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

Question

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?

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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...)).

Answer 1

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

Answer 2

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.