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I know this is an already answered question, but I couldn't make head or tail of it, and it's bugging me. I know I'm probably asking a silly question, but please bear with me as I'm 14 and this is my first post.

I saw this perturbation while studying up a bit of linearised gravity:

$$g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu} \tag{1}$$

and then the next line says: " you raise the indices to get":

$$g^{\mu\nu} = \eta^{\mu\nu} - h^{\mu\nu}. \tag{2}$$

I have done a bit of tensor calculus, but I still couldn't understand how the obvious fact $g_{\mu\nu}\, g^{\mu\rho} = \delta^{\rho}_{\nu}$ leads to the second equation. I tried everything I could, but all I get is the first equation with raised indices different from the given ones, and a persistent + sign. Please explain and derive the second equation step by step, as if to an idiot.

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Where did you see this question answered on the site? –  JamalS May 19 at 10:23

2 Answers 2

Note that: $$ h^{\mu \nu} = \eta^{\mu \rho}\eta^{\nu \lambda} h_{\rho \lambda} $$ Therefore, up to first order, we have: \begin{equation} \begin{aligned} g^{\mu \nu}g_{\nu \sigma} & = (\eta^{\mu \nu} - h^{\mu \nu})(\eta_{\nu \sigma} + h_{\nu \sigma}) \\& =\eta^{\mu \nu}\eta_{\nu \sigma} + \eta^{\mu \nu}h_{\nu \sigma} - \eta_{\nu \sigma} h^{\mu \nu} + \mathcal{O}(h^2) \\& = \delta^\mu_\sigma + \eta^{\mu \nu}h_{\nu \sigma} - \eta_{\nu \sigma} \eta^{\mu \rho}\eta^{\nu \lambda} h_{\rho \lambda} + \mathcal{O}(h^2) \\& = \delta^\mu_\sigma + \eta^{\mu \nu}h_{\nu \sigma} - \delta_\sigma^\lambda \eta^{\mu \rho} h_{\rho \lambda} + \mathcal{O}(h^2) \\& = \delta^\mu_\sigma + \eta^{\mu \nu}h_{\nu \sigma} - \eta^{\mu \rho} h_{\rho \sigma} + \mathcal{O}(h^2) \\& = \delta^\mu_\sigma + \eta^{\mu \nu}h_{\nu \sigma} - \eta^{\mu \nu} h_{\nu \sigma} + \mathcal{O}(h^2) \\& = \delta^\mu_\sigma + \mathcal{O}(h^2) \end{aligned} \end{equation} which is what we require.

Edit in response to comments. The inverse of $g_{\nu \sigma}$, which is denoted by $g^{\mu \nu}$, is defined to satisfy the following equation: $$ g^{\mu \nu}g_{\nu \sigma} = \delta^\mu_\sigma $$ In other words, we need to find an expression for $g^{\mu \nu}$ such that the following equations is satisfied: $$ g^{\mu \nu}(\eta_{\nu \sigma} + h_{\nu \sigma}) = \delta^\mu_\sigma $$ As I have shown above, the function that obeys the above equation up to first order in $h_{\mu \nu}$ is: \begin{equation} g^{\mu \nu} = \eta^{\mu \nu} - h^{\mu \nu} \end{equation} That is really all that is going on.

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Hunter, I do understand what you mean, but that's not what I asked. I asked how do we get the second equation from the first one....you used equation 2 in the first step itself. I only wanted to know how HOW do you get that....But thanks anyway!! –  GRrocks May 19 at 10:44
    
@GRrocks that makes no sense to me. What source are you studying from? –  Hunter May 19 at 10:45
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@GRrocks It is appears to be the exact same situation as equations (4.13) and (4.14). –  Hunter May 19 at 10:57
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Ahh...I get it now..Thanks a lot @Hunter..I guess I was stupid...but you showed earlier that $\eta^{\mu\nu}$ -$h^{\mu\nu}$ is the required function, so its more like trial and error. Can we not take an arbitrary tensor and show that to get equation 2 of my question ( i just edited the numbering of equations), that arbitrary tensor has to be $\eta^{\mu\nu}$ -$h^{\mu\nu}$ ???apologies if im being foolish again.. –  GRrocks May 19 at 11:17
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@GRrocks: well, the only 100% rigorous way to do what you ask is to write down $h_{ab}$ in components, invert the matrix, and drop the second order terms. Then, you can extract out the equation you want. But that's obviously a ton of work. –  Jerry Schirmer May 19 at 16:32

If you have $g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu}$, with $|h_{\mu\nu}|\ll 1$ a perturbation, $\eta_{\mu\nu} = \text{diag}(-1,1,1,1)$, then you can simply perform a Taylor expansion to obtain the inverse, $$g^{\mu\nu} = \eta^{\mu\nu} - h^{\mu\nu}+ \mathcal{O}(h^2),$$ where the indices of $h^{\mu\nu}$ are raised by $\eta^{\mu\nu}$. This is easy to see if you take the diagonal elements first, the off-diagonal ones are of second order, negligible. You don't need to solve a matrix equation for this, because the inverse is linearly dependent on $\eta_{\mu\nu}=\eta^{\mu\nu}$, $h_{\mu\nu}$ and their products, hence it must be of the form $g^{\mu\nu} = a\eta^{\mu\nu} + b h^{\mu\nu} + \ldots$, where the ellipses imply linear dependencies of higher order and the coefficients are constants. Also, note that $h_{\mu\nu} \eta^{\mu\mu} \eta^{\nu\nu} = h^{\mu\nu}$, no summation implied, namely $h^{\mu\nu} \propto h_{\mu\nu}$.

Also, a straightforward way to calculate $\delta g^{\mu\nu}$ is by use of $\delta g^{\mu\nu} = - g^{\mu\rho} g^{\nu\sigma} \delta g_{\rho\sigma}$ with $\delta g_{\mu\nu} = h_{\mu\nu}$.

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In general relativity, the demand $|h_{\mu\nu}| \ll 1$ is not enough to guarantee that the perturbation is small. Normally you treat case by case to determine what 'small' means for your particular background. –  JamalS May 19 at 16:27
    
@JamalS If we have a background described by $\eta_{\mu\nu}$, doesn't $|h_{\mu\nu}|\ll 1$ define smallness unambiguously? –  auxsvr May 19 at 16:30
    
For the Minkowski metric, of course, but not for a general background. I'm sorry if it seemed I was challenging your answer; the comment was meant as an addendum for future viewers. –  JamalS May 19 at 16:34
    
I assume that the OP is attempting to derive the newtonian limit, otherwise none of the answers makes sense. –  auxsvr May 19 at 17:03

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