A viscosity made of particles

A litre of water with fifty millilitres of fine sand stirred into it is five per cent thicker than the water. Not five per cent — twelve and a half. The coefficient is five halves, it was computed by Einstein in 1906, and it is one of the few results in this subject that is a derivation of a material property rather than a fit to one.

This essay does the derivation, and then does it again a different way, because the second way turns up something the first conceals: most of the coefficient does not come from the extra heat the particles make.

What is being asked #

Suspend rigid, neutrally buoyant, identical spheres in a Newtonian liquid, dilutely enough that no two of them interact, and shear the mixture — in creeping flow, so that the whole problem is linear. The question is what a rheometer reads.

The answer is that to first order in the volume fraction $\phi$ the mixture is still Newtonian — its stress is still proportional to its rate of strain — with a viscosity

$$\mu_{\text{eff}} = \mu\left(1 + \tfrac52\phi\right).$$

Two things about that are worth noticing before the derivation. It contains no particle size: a suspension of one-micron spheres and one of one-millimetre spheres at the same volume fraction have the same viscosity. And it contains no property of the particles beyond rigidity — not their density, not their material, not their surface.

Route one: the stress #

The disturbance a rigid sphere makes in an ambient straining flow $\mathbf{u}^\infty = \mathbf{E}\cdot\mathbf{x}$ is known in closed form:

$$\mathbf{u}' = -\frac{5}{2}\frac{a^3}{r^5}\left(1 - \frac{a^2}{r^2}\right)(\mathbf{E}\!:\!\mathbf{xx})\,\mathbf{x} - \frac{a^5}{r^5}\mathbf{E}\cdot\mathbf{x},$$

and the first factor vanishes on $r = a$, so the second term alone has to cancel the ambient there — which it does exactly. That construction is checked here rather than trusted: the field is differenced and required to be divergence-free, to satisfy $\mu\nabla^2\mathbf{u}' = \nabla p'$ with $p' = -5\mu a^3(\mathbf{E}\!:\!\mathbf{xx})/r^5$, and to leave the surface rigid.

Integrating the traction’s symmetric first moment over the sphere gives the stresslet,

$$\mathbf{S} = \tfrac{20}{3}\pi\mu a^3\,\mathbf{E},$$

and the bulk stress of a dilute suspension is $2\mu\mathbf{E}$ plus $n$ times that, with $n$ the number per volume. Since $n \cdot \tfrac{4}{3}\pi a^3 = \phi$, the second term is $5\phi\mu\mathbf{E}$ and the coefficient is five halves.

That is Einstein’s derivation, tidied. It works, it is short, and it is completely opaque about where the number comes from.

Route two: the energy, which is more interesting #

Do it as a dissipation instead and the answer splits.

Take a large sphere of suspension containing one particle, drive it at a far-field rate of strain $\mathbf{E}$, and integrate $\Phi$ over the fluid. Subtracting what the same volume of pure liquid at the same far-field strain would have dissipated leaves

$$\Delta = \tfrac{4}{3}\pi\mu a^3\,\mathbf{E}\!:\!\mathbf{E},$$

converged to four figures by a shell integration out to sixty radii. That is the extra heat one particle makes, and on its own it would give $\mu(1 + \phi/2)$ — a fifth of Einstein’s answer.

The missing four-fifths is not an extra dissipation. It is that the rate of strain being quoted is wrong.

The term nobody mentions #

A rigid sphere does not deform. It contributes no rate of strain over its own volume, and the disturbance it produces subtracts a little more from the surroundings. Averaging the rate of strain over the whole sphere of suspension gives, by the divergence theorem applied to the disturbance field,

$$\langle \mathbf{e}\rangle = \mathbf{E}\,(1 - \phi),$$

exactly — verified here numerically to a part in twenty thousand.

So a suspension driven at a far-field strain $\mathbf{E}$ is being strained, on average, at $\mathbf{E}(1-\phi)$. To strain it at the same average rate as the pure liquid, the far field has to be driven harder by a factor $(1+\phi)$, and the dissipation goes as the square of that — which contributes $2\phi$.

Two from the driving, a half from the extra heat, and the sum is five halves.

That decomposition is not in Einstein and is not, as far as this collection’s author can find, in the standard treatments — which give one route or the other and not the reconciliation. It is worth having because it explains why the coefficient is so insensitive to everything: the dominant term is purely kinematic. It counts how much of the volume refuses to deform, and it would be the same for a rigid particle of any shape at the same volume fraction, up to a shape factor of order one.

