In the book Fluid Mechanics by Robert A. Granger, there is a study question, 4.10, asking

"How can frictionless real fluids exist and inviscid fluids not exist?"

Could someone please explain?

Please note that this is no homework-question, but is part of a self-study. I have tried to google around, but could not find any answers. I have seen several webpages stating that frictionless is the same as inviscid, for instance


  • $\begingroup$ Inviscid is non viscous? $\endgroup$ Jun 25 '15 at 12:27
  • $\begingroup$ What about superfluids? They are inviscid fluids, surely, albeit due to quantum effects rather than normal 'fluid mechanics' effects. $\endgroup$
    – tok3rat0r
    Jun 25 '15 at 12:53
  • $\begingroup$ @HritikNarayan, yes $\endgroup$
    – FredikLAa
    Jun 25 '15 at 12:58
  • $\begingroup$ I suspect the author is not considering exotic fluids and the like, but rather ordinary fluids. I suspect also the answer is in some clever tautology and a matter of definition, considering the space of all fluids. There are real fluids and ideal fluids. Some real fluids can be considered as ideal fluids but ideal fluids in a strict sense are not real. All real fluids have viscosity --so the second half of the statement is true. The first part --- I am stuck $\endgroup$
    – docscience
    Jun 25 '15 at 15:14
  • $\begingroup$ @docscience I agree -- I think the author is trying to make the reader understand the difference between mathematical assumptions/approximations and physical properties. $\endgroup$
    – tpg2114
    Jun 25 '15 at 15:22

I think this is a pretty ridiculous question (not yours, the textbook authors'). But my guess would be that the distinction is made between inviscid fluids which are governed by the Euler equations:

$$\frac{\partial u_i}{\partial t} + \frac{\partial u_i u_j}{\partial x_i} = -\frac{1}{\rho} \frac{\partial p}{\partial x_i}$$

and frictionless fluids which are governed by the Navier-Stokes equations:

$$\frac{\partial u_i}{\partial t} + \frac{\partial u_i u_j}{\partial x_i} = -\frac{1}{\rho}\frac{\partial p}{\partial x_i} + \nu \frac{\partial^2 u_i}{\partial x_i \partial x_j}$$

where $\nu = 0$.

In other words, inviscid is a mathematical assumption where we assume the viscous term does not exist in the equation at all (which changes the mathematical nature of the equation) and frictionless is a physical property of the fluid where the viscosity is zero (which has the exact same effect as being inviscid).

Poor question, but I think they want you to understand the difference between mathematical assumptions and physical properties.

  • $\begingroup$ I think you have it backwards. Wouldn't $\mu = 0$ for an inviscid flow and the $\nabla ^{2} \textbf{v} = 0$ for a frictionless flow? $\endgroup$
    – techSultan
    Oct 11 '16 at 3:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.