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I asked the question in the MathOverflow, but didn't get any response. I thought maybe better luck here.

I'm reading the following paper about Petrov type D space times called "Type D vacuum metrics":

http://dx.doi.org/10.1063/1.1664958

by Kinnersley. I have a question about his choice of gauge. Here is the description:

In particular, starting with the tetrad $\lbrace l^{\mu},n^{\mu},m^{\mu},\overline{m}^{\mu}\rbrace$ where $l^{\mu},n^{\mu}$ are null, he then makes an argument that we can pick a scaling $A$ such that $l^{\mu}\rightarrow Al^{\mu}$ and $n^{\mu}\rightarrow (1/A)n^{\mu}$, and make $\nabla_{l}l=0$ (i.e. $l^{\mu}$ a geodesic vector field). He then picks a coordinate system $(x^{1},r,x^{2},x^{3})$ such that $l^{\mu}=(0,1,0,0)$. That's what he explains at the beginning of his section 2, and it's clear to me.

What confuses me is what he writes at the beginning of section 3C: he says that there is still freedom left if we choose the scaling, call it $A^{0}$, to be independent of $r$. Now, I understand that such choice will not violate the condition $\nabla_{l}l=0$. However, my question is: why doesn't it violate the already chosen $l^{\mu}=(0,1,0,0)$?

What am I missing? Why is one allowed to scale once more?

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I only had a brief look at the paper, but to me it looks like that even after choosing $A^{0}$, in such a way that $\nabla_{l}l=0$ and $l^{\mu}=(0,1,0,0)$ then you still have some freedom in what they call $\theta^0$ which will affect $m^{\mu}$ and $\overline{m}^{\mu}$ but not the first two.

I will agree that there is no more freedom in $A^{0}$, only for $\theta^0$ after you demand $l^{\mu}=(0,1,0,0)$. Alternatively, the way I would read it is that they have not demanded the condition $l^{\mu}=(0,1,0,0)$ in section C.

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  • $\begingroup$ thank you for the answer. So, have I understood you correctly: apart from the story with $\theta^{0}$ freedom, you also agree that choosing $A^{0}$ violates $l^{\mu}=(0,1,0,0)$? @Heterotic $\endgroup$
    – user46446
    Commented Jun 30, 2015 at 18:14
  • $\begingroup$ Yes, you have understood correctly what I am trying to say, even though I think your phrasing is slightly misleading. There is a value for $A^0$ that respects $l^{\mu}=(0,1,0,0)$. Setting $A^0$ to a different value violates $l^{\mu}=(0,1,0,0)$. $\endgroup$
    – Heterotic
    Commented Jul 1, 2015 at 8:03
  • $\begingroup$ oh then I understood wrong. But which value of $A^{0}$ apart from $A^{0}=1$ respects $(0,1,0,0)$? I just don't see how that's possible. Could you please elaborate a bit on that? Many thanks! @Heterotic $\endgroup$
    – user46446
    Commented Jul 1, 2015 at 11:17
  • $\begingroup$ I think we are both saying the same thing. All I am trying to convey is that the first comment should have been: "apart from the story with $\theta^0$ freedom, you also agree that choosing $A^0\neq1$ violates $l^{\mu}=(0,1,0,0)$". $\endgroup$
    – Heterotic
    Commented Jul 1, 2015 at 15:20
  • $\begingroup$ ok, great, now I got it! $\endgroup$
    – user46446
    Commented Jul 1, 2015 at 16:39

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