Appendix C Derivation of the stress tensor for a newtonian fluid

We derive the stress tensor by considering the dissipation of a motionless fluid seen by a rotating observer. This derivation is different from what is found in the literature (eg. Greiner,W. and Stock, 1991) and therefore presented here.

It is generally assumed that friction between fluid elements is proportional to the area of their surfaces. So in general the frictional or viscous force on a fluid element can be expressed like

$F_{visc,i}={\displaystyle {\oint }_A}{\sigma }_{ij}^{\prime }{n}_j𝑑A={\displaystyle {\int }_V}{\displaystyle \frac{\partial }{\partial r_j}}{\sigma }_{ij}^{\prime }𝑑V.$

(C.1)

This force leads to an irreversible rise of temperature in the fluid or an irreversible decrease of kinetic energy expressed by the equation for the dissipation⁵⁴⁵⁴See Landau and Lifschitz (1991).

${\displaystyle \frac{\partial }{\partial t}}{E}_{kin,visc}=-{\displaystyle {\int }_V}{\sigma }_{ij}^{\prime }{\displaystyle \frac{\partial }{\partial r_j}}{v}_i𝑑V.$

(C.2)

For a motionless fluid ($v_i=0$) and for a fluid with constant velocity ($\frac{\partial v_i}{\partial r_j}=0$) this integral is zero. But also a rotating observer of a motionless fluid should not see a rise in the temperature of a fluid ⁵⁵⁵⁵We do not consider here the a rigidly rotating fluid as is often done in the literature, because a rigidly rotating fluid is unphysical. This is so, because a rigidly rotating fluid can never fulfill the boundary condition $v_i=0$. However, for a rotating observer, the boundary is also rotating, so that the boundary condition for a boundary at distance $R$ is $v_i={ϵ}_{ijk}{\omega }_j{R}_k$ and there is no contradiction to the velocity field (C.4). that means

${\displaystyle {\int }_V}{\sigma }_{ij}^{\prime }{\displaystyle \frac{\partial }{\partial r_j}}{v}_i𝑑V=0.$

(C.3)

A rotating observer of a motionless fluid sees a velocity field of the form

$v_i={ϵ}_{ijk}{\omega }_j{r}_k.$

(C.4)

where ${\omega }_j$ is the angular velocity vector and $r_k$ is the position vector. It can be shown that for such a velocity field the Jacobian is antisymmetric (Greiner,W. and Stock, 1991), that means

${\displaystyle \frac{\partial v_i}{\partial r_j}}=-{\displaystyle \frac{\partial v_j}{\partial r_i}}.$

(C.5)

Using this and equation (B.5) in equation (C.3) we get

${\displaystyle {\int }_V}{\displaystyle \frac{1}{2}}\left({\sigma }_{ij}^{\prime }-{\sigma }_{ji}^{\prime }\right){\displaystyle \frac{\partial }{\partial r_j}}{v}_i𝑑V=0.$

(C.6)

This relation can only be fulfilled if the stress tensor ${\sigma }_{ij}^{\prime }$ is symmetric

${\sigma }_{ij}^{\prime }={\sigma }_{ji}^{\prime }.$

(C.7)

For a newtonian fluid is it assumed that the stress tensor is proportional only to the first derivatives of the velocity field. Together with the requirement of symmetry the most general form of such a tensor is

${\sigma }_{ij}^{\prime }=a\left({\displaystyle \frac{\partial v_j}{\partial r_i}}+{\displaystyle \frac{\partial v_i}{\partial r_j}}\right)+b{\delta }_{ij}{\displaystyle \frac{\partial v_k}{\partial r_k}}.$

(C.8)

Usually the trace is split off the first term and added to the second term so

${\sigma }_{ij}^{\prime }=a\left({\displaystyle \frac{\partial v_j}{\partial r_i}}+{\displaystyle \frac{\partial v_i}{\partial r_j}}-{\displaystyle \frac{2}{3}}{\delta }_{ij}{\displaystyle \frac{\partial v_k}{\partial r_k}}\right)+\left({\displaystyle \frac{2a}{3}}+b\right){\delta }_{ij}{\displaystyle \frac{\partial v_k}{\partial r_k}}.$

(C.9)

Using the definitions $a=\eta$ and $\frac{2a}{3}+b=\zeta$ we get the form most common in literature

${\sigma }_{ij}^{\prime }=2\eta \left[{\displaystyle \frac{1}{2}}\left({\displaystyle \frac{\partial v_i}{\partial r_j}}+{\displaystyle \frac{\partial v_j}{\partial r_i}}\right)-{\displaystyle \frac{1}{3}}{\delta }_{ij}{\displaystyle \frac{\partial v_k}{\partial r_k}}\right]+\zeta {\delta }_{ij}{\displaystyle \frac{\partial v_k}{\partial r_k}}.$

(C.10)