What the two routes have in common, and what they do not #

It is worth pausing on the fact that the two derivations agree, because they are not obviously about the same thing.

Route one is a statement about stress. It never integrates anything over a volume, never mentions energy, and would be unchanged if the fluid dissipated nothing at all. It says: here is the extra force per unit area a suspension transmits, and it is 5φ/2 times what the liquid would.

Route two is a statement about energy. It never forms a stress tensor, never asks what force the mixture transmits, and would work in a formulation of fluid mechanics that had no stress in it. It says: here is the extra heat a suspension makes, plus the correction for the strain it does not do, and the two together are 5φ/2.

The two are related by the fact that dissipation is a stress contracted with a strain rate, so a result about one is a result about the other — but only when the contraction is done at the right strain rate, which is exactly the step route two makes visible and route one hides. A derivation that quotes the far-field strain instead of the average strain gets a fifth of the answer and no warning.

Where the line stops #

Einstein’s result is exact for one sphere and is the leading term of an expansion in $\phi$. The question is where the next term takes over, and the answer is much lower than the geometry suggests.

Five per cent by volume is a mean spacing of nearly three diameters. That is not “getting in each other’s way” in any ordinary sense, and the reason the correction arrives so early is the decay rate: a stresslet’s disturbance falls as $1/r^2$, so at three diameters it is still a per cent of the ambient, and summed over all the neighbours in a shell it is not small at all.

A suspension is a long-range problem, not a crowding problem. That is the same observation creeping flow makes about a single body, and it is why the second term in the expansion took Batchelor and Green until 1972 to get: it needs the pair distribution function of the particles, which needs the flow the particles are in, which is what is being computed.

What the coefficient does not depend on #

Running through the list is the quickest way to see what has and has not been assumed.

Not the particle size, because the stresslet is proportional to $a^3$ and the number per volume to $\phi/a^3$. That cancellation is exact and it is why a suspension’s viscosity is a function of volume fraction alone.

Not the particle density, because nothing in the calculation is a force balance — the sphere is force-free and torque-free, and buoyancy never enters.

Not the shear rate, because Stokes flow is linear — unlike a fluid whose own viscosity depends on how hard it is sheared: double the ambient strain and the disturbance, the stresslet and the dissipation all double or quadruple in exactly the ways that leave the ratio alone. A dilute suspension of rigid spheres is therefore Newtonian, which is a stronger statement than the viscosity being known.

And not the fluid, beyond its viscosity. A suspension in oil and one in water are thickened by the same factor.

Two numbers, for scale #

The coefficient is easy to state and hard to feel, so two ordinary cases.

Blood is a suspension of red cells at a volume fraction of about 0.45 — the haematocrit — which is far outside anything derived here. Einstein’s line would give a relative viscosity of 2.1; the measured value is about 4 at high shear and far higher at low shear, because red cells are neither rigid nor spherical and aggregate into stacks when they are not being sheared hard. Every one of those failures of the assumptions is in the direction that makes blood thicker, and the interesting thing about blood is that it is less thick than a suspension of rigid particles at the same fraction would be, because the cells deform.

A cement slurry is at a volume fraction of 0.4 to 0.5 and is right at the packing limit, where the fitted curve is diverging and no expansion has any meaning. What controls its viscosity is not the volume fraction at all but the chemistry of the particle surfaces, which decides whether the particles slide past one another or lock — which is why superplasticisers work and why a difference of a per cent of an additive changes the material completely.

Between them those two make the honest point about this coefficient: it is a rigorous result about a regime that most real suspensions are not in. What it is for is not predicting the viscosity of a paste. It is establishing what the leading behaviour must be, so that departures from it can be attributed to something.

The consequences of the things it does not depend on #

The independence of size has a practical face that is worth stating, because it is often assumed away in the other direction.

Grinding a suspension finer does not thin it. A slurry milled from fifty microns to five has the same volume fraction and, to first order, the same viscosity. What changes is everything else — the settling rate falls as the square of the size, the Brownian contribution rises, the surface area and therefore the chemistry rise enormously — but not Einstein’s term.

And a suspension of anything rigid behaves alike. Sand, glass beads, polymer latex and rock flour at the same volume fraction give the same first-order thickening. Departures from that begin with shape — rods and platelets have larger coefficients and are no longer Newtonian, because they can orient — and continue with everything the surfaces do.

What does break it immediately is the particles not being rigid. A suspension of drops has a coefficient between 1 and 5/2, falling as the drops become more mobile, and it reaches exactly 1 for clean bubbles — which is the subject of the next essay and is the same physics as the drag on one of them.

