Linearized gravity: When do we let the metric be $\eta_{\mu \nu} + h_{\mu \nu}$ and when does it reduce to $\eta_{\mu \nu}$? I am following a standard text on GR. In the chapter on linearized gravity, the metric $g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}$ reduces to $\eta_{\mu \nu}$ when the metric act on tensor components which are $O(h)$. That make sense since we ignore terms smaller than $O(h)$.
Suppose now that $a^\mu$ denote a vector from some arbitrary tensor that is not necessarily $O(h)$ and $T^{\mu \nu}$ the stress energy tensor. It seems to me that the author of the textbook I am reading uses the following prescription, but it could be a coincidence:


*

*$T^{\mu\nu} = 0 \implies g_{\mu\nu} a^\mu = \eta_{\mu\nu}a^\mu$ and 

*$T^{\mu\nu} \neq 0 \implies g_{\mu\nu} a^\mu = (\eta_{\mu\nu} + h_{\mu
\nu})a^\mu$


My questions then: 
(1) Is this a coincidence, theorem, or something else? 
(2) If the metric acts on a tensor of unknown magnitude, is it then still possible to deduce whether $g$ reduce to $\eta$ or not?
EDIT: the textbook is Carroll, spacetime and geometry. The most relevant sections are 7.1, 7.2, 7.3, 7.4. Notice how Carroll use of the metric changes as the stress energy tensor changes.
 A: The equations are saying if there is NO stress energy in T, the stress energy tensor, then the metric is described by n (i.e. flat space), but when you have T != 0, then you can describe the metric by the flat space one plus some added little bits which are described by h. 
Essentially expanding the metric into a flat metric + corrections is always wrong in an exact sense, but for 'weak' gravitational fields its a really nice way to simplify the math.
1 ? (Perhaps: Separating a metric into flat + h can only be done when it makes sense.) You can tell if it made sense when you get an answer and h < 0.001 or so. 
2) No. If you don't know the magnitude of the stress energy, it makes no sense to try to use flat + h as a metric. 
OR
2) Yes - one could always use a flat + h metric, but the math would become tricky and harder than a full non linear treatment. You can in theory use any coordinate system on any spacetime, its just that the math gets hard. 
The metric reduces to the flat metric if and only if there is no energy around. - Or better it reduces to the flat metric if you don't care about the errors that the flat assumption makes.
A: There are two things happening here.  The first is a mathematical approximation in that when the metric describes a space-time that is approximately flat with a small amplitude variation we only keep terms that are linear in the perturbation, $h_{\mu \nu}$.  This applies to working out the Christoffel symbols and the Curvature tensors.  When It comes to raising and lowering indices of other tensors you would use the combination $\eta_{\mu \nu} + h_{\mu \nu}$.  The second thing is physics.  In GR the non-trivial metric is created by the presence of mass and energy in space-time and that is expressed in the stress-energy-momentum tensor, $T_{\mu \nu}$.  Einstein's field equations relate the curvature of space-time to the presence of a non-zero $T_{\mu \nu}$ somewhere.  I had previously, and incorrectly, stated that if $T_{\mu \nu} = 0$ then the metric is necessarily flat.  Clearly the existence of plane wave solutions for gravitational waves violates this.  In much the same way the Maxwell's equations predict source free EM waves.  However, we typically uses these in idealized settings and attribute a cause and effect relationship to source and wave, at least classically.  In the case of a non-trivial metric in source free regions of space, the useful ones typically have at least a point source somewhere, so $T_{\mu \nu}$ is non-zero somewhere (it is not globally zero).
My question to you would be, is there missing context to this?  Is the author stating making a global statement about $T_{\mu \nu}$ or is he using different rules at different points in space-time?
A: It says in the book that $h_{\mu\nu}$ are taken everywhere to at most first order. This means for examples that Christoffel symbols derivatives all vanish by the assertion, because they carry first order derivatives of the metric already.
Furthermore the stress-energy tensor is taken to 0th order.
