The geoid has a curvature only if you say where you stopped

How many triangles it takes ended by naming the one surface here that has never had a curvature computed for it, and it is the one that matters most. Every orthometric height on every map is measured above the geoid — an equipotential surface of the Earth’s own gravity, chosen to run at mean sea level. Height above what is where that surface first appears and the ellipsoid is a level surface is where the difference between it and the reference figure becomes a number.

It is a surface, so it has a Gaussian curvature. Nothing here has computed it.

The reason that gap stayed open turns out to be the finding, and it is not that the computation is difficult. It is that the question does not have an answer until one more thing is said.

The geoid is a series, and a curvature is two derivatives #

The ellipsoid is a shape. It is stated by two numbers, its curvature at any latitude is the exact $1/(MN)$ that the radius of curvature is two numbers computes, and asking for it is a request with one answer.

The geoid is not stated that way and cannot be. It is the level surface of a gravity field that is known through its spherical harmonic coefficients, and every model anybody has ever used is that expansion truncated at a degree: 36 for a satellite-only field, 200 for GRACE and GOCE together, 360 for EGM96, 2190 for EGM2008. The geoid model stops at a degree is about what that truncation costs a height and a slope. This is the same question one derivative further on, and one derivative changes the answer’s kind.

Write the geoid as a small radial perturbation of a sphere, $r = R + N$. To first order in $N$ the Gaussian curvature is

$$K = \frac{1}{R^2} - \frac{1}{R^3}\left(2N + \Delta N\right)$$

with $\Delta$ the Laplace–Beltrami operator on the unit sphere, whose eigenvalue on a degree-$n$ harmonic is $-n(n+1)$. So a degree-$n$ component of amplitude $N_n$ moves the curvature by

$$\delta K_n = \frac{n(n+1) - 2}{R^3}\,N_n$$

and the whole of this essay is in that numerator. The curvature spectrum is the height spectrum multiplied by $n^2$: one power for each derivative. The slope spectrum carries one power. The height spectrum carries none.

The factor $n(n+1)-2$ has one exact zero and it is the control the whole argument rests on. At $n = 1$ it is zero. A degree-one term in a radial expansion is a displacement of the centre, and moving a sphere cannot change its curvature — so the formula is required to return exactly nothing there, and it does. That is the only place the perturbation formula can be checked without a second computation, and if it failed the extra powers of $n$ would be arithmetic rather than geometry.

Kaula’s rule, and what follows from it #

Everything numerical here rests on one empirical statement, taken exactly as the geoid model stops at a degree takes it: the root-mean-square of the Earth’s normalised gravity coefficients at degree $n$ is about $10^{-5}/n^2$. Kaula wrote that down in 1963 and it has survived sixty years of better data as a description of the spectrum’s shape.

With $2n+1$ coefficients at each degree, the per-degree geoid height signal is $R \cdot 10^{-5}\sqrt{2n+1}/n^2$, which falls as $n^{-3/2}$. The degree variance therefore falls as $n^{-3}$, and a sum of $n^{-3}$ converges: the tail above $N$ is proportional to $N^{-2}$, so the omitted height is a number and the geoid has a shape.

Multiply each degree by $n$ for a slope and the variance falls as $n^{-1}$. That sum is a logarithm — it grows without bound and it grows so slowly that no power describes it. Multiply by $n^2$ for a curvature and the variance rises as $n$. That sum goes as $N^2$ and its square root as $N$.

Extending the tail from degree 720 to degree 120,000 — a factor of a hundred and sixty-six in where the sum is cut off — moves the three by very different amounts:

The height is insensitive to the stopping point, which is what it means for a quantity to exist. The slope is mildly sensitive, which is the logarithm. The curvature is almost entirely the stopping point, and a quantity whose value is 192 times larger when an arbitrary bound is moved is not a property of the Earth.

What the models actually carry #

• Degree 36, a satellite-only field resolving 556 kilometres: 0.039 per cent.
• Degree 200, GRACE and GOCE combined, 100 kilometres: 0.203 per cent.
• Degree 360, EGM96, 56 kilometres: 0.363 per cent.
• Degree 2190, EGM2008, 9 kilometres: 2.193 per cent.
• Degree 20,000, a one-kilometre local model: 20.0 per cent.

Each step down in wavelength multiplies the curvature by very nearly the ratio of the degrees, and the fitted exponent says so precisely.

That number is the essay’s whole claim in one line. One power of the cutoff, with a residual scatter of four parts in $10^5$ over a five-hundred-fold sweep — so the divergence is not an artefact of the low degrees, where Kaula’s rule is least reliable, nor of the high ones, where it is an extrapolation.

