The drop that is not a tear

The teardrop is one of the most reproduced wrong pictures in science. It appears on weather symbols, in children’s books, in the logos of water companies and in an alarming number of textbooks, and there is no configuration of falling water that has that shape.

Two facts settle it, and they are both about symmetry.

This collection has already established the static half of the story: how big a drop is allowed to be before its own weight shows. What follows is the moving half, and the two share nothing but the surface tension.

A drop in free fall has no up. In the frame falling with it, gravity is exactly cancelled by the acceleration, so the internal hydrostatic gradient is gone and the only thing that can shape it is surface tension — which makes it a sphere. That is the state of a drop for the first few metres after it leaves a cloud, and it is why astronauts’ water floats in balls.

A drop at terminal speed is not in free fall, but its weight is being carried by the air on its outside rather than by fluid on its inside, so the internal gradient is still very nearly absent. What shapes it is the pressure distribution of the flow going round it, and that distribution is fore-and-aft symmetric to leading order because ideal flow past a sphere is.

So the drop is symmetric front to back. A teardrop is not.

The calculation, which is four lines #

Ideal flow past a sphere gives a pressure coefficient of $C_p = 1 - \tfrac{9}{4}\sin^2\theta$, which this collection computes rather than quotes. The useful step is to decompose it into Legendre modes, because a sphere’s curvature responds to each mode independently. Using $\sin^2\theta = \tfrac{2}{3}(1 - P_2(\cos\theta))$:

$$C_p = -\tfrac{1}{2} + \tfrac{3}{2}P_2(\cos\theta).$$

A constant and a single second mode — nothing else. The constant shifts the internal pressure and does not deform anything.

Now perturb the sphere as $r = a(1 + \epsilon P_2)$. Its mean curvature changes by $(4/a)\epsilon P_2$ for that mode, so Young–Laplace requires the $P_2$ components to match:

$$\frac{4\sigma}{a}\epsilon = -\frac{3}{2}\cdot\frac{1}{2}\rho U^2 \qquad\Longrightarrow\qquad \epsilon = -\frac{3}{32}\,\mathrm{We},$$

with $\mathrm{We} = \rho U^2 d/\sigma$ formed on the diameter. The sign is the answer: $\epsilon$ is negative, $P_2$ is $+1$ at the poles and $-\tfrac12$ at the equator, so the poles move in and the equator moves out. Oblate. The width-to-height ratio is $1 + \tfrac{9}{64}\mathrm{We}$ to first order.

What Weber number a raindrop actually has #

The shape depends on the speed, the speed depends on the size through a drag balance, and the drag coefficient is the one quantity here that has to be imported — a measurement rather than a calculation. Balancing weight against drag and solving for the terminal speed with the Reynolds number inside the drag law:

A half-millimetre drop is a sphere to a part in two hundred. A two-millimetre one is a quarter wider than it is tall, which is plainly visible in a high-speed photograph and is what those photographs show. And a four-millimetre drop is past the range where a first-order shape means anything — real ones at that size are flattened buns with a dimple in the underside, and they break up shortly afterwards.

The theory announces its own death #

The linear shape has $r_{\text{pole}} = a(1 + \epsilon) = a(1 - \tfrac{3}{32}\mathrm{We})$, which reaches zero at $\mathrm{We} = 32/3 = 10.67$. At that Weber number the linearised drop has its two poles meeting at the centre and there is nothing left of it.

Measurement puts the critical Weber number for a drop breaking up in a stream at about 12.

Those are not the same calculation and the agreement is not a derivation — a first-order shape has no business predicting a break-up, and the linear theory is quantitatively wrong well before it gets there. But it is the right order for the right reason: the mechanism of break-up is the pressure distribution winning against the tension, and the linear theory fails at the point where the distortion becomes comparable with the drop. A theory whose breakdown coincides with the physical event it cannot describe is being honest about its own range.

Two mechanisms, and which one is acting #

A drop can be flattened by gravity through the Bond number or by aerodynamic pressure through the Weber number, and it is worth being able to say which. Their ratio is

$$\frac{\mathrm{We}}{\mathrm{Bo}} = \frac{\rho_a U^2 d/\sigma}{\rho_w g d^2/4\sigma} = \frac{4\rho_a U^2}{\rho_w g d},$$

which for a drop at terminal speed rises steeply with size because $U$ does. It passes one at about 0.7 mm and reaches five by two millimetres, so:

Small drops are Bond-number objects, held in shape against their own weight, and they are nearly spherical because their Bond number is tiny — a drop half a millimetre across is one per cent from a ball.

