The tides are caused by the gravitational field of the moon and sun. For simplicity, let’s consider a day when the sun and moon are aligned, as seen from the earth (i.e. new moon). The sun rises and sets only once per day. So why do most places get two high tides and two low tides on this day (and every other day)? This is discussed in the following sections.
Sometimes we do get only one tide cycle per day; the reasons for this are discussed in section 7.
People often miss the following distinction: You need to distinguish
This distinction can be explained in a number of ways. Let’s start with the pictorial approach:
inward
force
\ /
.
eee
eeeeeeeee
eeeeeeeeeeeee
(moon) <==== eeeeeeeeeeeee <==
large eeeeeeeeeeeee small
leftward eeeeeeeee leftward
force eee force
.
/ \
inward
force
And, about half a month later:
inward
force
\ /
.
eee
eeeeeeeee
eeeeeeeeeeeee
==> eeeeeeeeeeeee ====> (moon)
small eeeeeeeeeeeee large
rightward eeeeeeeee rightward
force eee force
.
/ \
inward
force
Key idea: A planet subjected to a truly uniform pull would feel no tidal effects whatsoever.
The average pull toward the moon is balanced by the earth’s orbital motion. That’s a solved problem, and is of no further interest when calculating the tides. The only thing that matters for the tides is the deviation from the average. It turns out
Larger-than-average to the right is the same as lower-than-average to the left. Both produce an outward force.
This explains why there is a large twice-a-day (semi-diurnal) component to the tide-producing force.
If you have trouble visualizing a saddle-shaped potential, here’s something that may help:
1) Take a piece of paper. Ordinary 8x11.5 copier paper will do.
2) Fold it in half. Cut it for about 1/2 inch perpendicular to the middle of the fold, as shown by the Xs.
+++++++++++++++++++++X+++++++++++++++++++++ <-- fold
| . X . |
| .X. |
| |
| |
| |
| |
| |
|_________________________________________| <-- free edges
3) Make darts as follows. (These are darts as in sewing, not darts as in throwing.) Fold the paper on each side of the cut, so that a crease runs from the end of the cut to the outside corner of the fold, along the path suggested by the dots in the diagram. Tape down the triangular flaps so they lie flat permanently.
4) Unfold the thing. Fold it in half along the perpendicular axis and make another set of darts.
5) Unfold it again. It will probably fall into a saddle shape of its own accord, but you can help it a little if necessary.
Hold the paper model edge-on, and observe the downward curvature of the nearest edge, and of the bottom of the saddle, and of the far-side edge.
Hold it edge-on the other way, and observe the upward curvature of the near edge, the saddle, and the far edge.
Imagine some particles moving on this surface under the influence of the potential. In one direction, the particles will be squashed toward each other by the concave-upward potential. In the other direction, the particles will be pulled apart by the concave-downward potential.
You know that any function y(x), if it’s not too wild, can be approximated by a polynomial. The same is true in higher dimensions. In this case, we want to represent the potential as a function of two position-variables. (We are forced to omit the third spatial dimension from our model.) The polynomial will have the form
| (1) |
The constant piece is uninteresting. A constant potential no forces.
The planar piece produces a uniform gravitational pull. It holds earth in its orbit around the moon. This is the easy part. You can illustrate this with a flat piece of cardboard. Let a marble roll down the tilted plane, to illustrate the relationship between tilted potential and force.
The saddle-shaped piece produces tides.
By taping together about 6 saddles you can create a model of the funnel-shaped 1/|R| potential.
So far, we have only discussed the field that drives the tides. Now we must ask how the water responds to this driving field. This is a different question entirely.
Imagine a tray containing a thin layer of water. Tilt the tray briefly, then return it to level. The water won’t move very much. But if you tilt it and hold it that way, all the water will move to the low end. So it is with the tides. The lunar and solar gravitational fields effectively tilt the earth’s gravitational potential, changing the local notion of what "horizontal" means.
