IFST · Parabolic ↔ Hyperbolic · Mechanics

A Maxwell body: fluid or solid?

τ ∂²u∂t² + ∂u∂t = ν ∂²u∂x²

τ = η/μ (Maxwell time) τ = 1.00

Time t t = 3.00

Newtonian fluid reference (τ = 0)

TypeHYPERBOLIC

b² − 4ac = 4ντ4.00

Shear-wave speed c = √(μ/ρ) = √(ν/τ)1.000

Regime at this tt/τ = 4.0

Shear velocity u(x, t)layer of length 40 between fixed walls (no slip, u = 0)

Maxwell body Newtonian fluid (τ = 0) fronts x₀ ± (w + ct)

Space–time mapcolour ∝ √|u|, so weak tails show

characteristics dx/dt = ±c current t

τ = 0: a Newtonian fluid

With a rigid spring only the dashpot is left. Momentum diffuses with kinematic viscosity ν = η/ρ, as in Stokes’ problems for a plate set in motion. The initial pulse is zero for |x − x₀| > w, yet at t = 0⁺ the whole layer moves.

τ > 0: elastic shear waves

A finite μ lets the body store elastic stress. Two real characteristics dx/dt = ±c appear, and outside the cone |x − x₀| < w + ct the layer has not moved yet. Since the initial stress is zero, u_t(x,0) = 0 and the pulse splits into two halves.

t ≫ τ: the solid flows

Stress relaxes as e^−t/τ and the waves damp as e^−t/2τ. Over times long compared with the Maxwell time the body behaves as a viscous fluid. This is how the mantle carries S waves and still flows during postglacial rebound.

How to use it in class

One knob, τ = η/μ, takes the same material from a Newtonian fluid (parabolic) to a nearly elastic solid (hyperbolic). The viscosity, the initial velocity pulse and the walls never change; only the stiffness of the spring does. The Modes tab follows one wavelength as a single oscillator. Each “Show this” button sets the demo to the state described and switches back to it.

Start with a Newtonian fluid

At τ = 0 the spring is rigid and only the dashpot remains. A sudden shear-velocity pulse spreads by viscous diffusion and is felt immediately across the whole layer.
Make the spring finite but stiff

At τ = 0.1 the type becomes hyperbolic and shear waves appear with c = √(ν/τ) ≈ 3.2, yet at t = 3 the yellow curve sits almost on the Newtonian one. The material is a solid on paper and a fluid in practice.
Watch a solid

At τ = 10, press Play from t = 0. The whole window is short compared with the Maxwell time, so the pulse splits into two shear waves at c ≈ 0.32 that damp slowly as e^−t/2τ. The Newtonian reference has already spread everywhere.
Reflections at the walls

At τ = 0.1 the shear waves reach the fixed walls near t ≈ 6 and come back with inverted velocity, as on a string with fixed ends.
Wait longer than the Maxwell time

Set τ = 1 and drag t well past τ. The waves die out and the profile settles onto the Newtonian one: given enough time, the solid flows.

Points to draw out

Type is about characteristics. The discriminant b² − 4ac = 4ντ jumps from 0 to positive as soon as τ > 0, while the solution changes continuously.
Time scale decides, not the material. A disturbance of frequency ω sees an elastic solid when ωτ ≫ 1 and a viscous fluid when ωτ ≪ 1. The same rock is a solid for a 10 s S wave and a fluid for ice-age loading.
Earth numbers. With η ≈ 10²¹ Pa·s and μ ≈ 7 × 10¹⁰ Pa, τ = η/μ ≈ 1.4 × 10¹⁰ s, roughly 450 years. Seismic periods are shorter by a factor of 10⁸ to 10¹⁰; postglacial rebound runs over thousands of years.
A caveat on attenuation. A pure Maxwell body has Q ≈ ωτ when ωτ ≫ 1, about 10¹⁰ for a 10 s wave with the numbers above. The observed mantle Q of a few hundred needs other mechanisms, which is why Burgers and Andrade models are used for seismic attenuation.
Reading the panels. Left: the Maxwell body (yellow), the Newtonian fluid (blue dashed) and the fronts x₀ ± (w + ct) (coral). Right: u over (x, t), coloured by √|u|, with the characteristics dx/dt = ±c.

Where the Maxwell body comes from

A spring μ and a dashpot η in series carry the same stress σ, and their strain rates add: ε̇ = σ̇/μ + σ/η. For a shear wave along x, with particle velocity u:

ρ∂u∂t=∂σ∂x, ∂σ∂t+στ=μ∂u∂x

Eliminating σ gives exactly the equation above, with

τ = η/μν = η/ρc = √(μ/ρ)

So the slider holds the viscosity fixed and changes the stiffness of the spring.

