But some will ask, in regard to the vertical movement, is it only in the descent of the water from P′ to P that work is performed? Water cannot descend from P′ to P, it will be urged, unless the entire column P O underneath descend also. But the column P O descends by means of gravity. Why, then, it will be asked, is not the descent of the column a motive power as real as the descent of the mass of water P′ P?
That neither force nor energy can be derived from the mere descent of the polar column P O is demonstrable thus:—The reason why the column P O descends is because, in consequence of the mass of water P′ P resting on it, its weight is in excess of the equatorial column E Q. But the force with which the column descends is equal, not to the weight of the column, but to the weight of the mass P′ P; consequently as much work would be performed by gravity in the descent of the mass P′ P (the one foot of water) alone as in the descent of the entire column P′ O, 10,000 feet in height. Suppose a ton weight is placed in each scale of a balance: the two scales balance each other. Place a pound weight in one of the scales along with the ton weight and the scale will descend. But it descends, not with the pressure of a ton and a pound, but with the pressure of the pound weight only. In the descent of the scale, say, one foot, gravity can perform only one foot-pound of work. In like manner, in the descent of the polar column, the only work available is the work of the mass P′ P laid on the top of the column. But it must be observed that in the descent of the column from P′ to P, a distance of one foot, each pound of water of the mass P′ P does not perform one foot-pound of work; for the moment that a molecule of water reaches P, it then ceases to perform further work. The molecules at the surface P′ descend one foot before reaching P; the molecules midway between P′ and P descend only half a foot before reaching P, and the molecules at the bottom of the mass are already at P, and therefore cannot perform any work. The mean distance through which the entire mass performs work is therefore half a foot. One foot-pound per pound of water represents in this case the amount of work derived from the vertical movement.
That such is the case is further evident from the following considerations. Before the polar column begins to descend, it is heavier than the equatorial by the weight of one foot of water; but when the column has descended half a foot, the polar column is heavier than the equatorial by the weight of only half a foot of water; and, as the column continues to descend, the force with which it descends continues to diminish, and when it has sunk to P the force is zero. Consequently the mean pressure or weight with which the one foot of water P′ P descended was equal to that of a layer of half a foot of water; in other words, each pound of water, taking the mass as a whole, descended with the pressure or weight of half a pound. But a half pound descending one foot performs half a foot-pound; so that whether we consider the full pressure acting through the mean distance, or the mean pressure acting through the full distance, we get the same result, viz. a half foot-pound as the work of vertical descent.
Now it will be found, as we shall presently see, that if we calculate the mean amount of work performed in descending the slope from the equator to the pole, 3½ foot-pounds per pound of water is the amount. The water at the bottom of the mass P P′ moved, of course, down the full slope E P 4 feet. The water at the top of the mass which descended from E to P′ descended a slope of only 3 feet. The mean descent of the whole mass is therefore 3½ feet. And this gives 3½ foot-pounds as the mean amount of work per pound of water in descending the slope; this, added to the half foot-pound derived from vertical descent, gives 4 foot-pounds as the total amount of work per pound of the mass.
I have in the above reasoning supposed one foot of water accumulated on the polar column before any vertical descent takes place. It is needless to remark that the same conclusion would have been arrived at, viz., that the total amount of work performed is 4 foot-pounds per pound of water, supposing we had considered 2 feet, or 3 feet, or even 4 feet of water to have accumulated on the polar column before vertical motion took place.
I have also, in agreement with Dr. Carpenter’s mode of representing the operation, been considering the two effects, viz., the flowing of the water down the slope and the vertical descent of the polar column as taking place alternately. In nature, however, the two effects take place simultaneously; but it is needless to add that the amount of work performed would be the same whether the effects took place alternately or simultaneously.
I have also represented the level of the ocean at the equator as remaining permanent while the alterations of level were taking place at the pole. But in representing the operation as it would actually take place in nature, we should consider the equatorial column to be lowered as the polar one is being raised. We should, for example, consider the one foot of water P′ P put upon the polar column as so much taken off the equatorial column. But in viewing the problem thus we arrive at exactly the same results as before.
Let P (Fig. 2), as in Fig. 1, be the surface of the ocean at the pole, and E the surface at the equator, there being a slope of 4 feet from E to P. Suppose now a quantity of water, E E′, say, one foot thick, to flow from off the equatorial regions down upon the polar. It will thus lower the level of the equatorial column by one foot, and raise the level of the polar column by the same amount. I may, however, observe that the one foot of water in passing from E to P would have its temperature reduced from 80° to 32°, and this would produce a slight contraction. But as the weight of the mass would not be affected, in order to simplify our reasoning we may leave this contraction out of consideration. Any one can easily satisfy himself that the assumption that E E′ is equal to P′ P does not in any way affect the question at issue—the only effect of the contraction being to increase by an infinitesimal amount the work done in descending the slope, and to diminish by an equally infinitesimal amount the work done in the vertical descent. If, for example, 3 foot-pounds represent the amount of work performed in descending the slope, and one foot-pound the amount performed in the vertical descent, on the supposition that E′ E does not contract in passing to the pole, then 3·0024 foot-pounds will represent the work of the slope, and 0·9976 foot-pounds the work of vertical descent when allowance is made for the contraction. But the total amount of work performed is the same in both cases. Consequently, to simplify our reasoning, we may be allowed to assume P′ P to be equal to E E′.
Fig. 2.
