# Will a propeller work in a superfluid?

Will a propeller work in a superfluid? Opinions differ.

• I see what you did there, very smart sir. Well done. – Mindwin Feb 10 '17 at 19:27
• What did I do? (Other than ask about something from a WB question) – JDługosz Feb 11 '17 at 1:27
• @Mindwin are you able to enlighten us as to what OP did? – Criggie Feb 12 '17 at 5:58
• Two things: Picked a popular post on Worldbuilding SE without an accepted answer, then cross-referenced the above post to the former, to drive traffic here. Smart, yes. Just to be clear, I see nothing wrong with this. – Pirx Feb 12 '17 at 13:58

No.

I actually tried this in an undergraduate physics lab long ago. We put two fan blades right up against each other in a glass dewar. One was driven and the other was free to spin. We filled the dewar with liquid He and spun the fan. The other spun just fine.

We pumped on the LHe until it transitioned to a superfluid. We spun the fan. The other just sat there and then slowly slowly started to turn.

Edit as requested

So yes, it worked a little. But so poorly that the best answer is no. As Does every superfluid have a normal and a superfluid component? says, it has two components. The normal component was responsible for the residual viscosity. If we had reduced the temperature, there would be a smaller fraction of normal component, and less viscosity. It would work even more poorly.

And I must include the obligitory XKCD.

• Your answer is no; but your anecdote is yes. If it slowly started to turn, it was working to some extent; although the power transmission may have been incredibly small. – JMac Feb 10 '17 at 14:07
• You are right. It worked much less. So poorly that the best answer is no it didn't work. As physics.stackexchange.com/q/311115/37364 says, it has two components. The normal component was responsible for the residual viscosity. If we had reduced the temperature, there would be a smaller fraction of normal component, and less viscosity. – mmesser314 Feb 10 '17 at 14:10
• It would be good if you clarified that in the answer then. The way you have it worded now contradicts itself unless you mention that it wasn't in a pure superfluid. – JMac Feb 10 '17 at 14:12
• @Davidmh - Just once. I am sure it was an expensive lab, but it was a great day. We made nitrogen and oxygen snow out of air. We put LHe in a porous clay cup. The holes were microscopic, so surface tension prevented it from leaking out the bottom. Below the critical temperature, it just poured through. We did it again with a non porous cup. Below the critical temperature, it climbed over the top and just poured off the bottom. These days you would just watch it on the internet. – mmesser314 Feb 11 '17 at 13:51

Interesting question. First of all, on the more general question, it is certainly possible to design devices that provide thrust in a superfluid, and some examples were given in the thread you have linked to. A simple propeller, on the other hand, will not work, if we can assume that the viscosity is exactly zero. If it's just very small, then the propeller should still work fine.

Now, we could still ask if there was a way to make a propeller work in a zero-viscosity fluid, using some auxiliary devices. The critical issue is that we need to create circulation in the fluid, without the help of viscosity. There are ways to create such circulation (using tangential wall jets, say), but I am not sure if effective devices that create forces in such fluids can be designed this way. My gut feeling is that simply using jet thrusters driven by some type of positive-displacement pumps might be the way to go.

To make this argument more rigorous, we will consider the fundamental equations describing this fluid flow problem, which are the incompressible Navier-Stokes equations for an incompressible fluid,

$$\rho\left(\frac{\partial\mathbf u}{\partial t} +{\mathbf u}\cdot{\mathbf\nabla}{\mathbf u}\right)=-{\mathbf\nabla}p+\nu\Delta{\mathbf u},$$

where we are ignoring potential volume forces (such as gravitational forces), plus the divergence-free condition,

$$\nabla\cdot {\mathbf u}=0,$$

within some closed two-dimensional domain $\Omega$ (in general multiply connected; see comments below for the 3-D case).

We will consider problems characterized by boundary conditions that prescribe the flow as being either normal to the boundary, ${\mathbf u}\cdot{\mathbf t}=0$ on part of the boundary $\partial\Omega$ (at inflow/outflow boundaries), or tangential to the boundary, ${\mathbf u}\cdot{\mathbf n}=0$ on other parts of the boundary (along solid walls). Here $\mathbf n$ and $\mathbf t$ are the unit normal and tangent vectors on the boundary, respectively. As our second boundary condition we prescribe a "traction-free condition", $\nu\,\Delta{\mathbf u}\cdot {\mathbf t}=0$, meaning the fluid slides across the surface without friction.

