Continuum mechanics: the stress tensor, and conservation laws in local form

Idea. Cauchy's theorem reduces every question about the traction on any plane to linear algebra with one symmetric matrix.

1. \(\vec t = \sigma\vec n = \tfrac13(130, 40, 20) = (43.33,\ 13.33,\ 6.67)\) MPa, with \(|\vec t|^2 = (16\,900 + 1600 + 400)/9 = 2100\), \(|\vec t| = 45.83\) MPa. Normal stress \(\vec n\cdot\vec t = \tfrac19(260 + 40 + 40) = \tfrac{340}{9} = 37.78\) MPa, a tension. Shear \(\sqrt{2100 - 37.78^2} = \sqrt{672.8} = 25.94\) MPa. For \(-\vec n\), \(\sigma(-\vec n) = -\sigma\vec n\) by linearity: \((-43.33, -13.33, -6.67)\), the third-law lemma ✓.

2. \(\operatorname{tr}\sigma = 50 - 20 + 10 = 40\), so \(p = -13.33\) MPa. Negative pressure: on average the point is in tension.

3. \(\vec e_3\) is an eigenvector with eigenvalue \(10\). The block has eigenvalues \(15 \pm\sqrt{225 + 1900} = 15 \pm 46.10\), i.e. \(61.10\) and \(-31.10\) MPa. The eigenvector for \(61.10\) satisfies \((50 - 61.10)x + 30y = 0\), so \(y/x = 0.370\): the direction \((0.938, 0.347, 0)\), at \(20.3^\circ\) from \(\vec e_1\) (check: \(\tan2\theta = 2\cdot30/(50 + 20) = 0.857\), \(2\theta = 40.6^\circ\) ✓). The principal stresses are \(61.10,\ 10,\ -31.10\) MPa and the largest shear is \((61.10 + 31.10)/2 = 46.10\) MPa, on the planes bisecting the first and third principal directions.

4. The matrix \(s\begin{pmatrix}0&1&0\\1&0&0\\0&0&0\end{pmatrix}\) has eigenvalues \(s, 0, -s\) with eigenvectors \((1,1,0)/\sqrt2\), \(\vec e_3\), \((1,-1,0)/\sqrt2\). Pure shear is, in the basis rotated by \(45^\circ\), an equal tension and compression. A brittle material cracks across the direction of maximum tension, \((1,1,0)/\sqrt2\): on planes at \(45^\circ\) to the stick's axis — the helical fracture of twisted chalk. The shear planes themselves, \(\vec n = \vec e_1\) and \(\vec e_2\), feel no normal stress at all.

Check. Dimensions: every entry in MPa ✓. The sum of principal stresses \(61.10 + 10 - 31.10 = 40 = \operatorname{tr}\sigma\) ✓, and their product \(61.10\times10\times(-31.10) = -19\,000 = \det\sigma\) (\(10\times(-1900)\)) ✓. The normal stress on any plane lies between the smallest and largest principal stresses: \(-31.10 \le 37.78 \le 61.10\) ✓.

Idea. Tractions act on areas and inertia on volumes. On a small enough region the surface forces must therefore balance by themselves, and "balance on every tetrahedron" is the statement that the traction is linear in the normal. Angular momentum then adds one more balance, whose only surviving term is the antisymmetric part of \(\sigma\).

1. Areas. Let \(T\) be the tetrahedron and \(\vec e_i\) a constant field. The divergence theorem: \(0 = \int_T\nabla\cdot\vec e_i\,\mathrm dV = \oint_{\partial T}\vec e_i\cdot\vec n\,\mathrm dA\). On the oblique face \(\vec e_i\cdot\vec n = n_i\); on the face in the plane \(x_i = \text{const}\) the outward normal is \(-\vec e_i\), giving \(-A_i\); on the other two coordinate faces \(\vec e_i\cdot\vec n = 0\). So \(An_i - A_i = 0\). \(\square\)

2a. The volume estimate. From the balance law, \(|\oint_{\partial P}\vec t\,\mathrm dA| \le C\,\mathrm{vol}(P)\) for small \(P\) near \(\vec x_0\), with \(C\) built from the bounds on \(\rho\), the acceleration and \(\vec b\).

