I'm interested in the paragraph under equation (38) of this reference:


To me, this is saying that given a null vector k^a, this defines a null direction (obviously) and then something about a null direction uniquely defining a null geodesic. I don't understand this statement - can someone explain it please?

And how would the situation be different for timelike geodesics? For example, surely if I decide to travel in a particular timelike direction, there is a unique timelike geodesic, no?


  • 1
    $\begingroup$ This site does have MathJax enabled, so you can copy the paragraph & equation directly. $\endgroup$
    – Kyle Kanos
    Oct 16, 2015 at 10:36
  • 2
    $\begingroup$ Minor comment to the post (v1): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. $\endgroup$
    – Qmechanic
    Oct 16, 2015 at 15:25

1 Answer 1


First of all, time-like geodesics is only for particles with non zero rest-mass. You are working with light rays or more specific a null congruence (which is basically a family of null curves or light rays).

That $k_a\,k^a=0$ means that $k_a$ is a null vector field, that is, at every point of the manifold, $k_a$ is tangent to a generator of the light cone at this point. There is an infinity null vector fields defining curves on the manifold, but not everyone is a geodesic.

If you don't have any background in general relativity... you are gonna love what is next (sarcasm detected) ;)

Now, in the Newman-Penrose formalism, it is possible, if you already have a null vector field $k_a$ to adapt a null tetrad, $\{k_a,\,l_a,\,m_a,\,\bar m_a\}$, to it. Then, one of the directional derivatives of the tetrad (in this formalism) would be:

\begin{equation} k^b\nabla_b k_a=(\epsilon+\bar\epsilon)k_a-\bar k \,m_a-k\,\bar m_a, \end{equation} where $\epsilon$ and $k$ are spin coefficients (which are part of the full set of the components of the Ricci rotation coefficients). Do you recognize the LHS of this equation?

Well, the Goldberg—Sachs theorem states that if you have a congruence of null geodesics which are shear-free (which this is basically a way of saying that the family of null rays, when evolve in time, they don't ''distort''), then $k=0$ (and other spin coefficient also is zero, but it is not here). So, the null geodesic equation can be put as \begin{equation} k^b\nabla_b k_a=(\epsilon+\bar\epsilon)k_a, \end{equation}

If besides of this, $\epsilon=0$, you have a congruence of null geodesics that have an affine parameter.

This is your case!!! you have a beautiful (and boring) family of shear-free, null geodesic light rays


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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