Let us imagine, temporarily, a planet that is earth-like except that it spins very slowly: slowly compared to the sloshing resonance frequency of all the oceans and bays, and also slowly compared to any viscous relaxation times. On such a planet, the water would come to equilibrium and would map out the potential for us.
However, the slow-spin approximation is a terrible approximation for the real earth. The earth spins quickly compared to oceanic slosh-times.
It is fairly easy to calculate how the water is being driven. Figuring out how a given body of water will respond is a much more complicated task. It depends on
It is fairly common for people to think that there should only be one tide cycle per day. The fallacious argument goes something like this: the water is attracted to the moon, so there should be a bulge on the side facing the moon, and a deficit on the opposite side of the earth.
This is fallacious at least two times over. For one thing, it fails to distinguish between the tide-producing potential and the water’s reaction to that potential, as discussed in section 4.1.
But there is a more fundamental fallacy here. Even if we were dealing with a slowly-rotating planet, on which the water could exactly follow the potential, it would be completely fallacious to think that the water bulges up on the moon-facing side and thins out on the opposite side. This is just completely wrong physics.
It is easy to fall into this trap if you think of the earth as being fixed in space, or if you think that the water responds to the moon’s gravitational "force" while the earth is so massive that it cannot respond. But this is completely wrong physics, too.
In fact, the moon’s gravitational field (like any other gravitational field) is not a "force" field at all. It is a force per unit mass, i.e. an acceleration field. At any given point, the gravitational field imparts an equal acceleration to all objects in that vicinity. Massive objects get a large force, and less-massive objects get a lesser force, so it all works out to a uniform acceleration in each vicinity. This is how gravity always works. This has been known for over 300 years.
To make a non-fallacious version of this argument, we can begin along the following lines: The water on the moon-facing side of the earth is attracted toward the moon more than the earth is. But the water on the far side is attracted less than the earth is. So (to a first approximation) the far-side water gets left behind as the earth accelerates away from it, leaving a bulge on the far side, just as surely as there is a bulge on the near side.
Another thing that sometimes mystifies people is why there is an inward force, a pinch, at all points 90 degrees away from the earth-moon line in figure 1 and figure 2. This will be mysterious indeed if you think only about the magnitude of the g field. But in fact g is a vector. The vector changes in magnitude as you move along the earth-moon axis, but it changes in direction as you move off that line. These changes in direction are perhaps less obvious, but no less important.
Again, it is crucial to understand that an object subjected to a uniform gravitational force would feel no tidal effects whatsoever.
Whereas ordinary gravity is an acceleration, the tides are driven by an acceleration per unit distance. The tide-producing field has different units (different dimensions) than the gravitational field.
Imagine a group of positively-charged particles held together by massless springs. If we suddenly apply a uniform electric field, the whole group will accelerate together. This acceleration will not induce any stress in the springs. (The analogy to gravitation is obvious; I mention electrical fields only because it might be hard to visualize suddenly applying a uniform gravitational field.)
The moon’s contribution to the earth’s gravitational potential has quadrupole symmetry: outward at a pair of antipodal points, and inward on the ring of points 90 degrees away from that pair.
(As discussed below, the quadrupole term is just the lowest term in a series, but the higher-order terms are quite tiny in comparison.)
So we see that this quadrupole term doesn’t even have the same symmetry as the plain old gravitational-pull term. Different symmetry, different dimensions – how different can you get?
The total tide-producing field is a superposition of the moon’s field and the sun’s field. They reinforce each other ∼2 times a month (specifically, at new moon and full moon). This is called a spring tide. They tend to cancel each other when the moon is halfway between new and full. This is called a neap tide. The sun’s tide-producing field is only half as large as the moon’s tide-producing field, so the cancellation is far from perfect.
It is instructive to write out the moon’s gravitational potential φ(X) as a Taylor series.
Let’s write the coordinates of the point of interest as:
| X = R + x (2) |
where R is the vector from the moon to the center of the earth, and x is the vector from the center of the earth to the point of interest, the point where we want to evaluate g(X). Do not expand the Taylor series around g(0), but rather around g(R+0); that is, around x=0, the center of the earth.
