Bubbles in a falling glass of champagne

[Grasshopper]

Suppose you just poured a glass of champagne, then you drop the glass straight down (so that there is no tilting).

Do the bubbles:

(1) Continue to rise with respect to the glass.
(2) Remain in place with respect to the glass.
(3) Sink with respect to the glass.My intuition is telling me that they remain in place because when the glass is in free fall, it should feel weightlessness, and I'm assuming that bubbles "want" to rise in a stationary (with respect to the ground) glass because of a pressure differential. I'm further assuming that in free fall, none of the champagne will be "feeling" gravity, and neither will the bubbles inside. As a result, the bubbles will remain stationary during the fall.

One caveat: I'm assuming if the glass fell long enough the bubbles might come a tiny bit closer together due to tidal forces.

So what would happen? I really don't know anything about fluid dynamics (I'm assuming this is the branch of physics relevant), but by all means, in addition to an answer, please feel free to post any math or physics related to the topic. Thanks for responding!

 

[TeethWhitener]

Relevant:
https://www.nasa.gov/audience/fored...carbon dioxide,foamy mess during space travel.

 

[ishika17]

I agree with you.
I think when the glass is released,the bubbles will remain stationary until the glass hits the floor .due to inertia,the bubbles will remain stationary and will not even rise with respect to the glass.

 

[jbriggs444]

Yes. Buoyancy is just another name for the effects of pressure gradients due to gravity. Cancel gravity (by being in free fall) and you cancel buoyancy.

If the wine were confined in a container on all sides (rather than being loosely held in a glass) then it would be subject to tidal tension in the vertical direction -- the bubbles would congregate in the middle vertically. But it would be in tidal compression horizontally -- the bubbles would migrate toward the edges horizontally. Tidal forces on an object as small as a wine glass and a duration as short as a fall from hand to floor are entirely negligible, so you are not likely to witness any such behavior.

 

[kuruman]

Relevant:
https://www.nasa.gov/audience/foreducators/5-8/features/F_Carbonated_Beverages_Space.html#:~:text=The bubbles of carbon dioxide,foamy mess during space travel.

Follow up question: Suppose an astronaut in orbit ingested a carbonated drink. Would (s)he be able to eructate the CO₂? I think not. Would that create discomfort? Probably.

 

[tech99]

I suggested in a thread some years ago that a bubble can never rise with an acceleration greater than -g, but I was assured this was nonsense.

 

[JT Smith]

Fortunately there is at least one current beer style that is non carbonated. Well, I like that style anyway.

The vertical movement question is easy enough but what else happens with the bubbles? Do more continue to come out of solution? Do they continue to grow in size or stabilize? Do they coalesce?

 

[kuruman]

I suggested in a thread some years ago that a bubble can never rise with an acceleration greater than -g, but I was assured this was nonsense.

I don't even know what "acceleration greater than -g" means. Acceleration is a vector that has magnitude and direction. The acceleration of a free falling object near the Earth's surface is g = 9.80 m/s² and its direction is down. What, in terms of magnitude and direction, would be the acceleration of an object accelerating with an "acceleration greater than -g" ?

 

[tech99]

I don't even know what "acceleration greater than -g" means. Acceleration is a vector that has magnitude and direction. The acceleration of a free falling object near the Earth's surface is g = 9.80 m/s² and its direction is down. What, in terms of magnitude and direction, would be the acceleration of an object accelerating with an "acceleration greater than -g" ?

Well I was thinking of a bubble accelerating upwards. So I thought the accelerating force in the opposite direction to gravity, namely upwards, so force is negative, so acc = f/m is negative.

 

[kuruman]

Well I was thinking of a bubble accelerating upwards. So I thought the accelerating force in the opposite direction to gravity, namely upwards, so force is negative, so acc = f/m is negative.

That works as an explanation of what you meant, but I don't think it explains why a bubble in a free-falling fluid does not rise. To begin with, rising bubbles do not accelerate up with acceleration 9.8 m/s² because of the fluid resistance that cannot be ignored. Most likely bubbles reach terminal velocity before reaching the top. A proper explanation must be based on dynamics and include why buoyancy goes away in free fall as @jbriggs444 pointed out.

