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?


3 Answers 3


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.

  • $\begingroup$ Right. p is not a vector field... By the way I still don't understand the last equality... $\endgroup$
    – user11543
    Dec 1, 2012 at 23:27
  • $\begingroup$ Here you can check a simple explanation of your last equality. $\endgroup$
    – Ana S. H.
    Dec 1, 2012 at 23:30

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.

  • $\begingroup$ It is simpler and more general considering it as a special case of the Gauss-Ostrogradsky theorem. $\endgroup$
    – Ana S. H.
    Dec 2, 2012 at 5:58
  • $\begingroup$ @Anuar: Except it's so simple I don't think it needs Gauss or Ostrogradsky's names attached to it. $\endgroup$
    – Ron Maimon
    Dec 2, 2012 at 6:04
  • $\begingroup$ Well, it's just matter of likes. $\endgroup$
    – Ana S. H.
    Dec 2, 2012 at 6:09

I'm 9 years late, but I'll answer anyway hoping to be useful to someone else. Before reading this answer it could be very useful (if not essential) reading the answer I wrote on the Cauchy equation. Note: I will use no bar for scalars, one bar for vectors (hat for versors) and two bars for tensors (for brevity, when I will say "tensor" I will always mean "rank 2 tensor", but of course scalars and vectors too are tensors).


The Euler equation is a differential equation that, with continuity equation, allows us to describe the motion of incompressible (density constant) and inviscid (pressure is the only responsible for surface forces) fluid. Indeed it also describes the motion of a compressible fluid but in that case we have too many variables, as we will see. The equation is \begin{equation} \frac{D \bar{v}}{Dt}= \bar{f} - \frac{\nabla p}{\rho} \tag{1} \end{equation} The meaning of terms $\frac{D \bar{v}}{Dt}$ and $\bar{f}$ are explained in my S.E. answer about Cauchy equation, $\rho$ is density and $p$, as we will see, is pressure. The necessity of Euler equation has origin by the fact that momentum conservation is expressed by Cauchy equation \begin{equation} \nabla \cdot \bar{\bar{\sigma}} + \rho \bar{f} = \frac{\partial (\rho \bar{v})}{\partial t} + \nabla \cdot \bar{\bar{F}} \end{equation} but this equation alone is not very useful, even if combined to continuity equation: stress tensor has nine components, six of these are independent if we assume it is symmetric, but unknowns are decidedly too much: considering also $\bar{v}$ (3 unknowns) and $\rho$, we have in total 10 unknowns in front of only 4 equations (continuity equation and Cauchy equation, that consists in three equation because is a vector equation). The problem can be made mathematically solvable by making these two assumptions

Assumption 1) the fluid is incompressible

Assumption 2) For every surface $d\bar{A}$ inside the fluid, the stress $\frac{d\bar{F}}{dA}$ is orthogonal to the surface.

I will show you how these assumption leads to the solvable (in principle) Euler equation.

On assumption 1

Assumption 1 doesn't need big arguments and we know it's very good for liquids. With this assumption $\rho$ became a constant (we remove an unknown, that anyway will be added with assumption 2 introducing $p$) and equations are simplified, for example continuity equation became $\nabla \cdot \bar{v}=0$ (this is not exploited in deriving Euler equation, but this is exploited in deriving Navier-Stokes equation).

On assumption 2

This is a more delicate point than assumption 1. First of all we must show that the assumption 2 is valid if and only if the stress tensor is multiple of identity tensor: \begin{equation} d\bar{F} \parallel d\bar{A} \qquad \Leftrightarrow \qquad \bar{\bar{\sigma}} = k \bar{\bar{I}} \end{equation} where $\bar{\bar{I}}$ is the identity vector and $k$ is a scalar field with dimensions of pressure. I did this proof.

