A liquid with one relaxation time rings in a tube, and a spectrum of times silences it

A solid, if it is not given time took the Deborah number at its word — a material’s relaxation time over the time it is watched for — and found that the word holds only for a material with one relaxation time. A real polymer has a spectrum of them, and the clean crossover between solid and liquid at $\lambda\omega = 1$ becomes a band decades wide in which the material is both. That was a statement about the material in a rheometer, where the deformation is imposed and the stress is read. This essay puts the material in a flow, where neither is imposed, and asks what the difference between one time and many does to something that can be measured with a flow meter.

The flow is the oldest oscillating flow there is: a pressure gradient pushed back and forth along a rigid tube, the problem that Womersley solved for blood and that has since been followed into elastic arteries and branching trees. In every one of those the liquid was Newtonian and the only elasticity was in the wall. Here the wall is rigid and the elasticity is in the liquid. A liquid that remembers its strain can carry a shear wave — a transverse wave travelling across the flow rather than along it — and a tube is a cavity for such waves. Shaken at the right frequency, the cavity rings.

The flow, in one line of Bessel functions #

The gradient is $-\partial p/\partial z = P e^{i\omega t}$ in a tube of radius $a$. The liquid is linear viscoelastic: its shear stress is a complex modulus $G^*(\omega)$ times the strain, so the axial momentum equation for the velocity amplitude $u(r)$ is

$$i\omega\rho\,u = P + \frac{G^*}{i\omega}\left(u'' + \frac{u'}{r}\right), \qquad u(a) = 0.$$

For a Newtonian liquid $G^* = i\omega\eta$ and this is Womersley’s equation. In general its solution is

$$u(r) = \frac{P}{i\omega\rho}\left[1 - \frac{J_0(kr)}{J_0(ka)}\right], \qquad k^2 = \frac{\omega^2\rho}{G^*},$$

and the flow rate is $Q = (\pi a^2 P/i\omega\rho)\,[1 - 2J_1(ka)/(ka\,J_0(ka))]$. The whole of the liquid’s character enters through one complex number, $k$, and $k$ is the wavenumber of a shear wave of frequency $\omega$ in that liquid. For a Newtonian liquid it is $(1-i)/\delta$ with $\delta$ the Stokes depth of the wall that shakes: a wave that dies within its own wavelength. For an elastic solid it is real, $\omega\sqrt{\rho/G}$, and the wave travels undamped at $\sqrt{G/\rho}$.

The smallest liquid with a relaxation time is Maxwell’s, $G^* = G\,i\omega\lambda/(1 + i\omega\lambda)$, viscosity $\eta_0 = G\lambda$. Two numbers then govern the flow: the Womersley number $\alpha = a\sqrt{\omega\rho/\eta_0}$ and the Deborah number $\mathrm{De} = \omega\lambda$, with $(ka)^2 = \alpha^2(\mathrm{De} - i)$. Their ratio,

$$E = \frac{\mathrm{De}}{\alpha^2} = \frac{\lambda\eta_0}{\rho a^2},$$

contains the liquid and the tube but not the frequency. It is the elasticity number, and it says that a given liquid in a given tube does not wander over the $(\alpha, \mathrm{De})$ plane as the frequency is changed; it moves along the line $\mathrm{De} = E\alpha^2$.

A cavity for shear waves #

When $\mathrm{De}$ is large the liquid’s modulus is nearly real and $ka \approx \alpha\sqrt{\mathrm{De}}\,(1 - i/2\mathrm{De})$. The flow is then large wherever $J_0(ka)$ is small — where the real part of $ka$ sits on a zero of $J_0$, at $2.405$, $5.520$, $8.654$ and so on. Those are the radial modes of a cylinder of elastic material clamped at its rim: the first is a core swinging one way, the second a core and an outer ring swinging against each other.

Along the line $\mathrm{De} = E\alpha^2$ the condition $\alpha\sqrt{\mathrm{De}} = j_{01}$ falls at $\mathrm{De} = j_{01}\sqrt{E}$. For $E = 300$ that is $41.65$; the peak of the flow found by searching the full solution sits at $41.652$. The resonance’s sharpness is set by the imaginary part of $ka$, $-\alpha/(2\sqrt{\mathrm{De}}) = -1/(2\sqrt E)$, so a more elastic combination of liquid and tube rings more sharply as well as at a higher Deborah number.

