# Distribution of bubble radius in carbonated water (soda, coke)?

After opening a cold sprite into a cup (operation under room temperature), numerous bubbles of various sizes are visible. What is the distribution of the size of these bubbles by radius?

It is obvious that this distribution is affected by temperature, the components of the solution, air pressure (altitude) and the butterflies in Texas. Thus, I am basically asking the family of the distribution rather that a specific distribution.

• What research have you done? What ideas do you have? Apart from listing possible factors. What kind of distribution are you expecting? Maxwell-Boltzmann? Poisson? Gaussian? – sammy gerbil Aug 29 '16 at 21:47
• I haven't done any research on it yet as it is mostly hobby, bit I feel it must follow either gamma or over of three hazard distribution families. – hyiltiz Aug 30 '16 at 23:29
• I was initially sceptical about the motivation for and value of your question, but it seems you have done some thinking about it. Getting a good answer depends on demonstrating that it is a worthwhile question and demonstrating that you have made an effort. So I recommend that you update your question to include your thinking - ie what distribution you think it is and why, what calculations you have made, what you have found from research etc. – sammy gerbil Aug 31 '16 at 12:38

Generally, when the outcome of some 'experiment' depends on many independent (and ideally identically distributed) random factors, the distribution of outcomes tends to be Gaussian. This result is known as the Central Limit Theorem. It is likely to be the case here.

My own observation is that bubble size under the same conditions (eg same depth in the liquid) tends to be quite uniform with very little variation.

On page 2 of The Quasi-Static Growth of CO2 Bubbles it is stated that bubble radius $R(t)$ during nucleation is observed to grow in proportion to $\sqrt{t}$ where $t$ is time. When bubbles reach a certain radius $R_0$ they have enough buoyancy to detach from the container and rise to the surface, continuing to expand as they do so. If we look only at attached bubbles, then $dt\propto RdR$. That is, the amount of time $dt$ which a bubble spends with radius $R$ to $R+dR$ is proportional to $R$. The probability of finding a bubble within $dR$ of radius $R$ is proportional to $dt$, therefore the distribution is expected to be proportional to $R$ from $R=0$ to $R_0$.

In the above chart I have measured the diameters (in pix) of all 59 bubbles in sufficient focus in the previous image, and averaged and plotted frequency against radius. The distribution is very approximately triangular, as predicted.

Raw data : diameters of bubbles in pix
18 18 22 17 16 16 15 6 7 23 17 18 20 23 21 17 19 19 21 9 22 24 22 9 17 21 20 6 20 19 5 28 25 22 18 23 18 22 18 12 9 11 12 24 23 26 23 20 22 14 12 13 22 14 19 22 8 6 11

If fitted with Weibull, then we have the following fit (using the data above).

• Using this proportionality, a simply linear f(R)=kR ∀R≤Ro where k is restricted by the laws of probability and is therefore k=2/R² should model the probability distribution. Intuitively, this doesn't feel right, though. – hyiltiz Sep 2 '16 at 15:53
• If $k=2/R^2$ it cannot be a constant of proportionality. What "intuitive" distribution are you expecting (gamma?), and why? $f(R)\propto R$ does give higher probability for large radius near $R_0$, which is what I seem to observe. – sammy gerbil Sep 2 '16 at 15:57
• I would suggest trying to fit your estimate data using Weibull distribution $f(x) = \frac{k}{\lambda} \left( \frac{x}{\lambda}\right)^{k-1}e^{-\left(\frac{x}{\lambda}\right)^k}, \forall x \ge 0$, which is one of the families in the Extreme Value theory; it makes sense in that a bubble tends to merge nearby bubbles around when formulating and only the largest bubble (local maxima) is observed. From the plots, it seems the shape parameter k has to be much larger than the location parameter. – hyiltiz Sep 3 '16 at 21:16
• @sammy_gebril, Can you please upload your measurements (maybe to Gist?) so I could fit a Weibull distribution? Weibull seems much more plausible than the triangular cutoff model. – hyiltiz Sep 27 '16 at 18:47
• Sorry I don't know how to use Gist. I have added the data to my answer. – sammy gerbil Sep 27 '16 at 22:57