Continuum Mechanics and Elasticity

Derives Proposition 30.7. The displacement is a vector, \(u'_{k}=a_{km}u_{m}\), and for an orthogonal transformation with constant coefficients the differential operator carries a vector index as well, \(\pp'_{i}=a_{il}\pp_{l}\). The product \(\pp_{i}u_{k}\) is therefore a rank-two tensor by the product rule of Section 13.2.2, its symmetric and antisymmetric parts are tensors by the addition and index-permutation rules there, and the trace is a rank-zero tensor by the contraction rule. That the trace is a scalar is the statement that \(\vect{\nabla}\cdot\vect{u}\) does not depend on the frame in which it is computed, which is what makes Proposition 30.8 a statement about the body rather than about the observer.

Derives Proposition 30.8. The map \(\vect{X}\mapsto\vect{X}+\vect{u}(\vect{X})\) has Jacobian matrix \(F_{ik}=\delta_{ik}+\pp_{k}u_{i}\), and a volume element is multiplied by \(\det F\). For any matrix \(A\), \(\det\left(\identity+A\right)=1+\operatorname{tr}A+O(A^{2})\), since the terms of the determinant that are not the product of diagonal entries contain at least two off-diagonal factors. Hence \(\det F=1+\pp_{i}u_{i}+O\left((\pp u)^{2}\right)\), and \(\pp_{i}u_{i}=u_{ii}=u_{ll}\) because the antisymmetric part of \(\pp_{i}u_{k}\) has zero trace. The deviatoric part is traceless by construction, so it changes no volumes to this order.

Derives Proposition 30.9. \(u_{ik}\) is a real symmetric array, hence the matrix of a self-adjoint operator on \(\R^{3}\) with the Euclidean inner product; by Theorem 5.75 it has an orthonormal eigenbasis and real eigenvalues. In that basis a material line element along the \(j\)-th eigenvector of length \(\dd l\) becomes, to first order, one of length \(\left(1+u^{(j)}\right)\dd l\), since the antisymmetric part contributes a rotation and not a length change. The trace is basis independent by Proposition 30.7.

Derives Proposition 30.12. With \(F_{ik}=\pp_{k}\chi_{i}=\delta_{ik}+\pp_{k}u_{i}\) one has \(\dd x_{i}=F_{ik}\dd X_{k}\), so

and subtracting \(\dd l^{2}=\delta_{lk}\dd X_{l}\dd X_{k}\) gives Equation (30.8). If \(E_{ik}\equiv0\) then every material length is preserved, so the motion is an isometry of \(\R^{3}\) and hence (Proposition 13.121) a rotation followed by a translation; the converse is immediate because a rigid motion has \(F\transpose F=\identity\).

Derives Proposition 30.14. Substitute \(u_{ij}=\tfrac{1}{2}(\pp_{i}u_{j}+\pp_{j}u_{i})\) into the left-hand side. The first term contributes \(\tfrac{1}{2}(\pp_{l}\pp_{k}\pp_{i}u_{j}+\pp_{l}\pp_{k}\pp_{j}u_{i})\) and the other three the analogous expressions; every one of the eight resulting third derivatives appears twice with opposite signs once the partial derivatives are freely commuted, which Proposition 7.73 licenses for a \(C^{3}\) field. The sum therefore vanishes term by term. The left-hand side is antisymmetric under \(i\leftrightarrow l\) and under \(j\leftrightarrow k\) and symmetric under the exchange of the pairs, which in three dimensions leaves \(3\times3\) components of which the six symmetric ones are the independent conditions — exactly the index symmetries of a curvature tensor (Proposition 13.154).

Derives Theorem 30.15. The construction is Cesàro's: recover first the rotation that the strain does not record, then the displacement. It uses no mathematics beyond Proposition 7.104(3), that a curl-free \(C^{1}\) field on a simply connected open set is a gradient.

Step 1: the rotation field. Set

which is \(C^{1}\) and antisymmetric in \(ij\). Were \(u_{ik}\) already the strain of a displacement, \(\Phi_{ijk}\) would be \(\pp_{k}\omega_{ij}\) with \(\omega_{ij}=\tfrac{1}{2}(\pp_{i}u_{j}-\pp_{j}u_{i})\) the infinitesimal rotation: substituting \(u_{jk}=\tfrac{1}{2}(\pp_{j}u_{k}+\pp_{k}u_{j})\) into Equation (30.10) gives \(\tfrac{1}{2}(\pp_{i}\pp_{k}u_{j}-\pp_{j}\pp_{k}u_{i})\), which is that derivative. Take this as the equation to be solved. For each fixed pair \(ij\) the field \(k\longmapsto\Phi_{ijk}\) is curl-free exactly when

vanishes, and the right-hand side is the compatibility expression Equation (30.9) with its four indices taken in the order \((i,k,l,j)\) and the sign reversed. It vanishes by hypothesis. By Proposition 7.104(3) on the simply connected \(\Omega\) there are therefore functions \(\omega_{ij}\) with \(\pp_{k}\omega_{ij}=\Phi_{ijk}\), each fixed up to a constant. Since \(\Phi_{ijk}=-\Phi_{jik}\), the constants may be chosen so that \(\omega_{ij}=-\omega_{ji}\), and \(\omega_{ii}=0\).

Step 2: the displacement. For each fixed \(i\) consider \(k\longmapsto u_{ik}-\omega_{ik}\), which is \(C^{1}\). Its curl is

every term cancelling in pairs, the last two because \(u_{kl}=u_{lk}\). Proposition 7.104(3) again supplies functions \(u_{i}\) with \(\pp_{k}u_{i}=u_{ik}-\omega_{ik}\). Symmetrising and using the symmetry of \(u_{ik}\) with the antisymmetry of \(\omega_{ik}\),

so \(\vect{u}\) has the prescribed strain.