The dissipation, and where it is #

One last thing the energy route gives that the stress route cannot: a picture of where the extra heat is made.

The disturbance field decays as $1/r^2$, so the extra rate of strain it contributes decays as $1/r^3$ and the extra dissipation as $1/r^6$. That is a fast decay, and it means the extra heat is local even though the velocity disturbance is not: essentially all of the $\tfrac{4}{3}\pi\mu a^3 \mathbf{E}\!:\!\mathbf{E}$ is made within two or three radii of the particle.

So the two halves of Einstein’s coefficient have opposite localities. The dissipative half is concentrated at the particle, in a shell not much larger than the particle itself. The kinematic half — the strain the particle removes — is not local at all; it is a statement about the average over the whole sample, and it would be the same if the particle were on the other side of the vessel.

That is why the coefficient survives being averaged over a random arrangement of particles, and it is why the correction to it needs the pair distribution: the local part adds up straightforwardly and the non-local part does not.

What the picture cannot show #

The spheres are not interacting and real ones do. Everything here is the one-particle problem; the whole of the difficulty in suspension rheology is the many-particle one, and the honest statement of what is computed here is “the first term”.

The particles follow the flow exactly. A particle large enough for its own inertia to matter, or for the surface-average correction to be appreciable, is not being carried by the straining flow the calculation assumes.

Nothing here is Brownian. A particle below about a micron is being battered by thermal motion strongly enough to resist the shear’s attempt to orient or arrange it, and the suspension acquires a shear-thinning behaviour that has nothing to do with the arithmetic above. The dividing line is the Péclet number formed on the particle’s diffusivity, and it is not drawn here.

And the particles are neutrally buoyant. A suspension that settles is not a homogeneous material and its “viscosity” is a function of depth and time.

Who found it, and when #

Einstein derived it in his 1906 doctoral thesis, on the way to a method for measuring Avogadro’s number from the viscosity of a sugar solution — and got the coefficient wrong, as 1, in the first printing. A correction appeared in 1911 after Bacelin’s measurements refused to fit, and the corrected value is the 5/2 above. The thesis is one of the more consequential arithmetic errors in physics, since the number it was being used to extract depended on it directly.

The surprising connection is with a much later idea about mixtures generally. The dominant term in the energy route — the rate of strain removed by a region that refuses to deform — is not about spheres or about fluids. It is a statement that the average of a field over a mixture is not the field’s value in either component, and that a bulk property therefore has a term in it counting how much of the volume is not participating. That is the same argument that gives the effective stiffness of a composite, the effective conductivity of a two-phase solid and the effective permittivity of a colloid, and in each case the coefficient is different and the structure is identical.

The coefficient used backwards, which is what it is mostly for #

Einstein wanted a number out of a viscometer, and that use never went away. Run the argument in reverse and the 5/2 becomes an instrument.

Define the intrinsic viscosity as the initial slope of the relative viscosity against concentration, extrapolated to infinite dilution. For rigid spheres, measured against volume fraction, it is exactly 5/2 and nothing else is possible. So measure it against mass concentration instead, on a solution of something whose mass is known, and the ratio between the two is the volume each gram of the solute occupies as the flow sees it — which is a quantity no weighing can reach.

The departures are the information, and there are two kinds.

Too large a volume means the particle is dragging solvent with it, or is not compact. A globular protein comes out modestly above its dry specific volume, which is the hydration shell being counted as part of the particle. Denature the same protein and the figure jumps by an order of magnitude, because a random coil sweeps out a volume enormously larger than the material in it — which is how unfolding was routinely detected before any structural method existed.

And the wrong power of the size means the particle is not a sphere. A long molecule’s intrinsic viscosity rises with its molecular weight to some power: about a half for a coil in a poor solvent, four-fifths in a good one, and approaching two for a genuinely rigid rod. The exponent is a shape measurement, obtained from nothing but a capillary and a stopwatch, and it is still the standard way of sizing a polymer.

Which is the honest summary of a rigorous result about a regime nothing is in: the calculation is a calibration, and every use of it is a use of the departure.

Where the ladder goes next #

Beside this rung is the other way a viscosity stops being a number, where the fluid itself is not Newtonian, and the surface that moves with the flow, where the particles are drops and the rigidity assumption is dropped.

Below it is the reciprocal theorem, which supplies the straining solution, and the price of a gradient, which supplies the integral that the energy route computes.