# Which of the following expressions for Christoffel symbols is correct?

As shown in the figure, after adding a small perturbation term $$h$$ to the metric $$g$$, the expression of Christoffel symbols in an article is Eq. (10)

But I think the result should be in the following form. Is my idea correct?

References: https://doi.org/10.1103/PhysRev.108.1063 eq. (5).

• you are not wrong, but remember the assumptions. When all of h is small compared to g, then the derivatives of g will also be tiny, and so your red terms are closer to 2nd order than first order. They are thus negligible compared to the simpler form. Commented Nov 6, 2023 at 13:58
• That means the result of Equation （10） ignores higher-order infinitesimals, right?
– math
Commented Nov 6, 2023 at 14:05
• Isn't that obviously the case? If you see an application of perturbation theory and all the terms are only linear in the perturbation, surely that is an expansion only valid up to first order. Commented Nov 6, 2023 at 14:22
• This might be similar to a convention used in Wald's treatment of linearized gravity: $g_{ab} = \eta_{ab} + \gamma_{ab}$ "In order not to have $\gamma_{ab}$ hidden in a raised or lowered index, it is convenient to raise and lower tensor indices with $\eta_{ab}$ and $\eta^{ab}$ rather than $g_{ab}$ and $g^{ab}$". In your case that would be raising and lowering with $\bar g_{\mu\nu}$ which would not give the 2nd term in your result Commented Nov 6, 2023 at 17:23
– math
Commented Nov 7, 2023 at 1:46

You're technically correct here, but you also have to remember that $$\bar{g}_{\alpha \beta;\nu} = 0$$ by the definition of the covariant derivative. So all of your red terms vanish.

(I'm assuming here that the $${}_{;\nu}$$ subscript notation refers to the covariant derivative associated with the background metric, which is how these calculations normally proceed. If it somehow refers to the covariant derivative associated with the perturbed metric, then $$\bar{g}_{\alpha \beta;\nu}$$ is not zero, but it is of $$\mathcal{O}(h)$$ and so it can be ignored at linear order in perturbation theory.)

I think your confusion lies in the distinction between "," and ";", i.e. partial and covariant derivative. First of all, by the product rule, we have that

$$\delta\Gamma^{\alpha}_{\mu\nu}=\delta\bigg(\frac{1}{2}g^{\alpha\delta}(\partial_{\mu}g_{\nu\delta}+\partial_{\nu}g_{\mu\delta}-\partial_{\delta}g_{\mu\nu})\bigg)=-\frac{1}{2}h^{\alpha\delta}(\partial_{\mu}\overline{g}_{\nu\delta}+\partial_{\nu}\overline{g}_{\mu\delta}-\partial_{\delta}\overline{g}_{\mu\nu})+\frac{1}{2}\overline{g}^{\alpha\delta}(\partial_{\mu}h_{\nu\delta}+\partial_{\nu}h_{\mu\delta}-\partial_{\delta}h_{\mu\nu})=-\frac{1}{2}h^{\alpha\delta}(\overline{g}_{\nu\delta,\mu}+\overline{g}_{\mu\delta,\nu}-\overline{g}_{\mu\nu,\delta})+\frac{1}{2}\overline{g}^{\alpha\delta}(h_{\nu\delta,\mu}+h_{\mu\delta,\nu}-h_{\mu\nu,\delta})$$

Note that the linearization of $$g$$ with upper indices is $$g^{\mu\nu}=\overline{g}^{\mu\nu}-h^{\mu\nu}$$. This is where the initial minus come from. Next, observe that

$$\delta\Gamma^{\alpha}_{\mu\nu}=-\frac{1}{2}h^{\alpha\delta}(\partial_{\mu}\overline{g}_{\nu\delta}+\partial_{\nu}\overline{g}_{\mu\delta}-\partial_{\delta}\overline{g}_{\mu\nu})+\frac{1}{2}\overline{g}^{\alpha\delta}(\partial_{\mu}h_{\nu\delta}+\partial_{\nu}h_{\mu\delta}-\partial_{\delta}h_{\mu\nu})=\frac{1}{2}\overline{g}^{\alpha\delta}(\nabla_{\mu}h_{\nu\delta}+\nabla_{\nu}h_{\mu\delta}-\nabla_{\delta}h_{\mu\nu})=\frac{1}{2}\overline{g}^{\alpha\delta}(h_{\nu\delta;\mu}+h_{\mu\delta;\nu}-h_{\mu\nu;\delta})$$

where the covariant derivatives are with respect to the background metric $$g$$. To see the last to last equality, just spell out the terms $$\nabla_{\mu}h_{\nu\delta}$$ etc. explicitely using Christoffel symbols.

• In other words, no red parts appear under linear approximation?@G. Blaickner
– math
Commented Nov 10, 2023 at 12:48
• Exactly. More rigorously, linearization is like taking derivatives: Take a curve of metric $g(\lambda)$ and set $g(0)=:\overline{g}$ as well as $\partial_{\lambda}g(\lambda)\vert_{\lambda=0}=:h$. Then $\delta\Gamma^{\lambda}_{\mu\nu}:=\partial_{\lambda}\Gamma(g(\lambda))^{\lambda}_{\mu\nu}\vert_{\lambda=0}$. In this case, you see that only linear terms survive. Commented Nov 10, 2023 at 14:44