# Euler equation of fluid dynamics

I'm trying to obtain Euler equation for a perfect fluid in laminar or stationary flow. A particle fluid is submitted at volume forces and surface force. The fist, in my case, is giving only by gravity and the second by pressure. By Newton's second law I obtain:

$$\vec{F}_V + \vec{F}_s = m\frac{d\vec{v}}{dt}.$$

An element of volume force is given by $$d\vec{F}_V = dm\vec{g}=\rho d\omega\vec{g}$$ and an element of surface force is given by $$d\vec{F}_S = -pd\vec{S}.$$

Integrating I obtain

$$\int_V \rho \,d\omega\vec{g} - \int_S p\,d\vec{S} = \frac{d\vec{v}}{dt}\int_V \rho\, d\omega$$.

Now Euler equation is written in local form as $$\rho\vec{g} - \nabla p = \rho \frac{d\vec{v}}{dt}.$$

My question is this: where the gradient of $p$ comes from? I must have the following identity $$-\int_S pd\vec{S} = -\int_V \nabla p\,d\omega.$$

Why the transformation from a surface integral to a volume integral is given by the gradient and not by the divergence? I'm doing something wrong in the previous calculations?

This confusion is caused by vector calculus. You should treat each component separately, and then it is obvious. For example, for the x component:

$$\int_V \partial_x p dx dy dz = \int_{\partial V} p(x) dy dz = \int_{\partial V} p dS_x$$

by the fundamental theorem of calculus (do the x integral first). Likewise for the other components. You can make up a proof for this from the divergence theorem by introducing the fictitious vector field

$$Q = (p,0,0)$$

And then the divergence of Q is the left hand side, while the right hand side is $Q\cdot dS$. But it's really just the fundamental theorem of calculus.

• It is simpler and more general considering it as a special case of the Gauss-Ostrogradsky theorem. – Ana S. H. Dec 2 '12 at 5:58
• @Anuar: Except it's so simple I don't think it needs Gauss or Ostrogradsky's names attached to it. – Ron Maimon Dec 2 '12 at 6:04
• Well, it's just matter of likes. – Ana S. H. Dec 2 '12 at 6:09

Because the Gauss-Ostrogradski theorem says that $$\iiint_{V}\nabla\cdot\mathbf{C}dv=\iint_{\partial V}\mathbf{C}\cdot\mathbf{n}da$$ Where $\mathbf{C}$ is a vector field. Here you don't have a vector field inside the integral. So, why do you expect that the G-O theorem is applied in this case?? By the way, the last equality that you wrote is correct. I don't know how to prove it, but I'm pretty sure that I saw that kind of theorem in Jackson's book of Electrodynamics.

• Right. p is not a vector field... By the way I still don't understand the last equality... – user11543 Dec 1 '12 at 23:27
• Here you can check a simple explanation of your last equality. – Ana S. H. Dec 1 '12 at 23:30