Earth’s Interior, Fluids, Elasticity and Seismic Waves
2. Temperature, electrical and magnetic properties of the interior
Temperature is steep and conductive in the lithosphere, then follows a mantle adiabat of about 0.3 to 0.5 K/km with a hot boundary layer at D″ (see the Heat Flow chapter). Seismic waves do not read temperature directly. It is constrained by the depths at which laboratory phase changes occur (410 and 660 km) and by the melting of iron alloy at the ICB, so the ICB temperature, of order 5000 K, is a melting temperature at about 330 GPa. The CMB lies at several thousand kelvin.
Electrical conductivity σ rises strongly with depth because conduction in minerals is thermally activated, σ = σ₀ exp(−E/kT). Dry crustal rock is a poor conductor, the mantle goes from about 10⁻³ to 10⁻² S/m in the shallow upper mantle to the order of 1 S/m or more in the lower mantle, and the liquid-iron outer core is a metal at 10⁵ to 10⁶ S/m. It is sounded by magnetotellurics and geomagnetic depth sounding, using natural field variations whose penetration is the skin depth δ = √(2/(ωμ₀σ)) ≈ 503 √(ρ/f) metres, with resistivity ρ = 1/σ in Ω·m and frequency f in Hz. Example: ρ = 100 Ω·m and f = 0.01 Hz give 503 × 100 ≈ 50 km, and lower frequencies sound deeper.
Magnetic properties. Ferromagnetic minerals such as magnetite lose their spontaneous magnetisation above the Curie temperature (about 580 °C for magnetite), which is reached at some tens of kilometres depth. The crust can therefore carry remanent magnetisation, producing short-wavelength anomalies, but the mantle cannot. The main geomagnetic field, mostly a dipole of a few tens of µT at the surface, is generated in the outer core by a dynamo, which needs a rotating, convecting, electrically conducting fluid. The conducting mantle screens out the fastest field variations.
3. Conservation laws, Newton’s law of viscosity, laminar and turbulent flow
Conservation laws underlie continuum mechanics. Mass: ∂ρ/∂t + ∇·(ρv) = 0, which for an incompressible fluid is ∇·v = 0. Momentum: ρ Dv/Dt = ∇·σ + ρg, which for a viscous incompressible fluid is the Navier–Stokes equation ρ Dv/Dt = −∇p + η∇²v + ρg. Energy: the first law, which gives the heat equation. For a solid with small displacement u the momentum law becomes ρ ∂²u/∂t² = ∇·σ + f, the starting point of seismic waves.
Newton’s law of viscosity says shear stress is proportional to the velocity gradient, τ = η dv/dy, where the dynamic viscosity η is in Pa·s (water about 10⁻³, the mantle 10²⁰ to 10²²). The kinematic viscosity is ν = η/ρ. Worked example. A film 2 mm thick between a fixed plate and a plate moving at 0.5 m/s, with η = 0.1 Pa·s, carries τ = 0.1 × 0.5 / 0.002 = 25 Pa. Mantle creep is not strictly Newtonian (strain rate rises faster than stress), so geophysicists quote an effective viscosity.
- Laminar flow moves in smooth layers. Couette flow between a fixed and a moving plate has a linear velocity profile and uniform shear stress ηU/h. Poiseuille flow in a pipe of radius R has a parabolic profile v(r) = (ΔP/4ηL)(R² − r²), a mean speed equal to half the maximum and discharge Q = πR⁴ΔP/(8ηL), so doubling the radius raises Q sixteen-fold. Between parallel walls a gap h gives Q = h³ΔP/(12ηL) per unit width, one reason wide dykes carry magma so well.
- Turbulent flow is chaotic, with eddies that carry momentum and heat far more effectively than molecular diffusion, a flatter mean velocity profile and drag that grows roughly as v².
