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The solution of the Einstein field equation with the negative cosmological constant $\Lambda = -3k^2$ or the Schwarzschild-anti-de-Sitter solution is: $$ds_4^2=-\left(1-\frac{2m}{r}+k^2r^2 \right)c^2dt^2+\left(1-\frac{2m}{r}+k^2r^2\right)^{-1} dr^2+r^2d\Omega^2.$$ At a fixed value of $r$, it then becomes: $$ds_4^2=-\left(1-\frac{2m}{r}+k^2r^2\right)c^2dt^2+r^2d\Omega^2.$$ I wonder why this metric becomes or is conformal to, $$ds_4^2=-c^2dt^2+\frac{1}{k^2}d\Omega^2$$ when $r\to\infty$. Could anyone explain it for me?

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For $r \to \infty$ we can neglect the term

$$1 - \frac{2m}{r} $$

since it is negligible in comparison to the $r^2$ term$^\dagger$. The metric becomes, approximately,

$$\mathrm{d}s^2 = -k^2 r^2 c^2 \mathrm{d}t^2 + r^2 \mathrm{d}\Omega^2 = k^2 r^2\left( -c^2 \mathrm{d}t^2 + \frac{1}{k^2} \mathrm{d}\Omega^2\right)\,.$$ Recall that two metrics $g,g'$ are conformal to one another if there is some positive scalar function $f(x)$, which in general can depend on position, such that $g' = f(x) g$.

$^\dagger$Physically this is just the statement that at large distances from the black hole, the effects of the black hole on the spacetime geometry become irrelevant, leaving us with pure AdS geometry.

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  • $\begingroup$ Thank you, and what is the conformal perfect fluid? I see something like the energy-momentum tensor is trace-free. If I consider a conformal perfect fluid on this metric, what will its equation of state be? $\endgroup$ – Lê Dũng Mar 14 '17 at 22:07
  • $\begingroup$ I suggest you ask that as another question! $\endgroup$ – gj255 Mar 14 '17 at 23:23

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