# Viscosity from DFT (VASP) using the Green-Kubo relation

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## Main Question or Discussion Point

Hello! In this paper https://pdfs.semanticscholar.org/e8a2/02f25555cd8c4f947bbbdff5a61a0ea0efd2.pdf the authors use VASP to determine MgSiO3 viscosity using the Green-Kubo relation

$\eta = \frac{V}{3k_{\rm{B}}T}\int_{0} \left<\sum_\limits{i<j}\sigma_{ij}(t+t_{0}).\sigma_{ij}(t_{0})\right>dt$ where $\sigma_{ij}$ (i and j = x, y, z) is the stress tensor, t is time and t0 is the time origin. But I've seen other papers use:

$\eta = \frac{V}{3k_{\rm{B}}T}\int_{0}^{\infty} dt \left< P_{xy}(t)P_{xy}(0)\right>$, where $P_{xy}$ is the off-diagonal component of the stress tensor $P_{αβ}$ ( α and β are Cartesian components).

OK, so clearly these are essentially exactly the same equation but the second uses only the xy component whereas the first seems to suggest a summation? So which is correct.

Also, VASP outputs the stress tensor components as XX YY ZZ XY YZ ZX. So which of these should I use to input into the Green-Kubo equation? And are there missing components? what about yx, zx, yx?

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## Answers and Replies

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OK, I've worked out that due to symmetry, XY=YX, YZ=ZY, ZX=XZ, so the tensor reduces from 9-components to 6. So that explains my second question (note typo when I wrote "what about yx, zx, yx?" should have said "what about yx, zy, xz?"

But I still don't understand why the first equation is suggesting a sum of the off-diagonal components whereas the second equation suggests only using one of the off-diagonal components $P_{xy}$.

TeethWhitener
Gold Member
But I've seen other papers
Can you give an example of these other papers? Were they looking at a 2D case? Without the proper context, it's really difficult to tell what' s going on.

TeethWhitener
Gold Member
Sure, I got the second equation from this paper: http://www.homepages.ucl.ac.uk/~ucfbdxa/pubblicazioni/PRL05161.pdf
I think this is just a schematic equation. Further into that paper (page 2), there's a paragraph starting with "There are five independent components of the traceless stress tensor..." The paper goes on to say that they find the shear viscosity by averaging the autocorrelation functions of these five components and integrating them from 0 to $t$, taking the limit as $t \to \infty$. (beginning of the paragraph that starts with "In Fig. 2") I haven't done the nitty gritty math, but it looks like this makes the two definitions in your OP equivalent.

Yes, looks like you're right. Thanks for spotting that.

DrDu
OK, I've worked out that due to symmetry, XY=YX, YZ=ZY, ZX=XZ, so the tensor reduces from 9-components to 6. So that explains my second question (note typo when I wrote "what about yx, zx, yx?" should have said "what about yx, zy, xz?"

But I still don't understand why the first equation is suggesting a sum of the off-diagonal components whereas the second equation suggests only using one of the off-diagonal components $P_{xy}$.
As liquids are isotropic, viscosity is a 4th order tensor which can be parametrized with only two parameters, $\eta$ and $\zeta$, $\eta$ is the same whether determined with xy or xz or whathever component of the deviator of the viscosity gradient or of their sums. On the other hand, the correlation of the Trace of the $\sigma_{ii}$ will probably yield the longitudinal viscosity $\zeta$.

Chestermiller
Mentor
As liquids are isotropic, viscosity is a 4th order tensor which can be parametrized with only two parameters, $\eta$ and $\zeta$, $\eta$ is the same whether determined with xy or xz or whathever component of the deviator of the viscosity gradient or of their sums. On the other hand, the correlation of the Trace of the $\sigma_{ii}$ will probably yield the longitudinal viscosity $\zeta$.
I think you mean the dilatational viscosity (proportionality constant between the trace of the stress tensor and the trace of the rate of deformation tensor).

DrDu