Tuesday, 28 June 2016

homework and exercises - Killing vector and one-form



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=Vt1,

from which you find Vt=4|b|2. Here we used Vxi=Vz=0.



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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