The Beautiful Result

I have just worked out a formula in celestial mechanics which is so beautiful, I’ve been stunned by it for a week.

It isn’t just the unexpected simplicity of the formula. It’s the fact that it’s one in the eye for the scientists. They are always contrasting how things look (to us) with how things really are (in the scientists’ eyes), and of course the idea is that Reality trumps Appearance every time.

Except now.

The beautiful one-in-the-eye-for-the-scientists formula says that the height of the tides depends on two things and two things only: what the Moon is made of (green cheese would make smaller tides than rock) and how big it looks when we look up at it. What about how big the Moon really is? And how far away it really is? Neither of these things matters in the slightest. Reality is irrelevant: it is Appearance that counts.

Beauty should be shared; and a beautiful result must have a proof that lives up to it. To me, this means a proof that is as pure and limpid and natural as the things it talks about, not line after line of algebra. Algebra does have beauty in some people’s eyes, but only because the beauty of what is underneath the algebra shines through the symbols. I’d rather try and communicate the beauty directly.

Here is my attempt.

The height of the tides

We all learn at school that the Moon makes the tides. This is true.

The Moon makes the tides by pulling the ocean up into a bulge beneath it, and, more obscurely, another bulge happens on the other side of the world, at the point furthest from the Moon. As the Moon goes round the sky, the bulges follow it. The Moon takes just under 25 hours to make a complete circuit, so the bulges have to travel at a thousand miles an hour or so, and one of them goes past us every 12½ hours. Broadly speaking, when a bulge goes past, it’s high tide, and half way between the bulges it’s low tide. I say ‘broadly speaking’ because the Earth isn’t covered by a pure, uniform, unending ocean. The seas have deeps and shallows and at the edge of them there is the land. Instead of a smoothly moving bulge there are sloshes and swirls. It’s like carrying a large shallow plate of soup.

The Sun is up there too, and the Sun makes tides as well, but they are weaker ones because it’s further away [so they tell us; but, as it happens, this is completely false]. The Sun goes round every 24 hours rather than 25 or so, and solar tides happen every 12 hours. When they coincide with the lunar ones we get an extra high tide: the spring tides. This happens twice a month, at full moon and new moon.

And that is about all there is to be said about tides, at school. One can’t spend long on them because there is so much more of the world to learn about. But it occurred to me recently to try calculating from first principles how high the tides ought to be.

What might the height of the tides depend on? You’d expect it to depend on how hard the Moon pulls the waters, which would mean the mass of the Moon (big Moon, big pull) and its distance from us (far Moon, weak pull). Equivalently, you could say that it depends on the volume of the Moon and its density and its distance (mass is volume times density) – or, come to that, the diameter of the Moon and its density and its distance (the volume is proportional to the cube of the diameter). And yes, it is all true. Whichever of these groups of three measurements one looks at, the height of the tides depends on all three of them.

Let’s do one more mutation. Think of the apparent size of the moon in the sky – how many degrees wide it is (it’s about half a degree, actually). The further away a big thing is, the smaller it looks. The further away a thing of a given apparent size is, the bigger it has to be (real diameter is apparent diameter times distance).
So yet another possible bunch of factors is the apparent size of the Moon, its density, and its distance from the Earth. On the face of it, this is just yet another reshuffle of the parameters and it doesn’t accomplish much. But in fact this is the moment when something special happens. It’s the reason why I’m writing this at all. Yes, the height of the tides depends on the apparent size of the Moon (big Moon, big tides). Yes, the height of the tides depends on the density of the Moon (heavy Moon, big tides). But – no, the height of the tides does not depend on the distance of the Moon at all.

(This also has exciting implications for the Sun, but I’ll come to those later).

At first sight it sounds absurd to say that the height of the tides doesn’t depend on the distance, but if you think about it, it isn’t absurd. If the Moon were twice the distance from us that it is, but was still half a degree wide when we looked up at it in the sky, it would have to be twice the physical size that it is now – and as we know, big moon, big tides. On the other hand, it would also be twice as far away – and as we know, far moon, small tides. There is nothing inherently absurd in supposing that these two effects could cancel each other out exactly. It remains to be proved that they do cancel each other out and that distance has no effect.

I could prove this by chucking formulae about and using some algebra, but one bit of algebra really looks much like another. I believe in simple results having limpid explanations, so I’ll do it in words and try to convince you without using a single line of mathematics.

How tides happen

First of all we need to revisit the school explanation of how tides happen. It’s not that what we were taught at school is wrong as such, but it’s over-simplified and consequently deceptive.