2b. Third law. Take the cylinder of radius \(r\) and height \(h\) with axis \(\vec n\) through \(\vec x_0\). The side has area \(2\pi rh\) and \(\vec t\) is bounded near \(\vec x_0\) (joint continuity on a compact set of points and unit normals), so as \(h\to0\) the side term and the volume bound vanish and \(\int_{D_r}(\vec t(\vec x,\vec n) + \vec t(\vec x,-\vec n))\,\mathrm dA = 0\) over the disc. Divide by \(\pi r^2\) and let \(r\to0\): by continuity, \(\vec t(\vec x_0,\vec n) + \vec t(\vec x_0,-\vec n) = 0\).

2c. Linearity. Scale the tetrahedron by \(\varepsilon\), so the oblique face has area \(a\varepsilon^2\) and the volume is \(a\varepsilon^3/3\). By continuity each face integral is its area times the traction at \(\vec x_0\), plus area \(\times\,o(1)\). The volume estimate gives

\[ \Big|a\varepsilon^2\Big(\vec t(\vec n) + \sum_in_i\,\vec t(-\vec e_i)\Big)\Big| \le C\frac{a\varepsilon^3}{3} + a\varepsilon^2\,o(1). \]

Divide by \(a\varepsilon^2\), let \(\varepsilon\to0\), and use 2b: \(\vec t(\vec n) = \sum_in_i\,\vec t(\vec e_i)\). For normals in other octants, the coordinate face in the plane \(x_i = x_{0,i}\) has outward normal \(-\operatorname{sgn}(n_i)\vec e_i\) and area \(A|n_i|\), contributing \(A|n_i|\,\vec t(-\operatorname{sgn}(n_i)\vec e_i) = -An_i\,\vec t(\vec e_i)\); the same formula results, and zero components follow by continuity in \(\vec n\).

2d. The tensor. Define \(\sigma(\vec m,\vec n) = \sum_{i,j}m_in_j\,\vec e_i\cdot\vec t(\vec e_j)\). This is bilinear by construction, and for a unit \(\vec n\) the formula of 2c gives \(\vec m\cdot\vec t(\vec n) = \sum_jn_j\,\vec m\cdot\vec t(\vec e_j) = \sigma(\vec m,\vec n)\). \(\square\)

3. Symmetry. For a material region \(\Omega\) and fixed \(i\ne j\), the angular momentum balance is

\[ \frac{\mathrm d}{\mathrm dt}\int_\Omega\rho(x_iv_j - x_jv_i) = \oint_{\partial\Omega}(x_it_j - x_jt_i)\,\mathrm dA + \int_\Omega\rho(x_ib_j - x_jb_i). \]

Left. By the corollary, \(\int\rho\,\frac{\mathrm D}{\mathrm Dt}(x_iv_j - x_jv_i)\), and since \(\mathrm Dx_i/\mathrm Dt = v_i\), the integrand is \(\rho(v_iv_j + x_ia_j - v_jv_i - x_ja_i) = \rho(x_ia_j - x_ja_i)\). Surface term. With \(t_j = \sigma_{jk}n_k\) it is the flux of \(w_k = x_i\sigma_{jk} - x_j\sigma_{ik}\), and \(\partial_kw_k = \delta_{ik}\sigma_{jk} - \delta_{jk}\sigma_{ik} + x_i\partial_k\sigma_{jk} - x_j\partial_k\sigma_{ik} = \sigma_{ji} - \sigma_{ij} + x_i(\operatorname{div}\sigma)_j - x_j(\operatorname{div}\sigma)_i\). Combine. Moving everything to one side,

\[ \int_\Omega\Big[x_i\big(\rho a_j - (\operatorname{div}\sigma)_j - \rho b_j\big) - x_j\big(\rho a_i - (\operatorname{div}\sigma)_i - \rho b_i\big)\Big] = \int_\Omega(\sigma_{ji} - \sigma_{ij}). \]

Cauchy's equation \(\rho\vec a = \operatorname{div}\sigma + \rho\vec b\) makes the left side zero. So \(\int_\Omega(\sigma_{ji} - \sigma_{ij}) = 0\) for every region (every ball is a material region at the instant considered), and the localisation lemma gives \(\sigma_{ij} = \sigma_{ji}\) everywhere. \(\square\)

4. The hidden hypothesis. That the only torques are moments of forces: no body couples, and no couple stresses (torques transmitted across surfaces independently of the forces). Newton's laws for point particles say nothing about this, because a point particle has no orientation for a torque to act on; the assumption fails for materials whose constituents carry their own angular momentum or are twisted by a field, such as a ferrofluid in a magnetic field. With a body couple, the right side of the balance acquires \(\int_\Omega m_{ij}\), and the same calculation ends in \(\int_\Omega(\sigma_{ji} - \sigma_{ij} + m_{ij}) = 0\), hence \(\sigma_{ij} - \sigma_{ji} = m_{ij}\).