Large drops are Weber-number objects, and the weight has stopped mattering because the air is carrying it. Their shape is set entirely by the outside.

That handover is why the two essays exist separately and why the pictures look so similar: an oblate spheroid produced by an internal hydrostatic gradient and an oblate spheroid produced by an external pressure field are hard to tell apart, and the physics is unrelated.

Where the teardrop comes from #

The picture is not invented from nothing. A drop pulled off a surface has that shape: as it detaches from a tap or a leaf, a neck forms and the liquid above the neck is drawn into a point. Photographs of dripping taps show it clearly, and it lasts for a few milliseconds before the neck pinches and the drop retracts into a sphere.

So the teardrop is a picture of a drop forming, reproduced as a picture of a drop falling, and the two states are separated by about a hundredth of a second and by the entire mechanism.

This is the same class of error as reading a photograph of a flow: the image is real, it was taken of the thing it claims to show, and the interpretation attaches it to the wrong moment. That is a harder error to correct than an invented picture, because there is always a photograph to point at.

The pressure field is the whole of it #

It is worth dwelling on how little is needed. The entire calculation used one number from the flow: the coefficient of the second Legendre mode in the surface pressure, which is 3/2 for a sphere in ideal flow. Nothing else about the flow enters the shape at all.

That is not an accident of the sphere. A pressure distribution on a nearly spherical surface can always be written as a sum of Legendre modes, the curvature responds to each one independently with a factor $(n-1)(n+2)$, and the shape is the mode-by-mode quotient. So any pressure field’s effect on a drop is a list of coefficients, and the shape it produces is that list divided by the curvature factors.

For ideal flow past a sphere the list has exactly one entry, because $1 - \tfrac94\sin^2\theta$ is exactly a constant plus $P_2$. A measured pressure distribution, with a separated wake behind, would have entries in $P_1$ (which is what a fore-and-aft asymmetry is), $P_3$ and upward, and the drop would acquire a small $P_1$ distortion — a genuine asymmetry, and one whose sign puts the flat side at the back.

A teardrop is a negative $P_1$ distortion of the wrong sign and about ten times the wrong magnitude. Which is a precise way of saying that it is not a small error in a right picture.

What a drop this size is doing to the light #

Worth a paragraph because it is the one consequence of the shape that everybody has seen.

A rainbow is formed by light refracting into a drop, reflecting once inside and refracting out, and the angle at which the emerging rays pile up — 42 degrees for red — depends on the drop being spherical. A flattened drop has a different geometry in the vertical plane from the horizontal one, so its rainbow angle differs between them.

The consequence is that large drops make a rainbow that is bright at the sides and weak at the top, and small drops make one that is even all the way round. Since large drops fall fast and small ones slowly, the shape of the bow carries information about the drop-size distribution in the shower — and the reason it does is the Weber number of the drops, through the axis ratios computed above.

The tolerance that matters there is a few per cent of axis ratio, which puts it between the one-millimetre and two-millimetre rows of the table — exactly where the shower’s drop distribution usually straddles.

The shape is measured operationally, every day #

The rainbow is the consequence of the flattening that everybody has seen. There is a second one that nobody sees and that is used continuously, and it is the strongest practical evidence that the shape computed here is the right one.

A dual-polarisation weather radar transmits and receives in two polarisations, horizontal and vertical, and reports the ratio of the two returned powers as the differential reflectivity. A sphere returns equally in both and gives zero. An oblate drop, falling with its flattened axis vertical, presents a larger horizontal extent than vertical, returns more in the horizontal channel, and gives a positive value — and the value grows with the drop’s axis ratio, which by the arithmetic of this essay grows with the drop’s size.

So a radar measuring that ratio is measuring the Weber number of the raindrops in a cloud, at range, in real time. The conversion from the measured ratio to a drop size uses an axis-ratio relation fitted to exactly the measurements this essay cites, and it is one of the standard operational products of every modern weather radar: the reflectivity alone gives a rainfall rate that is ambiguous between many small drops and few large ones, and the differential reflectivity separates them.

A teardrop would give nothing. A shape whose asymmetry is fore-and-aft has the same horizontal and vertical extent, so it would return equally in both channels; and a population of such drops, if they tumbled at all, would average to zero. What makes the measurement possible is precisely the two features this essay establishes — that the distortion is oblate rather than pointed, and that it is systematically oriented, because the flattening is caused by the flow the drop is falling through and therefore has its axis aligned with the fall.

