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p. 21 in this paper (http://arxiv.org/abs/0704.0247)

$V$ is Killing vector, where $V^2 = −4b\bar{b}$, which means it is timelike Killing vector.

The authors say:

From $V^2 = −4|b|^2$ and $V = ∂_t$ as a vector we get $V_t = −4|b|^2$,

My question here is how did the authors set $V_t$ equal to this value?

They add,

From $V^2 = −4|b|^2$ and $V = ∂_t$ as a vector we get $V_t = −4|b|^2$, so that $V = −4|b|^2(dt+σ)$ as a one-form, with $σ_t = 0$.

Why did they assume that?

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closed as unclear what you're asking by ACuriousMind, Gert, user36790, Sebastian Riese, yuggib Jan 7 '16 at 14:09

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ 1. Please include all relevant information into the question. What is $b$, what is $\sigma$, why is $V$ timelike, what manifold are we even on? 2. If $V$ is an ordinary vector field, $V^2 = -4\lvert b \rvert^2$, $V = \partial_t$ and $V_t = -4\lvert b\rvert^2$ do not make any sense if $V_t$ is meant to be the temporal component of $V$. By definition, $V_t = 1$ if $V = \partial_t$. $\endgroup$ – ACuriousMind Jan 5 '16 at 1:34
  • $\begingroup$ @ACuriousMind - If $V = \partial_t$ then $V^t = 1$ not $V_t$. $\endgroup$ – Prahar Jan 5 '16 at 1:48
  • $\begingroup$ @Prahar you're absolutely right since we're talking about dual maps. Maybe it is a typo from the authors or something deeper than that :S. $\endgroup$ – Beyond-formulas Jan 5 '16 at 1:53
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    $\begingroup$ @Prahar: Ah, yes, sorry. Beyond-formulas: It's a bit weird in the first place to claim to be able to write a vector field as a 1-form, since 1-form are the duals to vector fields. But if they mean that the dual form to $V$ can be written as such, then the claim follows directly from $V_t = -4|b|^2$: $V^\flat = V_\mu\mathrm{d}^\mu = V_t \mathrm{d}t + V_i\mathrm{d}x^i = V_t (\mathrm{d}t + \frac{V_i}{V_t}\mathrm{d}x^i)$ and defining $\sigma = \frac{V_i}{V_t}\mathrm{d}x^i$. $\endgroup$ – ACuriousMind Jan 5 '16 at 1:57
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My question here is how did the authors set $V_t$ equal to this value?

On page 21 the authors say: "Let us choose coordinates $(t, z, x_i)$ such that $V = \partial_t$ and $i = 1, 2$."

So they chose the coordinates such that $V=\partial_t$, which means $V^t=1$. Note that the other components of $V^\mu$ are zeros. Next we have $$V^2=V_\mu V^\mu=-4|b|^2=V_t V^t+V_{x_1}V^{x_1}+V_{x_2}V^{x_2}+V_{z}V^{z}=V_t*1,$$ from which you find $V_t=-4|b|^2$. Here we used $V^{x_i}=V^z=0$.

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?

$\sigma$ is a general one form on coordinates $x^i, z$, which means $\sigma=\sigma_{1}dx^1+\sigma_2 dx^2+\sigma_3 dz$, note that later they use the gauge freedom to set $\sigma_z=0$. They chose the coordinates to fix $V_t$ and the rest it is the most general one form on $x^i$. For example, the most general one form on coordinates $t,x^1, x^2,z$ is $\alpha=\alpha_0 dt+\alpha_1 dx^1+\alpha_2 dx^2+\alpha_3 dz$. 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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  • $\begingroup$ Thanks for your answer, but I'm afraid this doesn't answer my question. Maybe what ACuriousMnid mentioned above solved my issue. I can't see if your answer is adding anything to ACuriousMind's answer. Is it? $\endgroup$ – Beyond-formulas Jan 5 '16 at 2:42
  • $\begingroup$ @Beyond-formulas Well, he made several comments from the mathematical point of view and, I think, he did not answered your questions directly. My answer is direct and simple, also it does not contradict his :) But everything he said is right. $\endgroup$ – Yuri Jan 5 '16 at 2:47
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    $\begingroup$ $V=\partial_t$ means $V^t=1$, not $V_t=1$! To get the object with lower index you need to use the metric $V_t=g_{t\alpha}V^\alpha$. $\endgroup$ – Yuri Jan 5 '16 at 2:57
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    $\begingroup$ @PhilosophicalPhysics $V=\partial_t$ means that $V$ is a Killing vector along $t$, which means that the metric is independent of $t$. In general, if the Killing vector has the form $V^{y}=const$ it means that the metric is independent of $y$. $\endgroup$ – Yuri Jan 5 '16 at 4:14
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    $\begingroup$ @PhilosophicalPhysics I will give you an example, when I write $X=A dt+ B dz$ it means nothing but $X_t=A, X_z=B$. It is just the compact way to write vectors, instead of using smth like $X=(A,B)$. $\endgroup$ – Yuri Jan 5 '16 at 4:32

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