Check. Dimensions in part 4: \(m_{ij}\) is torque per volume, \(\mathrm{N\,m\,m^{-3}} = \mathrm{Pa}\), the same as a stress ✓. Special case: for a fluid at rest, \(\sigma = -pI\) is symmetric, consistent with part 3 ✓. The two powers of \(\varepsilon\) are exactly what the checkpoint asks you to name: \(\varepsilon^2\) for tractions, \(\varepsilon^3\) for inertia and body forces.

Another way

For part 3, the small-cube argument: shears on a cube of side \(\ell\) give torque \((\sigma_{21} - \sigma_{12})\ell^3\) about \(\vec e_3\), while the moment of inertia is \(\rho\ell^5/6\) and the other torques are \(O(\ell^4)\). The angular acceleration would be of order \(\ell^{-2}\), unbounded as \(\ell\to0\), unless \(\sigma_{21} = \sigma_{12}\). This is quicker and persuasive, but it assumes uniform stress over the cube; the integral proof does not.

Idea. The three moves of the deck: budget, divergence theorem, localisation.

1. The budget is the first line of rung 3: the amount in \(V\) changes by what flows out through \(\partial V\) (with the outward normal, hence the minus sign), what decays inside, and what the source adds. Leibniz's rule (unit 18, extended to a bounded region) moves \(\mathrm d/\mathrm dt\) inside as \(\partial_t\); the divergence theorem (unit 05) turns the boundary term into \(-\int_V\nabla\cdot(C\vec v - D\nabla C)\). Everything is now a volume integral over the arbitrary \(V\):

\[ \int_V\big[\partial_tC + \nabla\cdot(C\vec v) - D\nabla^2C + kC - s\big]\,\mathrm dV = 0 . \]

If the integrand is continuous, the localisation lemma gives

\[ \partial_tC + \nabla\cdot(C\vec v) = D\nabla^2C - kC + s . \]

2. \(\nabla\cdot(C\vec v) = \vec v\cdot\nabla C + C\,\nabla\cdot\vec v\), and the last term vanishes for incompressible flow. So \(\partial_tC + \vec v\cdot\nabla C = \mathrm DC/\mathrm Dt = D\nabla^2C - kC + s\): following a water particle, the concentration changes only by diffusion, decay and the source.

3. Continuity reads \(\mathrm D\rho/\mathrm Dt + \rho\,\nabla\cdot\vec v = 0\). With \(\rho \gt 0\), \(\mathrm D\rho/\mathrm Dt = 0\) if and only if \(\nabla\cdot\vec v = 0\). Example: a lake stratified in density, \(\rho = \rho(z)\), with a horizontal current \(\vec v = (U(z), 0, 0)\). Then \(\nabla\cdot\vec v = \partial_xU(z) = 0\), and \(\mathrm D\rho/\mathrm Dt = \partial_t\rho + U\partial_x\rho = 0\), though \(\rho\) varies with depth.

Check. Dimensions: \(\partial_tC\) is \(\mathrm{kg\,m^{-3}s^{-1}}\); \(D\nabla^2C\) needs \(D\) in \(\mathrm{m^2s^{-1}}\), as a diffusivity must be; \(kC\) needs \(k\) in \(\mathrm{s^{-1}}\) ✓. Limits: with \(D = k = s = 0\) and \(C = \rho\) the law is the continuity equation ✓; with \(\vec v = 0\), \(k = s = 0\) it is the diffusion equation, unit 18's parabolic example ✓. A number: a pollutant with \(k = 10^{-5}\ \mathrm{s^{-1}}\) in a river flowing at \(1\ \mathrm{m\,s^{-1}}\), with diffusion negligible, decays by a factor \(e\) over \(1/k = 10^5\) s, i.e. \(100\) km downstream.

