# Are third derivatives of metric perturbations zero?

I'm working on a problem related to fluid perturbations of stars. I'm following this paper. They start with the Einstein equation:

$$G_{\alpha \beta} = 8 \pi G T_{\alpha \beta}$$ and then perturb the metric, and the stress-energy of the matter (which they assume to be a perfect fluid). The metric is basically rewritten as

$$g_{\alpha \beta} = g^0_{\alpha \beta} + h_{\alpha \beta}$$ where $h_{\alpha \beta}$ represents the perturbation. After working through the algebra, they arrive at a coupled set of differential equations that relate the metric perturbation to the fluid perturbations.

My questions is: Is there any generic reason why $h'''_{\alpha \beta}$ would be zero? Assuming that $h_{\alpha \beta}$ only depends on $r$ and the primes represent derivatives with respect to $r$.

If there is no general rule about this, does anyone know of specific situations where the third derivative of a perturbed metric does go to zero?

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Since such a derivative can't be tensorial, there is no reason to expect to be able to say anything meaningful about whether or not it's zero, even for a specific, given spacetime. A tensor that vanishes in one set of coordinates vanishes in all other sets of coordinates, but this is not true for a non-tensorial quantity. So even if this particular non-covariant derivative $h'''$ did vanish in some set of coordinates, there would be other coordinates for which it didn't vanish.
Thanks Ben. I'm starting to think this was a poorly worded question. But to address some of the points you brought up. No, I do not believe the paper ever mentions any approximations it invokes about $h''' =0$. Yes, you are absolutely right about the tensorial nature of the metric, and I feel really dumb for not remembering it. I think the major issue is that I need to work out all of the algebra they apply to get to the perturbation equations before I can ask the correct question. Thanks again. –  tau1777 Jul 18 '13 at 5:26