I'm reading an article which includes the following equation involving a perturbed metric:

$$G_{AB} = \eta_{AB} + \overset{1}{\gamma}_{AB} + 2\overset{1}{\chi}_{(A,B)}\tag{4.1}$$

I don't understand how this equation was obtained; in particular, I don't understand how was obtained the third summand. Is there some literature explaining how this was obtained, or can you explain where it came from?


1 Answer 1


Given a arbitrary metric $g_{\mu\nu}$ you can introduce a reference (background) metric $\bar{g}_{\mu\nu}$ (in the paper notation it is just the Minkowski metric $\eta_{\mu\nu}$) in a way that $\delta{}g_{\mu\nu} = g_{\mu\nu} - \bar{g}_{\mu\nu}$ is small (in some sense). You can reintroduce the background metric in a way that the perturbation remains small, this action is parametrized by an small diffeomorphism (you can see why in the Mukhanov's review of perturbations: dx.doi.org/10.1016/0370-1573(92)90044-Z) generated by an arbitrary vector field $-\xi^\alpha$, thus, the background change by $$\bar{g}_{\mu\nu}\rightarrow\bar{g}_{\mu\nu}-\mathcal{L}_\xi\bar{g}_{\mu\nu},$$ where $\mathcal{L}_\xi$ is the Lie derivative with respect to $\xi^\alpha$. Given this transformation, the perturbation transform as $$\delta{}g_{\mu\nu}\rightarrow\delta{}g_{\mu\nu} + \mathcal{L}_\xi\bar{g}_{\mu\nu}.$$

If you choose a covariant derivative compatible with $\bar{g}_{\mu\nu}$, say $\bar{\nabla}_\alpha\bar{g}_{\mu\nu} = 0$, the Lie derivative can be written as $$\mathcal{L}_\xi\bar{g}_{\mu\nu} = 2\bar{\nabla}_{(\mu}\xi_{\nu)}.$$ In the paper the background metric is just the Minkowski metric, then in Cartesian coordinates $$\mathcal{L}_\xi\eta_{\mu\nu} = 2\partial_{(\mu}\xi_{\nu)} = 2\xi_{(\mu,\nu)},$$ where the comma represent the partial derivative.

So in general you can write a metric in terms of the background, perturbation and a gauge transformation as $$g_{\mu\nu} = \bar{g}_{\mu\nu} + \delta{}g_{\mu\nu} + 2\bar{\nabla}_{(\mu}\xi_{\nu)}.$$


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