Step 3: uniqueness. Let \(\vect{v}\) have zero strain, \(\pp_{i}v_{k}+\pp_{k}v_{i}=0\). The identity of step 1, applied with \(u_{ik}=0\), gives \(\pp_{k}w_{ij}=\Phi_{ijk}=0\) for \(w_{ij}=\tfrac{1}{2}(\pp_{i}v_{j}-\pp_{j}v_{i})\), so \(w\) is a constant antisymmetric array on the connected \(\Omega\). The symmetric part of \(\pp_{j}v_{k}\) vanishes by hypothesis, so \(\pp_{j}v_{k}=w_{jk}\) is constant and \(v_{k}=a_{k}+w_{jk}x_{j}\) with \(a_{k}\) constant; writing the antisymmetric \(w_{jk}\) as \(\varepsilon_{jkm}b_{m}\) and using the cyclic symmetry of \(\varepsilon\) puts this in the form \(\vect{v}=\vect{a}+\vect{b}\times\vect{x}\). Conversely every such field has zero strain. Two displacements with the same strain differ by one of them, which is the stated uniqueness.

Derives Theorem 30.18. First, \(\vect{t}(\vect{x},-\vect{n})=-\vect{t}(\vect{x},\vect{n})\). Apply Newton's second law to a flat pillbox of face area \(A\) and thickness \(h\) straddling the cut: the mass and hence the inertia and the body force are \(O(hA)\), while the two faces contribute \(\left[\vect{t}(\vect{n})+\vect{t}(-\vect{n})\right]A\) and the rim \(O\left(hA^{1/2}\right)\). Dividing by \(A\) and letting \(h\to0\) leaves the stated relation.

Now take a tetrahedron with three faces in the coordinate planes, meeting at \(\vect{x}\), and a fourth face of area \(A\) with outward unit normal \(\vect{n}\) at perpendicular distance \(h\) from \(\vect{x}\). The coordinate faces have areas \(An_{k}\) and outward normals \(-\hat{\vect{e}}_{k}\), and the volume is \(hA/3\). Newton's second law for the tetrahedron reads

where \(\vect{b}\) is the body force per unit mass and the mean values of the continuous fields over the faces have been used. Divide by \(A\) and let \(h\to0\): the left side vanishes, and

This is Equation (30.15). The nine numbers \(\sigma_{ik}\) so defined are a tensor by the quotient law of Section 13.2.2: \(t_{i}\) and \(n_{k}\) are vectors and \(\vect{n}\) is arbitrary, so the array contracting one into the other must transform as a rank-two tensor. Note that no constitutive assumption of any kind has been made — the argument is pure balance of momentum, and Equation (30.15) therefore holds for fluids, plastic solids and viscoelastic media as well.

Derives Proposition 30.19. Apply the balance of angular momentum about its own centre to a cube of side \(\ell\) centred at \(\vect{x}\). The torque about the \(z\) axis from the tractions on the four vertical faces is, to leading order, \(\left(\sigma_{yx}-\sigma_{xy}\right)\ell^{3}\); the torque from the body force is \(O(\ell^{4})\), because \(\int(\vect{r}\times\vect{f})\dd V\) over a region of size \(\ell\) centred on the reference point carries one factor of \(\ell\) from \(\vect{r}\) and three from \(\dd V\); and the moment of inertia is \(\rho\ell^{5}/6\). Hence the angular acceleration is

which diverges as \(\ell\to0\) unless \(\sigma_{yx}=\sigma_{xy}\). The same argument on the other two axes completes the proof. The hypothesis is essential and is not vacuous: a medium carrying an intrinsic torque per unit volume — a magnetised body in a field (Magnetism in Matter), a liquid crystal — adds a term of order \(\ell^{3}\) to the numerator and does support an antisymmetric part.

Derives Theorem 30.21. Let \(V\) be any fixed material region with boundary \(S\). Newton's second law for the material in \(V\) balances the rate of change of its momentum against the surface tractions and the body force:

using Equation (30.15). By the divergence theorem (Theorem 7.101) applied to each of the three vector fields \(k\mapsto\sigma_{ik}\),

The region \(V\) was arbitrary and the integrand is continuous, so it vanishes pointwise — the argument of the fundamental lemma Lemma 16.18, applied here with arbitrary subregions rather than arbitrary test functions.

Derives Proposition 30.23. Because \(\sigma_{ik}\) is symmetric, \(\sigma_{ik}\pp_{k}w_{i} =\sigma_{ik}w_{ik}\). Integrate the identity \(\pp_{k}\left(\sigma_{ik}w_{i}\right) =\left(\pp_{k}\sigma_{ik}\right)w_{i}+\sigma_{ik}\pp_{k}w_{i}\) over \(\Omega\) and apply Theorem 7.101:

On \(\pp\Omega_{\mathrm{u}}\) the field \(w_{i}\) vanishes and on \(\pp\Omega_{\mathrm{t}}\) the traction is \(\bar{t}_{i}\); substituting \(\pp_{k}\sigma_{ik}=-f_{i}\) rearranges the display into Equation (30.18). For the converse, run the computation backwards: Equation (30.18) becomes \(\int_{\Omega}\left(\pp_{k}\sigma_{ik}+f_{i}\right)w_{i}\dd V =\int_{\pp\Omega_{\mathrm{t}}}\left(\bar{t}_{i} -\sigma_{ik}n_{k}\right)w_{i}\dd S\) for all admissible \(w_{i}\), and choosing \(w_{i}\) supported in the interior and then near the traction boundary kills the two brackets in turn by Lemma 16.18.

Derives Proposition 30.25. Only the part of \(C_{iklm}\) symmetric in \((ik)\) and in \((lm)\) can contribute to \(C_{iklm}u_{ik}u_{lm}\), since \(u_{ik}\) is symmetric; this is the minor symmetry, and it reduces each index pair to the \(6\) values \(\{11,22,33,23,13,12\}\). The major symmetry is \(\pp^{2}F/\pp u_{ik}\pp u_{lm}=\pp^{2}F/\pp u_{lm}\pp u_{ik}\), which is Proposition 7.73 for the smooth function \(F\); without an energy function there would be no reason for it. The array is therefore a symmetric \(6\times6\) matrix, with \(6\times7/2=21\) independent entries.

