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When considering an AdS spacetime including a black hole, matter field and gauge field, the value of temporal component $A_t$ of the gauge potential $A_\mu$ on horizon always is set be zero, even the matter field is nonzero outsider black hole.

The explanation in paper Holographic phase transitions at finite baryon density (Page 6) is that the Killing vector on horizon $\partial_t$ vanishes, if the potential $A_\mu$ as a one-form is to be well-defined, then the temporal component $A_t(r_h)$ must vanish on horizon ($r_h$ is the horizon).

Question:

  1. What the relation between Killing vector and gauge potential lead to the vanishing condition($A_t(r_h)=0$) ?

  2. Is that vanishing value for $A_\mu$ on horizon dependent on a special gauge?

  3. What about the SU(N) gauge theory in above case?

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  • $\begingroup$ Not quite sure but: The vector $\partial_t$ is the "basis vector" for the ${}_t$ component of every vector. So if it vanishes, the temporal component of every vector must vanish. But having a $\partial_x$ vanish for any coordinate would indicate to me that you've chosen your coordinate system badly, anyway... $\endgroup$ – ACuriousMind May 25 '15 at 13:50
  • $\begingroup$ @ACuriousMind By"every vector must vanish" , do you mean both covariant vector and contravariant vector (0ne-form)? $\endgroup$ – Zoe Rowa May 25 '15 at 14:19
  • $\begingroup$ Well, depends on your metric - you know that the temporal components of vectors vanishes, just lower the index to see what that means for covectors. $\endgroup$ – ACuriousMind May 25 '15 at 14:21
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    $\begingroup$ @ZoeRowa The answer is in a paper by Racz and Wald and essentially relies on the existence of a bifurcate Killing horizon and linearity of the gauge field acting on the Killing vector. $\endgroup$ – user11128 Jan 17 '16 at 12:43

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