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In linearized general relativity indices are raised and lowerd by contracion with the flat space metric tensor $\eta_{\mu \nu}$. I don't really understand why we can do that. In the book gravitational waves by Michele Maggiore this is just called a "convention". That seems very weird to me, because raised and lowered indices have a geometrical meaning and I feel like such a convention would have consequences.

In other sources I found the short explanation, that using $\eta_{\mu \nu}$ instead of $g_{\mu \nu}(x)$ is an approximation which is correct to linear order in the perturbation $h_{\mu \nu}(x)$. This makes more sense to me, but nowhere was I provided some kind of calculation that proves this and trying myself, I failed to do it and came across some contradiction:

In linear theory the metric tensor is

$$g_{\mu \nu}(x) = \eta_{\mu \nu}+h_{\mu \nu}(x)~~~~~~~~~~\text{with}~|h_{\mu \nu}| \ll 1 $$

To find the linearized Christophel Symbols, one needs to find the inverse metric tensor $g^{\mu \nu}$ first. I found the following derivation, where raising indices via $\eta$ is used:

The Ansatz is $g^{\mu \nu}(x)=\eta^{\mu \nu} + \bar{h}^{\mu \nu}(x)~~~~~~~~~~\text{with}~|\bar{h}_{\mu \nu}| \ll 1$

then

$$g^{\mu \nu}g_{\nu \kappa}=\delta^\mu_\kappa$$

$$\Leftrightarrow~~~ \eta^{\mu \nu}\eta_{\nu \kappa}+\eta^{\mu \nu}h_{\nu \kappa}+\bar{h}^{\mu \nu}\eta_{\nu \kappa} + \bar{h}^{\mu\nu}h_{\nu\kappa} = \delta^\mu_\kappa $$

using $\eta^{\mu\nu}\eta_{\nu\kappa}=\delta^\mu_\kappa$ and ignoring the $\mathcal{O}(h^2)$ term we get

$$\eta^{\mu\nu}h_{\nu\kappa}=-\bar{h}^{\mu\nu}\eta_{\nu\kappa}$$

$$\Leftrightarrow~~~h^\mu_\kappa=-\bar{h}^\mu_\kappa$$

In the last step the index was raised with the flat space metric. So we end up with:

$$g^{\mu \nu}(x)=\eta^{\mu\nu}-h^{\mu\nu}(x)$$

Now my first problem is:

If I can just lower and raise indices of tensors with $\eta^{\mu\nu}$, why not just do that with $g_{\mu\nu}$ which is a tensor too? That would give:

$$g^{\mu\nu}(x)~=~\eta^{\mu\alpha}\eta^{\nu\beta}g_{\alpha\beta}(x)~=~\eta^{\mu\alpha}\eta^{\nu\beta}\eta_{\alpha\beta}+\eta^{\mu\alpha}\eta^{\nu\beta}h_{\alpha\beta}(x)~=~\eta^{\mu\alpha}\delta^\nu_\alpha+h^{\mu\nu}(x)~=~\eta^{\mu\nu}+h^{\mu\nu}(x)$$

But that is not what the first calculation gives...

My second problem is, that I just don't see how to justify the usage of $\eta_{\mu\nu}$ for raising and lowering indices. In linearized GR there is some symmetry under coordinate transformations

$$x^\mu \rightarrow x'^\mu=x^\mu+\xi^\mu(x)~~~~~~~\text{with}~|\partial_\nu\xi^\mu|\ll 1~~~~~~(1)$$

I would expect that under such coordinate transformations the components of contravariant vectors $A^\mu$ and covariant vectors $A_\mu$ would transform (up to linear order) in the common way, i.e

$$A'^\mu = \frac{\partial x'^\mu}{\partial x^\nu}A^\nu~~~~~~~~~~\text{and}~~~~~~~~~~A'_\mu=\frac{\partial x^\nu}{\partial x'^\mu}A_\nu$$

But if I put this to a test I get:

$$A'_\mu~=~\eta_{\mu\nu}A'^\nu~=~\eta_{\mu\nu}\frac{\partial x'^\nu}{\partial x^\alpha}A^\alpha~=~\eta_{\mu\nu}\frac{\partial x'^\nu}{\partial x^\alpha}\eta^{\alpha\beta}A_\beta~~~~~~(2)$$

plugging (1) into (2) I get

$$A'_\mu=\eta_{\mu\nu}\eta^{\alpha\beta}\left(\delta^\nu_\alpha +\frac{\partial \xi^\nu}{\partial x^\alpha}\right)A_\beta~=~\left(\delta^\beta_\mu+\eta_{\mu\nu}\eta^{\alpha\beta}\frac{\partial\xi^\nu}{\partial x^\alpha}\right)A_\beta$$

But what I would want to get is

$$A'_\mu=\frac{\partial x^\beta}{\partial x'^\mu}A_\beta~=~\left(\delta^\beta_\mu-\frac{\partial\xi^\beta}{\partial x'^\mu}\right)A_\beta$$

So I don't know how I can justify lowering indices via $\eta_{\mu\nu}$, if by doing so I don't get a covariant vector that transforms as covariant vectors do... I would be thankful if anybody finds my mistakes or knows, where I can read up on this.

