Does anisotropic stress cancel at the center by symmetry?

Let's define :

$$\sigma_{ij}=\tilde{\sigma_{ij}} - P \delta_{ij}$$ with $$\sigma$$ the stress tensor, $$P$$ the pressure and $$\tilde{\sigma}$$ the anisotropic stress tensor.

The balance of force imposes : $$\partial_{i}\tilde{\sigma}_{ij} - \partial_jP=0$$.

My question is do anisotropic stresses cancel at the center because of the symmetries. For example, for a fluid in spherical coordinates and radial symmetry we have :

$$\tilde{\sigma}_{rr}\sim\tilde{\sigma}_{\theta\theta}\sim \tilde{\sigma}_{\phi\phi} \sim \partial_rv-v/r$$

Do we have at the center of the sphere : $$\partial_rv-v/r=0$$ ?

I'm asking the question because of some intuition that anisotropic stresses have some directions, and that's not consistant with the center of symmetry, I mean we will have a singularity at the center. Is it right ? And if not does anisotropic stress resect certain conditions at the center ?

• What is the "constraints tensor"? Jul 11 '19 at 15:01
• Sorry I used the French word by mistake :)
– J.A
Jul 11 '19 at 15:03

If you consider a problem with spherical symmetry the balance of forces is given by

$$\frac{\partial \sigma_{rr}}{\partial r} + \frac{2}{r}[\sigma_{rr} - \sigma_{\phi\phi}] + f_r = 0\, ,$$

when $$\sigma_{\phi\phi} = \sigma_{\theta\theta}$$ have been used.

We also have the following (non-trivial) deformation components

\begin{align} &\epsilon_{rr} =\frac{\mathrm{d} u_r}{\mathrm{d} r}\, \\ &\epsilon_{\phi\phi} = \epsilon_{\theta\theta} =\frac{ u_r}{r}\, . \end{align}

$$\sigma_{rr} = \frac{2 \lambda u_r}{r} + \left(\lambda + 2 \mu\right) \frac{d}{d r} u_r\, ,\\ \sigma_{\phi\phi} = \sigma_{\theta\theta} = \lambda \frac{d}{d r} u_r + \frac{\lambda u_r}{r} + \frac{\left(\lambda + 2 \mu\right) u_r}{r}\, .$$

I suppose that what you call "anisotropic stress" is the deviatoric part of the stress tensor, i.e.,

$$\tilde{\sigma} = \frac{2}{3} \mu \left(\frac{d}{d r} u_r - \frac{u_r}{r}\right) \begin{pmatrix} 2 &0 &0\\ 0 &-1 &0\\ 0 &0 &-1\end{pmatrix}\, .$$

And I don't see that clear what you suggest.

• What if u = kr ? Jul 12 '19 at 1:19
• @ChetMiller, in that case, the deviatoric would be zero everywhere. That case corresponds to a sphere with uniform pressure applied in the exterior. Jul 12 '19 at 15:34
• I think you're writing the stress for an elastic solid and I wrote it for a fluid, don't you ?
– J.A
Nov 23 '19 at 22:31
• @J.A, yes I wrote for an elastic solid. In your question it was never implied that you wanted it for a fluid (Newtonian?). Nov 23 '19 at 22:54
• I wrote the stress in function of a velocity, so I had in mind more a fluid, but actually I think the answer should be the same. So my question is $\tilde{\sigma}(r=0)=0$ ? I mean does the anisotropy implies something concerning the anisotropic stress at the center ?
– J.A
Nov 23 '19 at 22:58