can easily compute the null tetrads for your case.

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In general, for any metric $g_{\mu\nu}$ we can choose an orthonormal basis $e^{a}_\mu$ (where then $e^a = e^a_\mu dx^\mu$) such that the tangent space any point is spanned by these and the metric in this basis is,

$$ds^2 = \eta_{ab} e^a e^b$$

where $\eta_{ab} = \mathrm{diag}(-1, +1, +1, \dots)$. A complex tull tetrad $\{l_\mu, n_\mu, m_\mu, \bar m_\mu\}$ is constructed as,

$$l_\mu dx^\mu = \frac{1}{\sqrt{2}}(e_0 + e_1), \quad n_\mu dx^\mu = \frac{1}{\sqrt{2}}(e_0-e_1), \quad m_\mu dx^\mu = \frac{1}{\sqrt{2}}(e_2 + i e_3).$$

It can be checked that these satisfy the defining properties of the null tetrad, namely that all the vectors of the tetrad have zero magnitude, $l_\mu m^\mu = l_\mu \bar m_\mu = n_\mu m^\mu = n_\mu \bar m^\mu = 0$, $l_\mu n^\mu = -1$ and $m_\mu \bar m^\mu = 1$ as well as the fact that the metric can be expressed as,

$$g_{\mu\nu} = 2 \left[ m_{(\mu}\bar{m}_{\nu)} - l_{(\mu}n_{\nu)}\right].$$

In certain scenarios, it is the case that $\{l_\mu, n_\mu\}$ may be aligned with the tangent vector to geodesics on the manifold, whose tangent $T$ satifies $\nabla_T T = 0$

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