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First, a point of principle - the raising and lowering of indices is a notational convention which is, in principle, completely unnecessary. When we take a tensor with "natural" index placement $T^{\mu\nu}$ and then write the collection of symbols $T_{\mu\nu}$, what we are really doing is saving ourselves the bother of writing $g_{\mu\alpha}g_{\nu\beta}T^{\alpha\beta}$.

Conventionally, when we see an index which differs from its natural location, we know that it has been raised or lowered with the metric. However, it's not really necessary to use the metric for this; any non-degenerate bilinear form would do. When the text says that indices are raised or lowered with the Minkowski metric, that's a perfectly legal convention; it just tells you how to interpret indices which are not in their natural positions.


The general idea in linearized gravity is that you can either operate in the framework of general relativity while using a metric $g=\eta+h$, or you can operate in the framework of special relativity (so $g=\eta$) and treat $h$ as a dynamical field on a flat spacetime. The convention your text spoke of is to do the latter.

In order to adopt this viewpoint, we must first derive the equations of motion for $h$ from the linearized Einstein equations. The result is $$\square \bar h_{\mu\nu}+\eta_{\mu\nu}\partial^\rho\partial^\sigma \bar h_{\rho\sigma}-\partial^\rho\partial_\nu \bar h_{\mu\rho}-\partial^\rho\partial_\mu \bar h_{\nu\rho}= -\frac{16\pi G}{c^4}T_{\mu\nu}$$

where $\bar h$ is the trace-reversed perturbation $$\bar h_{\mu \nu} = h_{\mu\nu} -\frac{1}{2} \eta_{\mu\nu}h \ \ , \ \ h \equiv \eta^{\mu\nu} h_{\mu\nu}$$ In the above (and its derivation), it doesn't matter whether you raise and lower indices with $\eta$, or you raise them with $g$ but throw away higher order terms; the result is the same, because the only indices being raised or lowered are on terms which are already $\mathcal O(h)$.

Once we have this equation of motion, we are free to jump back into the framework of special relativity. The (non-dynamical) metric is now taken to be $\eta$, and $h$ is treated as (dynamical) field which propagates on a flat background spacetime with the equation of motion given above, very much like the electromagnetic field (see e.g. gravitoelectromagnetism).


If I can just lower and raise indices of tensors with $\eta_{\mu\nu}$, why not just do that with $g_{\mu\nu}$ which is a tensor too?

Note that $\eta^{\mu\alpha}\eta^{\nu\beta}g_{\alpha\beta}$ is a perfectly reasonable tensor. However, its components are not the matrix inverse of $g_{\alpha\beta}$, which we need to calculate the Christoffel symbols en route to deriving the linearized Einstein equations.

My second problem is, that I just don't see how to justify the usage of $\eta_{\mu\nu}$ for raising and lowering indices. In linearized GR there is some symmetry under coordinate transformations [...] I would expect that under such coordinate transformations the components of contravariant vectors $A^\mu$ and covariant vectors $A_\mu$ would transform (up to linear order) in the common way.

If you want tensors to transform properly under local coordinate transformations, you need to change the metric components along with them. If you want the metric to take the canonical form of the Minkowski metric $\eta_{\mu\nu}=\operatorname{diag}(-1,+1,+1,+1)$, then you are restricted in the coordinate transformations which you can perform. In particular, you are restricted to global Poincare transformations, which leave the Minkowski metric invariant.

The coordinate transformation $x\rightarrow x'=x + \xi(x)$ does not generally leave $\eta$ invariant, so you shouldn't expect it to be compatible with raising/lowering indices with $\eta$ unless you're willing to let $\eta$ change. It is in this sense that special relativity does not possess general coordinate invariance.


Addendum:

So there might be properties of lower und upper indice components that I am used to, that are not valid anymore when changing from one to the other convention.