Where the geoid overtakes the ellipsoid #

The curvature is not one number measured the ellipsoid’s own variation: $K$ runs from $2.4747 \times 10^{-14}\,\mathrm{m^{-2}}$ at the equator down to $2.4417 \times 10^{-14}$ at the pole — the equator is the more curved end, because both principal radii are shortest there — a range of 1.334 per cent. That variation is the subject of an entire essay, and it is the fixed quantity everything here is measured against, because the ellipsoid is a stated shape and its curvature does not depend on anybody’s cutoff.

The geoid crosses it at degree 1,332.

Fifteen kilometres is a usable number and it deserves saying plainly. A geoid model resolved more finely than about fifteen kilometres has more curvature variation in its own undulations than the ellipsoid has between its equator and its pole. A model resolved more coarsely has less. EGM96 at 56 kilometres sits comfortably on the ellipsoid’s side of that line; EGM2008 at 9 kilometres sits on the other, carrying 1.6 times the ellipsoid’s whole flattening signal in bumps that are, individually, a few metres high.

That last comparison is the one where the surface curves the other way set up without being able to complete. It put the ellipsoid’s 1.35 per cent beside relief exceeding it by four orders of magnitude and observed that a great deal of care had gone into the smaller of the two. The geoid now sits between them and is worse than either, because the other two at least have values.

The total is exact, and that is not a consolation #

There is one curvature of the geoid that every model agrees on, to the last digit, and it is worth putting beside the divergence rather than leaving it to be discovered as an objection.

The integral of $K$ over the whole surface is exactly $4\pi$, whatever the cutoff. That is Gauss–Bonnet on a closed surface of genus zero, and it is a statement about topology rather than about gravity: the geoid is a deformed sphere, so its total curvature is the sphere’s total curvature and no amount of undulation changes it. The perturbation formula agrees, and it agrees for a reason that can be read off it. Every term $\delta K_n$ is a spherical harmonic of degree $n \ge 1$ multiplied by a constant, and a harmonic of positive degree integrates to zero over the sphere. So the bumps add curvature in some places and remove exactly as much in others, at every degree independently, and the books balance term by term.

Total curvature and the scale rule is where that fact is first used and how many times, not whether is where the same accounting is done by counting instead of by integrating. The point here is the contrast, and it is sharp: the geoid’s mean curvature is $4\pi$ over its area, which converges; its curvature at a place has no value at all. One number is a topological certainty and the other is a property of somebody’s truncation, and they are the integral and the integrand of the same function.

That kills one obvious escape. Faced with a divergent local quantity, the natural move is to average it over a patch — over a country, over a degree square — and quote the average. But averaging is integrating, and integrating is exactly the operation that makes the divergence disappear by discarding the thing that diverges. An average of $K$ over a region reduces, by Gauss–Bonnet, to the holonomy around that region’s boundary, which depends only on wavelengths longer than the boundary is. So a patch average is not a coarse measurement of a fine quantity; it is a different quantity, and it is the one the boundary can see.

Why it stayed open for thirteen measurements #

Nothing forced this question earlier, and the reason is that a height does not need a curvature.

An orthometric height is a distance measured along a plumb line to a surface. Producing one needs the geoid’s position — its height above the ellipsoid, which converges — and correcting a levelling run needs its slope, which diverges so slowly that a number can be quoted with a cutoff attached and be useful. A levelled height is not a distance and the line a height is measured along are both about the first derivative and neither reaches the second.

The second derivative is wanted only by somebody asking about the geoid as a surface in its own right: what its intrinsic geometry is, whether a patch of it can be flattened, how far a triangle on it departs from a plane one. Those are this collection’s questions rather than a levelling network’s, and the answer is that the geoid does not support them. It is a perfectly good surface to measure heights above and not a surface to do geometry on, which is a distinction none of the other surfaces here has needed.

A smaller triangle measures a different geoid #

How big a triangle it takes established the shape of the intrinsic measurement: a triangle’s excess is proportional to its area, so the curvature signal grows as $s^2$ while the angular noise stays put, and there is a best size set by the instrument. How many triangles it takes added the count. Both arguments assume the surface is given and the only question is whether the instrument can see it.

