$[\gamma_0,\gamma_r]_-=-2 \gamma_r \gamma_0$ and further that $\gamma^r \gamma_r=1$. So we find that in this representation the Dirac equation takes the following form

$$(i\gamma^\mu\partial_\mu-m-i\gamma^0\left(\frac{2M}{r}\right)^\frac{1}{2}(\partial_r+\frac{3}{4r}))\psi=0.
\label{61}$$

(This result for the Dirac equation in a Schwarzschild background with our special choice of gauging corresponds exactly to the one obtained in [@Lasenby:2002mc]).

Reformulation to a system of differential equations
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In the previous section we have derived the Dirac equation on a Schwarzschild background in a suitable representation. It has the form of the free Dirac equation with an additive radial dependant perturbation term. Now the task remains to reformulate the free part of the equation and to express it in spherical coordinates and then to separate the radial part of the state from the spherical symmetric one.

First we rewrite Eq. $(\ref{61})$ by expressing the $\gamma$-matrices by the well-known $\alpha$- and $\beta$-matrices which are defined by the relations $\gamma_0=\beta$ and $\gamma_k=\beta\alpha_k$ for $k=1,2,3$ and obtain

$$(i\beta\partial_t+i\beta\alpha^k\partial_k-m-i\beta\left(\frac{2M}{r}\right)^\frac{1}{2}(\partial_r+\frac{3}{4r}))\psi=0.
\label{62}$$

Rearranging this equation and multiplying it with $\beta$ we get

$$(i\alpha^k\partial_k)\psi=(-i\partial_t+{\beta}m+i\left(\frac{2M}{r}\right)^\frac{1}{2}(\partial_r+\frac{3}{4r}))\psi=0.
\label{63}$$

Since we are finally interested in stable orbits (the energy eigenstates) we separate the time to get the time independent Dirac equation.

Choosing the following ansatz

$$\psi=e^{-iEt}\Psi,
\label{64}$$

leads to

$$(i\alpha^k\partial_k)\Psi=(-E+{\beta}m+i\left(\frac{2M}{r}\right)^\frac{1}{2}(\partial_r+\frac{3}{4r}))\Psi.
\label{65}$$

We can 