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In linearized general relativity, we have the unperturbed metric and the perturbed metric. In all textbook treatments, they say that they are going to raise and lower indices with the unperturbed metric. Wald says that this is a matter of convenience. I'm confused as to how usually rigorous Wald can be so glib about this. Here are possibilities I would have considered a priori:

  1. Raising/lowering with the perturbed metric is logically/mathematically required;
  2. Raising/lowering with the unperturbed metric is logically/mathematically required;
  3. Raising/lowering with the perturbed metric is logically/mathematically required but we can use the unperturbed metric as an approximation;

I would have thought that 1) was the correct way to proceed.

Other authors besides Wald are similarly glib. Physicists do like to hand wave in perfect synchronicity.

Can somebody explain why it isn't required to use the perturbed metric?

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Say $$ g_{ab} = \eta_{ab} + h_{ab} $$

When acting on a field which is itself first order in $h_{ab}$ then raising/lowering with $\eta^{ab}$ and $\eta_{ab}$ produce the same answer as $g_{ab}$ and $g^{ab}$ at first order in the perturbation. For other cases you have to think about it and use $g_{ab}$, $g^{ab}$ where necessary to keep the degree of approximation consistent.

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  • $\begingroup$ OK, so then just to make sure I understand. The perturbed metric is what is logically/mathematically correct, but we can approximate it with the unperturbed metric. As always, we need to make sure that the approximation is appropriate for the order to which we are doing our calculation. If we needed to do a calculation correct to nth order then we would need to use the perturbed metric or retain up to n orders in h. $\endgroup$
    – mster8390
    Commented Sep 3, 2021 at 9:15
  • $\begingroup$ @mster8390 yes. $\endgroup$ Commented Sep 3, 2021 at 9:15
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The remark is simply establishing a notational convention. Wald is saying that (in the follow discussion) whenever you encounter a tensor

$$ T^{\mu\nu\lambda} $$

this is by definition equal to

$$ {T^{\mu\nu}}_\alpha \eta^{\alpha\lambda}$$

not

$$ {T^{\mu\nu}}_\alpha (\eta^{\alpha\lambda} - h^{\alpha\lambda}).$$

One could have chosen the opposite convention, but this convention is convenient (for the reasons mentioned by Andrew Steane in his answer).

Mathematically, this corresponds to considering all tensors in the expressions as tensors on the background spacetime. We could have formulated everything in terms of tensors of the full physical spacetime, but since the background spacetime is the one we control mathematically, this would be much more awkward.

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  • $\begingroup$ So you disagree that it is an approximation? A convention and an approximation are not the same thing. $\endgroup$
    – mster8390
    Commented Sep 3, 2021 at 12:13
  • $\begingroup$ The approximation is in how you identify tensors on the background manifold to tensors on the full (physical) manifold. It is a convention whether the objects you write with indices are the one or the other. (An it is the latter convention that determines whether indices are raised and lower by the background or full metric.) Whichever of the two you pick, you still need to make sure your equations are consistent. $\endgroup$
    – TimRias
    Commented Sep 3, 2021 at 14:28

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