More about tidal forces today (see also yesterday’s post) and the conceptual point underlying Earth’s ocean tides.
- (Quote) Because gravity dwindles at greater distances, the Moon’s pull is stronger on the near side of the Earth and weaker on the far side than it is on the Earth’s center. This uneven pull stretches our planet’s oceans slightly, resulting in a small bulge of water, not much taller than a human, both on the Earth’s side facing the Moon and on the opposite side, too.
- (Endnote) To explain why gravity leads to a water bulge on both sides of the Earth is too complex for a footnote, and I’d rather not repeat the most commonly heard explanations, which are misleading. One can see a hint of the cause as follows: if one drops a water balloon in constant gravity, it will fall as a sphere, whereas if it is pulled more strongly at the bottom than at the top, it will stretch into an oval as it falls.
Here I’ll explain this last observation more carefully.
Mere Gravity Isn’t Enough
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.
a SpherE of Water
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.)
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.
A Water-Covered Ball
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.
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.
The Earth’s Tides
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 the heavens govern the Earth, too.
27 Responses
I’m thinking the solid shell of the earth is being accelerated towards the moon as a more or less rigid body. Its oceans on the moonward side are drawn ahead. Its oceans on the other side lag behind. Is this a valid concept?
It isn’t invalid, but I find that it’s easy to get yourself confused that way. If you run that argument all the way, you find yourself implicitly assuming that the Earth rotates once an orbit [the way the Moon actually does], with the near side traversing a smaller circle than the far side; but that then gives the wrong answer for the tides. A better model is to first assume that it does not rotate at all, so that all parts of the Earth traverse a circle of exactly the same radius. (See for instance here.) Then later, as a last step, add its relatively rapid rotation. This makes it clearer that its rotation plays no role in generating the tides.
I find the description almost as thoroughly confusing as the “gravity pull” descriptions, especially as the comments add the notion of it being a local effect of almost stationary water. The usual rejoinder is that a glass of water has no tides.
For quite some years there has been descriptions that show how tides are an integrative effect of small force differences (as sketched in figure 2) over large bodies of water, with the details more elaborated than Figure 2 shows. I prefer the videos such as “What Physics Teachers Get Wrong About Tides! | Space Time | PBS Digital Studios” but Wikipedia has now a fair model as well:
“The tidal force produced by a massive object (Moon, hereafter) on a small particle located on or in an extensive body (Earth, hereafter) is the vector difference between the gravitational force exerted by the Moon on the particle, and the gravitational force that would be exerted on the particle if it were located at the Earth’s center of mass. … The lunar gravity residual field at the Earth’s surface is known as the tide-generating force. … The equilibrium tide is the idealized tide assuming a landless Earth.[49] It would produce a tidal bulge in the ocean, elongated towards the attracting body (Moon or Sun). It is not caused by the vertical pull nearest or farthest from the body, which is very weak; rather, it is caused by the tangential or tractive tidal force, which is strongest at about 45 degrees from the body, resulting in a horizontal tidal current.”
Your concerns are fair; after all, what is explained in this post is only why the tidal force is the “lunar gravity residual field.” What is not explained is fluid dynamics, which determines how the bulge actually forms. This, however, is another layer of complexity (and one can only fit so much into one post.) I will try to get the more accurate description into a later post.
I had understood that tides were caused by residual gravitational forces tangential to the Earth’s surface. i.e. the water on the left of the diagram is being pulled to the right, and the water on the right is being pulled to the left. Thus the tides are less of a “pull” and more of a “squeeze”. Was I wrong, or is it a different way of looking at the same phenomenon?
Well, it is certainly not a pull, as my diagrams show. The tidal effect is mainly a stretching toward and away from the Moon. It also involves a slight squeezing along the directions perpendicular to the Moon. You don’t see that in the water balloon example because (to make the problem simpler) I assumed gravity pulls in the same direction across the elevator. This is not true of the Moon’s pull across the Earth, which pulls toward the Moon’s center and thus is not always in the same direction as far as the Earth is concerned.
