# Deriving shallow water equations: why is the vertical velocity equal to the material derivative of the surface level?

I'm reading about the derivation of shallow water equations, and I have a hard time understanding why $$w(\eta) = \frac{D\eta}{Dt}$$ where $$w$$ is the vertical velocity, $$\eta$$ is the (top or bottom) vertical boundary of the layer and $$D/Dt$$ is the material derivative.

In Atmospheric and Oceanic Fluid Dynamics (2nd ed. 2017, chapter 3.1.2), Vallis gives the following explanation:

At the top the vertical velocity is the material derivative of the position of a particular fluid element. But the position of the fluid at the top is just $$\eta$$, and therefore $$w(\eta) = \frac{D\eta}{Dt}$$

Why is the vertical velocity at the top the material derivative?

Edit: Added a schematic illustration of the shallow layer. As you see, $$\eta$$ is not assumed to be constant. This is a statement of the kinematic free surface boundary condition: there can be no normal flow through the boundary, only tangential flow along it. Equivalently, the normal velocity (relative to the interface) of the free surface position is the same as the velocity of the fluid.

Defining the position of the free surface as $$z = \eta(x,y,t)$$ then taking the material derivative

$$\frac{\partial z}{\partial t} + u\frac{\partial z}{\partial x} + v\frac{\partial z}{\partial y} + w\frac{\partial z}{\partial z} = \frac{D\eta}{Dt}$$

Every term on the LHS besides $$\frac{\partial z}{\partial z}=1$$ goes to $$0$$ due to the fact that $$z$$ is an independent coordinate (see this answer for reference), leaving us with

$$w(z=\eta) = \frac{D\eta}{Dt}$$

• I made a small edit. Please let me know your thoughts). – gansub Jan 22 at 12:10
• looks great thanks! It is important to specify z as an independent coordinate to justify the simplification. I incorporated it with some minor changes to the text. – wingtorres Jan 22 at 14:57

As a refresher, let's refer back to the definition of the material derivative: $$\frac{D}{Dt}=\frac{\partial}{\partial t}+u\frac{\partial}{\partial x}+v\frac{\partial}{\partial y}+w\frac{\partial}{\partial z}$$

If we apply it to $$\eta$$, then we get: $$\frac{D \eta}{Dt}=\frac{\partial \eta}{\partial t}+u\frac{\partial\eta}{\partial x}+v\frac{\partial\eta}{\partial y}+w\frac{\partial\eta}{\partial z}$$

Let's, for a second, consider that $$\eta$$ is the height of the fluid. We can replace $$\eta$$ with $$z$$ evaluated at the fluid level. Thus, we get $$\frac{Dz}{Dt}|^{fluid}=\frac{\partial z}{\partial t}|^{fluid}+u\frac{\partial z}{\partial x}|^{fluid}+v\frac{\partial z}{\partial y}|^{fluid}+w\frac{\partial z}{\partial z}|^{fluid}$$ Since there is only one Eulerian derivative that is dependent on $$z$$, it follows $$\frac{D\eta}{Dt}=\frac{Dz}{Dt}|^{fluid}=w\frac{\partial z}{\partial z}|^{fluid}=w|^{fluid}=w(\eta)$$

• I still don't get it. If I understand correctly, you're saying that d eta/dt = d eta/dx = d eta/dy = 0. But eta is a function of t,x,y and is not assumed to be constant, so this is obviously not true? – blupp Jan 21 at 7:16
• @blupp eta is not a function of t,x, and y. – gansub Jan 21 at 13:48
• The vertical coordinate is orthogonal one (not curvilinear) – gansub Jan 21 at 14:16
• @gansub eta definitely varies with x in the derivation from Vallis. I'll update the question with a figure if that helps. – blupp Jan 21 at 16:53
• @blupp OK I saw that in the book. You may want to clarify what is the relationship between $\eta$ and z – gansub Jan 21 at 17:09