The result is:
| (3) |
where we are using the traditional albeit inelegant notation for evaluating derivatives, namely:
| ∇i φ(R) means calculate (d/dxi) φ(R + x) then evaluate it at x=0 (4) |
because we are taking R to be a constant and it wouldn’t make much sense to set X=R before taking the derivative.
We are summing over repeated indices. For instance term (3b) is just (x dot grad φ).
Homework: Show that equation 3 is the correct multi-dimensional Taylor series. Grind it out component by component if you have to.
We know φ(X) is given by the law of gravitation, proportional to 1/|X|, so you can in fact evaluate the RHS of equation 3 without any great brain strain. Homework: Grind it out. This is very good exercise.
OK, enough math; let’s do a little physics.
Why mention the potential? Why not just calculate the gravitational field directly?
We can understand the terms in equation 3 individually:
Term (3a) is a constant contribution to the potential. It is not interesting. We can make it go away by exercise of gauge invariance.
Term (3b) is a linear contribution to the potential. It represents a plane, tangent to the exact lunar potential, tangent at the earth’s location. It represents a steady uniform gravitational pull. It is not very interesting, because it is balanced by the earth’s orbital motion. Another way of saying almost the same thing: We can make this term go away by exercise of Einstein’s principle of relativity: it affects our reference frame AND everything we would like to measure, affecting them all equally.
Term (3c) is the lowest-order interesting term, the lowest-order term that we can’t get rid of. It is what drives the usual tides. It is a saddle-shaped contribution to the potential. It curves downward in both directions (+-) along the axis passing through earth and the moon, representing tidal stretching. It curves upward in both directions (+-) along the two perpendicular axes, representing tidal compression.
This term allows us to understand why small bodies of water have smaller tides: This tide-producing term is a second-order shape. Near the middle of the saddle, things are pretty flat. As you go farther out, things get more interesting.
In elementary physics, we like to use the lowest-order approximation to everything. Well, folks, in this problem there is no linear approximation that is of any use. The lowest-order interesting term is nonlinear.
Term (3d) is there to address a question that has come up a couple of times. Because the moon is far away, this term and all higher-order terms are small. But what would happen if a large asteroid passed close to the earth, so that its distance (|R| value) was only slightly greater than the radius of the earth (typical |x| value)? When in doubt, calculate this term and see if it is big enough to be interesting.
The gravitational field is just the gradient of the potential:
| gi(X) = ∇i φ(X) (5) |
It is also amusing to expand g(X) as a Taylor series of its own. You could write the series for g directly. Equivalently, you could take the gradient of equation 3 term by term – but remember that on the RHS of equation 3, the factors of the form ∇i ∇j φ(R) should be considered just constant tensors, having been evaluated at the constant point R; the gradient acts only on the xi xj factors.
The lowest-order term in the expansion for g is the steady gravitational pull. This is unobservable or at best uninteresting because we (the observers) and our frame of reference (the earth) are both freely falling under the influence of this steady pull. This term scales like
| |x|0 |R|−2 (6) |
The next term, the first-order term, is what drives the usual tides. This term scales like
| |x|1 |R|−3 (7) |
We can keep playing this game. There will be a third term, which we can call the hyper-tide driving term. It scales like
| |x|2 |R|−4 (8) |
This term will be smaller than the previous term by a factor of |x|/|R|. As mentioned above, this is small unless we are considering the case of an asteroid that comes very close to the earth.
There will be yet-higher-order terms, but they will be even smaller.
You may be wondering whether the tidal stretching lowers the density of the earth, or whether the pinch raises the density of the earth. The answer is neither one. It is easy to show that when averaged over the whole earth, the amount of stretching exactly balances the amount of pinching. Tides tend to re-arrange stuff, with no change in density. That is:
| (9) |
The tide-producing piece is the next term in the Taylor series following the steady-pull term. Call it h.
| (10) |
The average outward component of this, averaged over the surface of the earth S is:
| (11) |
where n is the unit outward normal. By the divergence theorem this is can be expressed as an integral over the volume of the earth V, namely:
| (12) |
which is zero because ∇ · ∇ φ = 0 everywhere outside the object (the moon) that creates the potential φ.
