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I'm puzzled by the statement below:

Consider the Einstein equation expanded to the linear order around the Schwarzschild background. This describes perturbation of black hole solution $$g_{\mu\nu}=g_{\mu\nu}^{\text{background}}+h_{\mu\nu}^{\text{perturbation}}.$$ We are interested in perturbation sourced by an external field. Thus, the boundary condition for $h_{\mu\nu}^{\text{perturbation}}$ is that it is finite at the horizon (and diverges at $r\to\infty$)

We are interested in the $g_{tt}$ component $$g_{tt}=\underbrace{-1+\frac{2GM}{r}}_{\text{Schwarzschild}}+\{\text{correction}\}.\tag{*}$$ The correction term can have the form $$\sum_{\ell=2}^\infty\frac{K_\ell}{r^{\ell+1}}+P_\ell r^\ell,$$ where $K_\ell$ are the tidal numbers.

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My question is how the equation ($*$) came about. In this equation, is the metric $g$ of pure Schwarzschild? Actually, I can't recognize the $P_\ell$ terms in the meantime. Any advice is welcome.

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The correction term is, by definition, $$ h_{\mu\nu} \equiv g_{\mu\nu} - g_{\mu\nu}^{\rm background} $$ and you have $$ g_{tt}^{\rm background} = -1 + \frac{GM}{r}. $$ Thus the equation comes about by a mathematical choice or definition: we choose to explore the difference between the actual metric and the Schwarzschild metric.

It is not clear in the question whether you mean that $K_l$ and $P_l$ are functions of angle here. But in any case, any function of $r$ can be expanded in a power series. By including both negative and positive powers I get the impression you do intend that the coefficients are functions of angle, so as to produce spherical harmonic functions are something like that, but in any case, to express one function in terms of others, any complete set of functions can be used.

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  • $\begingroup$ If I am not mistaken, did you mean that in ($*$), the correction is $h_{tt}^{\text{perturbation}}\quad$? Thank you. $\endgroup$
    – Boar
    Commented Jun 24, 2021 at 23:22
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    $\begingroup$ @Steve yes that's right $\endgroup$ Commented Jun 25, 2021 at 7:51

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