<p>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.</p>
<p>Defining the position of the free surface as
<span class="math-container">$$ z = \eta(x,y,t) $$</span>
then taking the material derivative</p>
<p><span class="math-container">$$\frac{\partial z}{\partial t} + 
u\frac{\partial z}{\partial x} +
v\frac{\partial z}{\partial y} +
w\frac{\partial z}{\partial z} =
\frac{D\eta}{Dt} 
$$</span></p>
<p>Every term on the LHS besides <span class="math-container">$\frac{\partial z}{\partial z}=1$</span> goes to <span class="math-container">$0$</span> due to the fact that <span class="math-container">$z$</span> is an independent coordinate (see this <a href="https://math.stackexchange.com/a/3067984">answer</a> for reference), leaving us with</p>
<p><span class="math-container">$$w(z=\eta) = \frac{D\eta}{Dt}$$</span></p>

CC BY-SA 4.0
2021-01-22T14:57:24.620Z
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
CC BY-SA 4.0
2021-01-22T09:27:20.120Z
2021-01-22T14:49:56.833Z
2021-01-22T14:49:56.833Z
wingtorres
<p>As a refresher, let's refer back to the definition of the material derivative: <span class="math-container">$$\frac{D}{Dt}=\frac{\partial}{\partial t}+u\frac{\partial}{\partial x}+v\frac{\partial}{\partial y}+w\frac{\partial}{\partial z}$$</span></p>
<p>If we apply it to <span class="math-container">$\eta$</span>, then we get:
<span class="math-container">$$\frac{D \eta}{Dt}=\frac{\partial \eta}{\partial t}+u\frac{\partial\eta}{\partial x}+v\frac{\partial\eta}{\partial y}+w\frac{\partial\eta}{\partial z}$$</span></p>
<p>Let's, for a second, consider that <span class="math-container">$\eta$</span> is the height of the fluid. We can replace <span class="math-container">$\eta$</span> with <span class="math-container">$z$</span> evaluated at the fluid level. Thus, we get <span class="math-container">$$\frac{Dz}{Dt}|^{fluid}=\frac{\partial z}{\partial t}|^{fluid}+u\frac{\partial z}{\partial x}|^{fluid}+v\frac{\partial z}{\partial y}|^{fluid}+w\frac{\partial z}{\partial z}|^{fluid}$$</span>
Since there is only one Eulerian derivative that is dependent on <span class="math-container">$z$</span>, it follows
<span class="math-container">$$\frac{D\eta}{Dt}=\frac{Dz}{Dt}|^{fluid}=w\frac{\partial z}{\partial z}|^{fluid}=w|^{fluid}=w(\eta)$$</span></p>

CC BY-SA 4.0
2021-01-21T07:16:21.583Z
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
CC BY-SA 4.0
2021-01-21T16:53:08.873Z
@gansub eta definitely varies with x in the derivation from Vallis. I'll update the question with a figure if that helps.
blupp
CC BY-SA 4.0
2021-01-21T01:11:35.973Z
2021-01-21T01:11:35.973Z
BarocliniCplusplus
<p>I'm reading about the derivation of shallow water equations, and I have a hard time understanding why
<span class="math-container">$$ w(\eta) = \frac{D\eta}{Dt}$$</span>
where <span class="math-container">$w$</span> is the vertical velocity, <span class="math-container">$\eta$</span> is the (top or bottom) vertical boundary of the layer and <span class="math-container">$D/Dt$</span> is the material derivative.</p>
<p>In Atmospheric and Oceanic Fluid Dynamics (2nd ed. 2017, chapter 3.1.2), Vallis gives the following explanation:</p>
<blockquote>
<p>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 <span class="math-container">$\eta$</span>, and therefore
<span class="math-container">$$ w(\eta) = \frac{D\eta}{Dt}$$</span></p>
</blockquote>
<p>Why is the vertical velocity at the top the material derivative?</p>
<p><em>Edit:</em> Added a schematic illustration of the shallow layer. As you see, <span class="math-container">$\eta$</span> is not assumed to be constant. <a href="https://i.sstatic.net/WyYWR.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/WyYWR.png" alt="Shallow layer" /></a></p>

<oceanography><fluid-dynamics>
Deriving shallow water equations: why is the vertical velocity equal to the material derivative of the surface level?
blupp
# Deriving shallow water equations: why is the vertical velocity equal to the material derivative of the surface level?

(Asked by: username_0 on 2021-01-20)

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. [![Shallow layer](https://i.sstatic.net/WyYWR.png)](https://i.sstatic.net/WyYWR.png)

**Tags:** oceanography, fluid-dynamics


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## 2 answers

### Answer by username_1 on 2021-01-22. Score: 4 | Accepted answer

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](https://math.stackexchange.com/a/3067984) for reference), leaving us with

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

> **Comments:**
> **username_1**: 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.  

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### Answer by username_2 on 2021-01-21. Score: 2

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
$$\frac{D\eta}{Dt}=\frac{Dz}{Dt}|^{fluid}=w\frac{\partial z}{\partial z}|^{fluid}=w|^{fluid}=w(\eta)$$

> **Comments:**
> **username_0**: 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?  
> **username_0**: @gansub eta definitely varies with x in the derivation from Vallis. I'll update the question with a figure if that helps.  

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