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Today’s topics:
- Relationship between stress and strain
- Equation of motion
- Wave Equation
Review from Last Lecture
The stress tensor is symmetric and linearly related to traction T: (1) σ = σij = σji = Ti(j) (2) Ti = σijn (^) j
For all Ti , there are 3 corresponding stress elements on the i-surface working along all three axes:
T
1
3
2
σ
σ (^) σ σ
σ
σ
σ
σ
σ
n
Stress components on the faces of a tetrahedron.
dS
dS
dS
dS
22
23 32 31
33
13
11
12
21
Figure by MIT OCW.
Adapted from Stein & Wysession (2003), An Introduction to Seismology, Earthquakes, and Earth Structure) , p. 40, Blackwell Publishing.
The strain tensor ε is also symmetric and can be written as the gradient of the displacement: (3) e = e ij = e i = ½ (δui /δu (^) j + δu (^) j/δui ) The trace of the strain tensor is the relative change in volume, or the cubic dilatation : (4) Tr( e ) = δui /δxj = V.^ u = ∆V/V
Relating Stress and Strain
To relate stress and strain, we are going to assume a medium with perfect linear elasticity.
Aside: If we had a viscous medium, we could relate the stress and strain rate using viscosity μ: ( 5 ) σ ∼ μ ė , where the strain rate Ė is given by (6) ė = δ e /δt
In one dimension, stress and strain can be related using Young’s Modulus E is a proportionality constant: (7) σ = E e , where Young’s Modulus is equal to the ratio between σx and E x : (8) E = σx / e x
σx e x :
From σx and e x
σx σx ½ e x
y
e.g. Take a rubber band and apply a stress in the -direction to get
, E can be found.
½ e
Note, however, that e y ≠0 even though σy =0; therefore, this simple skalar relationship does not hold in 2-D or 3-D. Instead, we need to be able to relate elements of the stress tensor to (combinations of) elements of the strain tensor. Because the stress tensor and the strain tensor are both 2nd-rank, we will need a 4th-rank tensor to do so, which gives us the generalized Hooke’s Law , (9) σij = C (^) ijkl e kl , also called the constitutive relationship between stress and strain, where Cijkl is known as the stiffness tensor or the elasticity tensor. Note that the generalized Hooke’s Law assumes perfect linear elasticity, so there are no effects from attenuation, etc.
Using equation (13), Young’s modulus can be written in terms of Lamé’s parameters for an isotropic medium: (17) E = σxx / E xx = (3λ+ 2 μ)μ / (λ+μ) Additionally, we often see Poisson’s ratio ν, where ( 18 ) ν = λ / 2 (λ+μ) Poisson’s ratio is often used to characterize the elastic properties of a medium, for example, for a fluid with μ=0, ν=0.5. As the rigidity of a material increases to infinity, Poisson’s ratio approaches 0.
A Poisson’s medium is an isotropic material with Lamé parameters such that λ = μ, giving ν=0.25. This value of Poisson’s ratio is reasonable for many crustal and mantle rocks, so it is often assumed in calculations. In the inner core, ν = 0.4, suggesting that the inner core is more “mushy” and sponge-like than the mantle, but still maintains some rigidity.
When considering P- and S-waves in the crust and mantle, we can assume a Poisson’s medium in order to relate them, such that (19) V (^) p ≅ Vs√ 3
Now, expanding σij for an isotropic medium with perfect linear elasticity, we get ┌ λ∆ + 2 μ e 11 2 μ e 12 2 μ e 13 ┐ (20) σij = C (^) ijkl e kl = │ 2 μ e 12 λ∆ + 2 μ e 22 2 μ e 23 │ └ 2 μ e 13 2 μ e 23 λ∆ + 2 μ e 33 ┘ Notice that the off-diagonals are pure shear stresses, and they are only dependent on μ. The diagonals refer to the normal stress and depend on both μ, λ, and ∆ (i.e, change in volume).
Aside: Generic anisotropy takes us back to 21 unknowns in Cijkl , so we have to make assumptions about the symmetry of the medium. For instance, in an olivine-rich medium, such as the mantle, the olivine crystals tend to align themselves in a constant direction. Thus, seismic waves will propagate faster along the crystal alignment than in other directions, creating anisotropy and necessitating 5 independent elements in Cijkl (hexagonal symmetry; transverse isotropy).
Equation of Motion
Let us revisit the stress tetrahedron. We have a traction T that can be broken up into 3 components, (T 1 , T 2 , T 3 ). We also know, using Newton’s 2nd^ Law of Motion and a force balance on the tetrahedron, that (21) ∑F = TiδS – (σi1 n 1 δS + σi2 n 2 δS + σi3 n 3 δS) + f i dV = ma = ρ (δ^2 ui /δt^2 ) dV where f i dV represents the body forces on the tetrahedron. Ignoring the body forces and assuming a=0, this gives us (22) Ti = σijn (^) j, which is true for pure equilibrium.
