p. 21 in this paper (http://arxiv.org/abs/0704.0247)
V is Killing vector, where V2=−4bˉb, which means it is timelike Killing vector.
The authors say:
From V2=−4|b|2 and V=∂t as a vector we get Vt=−4|b|2,
My question here is how did the authors set Vt equal to this value?
They add,
From V2=−4|b|2 and V=∂t as a vector we get Vt=−4|b|2, so that V=−4|b|2(dt+σ) as a one-form, with σt=0.
Why did they assume that?
Answer
My question here is how did the authors set Vt equal to this value?
On page 21 the authors say: "Let us choose coordinates (t,z,xi) such that V=∂t and i=1,2."
So they chose the coordinates such that V=∂t, which means Vt=1. Note that the other components of Vμ are zeros. Next we have V2=VμVμ=−4|b|2=VtVt+Vx1Vx1+Vx2Vx2+VzVz=Vt∗1,
Though what I don't get is their requirement that σt be equal to zero and why did they place a dt next to it. Why did they assume that?
σ is a general one form on coordinates xi,z, which means σ=σ1dx1+σ2dx2+σ3dz, note that later they use the gauge freedom to set σz=0. They chose the coordinates to fix Vt and the rest it is the most general one form on xi. For example, the most general one form on coordinates t,x1,x2,z is α=α0dt+α1dx1+α2dx2+α3dz. Compare it to their expression for V (after (4.7)) and you will see that they chose only the first components, the rest is arbitrary.
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