\label{57}$$

It is left to the reader to show that these choices fulfil the conditions $(\ref{56})$. The inverse components of the vierbein are determined by Eq. $(\ref{29})$ leading to

$$e^{\bar 0}_0=1\quad ,\quad  e^{\bar r}_0=\left(\frac{2M}{r}\right)^{\frac{1}{2}}\quad ,\quad e^{\bar r}_r=1\quad ,\quad e^{\bar 0}_r=0.
\label{58}$$

By the special choice of the vierbein in Eq. $(\ref{57})$, the special form of the metric is completely determined by Eq. $(\ref{36})$ and Eq. $(\ref{55})$. To find the specific Dirac equation we have to specify the connection coefficients $\omega_\mu^{mn}$ by inserting the above choice of the vierbein to Eq. $(\ref{48})$. In the expression arising from this most terms vanish and one obtains the following coefficients

$$\omega_0^{\bar 0 \bar r}=-\omega_0^{\bar r \bar 0}=-\frac{3M}{r^2} \quad, \quad  
\omega_r^{\bar 0 \bar r}=-\omega_r^{\bar r \bar 0}=-\frac{3}{2}\left(\frac{2M}{r}\right)^{\frac{1}{2}}\frac{1}{r}.
\label{59}$$

The coefficients with equal upper indices vanish because of the antisymmetry. By inserting Eqs. $(\ref{57})$ and $(\ref{59})$ to the general Dirac Eq. $(\ref{35})$ and using the antisymmetry of the generators and the cooresponding coefficients leading to $\omega_\mu^{mn}\Sigma_{mn}=2 \omega_\mu^{\bar 0 \bar r}\Sigma_{\bar 0 \bar r}$ we get the following expression

$$(i\gamma^\mu\partial_\mu-m-i\gamma^0\left(\frac{2M}{r}\right)^\frac{1}{2}\partial_r
-\gamma^r\left(-\frac{3}{2}\left(\frac{2M}{r}\right)^\frac{1}{2}\frac{1}{r}\right)\frac{i}{4}[\gamma_0,\gamma_r]_-)\psi=0.
\label{60}
$$

$\gamma^0$ and $\gamma^r$ obey the same anticommutation relation as $\gamma^0$ and $\gamma^i$ where $i$ runs from 1 to 3. From this we see that 