Null geodesic congruence

I came across a statement in Chandrasekhar's "Mathematical Theory of Black Holes" that I don't understand (rather say disagree):

Assume we have a Newman Penrose tetrad $\lbrace l, n,m,\overline{m}\rbrace$. We then arrive at the equation

$l_{i;j}l^{j}=(\epsilon+\overline{\epsilon})l_{i}-\kappa \overline{m_{i}}-\overline{\kappa}m_{i}$

($\kappa$ and $\epsilon$ are the spin coefficients). Then he says that the $l$-vectors form a congruence of null geodesics if and only if $\kappa=0$; and further, they are affinely parametrized if and only if in addition $\epsilon=0$.

Now, I don't agree with the second part of the statement. Shouldn't it be when $\epsilon+\overline{\epsilon}=0$, i.e. $\Re(\epsilon)=0$? Of course, if $\epsilon=0$ the affine parametrization holds, but it's a bigger constraint than just $\Re(\epsilon)=0$.

Can anyone tell me what's wrong with my reasoning?

Thank you

1. If you are giving a null vector field $$l$$, and make an arbitrary choice at every point to complete it to a tetrad $$\{l,n,m, \bar{m}\}$$, then indeed as you wrote that $$l$$ is geodesic if and only if for any tetrad you choose, $$\kappa$$ and $$\Re(\epsilon)$$ both vanish.
2. But the tetrad can, in many cases, be chosen with some degrees of freedom. In particular, when working with a "god-given" congruence $$\gamma$$, it is often very convenient to ask for your tetrad to be parallel transported by $$\gamma$$. In the case where you start off with a null vector field $$l$$, this convenient choice of tetrad will require $$l^j m_{i;j} = 0$$ and hence $$\epsilon = \Re(\epsilon)$$.
Indeed, a congruence generated by a null vector field is geodesic if and only if there exists a tetrad $$\{l,n,m,\bar{m}\}$$ such that $$l$$ is tangent to the congruence and $$\kappa = \epsilon = \pi = 0$$ (this can be achieved by making $$n$$ and $$m$$ parallel-transported along the congruence).
• Thank you for a great answer! @Willie Wong. I found now later in the chapter he says that with the choice $\epsilon=0$ the basis vectors $l,n,m,\overline{m}$ will remain unchanged as they are transported along $l$. So everything is clear now. May 5, 2015 at 8:46