# Does the potential energy of fluid rising on a string change?

Lets say I have a glass of water at rest. Then I go and hang a string above the water (vertically), such as the end of the string is immersed in the water. Over time some of the water is going to migrate upwards and climb the rope. The water level is going to drop slightly, as the rope gets "wet". Does the total potential energy of the water change, and if yes what provides the energy for the water to do work against gravity?

If the length of the string a really long, does the water stop climbing at some point? Is all this a result of surface tension on the water?

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Your question is very much like the problem of the ascent of sap in plants. –  Vijay Murthy Jan 3 '12 at 17:30

tl;dr: yes, the potential energy increases. Surface effects (surface tension leading to capillary action) provide the requisite energy, but the water will eventually stop rising as the weight of the water in the thread balances the surface tension.

Rather than discussing threads, I'm going to do as Lubos suggests and consider a capillary tube---a very small tube up which water will rise. This preserves the essential physics (contact between water and a surface) while simplifying matters enough that we can perform some explicit calculations.

Suppose, for the sake of concreteness, that the glass is a cylinder of radius $r$ and that the water has an initial height (before you put the capillary tube in) of $h$. Then the center of mass of the water is at $h_{cm} = h/2$, and the initial gravitational potential energy is $$U_{grav} = \rho V g h_{cm} = \frac{1}{2}\rho \pi r^2 h^2$$ (where I've written the $\rho$ for the density of water and put the zero of gravitational potential energy at the bottom of the glass).

If, then, you put a cylindrical capillary tube of radius $\delta r \ll r$ into the glass of water, and the water rises to a level $l$ above the new surface of the water in the glass, then we have a volume $$V_{tube}' = \pi \delta r^2 l$$ of water in the tube, and $$V_{glass}' = \pi r^2 h - V_{tube}' = \pi r^2 h \left(1 - \frac{\delta r^2}{r^2} \frac{l}{h}\right)$$ in the glass proper. The surface of the water in the glass is now at $$h' = V_{glass}'/(\pi r^2) = h \left(1 - \frac{\delta r^2}{r^2} \frac{l}{h}\right)$$ so the center of mass of the water in the glass is at $$h_{cm}' = h'/2 = h_{cm}\left(1 - \frac{\delta r^2}{r^2} \frac{l}{h}\right)$$ and the gravitational potential energy of the water in the glass is $$U_{grav, glass}' = \rho gV_{glass}'h_{cm}' = \rho g\pi r^2 h^2/2 \left(1 - \frac{\delta r^2}{r^2}\frac{l}{h}\right)^2$$ To next-to-leading order in $\delta r$ (since $\delta r \ll r$, $\frac{\delta r^2}{r^2}$ is small and $\frac{\delta r^4}{r^4}$ is very tiny indeed, so I can ignore it), this is $$U_{grav, glass}' \approx g\rho \pi r^2 h^2/2 \left(1 - 2\frac{\delta r^2}{r^2}\frac{l}{h}\right) = U_{grav} - g\rho\pi r^2 h^2 \frac{\delta r^2 l}{r^2 h}$$ The height of the center of mass of the water in the capillary tube is $$l/2 + h' = l/2 + h \left(1 - \frac{\delta r^2}{r^2} \frac{l}{h}\right),$$ so the potential energy of that water is $$U_{grav, tube}' = \rho V_{tube}' g (l/2 + h') = \rho g \pi \delta r^2l \left[l/2 + h \left(1 - \frac{\delta r^2}{r^2} \frac{l}{h}\right)\right];$$ to next-to-leading order in $\frac{\delta r}{r}$, this is $$U_{grav, tube}' \approx \rho V_{tube}' g (l/2 + h') = \rho g \pi\delta r^2l(l/2 + h).$$ At last, we can write down the total gravitational potential energy of all the water: $$U_{grav}' \approx U_{grav} + g\rho\pi \delta r^2l^2/2$$

To answer the first part of your question, the gravitational potential energy of the water does change. As you mentioned in your reply to Lubos, the drop in the level of the water in the glass does offset the increase in (gravitational) potential energy, but not totally.

So far, so good: I've been able to make a reasonably straightforward calculation of the gravitational potential energy of the two configurations, and compare them. The concepts are from freshman physics, and the techniques aren't hard to follow.

The thing is, the gravitational potential energy is not the only energy in this situation, and that's where the question gets complicated. There's also an energy associated with every interface---every surface of contact between two different materials.