Tides don’t happen purely because the Moon pulls on the waters beneath it. Think about it. Suppose that the Moon pulled on the waters beneath it, and on the Earth, and on the waters on the other side of the Earth, and it pulled on them all equally. With an equal pull everywhere, there would be nothing to affect the relative conformation of the seas and the Earth. There would be no tides. Since we know that there are tides, something must be making the pull different in different places. What is it?

The important missing point is that according to Newton’s law of universal gravitation the pull of the Moon gets less with distance. So the parts of the globe directly beneath the Moon, which are closest to it, get pulled towards it most. A point at the centre of the Earth will be further away and feel the Moon’s pull less; and the result of ‘greater pull minus lesser pull’ is a stretching force, pulling the surface of the Earth away from its centre. The Earth itself is rigid (or as near rigid as makes no difference), so nothing much happens to the rocks of the sea bed; but the sea itself, being made of water, is not rigid at all, and so the waters pile up.

At the antipodes it looks as if oughtn’t to work the same way, but it does really. The centre of the Earth is pulled towards the Moon as usual, while the surface (including any oceans) is pulled less. So again everything is being stretched apart, so again there is a bulge. Again the waters pile up. This is why there are two tidal bulges, and so two tides per 25 hours and not just one.

(That antipodal bulge trips a lot of people up. It tripped up our geography master at school when he was teaching us tides, to everybody’s delight. If it trips you up, to your distress, don’t worry, you’re not alone. I’m not going to say anything more about the antipodes in any case, because it’s tedious to have to tell each step of the story twice, once about the near-side bulge and once about the far-side one.)

The bottom line is that the tidal force doesn’t depend on the absolute strength of the Moon’s gravity but on how that gravity changes over distance – specifically, over the distance between the parts of the Earth nearest the Moon and the centre of the Earth. This revision of the school picture of things will make all the difference to the argument.

Apparent size

Apparent size counts. If we had a Moon made of the same stuff as ours and at the same distance as ours, the only way of increasing its apparent size would be to make it genuinely bigger. The bigger the Moon is, the more massive it is, and the harder its gravity pulls. Everything about its gravitational pull will be bigger in proportion, including the amount by which the pull varies between the nearest parts of the Earth and the centre. Since that is what makes the tides, the conclusion is: big Moon, big tides.

Density

If the Moon were made of gold instead of rock, it would be more massive. We already know that this means higher tides. So it’s exactly the same story: dense Moon, big tides.

Distance

Now for the core of the argument, where I’m claiming that distance in itself makes no difference at all.

Let’s change the distance of the Moon and change nothing else. To keep the sums simple, let’s move the Moon out to twice its present distance from us but let’s keep its apparent size (in degrees) unchanged, and let’s not change what it is made of.

To make this happen, we have to make the new Moon twice the physical diameter of the old. By straightforward geometry, that means that the new Moon will have eight times the volume of the old Moon – which, since the density hasn’t changed, means that the new Moon will be eight times as massive as the old.

So we have eight times the gravitational pull. If I’m to prove that it all makes no difference, I have to get rid of that factor of eight somehow.

Newton’s law of gravity is an inverse square law. The further away an object is, the less its gravitational pull becomes, and this depends not on the distance but on the square of the distance. So an object twice as far away pulls only a quarter as hard. If the new Moon, eight times as massive as the old, were in the same place where the old one used to be, its pull on us would be eight times the old Moon’s pull and the tides would be eight times as high. But the new Moon is twice as far away as the old, so it pulls only a quarter as hard: ¼ × 8 = 2 times the pull of the old Moon.

A factor of 2 is better than a factor of 8, but to be able to say ‘distance doesn’t matter’ I need a factor of 1. So let’s carry on.

We have a lunar gravitational pull that declines gently with increasing distance. What matters for the tides is not the absolute strength of the pull but how much it declines as it crosses the Earth: how much the Moon’s pull differs between the Earth’s surface and its centre.

First, consider the way things are now. The distance from the Earth to the Moon is about 60 times the radius of the Earth. So if we call the distance from the Moon to the nearest point of the Earth ‘1’, the distance from the Moon to the centre of the Earth will be ‘11/60’. Because gravity follows an inverse square law, this means that the Moon’s pull will get weaker by about 1/30 of its original value.

Now look at what happens when we have the Moon twice as far away: a distance of 120 times the radius of the Earth. Now the relative distance from the Moon to the centre of the Earth is not ‘11/60’ but ‘11/120’. That difference of 1/120 represents the space available for the Moon’s gravity to decline in, and it is half what it was with the old Moon. Over that distance the Moon’s pull will get weaker by only about 1/60 of its original value, not 1/30. It will decline, proportionately, half as much.

Thus although the absolute value of the Moon’s gravitational pull in this ‘new, bigger Moon’ picture is twice what it was, ‘new, bigger Moon’ means ‘new, more distant Moon’ and the relative amount of change in its pull from the surface of the Earth to the centre is half what it was. So with the new, bigger Moon, the stretching force that makes the tides is 2 × ½ = 1 times what it was before we pushed the Moon away and so made it bigger.

