5
$\begingroup$

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?

Improve this question
$\endgroup$
2
  • $\begingroup$ 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 $\endgroup$
    – Qmechanic ♦
    Commented Mar 11, 2014 at 19:43
  • $\begingroup$ @Qmechanic I'll make sure to do so in future posts, thank you Qmechanic. $\endgroup$
    – user32361
    Commented Mar 11, 2014 at 21:30

1 Answer 1

Reset to default
5
$\begingroup$

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.

Improve this answer
$\endgroup$
4
  • $\begingroup$ 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 $h$? $\endgroup$
    – user32361
    Commented Mar 11, 2014 at 18:24
  • 1
    $\begingroup$ In that case, you should expand the covariant derivative using the expansion given above and keep only the terms of first order in $h_{\mu\nu}$. $\endgroup$
    – Hrodelbert
    Commented Mar 11, 2014 at 20:03
  • $\begingroup$ Although I am not really familiar with the Lichnerowicz operator, I am quite sure that this is the way to go, since this is the only way to do perturbation theory consistently. It doesn't matter which operators appear in your Einstein equations (or in the stress-energy tensor. $\endgroup$
    – Hrodelbert
    Commented Mar 12, 2014 at 8:19
  • $\begingroup$ @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 $h$ $\endgroup$
    – Zo the Relativist
    Commented Aug 31, 2021 at 17:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.