Why do the bubbles in a champagne accelerate?

Question

The bubbles in champagne are familiar to most people. They form almost exclusively at the specific points in the champagne glass, and from these they rise faster and faster. Why do the bubbles in champagne accelerate?

Answer 1

As the other answers point out, the force that accelerates the bubble is buoyancy counteracted by drag and the weight of the bubble. The process that leads to perceptible acceleration of the bubble all the way while it rises, however, is that the bubble grows as gas diffuses into it, which increases the terminal velocity.

To show this, let's first solve the problem for a bubble of constant size.

Velocity of a Bubble of Constant Size

The drag for a small sphere at low Reynolds numbers is the Stokes drag $$ F_D = -6\pi \eta r v. $$ The net buoyancy (buoyancy plus the weight of the bubble) is given by $$ F_B = \frac 4 3 \pi r^3 (\rho_l - \rho_g) g. $$ The mass of the bubble is $$ m = \frac 4 3 \pi r^3 \rho_g. $$

The equation governing the dynamics of a bubble of constant diameter is: $$ m\dot v = F_D + F_B $$

The solution to that equation, with the initial condition $v(0) = v_0$, is $$ v(t) = v_\infty + (v_0 - v_\infty) e^{-t/\tau}, $$ with the time constant $\tau$ $$ \tau = \frac{2 r^2 \rho_g}{9 \eta} $$ and the terminal velocity $v_\infty$ $$ v_\infty = \frac{2 r^2 (\rho_l - \rho_g) g}{9 \eta}. $$

Of course this is just an approximation, as a bubble rising in water is not exactly spherical, etc. But it is a sufficient approximation.

Using the viscosity of water and the density of Carbon dioxide at 10 °C we can estimate time constant $\tau$ for a bubble 2 mm in diameter: $$ \rho_g = 2\,\mathrm{\frac{kg}{m^3}} $$ $$ \eta = 1.3\,\mathrm{mPa \cdot s} $$ $$ \tau \approx 0.3\,\mathrm{ms} $$

(Yes I know that viscosity of champagne is different from the one of water, but just need the order of magnitude here. Which is milliseconds.)

This means that the terminal velocity would be reached within a few milliseconds for a 2 mm diameter bubble and much faster for smaller ones, and in turn, that the acceleration you see is not caused by the bubble reaching its terminal velocity over the height of the glass.

What happens for a real bubble in champagne

So the conclusion is that the acceleration you see is because the bubble does not have a constant radius! (Which you can easily observe.)

The effect of the change of hydrostatic pressure is minute, the atmospheric pressure corresponds to about 10 m of water column, so the relative change in hydrostatic pressure, as the bubble rises in a glass (with a water column of perhaps 10 cm) is only 1%.

The relevant process is that gas is diffusing from the champagne to the bubble. There is a net diffusion into the bubble, because champagne is an over-saturated solution of carbon dioxide. The rate of diffusion approximately proportional to the surface area of the bubble, thereby we have $r \propto t$ for the bubble radius $r$. This increased bubble size means an increased terminal velocity and the bubble accelerates to match that. Since the time constant for reaching the terminal velocity is so small, we can say, as a not too bad estimate, that $$ v(t) = v_\infty(r(t)) \propto r(t)^2 \propto t^2. $$

And this matches the observation of the bubble accelerating more and more as it rises.

Answer 2

Basic forces which compete are buoyancy force and body weight. From here we can write net force equation : $$ ma = mg - \rho_{_f} gV $$ where $\rho_{_f}$ is fluid density, $V$ - body volume, or body's expelled liquid volume to be correct.

Solving equation above for body acceleration, gives : $$ a = g\left(1-\frac{\rho_{_f}}{\rho_{_b}}\right) $$ where $\rho_{_b}$ body density.

When $\rho_{_f} \lt\rho_{_b}$, then body goes down with positive acceleration and when $\rho_{_f} \gt \rho_{_b}$ - then body goes up with negative acceleration. $\rho_{_f} = \rho_{_b}$ - body stays at rest. In your case bubbles are filled with $\text{CO}_2$ gases which has relatively low density compared to champagne fluid. So in the parenthesized expression we get something $\lt 0$ and thus negative acceleration with which bubbles goes up.

It's worth to mention that this acceleration is not ever-lasting, because actually there's drag force too which comes into play at high speeds (fluid resistance to movement). So due to drag force at some point in time body should reach terminal speed, over which it can't speed-up anymore because any speedup will be canceled by equivalent increase of drag force. Does CO2 bubbles are able to reach terminal speed until they catches fluid surface - is another question out of scope here.

Answer 3

A constant buoyancy force on a bubble will accelerate it to some extent, but as it speeds up the drag force from the liquid increases, and the bubble will quickly reach an equilibrium speed at which the buoyancy force and drag force are equal. However, there is another factor at play here.

As a bubble rises the pressure in the surrounding liquid decreases, so the bubble expands. A larger bubble displaces a greater mass of liquid so the buoyancy force on it increases. A greater buoyancy force increases the equilibrium speed of the bubble, so it goes faster as it rises.

The effect can be seen in this YouTube video - if you look at the chains of small bubbles you will see the bubbles become larger and more spaced out as they rise.

Answer 4

A bubble rises because there is lower pressure (in this case champagne pressure) above the bubble than below it. As the bubble gets higher, the pressure continues to decrease and this allows for the acceleration. This is exactly the same as with a cork that has been pulled underwater.

What is interesting in this physics forum is that this is completely analogous to the way that gravity causes an object to fall with increasing acceleration. In fact, a race of people living beneath the sea would say that a rock experiences gravity and a cork experiences antigravity. Both behave in exactly the same manner. I theorize that this is because gravity is actually a low pressure system in spacetime.