Usually, we see a two high tides and two low tides per day. However, under some conditions we see that one of the high tides is higher than the other, and indeed in extreme cases there is just one high tide and one low tide per day.
How can this be?
Again, let’s start by discussing what the gravitational potential looks like. This is pretty simple. In the absence of tides, the earth’s potential is more-or-less spherical. And unchanging.
Now let’s add in the tides. To make things as simple as possible, let’s start with the case where the sun and moon are collinear with the earth, i.e. new moon or full moon. The tides squash the potential into a prolate ellipsoid. The potential bulges up at the point right under the moon and also bulges up at the antipodal point. It "tightens its belt" on the ring of points 90 degrees away.
As mentioned in section 4.1, the actual height of the tide is not a simple function of the size of the tide-producing potential.
The bulge in the potential stays in one place (dictated by the moon and sun) while the earth rotates within it. The water generally doesn’t have enough time to come into equilibrium with the potential.
To make things really messy, the spin axis isn’t aligned with the axis the aforementioned prolate spheroid. That’s a point that people tend to forget. And they forget that not everybody lives on the equator.
To be specific, suppose Moe is standing on Hainan island at noon on midsummer’s day. The sun is directly overhead. The sun is in the constellation Gemini, right near the Taurus border, near the star cluster M35. (Moe can’t confirm that directly; it’s hard to see stars in the daytime.) At the very same instant, Joe is standing at the antipodal point, namely Antofagasta, Chile. It is midnight there, and the full moon is directly overhead, in the constellation Sagittarius.
(Draco) Cat’s
eye
Eltanin o Polaris
* *
95 Her * * 36 Cam
NGC6535 ** / * Menkalinan
pinch/
|/
(Sag) M21 ** Moon bulge-Earth-bulge Sun ** M35 (Gemini)
/|
/pinch
/
Now, the scary thing about figure 4 is that the vertical axis is not the north/south axis. The usual definition of north/south is relative to the earth’s axis, which points to Polaris (roughly). In the diagram, the point directly up the page from the center of the earth is the North Ecliptic Pole (NEP) which is beautifully marked for us by the Cat’s Eye nebula, NGC6543 (an unforgettable number).
So let’s see how things look 12 hours later, when it is midnight at Hainan. The moon will not be directly overhead. Not even close! It will be in Sagittarius, about 45 degrees south of overhead.
The earth has rotated around its tilted axis. Hainan will be directly underneath the double star 95 Herculis.
This will not be associated with the bulge in the potential. It will sit in the more-or-less neutral regime between the bulge and the pinch. So Hainan will get a diurnal component to the potential: large at noon, neutral at midnight. The situation is similar at Antofagasta or any other location at comparable latitudes, north or south. If you find some location where the local geography allows the water to slosh with a resonant frequency near 24 hours, you will have a huge diurnal tide.
This is an example, i.e. a proof by construction that diurnal tides can exist. The general case is, of course, quite a bit more complicated.
In particular, there will always be a semi-diurnal component to the potential, in addition to (and larger than) whatever diurnal component there may be. So the usual case is that there will be two high-tide potentials and two low-tide potentials per day, but one of the high-tide potentials will be higher than the other.
Even that’s not the end of the story. The sun’s place in the sky moves north and south with the seasons. The moon’s orbit is not aligned with the earth’s orbit. And so on. Professional tide forecasters keep track of dozens of different contributions to the tide-producing potential. A list of some of the most-significant components (including period, strength, and conventional name) may be found in reference 1. A tutorial, including a discussion of how these components are used, may be found in reference 2. Some discussion of the physical origin of the largest components may be found in reference 3.
And as mentioned before, the way that the water responds to this driving force is very complex, depending on quirks of geography. Resonances and all that.
Finally, there are many non-tidal effects on the water that are as large or larger than the tidal effects. Weather systems produce wind and low pressure that move water around. Earthquakes can move spectacular amounts of water around.