The discrimination that follows is the part meteorologists care about most. Hailstones are large, return an enormous amount of power, and tumble — they have no aerodynamic reason to keep one axis vertical — so their differential reflectivity averages to nearly zero. Heavy rain and hail look alike in reflectivity alone and are told apart at a glance by the second channel: high reflectivity with a strong polarisation signal is rain, high reflectivity with none is hail.

Which is a satisfying place for the argument to end. The shape of a raindrop is set by the pressure field of the air going round it, it is oblate and aligned for that reason, and a national network of radars depends on both facts being true. The picture on the weather symbol is refuted by the instrument that drew the map underneath it.

What breaks the symmetry, since something must #

The fore-and-aft symmetry is a leading-order statement and real drops are not exactly symmetric. Two things break it and both are second order.

The wake. Ideal flow past a sphere has no wake and its pressure field is symmetric — which is d’Alembert’s paradox seen from the surface rather than from the force; a real flow at $\mathrm{Re}$ of several hundred has a separated wake behind the drop, and the base pressure there is lower than the front stagnation pressure. That pulls the rear surface outward slightly relative to the front, which is a flattening of the underside rather than a point — the dimple that appears in photographs of large drops, and the opposite of a teardrop.

The internal circulation. The air’s shear drives a slow circulation inside the drop, which changes the internal pressure distribution slightly. It is a small effect for water in air because the viscosity ratio is fifty to one, and it is not small for a drop of one liquid in another.

Both make the drop less like a teardrop rather than more.

What the picture cannot show #

The shape is first order in the Weber number. It is quantitative below about $\mathrm{We} = 1$ and indicative above it, and the figure stops drawing at $\mathrm{We} = 32/3$ where the linear shape stops being a shape. At $\mathrm{We} = 1.44$ — a two-millimetre drop — the linear width ratio is 1.20 and measurement gives about 1.10, so the theory over-predicts by a tenth.

The drag coefficient is measured, not computed. It is passed into the terminal-speed calculation rather than assumed, and the figures say so. A deformed drop has more drag than a sphere of the same volume, so the large end of the speed curve runs fast — a four-millimetre drop really falls at about 8.8 m/s rather than the 10.2 computed here.

The pressure field is ideal. It has no wake in it, so the base pressure is wrong, and the symmetry it produces is exact where the real one is only approximate. Using a measured pressure distribution instead would give the dimple and would need a measured input where this calculation has none.

And nothing here oscillates. Real raindrops of a millimetre and above oscillate continuously between oblate and prolate at tens of hertz, driven by their own shedding wake, and the shape drawn here is the mean of that oscillation rather than a state any individual drop holds.

Who found it, and when #

This is the same order of error as reading a Reynolds number off the wrong speed, and it survives for the same reason: nobody checks a picture they have seen a thousand times.

The physics is old and the correction is not. Rayleigh’s work on drop oscillations in 1879 has the Legendre modes and the curvature operator that this calculation uses; the aerodynamic flattening was described qualitatively for most of the twentieth century and measured properly by Pruppacher and Beard in 1970, whose axis-ratio correlation is still the standard.

The teardrop, meanwhile, has been corrected in print roughly once a decade since, in tones ranging from patient to exasperated, and it has not moved. It survives because it is a good picture of something else and because nobody looks at a falling raindrop closely enough to be surprised.

The surprising connection is that the shape is determined by the flow outside the drop and not by anything inside it. A drop has no idea it is falling — it feels no gravity in its own frame — and what it responds to is a pressure field it would feel equally if it were held stationary in a wind tunnel. Which is exactly the experiment that measured it, and is why the same axis ratios appear for drops in a vertical tunnel and for drops falling through still air — the same equivalence that makes a tunnel measurement mean anything at all.

Where the ladder goes next #

Above this rung is the oscillation: a drop is an oscillator with a spectrum of Legendre modes, the lowest of them at $\omega^2 = 8\sigma/\rho a^3$, and a falling drop is being driven by its own wake at a frequency that can come close to it. That resonance is one of the mechanisms proposed for raindrop break-up and it needs the unsteady problem this collection has not built.

Beside it sits the static drop, flattened by the other mechanism, and the jet, where the same Weber number decides break-up rather than shape. Below it is the pressure field that does all the work here, and which was computed for a different purpose long before this essay needed it.