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If I make no mistake, dynamic viscosity $\eta$ is measured in $ \rm kg /(s\, m) $. And volume is measured in $\rm m^3$.

The product of the two has the same units of Planck's quantum of action $\hbar$, namely $ \rm kg\, m^2 /s $. Does this mean that there is a uncertainty relation between volume and viscosity?

I have not found anything of the kind in the literature; however, usually, two quantities whose product has the units of $\hbar$ follow a uncertainty relation, such as position and momentum, phase and angular momentum, or energy and time.

So the question is: is there a relation of the type

$$ \Delta \eta \, \Delta V \geq \hbar \ \ ?$$

This is a question about the measurement precision of physical observables: can the simultaneous measurement of the viscosity and the volume of a fluid have measurement errors that are independent of each other and can be a small as desired?

If no such relation existed, then the right hand would be zero, and then it seems that it should be possible to circumvent the usual uncertainty relation between momentum and position.

tpg2114
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frauke
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    Why do you think an uncertainty relation has anything to do with the units of the product of these quantities? – ACuriousMind Jun 02 '19 at 17:36
  • The question is on hold because it is not clear what you're asking. When you ask about a quantum uncertainty relation, you first need to exhibit a theory in which the two quantities you are asking about are actually quantum observables. What theory are you using in which the volume and viscosity are quantum operators of a system? – ACuriousMind Jun 02 '19 at 17:54
  • A measurement error is a well-defined quantity. There is no need for a theory to define them. – frauke Jun 02 '19 at 18:06
  • A quantum uncertainty is not about measurement errors, cf. e.g. https://physics.stackexchange.com/q/54184/50583 and its many linked questions. – ACuriousMind Jun 02 '19 at 18:08
  • There are plenty of examples where two entirely different physical entities have the same units. One that jumps to mind is torque and energy. – Chet Miller Jun 02 '19 at 18:35
  • True, and there is an uncertainty relation for it as well! – frauke Jun 02 '19 at 19:11
  • You haven't shown that these, when interpreted as operators on some quantum state (which you'd have to then define what the viscosity and volume operators are and how they work on states, which needs a theory to do so), that their commutator is non-zero. As an example the units of $p_y \cdot x $ also have units of action however there is no uncertainty relation between them as they commute. – Triatticus Jun 02 '19 at 21:05
  • Hi Frauke -- I cleaned up some of the edits to the post. We appreciate the attempt to clarify your question and there are some reopen votes pending, so if a few more people agree it is clear enough, it will be reopened. Since this is a Q&A site, we don't want questions to generate discussions -- it should be clearly answerable. If you no longer want the question to be answered here, you can delete the post. – tpg2114 Jun 02 '19 at 23:02
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    @ACuriousMind OP seems to be asking whether such a theory exists. Still, the question is bad enough (and so completely sunk by its final paragraph as of v11) that I'm not particularly keen to vote to reopen in its current form. – Emilio Pisanty Jun 02 '19 at 23:07
  • @ACuriousMind The question is obviously not very well stated, but in the end this is what viscosity bounds $\eta/s>\hbar/(4\pi k_B)$, and recent ideas about quantum thermalization and chaos amount to. – Thomas Jun 03 '19 at 14:10

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The fact that the dimensions of two given variables multiply into those of action is by no means sufficient for there to exist a theory where they are canonical conjugates.

Viscosity, in particular, is extremely unlikely to find any sort of a home in a quantum mechanical framework. For one, viscosity reflects a dissipative process, and those are extremely hard to fit into QM. For another, quantum fluids are, generically, superfluids, with no viscosity.

There's nothing specific that says that this coupling is forbidden, but there's no requirement for it and you're unlikely to find one.

As for this,

If no such relation existed, then the right hand would be zero, and then it seems that it should be possible to circumvent the usual uncertainty relation between momentum and position.

That's a remarkable leap of logic with nothing to back it up.

Emilio Pisanty
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