Derivation. Derives Phenomenon 30.28. The undeformed state is an equilibrium and carries no stress, so the expansion of the energy density about \(u_{ik}=0\) has neither a constant nor a linear term:

with the constant coefficients of Definition 30.24. By Equation (30.19) the stress is then linear in the strain up to the cubic remainder, \(\sigma_{ik}=C_{iklm}u_{lm}\). That is Hooke's law, and its observed domain of validity is precisely the range of strains over which the remainder is negligible — which is why the law is a small-strain statement and not a property of any particular material.

If the body is isotropic, \(C_{iklm}\) is invariant under every rotation of the frame, so by Lemma 30.26 it has the form Equation (30.20). The minor symmetries of Proposition 30.25 force \(b=c\), since exchanging \(i\) and \(k\) in Equation (30.20) exchanges the coefficients of \(\delta_{il}\delta_{km}\) and \(\delta_{im}\delta_{kl}\). Writing \(a=\lambda\) and \(b=c=\mu\) and contracting with \(u_{lm}\) gives \(\sigma_{ik}=\lambda u_{ll}\delta_{ik}+2\mu u_{ik}\), which is Equation (30.21). That two constants suffice for an isotropic solid is a consequence of symmetry, not a further approximation [Landau:1986].

Derives Proposition 30.30. The second form of Equation (30.23) is a sum of two squares in variables that are independent — the trace \(u_{ll}\) and the five components of the traceless \(d_{ik}\) — so the quadratic form is positive definite exactly when both coefficients are positive, which is Theorem 5.112 read on a form already in normal shape. Each condition is realised by an admissible strain: a pure dilatation \(u_{ik}=\epsilon\delta_{ik}\) has \(F=\tfrac{9}{2}K\epsilon^{2}\) and no shear, and a pure shear as in Example 30.10 has \(F=\mu\gamma^{2}/2\) and no dilatation. A body violating either bound would gain energy by deforming and would not be found in the unstrained state at all.

Derivation. Derives Phenomenon 30.32. Take the bar along \(z\) and impose the uniaxial stress \(\sigma_{zz}=\sigma\) with all other stress components zero. By isotropy the strain is diagonal, with \(u_{xx}=u_{yy}\equiv u_{\perp}\) and \(u_{zz}\equiv u_{\parallel}\). The transverse components of Equation (30.21) read \(0=\lambda(2u_{\perp}+u_{\parallel})+2\mu u_{\perp}\), so

which is positive whenever \(\lambda>0\): the transverse strain has the sign opposite to the axial one, as observed. Substituting back into the axial component gives \(\sigma=\mu(3\lambda+2\mu)u_{\parallel}/(\lambda+\mu)\), so that \(E=\mu(3\lambda+2\mu)/(\lambda+\mu)\) — the modulus is not an independent constant but a combination of the two in Equation (30.21). For the bounds, use Proposition 30.30: \(\mu>0\) and \(K=\lambda+\tfrac{2}{3}\mu>0\). Writing \(x=\mu/K\), so that \(x\) ranges over \((0,\infty)\), Equation (30.26) becomes

a strictly decreasing function of \(x\) with limits \(\tfrac{1}{2}\) as \(x\to0\) and \(-1\) as \(x\to\infty\). Its range is therefore the open interval of Equation (30.25). A solid far harder to compress than to shear sits at the top of that interval: it is effectively incompressible and contracts transversely by half of what it extends.

Derives Proposition 30.33. The derivation of Phenomenon 30.32 gave \(\nu=\lambda/\left[2(\lambda+\mu)\right]\) and \(E=\mu(3\lambda+2\mu)/(\lambda+\mu)\). Solving the first for \(\lambda\) gives \(\lambda(1-2\nu)=2\mu\nu\), the third relation of Equation (30.28); substituting it into the second gives \(E=\mu\left[3\cdot2\mu\nu/(1-2\nu)+2\mu\right] \big/\left[2\mu\nu/(1-2\nu)+\mu\right]=2\mu(1+\nu)\), the first. Then \(K=\lambda+\tfrac{2}{3}\mu =2\mu\nu/(1-2\nu)+\tfrac{2}{3}\mu =\tfrac{2}{3}\mu(1+\nu)/(1-2\nu)=E/\left[3(1-2\nu)\right]\), the second. Eliminating \(\nu\) between the first two of Equation (30.28) gives \(3K(1-2\nu)=2\mu(1+\nu)\), hence \(\nu=(3K-2\mu)/\left[2(3K+\mu)\right]\) and, back-substituting, \(E=2\mu(1+\nu)=9K\mu/(3K+\mu)\). Finally \(M=\lambda+2\mu=\left(K-\tfrac{2}{3}\mu\right)+2\mu =K+\tfrac{4}{3}\mu\).

Derives Proposition 30.34. Let \(\vect{u}\) be the equilibrium field and \(\vect{v}=\vect{u}+ \vect{w}\) any admissible competitor, so that \(\vect{w}\) vanishes on \(\pp\Omega_{\mathrm{u}}\). Because \(F\) is the quadratic form Equation (30.22), \(F(u_{ik}+w_{ik})=F(u_{ik})+\sigma_{ik}w_{ik}+F(w_{ik})\) exactly, the cross term being \(C_{iklm}u_{lm}w_{ik}=\sigma_{ik}w_{ik}\) by the major symmetry of Proposition 30.25. Hence

The first three terms cancel by Proposition 30.23, and the last is non-negative by Proposition 30.30, vanishing only if \(w_{ik}\equiv0\), i.e. only if \(\vect{w}\) is a rigid displacement (Proposition 30.12). This is the Dirichlet principle of Proposition 10.91 for the elastic operator, and the same convexity underwrites the direct method Theorem 16.70.

