The surface a spin decides

Put a dish of water on a turntable and start it spinning. The water climbs the outside, dips in the middle, and after a minute or two settles into a shape and stays there. The shape is a paraboloid, and it is a paraboloid exactly — not to first order, not for small rotation rates, but at every speed at which the liquid stays in the dish.

Two ways to get the same surface #

The quickest derivation is in the rotating frame, where the liquid is at rest and the centrifugal term appears as a force. Its potential energy per unit mass is $-\tfrac12\omega^2 r^2$, so the total effective potential is

$$\Phi = gz - \tfrac12 \omega^2 r^2.$$

A free surface of a liquid at rest is a surface of constant $\Phi$: if it were not, there would be a component of effective gravity along the surface, and the liquid would flow until there was not. That is the same argument that makes the surface of a still pond flat, with one extra term in the potential. Setting $\Phi$ constant,

$$z(r) = z_0 + \frac{\omega^2 r^2}{2g}.$$

A parabola, with no small quantity dropped anywhere.

The second route uses no rotating frame and is worth doing, because it makes clear that the answer is not an artefact of one. In the laboratory, a parcel of liquid at radius $r$ is going in a circle and therefore accelerating inward at $\omega^2 r$. The only horizontal force available is the pressure gradient, so

$$\frac{\partial p}{\partial r} = \rho\,\omega^2 r, \qquad \frac{\partial p}{\partial z} = -\rho g.$$

Integrating gives $p = \tfrac12\rho\omega^2 r^2 - \rho g z + \text{const}$, and the free surface is where $p$ equals atmospheric — which is the same parabola, with $\rho$ cancelling out of the shape while remaining in the pressure.

Half up and half down #

The parabola fixes the surface’s shape and not its position, and where it sits is decided by the one thing the derivation has not used: the liquid has to go somewhere.

The volume is conserved, so the mean height of the surface over the dish is unchanged by spinning. For a paraboloid over a circular dish that mean is easy to take — the average of $r^2$ over a disc, weighted by the area at each radius, is half the value at the rim — so the surface’s mean height is exactly half its rim height above its vertex.

Which gives an exact and rather satisfying statement: the rim rises by $\omega^2R^2/4g$ and the centre falls by the same amount. Half of the excursion goes up and half goes down, whatever the spin rate, the dish size or the depth. For the half-metre dish at 20 rpm the rim is 56 mm above the vertex, so the rim has climbed 28 mm and the centre has dropped 28 mm from where the water stood still.

That is what sets the practical limit on how fast a dish may be turned. Spinning stops working when the vertex reaches the floor of the dish — the liquid parts in the middle and the surface is no longer a complete paraboloid — and that happens when the depth at rest is less than $\omega^2R^2/4g$. A shallow layer is therefore a slow one, which is exactly the wrong way round for an instrument that wants a short focal length and a thin layer of mercury, and the compromise between the two is what fixes the focal ratio.

A mirror figured by a clock #

A parabola written $z = r^2/4f$ has focal length $f$. Comparing with the surface above,

$$f = \frac{g}{2\omega^2}.$$

The focal length of a spinning dish of liquid depends on the spin rate and on nothing else. Not on the liquid, not on the size of the dish, not on how much is in it. Doubling the rotation rate quarters the focal length, and there is no other adjustment available.

That the focusing is exact rather than approximate is the claim worth checking rather than asserting, and it can be checked on the drawn curve. A ray coming straight down hits the surface where the slope is $dz/dr = \omega^2 r/g$, so the surface is tilted by $\alpha = \arctan(\omega^2 r/g)$ from horizontal, and the reflected ray makes an angle $2\alpha$ with the vertical. It crosses the axis at a height

$$z(r) + \frac{r}{\tan 2\alpha} = \frac{r^2}{4f} + f - \frac{r^2}{4f} = f,$$

independently of $r$. The $r^2$ terms cancel identically, which is what “no spherical aberration” means.

