Viscosity coefficients

Question

I'm using the 2nd edition of "Transport Phenomena" by Bird and Stewart.

I am having trouble with one of the equations: $$\tau_{ij} = \sum_k \sum_l \mu_{ijkl} \frac{\partial v_k}{\partial x_l} $$ Here, $i,j,k=1,2,\:or\:3$. And $x_1,x_2,x_3=x,y,z$ respectively. Similarly $v_1,v_2,v_3=v_x,v_y,v_z$. The derivative is meant to express that the velocity of the fluid towards the $k$ direction varies as you move along the $l$ coordinate. The $\tau_{ij}$ is the viscous stress tensor. According to the statement, the viscosity tensor can be expressed as a linear combination of all the velocity gradients. So far I am good. Then it goes on and puts the double series. Which I suppose it is there to express that the velocity of the molecules moving in the $x,y,z$ direction can vary as one moves along the $x,y,\:or\:z$ coordinates. However, I do not understand why there are 81 viscosity coefficients. I cannot imagine a physical interpretation to it. Shouldn't the viscosity coefficients be 9, $\mu_{kl}$? One way that I thought I could rationalize the reason for 81 coefficients instead of 9 was that it is possible that the the direction of the tensor matters. But isn't this information already included on the linear combination of the velocity gradients (without the viscosity coefficient)? Considering this, how and why would the direction of the tensor matter? Why are there 81 coefficients?

Answer 1

The formula that you write out is just a consequence of linearity without any additional requirements. Suppose you had a vector $\mathbf{D}$ that was linearly dependent on another vector $\mathbf{E}$. Then, one would write $D_i =\sum_j \epsilon_{ij}\ E_j$. With no further conditions, $\epsilon_{ij}$ would have $9=3\times3$ coefficients. In your formula, the viscosity tensor $\tau_{ij}$ has $9$ components (assuming no symmetry). If it depends linearly on the velocity gradients, $\partial_i v_j$, which again has $9$ components. Then, the most general formula compatible with linearity would be the one that you have written out. The tensor $\mu_{ijkl}$ has $9\times9=81$ components assuming no symmetry.