Vectors are still vectors and covectors are still covectors, regardless of which bilinear form you use to map between them. That is, given some vector $\mathbf X$ and two nondegenerate bilinear forms $\mathbf g$ and $\mathbf B$, the quantities $Y_\mu = g_{\mu\nu} X^\nu$ and $Z_\mu = B_{\mu\nu}X^\nu$ are both covectors.

Put differently, a choice of form to do the raising and lowering amounts to a choice of a unique covector partner for each vector. A different form means different partners, but the geometric properties of the vector/dual space remain the same.

It looks like the convention is inconsistent if the inverse metric $g^{\mu\nu}$ is not the same as $\eta^{\mu\alpha}\eta^{\nu\beta}g_{\alpha\beta}$, but still labeled with upper indices. That also leads to the question if there are more tensors the new convention does not apply to?

A (2,0)-tensor $T^{\mu\nu}$ and its corresponding, index-lowered (0,2)-tensor $T_{\mu\nu}$ are not matrix inverses of one another. The only tensor for which this is true is the one you've chosen to do the raising and lowering. There's nothing inconsistent about this.

The inverse metric, by definition, is a (2,0)-tensor whose components $(g^{-1})^{\mu\nu}$ are the matrix inverse of $g_{\mu\nu}$. From here,

$$g^{\mu\nu} \equiv (g^{-1})^{\mu\alpha}(g^{-1})^{\nu\beta}g_{\alpha\beta}=(g^{-1})^{\mu\alpha}\delta^\nu_\alpha = (g^{-1})^{\mu\nu}$$

But this is a consequence of the defintion of the inverse metric, not the definition itself. In particular, the inverse metric is not defined to be the index-raised version of the metric, because what would that even mean without a tensor to do the raising?

I always thought that the whole point of using 4-vectors and tensors to write down equations in relativity was that the components behave in a certain way under coordinate transformations. If that is not true anymore, what is the point then of raising and lowering indices? Is a $A_{\mu}B^\mu$ even a scaler under coordinate transformations? Are the equations still form invariant?

It is still true. But when you perform a coordinate transformation, you need to transform everything - that means the metric as well. Explicitly, if you start with the metric $\eta$ and perform the coordinate transformation $x\rightarrow x'=x+\xi(x)$, then the components of the metric become $$\eta_{\mu\nu}\rightarrow \eta'_{\mu\nu} =\eta_{\mu\nu} -2 \partial_{(\mu}\xi_{\nu)}+\mathcal O(\partial \xi^2)$$ where the index on $\xi$ has been lowered with $\eta$. If you plug this in to your computation

$$A'_\mu=\eta'_{\mu\nu}A'^\nu = \eta'_{\mu\nu} \frac{\partial x'^\nu}{\partial x^\alpha}A^\alpha = \eta'_{\mu\nu} \frac{\partial x'^\nu}{\partial x^\alpha}\eta^{\alpha\beta}A_\beta$$

Then everything works just fine. However, you are no longer working with the canonical form of the Minkowski metric, $\eta=\operatorname{diag}(-1,+1,+1,+1)$. If you want to preserve the form of $\eta$, then you must restrict yourself to global Poincaré transformations, as we generally do in special relativity.

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  • $\begingroup$ Thanks for the answer. I had to think for a day but now I see that I'm still confused about this whole thing. To the first point: Of course anybody is in his legal right to to raise and lower indices to according to his or her own rules. But when you change a convention normally you also have to be aware of changes in the meaning of things. So there might be properties of lower und upper indice components that I am used to, that are not valid anymore when changing from one to the other convention. The two points I made in the original post are examples for that: ... $\endgroup$ Jul 8, 2020 at 6:45
  • $\begingroup$ It looks like the convention is inconsistent if the inverse metric $g^{\mu \nu}$ is not the same as $\eta^{\mu\alpha}\eta^{\nu\beta}g_{\alpha\beta}$, but still labeled with upper indices. That also leads to the question if there are more tensors the new convention does not apply to? $\endgroup$ Jul 8, 2020 at 6:46
  • $\begingroup$ With the second point I have my biggest problem. I always thought that the whole point of using 4-vectors and tensors to write down equations in relativity was that the components behave in a certain way under coordinate transformations. If that is not true anymore, what is the point then of raising and lowering indices? Is a $A_\mu B^\mu$ even a scaler under coordinate transformations? Are the equations still form invariant? $\endgroup$ Jul 8, 2020 at 6:49
  • $\begingroup$ @BenitoMcLanbeck I have added a substantial section to my answer to address your additional questions. $\endgroup$
    – J. Murray
    Jul 8, 2020 at 14:07

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