Derives Proposition 30.38. Write \(\eta_{b}=u_{ik}n_{i}n_{k}\) for the bond of length \(r_{b}\) and direction \(\hat{\vect{n}}\). By Equation (30.8) the strained length is \(r_{b}'=r_{b}\left(1+2\eta_{b}\right)^{1/2} =r_{b}\left(1+\eta_{b}-\tfrac{1}{2}\eta_{b}^{2}+\cdots\right)\). Because every atom sits at a centre of symmetry the bonds come in antiparallel pairs, so no internal relaxation of the sublattices is induced by a homogeneous strain and the energy per unit volume is simply \(\sum_{b}\varphi(r_{b}')\) over the bonds of one cell divided by its volume. Expanding,

The coefficient of the term linear in \(u_{ik}\) is the residual stress, which vanishes by the zero-stress hypothesis. The quadratic term gives

since \(\eta_{b}^{2}=u_{ik}u_{lm}n_{i}n_{k}n_{l}n_{m}\); the weights \(w_{b}\) depend on the bond length alone, and their exact normalisation — how the bond sum is counted — does not matter for what follows. Every summand is a product of four components of one and the same unit vector, hence invariant under any permutation of \(iklm\); so is \(C_{iklm}\). In particular \(C_{1122}=C_{1212}\), which in the notation of Definition 30.37 reads \(C_{12}=C_{66}\), and \(C_{66}=C_{44}\) for a cubic crystal: that is Equation (30.32). The other five relations are the remaining independent identifications among the \(21\) constants.

Derives Theorem 30.40. Substitute Equation (30.21), with \(u_{ik}=\tfrac{1}{2}(\pp_{i}u_{k}+\pp_{k}u_{i})\), into Cauchy's equation Equation (30.17). Since \(\lambda\) and \(\mu\) are constants,

where \(\pp_{k}\pp_{i}u_{k}=\pp_{i}\pp_{k}u_{k} =\pp_{i}(\vect{\nabla}\cdot\vect{u})\) by Proposition 7.73. This is Equation (30.33). Navier reached the same equation from a molecular model in which the constants were not independent [Navier:1827]; Cauchy reached it from the stress tensor, in which they are [Cauchy:1823], and Remark 30.39 says which of the two the measurements support.

Derives Proposition 30.41. Differentiate: \(\pp_{t}e=\rho\dot{u}_{i}\ddot{u}_{i} +\sigma_{ik}\dot{u}_{ik}\), using Equation (30.19) for the second term. Because \(\sigma_{ik}\) is symmetric, \(\sigma_{ik}\dot{u}_{ik} =\sigma_{ik}\pp_{k}\dot{u}_{i}\). Substituting \(\rho\ddot{u}_{i}=\pp_{k}\sigma_{ik}+f_{i}\) from Equation (30.17),

which is the stated balance with \(q_{k}\) as defined.

Derives Theorem 30.42. Let \(\vect{w}\) be the difference of two solutions. It solves Equation (30.33) with \(\vect{f}=0\), with zero initial data, and with \(\vect{w}=0\) or \(\sigma_{ik}[\vect{w}]n_{k}=0\) at each boundary point. By Proposition 30.41 its total energy \(\mathcal{E}(t)=\int_{\Omega}e\,\dd V\) satisfies

by Theorem 7.101, since on each part of the boundary one factor of the integrand vanishes — the traction where the traction is prescribed, the velocity where the displacement is. Hence \(\mathcal{E}(t)=\mathcal{E}(0)=0\) for all \(t\). But \(\mathcal{E}\) is an integral of a sum of two non-negative terms, the second positive definite by Proposition 30.30, so \(\dot{\vect{w}}\equiv0\) and \(w_{ik}\equiv0\); by Proposition 30.12 \(\vect{w}\) is a rigid displacement, constant in time, and it vanishes if the displacement is prescribed anywhere. This is what makes the problem well posed in the sense of Definition 10.24, and the same energy identity gives the continuous dependence on the data.

Derivation. Derives Phenomenon 30.43. Look for plane-wave solutions of the source-free Equation (30.33), \(\vect{u}=\vect{A}\exp\left[\ii\left(\vect{k}\cdot\vect{x} -\omega t\right)\right]\), with \(\vect{A}\) a constant polarization vector. Each spatial derivative brings down \(\ii k_{j}\) and each time derivative \(-\ii\omega\), so \(\vect{\nabla}(\vect{\nabla}\cdot\vect{u})\mapsto -\left(\vect{k}\cdot\vect{A}\right)\vect{k}\) and \(\nabla^{2}\vect{u}\mapsto-k^{2}\vect{A}\), giving

A plane wave exists exactly when \(\vect{A}\) is an eigenvector of the acoustic tensor \(\Gamma_{il}\), with \(\rho\omega^{2}\) the eigenvalue. \(\Gamma\) is real and symmetric, so Theorem 5.75 supplies an orthonormal eigenbasis, and it can be read off: \(\hat{\vect{k}}\) is an eigenvector with eigenvalue \((\lambda+\mu)k^{2}+\mu k^{2}=(\lambda+2\mu)k^{2}\), and every vector orthogonal to \(\vect{k}\) is an eigenvector with eigenvalue \(\mu k^{2}\). Those three exhaust \(\R^{3}\), so there are no other polarizations. Hence

which are the two speeds Equation (30.35). Both branches are non-dispersive: \(\omega\) is proportional to \(k\), so a pulse travels undistorted, which is what makes an arrival time a meaningful observable at all.

The physical labels follow from the polarizations. The longitudinal branch has \(\vect{\nabla}\times\vect{u}=0\) and carries the dilatation \(\vect{\nabla}\cdot\vect{u}\); the transverse branch has \(\vect{\nabla}\cdot\vect{u}=0\) and carries the rotation, deforming without changing any volume. Finally

by Proposition 30.30: the ordering of the two speeds is not an accident of the materials that happen to exist but follows from the existence of a stable unstrained state. Setting \(\mu=0\) annihilates the transverse branch and leaves \(v_{\mathrm{P}}=\sqrt{K/\rho}\), the sound speed of Fluid Dynamics.