The Reynolds number Re = ρ v L/η = vL/ν is the ratio of inertial to viscous forces. In a pipe, flow is laminar below Re of about 2000 and turbulent above about 4000, with a transitional range between. Lava of ρ = 2700 kg/m³, speed 0.5 m/s, depth 2 m and η = 100 Pa·s has Re = 27, laminar; water at 0.5 m/s in a 0.1 m pipe has Re = 5×10⁴, turbulent. In the mantle Re is about 10⁻²⁰, so inertia is negligible and the flow is a slow, creeping one; in the outer core Re is enormous.
4. Elasticity: stress and strain tensors and the generalised Hooke’s law
Stress at a point is a symmetric second-rank tensor σ_ij. The traction on a surface of unit normal n is t_i = σ_ij n_j. Diagonal terms are normal stresses and off-diagonal terms are shears; σ_ij = σ_ji (rotational equilibrium) leaves six independent components, and three principal stresses act on axes where the shear vanishes. Seismology counts tension positive, whereas much of geology counts compression positive, so check the convention. Strain for small displacement u is ε_ij = ½(∂u_i/∂x_j + ∂u_j/∂x_i), again symmetric with six components. Its trace ε_kk = ∇·u is the volume change per unit volume (the dilatation), and the antisymmetric part ½(∂u_i/∂x_j − ∂u_j/∂x_i) is a rigid rotation that produces no stress.
The generalised Hooke’s law is linear: σ_ij = c_ijkl ε_kl. Its 81 constants fall to 36 because σ and ε are symmetric, and to 21 because a strain-energy function exists, which is the most general anisotropic solid. Transverse isotropy (aligned olivine, fine layering) needs 5, and an isotropic solid needs only 2: σ_ij = λ δ_ij ε_kk + 2μ ε_ij, with Lamé constant λ and shear modulus (rigidity) μ. In a fluid μ = 0 and σ_ij = −p δ_ij.
| Quantity | Meaning | In terms of λ and μ |
|---|---|---|
| Shear modulus μ | Shear stress ÷ shear strain | μ |
| Bulk modulus K | Pressure ÷ fractional volume change | λ + 2μ/3 |
| Young’s modulus E | Axial stress ÷ axial strain | μ(3λ + 2μ)/(λ + μ) |
| Poisson’s ratio ν | − lateral strain ÷ axial strain | λ / (2(λ + μ)) |
Poisson’s ratio lies between 0 and 0.5 for ordinary materials (0.5 for an incompressible fluid) and between about 0.2 and 0.3 for most rocks. A Poisson solid has λ = μ and ν = 0.25. Worked example. λ = 20 GPa and μ = 15 GPa give K = 20 + 10 = 30 GPa, E = 15 × 90 / 35 = 38.6 GPa and ν = 20/70 = 0.286.
5. The wave equation, body and surface waves and the four names
Putting Hooke’s law into the momentum law for a homogeneous isotropic medium without body force gives the elastic wave equation ρ ∂²u/∂t² = (λ + 2μ) ∇(∇·u) − μ ∇×(∇×u). Writing u = ∇φ + ∇×ψ (the Helmholtz decomposition, with scalar potential φ and vector potential ψ) splits it into two wave equations, ∇²φ = (1/α²) ∂²φ/∂t² and ∇²ψ = (1/β²) ∂²ψ/∂t², where α = √((λ + 2μ)/ρ) and β = √(μ/ρ). A plane harmonic wave u = A exp[i(k·x − ωt)] has phase speed c = ω/k and wavelength c/f; a 5 Hz wave at 5 km/s has a wavelength of 1 km.
- P wave: u = ∇φ, so ∇×u = 0 (irrotational) and ∇·u ≠ 0 (dilatational, it changes volume). Motion is along the propagation direction (longitudinal, compressional). It is the faster wave and travels in solids and fluids.
