# Change in appearance of liquid drop due to gravity

A liquid drop is spherical in shape due to surface tension. But why does it appear as a vertical line under the free-fall due to gravity? (E.g. During a rain - falling raindrop) Is there a specified length for the line or does it vary with the size of drops?

• Your original question was better--- they appear long because they are moving fast, and so streak in your eye. Commented Aug 9, 2012 at 8:32
• I'm not sure I exactly understand your question but maybe that can be interesting ems.psu.edu/~fraser/Bad/BadRain.html Commented Nov 2, 2012 at 14:51

A drop that is free falling in vacuum is spherical. This is because free falling in a gravitational field is the same thing as being at rest with no gravitational field present: the gravitational field and the acceleration cancel each other out.

Rain drops falling to the earth can have various shapes depending on their size, although I am not aware that they can become elongated (can you provide a source?). These shapes are due to the air flowing past them, in a rather intuitive way: roughly, air flow causes the bottom to become flatter.

Edit: Regarding the appearance of raindrops (as opposed to their physical shape), consider taking a photograph of falling rain. The camera integrates the incoming light over the exposure time $$t$$, during which the drop travels a distance of $$v t$$, where $$v$$ is the velocity. If we are close to the ground this will be the terminal velocity, which is about $$2_{m/s}$$. If we use an exposure time of $$t=1/60_s$$ (say we are using a flash), the drop will trace a line of length $$\sim 3_{cm}$$. The apparent line on the photograph then has to account for distance from the drop, etc.

• They don't become flat, they become elongated, OP is asking why. Commented Aug 9, 2012 at 3:53
• Their shape depends on their size. Small drops are spherical, larger ones look sort of like pancakes (this is what I mean by 'flat', perhaps this was not clear), yet larger ones look like parachutes. I don't know at which size (if any) they become elongated, but the shape is due to the flow of air and not gravity. I will try to clarify... Commented Aug 9, 2012 at 4:38
• I mistook a drawing for a photo! You are right--- raindrops are flat on the bottom. I assumed this would be unstable because the raindrop weight would be greater at the center. I found this paper: weather.about.com/od/cloudsandprecipitation/a/rainburgers.htm (although the wikipedia link you provide is fine). Commented Aug 9, 2012 at 5:06
• @CrazyBuddy then that's the question about optical illusions Commented Aug 9, 2012 at 7:53
• @CrazyBuddy What confuses me at least is that you call them 'horizontal' lines -- I never saw a raindrop that looks like that. That is why I thought you meant this flattening. Do you mean vertical lines? Commented Aug 9, 2012 at 13:29

It is not an answer but an unfinished draft of how to get the shape :

Let $$y(x)$$ be the profile function of this axisymmetric shape. Supposing a laminar regime friction implies the total counterforce to weight is : $$F=2\pi av\int_0^L\frac{y'y}{\sqrt{1+y'^2}}dx$$ because the friction force is along the normal, while the projection along the axis is needed.

This is a variational problem, the minimum of F is seeked under the constraint of constant volume, given by : $$V=\pi\int_0^L y^2dx$$ and $$y(0)=y(L)=0$$.

Using Lagrange multipliers we get the functional : $$I[y]=\int_0^L \frac{\pi a v y y'}{\sqrt{1+y'^2}}-\lambda( \frac{V}{L}-\pi y^2)dx$$. The Euler-Lagrange equation permits to express y as a function of the multiplier and the derivatives, but I stopped there because it becomes a bit tedious and it is late here.

Then should follow to derivate towards $$\lambda$$ getting a final ODE for y, nevertheless it is nonlinear and of high order.

The latter should imply in particular that there are families of shapes, thus maybe there is no agreement on it.