We choose an initial condition for a velocity field ${\mathbf u}_0={\mathbf u}(t=0)$ which satisfies $\nabla\cdot{\mathbf u}=0$ (incompressibility condition) as well as $\nabla\times{\mathbf u}=0$ (irrotationality) in the interior of $\Omega$. Finally, we require that the initial velocity field ${\mathbf u}_0$ is at least $C^0$-continuous on the closed domain $\Omega$. We note in passing that this condition is non-trivial and often violated in many practical problems (meaning, numerical solutions). It plays a crucial role in the mathematics of the Navier-Stokes problem, however: If this condition is violated then the Navier-Stokes problem is in fact ill-posed. To avoid unnecessary complications I will require the slightly stronger condition of $C^1$-continuity below. The difference is mathematically non-trivial, but should be of no physical consequence.

Now, with these preparations it is possible to prove that the space of potential flow solutions of this problem is an invariant subspace of the Navier-Stokes equations, meaning that if our initial condition represent an irrotational, incompressible flow, the flow must remain so at all future times. One specific consequence of this situation is that the circulation $\Gamma$ satisfies

$$\Gamma=\oint_C {\mathbf u}\cdot\mbox{d}{\mathbf s}=\iint_\Omega \nabla\times{\mathbf u}\,\mbox{d}{\mathbf x}=0$$

at all times. I will note that I'm leaving out the proof. The technical details are somewhat severe; for details see Ladyzhenskaya's Mathematical Theory of Viscous Incompressible Flow.

At this point the only additional ingredient we need is the Joukowski theorem which says that forces in potential flows around closed contours are proportional to the circulation $\Gamma$ around the contour. Since we have shown above that the circulation remains zero, there can be no forces.

I remind the reader that the above argument assumes a two-dimensional domain. I'll just state without proof that it can be extended to the three-dimensional case, at the cost of considerable mathematical complexity...

Finally, it is worth pointing out that the above argument addresses the mathematics of this problem for the case of an ideal inviscid fluid. If you preform an experiment in such a flow, a number of real-world effects may change the outcome significantly. For example, the potential flow around sharp trailing edges of the propeller's airfoils requires extremely strong pressure gradients. I am fairly certain that any real fluid would be subject to cavitation under these conditions, and who knows what might happen then. Most certainly the model of the ideal incompressible fluid will not apply to the situation anymore.

• "The critical issue is that we need to create circulation in the fluid": is this essentially the same reasoning that shows one cannot have lift from a wing in a zero viscosity 2D potential flow unless one introduces circulation? Or is this too simple minded? – Selene Routley Feb 10 '17 at 12:58
• Yes, same thing. – Pirx Feb 10 '17 at 13:30
• I'd love to hear any reasoning behind statement, that propeller can't work in exactly zero viscosity. If we start moving propeller doesn't it apply some force to liquid particles? If yes, then it does generate non-zero thrust. If no, well, i'd like to imagine it, so share your thoughts. – nnovich-OK Feb 10 '17 at 14:07
• That's a fair question to ask. The answer boils down to what I said, but the details are anything but simple and won't fit into a comment. I'll see if I can find the time to expand on my original answer later today. – Pirx Feb 10 '17 at 14:53
• @slebetman The answer is viscosity - basically liquid inertia. No, that answer has no connection to physical reality. But you are correct that the question of how it is that propellers (or wings for that matter) can produce lift is indeed quite interesting. As Jan has said, viscosity has nothing to do with inertia, and the role of viscosity in the generation of lift is much more subtle: In a sense viscosity plays a purely auxiliary role, in that it is needed to set up a flow that can produce lift, but this exact same flow will produce lift just as nicely in the complete absence of viscosity. – Pirx Feb 12 '17 at 13:49

Fluids with viscosity zero (superfluids) are unable to exchange energy with a moving object submerged in it. So if the object is a rotating propeller, the fluid can't transfer energy to it, which is the same as saying that if the propeller is attached to a submarine, the submarine's kinetic energy can't increase, so the propeller is of no use. Even a square piece of metal moving through a superfluid doesn't experience a change in velocity (which is rather counter-intuitive).

• Does that same argument imply that peristaltic pumps wouldn't work, either? – ruakh Feb 11 '17 at 6:05
• Can you include some reference for that? E.g. how do I determine whether the assertions made by this post or peterh’s is correct? – JDługosz Feb 11 '17 at 10:52
• @ruakh-No. A peristaltic pump is not an object that is in its entirety submerged in the superfluid. It's completely around the fluid. There is no energy exchange between the fluid and the mass of the pump, though. You can compare it to a superfluid in a plastic ball. As you throw the ball away then, of course, the fluid in the ball will acquire kinetic energy. – descheleschilder Feb 11 '17 at 12:49
• @JDlugosz-I asked a same kind of question about the movement of a piece of metal through a fluid without viscosity (a superfluid). I thought at first the metal would be stopped by the fluid, which turned out not to be true. No momentum is transferred to moving objects, wich is not to say that the laws of Newton don't work in a superfluid. It's because the laws of Newton hold that there is no driving force. It's like a rotating propeller in outer space. – descheleschilder Feb 11 '17 at 13:05