Graduate Fluids and the no-penetration condition

[member 428835]

Hi PF!

Denote the velocity of a fluid $\vec u$ and define a potential $\vec u = -\nabla \phi$. Let $\hat n$ be an outward-oriented surface normal to a solid boundary. I would express no penetration at the boundary as $$ u \cdot \hat n = 0 \implies \nabla \phi \cdot \hat n = 0.$$
However, the text writes $$\frac{\partial \phi}{\partial n} = 0$$ where $n$ is the direction of the outward-oriented normal. Can someone explain this result?

 

[NFuller]

The gradient operator is defined in cartesian coordinates as
$$\nabla=\frac{\partial}{\partial x}\hat{x}+\frac{\partial}{\partial y}\hat{y}+\frac{\partial}{\partial z}\hat{z}$$
Looking at a particular example, if $\hat{n}=\hat{x}$ then
$$\nabla\phi\cdot\hat{n}=\nabla\phi\cdot\hat{x}=\frac{\partial\phi}{\partial x}\hat{x}\cdot\hat{x}+\frac{\partial\phi}{\partial y}\hat{y}\cdot\hat{x}+\frac{\partial\phi}{\partial z}\hat{z}\cdot\hat{x}=\frac{\partial\phi}{\partial x}=\frac{\partial\phi}{\partial n}$$

 

[boneh3ad]

The above answer has some math errors when carrying out the dot product.

Really, though, you don't need any fancy math to figure this out if you just think about the meaning of a dot product. If you just remember that $\vec{a}\cdot\vec{b}$ is the projection of $\vec{a}$ onto $\vec{b}$ (i.e. the size of $\vec{a}$ in the direction of $\vec{b}$ if $\vec{b} = \hat{b}$ is of unit length), you can apply that here. In this case, you've got $\nabla\phi \cdot \hat{n}$. Since $\nabla \phi$ is the gradient and $\hat{n}$ is a unit vector normal to the surface, the dot product represents the magnitude of the gradient in the direction of the unit normal, or $\frac{\partial \phi}{\partial n}$.

 

[member 428835]

In this case, you've got $\nabla\phi \cdot \hat{n}$. Since $\nabla \phi$ is the gradient and $\hat{n}$ is a unit vector normal to the surface, the dot product represents the magnitude of the gradient in the direction of the unit normal, or $\frac{\partial \phi}{\partial n}$.

Riiiiiight, just like the directional derivative! Got it!

 

[NFuller]

The above answer has some math errors when carrying out the dot product.

Can you be more specific?

 

[boneh3ad]

Your assumed form of your normal vector had one nonzero component yet somehow it was magically distributed to all of the gradient components.

 

[NFuller]

Yes, that's because a dot product distributes over addition.
$$\nabla\phi\cdot\hat{x}=\left(\frac{\partial\phi}{\partial x}\hat{x}+\frac{\partial\phi}{\partial y}\hat{y}+\frac{\partial\phi}{\partial z}\hat{z}\right)\cdot\hat{x}=\frac{\partial\phi}{\partial x}\hat{x}\cdot\hat{x}+\frac{\partial\phi}{\partial y}\hat{y}\cdot\hat{x}+\frac{\partial\phi}{\partial z}\hat{z}\cdot\hat{x}=\frac{\partial\phi}{\partial x}$$

 

[boneh3ad]

It certainly is distributive over addition, and I suppose that, while the way you've done it is technically correct, you've turned one dot product into three dot products. It would be much simpler to just compute the first one directly. Otherwise, what happens when you have to generalize it to a situation where $\hat{n}$ is not parallel to one of the Cartesian directions?

In your example (which is a special case, not general), you would do it directly by:
\nabla\phi \cdot \hat{n} = \left( \hat{\imath}\dfrac{\partial \phi}{\partial x} + \hat{\jmath}\dfrac{\partial \phi}{\partial y} + \hat{k}\dfrac{\partial \phi}{\partial z} \right) \cdot (\hat{\imath}) = \left( \hat{\imath}\dfrac{\partial \phi}{\partial x} + \hat{\jmath}\dfrac{\partial \phi}{\partial y} + \hat{k}\dfrac{\partial \phi}{\partial z} \right) \cdot (1\hat{\imath} +0\hat{\jmath} + 0\hat{k})
= \left[(1) \left(\dfrac{\partial \phi}{\partial x} \right) + (0) \left(\dfrac{\partial \phi}{\partial y} \right) + (0) \left(\dfrac{\partial \phi}{\partial z}\right) \right] = \dfrac{\partial \phi}{\partial x}

 

[NFuller]

Otherwise, what happens when you have to generalize it to a situation where ^nn^\hat{n} is not parallel to one of the Cartesian directions?

The same logic still applies. Let's say $\hat{n}=2\hat{x}+1\hat{y}$ then
$$\nabla\phi\cdot\left(2\hat{x}+1\hat{y}\right)=\left(\frac{\partial\phi}{\partial x}\hat{x}+\frac{\partial\phi}{\partial y}\hat{y}+\frac{\partial\phi}{\partial z}\hat{z}\right)\cdot\left(2\hat{x}+1\hat{y}\right)=\frac{\partial\phi}{\partial x}\left(\hat{x}\cdot2\hat{x}+\hat{x}\cdot\hat{y}\right)+\frac{\partial\phi}{\partial y}\left(\hat{y}\cdot2\hat{x}+\hat{y}\cdot\hat{y}\right)+\frac{\partial\phi}{\partial z}\left(\hat{z}\cdot2\hat{x}+\hat{z}\cdot\hat{y}\right)=2\frac{\partial\phi}{\partial x}+\frac{\partial\phi}{\partial y}$$
This is certainly pedantic for the simple case here but this process is useful when working with a non-Cartesian system or when $\hat{n}$ is defined on an irregular boundary.