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Plane elastic waves in the adopted model: the dispersion condition and the longitudinal and transverse speeds

Physical scope

Conditional on the small-strain isotropic linear elastic branch: a plane displacement field a cos(<k,x> - omega t) satisfies the model's displacement equation with zero body force on all of space-time if and only if its dispersion relation holds; only solutions of that equation have the stated longitudinal or transverse phase speeds, which are real and positive under the branch's positivity hypotheses. The theorem asserts no completeness of plane waves, no measured modulus and no empirical value.

Statement

Adopt the small-displacement isotropic linear elastic branch of Adopted isotropic linear elastic constitutive model with its small-strain regime and strain energy with reference density ρ0>0 of dimension μℓ−3 and Lamé parameters λ,μs of dimension μℓ−1τ−2, satisfying μs>0 and λ+2μs3>0 (Continuum body data, the fixed inertial frame and the dimensions of the fields). Let (E) be the displacement equation of The displacement equation of the linear elastic branch: rho_0 partial_t^2 xi = rho_0 b + mu_s Laplacian xi + (lambda + mu_s) grad div xi with body force b≡0. Let x∈R3 denote position of dimension ℓ and t∈R time of dimension τ. Let k∈R3∖{0} be a wave vector of dimension ℓ−1, a∈R3 a polarization amplitude of dimension ℓ, and ω∈R∖{0} an angular frequency of dimension τ−1. The phase is dimensionless. Define the plane displacement field

ξ(t,x):=acos⁡(⟨k,x⟩2−ωt),

with the Euclidean inner product ⟨⋅,⋅⟩2 (The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn). Then ξ satisfies (E) on all of R×R3 if and only if

ρ0ω2a=μs∣k∣22 a+(λ+μs) ⟨k,a⟩2 k.(D)

Moreover, if this plane wave satisfies (E) (equivalently, if (D) holds), then for a≠0: if a is parallel to k (longitudinal), then ρ0ω2=(λ+2μs)∣k∣22 and its phase speed is ∣ω∣/∣k∣2=cL, where cL:=(λ+2μs)/ρ0>0; if a is orthogonal to k (transverse), then ρ0ω2=μs∣k∣22 and its phase speed is ∣ω∣/∣k∣2=cT, where cT:=μs/ρ0>0. Positivity follows from the branch hypotheses, and [cL]=[cT]=ℓτ−1 (Continuum body data, the fixed inertial frame and the dimensions of the fields, Dimensions and SI unit conventions of the non-relativistic classical mechanics framework). No assertion is made that every solution of (E) is a superposition of plane waves, and no measured modulus or speed is claimed.

Facts & Assumptions

Given: the linear elastic branch data (ρ0,λ,μs); the vectors k,a and the frequency ω; the plane wave ξ.

[F1]

Displacement equation (The displacement equation of the linear elastic branch: rho_0 partial_t^2 xi = rho_0 b + mu_s Laplacian xi + (lambda + mu_s) grad div xi): in the branch's description the momentum equation is ρ0∂t2ξ=ρ0b+μsΔξ+(λ+μs)∇(div⁡ξ); with b≡0 the right-hand side is the displayed operator applied to ξ.

[F2]

Plane-wave calculation (Exact plane-wave solutions of the isotropic linear elastic model, with longitudinal and transverse speeds, The small-strain isotropic linear elastic constitutive law and its strain energy): for the displayed field and stress, the identity ρ0∂t2ξi=∑j∂jσij on R×R3 holds if and only if (D) holds.

[F3]

Positivity hypotheses (Adopted isotropic linear elastic constitutive model with its small-strain regime and strain energy clause (ii)): the branch's moduli satisfy μs>0 and λ+2μs3>0.

[F4]

Dimensions (Continuum body data, the fixed inertial frame and the dimensions of the fields, Dimensions and SI unit conventions of the non-relativistic classical mechanics framework): [ρ0]=μℓ−3, [λ]=[μs]=μℓ−1τ−2, [k]=ℓ−1, [a]=ℓ, and [ω]=τ−1.

Proof

technique · direct; identify the branch's displacement equation with the stress-divergence identity of the cited plane-wave theorem, then read off the two polarization branches and their positivity
1.1givenF1F2

The field ξ is C∞ on R×R3 and the branch's displacement equation (E) of [F1] with b≡0 has right-hand side μsΔξ+(λ+μs)∇(div⁡ξ), which is exactly the divergence of the branch's stress σ=λ(tr⁡e)I3+2μse as computed by [F1]; the cited plane-wave theorem [F2] therefore gives that (E) holds on all of R×R3 if and only if (D) holds.

2.1step 1.1F2algebra

Polarization branches: suppose the plane wave satisfies (E), so (D) holds by step 1.1, and let a≠0. If a∥k, then a=αk with α≠0, and substituting into (D) gives ρ0ω2αk=μs∣k∣2αk+(λ+μs)∣k∣2αk; dividing by α≠0 and using k≠0 gives ρ0ω2=(λ+2μs)∣k∣2, hence ∣ω∣/∣k∣2=cL. If a⊥k, the inner product term of (D) vanishes, so ρ0ω2a=μs∣k∣2a; dividing by the nonzero vector a gives ρ0ω2=μs∣k∣2, hence ∣ω∣/∣k∣2=cT.

3.1step 2.1F3F4algebra∎

Positivity and dimensions: by [F3], μs>0 gives cT2=μs/ρ0>0, and λ+2μs=(λ+2μs3)+4μs3>0 gives cL2=(λ+2μs)/ρ0>0, so both squared speeds are real and positive. By [F4], both squared speeds have dimension ℓ2τ−2, hence [cL]=[cT]=ℓτ−1. No completeness, boundary-value or empirical claim is made.

Remarks

  • Exact solutions of the model's linear equation. The waves displayed are exact solutions of (E) with b≡0, not approximations, but they are special solutions: a general initial-value problem must be solved by other means, and the smooth fixed-end and chain items below supply existence results for particular geometries. Existence for general data in R3 is not asserted on this page.

  • Reading of the speeds. For a material with measured moduli the speeds cL=(λ+2μs)/ρ0 and cT=μs/ρ0 are model predictions; comparing them with measured sound speeds tests the model and its idealizations, and any such comparison uses the moduli as qualified empirical premises. No measured value is asserted here.

Depends on

Dependency tree · 2 levels

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