Idea. A steady velocity field can still accelerate every particle, because the particle moves to where the velocity is different. The material derivative is the bookkeeping for exactly that.

1. \(\mathrm D\vec v/\mathrm Dt = -\Omega^2(x,y,0)\) (rung 3; the deck's rotation example). Euler: \(\rho\,\mathrm D\vec v/\mathrm Dt = -\nabla p - \rho g\vec e_3\), so \(\nabla p = \rho(\Omega^2x, \Omega^2y, -g)\). Integrating, \(p = p_c + \tfrac12\rho\Omega^2(x^2 + y^2) - \rho g z\), where \(p_c\) is the pressure on the axis at \(z = 0\). This is a gradient field, as it must be for a solution to exist. The free surface \(p = p_{\text{atm}}\), taking the surface's lowest point at \(z = 0\) (so \(p_c = p_{\text{atm}}\)), is \(z = \Omega^2r^2/2g\), a paraboloid. With \(\Omega = 2\pi\ \mathrm{s^{-1}}\), \(r = 0.1\) m: \(z = (39.48\times0.01)/(2\times9.807) = 0.0201\) m, about \(2.0\) cm.

2. Nothing depends on \(t\), so \(\partial_t\vec v = 0\). \(\vec v\cdot\nabla = \alpha(x\partial_x - y\partial_y)\), and \((\vec v\cdot\nabla)\vec v = \alpha(x\cdot\alpha,\ -y\cdot(-\alpha),\ 0) = \alpha^2(x, y, 0)\ne0\). Particles speed up along \(x\) and slow down along \(y\) as they pass the stagnation point. \(\nabla\cdot\vec v = \alpha - \alpha = 0\): incompressible.

3. \(\nabla\cdot\vec v = 3\alpha\), \(\nabla\rho = 0\). Continuity: \(\dot\rho + 3\alpha\rho = 0\), so \(\rho = \rho_0e^{-3\alpha t}\). Transport theorem with \(f = 1\): \(\dot V = \int_{\Omega_t}3\alpha\,\mathrm dV = 3\alpha V\), so \(V = V_0e^{3\alpha t}\). The mass \(\rho V = \rho_0V_0\) is constant ✓. Directly: each particle moves as \(\vec x = \vec X e^{\alpha t}\), so lengths grow as \(e^{\alpha t}\) and volumes as \(e^{3\alpha t}\) ✓; and \(J = \det(e^{\alpha t}I) = e^{3\alpha t}\) satisfies \(\dot J = 3\alpha J\), Euler's expansion formula ✓.

Check. Part 1 dimensions: \(\Omega^2r^2/g\) is \(\mathrm{s^{-2}m^2/(m\,s^{-2})} = \mathrm m\) ✓; limit \(\Omega\to0\): a flat surface ✓. The paraboloid is also what the rotating-frame picture of unit 03 gives, with the centrifugal potential \(-\tfrac12\Omega^2r^2\) per unit mass added to gravity's \(gz\) ✓.

Idea. Statics is Cauchy's equation with no acceleration; a closure — isothermal or adiabatic — turns it into an ODE. The same closure question as for sound, with a different observable.

1. \(0 = \operatorname{div}(-pI) + \rho\vec g = -\nabla p - \rho g\vec e_3\): \(p\) depends on \(z\) alone and \(\mathrm dp/\mathrm dz = -\rho g\).

2. \(p' = -(Mg/RT_0)p\), so \(p = p_0e^{-z/H}\) with \(H = RT_0/Mg = 8.314\,462\times293.15/(0.028\,965\times9.806\,65) = 8581\) m \(= 8.58\) km. At \(5\) km, \(p/p_0 = e^{-5/8.581} = 0.558\).

3. Rung 3 gives \(T' = -\Gamma\), constant, with \(\Gamma = \frac{0.4}{1.4}\times\frac{0.028\,965\times9.806\,65}{8.314\,462} = 9.76\times10^{-3}\ \mathrm{K\,m^{-1}}\), i.e. \(9.76\) K per km — the dry adiabatic lapse rate. So \(T = T_0 - \Gamma z\), zero at \(z = T_0/\Gamma = 293.15/9.76 = 30.0\) km, where the model's atmosphere ends. At \(5\) km, \(T = 244.3\) K and \(p/p_0 = (T/T_0)^{\gamma/(\gamma - 1)} = (0.8335)^{3.5} = 0.529\).

