I'm trying to understand how to compute the null future-directed vector fields if I have a given (Bondi) metric


with $d\Omega$-standard metric on the unit sphere. I was reading this paper http://www.jstor.org/stable/2118619 and there came the statement (no proof) that the null future directed vector fields, called $n$ and $l$, are given by

$n=2e^{-\nu}\frac{\partial}{\partial u}-e^{-\lambda}\frac{\partial}{\partial r}$

$l=e^{-\lambda}\frac{\partial}{\partial r}$

with $g(n,l)=-2$.

I do not understand where these expressions for the null vector fields came from, i.e. what kind of formulas did one use? I know that null vector fields are composed of null vectors, and a null vector $v$ is such that $g(v,v)=0$, but this doesn't really bring me a way to calculate $n$ and $l$.

Attempt at the solution: Let $v=v_{1}\frac{\partial}{\partial u}+v_{2}\frac{\partial}{\partial r}$ be a null vector field. Then should hold:


this gives us two possible solutions, let call them $l$ and $r$, so as to be consistent with the paper. The first solution gives $v_{1}=0$ so one can write

$l=\mathrm{something}\cdot\frac{\partial}{\partial r}$

and the second solution gives $v_{1}=-2e^{\lambda-\nu}v_{2}$, so that we can write

$n=(-2e^{\lambda-\nu}\cdot\mathrm{somethingElse})\cdot\frac{\partial}{\partial u}+\mathrm{somethingElse}\cdot\frac{\partial}{\partial r}$.

So I seem to get the general form of these null fields, but still have two open questions:

  1. where do I see that they are "future-directed"?

  2. how was the choice of "something" and "somethingElse" made? (in the paper, rather specific functions occur there without any explanation)

  • $\begingroup$ In order to determine if they are future pointing you will need to take their inner product with a vector field which is already known to be future pointing. Such a choice must have been made to define future pointing in the first place. Often this choice is implicit in the choice of coordinates, so have a careful read over the paper. $\endgroup$ – Ben Whale Nov 12 '13 at 21:34
  • $\begingroup$ The choice of the coefficients is, to a certain extent, free. Remember that the condition of being a null vector is independent of scaling. Hence you need other conditions to fix the coefficients. These are arbitrary choices, though sometimes there are good reasons for making a particular choice. Your normalization condition on n and l are an example. I have, for example, seen other authors take g(n,l)=-1 rather than -2. $\endgroup$ – Ben Whale Nov 12 '13 at 21:39

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