 

[kuruman]

If the wine were confined in a container on all sides (rather than being loosely held in a glass) then it would be subject to tidal tension in the vertical direction -- the bubbles would congregate in the middle vertically. But it would be in tidal compression horizontally -- the bubbles would migrate toward the edges horizontally. Tidal forces on an object as small as a wine glass and a duration as short as a fall from hand to floor are entirely negligible, so you are not likely to witness any such behavior.

I don't understand why there would be bubbles in what you describe. There are no bubbles in corked bubbly wine. When you uncork and release the pressure, bubbles are formed. When you put the cork back on tightly, the air space eventually gets pressurized and no more bubbles are formed; we are back to where we started except that the wine has lost some of its CO₂.

What is the difference between a corked wine bottle and wine "confined on all sides"? If it's the air space, that can be eliminated: drill a small hole in the cork, push it all the way down to the fluid level then plug the hole.

 

[jbriggs444]

I don't understand why there would be bubbles in what you describe. There are no bubbles in corked bubbly wine. When you uncork and release the pressure, bubbles are formed. When you put the cork back on tightly, the air space eventually gets pressurized and no more bubbles are formed; we are back to where we started except that the wine has lost some of its CO₂.

What is the difference between a corked wine bottle and wine "confined on all sides"? If it's the air space, that can be eliminated: drill a small hole in the cork, push it all the way down to the fluid level then plug the hole.

I would envision an enclosure with enough head space and low enough pressure to allow for bubbles to form. The container is there to keep the champagne from separating into globs under tidal tension. And to provide a boundary against which a pressure gradient can form. The bubbles are there (at least for a while) because the container is not pressurized high enough to force them back into solution.

 

[tech99]

That works as an explanation of what you meant, but I don't think it explains why a bubble in a free-falling fluid does not rise. To begin with, rising bubbles do not accelerate up with acceleration 9.8 m/s² because of the fluid resistance that cannot be ignored. Most likely bubbles reach terminal velocity before reaching the top. A proper explanation must be based on dynamics and include why buoyancy goes away in free fall as @jbriggs444 pointed out.

I think buoyancy goes away in free fall because the liquid cannot fall fast enough to fill space beneath the bubble, and so cannot exert any upward pressure. For the same reason, a buoyant object in a static liquid, even if totally streamlined, could not accelerate upwards faster than -g. Further, it cannot leap out of the liquid higher than its depth of immersion.

 

[jbriggs444]

I think buoyancy goes away in free fall because the liquid cannot fall fast enough to fill space beneath the bubble, and so cannot exert any upward pressure. For the same reason, a buoyant object in a static liquid, even if totally streamlined, could not accelerate upwards faster than -g. Further, it cannot leap out of the liquid higher than its depth of immersion.

I am pretty sure we can perform an experiment to prove this wrong.

The buoyant force on a bubble has to do with the weight of the water displaced. The mass of a bubble can be negligible. The acceleration would, naively, be near infinite.

However, @kuruman makes the important point that it is not just the bubble that is moving. The water around the bubble also moves. The acceleration of the water need not match the acceleration of the bubble if there is more water than bubble.

This is a complicated problem for which I do not have a solution at hand. My recollection is that an engineering approximation is that for a particular configuration, one can associate an "effective mass" with the bubble to account for the motion of the water.

Edit: Google, however, knows much.

https://journals.sagepub.com/doi/pdf/10.1260/1757-482X.7.3.129#:~:text=The initial acceleration of a,based on potential- flow theory.

The abstract claims initial accelerations of 3.3 g.