The case drawn is a wormlike-micelle solution, a surfactant in water whose molecules assemble into long, flexible, breakable chains — cetylpyridinium chloride with sodium salicylate is the standard one — with a relaxation time of about a second and a zero-shear viscosity of tens of pascal-seconds, in a tube one centimetre in radius. $E = (1\,\mathrm{s})(30\,\mathrm{Pa\,s})/ (1000\,\mathrm{kg\,m^{-3}}\times 10^{-4}\,\mathrm{m^2}) = 300$. The shear-wave speed $\sqrt{G/\rho}$ is 17 centimetres a second, and the first resonance, $j_{01}\sqrt{G/\rho}/(2\pi a)$, falls at 6.6 Hz.

Sixteen hundred times, for the same push #

At that frequency the flow carries 1661 times what a Newtonian liquid of the same zero-shear viscosity carries at the same frequency, and 1660 times the steady Poiseuille flow of the same gradient. The tube is small enough, $\alpha = 0.37$, that the Newtonian liquid hardly notices it is being oscillated at all; its flow at 6.6 Hz is its steady flow. The Maxwell liquid’s is not remotely, and the reason is not that it has become less viscous. It is that the gradient is driving a resonance of the liquid itself, and a resonance’s amplitude is limited only by its losses.

The profile at the first resonance is the mode itself, $J_0(j_{01}r/a)$: largest on the axis, zero at the wall, one sign everywhere. At the second resonance the centreline velocity is still 1276 times the steady value, but the flow rate is only 317 times, because the ring beyond the node is moving the other way and cancels most of the core’s flux. A flow meter would read the second resonance as a fifth of the first; a velocimeter on the axis would read them as nearly equal.

The phase says the same thing in time. Below the first resonance the flow leads the gradient by nearly a quarter of a cycle, as a spring’s velocity leads the force on it — the liquid is being bent elastically, not pushed viscously. Through the resonance the phase swings by half a cycle and the flow lags, as a mass’s velocity lags a force above the natural frequency.

The gain grows with the elasticity number #

The peak gain over the Newtonian liquid, computed at the first resonance for elasticity numbers from one to three thousand, grows in proportion to $E$: 56 at $E = 10$, 554 at 100, 5534 at 1000 — about $5.5E$ throughout. The constant is not a fit, and it is worth having because it shows where the gain comes from. Near the first resonance $ka = j_{01} + i\varepsilon$ with $\varepsilon = -1/(2\sqrt E)$, so $J_0(ka) \approx -J_1(j_{01})\,i\varepsilon$ and the Bessel ratio in the flow rate becomes $2J_1/(ka J_0) \approx 2i/(j_{01}\varepsilon)$, of size $4\sqrt E/j_{01}$. The prefactor $8/\alpha^2$ is evaluated where the line $\mathrm{De} = E\alpha^2$ meets $\alpha^2\mathrm{De} = j_{01}^2$, at $\alpha^2 = j_{01}/\sqrt E$. Together,

$$\left|\frac{Q}{Q_s}\right|_{\text{peak}} \approx \frac{8\sqrt E}{j_{01}}\cdot\frac{4\sqrt E}{j_{01}} = \frac{32E}{j_{01}^2} = 5.53\,E.$$

Half of the $E$ comes from the resonance’s sharpness: the shear wave’s loss per crossing falls as $1/\sqrt E$, so the cavity stores more cycles of motion before it gives them up. The other half comes from where the resonance sits: a more elastic liquid reaches it at a smaller Womersley number, where inertia, which is what the factor $1/i\omega\rho$ represents, resists the push less. The same estimate for the $n$-th resonance gives $32E/j_{0n}^2$ — 315 at the second for $E = 300$, against 317 found — so the higher modes are weaker as the square of their order, which is the rule a drum skin obeys too.