(proof $\Rightarrow$)

Because of stress tensor definition $d\bar{F}= \bar{\bar{\sigma}} \cdot d \bar{A}$ we have that our hypothesis can be written in this way \begin{equation} \begin{pmatrix} \sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\ \sigma_{yx} & \sigma_{yy} & \sigma_{yz} \\ \sigma_{zx} & \sigma_{zy} & \sigma_{zz} \end{pmatrix} \begin{pmatrix} dA_x \\ dA_y \\ dA_z \end{pmatrix} = k \begin{pmatrix} dA_x \\ dA_y \\ dA_z \end{pmatrix} \end{equation} So we have $\sigma_{xx} dA_x + \sigma_{xy} dA_y + \sigma_{xz} dA_z = k dA_x$, $\sigma_{yx} dA_x + \sigma_{yy} dA_y + \sigma_{yz} dA_z = k dA_y$ and $\sigma_{zx} dA_x + \sigma_{zy} dA_y + \sigma_{zz} dA_z = k dA_z$. By hypothesis these equations must be true for every $d\bar{A}$ so we conclude that $$ \sigma_{xx}=\sigma_{yy}=\sigma_{zz}=k $$ $$ \sigma_{xy}=\sigma_{xz}=\sigma_{yx}=\sigma_{yz}=\sigma_{zx}=\sigma_{zy}=0 $$ In other words we have the thesis: $\sigma_{ij}=0$ if $i\neq j$, and $\sigma_{ii}=k$.

(proof $\Leftarrow$)

By definition of stress tensor $d\bar{F}= \bar{\bar{\sigma}} \cdot d \bar{A}$, so exploiting the hypothesis $\bar{\bar{\sigma}} = k \bar{\bar{I}}$ we have \begin{equation} d\bar{F} = k \bar{\bar{I}} \cdot d \bar{A} = k d \bar{A} \tag{2} \end{equation} and so $d\bar{F} \parallel d\bar{A}$.

Again on assumption 2

Now we will show that if assumption 2 is valid then stress tensor is given by \begin{equation} \bar{\bar{\sigma}} = -p\bar{\bar{I}} \tag{3} \end{equation} where $p$ is the positive scalar field that we can call pressure, and $\bar{\bar{I}}$ is as always the identity tensor. To see why, let's consider that we have already seen that with assumption 2 the stress tensor must assume the form $\bar{\bar{\sigma}} = k \bar{\bar{I}}$. By definition of stress tensor (see my S.E. answer on Cauchy equation) this means that the "upper" side (the one containing the vector $d\bar{A}$) exert on the "lower" part a force given by (2): $d\bar{F} =k d \bar{A}$. The forces that portions of fluid exert on other portions of fluid are not tensile, but compressive forces, so $d\bar{F}$ must always have the opposite direction of $d\bar{A}$, and so $k<0$. By naming $p$ the positive scalar field that describe these forces, I get (3). Conclusion: assumption 2 allows us to describe the tensor field $\bar{\bar{\sigma}}$ exploiting a scalar field, reducing by five the number of unknowns. By the way I used the letter $p$ because we can see pressure as the magnitude of normal stress not given by viscosity (as can be seen deriving Navier-Stokes equation, if we assume a non zero viscosity there can be normal stresses that have a different origin than that, so if fluid is not ideal it would be wrong to say that pressure is the magnitude of normal stress, but it is true in static case and/or in absence of viscosity: in both cases the tensor $2\eta \bar{\bar{\epsilon}}$ that we have in Navier-Stokes equation became $\bar{\bar{0}}$).

Proof of Euler equation

The shorter version of the Cauchy equation (see my S.E. answer about it) is \begin{equation} \frac{D \bar{v}}{Dt} =\frac{1}{\rho} \nabla \cdot \bar{\bar{\sigma}} + \bar{f} \end{equation} We have seen that with our assumption stress tensor is writable as $\bar{\bar{\sigma}} = -p\bar{\bar{I}}$. So we immediately get the Euler equation by observing that \begin{equation} \nabla \cdot (-p\bar{\bar{I}}) = \left( \frac{\partial}{\partial x} \quad \frac{\partial}{\partial y} \quad \frac{\partial}{\partial z} \right) \begin{pmatrix} -p & 0 & 0 \\ 0 & -p & 0 \\ 0 & 0 & -p \end{pmatrix} = - \nabla p \end{equation}

Final note

To prove the Cauchy equation we didn't exploit the hypothesis of incompressibility, so the equation works for compressible fluid too (but inviscid). At first glance it may seem very ugly assume a limitanting and not exploited hypothesis, and in some way it is true, but note that this assumption reduces the unknowns and so it makes the system Euler+continuity solvable, so it makes sense assuming $\rho=$constant.


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