Bulk Mass Coordinates

These are preliminary notes explaining the difference between what I'm terming "specific-mass coordinates" and "bulk-mass coordinates".

The typical hybrid mass coordinate is defined by $\partial_{\eta} p_h \equiv \partial_{\eta} F_{m} = p_s\partial_{\eta} [A] + p_0 \partial_{\eta} [B]$ where I am going to use $F_{m}$ to refer to a fictitious force that in previous HOMME versions was called "hydrostatic pressure".

The well-posedness of this coordinate results from the fact $\int_{z}^{\textrm{top}} g \rho \intd{z}$ is monotonically increasing in $z$. Note that the monotonicity holds regardless of whether the atmosphere is in hydrostatic balance. However, the form of this equation does indicate two important things about this equation. Firstly, the use of a notion of "pressure" in the definition of our mass coordinate is largely a misnomer. The notion of "hydrostatic pressure" in the non-hydrostatic HOMME model is misleading and should be abandoned. Secondly, this equation does allow us to precisely explain where quantities such as $\Delta p$ which show up in the legacy coordinates we use come from.

We should probably retain legacy $\eta$ model coordinates. To my knowledge, they are the best understood way to combine terrain following and pure-mass coordinates in one concise coordinate system.

Assume (incorrectly) that the atmosphere is in hydrostatic balance between $z_1, z_2$. Then $$\begin{align*} \int_{z_1}^{z_2} \rho g \intd{z} = \int_{p_2}^{p_1} \intd{p} = \Delta p \end{align*}$$

Ignore variation in $g$ as it is merely a nuissance for the moment. Then $\frac{\Delta p}{g} = \Delta \hat{m}$, where $\hat{m}$ is the columnar mass within a certain level. It has units $\textrm{kg m}^{-2}$.

There are two ways of recovering the correct mass when we go to the deep atmosphere in spherical coordinates. Either one recovers mass-weighted integrals as $$\int \int [\rho X] \hat{r}^2 \intd{s} \intd{A} \textrm{ or } \int \int [\rho \hat{r}^2] X \intd{s} \intd{A}$$ where the integral on the left corresponds to introducing time dependence to the metric that is used to compute weak differential operators. (Strictly speaking, the $\mathrm{d}A$ here contains a $a$ factor that is the mean radius of the earth.) Under this interpretation, $\rho$ corresponds to specific density. What I mean by this is that $\rho$ is the mass per unit volume at a gridpoint. This is an unproblematic interpretation with either a shallow metric, or a modified deep metric.

We have elected to follow the strategy that corresponds to the integral on the right. This means that we then find $\frac{1}{g_0} \int_{z_1}^{z_2} [\hat{r}^2 \rho] g_0 \intd{z} = \Delta \hat{m}$

implications for initialization

Assuming we are in hydrostatic balance, $$\frac{\int_{z_1}^{z_2} [\hat{r}^2 \rho] g_0 \intd{z} }{g_0} = \frac{\Delta p}{g_0} = \Delta \hat{m}.$$

If we are given a "pressure-based" initialization routine, it expects that $\frac{\Delta p}{\hat{g}} = \int \rho \hat{g} \intd{z}$ (we neglect variation in g in the integral and let $\hat{g}$ be the midpoint value.)