A thinner tube, a more viscous liquid or a longer relaxation time all raise $E$, and all raise the gain in proportion. The tube radius enters squared, so the same micelle solution in a tube of two millimetres radius has $E = 7500$ and a first resonance at 33 Hz with a gain of about forty thousand — linear theory’s number, and one the strain limit below will not let a real liquid reach.

The proportionality is a claim about a liquid with one relaxation time. The next figure shows what a spectrum does to it.

Across the tube, not along it #

The arteries in those essays had elasticity too, and it produced waves, but of a different kind. An elastic wall lets the pressure travel along the tube as a longitudinal pulse whose wavelength, at a heart rate, is metres; the tube becomes a transmission line, and its resonances are set by its length and its branching, as the transparent tree showed. An elastic liquid in a rigid tube carries a transverse wave across the flow, from the wall to the axis and back, and the resonance is set by the radius. The first kind of tube rings when its length is a few wavelengths; this kind rings when its radius is a little under half of one.

That difference is why the effect is large here and modest there. A pulse travelling along a tube is fed at one end and loses energy everywhere along its length; a standing shear wave across the tube is fed uniformly, since the pressure gradient pushes every annulus of liquid at once, and loses energy only through the liquid’s own small viscous part. The pressure gradient is the ideal drive for the radial mode — it has the mode’s sign everywhere — which is why the first resonance carries nearly all of the response and the second, whose outer ring the uniform push works against, five times less.

There is a closer relative. A damper that turns into a spring found a gas film that stops resisting as a dashpot and starts resisting as a spring when it is squeezed faster than it can escape, and a drop rings like a bell found a resonance where surface tension supplies the spring. In every case the question a number answers is whether the spring is there at the frequency of the forcing; whether it then rings is a question about what it loses. Bubbly water is the case where the answer is reversed: above their resonance, bubbles make water faster, because a spring driven above its natural frequency stops yielding at all. The micelle solution between its resonances is the other way about: at 12 Hz it still carries 24 times the Newtonian liquid’s flow, because the spring is there at every frequency above $\mathrm{De} \approx 1$ and the resonances are only where it is driven at its own pitch.

A resonance is a rheometer #

Read backwards, the peak is a measurement. Its frequency fixes the shear-wave speed, $\sqrt{G/\rho} = 2\pi f a/j_{01}$, and so the modulus; its height fixes $E$ and so, with the modulus, the relaxation time. For the micelle solution, a peak at 6.63 Hz in a tube 1 cm in radius gives a wave speed of 0.173 m/s and $G = 30$ Pa, the number that went in. A rheometer measures the same modulus by imposing a strain on a thin gap, where inertia is negligible by design; the tube measures it through inertia, which is what makes it a cavity. The two agree for a liquid with one relaxation time, and the disagreement for a real one — a peak lower and broader than the Maxwell fit of its rheometer data predicts — is a measurement of how much spectrum lies between the cycle and the longest time.

A spectrum silences it #

The comparison liquid is the Rouse spectrum of the essay below: forty modes of equal modulus with times $\lambda_1/p^2$, the relaxation of a flexible chain whose $p$-th mode relaxes $1/p$ of the molecule. It is given the same longest time, one second, and the same zero-shear viscosity, 30 Pa·s, so $E$ is the same 300 and the two liquids move along the same line in the plane above. Nothing in the Deborah number tells them apart.

The flow does. The Rouse liquid’s response rises gently to 20.2 at 19.8 Hz and falls; there is no cavity mode to speak of, and its peak grows only as about the square root of $E$, from 5.4 at $E = 10$ to 64 at 3000. The reason is in its loss tangent. Between its longest time and its shortest — here between $\mathrm{De} = 1$ and $1600$ — a Rouse liquid’s two moduli both grow as the square root of frequency, and its loss tangent sits near one. A shear wave in such a material loses $e^{-2\pi\tan(\delta/2)}$ of its amplitude per wavelength, and with $\tan\delta$ near one that is most of it.

So the Rouse liquid at $\mathrm{De} = 100$ is “elastic” by the Deborah number’s measure — its longest time is a hundred times the cycle — and it cannot carry a wave across a tube. The surprise is which side of the comparison is unusual. A Maxwell liquid’s loss tangent falls as $1/\mathrm{De}$ without limit, which no polymer melt does; the wormlike micelle comes close because its chains break and re-form faster than they can reptate, and Cates showed in 1987 that breaking averages every chain’s relaxation into a single exponential. The liquid that rings is the one whose molecules fall apart. The intact polymer, whose spectrum records every length of chain, cannot.

