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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?

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In a situation like this it is useful to explain the context and define terms as not every presentation makes the same underling assumptions or uses the same symbols. In this case you appear to be working the conservative field case, but using different symbol for the potential than, say, the wikipedia presentation. All that said, this is just a conservation of energy argument; you just have to correctly identify all the contributions to the energy in your case. – dmckee Mar 8 '13 at 23:38
@dmckee, in (2.13) you can find the definitions. thank you! – Alon Shmiel Mar 8 '13 at 23:45
Alon, (2.13) of what? There is no (2.13) in you post nor any links to other resources. – dmckee Mar 8 '13 at 23:49
sorry, I forgot to paste the link: – Alon Shmiel Mar 8 '13 at 23:50
up vote 2 down vote accepted

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!

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