# 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

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

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 $$\frac{D\eta}{Dt}=\frac{Dz}{Dt}|^{fluid}=w\frac{\partial z}{\partial z}|^{fluid}=w|^{fluid}=w(\eta)$$
• @blupp OK I saw that in the book. You may want to clarify what is the relationship between $\eta$ and z Jan 21 at 17:09