When deriving Stokes law one uses the Navier Stokes equation with the assumptions:

  1. low Reynolds number

  2. stationary flow

  3. in compressible flow

leading to this version of the N.S : $$\nabla p = \eta \Delta v $$

In my book then the force on a sphere is calculated using the stress tensor for isotropic incompressible fluids:

$$\sigma_{ij}= -\delta_{ij} p + \eta ({\partial v_i \over \partial x_j}+{\partial v_j \over \partial x_i})$$

the force than can be calculated by:

$$F=\int_{\partial V} \sigma_{ij} \vec n dS=\int_V {\partial \sigma_{ij} \over \partial x_j} dV$$

This is the cause of my strong confusion using the expression ${\partial \sigma_{ij} \over \partial x_j}$ and comparing it with the N.S we see the following: $${\partial \sigma_{ij} \over \partial x_j}=-\delta_{ij} \partial_j p + \eta \Delta v_i=0$$

were i used $ {\partial \over \partial x_ j} ({\partial v_i \over \partial x_j}+{\partial v_j \over \partial x_i})= ({\partial^2 v_i \over \partial x_ j \partial x_j}+\underbrace{{\partial^2 v_j \over \partial x_i \partial x_ j}}_{=0 })=\Delta v_i$ due to incompressibility.

What in hell am i doing wrong? It appears the second term in the solution for $v=\nabla \Phi + v_2$ i.e. $v_2={a\over r}$ is the key but should it not vanish aswell? In the equation for $F$?

Update: The equation $${\partial \sigma_{ij} \over \partial x_j}=0$$ is definetly an equivalent for the Navier srokes equation used above. (Source: Guyon, Hulin, Petit, Mitescu; Physical Hydrodynamics) So how can you even use this to calculate the Stokes friction?

  • $\begingroup$ the force density on a surface is given by $\sigma \cdot \vec{n}$ where $n$ is the surface normal. Use this + divergence theorem $\endgroup$ – Bort Feb 12 '16 at 12:49
  • $\begingroup$ it does. $F=\int \sigma \cdot \vec{n} \mathrm{d}A$ is the definition of the stress tensor (go back to cauchy momentum equation if you do not believe this). (this will be my last comment on this, as I really struggle to remain polite) $\endgroup$ – Bort Feb 12 '16 at 13:18
  • $\begingroup$ Maybe you can correct the spelling of "friction" in the title. $\endgroup$ – Thomas Feb 12 '16 at 16:51
  • $\begingroup$ For another treatment of this in depth check out Transport Phenomena by Bird, Stewart and Lightfoot $\endgroup$ – nluigi Feb 12 '16 at 23:09
  • $\begingroup$ thanks for the tipp I'll keep it in mind to check it out ^^ $\endgroup$ – pindakaas Feb 13 '16 at 22:39

The problem is that you are not allowed to use Gauss law in this way. The force is given by $$ F_i= \int_{S} n_j\sigma_{ij} dS $$ where $S$ is the surface of the sphere. In order to use Gauss law you have to interpret $S$ as the boundary of some volume. You cannot use the interior of the sphere, because the velocity field is not defined there. But you cannot use the exterior either, because then you have to include a surface at infinity. Indeed, this surface cannot be neglected because the pressure asymmetry falls as $1/r^2$. Physically, this is clear: The force on the sphere arises from somebody pushing the fluid at infinity.

In practice this demonstrates that you can either compute $F$ from the original surface integral (this is what is usually done in text books), or you can compute it from a surface integral at infinity.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.