- S wave: u = ∇×ψ, so ∇·u = 0 (equivoluminal, no volume change) and ∇×u ≠ 0 (rotational). Motion is transverse, resolved into SV (in the vertical plane of propagation) and SH (horizontal). It cannot travel in a fluid because μ = 0.
| Wave | Particle motion | Speed | Notes |
|---|---|---|---|
| P | Along the ray | α = √((K + 4μ/3)/ρ) | Fastest; solids and fluids |
| S | Perpendicular to the ray: SV, SH | β = √(μ/ρ) | Solids only |
| Rayleigh | Retrograde ellipse in the vertical plane | About 0.92 β | Free surface; confined to about a wavelength |
| Love | Horizontal, transverse (SH type) | Between β of layer and of half-space | Needs a slow surface layer |
6. Wave speeds, moduli and Poisson’s ratio
The key relations are Vp = √((λ + 2μ)/ρ) = √((K + 4μ/3)/ρ) and Vs = √(μ/ρ), so (Vp/Vs)² = (λ + 2μ)/μ = 2(1 − ν)/(1 − 2ν) and ν = (Vp² − 2Vs²) / (2(Vp² − Vs²)). For a Poisson solid (λ = μ, ν = 0.25) Vp/Vs = √3 ≈ 1.73. For a fluid μ = 0, so Vs = 0, ν = 0.5 and Vp = √(K/ρ). Because K > 0, Vp must exceed (2/√3) Vs ≈ 1.155 Vs in any solid, and crustal rocks have Vp/Vs of about 1.7 to 1.8.
Worked example. ρ = 2500 kg/m³, λ = 20 GPa, μ = 15 GPa. Vp = √(50×10⁹/2500) = 4472 m/s ≈ 4.47 km/s, Vs = √(15×10⁹/2500) = 2449 m/s ≈ 2.45 km/s, so Vp/Vs = 1.83. Check: ν = (20 − 12)/(2 × (20 − 6)) = 0.286, which equals λ/(2(λ + μ)) = 20/70. If only the ratio Vp/Vs = 1.8 is known, ν = (3.24 − 2)/(2 × 2.24) = 0.277.
Key takeaways
- The Earth has a crust, a mantle with jumps at about 410 and 660 km, a liquid outer core (Vs = 0, Vp falls from about 13.7 to 8 km/s at the CMB) and a solid inner core. Pressure follows dP/dz = ρg (about 136 GPa at the CMB) and density gradients follow dρ/dz = ρg/φ for a homogeneous adiabatic layer.
- Electrical conductivity rises with depth (mantle about 1 S/m, outer core 10⁵ to 10⁶ S/m) and sets the skin depth 503√(ρ/f) m. Magnetisation needs temperatures below the Curie point, so only the crust is magnetised and the main field comes from the outer-core dynamo.
- Newton’s law of viscosity is τ = η dv/dy. Poiseuille discharge is πR⁴ΔP/(8ηL), and Re = ρvL/η separates laminar flow (below about 2000 in a pipe) from turbulent flow (above about 4000); the mantle has Re of about 10⁻²⁰.
- Stress and strain are symmetric tensors with six components each; a general anisotropic solid has 21 elastic constants, an isotropic one has 2 (λ, μ), and ε_kk = ∇·u is the dilatation. K = λ + 2μ/3, E = μ(3λ + 2μ)/(λ + μ), ν = λ/(2(λ + μ)).
- The wave equation splits into P (dilatational, irrotational, α = √((λ + 2μ)/ρ)) and S (rotational, equivoluminal, β = √(μ/ρ), solids only). Rayleigh waves are about 0.92β and Love waves need a slow surface layer; surface waves fall off as 1/√r, body waves as 1/r.
- ν = (Vp² − 2Vs²)/(2(Vp² − Vs²)); a Poisson solid has λ = μ, ν = 0.25 and Vp/Vs = √3; a fluid has ν = 0.5 and Vs = 0. With moduli in GPa and density in g/cm³ the speeds come out in km/s.
Practice questions (17)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
Which observation gives direct evidence that the outer core is liquid?