To summarize, doubling the distance of the Moon does this:

Distance from the Earth to the Moon: × 2.
Diameter of the Moon: × 2 (to keep the same apparent size).
Volume of the Moon: × 8 (geometry).
Moon’s mass: × 8 (no change in density).
Moon’s gravity when it reaches us: × 8 ÷ 2 ÷ 2. That is, × 2 (Newton’s law of gravitation).
Size of the Earth relative to the Earth-Moon distance: ÷ 2 (size didn’t change, distance did).
Change in Moon’s gravity when crossing the Earth: × 2 ÷ 2. That is, × 1.
Height of the tides: × 1.

The Sun

And the Sun? Ah yes, the Sun. The Sun raises smaller tides: but why?

Despite what we remember being told at school, the weakness of the solar tides can’t be because the Sun is far away, because we’ve just proved that distance doesn’t make any difference. The weakness of the solar tides can’t be because the Sun is smaller in the sky than the Moon, because it isn’t, it’s the same size (a mysterious and interesting fact). So the only remaining difference is density. If the Sun were made of moon rock, its tides would be as strong as the Moon’s. They aren’t, so the Sun must be less dense than the Moon.

One could imagine a mad experiment of measuring the Sun’s density without ever looking up at it, by measuring the difference in height between solar and lunar tides or, what is equivalent, by measuring the difference between spring and neap tides. It’s one of those things that is all right in principle but nearly impossible in practice. The trouble is that with all the swirling and sloshing that goes on, an unambiguous measurement of the ‘real’ height of the tidal bulge is hard to get.

Unless one tamed it. Take an Olympic-sized swimming pool. The tides in the pool would only be a few microns high, but with lasers and sensitive measuring cameras it should be measurable. There will be many influences at the micron level – tiny changes in temperature, cars driving past outside, people talking, even – but they could be averaged out. It would make a nice challenge for a student in need of a dissertation.

The distant future

The tidal bulge races across the sea from east to west, following the Moon. But it always lags a little bit behind because friction (especially against the beds of shallow seas) holds the sea back as it tries to follow the Moon. So the tidal bulge is never exactly beneath the Moon but always a bit to one side. The result is that the Earth (with the sea) is a little lop-sided viewed from the Moon, which accordingly – following the laws of gravitation – pulls on one side of it more than it does on the other side.

The resultant asymmetrical force acts to pull the Earth ‘backwards’ in its rotation and to pull the Moon ‘forwards’ in its orbit.

As far as the Earth is concerned, slowing down its rotation makes the day longer. Six hundred million years ago a day was 1/400 of a year (the days have been counted by looking at daily and annual variations in the shells of molluscs). Today a day is 1/365.2425 of a year and it gets longer by a couple of thousandths of a second per century.* In a couple of million years’ time the day will have lengthened to 1/365 of a year and there will be no more February 29.

As for the Moon, the lopsided tidal bulge pulls the Moon forward and, by making it go faster, swings it further out from the Earth by about an inch and a half a year. The further out the Moon is, the smaller it looks. In time the tides will be a little less, and one day (in a thousand million years or so) the Moon will be too small to cause total eclipses of the Sun.

The mathematical answer

I promised no mathematics. Here are mathematics.

I have two excuses. First, verbal argument, though persuasive, can be seductive, and it is a good idea to check the result with mathematical equations, which can’t be seduced by fine words. Secondly, even if we know that the height of the tides depends only on the density of the Moon (or Sun or whatever is raising the tides) and on its apparent size in the sky, it would be good to know exactly how it depends on these two things.

The third excuse is that the mathematical result is indecently beautiful. Here it is:

The tidal range (high tide to low tide) is ρθ3 times the radius of the Earth, where ρ is the density of the Moon relative to the Earth and θ is the angular radius of the Moon as seen from the Earth, in radians.

And now for the proof. We assume a perfectly rigid and uniform Earth, inertialess and perfectly fluid oceans, and all the other idealisations that are common in calculations of this kind.

Let’s define some symbols:
Moon’s distance = d.
Moon’s apparent radius in the sky = θ radians.
Moon’s density = ρM.
Earth’s radius = R.
Earth’s density = ρE.
Moon’s density relative to the Earth = ρ = ρE / ρM.
Newton’s gravitational constant = G.

It follows that:

Moon’s actual radius = d tan θ, which for small θ is practically equal to dθ.
Moon’s mass = mM = 4/3πρMd3θ3.
Earth’s mass = mE = 4/3πρER3.
Force of the Earth’s gravity at the Earth’s surface = g
And Newton’s law of gravitation tells us that g = GmER−2 = 4/3GπρER.

