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When a drop of liquid splits into a number of drops, each drop tries to minimize its area but the overall surface area of drops increases. How does the overall surface area of the drops increase????

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    I just tried to mathematically split a drop of volume "V" into two drops of volume "V/2". When I did, the total surface area remained constant. Do you have an example to go along with your question? – David White May 28 '20 at 17:09
  • Try the derivation of the work done in splitting a bigger drop into n smaller drops in this topic it is given in the introductory paragraph that when a drop of radius R splits into smaller n dropswhere is drop tries to minimise its area but the overall surface area of the drop increases I want to know why does the surface area of the drops increase when each drop tries to minimise its area –  May 28 '20 at 17:18
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    As Monocerotis pointed out, the total area remains the same, but the surface-area/volume ratio increases. – David White May 28 '20 at 17:20
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    @DavidWhite the total surface area remained constant Can you doublecheck? I found that the surface area increases as the cube root of the number of droplets. – G. Smith May 28 '20 at 18:50
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    @G.Smith, I just checked the case of splitting one drop into two while holding the total volume constant. For an initial volume of $\frac{4}{3}\pi r^3$ with an area of $4 \pi r^2$, a drop half that volume has a volume of $\frac{2}{3} \pi r^3$. I took the derivative of that to arrive at an area of $2 \pi r^2$ for the half sized drop. Was that an error on my part? – David White May 28 '20 at 21:45
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    @G.Smith, never mind ... my assumed "quick" method proved to be incorrect. You are right. The area increases as the number of drops increase. – David White May 28 '20 at 22:01
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    @DavidWhite Agreed, but I can’t put my finger on why your derivative method is invalid. – G. Smith May 28 '20 at 22:03
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    @G.Smith, something apparently went wrong with the $r^3$ or $r^2$ term. I'll have to think about this issue for a bit to find out where it went wrong. Thanks for bringing it to my attention. – David White May 28 '20 at 22:08

2 Answers2

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It does not try to minimize it's total surface area,rather it minimizes it's $\frac{surface \\\ area}{volume} $ ratio to attain minimum potential and thus the droplets assume spherical shapes as it is the best way to do so.

Monocerotis
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  • What is the benefit of attaining a minimum potential –  May 28 '20 at 17:20
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    @HenaAnvar That's a different question altogether and is not meant to be answered in the comments section. Possible answer to your question: https://physics.stackexchange.com/questions/113092/why-does-a-system-try-to-minimize-potential-energy –  May 28 '20 at 17:22
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    that begs for a completely new question,but in gist minimum energy means lower internal energy of the constituent particles which means more stability – Monocerotis May 28 '20 at 17:24
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How does the overall surface area of the drops increase????

It’s just geometry.

The area of a sphere with radius $r$ is $A=4\pi r^2$ and the volume is $V=\frac43\pi r^3$, so the relationship between area and volume for one sphere is

$$A_1=4\pi\left(\frac{3V_1}{4\pi}\right)^{2/3}=CV_1^{2/3}$$

where $C$ is a numerical constant.

Now suppose one sphere of volume $V_1$ divides into $N$ spheres each of volume $V_1/N$. Their total area will be

$$A_N=NC\left(\frac{V_1}{N}\right)^{2/3}$$

or

$$A_N=N^{1/3}A_1.$$

Thus the area increases as the cube root of the number of spheres.

G. Smith
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