Idealized eta coordinates

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We're going to follow the strategy outlined in DCMIP 2008. We use the definition p(η)=A(η)p0+B(η)ps p(\eta) = A(\eta) p_0 + B(\eta) p_s

  1. Establishing a reference temperature profile T(z)T(z).
  2. Use this to determine pressure levels pint,k=p(zint,k)p_{\textrm{int}, k} = p(z_{\textrm{int}, k})
  3. Define A(η)=η−B(η)A(\eta) = \eta - B(\eta), and B=(η−ηtop1−ηtop)c B = \left(\frac{\eta - \eta_{\textrm{top}}}{1 - \eta_{\textrm{top}}}\right)^c, η(z)≡p(z)ps\eta(z) \equiv \frac{p(z)}{p_s}
  4. Calculate full model levels by averaging interfaces.

We will be using c≡1c \equiv 1.

special considerations:

Suppose we want Δz(p) \Delta z(p) to have a specific (approximate) height profile, e.g. in Skamarock, et al.. Suppose we have already found pk,k=K+1,…,k p_k, k=K+1, \dots, k (ηtop\eta_{\textrm{top}} is a free parameter, while bad things happend if ηs≠1 \eta_{s}\neq 1)

Assume the atmosphere is in hydrostatic balance. Under a constant lapse rate, p(z)=p0(T0−ΓzT0)gΓRd p(z) = p_0 \left(\frac{T_0 - \Gamma z}{T_0}\right)^{\frac{g}{\Gamma R_d}} The temperature is then T(p)=T0(pp0)−RdΓgT(p) = T_0 \left(\frac{p}{p_0} \right)^{-\frac{R_d \Gamma}{g}} , and the height is z(p)=T0Γ[1−(pp0)RdΓg]z(p) = \frac{T_0}{\Gamma} \left[1 - \left( \frac{p}{p_0} \right)^{\frac{R_d \Gamma}{g}} \right]

Suppose we have pint,k+1,zint,k+1p_{\textrm{int}, k+1}, z_{\textrm{int}, k+1}.

Note ηmid,k=12(pint,k+1ps+pint,kps) \eta_{\textrm{mid}, k} = \frac{1}{2}\left(\frac{p_{\textrm{int},k+1}}{p_s} + \frac{p_{\textrm{int}, k}}{p_s}\right)

The equation we want to solve is Δz(pint,k+1+pint, k2)=z(pint,k)−z(pint,k+1) \Delta z\left(\frac{p_{\textrm{int}, k+1} + p_{\textrm{int, k}}}{2} \right) = z(p_{\textrm{int}, k}) - z(p_{\textrm{int}, k+1})

The equation that's easy to solve is Δz(pint, k+1)=z(pint,k)−z(pint,k+1) \Delta z\left(p_{\textrm{int, k+1}} \right) = z(p_{\textrm{int}, k}) - z(p_{\textrm{int}, k+1}) , or pk+1=p(Δz(pint, k+1)+z(pint,k+1))p_{k+1} = p(\Delta z\left(p_{\textrm{int, k+1}}) +z(p_{\textrm{int}, k+1}) \right)

This is an approximation, but we are making several other approximations that won't hold since we have topography.

Brief digression:

Under the profile T(z)=T0T(z) = T_0, the pressure looks like p=p0exp⁡(zRdT0g−1) p = p_0 \exp\left(\frac{z}{R_d T_0 g^{-1}} \right) while under T(z)=T0−ΓzT(z) = T_0 - \Gamma z the pressure looks like p=p0(T0−ΓzT0)gRdΓp = p_0\left(\frac{T_0 - \Gamma z}{T_0} \right)^{\frac{g}{R_d \Gamma}} . The fact that these don't immediately agree when Γ=0\Gamma = 0 has always irritated me. Here's why that is:

An intermediate step in integrating the hydrostatic integral reads log⁡(p)=log⁡(p0)+log⁡([T0−ΓzT0]gRdΓ)=log⁡(p0)+log⁡([T0−ΓzT0]gRd)Γ \begin{align*} \log(p) &= \log(p_0) + \log\left(\left[\frac{T_0 - \Gamma z}{T_0} \right]^{\frac{g}{R_d \Gamma}} \right) \\ &= \log(p_0) + \frac{\log\left(\left[\frac{T_0 - \Gamma z}{T_0} \right]^{\frac{g}{R_d}} \right)}{\Gamma} \end{align*}

Noting that the numerator and denominator are both differentiable functions of Γ\Gamma, we then find lim⁡Γ→0+log⁡(p/p0)=log⁡(p/p0)=lim⁡Γ→0+log⁡([T0−ΓzT0]gRd)Γ=lim⁡Γ→0+∂Γ[gRdlog⁡(T0−ΓzT0)]∂ΓΓ=lim⁡Γ→0+gRd⋅−zT0⋅T0T0−Γz=−zRdT0g−1 \begin{align*} \lim_{\Gamma \to 0^+} \log(p/p_0) = \log(p/p_0) &= \lim_{\Gamma \to 0^+} \frac{\log\left(\left[\frac{T_0 - \Gamma z}{T_0} \right]^{\frac{g}{R_d}} \right)}{\Gamma} \\ &= \lim_{\Gamma \to 0^+} \frac{\partial_\Gamma \left[\frac{g}{R_d}\log\left(\frac{T_0 - \Gamma z}{T_0} \right)\right]}{\partial_\Gamma \Gamma} \\ &= \lim_{\Gamma \to 0^+} \frac{g}{R_d}\cdot -\frac{z}{T_0}\cdot\frac{T_0}{T_0 - \Gamma z} \\ &= -\frac{z}{R_d T_0 g^{-1}} \end{align*}

Unfortunately, this makes it very difficult to find, e.g., a unified expression for p(z)p(z) that agrees in both cases in floating point arithmetic :(.

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