From the post demonstration of expand perturbations of metric,

I understand the full demo below excepted for the last result, i.e :

$$ \sqrt{-g}=\sqrt{-\operatorname{det} b}\left(1+\frac{1}{2} \operatorname{tr}\left(b^{-1} h\right)-\frac{1}{4} \operatorname{tr}\left(b^{-1} h\right)^{2}+\frac{1}{8} \operatorname{tr}^{2}\left(b^{-1} h\right)\right)+\mathcal{O}\left(h^{3}\right) $$ In physics notation this reads $$ \sqrt{-g}=\sqrt{-b}\left(1+\frac{1}{2} h_{\mu}^{\mu}-\frac{1}{4} h^{\mu \nu} h_{\mu \nu}+\frac{1}{8}\left(h_{\mu}^{\mu}\right)^{2}\right)+\mathcal{O}\left(h^{3}\right) $$

I have problems of understand to derive for example $$\frac{1}{2} h_{\mu}^{\mu}\equiv \frac{1}{2} \operatorname{tr}\left(b^{-1} h\right)$$

Here the full demo :

\begin{align} \sqrt{-\det{g}} = \sqrt{-\det{(b+h)}} &= \exp\log\sqrt{-\det{(b+h)}}\\ &= \exp{\frac{1}{2}\log{(-\det{(b+h)})}}\\ &= \exp{\frac{1}{2}\log{(-\det{b}\det{(1+b^{-1}h)})}}\\ &= \exp{\left[\log{\sqrt{-\det{b}}} + \frac{1}{2}\log{\det{(1+b^{-1}h)}}\right]}\\ &= \sqrt{-\det{b}}\exp{\left[\frac{1}{2}\log{\det{(1+b^{-1}h)}}\right]}\\ &= \sqrt{-\det{b}}\exp{\left[\frac{1}{2}\operatorname{tr}{\log{(1+b^{-1}h)}}\right]}\\ &= \sqrt{-\det{b}}\exp{\left[\frac{1}{2}\operatorname{tr}{\left[b^{-1}h-\frac{1}{2}(b^{-1}h)^2+\dots\right]}\right]}\\ &= \sqrt{-\det{b}}\exp{\left[\frac{1}{2}\operatorname{tr}(b^{-1}h)-\frac{1}{4}\operatorname{tr}{(b^{-1}h)^2}+\dots\right]}\\ &= \sqrt{-\det{b}}\left(1 + \frac{1}{2}\operatorname{tr}(b^{-1}h)-\frac{1}{4}\operatorname{tr}{(b^{-1}h)^2} + \frac{1}{2}\left(\frac{1}{2}\operatorname{tr}(b^{-1}h)-\frac{1}{4}\operatorname{tr}{(b^{-1}h)^2}\right)^2\right) + \mathcal O(h^3)\\ &= \sqrt{-\det{b}}\left(1 + \frac{1}{2}\operatorname{tr}(b^{-1}h)-\frac{1}{4}\operatorname{tr}{(b^{-1}h)^2} + \frac{1}{8}\operatorname{tr}^2{(b^{-1}h)}\right) + \mathcal O(h^3)\\ \end{align}

In physics notation this reads $$ \sqrt{-g}=\sqrt{-b}\left(1+\frac{1}{2} h_{\mu}^{\mu}-\frac{1}{4} h^{\mu \nu} h_{\mu \nu}+\frac{1}{8}\left(h_{\mu}^{\mu}\right)^{2}\right)+\mathcal{O}\left(h^{3}\right) $$

How to make match these 2 terms : $\frac{1}{2} h_{\mu}^{\mu}$ and $\frac{1}{2} \operatorname{tr}\left(b^{-1} h\right)$ ?

Any help is welcome.


$b$ is their notation for the original metric $g_{\mu\nu}$ and $b^{-1}$ is just their notation for the inverse metric $g^{\mu\nu}$. So ${\rm tr} \{b^{-1}h\}=g^{\mu\alpha} h_{\alpha \mu}={h^\mu}_\mu$, and so on.

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  • $\begingroup$ Thanks for your quick answer. However, in the answer of the post cited, they have written : $$g_{\mu \nu}=b_{\mu \nu}+h_{\mu \nu}$$ , so $b_{\mu\nu}$ can't be equal to $g_{\mu\nu}$, can it be ? $\endgroup$ – youpilat13 Sep 16 at 17:57
  • 1
    $\begingroup$ Just notation. The standard notation is $g_{\mu\nu}\to g_{\mu\nu}+ h_{\mu\nu}$. I used the standard notation because who ever used the "b"notation was just being difficult. $\endgroup$ – mike stone Sep 16 at 22:48
  • $\begingroup$ Okay, I understand better this way. thanks $\endgroup$ – youpilat13 Sep 16 at 23:08

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