Are third derivatives of metric perturbations zero?

Question

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?

Answer 1

Is there somewhere in the paper that they say that the third derivative vanishes, or invoke its vanishing as an approximation?

In general, you can't make tensorial objects by differentiating the metric. To get a tensor by differentiating a tensor, you have to take a covariant derivative. But the covariant derivative of the metric vanishes identically.

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.