Let P (fig. 1) be the surface of the ocean at the pole, and E the surface at the equator; P O a column of water at the pole, and E Q a column at the equator. The two columns are of equal weight, and balance each other; but as the polar water is colder, and consequently denser than the equatorial, the polar column is shorter than the equatorial, the difference in the length of the two columns being 4 feet. The surface of the ocean at the equator E is 4 feet higher than the surface of the ocean at the pole P; there is therefore a slope of 4 feet from E to P. The molecules of water at E tend to flow down this slope towards P. The amount of work performed by gravity in the descent of a pound of water down this slope from E to P is therefore 4 foot-pounds. But of course there can be no permanent circulation while the full slope remains. In order to have circulation the polar column must be heavier than the equatorial. But any addition to the weight of the polar column is at the expense of the slope. In proportion as the weight of the polar column increases the less becomes the slope. This, however, makes no difference in the amount of work performed by gravity. Suppose now that water has flowed down till an addition of one foot of water is made to the polar column, and the difference of level, of course, diminished by one foot. The surface of the ocean in this case will now be represented by the dotted line P′ E, and the slope reduced from 4 feet to 3 feet. Let us then suppose a pound of water to leave E and flow down to P′; 3 foot-pounds will be the amount of work performed. The polar column being now too heavy by the extent of the mass of water P′ P one foot thick, its extra pressure causes a mass of water equal to P′ P to flow off laterally from the bottom of the column. The column therefore sinks down one foot till P′ reaches P. Now the pound of water in this vertical descent from P′ to P has one foot-pound of work performed on it by gravity; this added to the 3 foot-pounds derived from the slope, gives a total of 4 foot-pounds in passing from E to P′ and then from P′ to P. This is the same amount of work that would have been performed had it descended directly from E to P. In like manner it can be proved that 4 foot-pounds is the amount of work performed in the descent of every pound of water of the mass P′ P. The first pound which left E flowed down the slope directly to P, and performed 4 foot-pounds of work. The last pound flowed down the slope E P′, and performed only 3 foot-pounds; but in descending from P′ to P it performed the other one foot-pound. A pound leaving at a period exactly intermediate between the two flowed down 3½ feet of slope and descended vertically half a foot. Whatever path a pound of water might take, by the time that it reached P, 4 foot-pounds of work would be performed. But no further work can be performed after it reaches P.