Can Earth’s waters absorb gravitational change energy?
Your question is not entirely clear; but if you are asking, does the flow of water caused by the tides potentially absorb energy from the Earth-Moon system, the answer is yes. A similar but much more dramatic effect afflicts the moon Io of Jupiter; the tides on Io from its giant planetary host are so enormous that the rock of the little moon absorbs quite a bit of energy, heating it enough to drive volcanoes.
Yes. How can this absorption occur at the level of signal exchange? Do we need a theory of quantum gravity to explain it?
Dr.Stassler:
Another great explanation. I have always realized that it was the “gravity gradient” that caused the tides. Isn’t the gravity gradient of planets like Jupiter the cause of internal heating on orbiting moons? Also, the gravity gradient goes, I believe like 1/r^3 so if the moon was much further away from earth, relative to its current orbit, wouldn’t tides be a lot less in magnitude?
I remember a lecture some years back that gravity gradients can also cause elliptical orbits to become more circular…can you elaborate at all on that process?
You’re right about the internal heating of Jupiter’s moons (most notably Io, whose volcanism arises from this source) and you are also right that it falls like inverse-distance-cubed (rather than the Newtonian force’s own inverse-distance-squared law) — which follows from simple calculus by the way. As for tides’ effects on orbits, I’m afraid I’ve never thought hard about this subject, and so at the moment I wouldn’t be able to explain it any more simply than what is on Wikipedia, which looks reasonable: https://en.wikipedia.org/wiki/Tidal_circularization
but why the symmetrical shape? why doesn’t most of the water move to the moon-facing side of the earth, i.e. why is the volume of water ovoid rather than teardrop shaped? your water balloon analogy is helpful but assumes rather than explains the resulting ovoid distribution of water…
Look at the green arrows at the bottom right of Figure 1. It is not gravity’s pull but the *change* in gravity’s pull that causes the tidal effect… which, as the green arrows at bottom right show, is not a pull downward but a stretching upward and downward. [I did not assume anything here; everything here is the result of direct calculations.] For further insight, try yesterday’s post, which explains this point in more detail and perhaps in a way you may find easier to follow.
ah… right, that makes sense. the only remaining thing for me to get my head around is that fact that the earth’s rotation speed must be such that the volume of water has time to constantly rearrange itself into the ‘right’ shape so that your static analysis (i.e. ignoring rotation) is approximately right (and, yes, a great way to understand things intuitively). i’ve read that tsunami’s travel ~800 km/h so a tsunami would take about 12 hours to go 1/4 way around the earth. Since the high tide bulge ‘moves’ 1/4 way around the earth in only 6 hours it must be the case that the rotation speed is actually quite close to being too fast to justify the static analysis you’ve done. but, ex post, it must be slow enough (barely?) to allow it…
This is a very good question about a subtle point.
First, it’s crucial that the water can rearrange itself and move the bulges around very quickly without the water itself moving very far at all. This is similar to other phenomena involving waves. Your example of a tsunami is a good one, as it obviously travels much faster than the water itself does. Another is seen in a crowd “doing the wave” at a stadium — the wave is made of people, of course, but it travels much faster than a person could run. So the water doesn’t have to actually move far, just its rearrangement into bulges does.
The difference between a tsunami and a tide is that the tsunami is created at one moment in one location and then travels freely, while the tide is constantly being pushed/pulled by the Moon. Think about what happens when you drop a rock in a pond; the waves move outward at a certain speed set by the water itself. But compare this with a wave that you could make by pushing your hand rapidly through the water; that wave could move much faster than the ripples from a rock, if you move your hand fast enough, because the speed of the wave is set by the speed of your hand. The latter is closer to what we are dealing with here. The Earth’s rotation and the Moon’s slower-moving tidal effect set the location of the bulge, which is not a freely moving wave.
But it is a good question, because of course the ocean turns with the Earth to a good approximation, and so to understand how the water rearranges itself, while it is coping with the friction of the sea floor, containment by the continents, and the stretching from the Moon, is not simple. I will try to see if I can find a decent reference on this that might address your question more completely.