Derives Proposition 30.44. From Equation (30.35), \(v_{\mathrm{P}}^{2}/v_{\mathrm{S}}^{2}=(\lambda+2\mu)/\mu =\lambda/\mu+2\), and \(\lambda/\mu=2\nu/(1-2\nu)\) by Equation (30.28); adding gives \(\left(2\nu+2-4\nu\right)/(1-2\nu)=2(1-\nu)/(1-2\nu)\), which is Equation (30.37). Differentiating the square, \(\dd\left[2(1-\nu)/(1-2\nu)\right]/\dd\nu =2/(1-2\nu)^{2}>0\), so it increases; its values at the endpoints \(\nu=-1\) and \(\nu\to\tfrac{1}{2}\) are \(4/3\) and \(+\infty\). The density cancels between the two speeds, which is the practical point: an ultrasonic apparatus that measures only times and one length returns \(\nu\) directly.

Derivation. Derives Phenomenon 30.47. The order of the first two groups is Equation (30.38) with \(v_{\mathrm{P}}>v_{\mathrm{S}}\), established in Phenomenon 30.43. That the third group comes last is Proposition 30.49 and Proposition 30.50: both surface branches travel below \(v_{\mathrm{S}}\), the Rayleigh wave at \(0.919\,v_{\mathrm{S}}\) for a Poisson solid and the Love wave between the two shear speeds of the layered structure. That the third group is the largest at great range, though the source radiates mostly body waves, is geometric. A body wave from a point source spreads over a hemisphere of area \(2\pi r^{2}\), so its energy flux falls as \(r^{-2}\) and its amplitude as \(r^{-1}\); a surface wave is confined to a skin of fixed thickness of order one wavelength and spreads over a cylinder of area \(2\pi r\times\text{const}\), so its flux falls as \(r^{-1}\) and its amplitude only as \(r^{-1/2}\). At \(r=1000\,\mathrm{km}\) from a source of characteristic size \(10\,\mathrm{km}\) the surface wave is therefore favoured by a factor \(\left(r/r_{0}\right)^{1/2}=10\) in amplitude over the body waves, which is why it dominates a distant record.

That the train is long and its period varies through it — the observed dispersion — is Proposition 30.50: the Love speed depends on \(\omega H/\beta\), so different periods arrive at different times and a single impulsive source is drawn out into a train ordered in period. A homogeneous half-space would give a Rayleigh wave with no dispersion at all (Proposition 30.48 contains no length), so the observed dispersion is itself the evidence that the Earth is layered. Finally, the confinement to the surface: both branches are built from solutions decaying as \(\ee^{-qz}\) with \(q\) of order \(k\), so the motion is negligible below a depth of about one wavelength — \(35\,\mathrm{km}\) for a \(10\,\mathrm{s}\) Love wave travelling at \(3.5\,\mathrm{km}/\mathrm{s}\).

Derives Proposition 30.48. Let \(z\) measure depth from the free surface, so that the solid occupies \(z>0\), let the propagation be along \(x\), and seek a motion with \(u_{y}=0\) and no dependence on \(y\). Build it from two potentials,

This is an ansatz, not a decomposition theorem, and it is all the proposition needs: it asserts that such a wave exists, so it is enough to exhibit one. Since \(\vect{\nabla}\cdot\vect{u}=\nabla^{2}\phi\), substituting Equation (30.40) into the source-free Equation (30.33) and collecting gives

for the \(x\) component and the same two brackets differentiated by \(\pp_{z}\) and \(-\pp_{x}\) for the \(z\) component, so it suffices that each bracket vanish: \(\phi\) obeys the wave equation at \(v_{\mathrm{P}}\) and \(\psi\) at \(v_{\mathrm{S}}\), the two speeds Equation (30.35).

Take the surface wave

with \(k>0\). The two wave equations fix \(p^{2}=k^{2}-\omega^{2}/v_{\mathrm{P}}^{2}\) and \(q^{2}=k^{2}-\omega^{2}/v_{\mathrm{S}}^{2}\), so writing \(c=\omega/k=\xi v_{\mathrm{S}}\),

both real and positive — so that the motion decays with depth over a distance of order \(1/k\) — precisely when \(0<\xi<1\), since \(\gamma<1\) by \(v_{\mathrm{S}}<v_{\mathrm{P}}\) (Phenomenon 30.43).

The free surface \(z=0\) carries no traction (Definition 30.22), which for this motion is the two conditions \(\sigma_{zz}=\sigma_{xz}=0\). From Equation (30.21),

Insert Equation (30.41). In the first, the coefficient of \(A\) is \(\lambda\left(p^{2}-k^{2}\right)+2\mu p^{2} =(\lambda+2\mu)p^{2}-\lambda k^{2}=2\mu k^{2}-\rho\omega^{2} =\mu k^{2}\left(2-\xi^{2}\right)\), using \(\rho\omega^{2}=\mu k^{2}\xi^{2}\); in the second, the coefficient of \(B\) is \(\mu\left(q^{2}+k^{2}\right)=\mu k^{2}\left(2-\xi^{2}\right)\), the same combination. The cross terms are \(2\ii\mu kq\) in the first and \(-2\ii\mu kp\) in the second. Dividing throughout by \(\mu\), the two conditions at \(z=0\) are therefore

a homogeneous pair with a non-trivial solution only if its determinant \(k^{4}\left(2-\xi^{2}\right)^{2}-4k^{2}pq\) vanishes. Dividing by \(k^{4}\) and substituting Equation (30.42) gives Equation (30.39).