On the geoid that assumption fails, and it fails in a direction nothing else here has met. A triangle of side $s$ is blind to undulations much shorter than $s$ and sees those much longer than $s$ as a uniform tilt, so the curvature it can measure is the curvature contained in the model at that resolution — which grows as the triangle shrinks:

• a 100 km triangle can resolve 0.203 per cent;
• a 10 km triangle, 2.003 per cent;
• a 1 km triangle, 20.0 per cent.

So refining the survey does not converge on an answer. The signal-to-noise ratio improves in both directions at once — smaller triangles have larger relative noise from the angles and larger relative signal from the surface — and the quantity being measured changes with the instrument. That is not measurement error. It is the same situation how small is flat enough describes for a working tolerance, seen from the far side: there is no scale at which the surface settles down.

What each number was checked against #

The whole argument is a claim about how three quantities scale, so every control is about scaling rather than about agreement with a reference value.

A degree-one perturbation must move the curvature by exactly nothing. $n(n+1)-2$ is zero at $n = 1$, and a degree-one radial term is a displacement of the centre. This is the one exact zero the formula has, and it is checked to within $10^{-30}$.

The omitted height must barely depend on where the tail is stopped, required to move by less than two per cent over a factor of fifty-five in the cutoff. It moves by 1.4 per cent. If it did not, the geoid would not have a height either and the essay would be about something much larger.

The omitted slope must grow, but between 1.4 and 4 times over the same range. It grows by 1.79. A logarithm is bounded that way and a power is not, so this is what separates the middle case from the other two.

The omitted curvature must grow by more than twenty times over the same range. It grows by 56.

The contained curvature must grow as the first power of the cutoff, checked as an exponent within 0.08 of one at $R^2$ above 0.999. It gives 0.9913 at 0.999963.

And the crossing with the ellipsoid must fall inside the range of models anybody uses, between degree 36 and degree 36,000. It falls at 1,332. A crossing outside that range would mean the comparison was decided before it was made — either every real model beats the ellipsoid or none does — and the fifteen-kilometre figure would be a curiosity rather than a reading.

Where the model stops #

Kaula’s rule is an input and it is approximate. The real spectrum departs from $10^{-5}/n^2$ by a factor of two or three in places, and above degree 2190 nobody has global data at all, so the high-degree behaviour is an extrapolation. This matters less than it looks: the divergence follows from the exponent of the power law and survives any rule of the form $n^{-p}$ with $p < 5/2$, which every proposed spectrum satisfies. The specific numbers would move; the fact that there is no limit would not.

The perturbation is first order in $N$. Geoid undulations reach about a hundred metres against a radius of six and a half million, so the second-order terms are one part in $10^5$ of the first-order ones — far below the factor of 190 the argument is about.

The curvature computed is an RMS over the sphere, not a value at a place. A point value would need the actual coefficients rather than their statistics, and it would be larger where the field is rough and smaller where it is smooth. What is being measured here is the size of the quantity, and an RMS is the right measure of a size.

And the geoid is taken as a surface rather than as a boundary condition. It is defined by a potential, not by a shape, and a level surface of a field with a rough spectrum is rough for a reason that has nothing to do with topography being rough. A mountain is not a buried sphere and a deflection is the slope of a mass are where the local sources of that roughness are priced; nothing here separates the part of the spectrum that comes from topography from the part that comes from density.

Still open: whether any surface a survey uses has a curvature at all #

The result generalises further than it was asked to, and the generalisation is uncomfortable enough to be worth writing down rather than leaving implied.

Any surface given as a truncated expansion has this problem, and the threshold is the spectral exponent. A field whose degree variance falls faster than $n^{-5}$ has a convergent curvature; one that falls more slowly does not. Gravity falls as $n^{-3}$, topography more slowly still, and the only surfaces here with convergent curvature are the ones written as closed forms — the sphere, the ellipsoid, the triaxial body, the torus.

So the honest statement is narrower than “the geoid has no curvature” and worse. Every surface a survey actually meets is given by data at a resolution, and for every one of them the Gaussian curvature is a property of the resolution. The ellipsoid’s 1.35 per cent is not a small effect that better data would refine; it is a different kind of quantity from anything measurable, and the eleven essays that priced it were pricing the last surface in the subject that has a curvature to price.

What would settle the practical half of it is a measurement no spectrum can supply: whether any real computation ever reads a curvature off the geoid rather than off the ellipsoid. A levelling network propagates heights along a surface and its corrections use the ellipsoid’s radii; a geodetic solve uses the ellipsoid throughout. If nothing ever asks the geoid for a second derivative, then the quantity that does not exist is also the quantity nobody needed — which would be a satisfying end, and is not something a spectrum can be asked.