# Gauge freedom in the tetrad

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

http://scitation.aip.org/content/aip/journal/jmp/10/7/10.1063/1.1664958

by Kinnersley. I have a question about his choice of gauge.

In particular, starting with the tetrad $\lbrace l^{\mu},n^{\nu},m^{\mu},\overline{m}^{\mu}\rbrace$ 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?

Thanks for any help.

This post imported from StackExchange MathOverflow at 2015-07-05 20:45 (UTC), posted by SE-user GregVoit
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