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I am reading some lecture notes on general relativity, where the author talks about perturbation theory applied to GR. In the case of a weak gravitational field, one perturb about the Minkowski metric.

The author says that one can take small perturbations about a background spacetime apart from Minkowski, in which case the metric is

$g_{\mu \nu} = g_{\mu \nu}^{(0)} + h_{\mu \nu} $.

But in this case, what is the zeroth-order term $g_{\mu \nu}^{(0)}$, or does it depend on the exact metric, and then one just applies perturbation theory to it? I assume there are some differences from traditional perturbation theory when applied to metrics?

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You have to choose a background metric around which you do the perturbation theory. It could be Minkowski, for example, but it could also be Schwarzschild for another example. The former might be appropriate if you are "far" from all strong sources. The latter might be appropriate if you want to understand the relatively small deviations from Schwarzschild that you'd get if you had small objects orbiting a black hole.

As with other perturbation theory approaches, you need some insight into what makes a sensible background in order to get started. In GR, that corresponds to picking the form of what you labeled $g_{\mu\nu}^{(0)}$ based on the physics of your problem.

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  • $\begingroup$ So the zeroth-order term above could be the Schwarzchild metric, but depends on how you have formulated the problem as well as the physics of the problem? $\endgroup$ – Tom Aug 8 '19 at 14:16
  • $\begingroup$ No, if you want to perturb around the Schwarzschild metric then the zero-order term is the Schwarzschild metric. To have decided that you want to perturb around Schwarzschild, however, you should have had some insight into your system that suggests that the spacetime is "almost" Schwarzschild. $\endgroup$ – Brick Aug 8 '19 at 14:18
  • $\begingroup$ Yes, that's what I meant, sorry if I was not clear. The zeroth-order term is Schwarzchild, if the physical system can in some sense be regarded as close to Schwarzchild. Is there some asymptotic sense in which any system can be regarded as Schwarzchild or Minkowski as one goes out to spatial infinity? $\endgroup$ – Tom Aug 8 '19 at 14:21
  • $\begingroup$ Yes, any localized distribution of matter with no angular momentum and no charge will look like Schwarzschild far enough away. (With angular momentum or charge then like the other "standard" BH solutions.) Schwarzschild itself looks like Minkowski even further away. $\endgroup$ – Brick Aug 8 '19 at 14:54
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    $\begingroup$ Kerr for angular momentum and no charge. Reissner–Nordstrom with charge and no spin. $\endgroup$ – Brick Aug 8 '19 at 14:59
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As @Brick already said, $g_{\mu \nu}^{(0)}$ depends on the subject you want to apply perturbation theory to. It may be the Schwarzschild metric if you have small objects around a black hole. Another example are gravitational waves, where you have the minkowski metric $\eta$ for flat space with a perturbation for the waves.

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  • $\begingroup$ What does it imply for the spacetime 4-manifold if the metric with which it is equipped is a perturbation around some other metric, is there some asymptotic interpretation as one goes out to spatial infinity? $\endgroup$ – Tom Aug 8 '19 at 14:30
  • $\begingroup$ If i remember correctly, in the limit of an infinite distance the space time is always flat and therefore described by the minkowski metric. $\endgroup$ – Jan2103 Aug 8 '19 at 14:34
  • $\begingroup$ But surely if the spacetime contains a black hole, it must be described by the Schwarzchild metric as one goes to spatial infinity? $\endgroup$ – Tom Aug 8 '19 at 14:36
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    $\begingroup$ The gravitation is described by the Schwarzschild metric. $$ \mathrm{d} s^{2}=g_{\mu \nu} \mathrm{d} x^{\mu} \mathrm{d} x^{\nu}=-\left(1-\frac{r_{\mathrm{s}}}{r}\right) \mathrm{d} t^{2}+\frac{1}{1-\frac{r_{\mathrm{s}}}{r}} \mathrm{d} r^{2}+r^{2} \mathrm{d} \theta^{2}+r^{2} \sin ^{2}(\theta) \mathrm{d} \phi^{2} $$ In the limit $r \rightarrow \infty$ you will get the minkowski metric $$\text{d} s^2= -\text{d}t^2 + \text{d} r^2(\text{d}\theta^2 + \text{sin}^2 \text{d} \phi^2)$$ $\endgroup$ – Jan2103 Aug 8 '19 at 14:39

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