Now, consider an accelerating seismic wave. The equation of motion will be (23) (Ti - σijn (^) j)δS + f i dV = ρ (δ^2 u (^) i /δt^2 ) dV If the traction cancel, i.e., Ti - σijn (^) j , equation (22) would give (24) f i = ρ (δ^2 u (^) i /δt^2 ), that is acceleration would only be due to the body forces; but if there is a net change in stress, (Ti - σijn (^) j) can be considered as the non-lithostatic (deviatoric) stress, σij’. Therefore, (25) σij’u (^) jδS + f i dV = ρ (δ^2 ui /δt^2 ) dV From now on, σij will be used to refer to the deviatoric stress, σij’.
Now, to develop equation (25) further we want to get rid of either δS or dV. We will use Gauss’ Divergence Theorem to transform a surface integral into a volume integral. Gauss’ theorem uses flux to relate volume to surface area.
Consider a field a with a flux through a surface with area δS:
δS = ňdS
ň
a
The total flux of the field in and out of the surface is given by (26) ∫ a.dS = ∫ai n (^) i dS which is related to the amount of field generated or absorbed in the volume within dS. In other words (27) (^) S∫ai ni dS = ∫V.^ a dV, where V.^ a is the source (or sink) of the field. This is Gauss’ Divergence Theorem. If V.^ a = 0, the field is source/sink-free.
So, taking equation (25) and integrating both sides, we get (28) ∫σiju (^) jδS + ∫ f i dV = ∫ρ (δ^2 ui /δt^2 ) dV Now we apply Gauss’ Theorem and combine terms to get (29) ∫(δσij/δx (^) j + f i )dV = ∫ρ (δ^2 u (^) i /δt^2 ) dV, which leads to (30) ρ (δ^2 ui /δt^2 ) = δσij/δx (^) j + f i = σij, j + f i (Stein and Wysession not.) = δjσij + f i ( van der Hilst not.)
In vector form, we get the equation of motion,
(31) ρ ü = f + V.^ σ
Equation (30 or 31) is known as either Navier’s or Cauchy’s Equation of Motion , as they independently worked on developing it in the 19th^ century.
From equations (33) and (35), we can show that the 2nd^ derivative in time is balanced by the 2nd derivative in space, so that it begins to look like a wave equation: (36) ρ(δ^2 ui /δt^2 ) = (λ+ 2 μ)δ^2 uk /(δxjδxk ) + μV^2 ui Dividing both sides by ρ, we obtain the equation of motion under assumptions of a homogeneous, isotropic, linear perfect elastic medium, with no body forces, and undergoing infinitesimal strain:
(37) δ^2 ui /δt^2 = [(λ+ 2 μ)/ρ]δ^2 u (^) k /(δxjδxk ) + (μ/ρ)V^2 ui , or (38) ü = [(λ+ 2 μ)/ρ]V(V.^ u ) + (μ/ρ)V^2 u
Note that the general form of the wave equation is δ^2 ☺/δt^2 =c^2 V^2 ☺with c the propagation speed of disturbance☺, so that we can see that the terms (μ/ρ) and (λ+ 2 μ)/ρ are related to wave speed.
Some useful vector identities: The divergence of a curl is zero: (39) V.^ (Vx u ) = 0 The curl of a gradient is zero: (40) Vx (VΦ) = 0 The Laplacian of a vector u is given by (41) V^2 u = V(V.^ u ) - Vx (Vx u ) (NB V(V.^ u ) = V^2 u only if u is a conservative, rotation free field).
Combining equations (38) and (41), we get (42) ü = [(λ+ 2 μ)/ρ]V(V.^ u ) - (μ/ρ)[Vx (Vx u )]
Some remarks on equation (42): Remember that in an isotropic medium such that (43) σij = λδij∆ + 2μ E ij, the diagonals of σij have terms in both ∆ and μ. The off-diagonals are only in terms of μ. The same thing occurs with (μ/ρ)[Vx (Vx u )].
Also, V.^ u = ∆ = e kk , which shows that in a general elastic medium an accelerating system has a volume change (e.g. P-wave) component as well as a rotation (curl) component with no volume change.
Equation (43) can be separated into a pure volume change component (λδij∆) and a pure shear component (2μ E ij). However, λ depends on both κ and λ; that is, a pure volume change cannot occur without a change in shear because volume change is nonzero only when i=j, and when i=j, there still exists a shear component in equation (43). Thus, there must always be slip somewhere to account for a volume change in the system.
Notes: Lori Eich (Feb. 14, 2005)