At a point on the Earth beneath the Moon the distance to the moon is d−R, so Newton’s law of gravitation says that the pull of the moon is GmM/(d−R)2. At the centre of the Earth the pull of the moon is GmM/d2.

Thus the net gravitational effect of the Moon as far as things on Earth are concerned, the difference in the lunar pull between the centre of the Earth and the surface nearest the Moon, is GmM/(d−R)2 − GmM/d2.

Since d is large in comparison to R, this is practically equal to 2GmMR/d3.

Since mM = 4/3πρMd3θ3, the difference in pull is 8/3GπρMθ3R.

That is an ugly formula, and it isn’t interesting in itself. What is going to matter is not the absolute value of the difference but how big it is in comparison to the Earth’s own gravity g. Let’s say, therefore, that the difference in pull is δ times g.

What, then, is the value of δ?

δ is 8/3GπρMθ3R divided by 4/3GπρER. Cancelling out, this gives us δ = 2ρθ3.

To summarize: the Earth’s surface gravity without any lunar influence is g. Right below the Moon, the net surface gravity is (1−δ)g.

Now we know how the effective gravity felt at the Earth’s surface varies depending on where the Moon is in the sky – (1−δ)g under the moon and at the antipodes, g half way between. That is: (1−δ)g at high tide and g at low tide.

This is nice, but how does the gravity affect the height of the tides? In an earlier draft I thought I had the answer but I was wrong by a factor of 2. So if I take small steps now, it is in order to be sure.

Potential energy is the energy a thing has by virtue of where it is in a gravitational field. If I pick up a brick in the street and carry it up to my roof, it acquires potential energy by having been carried higher up, and the increase in energy is equal to the amount of work I put in to carrying it up there. If I throw the brick off the roof and it falls to the ground, some of the potential energy is converted to kinetic energy: energy of motion.

The application of all this in this case is that a liquid in a gravitational field will settle so that the potential energy at every point of its surface is the same. (Because if not, a drop of water at a point of higher potential energy would be able to flow to a point of lower potential energy).

The question ‘How high are the tides?’ is thus the same as the question ‘Given a gravitational field varying as we’ve described, what shape is the equipotential surface?’.

Suppose that instead of a brick I carry a drop of water, and instead of starting in the street I start at the centre of the Earth. As I climb up to the surface, how much force do I have to exert over each inch, and what is the total potential energy I have given to the water drop as a result? The answer isn’t as simple as you think, because the gravity will vary as I climb. At the centre of the Earth I and the water will weigh nothing because weight pulls things down and at the centre of the Earth there is no down.

To get back to the algebra, if I carry a drop of water up from the centre of the Earth to its surface (and lunar gravity isn’t interfering), the potential energy I’ve given it is ½gR. (This is half what it would have been if gravity had been constant all the way up).

So if we assume the height of the ocean at low tide to be exactly R from the centre of the Earth, the potential energy of the surface of the water will be ½gR.

As for high tide, exactly the same calculation applies but with different figures. This time the force of gravity is weakened by the stretching effect of the Moon’s pull, so it varies from 0 to (1−δ)g as we climb. This means that the potential energy at a distance R from the centre (the height of low tide) is ½(1−δ)gR. This is not enough. To match the potential energy of the rest of the sea, I have to carry my water higher until the potential energy reaches ½gR. That is, I need ½δgR of extra potential energy, which, since the gravity up here is (1−δ)g, can be obtained by climbing an extra distance of ½δR/(1−δ). Since δ is so small, 1−δ is practically 1, so the extra distance to be climbed is ½δR.

Remembering that δ is shorthand for 2ρθ3, here is the Beautiful Result:

The tidal range (high tide to low tide) is ρθ3 times the radius of the Earth, where ρ is the density of the Moon relative to the Earth and θ is the angular radius of the Moon as seen from the Earth, in radians.

Putting in some approximate figures: the Moon and the Earth are both made of rock, so let’s assume it’s the same kind of rock: ρ = 1. The moon is ½° wide, so its radius is ¼°, which is 0.00436 radians. The Earth’s radius is 6366 kilometres. Multiplying it all out, we get a tidal range of just over half a metre from low tide to high tide.

Half a metre is a lot less than you will see on the coast if you go out and look. This is because of resonance. If you give regular small pushes to an oscillatory system – such as a plate with soup in it, or a child on a swing, or an ocean – then if you do it at just the right intervals the system will have had time to respond to your push, swing out and swing back, and it will be swinging out again just as the next push comes, so the each push adds to the swing that is already there. With soup, swings, and seas the result is exactly the same: the small pushes add up to a very big movement indeed.