Does the free surface of a liquid in a rotating frame depend on the viscosity?

If we consider an incompressible, isothermal, stationary Newtonian flow (density ρ =const, viscosity μ =const), is it true that the form of the parabolic free surface in a rotating frame (constant angular velocity, of course) is the same as a flow with no viscosity?

I'm currently trying to understand the difference between these two surfaces, but it seems that in this case there is no difference between the Euler equation and the Navier-Stokes equation (as the second order partial derivitaves of the velocity are all zero.) Is this true?

• By intuition there must be some kind of difference in the surface.. where can I find the analytical reasoning? Feb 24 at 23:12
• Are you asking if the shape of the interface of a completely stationary liquid, but which is viewed from a rotating frame, is the same as the interface of the same stationary liquid (but with no viscosity) viewed from that same rotating frame? Feb 24 at 23:15
• No, it's just the shape of the free surface in rotation. By stationary I'm trying to say the velocity and pressure does not depend on time. Feb 24 at 23:18
• So in both cases, you're just considering a liquid that that is rotating with a uniform angular velocity on some container, and checking if changing the viscosity matters? Feb 24 at 23:21
• You can't initiate the rotation of an inviscid fluid in the situation you are describing because there is slip at the wall of the cylinder. However, if you were somehow able to establish the flow by some means, the shape of the free surface would be the same as for a viscous fluid. Feb 24 at 23:35

To observe this, consider a flow in cylindrical coordinates consistent with a uniform angular velocity $$\omega$$. The velocity field for this flow would just be: $$\vec{v} = \omega r \hat{\theta}$$
Taking the Laplacian in polar coordinates, one finds that: $$\nabla^2\vec{v} = \vec{0}$$
$$\mu\nabla^2\vec{v} = \vec{0}$$