I'm studying fluid mechanics and I have the following doubt: on the book, the author first deduces the differential form of the equation of balance of momentum. First he argues that if the fluid is ideal and is contained in a region $D\subset \mathbb{R}^3$ and if the pressure is $p : D\times \mathbb{R}\to \mathbb{R}$ then the stress force per unit volume is $-\nabla p$ and the body force per unit volume is $\rho \mathbf{b}$ (with $\mathbf{b}$ being the body force per unit mass). In that case the law becomes

$$\rho \dfrac{D\mathbf{u}}{Dt} = -\nabla p + \rho\mathbf{b}$$

Then with some manipulations he derives the integral form of the law. Basically he considers a fixed region $W$ and consider the total momentum contained in $W$ as

$$\int_W \rho \mathbf{u} \ dV$$

In that case the integral form can be obtained differentiating that. So, we gain the following law:

$$\dfrac{d}{dt}\int_W \rho \mathbf{u} \ dV = - \int_{\partial W} (p\mathbf{n}+\rho \mathbf{u}(\mathbf{u}\cdot \mathbf{n}))\ dV + \int_W \rho \mathbf{b} \ dV$$

This law is over a fixed region. So we choose a region on the fluid, compute the momentum inside it and see how it varies as the fluid flows through it.

Now, to assume as little differentiability as possible, the author proceeds to obtain one integral form of the law directly from basic principles. If $\varphi$ is the fluid flow map, $\varphi_t = \varphi (\cdot, t)$, then he states the following:

$$\dfrac{d}{dt} \int_{\varphi_t(W)} \rho \mathbf{u} \ dV = \mathbf{S}_{\partial \varphi_t(W)} + \int_{\varphi_t(W)} \rho \mathbf{b} \ dV$$

Where $\mathbf{S}_{\partial \varphi_t(W)}$ is the total force at time $t$ exterted on the fluid contained in $\varphi_t(W)$ by means of stress on its boundary $\partial \varphi_t(W)$.

This is quite clear: we pick a region, look at the mometum and the rate of change of momentum should be equal to the total force applied to it. Now this region is time-dependent. That is, instead of looking at a fixed region and at the fluid flowing through it, the author is following a chunk of fluid that moves with time.

My question is: why, when deriving the integral form of the balance of momentum law from the differential form, we consider a fixed region and why, when stating it from basic principles, we consider a region varying with time? Is there some connection between this and the differences between the Lagrangian and Eulerian points of view?

  • $\begingroup$ This might not be needed to answer the question, but what are $\mathbf{b}$ and $\mathbf{S}$ here? $\endgroup$ – user10851 Aug 29 '14 at 1:13
  • $\begingroup$ Thanks for pointing it out @ChrisWhite, I edited the question. $\mathbf{S}$ is the total stress force and $\mathbf{b}$ the body force per unit mass. $\endgroup$ – user1620696 Aug 29 '14 at 1:23

You are right, difference in the bounds of the integrals represents the difference between the Eulerian and Lagrangian perspectives. The Lagrangian frame of reference follows a volume of fluid, so Newton's laws can be directly applied. However, the bounding surfaces of the volume can change over time. This makes it difficult to actually apply the integral, as we will need some way of tracking the material volume.

On the other hand, the Eulerian perspective is fixed in space. An integral formulated in the Eulerian frame of reference can be much easier to use. For example, if we have opaque pipe with no source or sink of momentum, then we have:

$$ \int_{\partial W} \rho \mathbf{u}(\mathbf{u}\cdot \mathbf{n}))\ dA = - \int_{\partial W} p\mathbf{n}\ dA + \int_W \rho \mathbf{b} \ dV. $$ where $\partial W$ covers the two ends of the pipe. So if we can measure the momentum flux at the two ends, we can find the bulk force acting on the entire volume of fluid in the pipe.

While both integrals are technically correct, the Eulerian form can be much more practical to use. A more detailed connection between the two frames of reference can be found here.


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