Bernoulli's theorem: $\frac{p}{\rho}+\frac{1}{2}u^2+\phi$ is constant along a streamline I am trying to understand the Bernoulli's theorem:


*

*$\frac{p}{\rho}+\frac{1}{2}u^2+\phi$ is a constant along a streamline
I got that:
$\frac{\partial u}{\partial t}$ + ($\nabla \times u)\times u$ = $-\nabla(\frac{p}{\rho}+\frac{1}{2}u^2+\phi)$
For a steady flow: $\frac{\partial}{\partial t}$ = 0 and then:
($\nabla \times u)\times u$ = -$\nabla H$ with the scalar: $H$ = $\frac{p}{\rho}+\frac{1}{2}u^2+\phi$
now, I didn't understand the next steps:
Taking the “dot product” of ($\nabla \times u)\times u$ = -$\nabla H$ the left hand side vanishes, as ($\nabla \times u)\times u$  is perpendicular to  $u$ and we get:
$(u \cdot \nabla)H = 0$
This implies that $H$ is constant along a streamline
Can someone explain me this thing in other words please?
 A: Let $\mathbf u(t, \mathbf x)$ represent the velocity vector field of the fluid.  Let $\mathbf x(t)$ denote the position of a particle moving with the fluid, then the velocity $\dot{\mathbf x}(t)$ of the particle at a time $t$ will be equal to the velocity of the fluid flow at the point $(t, \mathbf x(t))$, namely
$$
  \mathbf u(t, \mathbf x(t)) = \dot{\mathbf x}(t)
$$
Now suppose that $\mathbf u(t, \mathbf x)\cdot\nabla H(t, \mathbf x) = 0$.  He want to show that this implies that $H$ is constant along the path of a particle moving with he fluid.  Notice that for any path $\mathbf x(t)$ we have
$$
  \frac{d}{dt}H(t, \mathbf x(t)) = \frac{\partial H}{\partial t}(t, \mathbf x(t))+\dot{\mathbf x}(t)\cdot\nabla H(t, \mathbf x(t))
$$
Assuming then that $\partial_t H = 0$, and assuming that the path $\mathbf x(t)$ is that of a particle moving with he fluid, the equations written above imply
$$
  \frac{d}{dt}H(t, \mathbf x(t)) = \mathbf u(t, \mathbf x(t))\cdot\nabla H(t, \mathbf x(t)) = 0
$$
so the quantity $H$ is constant along a flow line, as desired!