A 2.0 g figure which the abstract also mentions can be found here:

https://physics.stackexchange.com/q...-of-an-air-bubble-rising-in-a-liquid-increase:

Incidentally, for laminar flow it is possible to calculate the "inertia" of the spherical bubble. Even though the bubble is filled with gas, and you would expect it to have very low inertia (mass), it will not accelerate very rapidly. The reason is that as the bubble moves up, liquid has to flow down and around the bubble. If the bubble accelerates, the liquid around it must also accelerate. It turns out that there is an exact mathematical solution for the case of laminar flow around a sphere, and that the inertia of the sphere is equal to half the mass of the displaced liquid. And since the weight of the liquid is equal to the force of buoyancy, it follows that the initial acceleration of the bubble is 2g - it "falls up" at twice the acceleration of gravity. But it will reach a steady state velocity pretty quickly. And then, as the bubble gets closer to the surface and expands slightly, there will be a further acceleration (which is normally ignored).

 

[A.T.]

I think buoyancy goes away in free fall because the liquid cannot fall fast enough to fill space beneath the bubble, and so cannot exert any upward pressure. For the same reason, a buoyant object in a static liquid, even if totally streamlined, could not accelerate upwards faster than -g.

I'm not convinced by the above argument:
- The liquid doesn't have to accelerate downwards at same rate the buoyant object accelerates upwards, because the liquid can move in from the sides to fill the gap.
- The liquid is not in free fall, but also accelerated by pressure gradients, so I see no reason why no part of the liquid can ever accelerate more than g in some direction.

Further, it cannot leap out of the liquid higher than its depth of immersion.

Apparently you have never played with a ball in water. It can easily rise above the surface to multiples of the submersion depth at release, if you don't submerge it too deep.

Here a paper on this:
https://journals.aps.org/prfluids/abstract/10.1103/PhysRevFluids.1.074501

 

[etotheipi]

Apparently you have never played with a ball in water. It can easily rise above the surface to multiples of the submersion depth at release, if you don't submerge it too deep.

Yeah, I haven't read the paper yet but I would have thought that doing a back-of-the-envelope calculation, i.e. assuming incompressible, constant density fluid, no resistive/drag forces, then work by surface forces equals the increase in gravitational energy,$$\rho_l g V d = \rho V g(d+h) \implies \rho_l d = \rho(d+h) \implies h = d \left(\frac{\rho_l}{\rho} - 1 \right)$$where $d$ is the initial depth below the surface, and $h$ the maximum height above the surface reached. To rise to its initial submersion depth would require $\rho_l = 2\rho$, whilst rising above the initial submersion depth requires $\rho_l > 2\rho$.

 

[Grasshopper]

Sidenote: All of this ^^ is why I love this forum. Please continue!

 

[A.T.]

Further, it cannot leap out of the liquid higher than its depth of immersion.

Apparently you have never played with a ball in water. It can easily rise above the surface to multiples of the submersion depth at release, if you don't submerge it too deep.

Here a paper on this:
https://journals.aps.org/prfluids/abstract/10.1103/PhysRevFluids.1.074501

If you cannot dowload the paper from there, try here:
https://www.researchgate.net/publication/309628307_Water_exit_dynamics_of_buoyant_spheres

Look at figure 5(a) (data for a ping pong ball): When released from 1 diameter depth (center to surface) it jumps up to ~2.8 diameters above the surface.

 

[Straight6]

Surely the answer is 2. What makes the bubbles rise, buoyancy, the fact that the bubble is at a lower density than its surrounding, and so (when positively buoyant) will rise until it reaches equilibrium. The reason it is rising is because pressure increases with depth due to the force exerted by the mass of water above it, every scuba diver knows this. So, if the glass is dropped there is to pressure gradient in the fluid any more,there is no "up" to rise towards, therefore the bubble cannot move towards equilibrium so it will remain in place, although it may expand in volume minutely as the pressure in the surrounding liquid lowers as there is no longer a force exerted from the mass of liquid above it.

 

[DaveE]

It's not bubbles, but there's a good relevant demo somewhere in the middle of this clip. Water doesn't leak out of a container with holes in the bottom in free fall. Skip to the demo in the middle, the rest is a waste of time IMO.

 

[sophiecentaur]

No one has explicitly mentioned that the bubbles will increase in size and push a lot of the bubbly out of the glass. Judging what happens rapidly when a bottle of soda or champagne is shaken first and then opened, given a bit of time, the liquid would be pretty well all pushed out, displaced by more than its own volume of CO2. The event will form a 'uniform' foam.