# Interpreting perturbation theory in general relativity

In quantum mechanics, we start with a Hamiltonian $$H_0$$ for which we know the exact eigenstates and energy eigenvalues. We perturb it by a known term $$H$$, and then attempt to compute (approximately) the new eigenstates and eigenvalues.

In general relativity, my understanding is we start with a metric $$g_{\mu \nu}$$, and perturb it by a known $$h_{\mu \nu}.$$ But in my lecture notes (https://arxiv.org/abs/0804.2595), the lecturer shows how to compute $$h_{\mu \nu}$$. I thought we perturbed a system by a known quantity; can someone clarify the regular procedure of perturbation theory in general relativity, and what typical 'goals' are?

The only alternative I see is that we perturb a known solution $$g_{\mu \nu}$$ by an unknown perturbation $$h_{\mu \nu}$$, state how we would like the stress-energy $$T_{\mu \nu}$$ to change, and then try and compute $$h_{\mu \nu}$$ such that it does. Could this be the correct interpretation?

• Minor comment to the post (v1): In the future please link to abstract pages rather than pdf files, e.g., arxiv.org/abs/0804.2595 Mar 11 '14 at 19:43
• @Qmechanic I'll make sure to do so in future posts, thank you Qmechanic. Mar 11 '14 at 21:30

Yes, your second guess is more or less correct. In GR, perturbing the metric is the usual way of doing perturbation theory. One writes for the true metric $g_{\mu\nu}$ an expansion of the form $$g_{\mu\nu} = g^{(0)}_{\mu\nu}+h_{{\mu\nu}}+O(h^2),$$ where $g^{(0)}_{\mu\nu}$ is known and usually called background metric. One then substitutes this into the Einstein equations and find equations for $h_{\mu\nu}$. Solving those then gives you the first order correction to the background metric.
• A quick question: in perturbation theory, whenever I see a covariant derivative, e.g. $\nabla_{a}h^{ab}$, does it mean the covariant derivative featuring the Christoffel symbols of the full $g$, only $g^{(0)}$ or only $h$? Mar 11 '14 at 18:24
• In that case, you should expand the covariant derivative using the expansion given above and keep only the terms of first order in $h_{\mu\nu}$. Mar 11 '14 at 20:03
• @user32361 it will end up not mattering if you are computing a term like that using $g$ or $g^{(0)}$, because the christoffel symbols of the two operators only differ by terms involving $h$, and you already have a term of h in the argument of the derivative, so anything else will be at least second order in $h$ Aug 31 at 17:43