# Again shear stress tensor

Hello,

today a new problem has appeared.
Usually, the shear stress tensor Pi is calculated in the form

``````
Pi_ij = eta * (dv_i / dx_j + dv_j / dx_i)

``````

Obviously, this expression is symmetric. It can be calculated in the LB code in a local way,

``````
Pi_ij = (1 - 1 / (2 * tau)) * sum_k f^neq_k * (c_ik * c_jk - delta_ij / 3)

``````

in three dimensions. However, this quantity is not physical. Why is that so? If you have an area element in the xy-plane and you want to measure (in an experiment) the force due to shear on that area element PARALLEL to the xy-plane, then you measure

``````
Pi'_xz = eta * dv_z / dx

``````

and

``````
Pi'_yz = eta * dv_z / dy

``````

Those two quantities are not the same as Pi_xz and Pi_yz using the definition at the top. The shear tensor Pi_ij has five independent components, since it is symmetric and traceless. The shear tensor Pi’_ij has EIGHT independent components (if the flow is incompressible). The missing three components are the asymmetric parts, which have been removed.

Now, since I want to simulate the physical shear stress tensor, which can be measured in an experiment, I want to calculate the quantity Pi’ and not Pi. It is not possible to extract Pi’ from Pi, since the asymmetric information is gone.
But is it possible to find a local expression usable in a LB code, which directly produces Pi’?
My understanding is that this is not possible, since one cannot find a second moment of the velocities which is asymmetric.
My conclusion is that, if one is interested in Pi’, one has to use a finite difference scheme, which of course is not local.

Does anybody have an idea regarding this issue?

Thanks a lot,
Timm

Hi Timm,

Your understanding and your conclusion are right. The lattice Boltzmann variables contain only information on the symmetric part of the shear stress, but not on the antisymmetric part. The antisymmetric part can only be accessed through a non-local approach, by using for example a finite difference scheme.

However, claiming that the symmetric strain-rate tensor S is non-physical seems a too strong statement to me. The Navier-Stokes equations for example depend only on S, and not on the antisymmetric part. And they are physically meaningful equations.

Hi Jonas,

thanks for your reply. You are absolutely right. Perhaps my question was not precise enough.
If I have a particle in the fluid flow and I want to measure the force on the particle in the xy-plane, then I will need the asymmetric parts of the shear tensor. If one however is only interested in the NS equations, the symmetric part will suffice. In that sense the symmetric shear tensor S has a physical meaning, but it is not applicable to my special needs.
However, I am lucky at this point, because the z-component of my flow velocity is very small compared to the x- and y-components. This means that I can neglect the terms dv_z / dx and dv_z / dy and I am fine. But in more general flows this is not satisfied.

Timm