A sphere does something else, and the difference is the point. Rays striking a spherical mirror further from the axis cross it closer to the vertex, so there is no single focus and the blur cannot be got rid of by making the mirror better — it is the defect a telescope maker spends the effort to remove, by figuring the glass away from a sphere toward a paraboloid. A spinning liquid never has to have it removed, because a rotating fluid in equilibrium cannot settle into any shape but the right one.

What the surface then does is form an image. A mirror of focal length $f$ images an object at distance $u$ to a point the mirror equation fixes, and for an object at infinity — a star — that point is at $f$ exactly. The whole of a liquid mirror telescope is that construction with $f$ set by a motor speed: $f = g/2\omega^2$, so a dish turning once every eight seconds has a focal length of about four metres, and changing the focal length means changing the gearing.

The instrument #

The idea is Isaac Newton’s, in a note of 1680; the first working attempt was Ernesto Capocci’s in 1850 and it did not work. Robert Wood built one in Baltimore in 1908 that did, and the modern versions began in the 1980s once air bearings and speed control were good enough. The Large Zenith Telescope in British Columbia carried a six-metre dish of mercury, three millimetres deep, turning at 6 rpm, from 2003 to 2016.

Three things follow from the physics and constrain the instrument absolutely.

It can only point up. The surface is a level set of gravity plus rotation, and gravity points down. Tilting the dish gives a surface that is not a paraboloid about the tilted axis, and there is no arrangement that produces a tilted paraboloid. A liquid mirror observes the strip of sky that passes overhead, and nothing else.

The focal length is fixed by the speed and the speed must be very steady. A fractional wobble $\delta\omega/\omega$ shifts the focus by $2\delta\omega/\omega$ of a focal length, so parts in $10^5$ are needed. Air bearings and synchronous drives supply it, and the requirement is not on the average rate but on its steadiness, since a resonance in the drive at the right frequency would grow rather than merely wobble.

Surface waves are the enemy. Any disturbance — a draught, a vibration, an imperfectly balanced bearing — appears as a travelling ripple on the optical surface, and a ripple of a hundred nanometres is a quarter of a wavelength. The mercury is kept shallow partly so that waves damp fast.

The shallowness is doing more work than it looks. A three-millimetre layer means about four hundred kilograms of mercury for a six-metre dish rather than a hundred tonnes; it puts the whole layer within a few times the viscous penetration depth so that disturbances die in seconds rather than minutes; and it makes the spin-up time short. What it costs is that the layer is thinner than the depth of the parabola, so the dish itself has to be figured to roughly the right paraboloid first, and the mercury merely perfects it.

The settling takes a minute or two, and the reason is worth stating because it is the one part of the process that is not equilibrium. When the dish starts turning the liquid does not, and it is brought up to speed only by viscosity at the wall and the base — a diffusion of momentum inward whose timescale is the spin-up time, seconds for a shallow dish and minutes for a deep one. Until then the surface is not an equipotential of anything, and the flow seen in the frame of the dish is real motion rather than the fictitious kind.

Why it settles in a minute rather than in an hour #

The remark that viscosity brings the liquid up to speed hides an interesting failure of the obvious estimate, and the failure is the reason the experiment is possible at a bench at all.

The obvious estimate is diffusion. Momentum spreads through a fluid at the same rate as anything else carried by molecular motion, so the time for the wall’s rotation to reach the middle of a layer of depth $H$ should be about $H^2/\nu$. For three centimetres of water that is fifteen minutes, and for a dish half a metre across, if the momentum had to come in from the rim, it would be days.

It settles in a minute or two. What actually happens is not diffusion but a circulation. The layer right against the rotating floor — a thin one, of thickness $\sqrt{\nu/\Omega}$ — is dragged round with it, and being in rotation it is flung outward; that outflow has to be fed, so fluid is drawn down from the interior at the centre and returned outward along the bottom. Every parcel of liquid therefore passes through the thin fast layer at the floor and is spun up there, rather than waiting for momentum to reach it.

The timescale that governs it is $H/\sqrt{\nu\Omega}$ rather than $H^2/\nu$, and for the same three centimetres of water at a couple of radians per second that is about twenty seconds — some forty times faster. The thinner and faster the arrangement, the larger the advantage.