If you ponder a little while, that statement becomes pretty obvious. Think about a kid blowing a soap-bubble: he has to do some amount of work (call it $\delta W$) by blowing on that little thing shaped like a Quidditch goal in order to increase the surface area (inside and out) of the bubble. We define the surface tension $\gamma$ as $$\gamma = \frac{\delta W}{\delta A}$$ (more or less---in actuality, we have to pass to the derivative: consider very small $\delta W$ and $\delta A$. For a slightly less intuitive example in which the surface tension comes out a bit more nicely, without having to pass to a derivative, see https://en.wikipedia.org/wiki/Surface_tension#Two_definitions). I don't actually understand what causes surface tension; I assume it's inter-molecular forces (e.g. van der Waals or polar forces), but I'd like to see some calculations, or better yet experiments, bearing that out. So that's one part of the answer: the surface tension, and ultimately the forces between the water molecules, provide the work required to draw the water up into the tube.

But that's not the whole story. Naively, even when we take surface tension into account, the minimum energy condition would seem to be a perfectly flat surface. Keeping the level of the water thus at $h$, inside the tube or outside, would certainly minimize the surface area of the water. Nor does saying surface tension!!one!'' really constitute an explanation.

In order to come up with a satisfactory explanation, we have to shift from thinking about energies to thinking about forces. One can think of surface tension as a sort of analogue to pressure: it tells you how much force is exerted across a line of given length drawn along an interface between two substances. Surface tension is to the length of a line along an interface as pressure is to the area of the interface.

This has a surprising result: that at an ideal three-way interface, for example at (an idealized version of) the edge of a droplet sitting on a table or the place where the water meets the capillary tube, the liquid makes a specific angle with the solid. In this case, there are three relevant sets of interfaces, each with its own surface tension: liquid-gas ($\gamma_{LG}$, water and air), liquid-solid ($\gamma_{LS}$, water and the glass of the capillary tube), and gas-solid ($\gamma_{GS}$, air and glass).

If the whole situation is in equilibrium, we can do a force balance: the horizontal component of the force due to the solid-gas surface tension has to equal the sum of those due to solid-liquid and liquid-gas surface tensions. Otherwise, the interface would be moving. Writing $\theta_e$ for the contact angle, the angle between the solid and the liquid measured through the liquid (that is to say, on the liquid side of the interface), and dividing through by the length of the three-way interface, $$\gamma_{GS} = \gamma_{LS} + \gamma_{LG} \cos\theta_e.$$ This is Young's Equation (https://en.wikipedia.org/wiki/Wetting#Simplification_to_planar_geometry.2C_Young.27s_relation); solving for the contact $\theta_e$, $$\theta_e = \cos^{-1}\frac{\gamma_{GS} - \gamma_{LS}}{\gamma_{LG}}$$ This immediately rules out the all-horizontal case, which would require $\theta_e = 0$, or $$\frac{\gamma_{GS} - \gamma_{LS}}{\gamma_{LG}} = \frac{\pi}{2}$$ which is not the case for water.

It also give us a clue as to how to proceed: the vertical component of the force due to surface tension very near the three-way interface between water, air, and tube is $\gamma (2\pi r) \cos \theta_e$; this force must balance the weight of the water in the tube, which is $r (\pi r^2) l \rho$ in the notation from the first part of this answer. At last, we see that surface tension draws the water up to a height $$l = \frac{2\cos\theta_e}{rg \rho}$$

So: yes, the potential energy increases. Surface effects (surface tension leading to capillary action) provide the requisite energy, but the water will eventually stop rising as the weight of the water in the thread balances the surface tension.

There's Wikipedia, of course: look at the articles on surface tension, capillary action, and contact angle. I also used

• Landau, L.D. and E.M. Lifshitz. Fluid Mechanics. 2nd English edition, revised. Trans. J.B. Sykes and W.H. Reid. Course of Theoretical Physics Vol. 6. Oxford: Pergamon Press, 1986. \S 61, pp. 238-244.

Note in particular problems 2 and 3.This book leaves a great deal for the reader to figure out on his own. I'm coming to believe that that is usual in the Course of Theoretical Physics.

Much more explicit is

• Kundu, Pijush K. and Ira M. Cohen. Fluid Mechanics. 2nd edition. San Diego: Academic Press, 2002. pp. 9-12.

from which I drew (with modification) my calculation of the height to which capillary action draws the water. Note in particular example 1. This appears to be an undergrad engineers' textbook.

For more information on wetting---how liquids interact with solid surfaces in general---you might try Wikipedia, of course, but also

Though I've only read a little bit and can't vouch for its quality myself, it appears to be frequently cited, so that's something.

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Nice answer. Thanks for your help. –  ja72 Jan 4 '12 at 22:08

When the potential energy of a water molecule is defined as $h\rho g$, then it obviously increases when the water molecules get higher. The conservation of energy is guaranteed by properly including the surface tension and, even more precisely, cohesion into the equation, indeed:

There is a reduction of the total energy by a term that is proportional to the total area of the surface by which the water is in contact with another material. A proper calculation shows that the cohesion or surface tension may only lift the water molecule to a certain finite weight.