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 Answer 1


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.

  • $\begingroup$ 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, 2015 at 8:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.