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Answer: A — S waves from earthquakes are not transmitted through it, because a liquid has no shear modulus
A liquid has μ = 0 and so cannot carry shear waves; S waves vanish at the CMB. A drop in P speed, a large density jump or high conductivity can all occur in a solid with a different composition or temperature, so none of them proves liquidity by itself. The P-speed drop is the tempting option because it is what the CMB looks like on a velocity profile.The Adams–Williamson equation dρ/dz = ρg/φ describes the density rise with depth poorly across the 410 and 660 km discontinuities, mainly because it assumes
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Answer: C — a homogeneous, well-mixed adiabatic layer with no change of phase or composition
The equation integrates a compression-only density gradient through a layer of uniform composition and phase. A phase change such as olivine to a denser spinel-like form adds extra density that compression alone cannot give. The seismic parameter φ already contains Vs, and g enters the equation explicitly, so those are not the failing assumptions.A film of oil 2 mm thick (η = 0.1 Pa·s) lies between a fixed plate and a plate moving at 0.5 m/s, and the velocity varies linearly across the film. What is the shear stress in the film?
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Answer: D — 25 Pa
By Newton’s law τ = η U/h = 0.1 × 0.5 / 0.002 = 25 Pa. Using h = 2 without converting millimetres to metres gives the tempting 0.025 Pa, and the other values come from further slips of the decimal point.Mantle convection has a Reynolds number of about 10⁻²⁰. What does this imply?
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Answer: B — Inertial forces are negligible beside viscous forces, so the flow is a slow, creeping, laminar one
Re = ρvL/η is the ratio of inertial to viscous forces; a value of 10⁻²⁰ means viscosity dominates completely and the flow is creeping and laminar. A high viscosity lowers Re and so suppresses turbulence rather than causing it, and a small Re comes from the very slow speed (cm per year) together with the huge η.How many independent elastic constants describe the most general anisotropic elastic solid and an isotropic elastic solid, respectively?
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Answer: D — 21 and 2
The 81 components of c_ijkl fall to 36 by the symmetry of stress and strain and to 21 because a strain-energy function exists. Isotropy leaves two, λ and μ. The figure 36 omits the energy argument, and a third constant is not needed because E and ν are fixed by λ and μ.A seismic wave whose displacement field has zero curl but non-zero divergence is
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Answer: A — a P wave, which is irrotational and dilatational
Zero curl means irrotational and non-zero divergence means a change of volume (dilatational); that is the P wave, u = ∇φ. SV and SH waves are rotational and equivoluminal, and a Love wave is built from SH motion.For a Poisson solid, in which the Lamé constants satisfy λ = μ, which pair of values is correct?
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Answer: C — Vp/Vs = √3 and ν = 0.25
With λ = μ, (Vp/Vs)² = (λ + 2μ)/μ = 3, so Vp/Vs = √3, and ν = λ/(2(λ + μ)) = 1/4. The pair 2 and 0.33 is self-consistent but belongs to λ = 2μ, and ν = 0.5 would mean a fluid with Vs = 0.Why can the main geomagnetic field not be due to magnetised mantle rock?
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Answer: B — Temperatures tens of kilometres down exceed the Curie points of magnetic minerals, so the rock cannot stay magnetised
Ferromagnetic minerals lose spontaneous magnetisation above their Curie temperature (about 580 °C for magnetite), reached at some tens of kilometres depth. The field is therefore made by a dynamo in the liquid outer core, and only the crust carries remanent magnetisation. The mantle is neither a perfect conductor nor does it cancel the field.Going downward across the core–mantle boundary, which set of changes is correct?
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Answer: C — Vp falls from about 13.7 to about 8 km/s, Vs falls to zero and density jumps from about 5.6 to about 9.9 g/cm³
At the CMB Vp drops from about 13.7 to about 8 km/s, Vs goes to zero, and density increases from about 5.6 to about 9.9 g/cm³ because the iron-rich core is far denser than the silicate mantle. The statement that density falls confuses liquid with light, and Vs cannot stay at 7 km/s in a liquid.Which of the following statements about stress, strain and elastic constants are correct?
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Answer: A — The stress tensor is symmetric, so it has six independent components; B — The trace of the small-strain tensor, ε_kk = ∇·u, is the volume change per unit volume; C — An isotropic elastic solid is described completely by two independent constants such as λ and μ
Symmetry of stress, the meaning of the strain trace and the two-constant isotropic law are all standard. An ideal liquid has μ = 0, so it cannot support shear waves at all; S waves do not travel through the outer core.Which of the following statements about P and S waves are correct?
