Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Can a small amount of smoke be dense enough to stay in the air keeping its shape for a minute or so?
Or does it always dissipate quickly?
If not smoke, can anything else stay in the air for a minute while keeping its shape?

share|cite|improve this question
up vote 3 down vote accepted

Can a small amount of smoke be dense enough to stay in the air keeping its shape for a minute or so? Or does it always dissipate quickly?

I read this and think to myself "optimization problem". Firstly, you should know the following, which is the law of diffusion:

$$\frac{\partial \phi}{\partial t} = D\,\frac{\partial^2 \phi}{\partial x^2}$$

For clarity, $\phi$ is a function that represents the distribution of the concentration of the gas. It is a scalar function of 3 variables, which is to say that I could write it as $\phi(\vec{r})$, where $\vec{r}$ represents typical $x,y,z$ coordinates. If have an image in your mind of a cloud that is the shape of a snowman, that can be represented by that function, so can smoke rings or whatever you desire.

The clarity of the shape degrades over time, exactly per the above diffusion equation. Picture blurring an image in Photoshop. That is very similar to the process that happens.

  • Q: Can you reduce the rate at which this blurring happens?
  • A: Yes you can

The rate at which $\phi$ (your snowman) degrades in sharpness comes from the magnitude of $|d\phi/dt|$. This magnitude is proportional to the diffusion coefficient $D$ as well as that other derivative with respect to $dx^2$, but that term is representative of the sharpness itself, so we don't want to reduce that, we would rather reduce $D$. In order to reduce $D$, we need to first talk about mean free path and velocity of the gas molecules. I'll use this source and refer to the mean free path $\lambda$ (units of length) and average speed $\bar{c}$. In general D is proportional to those two.

$$D \propto \lambda \bar{c}$$

For a gas cloud the parameter $\lambda$ has mostly to do with the density of the gas, as well as some other things. Again, we would like to minimize $D$, but $\lambda$ might not have much design freedom. On the other hand, $\bar{c}$ could have great design freedom. This is also dependent on the temperature of the gas, but more specifically, the temperature is a measure of the kinetic energy of the molecules. I'll say fairly generally:

$$\frac{1}{2} m \bar{c}^2 = \frac{3}{2} k T$$

Never mind very much what $k$ is (it's just a physical constant), what matters is that this equation has temperature $T$ and $m$. I'm taking your question to be most likely concerned with normal air. That means that it is unlikely that we would have $T$ as a design variable. However, since you are not specifying the gas you are working with, it's possible we could choose that, and the selection of that gas determines $m$ which is a factor in determining $\bar{c}$ which is a factor in determining $D$, which determines the persistence of your cloud image.

Bottom line: Heavier gases will diffuse more slowly, meaning the image will persist longer.

An example of a high molecular weight gas is common refrigerant gases, like R-134a. If you released that into the air it will diffuse rather slowly compared to other examples. NOTE: don't do this, it would be dangerous and probably illegal.

share|cite|improve this answer
very good answer and very concise and easy to follow. But now the question becomes about feasibility. What gas/fog/smoke is legal to do such experiment with while alleviating the diffusion as much as possible? – BeemerGuy Aug 30 '11 at 17:57
@BeemerGuy Don't be so optimistic! Even the heavy gases like R-134a are only heavier than light gases by a factor of around 100. That means $\bar{c}$, as well as $D$, will only be smaller by a factor of 10. Plus, that diffusion equation is only valid in the absence of a density difference. The gas will fall quickly, eliminating the shape. Instead, you should have a transparent heavy gas (possible? I dunno) in a closed container, then introduce another gas of similar mass that is easily visible in your desired shape. Minimize any bulk flow and it might stick around... a little bit? – Alan Rominger Aug 30 '11 at 18:11
so the short answer to my original question is: it will never act the way I need it to (the stillness rather than just maintaining shape). Correct? – BeemerGuy Aug 30 '11 at 18:27
@Beemer you asked "does it always dissipate quickly?" but the shape will always be diffusing from the moment it is released unless there is a barrier. It just won't be very much very quickly. Obviously ordinary shapes in smoke have some amount of permanence, and I'm just saying you could create something that does a little better job at that. I can't define "stillness" well, all this time I'm assuming there is no wind. It remains but it gets fuzzier. – Alan Rominger Aug 30 '11 at 18:37

Quick answer: I believe due to Brownian Motion, a small cloud of smoke particles would dissipate in fairly short order. Indeed, smoke particles are used in experiments to confirm the particulate nature of Air.

However, if you were to wrap that smoke in a soap bubble, the surface tension could hold for longer than a minute…

share|cite|improve this answer

Would, say, clouds or fog count? In this case, the temperature differences between pockets of the air leads to accumulation of water vapor. Off the top of my head, it might be possible to imagine a Van der Waals stabilized gas, or a heterogenous electrostatically stabilized 'smoke'.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.