Linearized Gravity & Metric Perturbation when Indices Raised

[George Keeling]

TL;DR Summary

$g^{\mu\nu}=\eta^{\mu\nu} \pm h^{\mu\nu}$?

I have just met linearized gravity where we decompose the metric into a flat Minkowski plus a small perturbation$$g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu},\ \ \left|h_{\mu\nu}\ll1\right|$$from which we 'immediately' obtain $$g^{\mu\nu}=\eta^{\mu\nu}-h^{\mu\nu}$$I don't obtain that. In my rule book for tensor index manipulation it says you can always raise all the indices on an equation so I get$$g^{\mu\nu}=\eta^{\mu\nu}+h^{\mu\nu}$$Here's how to get a minus sign. For small changes $$\delta\left(g^{\mu\rho}g_{\rho\nu}\right)=g^{\mu\rho}\delta\left(g_{\rho\nu}\right)+\delta\left(g^{\mu\rho}\right)g_{\rho\nu}=0$$$$\Rightarrow g^{\mu\rho}\delta\left(g_{\rho\nu}\right)g^{\nu\sigma}=-\delta\left(g^{\mu\rho}\right)g_{\rho\nu}g^{\nu\sigma}$$now if I say $\delta g\rightarrow h$ I get $$g^{\mu\rho}g^{\nu\sigma}h_{\rho\nu}=-h^{\mu\sigma}$$So raising indices on $h_{\mu\nu}$ produces the minus sign! Why should I accept this exceptional result instead of the usual one that $g^{\mu\rho}g^{\nu\sigma}h_{\rho\nu}=h^{\mu\sigma}$?

@haushofer proved the minus sign a different way here but that doesn't tell me why the 'normal' result is wrong.

 

[Orodruin]

Summary:: $g^{\mu\nu}=\eta^{\mu\nu} \pm h^{\mu\nu}$?

I don't obtain that. In my rule book for tensor index manipulation it says you can always raise all the indices on an equation

This is wrong in this case because $\eta^{\mu\nu}$ has a predefined meaning that is different from the symbol you get when you raise the indices with the metric g. It is therefore not a tensor in this case.

 

[vanhees71]

It's because $g^{\mu \nu}$ is obtained from $g_{\mu \nu}$ as the inverse matrix, i.e., via
$$g^{\mu \nu} g_{\nu \rho}=\delta_{\nu}^{\rho}.$$
Now make
$$g_{\nu \rho}=\eta_{\nu \rho}+h_{\nu \rho}$$
and
$$g^{\mu \nu}=\eta^{\mu \rho}+\tilde{h}^{\mu \nu},$$
where
$$(\eta_{\mu \nu})=(\eta^{\mu \nu})=\mathrm{diag}(1,-1,-1,-1).$$
From this you get, neglecting the term in second order of $h_{\mu \nu}$,
$$g^{\mu \nu} g_{\nu \rho} = (\eta^{\mu \nu}+\tilde{h}^{\mu \nu})(\eta_{\nu \rho} + h_{\nu \rho})
=\delta_{\rho}^{\mu}+{h^{\mu}}_{\rho} + {\tilde{h}^{\mu}}_{\rho} +\ldots \stackrel{!}{=} \delta_{\rho}^{\mu}.$$
Note that here by definition the raising and lowering of indices is done with the "Minkowski metrix" $\eta_{\mu \nu}$ or $\eta^{\mu \nu}$ respectively. We get
$${\tilde{h}^{\mu}}_{\rho} = -{h^{\mu}}_{\rho}$$
or raising the 2nd index again with $\eta^{\rho \nu}$:
$$\tilde{h}^{\mu \nu}=-h^{\mu \rho}.$$
So the additional - comes from the extra convention to lower and raise indices with $\eta_{\mu \nu}$ and $\eta^{\mu \nu}$ respectively when working within "linearized GR".

 

[George Keeling]

It is therefore not a tensor in this case.

Raising indices on $g_{\mu\nu}$ and $\eta_{\mu\nu}$ worked but on $h_{\mu\nu}$ didn't. Is $h_{\mu\nu}$ the It you refer to? (Edit: No, it's $\eta_{\mu\nu}$ or ${\widetilde{\eta}}^{\mu\nu},$ see below)

 

[George Keeling]

Note that here by definition the raising and lowering of indices is done with the "Minkowski metrix"

When raising indices on $h$ I don't think it makes any difference whether you use $g$ or $\eta$ or whether there's a + or a - in the $h_{μν}$ because nearly everything is second order or smaller: $$g^{\mu\sigma}g^{\nu\rho}h_{\sigma\rho}=\left(\eta^{\mu\sigma}\pm h^{\mu\sigma}\right)\left(\eta^{\nu\rho}\pm h^{\nu\rho}\right)h_{\sigma\rho}=\eta^{\mu\sigma}\eta^{\nu\rho}h_{\sigma\rho}\pm\eta^{\mu\sigma}h^{\nu\rho}h_{\sigma\rho}\ldots$$
The proof you give is pretty like the one I gave. Yours raises and lower indices on ${\widetilde{h}}^{\mu\nu},h_{\mu\nu}$ in the normal way. Why is it suddenly OK to do that?
I'm still confused.

