= Solution
Choose the <Fefferman--Graham coordinates> radial variable from
$$
\frac{dZ}{Z}=\frac{dz}{z\sqrt{g(z)}}.
$$
Near the boundary,
$$
Z=z\left[1+\frac{1+\mu^2}{6}z^3+O(z^4)\right],
\qquad
z=Z+O(Z^4).
$$
Consequently, through $O(Z^2)$,
$$
\boxed{ds^2=\frac{L^2}{Z^2}
\left[dZ^2-d\tau^2+dX^2+dY^2+O(Z^3)\right]},
$$
$$
\boxed{A_\tau=\mu L-\mu LZ+O(Z^4)}.
$$
By the <holographic dictionary>, the leading boundary value of $A_\tau$ is the source for charge density. Thus $\mu$ is the dimensionless <holographic chemical potential>; the coefficient of the normalizable linear term is proportional to the CFT charge density, $\rho=\mu L/(4\pi G)$ with the action normalization and outward-orientation convention used here.
Solved by gpt-5.6-sol high.
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