I have been going through the Chandrasekhar's "The Mathematical Theory of Black holes", in particular the chapter on Newman Penrose formalism.


I have a question about what he calls a "class III transformation", where, given a null tetrad $\lbrace l,n,m,\overline{m} \rbrace$ we can rescale the null directions and rotate the other as


$\lbrace Al,A^{-1}n,e^{i\phi}m,e^{-i\phi}\overline{m} \rbrace$


where $A$ is a positive function. Such transformation indeed preserves the underlying orthogonality and normalization conditions. Furthermore, the direct consequence (I also don't know how to prove that) of such choice of scaling is that some spin coefficients vanish, in particular $\epsilon+\overline{\epsilon}=0$.


My question is: how do we know that the function $A$ *exists*? 


**Attempt at the solution**:


I started by writing out the covariant derivative


$\nabla\_{l}l=Dl^{a}=(\epsilon+\overline{\epsilon})l^{a}-\overline{\kappa}m^{a}-\kappa\overline{m}^{a}$.


What I want to show is that there exists a choice of scaling such that the right hand side vanishes. I then substituted $l\rightarrow Al$ but it didn't get me anywhere. I suppose the real solution is more complicated than that because I have to show the existence of a solution to a PDE but I still have no idea how to start.


This is not homework. 



Null tetrad transformation
From the equation you wrote down you have that the scalar product 
$$(\nabla\_\ell \ell, n) = \pm (\epsilon + \bar{\epsilon})$$ 
(the $\pm$ is from the sign convention; I don't remember which one Chandrasekhar uses). So rescaling both $\ell$ and $n$ you get
$$ \nabla\_{A\ell} (A\ell) = A^2 \nabla\_\ell \ell + A (D A) \ell $$
and hence
$$ (\nabla\_{A\ell} (A\ell), A^{-1} e) = \pm A (\epsilon + \bar{\epsilon}) \pm D(A) $$


This means that $A$ solves the ODE 
$$ D (\ln A) + \epsilon + \bar{\epsilon} = 0 $$
along the integral curves of $\ell$. 


So now just set any hypersurface transverse to these integral curves, on this hypersurface prescribe $A \equiv 1$, and solve the transport equation alone integral curves. 




---


Choosing the right hand side to always vanish, however, is not always possible through only tetrad transformations. For the right hand side to vanish is the statement that $\ell$ is geodesic. Since your tetrad transformation only changes $\ell$ by rescaling, if you start out with $\ell$ such that its integral curves are not geodesics, you can never through tetrad transformations make it a geodesic vector field. 


An example is the following: consider in Minkowski space the tetrad


$$ \sqrt{2} \ell = \partial\_t + \cos(x\_3) \partial\_1 + \sin(x\_3) \partial\_3 $$


$$ \sqrt{2} n = \partial\_t - (\cos(x\_3) \partial\_1 - \sin(x\_3) \partial\_3) $$


$$ \sqrt{2} m = - \sin(x\_3) \partial\_1 + \cos(x\_3) \partial\_3 + i \partial\_2 $$




---


Finally, note that the tetrad transformations that can be used to kill some degrees of (gauge) freedom are usually only available "locally". For example: depending on the global structure of your spacetime, starting from a (global) tetrad it may be impossible to find a global $A$ such that $(\epsilon + \bar{\epsilon})$ vanish everywhere. (This is not unrelated to the fact that it may be impossible to define a tetrad globally due to topological obstructions.) 