The rearrangement is demonstrated by tidal races.
Would you care to cite your sources on It is not gravity’s pull but the *change* in gravity’s pull that causes the tidal effect…
Well, if you’re a professional physicist, you just calculate it. That’s the whole point about science, especially theoretical physics; if you understand it, you don’t need to cite sources, since you can directly check it yourself.
However, it’s easy to find sources. Here’s are some examples of other people who’ve also shown how it’s done.
A general introduction is here:
https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/13-7-tidal-forces/
Here’s a more technical calculation:
https://www.sfu.ca/~boal/390lecs/390lec8.pdf
Here’s a book that illustrates the effects without calculation — see page 12, especially figure 1.3:
https://www.worldscientific.com/doi/pdf/10.1142/9789813234956_0001
I could go on and on… that’s the great thing about established science. The debates are over.
First, many thanks for actually citing sources, one gets depressed by all of the facts that experts tell us. I obviously don’t want to negate scientific milestones, I refer only to the annoying tendency of main stream media to make claims from Experts without telling us who they are. I believe that Peer review is the thing here.
That said, other, very simple propositions in Physics seem to be open to suspicion.
Without wishing to be accused of being unnecessarily adversarial…
How do we explain the very odd behavior of light waves in media.
Any fan of Pink Floyd will regard the Prism effect as a lesson in Physics
White light enters the prism and exits in the form of rainbow light, deflected by the slowing down of the light in the prism.
Question 1 Why are the different “wavelengths” shifted in different directions? I mention this to undermine the established answer to…
Question 2 How is it possible for light to “slow down” in the Glass but then resume it’s previous speed outside in the air (where it must be slower than C?)
I’ve seen the (theoretical) answer to 2 but it does not coincide to any reasonable answer to 1.
Wishing you well
MTA
Great questions.
(1) is easy: the speeds of the different wavelength are slightly different in the material (“dispersion”), and this difference in speed directly translates into slightly different angles of refraction [well-enough explained in https://en.wikipedia.org/wiki/Snell%27s_law ]
(2) is trickier. Really good explanations are hard to come by, and in fact I’m planning to try to write a decent one soon — not sure I will succeed, because the material science isn’t trivial. Let me, for now, give you an instructive but incomplete model of the phenomenon that can at least explain how the behavior could be logical.
Suppose that what happened to the light, once in the material, is that each photon has a certain probability of being absorbed by an atom, and then that atom emits a photon along the same direction and with the same wavelength as the one that came in. (That’s less crazy than it sounds: see the last sentence on stimulated emission in https://eng.libretexts.org/Bookshelves/Electrical_Engineering/Electro-Optics/Direct_Energy_(Mitofsky)/07%3A_Lamps%2C_LEDs%2C_and_Lasers/7.01%3A_Absorption%2C_Spontaneous_Emission%2C_Stimulated_Emission ) Now, in this model, every photon travels always at the speed c; but because there is a slight delay between when an atom absorbs a photon and when it emits a photon, the average speed of the light is reduced. To say that again: photon by photon, the light travels at c; but the light beam, with its photons being absorbed and reemitted over and over again, traverses the material at a speed below c. Then, when the beam reaches the edge of the material, the last photons emitted by the atoms enter the vacuum, all traveling at c; and since the absorption and reemission ceases there, the beam now continues on at the speed c.
Again, this is a crude model that does not capture the complexity. But it shows you how it can be logical for the light beam to slow inside the material and instantly return to its initial speed once it reenters the vacuum (or air, since the speed of light in air is only slightly reduced.)
Keep an eye on this channel. I do sincerely hope to give a more accurate answer that would explain why all frequencies of light would slow down and why their speeds would not be equal. But it’s not trivial to both get it straight and make it comprehensible; materials are complicated!!
Many thanks again, I’m aware that my topic is probably not a part of the current discussion, could you point me to another area of your excellent blog or should I continue here as I’d love to discuss the boundary between Newtonian Physics and Quantum Mechanics as these two questions are in those two camps (I believe)
Best regards,
MW
How about here?
