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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

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Yes and no.

  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).

Since you didn't provide the context, it is unclear whether what you quote is an actual error in the book, a spot where he is implicitly using the freedom to adjust tetrads, or a spot where he has already explicitly chosen the tetrad so 2 holds and in which case there is nothing amiss at all.

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  • $\begingroup$ 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. $\endgroup$ – GregVoit May 5 '15 at 8:46

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