# 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:

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

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...)).

## 2 Answers

$$\mathbf{v} \cdot \nabla H$$ 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.

• Thanks for making the answer look cleaner! – JG123 Aug 9 '19 at 19:20

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.