Two things worth carrying from that. A boundary layer can be an enormously more effective transport route than the bulk, because it moves fluid rather than momentum, and any estimate that reaches for a diffusion time in a rotating system is likely to be wrong by orders of magnitude. And the spin-down after the motor stops takes the same time for the same reason, which is why a dish left to slow on its own comes to rest far sooner than its own viscosity would suggest.

What “exactly” is worth here #

It is worth pausing on how unusual the exactness is, because most shapes in fluid mechanics are approximations.

A hanging chain is a catenary exactly, and a hanging chain loaded uniformly along the horizontal is a parabola exactly, and neither of those is the shape of a real bridge cable, which is somewhere between. A soap film is a minimal surface exactly and a soap bubble a sphere exactly, both because the surface tension is uniform. The rotating liquid joins that short list, and it joins it for the same reason: the condition defining the surface is an equality between two scalar fields, with no length scale in it that could be compared against another one.

The moment a second length enters, the exactness goes. Surface tension supplies one — the capillary length — and the shape near the rim stops being a parabola. A finite depth supplies another once the vertex would fall below the floor of the dish. Neither of those is a correction to the parabola; each replaces the problem with a different one that has a boundary condition the parabola cannot satisfy.

Where else the same surface appears #

The condition — a free surface is an equipotential of the effective potential — does not care what makes the potential.

A tanker’s baffles. Liquid in an accelerating vehicle settles with its surface tilted at $\arctan(a/g)$, because the effective gravity has been tilted. That is the same statement with a uniform acceleration instead of a centrifugal one, and it is why an aircraft fuel gauge has to know the attitude.

A centrifuge. At high $\omega$ the term $\tfrac12\omega^2r^2$ dominates $gz$ entirely and the surface becomes a cylinder — the liquid climbs the wall and the paraboloid’s vertex drops below the bottom of the tube. The crossover is where $\omega^2 R^2/2g$ exceeds the depth.

A planet. A self-gravitating rotating fluid body settles into an oblate figure for the same reason, though the potential is no longer external and the problem becomes self-consistent — the shape and the field decide each other. Jupiter, which turns once in ten hours, is visibly flattened: its equatorial radius exceeds its polar one by 6.5%, and the whole of that is the term added to the potential here.

The same tilt of effective gravity is what a banked road is built out of. A bank tilts the normal force so that its horizontal component supplies the turning, and a liquid surface tilts itself for the same reason and to the same angle. One is designed and the other settles; the arithmetic is identical, and in both cases the surface is perpendicular to the effective gravity at every point.

The pressure underneath, where the density comes back #

Nothing about the liquid appears in the shape, and a great deal of it appears one line further down.

Integrating the pressure gradients gave

$$p(r, z) = p_0 + \tfrac12\rho\,\omega^2 r^2 - \rho g z,$$

so at the bottom of the dish the pressure rises quadratically outward. For mercury in a three-metre dish at 8 rpm that is 1.9 kPa of extra pressure at the rim over the centre — small against atmospheric, and enough to matter for a dish whose figure must hold to a fraction of a wavelength, since the container itself deflects under it.

The apparent weight of a floating object changes with radius for the same reason: the effective gravity at radius $r$ is $\sqrt{g^2 + \omega^4 r^4}$ in magnitude and tilted from the vertical, so a small float placed off-centre sits slightly deeper and leans. Buoyancy is a pressure difference, and the pressure field has changed.

Buoyancy in the still case is the weight of the displaced fluid, and an object floats at the depth that makes those equal. In the spinning dish the same balance holds against the effective gravity, which is larger than $g$ everywhere but the centre and tilted as well — so a floating object drifts outward, and rides at a slightly different draught depending on where it has drifted to. The equilibrium is at the rim, which is why anything floating on a centrifuged liquid ends up against the wall.

The shape, frozen #

The constraint that a liquid mirror can only point up is a consequence of gravity being the potential, and there is a way round it that keeps everything else: make the shape while the material is liquid, and then stop it being liquid.

