Appendix J Derivation of the stress tensor for a newtonian fluid

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

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

(J.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 dissipation

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

(J.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 ³¹³¹We do not consider here the a rigidly rotating fluid as it os often done in the literature, because a rigidly rotating fluid is unphysical. that means

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

(J.3)

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

$v_i={ϵ}_{ijk}{\omega }_j{r}_k$

(J.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, that means

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

(J.5)

Using this and equation (H.5) in equation (J.3) we get

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

(J.6)

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

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

(J.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 for of such a tensor is

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

(J.8)

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

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

(J.9)

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

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

(J.10)