Can we use flat space coordinates in linearized gravity?

Question

I am studying the production of gravitational waves in linearized general relativity. While it is not mentioned anywhere, I am convinced, that the coordinates used in the derivations I have seen, are cartesian.

For example the Energy-Momentum-Tensor is taylor-expanded in Maggiore page 106:

$$T_{kl}\Bigl(t-\frac{r}{c}+\frac{\boldsymbol{x'}\cdot \boldsymbol{n}}{c},\boldsymbol{x'}\Bigr)~\approx~T_{kl}\Bigl(t-\frac{r}{c},\boldsymbol{x'}\Bigr)+\frac {x'^in^i}{c}\partial_tT_{kl}+...$$

Where the derivative is evaluated at the point $(-r/c,\boldsymbol x')$, $\boldsymbol{x'}$ is a position vector and $\boldsymbol{n}$ is the unit normal vector pointing from the source to the observer. So the dot-product is $\boldsymbol{x'}\cdot\boldsymbol{n}=x'^in^i$ which suggests cartesian coordinates. But if spacetime is curved, there is no way to have cartesian coordinates. It already seems odd to me that the position can be described by a vector $\boldsymbol{x'}$ which is generally not possible in general relativity.

There are many other passages that make me think, that flat space coordinates are used. But this one seems to be very clear.

Now, the condition on which the equation above is based is that spacetime is described by

$$g_{\mu\nu}(x)=\eta_{\mu\nu}+h_{\mu\nu}(x)$$

which means that spacetime is nearly flat and the coordinates are nearly Lorentz coordinates i.e. nearly cartesian spatial coordinates. But nearly flat does not mean totally flat and I haven't seen any kind of argument explaining, why one can use a position vector or treat the coordinates just in the same way as one would treat cartesian ones.

So my questions are:

(1) Am I right in my interpretation, that the coordinates are treated just as cartesian ones here and in linearized theory in general (meaning linearized around $\eta_{\mu\nu}$ of course)?

(2) If so: Why can we do that? I tried to make an argument myself, checking for example if one could show that the Pythagorean Theorem is valid up to Order $\mathcal{O}(h^2)$ but I didn't have any success...

Answer 1

When linearizing general relativity you can use any set of coordinates for the background spacetime you like. In particular if the background is flat Minkowski space you can use Cartesian coordinates. (Other common choices include spherical and cylindrical coordinates.)

However, the association of coordinates on the background spacetime to coordinates on the full spacetime is not wholly unambiguous. You can always slightly change the coordinates on the background, and interpret the change in the background metric as a change in the metric perturbation. This is what gives rise to the coordinate gauge freedom in linearized gravity.

Note, that while the chosen coordinates may be Cartesian coordinates on the background, they cannot be expected to behave as Cartesian coordinates on the full spacetime. For example, $\tfrac{\partial}{\partial x}$ will not necessarily by orthogonal to $\tfrac{\partial}{\partial y}$ (their innerproduct will generally be $\mathcal{O}(h)$).