A furnace holding several tonnes of borosilicate glass, mounted on a turntable and rotated while the glass is molten, produces exactly the surface this essay computes. Cool it slowly, still turning, and the glass solidifies with a front face that is already very nearly the required paraboloid — after which it can be tilted, mounted and pointed anywhere at all, because it is no longer a fluid and no longer obliged to be an equipotential.

The saving is not marginal. A conventional blank starts as a flat disc and the whole of the parabolic sag has to be ground and polished away, which for a large fast mirror is tonnes of glass and years of work. Spin-cast, the sag is there from the start and what is left is figuring — removing the last micrometres rather than the first centimetres.

The spin rate is chosen by exactly the relation in this essay: the furnace turns at whatever rate gives the focal length wanted, and a mirror of a stated focal ratio is made by setting a speed. That several of the largest telescope mirrors in existence were figured in the first instance by an equipotential, and only afterwards by an optician, is the most consequential use this piece of hydrostatics has been put to.

What it does not solve is the shrinkage. Glass contracts as it cools, and it does not contract uniformly, so the frozen surface is close to the intended paraboloid rather than equal to it. The method removes most of the work and none of the last stage of it.

Where the model stops #

The liquid rotates as a solid body. The derivation assumes every parcel is at rest in the rotating frame, which is only true once viscosity has done its work. During spin-up, and for any liquid with a genuine circulation, the surface is not this shape.

Surface tension has been ignored. At the rim the liquid climbs the wall by a meniscus of order the capillary length — 2.7 mm for water, 1.9 mm for mercury — and near the axis the curvature term is negligible. For a dish tens of centimetres across the meniscus spoils the outermost few millimetres and nothing else, which is why the optical aperture of a liquid mirror is always quoted smaller than the dish.

Gravity is uniform. Over a six-metre dish it varies by parts in $10^{7}$, which is negligible. Over an ocean it is not, and the equipotential is the geoid rather than a paraboloid.

The reflectivity is the liquid’s problem, not the shape’s. Mercury reflects about 75% in the visible, against 92% for a good aluminium coating, and it oxidises. Nothing in the shape argument cares, and it is the reason liquid mirrors have never displaced glass for anything but survey work where cost per square metre dominates.

The shape is a paraboloid, and a telescope wants a paraboloid of a chosen focal ratio. The two constraints — the focal length must be $g/2\omega^2$ and the dish must not throw its contents out — put a floor under the focal ratio for a given aperture. At 6 rpm and six metres, $f/D$ is about 1.5, and going faster to shorten it raises the rim without limit.

What the pictures cannot show #

Every figure draws a cross-section, and the surface is a paraboloid of revolution. The distinction matters for the optics: a parabolic cylinder would focus to a line, and it is the rotational symmetry that makes the focus a point.

The rotating figures plot height against radius on axes with different scales, so the surfaces look far deeper than they are. At 20 rpm across half a metre the rim is 56 mm up over a 500 mm radius — a slope of about six degrees at the edge — and drawn to scale the curve would be nearly flat.

And nothing here shows the liquid moving. Every figure is the settled state in the rotating frame, in which the whole system is stationary; the fact that the mercury is travelling at two metres a second at the rim of a six-metre dish appears nowhere.

Where the ladder goes next #

The ladder has gone from pressure depending only on depth, through what that lets a hydraulic press do, to a case where the potential has a second term in it and the surface takes a shape that can be used.

The rungs above are the ones where the fluid is no longer at rest in any frame. A rotating fluid with a free surface and a disturbance supports waves whose restoring force is the effective gravity, and their behaviour is different from ordinary surface waves because the frame is rotating. Beyond that lies the self-consistent problem — a rotating self-gravitating mass — where the potential and the shape must be solved for together, and where the sequence of allowed figures stops being a single family.

The habit worth carrying: when a free surface is involved, do not solve for the flow. Write down the potential, set it constant, and read the shape off. It works whenever the fluid is at rest in some frame, and the trick is usually finding which frame that is.