It remains to know that an admissible root exists. Put \(t=\xi^{2}\) and \(f(t)=\left(2-t\right)^{2} -4\sqrt{\left(1-t\right)\left(1-\gamma t\right)}\). Then \(f(0)=0\) and \(f'(0)=-4+2\left(1+\gamma\right)=2\left(\gamma-1\right)<0\), so \(f\) is negative just to the right of the origin, while \(f(1)=1>0\); by the intermediate value theorem (Theorem 7.23) \(f\) has a root in \((0,1)\), and \(\xi=\sqrt{t}\) is the Rayleigh ratio. Finally, no length enters Equation (30.39): \(\xi\) depends on \(\gamma\) alone, so \(c=\xi v_{\mathrm{S}}\) is independent of \(k\) and the wave does not disperse. The depth of penetration, by contrast, is \(1/p\) and \(1/q\), both of order the wavelength.

Derives Proposition 30.49. Put \(t=\xi^{2}\) and \(\gamma=1/3\) and square Equation (30.39):

Expanding, \((2-t)^{4}=t^{4}-8t^{3}+24t^{2}-32t+16\) and the right side is \(16-\tfrac{64}{3}t+\tfrac{16}{3}t^{2}\), so the difference is \(t^{4}-8t^{3}+\tfrac{56}{3}t^{2}-\tfrac{32}{3}t =\tfrac{t}{3}\left(3t^{3}-24t^{2}+56t-32\right)\). The root \(t=0\) is the trivial static solution. The cubic has \(t=4\) as a root by inspection, and dividing out leaves \(3t^{2}-12t+8=0\) with roots \(t=2\pm2/\sqrt{3}\). Admissibility requires \(0<t<1\), since both radicals in Equation (30.39) must be real for the depth dependence to decay: this excludes \(t=4\) and \(t=2+2/\sqrt{3}=3.1547\) and leaves \(t=2-2/\sqrt{3}=0.845299\). Squaring can create roots, so verify in the unsquared equation: \((2-t)^{2}=1.33333\) and \(4\sqrt{1-t}\sqrt{1-t/3} =4(0.393320)(0.847487)=1.33333\), so it is genuine. Hence \(\xi=\sqrt{0.845299}=0.91940\). The Rayleigh wave is thus slower than the S wave by \(8\,\mathrm{\%}\), which is what puts it last on the seismogram of Phenomenon 30.47, and its speed is a fixed multiple of \(v_{\mathrm{S}}\) with no free length — so on a homogeneous half-space it does not disperse.

Derives Proposition 30.50. Let \(z\) measure depth from the free surface, and look for a motion purely along \(y\), \(\vect{u}=\left(0,v,0\right)\) with \(v=f(z)\exp\left[\ii(kx-\omega t)\right]\). Such a field has \(\vect{\nabla}\cdot\vect{u}=0\), so Equation (30.33) reduces in each homogeneous medium to \(\rho\,\pp_{t}^{2}v=\mu\nabla^{2}v\), and hence to

In the layer take \(c>\beta_{1}\), so \(s^{2}=\omega^{2}/\beta_{1}^{2}-k^{2}>0\) and \(f=A\cos sz+B\sin sz\). The free surface carries the traction \(\sigma_{yz}=\mu_{1}\pp_{z}v\), which must vanish at \(z=0\): hence \(B=0\) and \(f=A\cos sz\). In the half-space take \(c<\beta_{2}\), so \(q^{2}=k^{2}-\omega^{2}/\beta_{2}^{2}>0\) and the only solution bounded at depth is \(f=C\ee^{-q(z-H)}\). Continuity of the displacement and of the traction \(\mu\pp_{z}v\) across \(z=H\) gives

and dividing the second by the first, \(\mu_{1}s\tan sH=\mu_{2}q\), which is Equation (30.47) once \(s\) and \(q\) are written out. If \(\beta_{1}\geq\beta_{2}\) the two requirements \(c>\beta_{1}\) and \(c<\beta_{2}\) are incompatible and no solution exists — a homogeneous half-space, in particular, carries no Love wave, which is why Love had to introduce the layer [Love:1911]. Finally \(k\) and \(\omega\) enter Equation (30.47) separately and not only through \(c\), so \(c\) depends on frequency; as \(\omega\to\infty\) the wave concentrates in the layer and \(c\to\beta_{1}\), while near the cut-off of each higher branch \(c\to\beta_{2}\).

Derives Proposition 30.51. Under the kinematic hypothesis, a cross-section at \(x\) rotates through the angle \(w'(x)\), so the axial displacement of the fibre at height \(z\) is \(u_{x}=-z\,w'(x)\) and its axial strain is \(u_{xx}=\pp_{x}u_{x}=-z\,w''\). The lateral faces of the rod are free, so \(\sigma_{yy}=\sigma_{zz}=0\) and the state is uniaxial stress — whence \(\sigma_{xx}=E\,u_{xx}\) by Definition 30.31, with \(E\) and not \(M\) appearing. The moment of these stresses about the axis \(z=0\) is \(-\int_{S}\sigma_{xx}z\,\dd S=Ew''\int_{S}z^{2}\dd S=EIw''\). The axis \(z=0\) is fixed by requiring zero net axial force, \(\int_{S}\sigma_{xx}\dd S=0\), i.e. \(\int_{S}z\,\dd S=0\): the neutral axis passes through the centroid.

Derivation. Derives Phenomenon 30.52. Let the column occupy \(0\leq x\leq L\) under the axial compression \(P\) and let \(y(x)\) be a small lateral deflection, with the ends pinned so that \(y(0)=y(L)=0\) and no moment is transmitted there. By Proposition 30.51 the bending moment at \(x\) is \(EI\,y''(x)\), and the moment of the applied load about the same section is \(-P\,y(x)\); equilibrium of the deflected shape therefore requires

This is an eigenvalue problem. Writing \(k^{2}=P/(EI)\), the general solution is \(y=A\sin kx+B\cos kx\); the condition at \(x=0\) forces \(B=0\), and the condition at \(x=L\) then requires either \(A=0\) — the straight column, available at every load — or \(\sin kL=0\), i.e.\ \(kL=n\pi\) with \(n\) a positive integer. A deflected equilibrium therefore exists only at the discrete loads \(P_{n}=n^{2}\pi^{2}EI/L^{2}\), and the smallest of them, \(n=1\), is Equation (30.50). Below it the straight column is the only equilibrium; at it a one-parameter family of bent shapes appears, which is what is seen as the sudden onset of buckling. Note that the material enters only through \(E\) and the cross-section only through \(I\): the yield stress of Phenomenon 30.73 plays no part, so a slender column fails long before its material does.

