# Are the viscosity and the volume of a fluid related by an uncertainty relation?

+ 0 like - 0 dislike
225 views

Is there a relation of the type  $\Delta \eta \ \Delta V \geq \hbar$ ?

I have found nothing via Google Scholar on this topic.

There might be applications for the inequality in black hole physics.

edited Jun 2, 2019

There are many consistency reasons to answer no. But it's too broad to answer anything :) Perhaps were you looking for Dam Thanh Son articles around Hall viscosity like, but not only, Effective field theory for fluids: Hall viscosity from a Wess-Zumino-Witten term . I'm curious to know what you would answer today.

Thank you - I was indeed thinking about the various publications of Son, Starinets and Kovtun.

+ 1 like - 0 dislike

The free stress-energy tensor can be defined as

$$T^{\mu\nu}_{\rm free} = \int f(x^\mu,p_\mu) p^\mu p^\nu \frac{d^3 p}{p_0 m^2}$$

where $f$ is the distribution function, and I assume massive particles of mass $m$. The interaction stress-energy tensor can then be chosen so that $T^{\mu\nu}_{\;\;;\nu}$ is equivalent to the second moment of the collisional Boltzmann equation

$${f,H_{\rm free}} = \frac{\delta f}{\delta t}\Big|_{\rm coll.}$$

Generally, this is not possible. However, it is generically possible to obtain a volume-averaged version of the second moment that offers an asymptotic description of the dynamics in terms of an averaged stress-energy tensor $\bar{T}^{\mu\nu}$. This amounts to a free stress energy tensor with additional viscosity terms that scale as the gradients of stress energy times kinematic viscosity, $\sim \nu T^{\mu\nu}_{{\rm free};\kappa}$. However, the volume averaging itself adds terms

$$\bar{T^{\mu\nu}}_{\rm free} \equiv \frac{1}{\delta V}\int_{\delta V} T^{\mu\nu}_{\rm free} d^3 x = T^{\mu\nu}_{\rm free} + \propto T^{\mu\nu}_{\rm free;\kappa} \delta V^{1/3} + \propto T^{\mu\nu}_{\rm free;\kappa\gamma} \delta V^{2/3} + ...$$

In other words, the averaging volume itself adds kinematic viscosity, or $\Delta \nu/\Delta V^{1/3} \sim C$, where $C$ is a dimensionless geometrical-type constant. By going quantum, one can assume that $\delta V^{1/3}$ needs to be about as large as the De Broglie wavelength or smaller for convergence (note I assume massive particles), and assuming an ideal-gas-type behaviour, we have $\delta V^{1/3} \sim h/ \sqrt{m k_{\rm B} T}$, so I get

$$\Delta \nu \sqrt{m k_{\rm B} T} \lesssim h$$

This is really nothing like your relation. However, it can perhaps be useful for you to see that what kind of relation you get depends on the physical context you cite.

I think you might get interesting stuff from a gradient-type expansion of a conformal quantum fluid rather than massive non-relativistic gases as considered in this answer.

answered Jun 4, 2019 by (1,645 points)

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\varnothing$ysicsOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.