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Answer: A — In the same solid, P waves travel faster than S waves; B — S waves cannot propagate through a fluid because the shear modulus of a fluid is zero; D — S waves are equivoluminal and rotational, with motion perpendicular to the propagation direction
Vp exceeds Vs in any solid, S waves need shear rigidity, and S motion is transverse, rotational and equivoluminal. P waves are the opposite of equivoluminal: they compress and dilate the material.Which of the following statements about the interior of the Earth are correct?
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Answer: A — Electrical conductivity generally increases with depth through the mantle; B — The electrical conductivity of the liquid-iron outer core is many orders of magnitude greater than that of the mantle; C — The mantle transmits S waves like an elastic solid but flows like a very viscous fluid over millions of years
Conductivity rises with temperature and depth and the core is a metal, while the mantle shows elastic behaviour on seismic time scales and viscous creep over geological ones. Mantle viscosity is of order 10²⁰ to 10²² Pa·s, not 10³.A rock has density 2500 kg/m³ and Lamé constants λ = 20 GPa and μ = 15 GPa. Find its P-wave speed in km/s, to two decimal places.
Numerical answer — type the value.
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Answer: 4.47
Vp = √((λ + 2μ)/ρ) = √((20 + 30) × 10⁹ / 2500) = √(2.0 × 10⁷) = 4472 m/s, which is 4.47 km/s. Using λ + μ instead of λ + 2μ gives 3.74 km/s, and forgetting the GPa conversion gives an absurdly small speed.In a crustal layer the ratio of P to S wave speed is Vp/Vs = 1.8. Find Poisson’s ratio, to two decimal places.
Numerical answer — type the value.
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Answer: 0.28
ν = (Vp² − 2Vs²) / (2(Vp² − Vs²)). With Vs = 1 and Vp = 1.8, Vp² = 3.24, so ν = (3.24 − 2) / (2 × 2.24) = 1.24 / 4.48 = 0.2768, which is 0.28. A Poisson solid (Vp/Vs = 1.73) would give 0.25.Basaltic lava of density 2700 kg/m³ and viscosity 100 Pa·s flows at 0.5 m/s in a channel 2 m deep. Take the flow depth as the length scale. Find the Reynolds number.
Numerical answer — type the value.
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Answer: 27
Re = ρ v L / η = 2700 × 0.5 × 2 / 100 = 27. This is far below the pipe-flow threshold of about 2000, so the flow is laminar. Using the viscosity in poise or the depth in centimetres changes the answer by powers of ten.In a homogeneous adiabatic layer of the upper mantle, ρ = 3300 kg/m³, g = 9.9 m/s², Vp = 8.5 km/s and Vs = 4.7 km/s. Use the Adams–Williamson equation dρ/dz = ρg/φ with φ = Vp² − (4/3)Vs². Find dρ/dz in kg/m³ per km of depth, to two decimal places.
Numerical answer — type the value.
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Answer: 0.76
φ = 8.5² − (4/3) × 4.7² = 72.25 − 29.45 = 42.80 km²/s² = 4.280 × 10⁷ m²/s². dρ/dz = 3300 × 9.9 / 4.280 × 10⁷ = 7.63 × 10⁻⁴ kg/m⁴, which is 0.76 kg/m³ per km. Using Vp² alone for φ would give 0.45, and leaving φ in km²/s² gives a result 10⁶ times too large.Natural magnetotelluric fields of frequency 0.01 Hz sound a region of resistivity 100 Ω·m. Use the skin depth δ ≈ 503 √(ρ/f) metres, with ρ in Ω·m and f in Hz. Find δ in km, to the nearest km.
Numerical answer — type the value.
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Answer: 50
δ = 503 × √(100 / 0.01) = 503 × √10⁴ = 503 × 100 = 50 300 m, which is about 50 km. Quoting it in metres (50300) or using the conductivity in place of the resistivity are the usual slips.