1
$\begingroup$

Context

I am trying to follow a derivation of Bernoulli's equation from Acheson's Elementary Fluid Dynamics 1990.

After making some simplifying assumptions and doing some algebra I got to:

$$( \nabla \times \mathbf{v}) \times \mathbf{v} = - \nabla H$$

where:

$\mathbf{v}$ is a vector field representing the fluid velocity

$H$ is an expression (we are trying to show that $H$ is a constant).

Next we take the dot product with $\mathbf{v}$.

$$\mathbf{v}\cdot (( \nabla \times \mathbf{v}) \times \mathbf{v}) = -\mathbf{v} \cdot \nabla H$$

The left hand side is zero (this can be shown using vector identities). Hence we get to $-\mathbf{v} \cdot \nabla H =0$ which implies that $\mathbf{v} \cdot \nabla H =0$

Apparently it is now trivial to see that "$H$ is constant along the streamlines". But I do not understand why?

Question

Why does $\color{blue}{\mathbf{v} \cdot \nabla H =0}$ imply that $\color{blue}{H}$ is constant along a stream line?


Here is an example fluid velocity field with streamlines if anyone wants to refer to a concrete example:

$\mathbf{v} = x\hat{i}+(-y)\hat{j}+ 0\hat{k}$

enter image description here

here is the same field with the corresponding potential shown as contours (which I notice are perpendicular to the stream lines (the dot product of perpendicular vectors is zero...)).

enter image description here

$\endgroup$
1
3
$\begingroup$

$\mathbf{v} \cdot \nabla H$ is the directional derivative of $H$ in the direction of $\mathbf{v}$. Since $\mathbf{v} \cdot \nabla H =0$, we have that the instantaneous rate of change of $H$ in the direction of $\mathbf{v} $ is $0$. So $H$ is constant in the direction of $\mathbf{v}$ and thus $H$ is constant along the streamlines of the fluid.

$\endgroup$
1
  • 1
    $\begingroup$ Thanks for making the answer look cleaner! $\endgroup$ – JG123 Aug 9 '19 at 19:20
1
$\begingroup$

The chain rule also may shed some light.

$\vec{v}$ is the velocity vector so:

$\vec{v}=v_x\hat{i}+v_y\hat{j}+v_z\hat{k} =\frac{dx}{dt}\hat{i}+\frac{dy}{dt}\hat{j}+\frac{dz}{dt}\hat{k}$

$$\nabla H=\frac{\partial H}{\partial x}\hat{i}+\frac{\partial H}{\partial y}\hat{j}+\frac{\partial H}{\partial z}\hat{k}$$

By the chain rule we have:

$$\frac{dH}{dt}=\frac{\partial H}{\partial x}\frac{dx}{dt}+\frac{\partial H}{\partial y}\frac{dy}{dt}+\frac{\partial H}{\partial z}\frac{dz}{dt}=\frac{d\vec{s}\cdot \nabla H}{dt}$$

And keep in mind, total derivative of $f$ along a path is $df=\nabla f\cdot d\vec{s}$

where $d\vec{s}$ is infinitesimal length element in direction of change.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.