Derives Theorem 30.56. By Equation (30.48) the elastic energy density is \(\tfrac{1}{2}E\left(zw''\right)^{2}\); integrating over the cross-section gives \(\tfrac{1}{2}EI\left(w''\right)^{2}\) per unit length, so the action of the rod is \(\int\!\!\int\Lag\,\dd x\,\dd t\) with Lagrangian density

The density depends on a second derivative, so the stationarity condition is the Euler–Poisson equation Theorem 16.38,

which evaluates to \(q-\pp_{x}^{2}\left(EIw''\right) -\rho S\pp_{t}^{2}w=0\), i.e.\ Equation (30.53). The boundary terms produced by the integration by parts are \(\left[EIw''\,\delta w'\right]_{0}^{L}\) and \(-\left[\left(EIw''\right)'\delta w\right]_{0}^{L}\); where the end is free, \(\delta w\) and \(\delta w'\) are unconstrained and each bracket must vanish separately, which is Theorem 16.31. Where the end is clamped, \(\delta w\) and \(\delta w'\) vanish and the conditions are \(w=w'=0\) instead — two conditions at each end in every case, as a fourth-order equation requires.

Derives Proposition 30.57. Substitute \(w=\exp\left[\ii\left(kx-\omega t\right)\right]\) into the source-free Equation (30.53) with \(EI\) constant: \(-\rho S\omega^{2}+EIk^{4}=0\), which is Equation (30.54). Then \(v_{\mathrm{ph}}=\omega/k=k\sqrt{EI/\rho S}\) and \(v_{\mathrm{g}}=\dd\omega/\dd k=2k\sqrt{EI/\rho S}=2v_{\mathrm{ph}}\). For a finite beam, separating \(w=W(x)\ee^{-\ii\omega t}\) gives \(W''''=\beta^{4}W\) with \(\beta^{4}=\rho S\omega^{2}/(EI)\), whose general solution is a combination of \(\sin\beta x,\cos\beta x,\sinh\beta x,\cosh\beta x\); imposing \(W=W'=0\) at the clamped end and \(W''=W'''=0\) at the free one gives a \(4\times4\) homogeneous system whose determinant reduces to \(\cos\beta L\cosh\beta L+1=0\), and its smallest positive root is \(\beta_{1}L=1.8751\).

Derivation. Derives Phenomenon 30.60. Take a circular cylinder of radius \(a=d/2\) along \(z\) and twist it uniformly, so that the section at height \(z\) rotates through \(\theta'z\) with \(\theta'=\Phi/L\) the twist per unit length. In cylindrical coordinates the displacement is \(u_{\phi}=\theta'zr\), \(u_{r}=u_{z}=0\), whose only non-vanishing strain component is \(u_{\phi z}=\tfrac{1}{2}\theta'r\); the deformation is a pure shear, so no volume changes and only \(\mu\) can enter. By Equation (30.21) the stress is \(\sigma_{\phi z}=2\mu u_{\phi z}=\mu\theta'r\), and the torque transmitted across a section is

Setting \(\theta'=\Phi/L\) gives Equation (30.55). The \(d^{4}\) is the content of Coulomb's measurement: a factor \(d^{2}\) because the sheared area grows so, and a second factor \(d^{2}\) because the lever arm and the strain each grow with \(r\). The restoring torque is linear in \(\Phi\) because Hooke's law is linear in the strain, so the torsional pendulum is harmonic and its period is amplitude independent, giving \(T=2\pi\sqrt{I_{0}/\kappa}\) from Oscillations and Mechanical Waves.

Derivation. Derives Phenomenon 30.65. Free harmonic motion of the plate is \(w=W(x,y)\ee^{-\ii\omega t}\) in Equation (30.58) with \(q=0\), so \(W\) satisfies \(D\nabla^{4}W=\rho h\omega^{2}W\) with the plate's own boundary conditions. That is a self-adjoint eigenvalue problem of the kind of Definition 9.54: its spectrum is a discrete increasing sequence \(\omega_{1}<\omega_{2}<\cdots\), so a resonance can only occur at one of countably many frequencies and the mode shape is locked to the frequency. That is the observed discontinuity: sweeping the driving frequency, nothing changes until the next \(\omega_{n}\) is reached, and then the whole figure is replaced by the next eigenfunction.

The sand marks the nodal set \(\{W=0\}\) because a grain leaves the surface whenever the plate's peak downward acceleration exceeds \(g\): with local amplitude \(A\) the condition is \(\omega^{2}A>g\), so at \(f=500\,\mathrm{Hz}\) any region moving by more than \(A=g/\omega^{2}=1.0\,\mu\mathrm{m}\) throws its grains clear. They land elsewhere, are thrown again, and random-walk until they reach a region where \(A\) falls below the threshold — the neighbourhood of a node. The figure is therefore a picture of \(\{W=0\}\) broadened by the threshold, which is why the curves are sharp: near a nodal line \(W\) grows linearly, so the width of the deposit is the threshold amplitude divided by the slope of \(W\), and both are small.

The scaling is Proposition 30.66. Nothing in this argument is peculiar to a plate; the same three statements — discrete spectrum, nodal set, threshold — describe every visualisation of a standing wave in Experiment: Waves and Acoustics.