Many thanks indeed, now I can tell the wife that her moods can be predicted to six decimal places.
Best regards
The entire planet is falling towards the Moon at an acceleration which decreases with distance. (Yes, the Earth and Moon are in orbit, but the Earth is still accelerated toward the Moon.) That is a free-fall, and if you are orbiting with the Earth and accelerated (by the Moon) at that rate, you will feel no force (from the Moon). (The same is true for the Sun, but we can consider one body at a time).
So, if you are at on a place where the Moon was on the horizon, you’d be at the same distance from the Moon as the center of the Earth and so you’d feel nothing from the Moon.
But, if you are closer to the Moon, say, where the Moon is directly overhead, you are being pulled by the Moon more than the center of the Earth (your acceleration towards the Moon is more than the acceleration of the planet as a whole) and you feel that as a tidal force directed towards the Moon. A liquid ocean would have a bulge there, pointed at the Moon.
However, if you are further away from the Moon, say, at the lunar antipode where the Moon is directly underfoot, you are being pulled by the Moon less than the center of the Earth (your acceleration towards the Moon is less than the acceleration of the planet as a whole) and you feel that as a tidal force directed away from the Moon. A liquid ocean would have a bulge there as well, pointing away from the Moon.
A deeper way to put it is that, yes, you are still being pulled towards the Moon at the lunar antipode, but less than the planet you are sitting on, and so you experience that (relative to that planet) as a force directed away from the Moon.
And, of course, this two bulges per attracting body is exactly what we observe. If you do the math (which is easy) you can show that to linear order (i.e, a few per cent for the Moon) the bulges should have equal heights, again, basically what we observe.
I hope all of this helps.
yes, this is the point that matt made in his first response and i agree with both of you. it’s the difference in acceleration that matters.
in the case of a non-rotating earth, it makes the tidal bulges very intuitive.
also, if the earth rotates slowly enough that the water can ‘equilibrate’ to the the varying lunar gravitational force then the existence of 2 high tides per day is obvious (with bigger high-to-low tide differences at the points which are closest and farthest from the moon and minimal tides, etc.).
conversely, if the earth rotates very quickly then the water wouldn’t have time to rearrange itself into its equilibrium 2 bulge configuration before the earth had rotated enough to change that equilibrium configuration.
the issue is “slowly” or “quickly” relative to what? there’s a 24 hour timescale for rotation and there’s a timescale which i’m having trouble defining which governs how quickly the water in the oceans can react to changing tidal forces. it seems to me that the two timescales happen to be pretty close, which is weird.
to go totally off the rails and make up a story for why the timescales are similar: perhaps the earth’s rotation speed used to be faster and was slowed down over the ages by the friction from the shear forces of dragging the sea beds under the resisting water at the bottom? perhaps we have a 24 hour day now because that’s the highest rotation speed at which we no longer dissipate rotational kinetic energy as heat from having the seabed shearing under the water? probably nonsense, but i wonder if the fossil or geological record allows one to make any inferences about the length of earth’s day in early geological eras? probably not, but it would be cool…
“the issue is “slowly” or “quickly” relative to what? there’s a 24 hour timescale for rotation and there’s a timescale which i’m having trouble defining which governs how quickly the water in the oceans can react to changing tidal forces. it seems to me that the two timescales happen to be pretty close, which is weird.”
I don’t feel I fully understand this either [though I don’t think your final paragraph is anywhere near correct; the long-term future of Earth’s spinning rate and orientation is an area of research, about which I am not well-informed.] There is a known time-difference between the location of the bulge and the Moon being overhead, which is associated with the Earth’s rotation; I do not know how to calculate that time-difference and so I don’t know how it depends on the Earth’s rotation rate. Curious to know if anyone else here does.
i also don’t think the final paragraph is anywhere near correct – hence the “off the rails” and “nonsense” decorators :D. but it’s fun to posit something a little crazy that should be falsifiable