Why is the metric linearized to determine the mass spectrum of five dimensional Kaluza-Klein? In this review about Kaluza-Klein theories, (page 1115) in order to determine the mass spectrum of the 5 dimensional theory the metric is expanded to first order.
Why this? Why not retain the full metric? Why does it consider only small perturbations? Is it because doing considering the full metric is just too complicated or is something else behind this?
 A: The mass of a quantum of a field is defined from the second derivative of the potential term
$$ m^2 = \left. \frac{\partial^2 V(\phi)}{\partial \phi^2} \right|_{\phi=0} $$
and similar for fields with spin (fields that are not scalar fields). The general form of the potential – or the whole Lagrangian – is always more complicated but only this leading term determines how non-interacting wave packets and the particles of the field propagate. The higher-order terms only affect the interactions.
So the potential may be expanded as
$$ V = V_0 + 0\Phi +\frac{m^2}{2}\Phi^2+ O(\Phi^3) $$
Here, the constant term doesn't matter except for causing curvature in general relativity (the cosmological constant). The linear term may be set to zero by redefining the field $\Phi$ additively so that its vev is $\Phi=0$, the quadratic term is the first important nontrivial term, and the higher-order terms don't affect small "waves" i.e. masses of the particles at all because the equations of motion are of the form
$$\Box \Phi =m^2 \Phi + O(\Phi^2) $$
and the higher-order terms may be neglected for a small $\Phi$. The quantization of the field from which the particles' masses may be extracted effectively deals with the infinitesimal values of $\Phi$, too.
To calculate the leading non-trivial (non-constant and nonzero) term in the potential, it is enough to linearize the dependence of all other things on the fields at the relevant point.
So there is no inaccuracy introduced whatsoever.