Derives Proposition 30.66. Write \(\vect{x}=a\vect{\zeta}\) with \(\vect{\zeta}\) ranging over the fixed reference shape. Then \(\nabla^{4}=a^{-4}\nabla_{\zeta}^{4}\), and the eigenvalue problem \(D\nabla^{4}W=\rho h\omega^{2}W\) becomes \(\nabla_{\zeta}^{4}W=\Lambda W\) with \(\Lambda=\rho h\omega^{2}a^{4}/D\). The reduced problem involves only the reference shape and the boundary conditions, so its eigenvalues \(\Lambda_{n}\) are pure numbers. Solving for \(\omega\),

using Equation (30.57), and taking the square root gives Equation (30.59): proportional to \(h\), inversely proportional to \(a^{2}\), exactly as observed in Phenomenon 30.65. The same relation read backwards is a measurement of \(E/\left[\rho(1-\nu^{2})\right]\) from a frequency, a thickness and a diameter.

Derivation. Derives Phenomenon 30.70. The exponent follows from three geometric and dimensional statements and needs no solution of the elastic problem. Let \(a\) be the radius of the contact circle. First, the geometry of two nearly touching spheres gives \(\delta=a^{2}/R\) to leading order in \(a/R\), since the gap between the undeformed surfaces at distance \(a\) from the axis is \(a^{2}/2R_{1}+a^{2}/2R_{2}=a^{2}/2R\) and the two surfaces must meet there. Second, the only strain scale in the contact region is the indentation divided by its own size, \(\delta/a=a/R\), because \(a\) is the only length over which the displacement varies. Third, the stress is then \(E^{*}\) times that strain and acts over an area \(\pi a^{2}\), so

which is Equation (30.60) up to the pure number \(4/3\). The exponent \(3/2\) is exact and is purely geometric: it says that a Hertzian contact stiffens because the contact area grows with the load, not because any material stiffens. The combination \(E^{*}\) is the plane-strain modulus \(E/(1-\nu^{2})\) of Definition 30.63, appearing because the material just under a broad contact patch is laterally confined, and the two bodies act as compliances in series — hence the sum of reciprocals. The coefficient \(4/3\) and the pressure distribution \(p(r)=p_{0}\sqrt{1-r^{2}/a^{2}}\) require the full elastic solution for a half-space [Hertz:1882].

Derivation. Derives Phenomenon 30.73. The two halves are Proposition 30.74 and Proposition 30.76 respectively. What the pair says together is that “strength” is not one material property but two, governed by different parts of the stress tensor: ductile flow by the deviatoric part \(s_{ik}\) of Equation (30.16), which is why it is insensitive to pressure, and brittle fracture by the elastic energy available to be released, which is why it is sensitive to the flaw population and hence to the specimen's history rather than to its composition alone. That is also why the same material can do both: a steel that flows at room temperature cleaves below its transition temperature, where the shear stress needed for flow rises above the stress needed to run a crack.

Derives Proposition 30.74. Work in the principal frame of Definition 30.20, so \(\sigma=\diag\left(\sigma_{1},\sigma_{2},\sigma_{3}\right)\) with \(\sigma_{1}\geq\sigma_{2}\geq\sigma_{3}\). For a unit normal \(\vect{n}\) the traction is \(t_{i}=\sigma_{i}n_{i}\) (no sum), its normal component is \(\bar{\sigma}=\sum_{i}\sigma_{i}m_{i}\) with \(m_{i}=n_{i}^{2}\), and the shear component has magnitude

Since \(\sigma_{3}\leq\sigma_{i}\leq\sigma_{1}\) we have \(\left(\sigma_{i}-\sigma_{3}\right) \left(\sigma_{1}-\sigma_{i}\right)\geq0\) for each \(i\); averaging with the weights \(m_{i}\) gives \(\sum_{i}\sigma_{i}^{2}m_{i}\leq \left(\sigma_{1}+\sigma_{3}\right)\bar{\sigma}-\sigma_{1}\sigma_{3}\). Hence

the last step being the inequality between the geometric and arithmetic means of two non-negative numbers whose sum is \(\sigma_{1}-\sigma_{3}\). Both inequalities are equalities when \(m=\left(\tfrac{1}{2},0,\tfrac{1}{2}\right)\), i.e. for \(\vect{n}\) at \(45^\circ\) between the first and third principal directions, giving Equation (30.62). Adding \(-p_{0}\delta_{ik}\) shifts every principal stress by the same \(-p_{0}\) and leaves every difference, hence \(\tau_{\max}\), unchanged.

Derives Proposition 30.76. Let \(U(c)\) be the total energy of the plate at fixed remote displacement, measured from the uncracked state. Introducing the crack relaxes the material near it and releases stored elastic energy; the elastic solution for an elliptical hole degenerating to a crack, which Griffith took over for this purpose, gives that release as \(\pi\sigma^{2}c^{2}/E\) per unit thickness [Griffith:1921]. Creating the two new faces costs \(\gamma\) per unit area, and the two faces of a crack of length \(2c\) have total area \(4c\) per unit thickness. Hence

The derivative vanishes at \(c^{*}=2E\gamma/\left(\pi\sigma^{2}\right)\) and \(\dd^{2}U/\dd c^{2}=-2\pi\sigma^{2}/E<0\), so \(c^{*}\) is a maximum of the energy — an unstable equilibrium. A crack shorter than \(c^{*}\) can only grow by climbing the energy barrier and does not; one longer than \(c^{*}\) releases more energy than it costs at every subsequent length and accelerates without limit. Solving \(c=c^{*}\) for \(\sigma\) gives Equation (30.63). The criterion is an energy balance and not a stress criterion, which is exactly what was needed: Remark 30.72 showed that the stress at a crack tip is infinite for any load, so no stress criterion could ever have distinguished a safe crack from a dangerous one.

Derives Proposition 30.80. For a constant \(s_{ik}\) the first term of Equation (30.64) vanishes after the step and integration gives \(d_{ik}(t)=d_{ik}(0)+s_{ik}t/(2\eta)\) with the instantaneous elastic response \(d_{ik}(0)=s_{ik}/(2\mu)\); factoring out \(s_{ik}/2\mu\) and using Equation (30.65) gives the stated form. For constant \(d_{ik}\) the left side vanishes and Equation (30.64) becomes \(\dot{s}_{ik}=-\left(\mu/\eta\right)s_{ik}\), whose solution is the stated exponential.