# A problem of the 2-Layer hydrostatic model

The 2-layer hydrostatic model is like this And the pressure of the two layers and the top are and it is easy to find the horizontal gradient of $$p_1$$ and $$p_2$$ \begin{equation} \begin{aligned} \nabla p_1 &= \nabla p_H\\ \nabla p_2 &= \nabla p_H + g(\rho_2 - \rho_1)\nabla h_2\\ \end{aligned} \end{equation} Now my problem is that how to get the result $$\begin{equation} p_2 - p_1 = g(\rho_2 - \rho_1)\eta \end{equation}$$

from the equations above, where$$\eta = h_2 - H_2$$. I find this result in the book"Fundamentals of Geophysical Fluid Dynamics" at page 164.

• What have you tried? Is this the solution with the rigid layer (top) or with variable water level? Apr 14 at 13:31
• Is that equation in your reference? I can't find it Apr 14 at 13:58
• Just to help us verify... not a homework question, correct? Apr 15 at 1:40
• @JeopardyTempest no…I find this problem when I was reading the book
–  Hou
Apr 15 at 8:12
• I'm with @J.Fregin... I do not see your final result equation in the book??? Apr 15 at 10:25

Let's say our goal is to find $$\eta$$ which is the displacement of the fluid relative to its resting position at $$z = H_2$$.

We find $$\begin{equation} \begin{split} p_2 - p_1 &= \require{cancel} \bcancel{p_H} + \rho_1 g(\require{cancel} \bcancel{H}-h_2) + \rho_2g(h_2 - z) \require{cancel} \bcancel{-p_H} - \rho_1g(\require{cancel} \bcancel{H}-z) \\ &= (\rho_2 - \rho_1)gh_2 - (\rho_2 - \rho_1)gz \\ & = (\rho_2 - \rho_1)g(H_2 + \eta) - (\rho_2 - \rho_1)gz \end{split} \end{equation}$$

Dividing by $$(\rho_2 - \rho_1)g$$ yields

$$\begin{equation} \eta + H_2 - z = \frac{p_2 - p_1}{(\rho_2 - \rho_1)g}. \end{equation}$$ The equation above tells us the distance to the interface at some height $$z$$ (remember the coordinate origin is at the bottom of the domain). We are interested in the displacement relative to the mean interface height $$z = H_2$$, so we have $$\begin{equation} \eta = \frac{p_2 - p_1}{(\rho_2 - \rho_1)g}, \end{equation}$$ which is the result you are looking for. However, in the document they say that they use four more equations to derive the result. Maybe it's a mistake or maybe I'm missing something.

I was a little confused that you said we can find the result in the document - so for anyone wondering:

Using $$\phi_n = p_n / \rho_0$$ and $$g' = g(\rho_2 - \rho_1)/\rho_0$$, we find $$\begin{equation} \eta = \frac{(p_2 - p_1)\rho_0}{(\rho_2 - \rho_1)g\rho_0} = -\frac{\phi_1 - \phi_2}{g'}, \end{equation}$$ which corresponds to what's shown in the document.

• I’m little confused about the coordinate $z$ here, I mean, to compute the pressure difference between two locations in two layers respectively, why you only used one $z$ in your answer not $z_1$(corresponding to the first layer $/rho_1$) and $z_2$(corresponding to $/rho_2$)? I think the pressure difference is a function of $z_1$ and $z_2$.
–  Hou
Apr 17 at 4:22
• Good question - I have two for you too: What happens if you just use $z= H_2$ ? Does this capture the qualitative behaviour of the displacement $\eta$? I think we can't interpret the equations as a difference between two locations. There is just one. Apr 17 at 8:16
• But what is the physical meaning of the equation when you set $z=H_2$? $p_1$ and $p_2$ should be the pressure of the two layers respectively.
–  Hou
Apr 17 at 9:10
• The way I understand it is that you just assume that either layer 1 or layer 2 extends to $H_2$, which of course is unphysical in the case of one layer. I struggle to find another explanation for this using the simple rearrangement in my answer above. Notice that in practice you would solve the pde-system to find your unknowns instead of using this equation. Maybe someone else can help out here? Apr 17 at 11:11
• Well, I have no ideas now, actually I think it is no need to compute the pressure difference, since we actually use the gradient of the pressure in the momentum equations
–  Hou
Apr 17 at 11:23