1851
G. G. Stokes, “On the effect of the internal friction of fluids on the motion of pendulums”. The fluid next to a moving plane is set in motion by viscous diffusion with ν = η/ρ. This is τ = 0.
1867
J. C. Maxwell, “On the dynamical theory of gases”. Viscosity is elastic stress that relaxes in a time τ, so η = μτ. The same paper has a relaxation term for the heat flux (see the heat page).
1890s
Kelvin and Voigt use the other basic arrangement, spring and dashpot in parallel. It damps waves but never flows.
1935
N. A. Haskell estimates a mantle viscosity of about 10²¹ Pa·s from the postglacial uplift of Fennoscandia.
1946
J. Frenkel, Kinetic Theory of Liquids: a liquid supports shear waves at frequencies above about 1/τ.
1974
W. R. Peltier, “The impulse response of a Maxwell Earth”. Maxwell rheology becomes the standard model for glacial isostatic adjustment.

One wavelength, one oscillator

Each sine mode u = A(t) sin(qx) of the layer obeys τ Ä + Ȧ + ν q² A = 0. That is a unit mass on a Maxwell element whose spring and dashpot both scale with q², so τ is the same for every mode and only ω₀ = c q changes. Pick a wavelength and watch the mass after a kick (A(0) = 1, no initial force).

Wavelength λ = 2L/n n = 10 · λ = 8.00

τ = η/μ (Maxwell time) τ = 1.00

Same τ as the Demo tab.

This modeUNDERDAMPED

ω₀τ = q√(ντ)0.79

Critical λc = 4π√(ντ)12.6

Rates (1/time)–

Dispersion, log–logω = −i s for decaying modes

Re ω decay rates s ω = cq Newtonian νq² q_c

Mode amplitude after a kicktime window adapts to the mode

velocity A(t) Newtonian e^−νq²t displacement ∫A dt (scaled)

The mode as a mass on a Maxwell elementk_eff, c_eff ∝ q²

start final position, the same for every τ

Mode shape in the layeru(x, t) = A(t) sin(qx), walls at x = 0 and 40

u(x, t)

From the continuity equation

	Heat	Momentum (shear)	Cable
Conserved density	ρC_pT	ρu	CV
Flux	Q	−σ	I
Continuity	ρC_p ∂T/∂t + ∂Q/∂z = 0	ρ ∂u/∂t − ∂σ/∂x = 0	C ∂V/∂t + ∂I/∂x = 0
Instant flux law	Q = −k ∂T/∂z (Fourier)	σ = η ∂u/∂x (Newton)	RI = −∂V/∂x (Ohm)
Relaxing flux law	τ ∂Q/∂t + Q = −k ∂T/∂z (Cattaneo)	τ ∂σ/∂t + σ = η ∂u/∂x (Maxwell)	(L/R) ∂I/∂t + I = −(1/R) ∂V/∂x
Diffusivity	α = k/(ρC_p)	ν = η/ρ	1/(RC)
Front speed	√(α/τ)	√(μ/ρ) = √(ν/τ)	1/√(LC)

Here τ is the relaxation time and σ the shear stress. On the continuity slides τ denotes the stress tensor and σ a source term.

Two derivatives, two laws. Continuity supplies one ∂/∂x: the net flux through the two faces of the element. The flux law supplies the other: the flux depends on a gradient. Continuity plus an instant flux law gives diffusion; continuity plus a relaxing flux law gives the telegraph equation.
The same τ for every mode. On a chain of masses joined by Maxwell links, mode q feels k_eff = 4k sin²(qa/2) and c_eff = 4c sin²(qa/2). The factor is common, so τ = c/k does not depend on q; only ω₀ = c q does.
Short waves see a solid, long waves a fluid. Modes with λ < λ_c = 4π√(ντ) oscillate and propagate; longer ones only creep. At τ = 0, λ_c = 0 and every mode creeps.
The dashpot always wins in the end. Integrating the mode equation gives ∫A dt = 1/(νq²) for any τ. After the kick the mass ends at the same place; τ only changes the path, with overshoot and ringing when the mode is underdamped.

Model: ν = 1, layer 0 ≤ x ≤ 40 with u = 0 at both walls, initial velocity pulse u = cos²(π(x − x₀)/2w) for |x − x₀| < w (x₀ = 20, w = 1) and zero elsewhere, no initial stress (σ = 0, hence u_t(x,0) = 0). Exact solution by expansion in 400 sine modes; each mode is a damped oscillator τ a″ + a′ + ν k² a = 0. Keys: Space play/pause, ←→ on a slider to step.

τ = 0: a Newtonian fluid

τ > 0: elastic shear waves

t ≫ τ: the solid flows

How to use it in class

Start with a Newtonian fluid

Make the spring finite but stiff

Watch a solid

Reflections at the walls

Wait longer than the Maxwell time

Points to draw out

Where the Maxwell body comes from

One wavelength, one oscillator

From the continuity equation