In Landau & Lifshitz volume 6 on fluid mechanics, they derive the most general expression for the viscous stress tensor $\sigma_{ik}'$.

First they note that it must depend on the spatial derivatives of the fluid velocity. This is fair. Second, they assume the derivatives are small so that only linear terms contribute. I can accept this too.

Then, they state that since $\sigma_{ik}'$ must vanish for both constant velocity $\vec{v}$ and for uniform rotation $\vec{v} = \vec{\Omega} \times \vec{r}$ only symmetrical combinations of the form

$$\frac{\partial v_i}{\partial x_k} + \frac{\partial v_k}{\partial x_i}$$

can be contained in $\sigma_{ik}'$. I understand this as:

$$\sigma_{jl}' = \alpha_{ikjl} \left( \frac{\partial v_i}{\partial x_k} + \frac{\partial v_k}{\partial x_i} \right)$$

where $\alpha_{ikjl}$ are constants in the linear combination of derivatives. They then conclude (for an isotropic fluid) that

$$\sigma_{ik} = a \left( \frac{\partial v_i}{\partial x_k} + \frac{\partial v_k}{\partial x_i} \right) + b \frac{\partial v_l}{\partial x_l}. $$

I don't quite see how they get here. I have convinced myself that for an isotropic fluid all off-diagonal derivatives must contribute equally (no preferred direction in space) and all the diagonal terms must also be treated on an equal footing, though diagonal and off-diagonal terms can be different. But I don't see how they can eliminate all the other off-diagonal terms from $\sigma_{ik}'$. E.g. for $ik = xy$,

$$\sigma_{xy}' = a \left( \frac{\partial v_x}{\partial y} + \frac{\partial v_y}{\partial x} \right) + b \frac{\partial v_l}{\partial x_l}.$$

the viscous stress tensor only includes off-diagonal derivatives of x and y, but not of z. How does he know this? I suspect the answer might lie in the physics - that somehow I could convince myself that the z-terms shouldn't contribute to the x-y part of the tensor - but I haven't figured out the argument and couldn't find it anywhere.

I hope the question is clear.


1 Answer 1


For an isotropic system the viscosty tensor must be itself isotropic i.e. have the same numerical componets in all orthogonal frames. The most general 4-index isotropic tensor in 3d is
$$ \alpha_{ijkl}= a\delta_{ij}\delta_{kl}+ b \delta_{ik}\delta_{jl}+ c\delta_{il}\delta_{jk} $$ for some constants $a,b,c$. Also, I expect that $$ \alpha_{ijkl}=\alpha_{klij}, $$ so that cuts the number of constants down to two.

  • $\begingroup$ Thanks a lot! Your last condition (that resembles some sort of symmetry condition) - where does it come from? Physically $\endgroup$
    – Albert
    Mar 1, 2021 at 9:08
  • $\begingroup$ Also, I don't see how your last condition is a further constraint. Wouldn't it mean: $a \delta_{ij}\delta_{kl} + b \delta_{ik}\delta_{jl} + c\delta_{il}\delta_{jk} = a \delta_{kl}\delta_{ij} + b \delta_{ki}\delta_{lj} + c\delta_{kj}\delta_{li},$ which is always true? How does that cut down the number of constants? $\endgroup$
    – Albert
    Mar 1, 2021 at 9:37
  • $\begingroup$ The symmetry $\alpha_{ijkl}= \alpha_{klij}$ requires $b=c$. $\endgroup$
    – mike stone
    Mar 1, 2021 at 12:50

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