I'm studying Post-Newtonian theory on the book "Gravity" by Poisson and Will and I found a few formulas that I can't obtain by myself. I'm pretty sure it must be quite simple, but can't find the right method to get the result.

First of all, a really small introduction to the topic: instead of using the usual metric $g_{\alpha \beta}$, the inverse gothic metric $\mathfrak{g}^{\alpha \beta} = \sqrt{-g}\, g^{\alpha \beta}$ is used, where $g = \det (g_{\alpha \beta}) = \det (\mathfrak{g}^{\alpha \beta})$ (so that the relation can be easily reversed to obtain the usual metric). Then this inverse gothic metric is split in $\mathfrak{g}^{\alpha \beta} = \eta^{\alpha \beta} - h^{\alpha \beta}$, where $\eta^{\alpha \beta}$ is the usual Lorentzian metric $\operatorname{diag}(1, -1, -1, -1)$ (or with the opposite signs, if you prefer) and $h^{\alpha \beta}$ is a small correction of order $G$ (the gravitational constant, a placeholder for something a little bit more complicated and much more meaningful containing $G$).

It's easy to find that $g_{\alpha \beta} = \sqrt{-g}\, \mathfrak{g}_{\alpha \beta}$, but now it gets tricky. On pages 335-336 it is stated that we can perform a Post-Minkowskian expansion of a few quantities, i.e. they can be expressed as $\sum \limits_{n = 0}^{\infty} G^n k_{(n)}$, where $k_{(n)}$ are functions. The formulas are these (7.20a - 7.20d): $$g_{\alpha \beta} = \eta_{\alpha \beta} + h_{\alpha \beta} - \frac{1}{2} h \eta_{\alpha \beta} + h_{\alpha \mu} h^{\mu}{}_{\beta} - \frac{1}{2} h h_{\alpha \beta} + \left( \frac{1}{8} h^2 + \frac{1}{4} h^{\mu \nu} h_{\mu \nu} \right) \eta_{\alpha \beta} + O(G^3)$$ $$g^{\alpha \beta} = \eta^{\alpha \beta} - h^{\alpha \beta} + \frac{1}{2} h \eta^{\alpha \beta} - \frac{1}{2} h h^{\alpha \beta} + \left( \frac{1}{8} h^2 + \frac{1}{4} h^{\mu \nu} h_{\mu \nu} \right) \eta^{\alpha \beta} + O(G^3)$$ $$(-g) = 1 - h + \frac{1}{2} h^2 - \frac{1}{2} h^{\mu \nu} h_{\mu \nu} + O(G^3)$$ $$\sqrt{-g} = 1 - \frac{1}{2}h + \frac{1}{8} h^2 - \frac{1}{4} h^{\mu \nu} h_{\mu \nu} + O(G^3)$$

"Post-Minkowskian expansion" seems to have a vague meaning, as can be deduced from the definition, but I suppose here it's just a synonym of "Taylor expansion around $h^{\alpha \beta} = 0$".

If that's the case, I'm not sure what to expand and how, since it's not obvious or possible to isolate the quantity I'm looking for because of paired indices and products between tensors (and by tensor I mean - wrongly - everything that has indices). If that's not the case, then how can I calculate the expansion?

I tried to find the solution elsewhere, but it seems that the literature regarding Post-Newtonian theory is not so abundant and this specific topic isn't covered by anyone else, at least in the articles I found.

Edit: I forgot to specify that every index rasing/lowering and every contraction is made with respect to $\eta$. So for example $h = \eta^{\mu \nu} h_{\mu \nu}$ and $h^{\mu}{}_{\nu} = \eta^{\mu \sigma} h_{\sigma \nu}$.

  • $\begingroup$ Hey, I'm currently following the same book but I'm stuck with the problem 6.9 in the same chapter, could just tell how do you solve? or a hint at least because I don't find any sources for this particularly and If there are any, just tell me the names. $\endgroup$ May 13, 2021 at 15:37
  • $\begingroup$ I'm sorry @VikashKotteeswaran, I don't have the book anymore. I borrowed it in the university library at the time. $\endgroup$
    – GRB
    May 15, 2021 at 16:45
  • $\begingroup$ It's alright, no problem $\endgroup$ May 15, 2021 at 18:29

1 Answer 1


For eq3 and eq4 it is a simple exercise to show that in general $$ \text{det }(\mathbb{1}- \mathbb{M}) = 1 - \text{tr }\mathbb{M} + \frac{1}{2} \text{tr }^2\mathbb{M} - \frac{1}{2}\text{tr }\mathbb{M}^2 + \mathcal{O}\left(\mathbb{M}^3\right) $$ hint: use the Levi-civita identity for the determinant Applying this formula immediately gives you det$(-g)=$det$(-\mathfrak{g})=\ldots$, and the square root by taylor expanding again.

Then for eq2 just use $$ g^{\alpha\beta}=\frac{1}{\sqrt{\text{det}(-g)}}\mathfrak{g}^{\alpha\beta}=\frac{1}{\sqrt{\text{det}(-\mathfrak{g})}}\mathfrak{g}^{\alpha\beta} = \frac{\eta^{\alpha\beta}-h^{\alpha\beta}}{1-\frac{1}{2}h + \frac{1}{8}h^2-\frac{1}{4}h^{\alpha\beta}h_{\alpha\beta}+\ldots} $$ and just do a taylor expansion (denomenator is just scalars).

Finally you can deduce eq1 from eq2 by inverting the latter perturbatively

  • $\begingroup$ It seems this is the correct idea. There is a small error (it's better to expand $\dfrac{1}{\sqrt{-g}}$ directly instead of calculating a double expansion), but I managed to find easily equations 2, 3 and 4. Sadly I'm not sure of what "inverting perturbatively" means. I also don't understand where your first formula comes from, I just used it trusting you: it works, but I don't feel happy about blind trust. Do you have a hint to obtain the formula? $\endgroup$
    – GRB
    May 12, 2015 at 13:34
  • $\begingroup$ I found also equation 1, I managed to express $g_{\alpha \beta} = \eta_{\alpha \beta} + k^{(1)}_{\alpha \beta} + k^{(2)}_{\alpha \beta}$, where $k^{(1)}$ and $k^{(2)}$ are respectively of order $G$ and $G^2$, and then solve $g_{\alpha \beta} g^{\beta \gamma} - \delta_{\alpha}^{\gamma} = O(G^3)$. Is that what you meant? Still I can't understand your first formula, the founding stone of all the calculations. $\endgroup$
    – GRB
    May 12, 2015 at 14:38
  • $\begingroup$ I feel like an idiot... I managed to calculate your first formula to obtain the approximate determinant and then forgot completely about it. Thanks a lot for the help! $\endgroup$
    – GRB
    May 12, 2015 at 15:15

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