That is the argument the Deborah number could not make. It is a ratio of two times, and the resonance needs a statement about the whole distribution of times between the cycle and the longest one: whether anything is still relaxing at the frequency of the experiment. The rheometer’s measurement found that the spectrum turns a crossover into a band; here the band is where the resonance would have been, and it absorbs it.

Where the model stops #

The theory is linear, and at a resonance that is a strong restriction. The strain in the liquid scales with the velocity over the frequency, and at a gain of sixteen hundred a modest push is a large strain. With the micelle solution in the 1 cm tube, a gradient amplitude of about 100 Pa per metre — a pascal per centimetre — already strains the liquid by order one at the resonance, and wormlike micelles leave the linear range at strains of that order — they shear-thin and band, so that their viscosity depends on the question asked of them; real resonance peaks are therefore lower, and depend on how hard the liquid is driven, in a way this calculation does not contain. The tube is rigid and infinitely long, so there are no end effects and no wall elasticity of the kind the arterial essays needed. The liquid is incompressible and isothermal, and the energy the resonance dissipates is assumed to leave without warming it. And the Maxwell and Rouse liquids are two idealisations at the ends of a range: real micellar solutions depart from a single time at high frequency, where the breaking itself is resolved, and entangled melts have a plateau that neither model contains.

The convention matters to every number quoted. The comparison is at equal zero-shear viscosity, the property a steady flow measures; at equal plateau modulus, or equal longest time and modulus, the gains would differ. The “gain” is the amplitude of the flow rate against a Newtonian liquid at the same frequency, not against the same liquid’s steady flow, though at $\alpha = 0.37$ the two are within a tenth of a per cent.

Who found it #

The Maxwell liquid’s oscillating tube flow follows from Womersley’s in a line and has been written down many times since the 1950s. Its resonance was pointed out as an enhancement of the dynamic response by del Río, López de Haro and Whitaker in 1998, who located the peaks at the zeros of $J_0$ and traced them to the liquid’s elasticity. Castrejón-Pita, del Río and co-workers reported an experimental observation in 2003, with a micellar solution of the kind drawn here, of peaks near the predicted frequencies that a Newtonian liquid in the same apparatus did not show. The shear waves themselves are older: Ferry measured transverse waves in gels in the 1940s and used their speed and damping to read off the modulus, which is the same complex wavenumber this essay’s whole answer depends on.

How it was checked #

With the relaxation time set to zero the flow is Womersley’s, computed independently from his function $F_{10}$, to one part in $10^9$. At $\mathrm{De} = 10^{-6}$ both liquids carry their steady Poiseuille flow to two parts in $10^8$. The profile at a Womersley number of 1.2 and a Deborah number of 40 satisfies the momentum equation, evaluated with the stress from the complex modulus and the Laplacian by central differences, to four parts in $10^8$, and the velocity at the wall is zero to rounding. A one-mode Rouse spectrum reproduces the Maxwell modulus exactly. The first resonance located by a golden-section search on the full solution sits at $\mathrm{De} = 41.652$, against the asymptotic $j_{01}\sqrt E = 41.653$, and the ratio of the gains at $E = 1000$ and $E = 100$ is 9.98. The solver refuses a tube with no oscillation, a negative Deborah number, a liquid with no elasticity, a spectrum with no modes, and a Bessel argument past where its series keeps its digits.

Still open: the resonance a strong push can reach #

The gain here is a linear limit, and at a resonance the linear limit is reached first. The next calculation keeps the tube and the micelle solution and gives the liquid a nonlinear constitutive law that saturates its stress at large strain — a Giesekus or Johnson–Segalman liquid, the two that reproduce a micellar solution’s shear banding — and drives the tube at the resonance with growing gradient amplitudes, to find where the flow rate stops growing in proportion to the push, how far below the linear sixteen hundred the peak is when the strain is of order one, and whether the resonance frequency moves as the liquid softens.