What does the following sentence imply?

"Because of the existence of forces across any line in the surface of a liquid, the surface tends to shrink whenever it gets a chance to do so."

Why does the existence of forces cause tendency to shrink the surface? I understand that the surface molecules have extra energy than the molecules in the interior, so lesser surface area means lesser surface energy(the extra energy), which is thermodynamically favourable. But how does the existence of forces relate to this?


1 Answer 1


Consider a disk shaped element of the air/water interface:

Surface tension

The surface tension, $\gamma$, is a force per unit length i.e. you multiply it by a line length to get the total force acting normal to that line. In this case the force normal to the edge of the disk is $\gamma$ times the circumference:

$$ F = \gamma 2 \pi r $$

Suppose let the disk shrink by reducing the radius by an infinitesimal amount $dr$. The work done (by the surface tension) is:

$$ dW = F dr = \gamma 2 \pi r $$

and if we integrate this from $0$ to $r$ we get the total energy stored in the disk:

$$ E = \int_0^r \gamma 2 \pi r = \gamma \pi r^2 $$

So the energy is proportional to the area of the disk as we expect.

Conversely you can get the force by differentiating the surface energy:

$$ F = \frac{dE}{dr} = \gamma 2 \pi r $$

  • $\begingroup$ This derivation was also present in the book I read. The problem is that the statement I gave above was made before this derivation. Does it follow that, anywhere, the existence of forces(as in this case) means that the system has stored potential energy? $\endgroup$
    – Shubham
    Apr 8, 2014 at 9:10
  • $\begingroup$ Generally speaking, if the system has a potential energy the force will be the differential of the potential energy with length. $\endgroup$ Apr 8, 2014 at 9:11
  • $\begingroup$ That is because of their mathematical definitions, but does the converse follow? $\endgroup$
    – Shubham
    Apr 8, 2014 at 9:21
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    $\begingroup$ You say That is because of their mathematical definitions but this is a misrepresentation. It's because work is force times distance, which is a physical principle rather than a mathematical one. $\endgroup$ Apr 8, 2014 at 9:26
  • 1
    $\begingroup$ @Shubham: Yes, that's true; if there's a force and it's able to move in that direction, then it will (because there's a force) and so energy will be released upon doing so ($w=F\cdot d$). There's only one place that energy could have come from: the system. So, the system energy will have decreased. $\endgroup$ Apr 8, 2014 at 12:11

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