 

[Orodruin]

Raising indices on $g_{\mu\nu}$ and $\eta_{\mu\nu}$ worked but on $h_{\mu\nu}$ didn't. Is $h_{\mu\nu}$ the It you refer to?

Raising the indices on $g$ and $\eta$ does not use the same metric. That is the entire point. They differ by a quantity of order h.

 

[Orodruin]

Consider the one-dimensional manifold with metric $g= 1+h$. The inverse metric is then $1/g = 1/(1+h) = 1-h + \mathcal O(h^2)$. This already tells you that the + sign in the inverse is wrong.

 

[PeterDonis]

In my rule book for tensor index manipulation

What rule book? Please give a specific reference.

 

[George Keeling]

Raising the indices on g and η does not use the same metric. That is the entire point. They differ by a quantity of order h.

At last I get it:
we can raise indices on $$g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}$$and get$$g^{\mu\nu}=\eta^{\mu\nu}+h^{\mu\nu}$$but the $\eta^{\mu\nu}$ in that is the Minkowski metric in the $g^{\mu\nu}$ manifold. So its components are not the usual$${\widetilde{\eta}}^{\mu\nu}=\left(\begin{matrix}-1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\\\end{matrix}\right)$$but something slightly different. We want to use ${\widetilde{\eta}}^{\mu\nu}$ , not $\eta^{\mu\nu}$. So we'll also have to replace $h^{\mu\nu}$ by an unknown tensor ${\widetilde{h}}^{\mu\nu}$ and we get $$g^{\mu\nu}={\widetilde{\eta}}^{\mu\nu}+{\widetilde{h}}^{\mu\nu}$$which is the same as

$g^{\mu \nu}=\eta^{\mu \rho}+\tilde{h}^{\mu \nu},$

give or take a tilde.
I had also missed the significance of the following line

$(η_{μν})=(η^{μν})=diag(1,−1,−1,−1).$

and I am using the opposite sign convention and using up more screen space. Then I can just follow vanhees61.

Thank you both for your patience!

 

[George Keeling]

What rule book? Please give a specific reference.

It's my collected notes on various rules about tensors. You probably wouldn't call it a proper book!

 

[PeterDonis]

It's my collected notes on various rules about tensors.

Since you open with a claim about there being an error in Carroll, and your claim is erroneous, you should not be depending on these to be accurate. I would strongly suggest using a textbook on tensor analysis as a reference if you want to be sure you're using the right rules. I would also strongly suggest asking for input from others before treating any claim about an error in a published textbook as accurate.

You probably wouldn't call it a proper book!

Indeed not.

 

[vanhees71]

Well, I think it's a good attitude, not to blindly believe what's printed in textbooks or even peer-reviewed papers. If I'd had a recipe against typos, misconceptions, or other errors, I'd made a lot of money selling the method ;-)). Of course, it's likely that one misunderstands the textbook, but if you can't rederive some result in it, it's good to question whether the author of the textbook or the paper may be in error and look for other treatments of the subject.

 

[George Keeling]

Since you open with a claim about there being an error in Carroll,

Sorry if it looked like that. I should have emphasised that it was me being so thick that I couldn't understand.

 

[George Keeling]

One more thing

I would strongly suggest using a textbook on tensor analysis

Can you recommend such a textbook with lots of exercises?

 

[PeterDonis]

Can you recommend such a textbook with lots of exercises?

If you mean a math textbook, I can't recommend one since my knowledge of tensor analysis comes from physics textbooks. The one I first learned it from was MTW, but that might be too heavy (both in the literal and figurative sense--the book weighs more than any other book in my library by a fair margin, excepting only the unabridged dictionary and the CRC). Carroll's book (or his free online lecture notes) might be a better, lighter introduction from a physics viewpoint, but it might not meet the "lots of exercises" part.

 

[Orodruin]

I know which textbook I would suggest ... but I am quite biased on the matter ;)

 

[vanhees71]

That I can warmly suggest:

https://www.taylorfrancis.com/books...l-methods-physics-engineering-mattias-blennow

I learned GR first for myself from Landau&Lifshitz vol. 2. That provides the necessary math in a very efficient manner. It's however restricted to holonomous bases. Then I got MTW an also learned the more modern Cartan approach. It emphasizes the geometrical interpretation ("geometrodynamics"). If you want less geometry and a more physical approach, Weinberg's book (1971) is a gem (as all textbooks by Weinberg).

 

[George Keeling]

Strangely, I have three of the books mentioned! So I guess I have plenty to read and study. Thanks again.