Of Particular Significance

Chapter 4, Endnote 12


Tides are subtle, and not all explanations found in textbooks and websites are correct. The tides are caused by a double-bulge in the surface of the sea, at the longitudes that face toward and away from the Moon; these lead to twice-daily tides as the Earth rotates. Some websites say that this has to do with the gravitational pull of the Moon and the rotation of the Earth, but the rotation of the Earth has nothing to do with the creation of the bulges. Other websites say that one bulge is created by the Earth’s gravity and the other by some sort of centrifugal force — but centrifugal forces are fictitious and perspective-dependent, so that is not an ideal starting point when one is trying to gain some understanding.

The key point is that if gravity were a constant force that did not decrease with distance, then an Earth-Moon system in orbit, even with a rotating Earth, would have no tidal bulge in its ocean. Only because gravity varies with distance are there ocean tides. And this is most easily seen in an object that is falling down, rather than in orbit.


The role of varying gravity in the double-bulge effect is most easily understood in the context of a falling water balloon. In gravity that does not vary with distance, as is essentially the case on human scales near the Earth’s surface, all objects accelerate downward at the same rate, and that includes all parts of a water balloon. If the balloon is made perfectly spherical and is then released, all of the water in the balloon will fall in exactly the same way, and the balloon will remain spherical (Fig. 1 top left), as though there were no forces acting on it at all (Fig. 1 top right).

This is as required by Einstein’s Principle of Equivalence (discussed in this earlier note,) which states that constant gravity cannot be distinguished from constant acceleration. It implies also that constant gravity with matching acceleration — free fall — cannot be distinguished from being stationary and free of gravity.

If instead gravity varies substantially across the water balloon, then the lower parts of the balloon will accelerate downward faster than than the upper parts, causing the initially spherical balloon to become an oval as it falls (Fig. 1 at bottom left). The change in gravity from top to bottom creates a net tidal force (Fig. 1 at bottom right), which pulls the water into the shape of an ovoid (a three-dimensional oval) which, compared to a sphere, has a double-bulge. (See also Figs. 1 and 3 of the earlier note.)

Figure 1: (Top left) In uniform gravity, the water sphere falls uniformly and remains a sphere, just as it would if (top right) there were no gravity and motion at all. (Bottom left) In gravity that is stronger at lower altitude, the water at the sphere’s top will fall more slowly than that at the bottom, and the sphere will elongate until the balloon skin prevents it from stretching further. The resulting ovoid shape exhibits two bulges compared to a sphere (light blue). This shape is created (bottom right) by a real, not relative, tidal force (green arrows), which stretches the balloon vertically and also causes it to contract horizontally.

If the water were not enclosed in a balloon skin, it would continue to stretch indefinitely. But the balloon provides an additional inward force that keeps the now ovoid shape of the water intact as it falls.


Essentially the same idea would apply if the balloon instead enclosed a spherical wooden ball covered by a thin layer of water. Varying gravity would again pull more strongly on the lower parts of the water and wood compared to the upper parts, tending to cause them to stretch. But there are strong internal forces holding the wood together (after all, wood, unlike water, cannot be poured into a glass) and so the wood resists the stretching, remaining almost perfectly spherical. The watery layer, by contrast, easily flows, and it thereby reshapes itself into an oval shell, thicker on the upper and lower areas of the wooden ball, and thinner around the middle.

Figure 2: Same as Fig. 1, but with a wooden spherical core surrounded by a layer of water. This is somewhat analogous to a rocky planet covered with a watery ocean.

This case is more similar to the Earth; tides act on the Earth’s rock as well as its water, but because rock does not quickly flow and has strong internal forces that help it maintain its shape, it bulges less than the water of the ocean.


Now, to what extent does this example have anything to do with the Moon’s effect on the Earth? Let’s start by addressing two simple points.

First, the water is prevented from sliding away from the wooden sphere by the skin of the balloon. For the Earth, the role of the balloon skin is played by our planet’s gravity, which pulls the water inward almost equally around the Earth and keeps it from drifting away. Because of this, on a spherical non-rotating Earth without a Moon, the oceans would have a constant depth everywhere around the planet, as in Fig. 2, top right. [The rotation of the Earth does affect the oceans, but these effects are centered around the equator, i.e. at zero latitude, unlike the tidal bulges which are at definite longitudes’; thus they have nothing to do with our current focus.]

Second, the Earth isn’t perfectly spherical. But if we replaced our perfectly spherical wooden ball of Fig. 2 with an imperfect one, with its own raised and lowered areas, and even spinning on top of all that, the same types of watery bulges would appear, only now vastly more complicated. Tides on the real Earth are indeed complex and far beyond simple estimates and pictures. Nevertheless, the conceptual basis for the tides is as illustrated in Fig. 2.

More confusing is that the Earth isn’t falling straight down like a balloon in a gravitational field, it is co-rotating, with the Moon, around a point somewhat displaced from the Earth’s center. Yet the Earth’s actual motion is similar, in that its motion is caused only by the effects of gravity — it is, on average, in free fall. Gravity is very special. If gravity’s pull due to the Sun, Moon and everything else were perfectly constant across the Earth, then because gravity is universal, its effects on all parts of the Earth, no matter what they are made of, would be the same. The entire Earth, including its air and oceans, would move around the universe as a coherent unit, and would be neither squeezed nor stretched in any way. This again reflects the principle of equivalence that underlies Einstein’s view of gravity; an object subject to a perfectly constant gravitational field would act as though it is weightless and subject to no forces of any sort. (Einstein might say it is weightless, but we will not split such hairs here.) This is why astronauts in space stations act as though weightless… or rather, almost weightless. And this is why the picture at the top of Fig. 2 would apply to our planet if the gravitational forces it experienced were uniform.

But when gravity is not uniform (and it never really is), then the lack of uniformity ruins the weightlessness, creating net forces of squeezing or stretching — tidal forces — as the different parts of an extended object attempt to fall at different rates or in different directions. We’ve seen this in the simple intuitive setting of a water balloon, but the principle is much more general than that.

The non-constant gravitational effect of the Moon on the Earth causes such tidal forces. So does the Sun’s gravity on the Earth, with an impact slightly smaller than that of the Moon. These forces shift rock, water and air. The rock, with stronger internal forces that maintain its integrity, responds much less than does the water. The air bulges too, but not only is there no easy way for us to observe this, the Sun’s daily heating creates a more important once-a-day bulge in the atmosphere. And so the only effects we easily observe in ordinary life are those of the water’s double bulge — our original clue that the laws that governs


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A decay of a Higgs boson, as reconstructed by the CMS experiment at the LHC