# Linearising with a density fluctuation

I have some trouble understanding orders. Starting with the continuity equation $\partial_t\rho=-\nabla_r .(\rho \vec{u})$ and applying a peturbation to the density $\rho(\vec{r},t)=\bar{\rho}(\vec{r})+\delta\rho(\vec{r},t)$ I know that the result to linear order is $\partial_t\delta\rho=-\nabla_r .(\bar{\rho} \vec{u})$. I do not understand clearly why terms with $\nabla.\delta$ are treated as second order. Does a derivative imply higher order, because as far as I am aware higher order would be a product of flucuations i.e. $\delta\delta$ terms.

I'd wager the reason why the perturbed approximation looks like that is because you're actually perturbing both the density and the velocity in the following way:

$$\rho(\vec{r},t) = \bar{\rho}(x) + \delta\rho'(x,t)$$

$$\vec{u}(\vec{r},t) = \vec{0} + \delta\vec{u}'(x,t)$$

In this case, the only term that is inside the divergence in the final approximation is the only term that isn't quadratic in $\delta$, $\bar{\rho}(x)\delta\vec{u}'(x,t)$, justifying the approximation without needing to reference the magnitude of spatial gradients. Note the $\delta$ in the velocity perturbation is often omitted because you're perturbing off of a rest state.

Given the cosmology tag, I'll add some specific context; when considering the expansion of the universe, you'd have to consider Hubble flow and co-moving coordinate frames, but the process is the same as the one I detailed above in the correct reference frame.

Both of these approaches are discussed fairly well here.

• Indeed I have been using these very same notes, I was not careful in considering the velocity perturbation. Thanks. ( I still wonder about spatial gradients and magnitude arguments though)
– AAM
Jul 15 '18 at 16:29
• No problem! I’ll add that this approximation can fairly often fail to hold water, particularly when small length scales are considered (precisely because spatial gradients can be big even though the differentiated function can be tiny). Jul 15 '18 at 16:31