Check. Dimensions: \(RT/Mg\) is \(\mathrm{J\,mol^{-1}}/(\mathrm{kg\,mol^{-1}\,m\,s^{-2}}) = \mathrm m\) ✓; \(Mg/R\) is \(\mathrm{K\,m^{-1}}\) ✓. Limit \(\gamma\to1\): \(\Gamma\to0\), the isothermal atmosphere ✓. For small \(z\) the two agree: expanding \((1 - \Gamma z/T_0)^{\gamma/(\gamma - 1)}\) to first order gives \(1 - z/H\), as does \(e^{-z/H}\) ✓. The real troposphere cools at about \(6.5\) K per km, between the two, because rising moist air releases latent heat — neither of these idealised closures is exact for weather, as neither was exactly right for Newton.

Idea. Volume change and shape change are orthogonal, each has its own modulus, and stability says both moduli are positive. Every inequality between wave speeds follows from that.

1. \(\mathrm dW = \lambda(\operatorname{tr}\varepsilon)\operatorname{tr}(\mathrm d\varepsilon) + 2\mu\,\varepsilon:\mathrm d\varepsilon = \big(\lambda(\operatorname{tr}\varepsilon)I + 2\mu\varepsilon\big):\mathrm d\varepsilon = \sigma:\mathrm d\varepsilon\), using \(\operatorname{tr}(\mathrm d\varepsilon) = I:\mathrm d\varepsilon\) and the symmetry of \(\varepsilon\).

2. With rung 2's identity, \(W = \tfrac12\lambda(\operatorname{tr}\varepsilon)^2 + \tfrac13\mu(\operatorname{tr}\varepsilon)^2 + \mu|\varepsilon_{\text{dev}}|^2 = \tfrac12(\lambda + \tfrac23\mu)(\operatorname{tr}\varepsilon)^2 + \mu|\varepsilon_{\text{dev}}|^2\). If \(K, \mu \gt 0\), \(W\) is a sum of nonnegative terms, zero only when \(\operatorname{tr}\varepsilon = 0\) and \(\varepsilon_{\text{dev}} = 0\), i.e. \(\varepsilon = 0\). Only if: \(\varepsilon = I\) gives \(W = \tfrac92K\), so \(K \gt 0\); a nonzero trace-free \(\varepsilon\) gives \(W = \mu|\varepsilon|^2\), so \(\mu \gt 0\). \(\square\)

3. From rung 3, \(\rho(c_L^2 - c_T^2) = K + \mu/3 \gt 0\), and \(c_L^2/c_T^2 = K/\mu + 4/3 \gt 4/3\). Yes, \(\lambda = K - \tfrac23\mu\) can be negative when \(K \lt \tfrac23\mu\): nothing forbids it, and some foams and engineered "auxetic" materials realise it. What stability forbids is \(K\le0\) or \(\mu\le0\).

4. \(\lambda = \mu\): \(K = \tfrac53\mu\), and \(c_L^2/c_T^2 = 3\), \(c_L/c_T = \sqrt3 = 1.732\).

Check. Dimensions: \(W\) in \(\mathrm{J\,m^{-3}} = \mathrm{Pa}\), moduli in Pa, strain dimensionless ✓. Limit \(\mu\to0\): \(W = \tfrac12K(\operatorname{tr}\varepsilon)^2\), a fluid storing energy only in compression, and \(c_L^2/c_T^2\to\infty\) ✓. Number: steel with \(\lambda = 115\), \(\mu = 80\) GPa gives \(K = 168.3\) GPa and \(c_L/c_T = 1.854 \gt 1.155\) ✓.

Idea. In this geometry the fluid particles never change speed, so inertia drops out entirely and the pressure push is balanced by viscous drag alone.

1. Rung 2 kills the nonlinear term; the \(y\) and \(z\) components give \(\partial_yp = \partial_zp = 0\). The \(x\) component, \(0 = -\partial_xp + \eta u''\), gives \(u'' = -G/\eta\). With \(u(0) = u(h) = 0\):

\[ u(y) = \frac{G}{2\eta}\,y\,(h - y). \]

2. Maximum at \(y = h/2\): \(u_{\max} = Gh^2/8\eta = 100\times10^{-6}/(8\times1.002\times10^{-3}) = 0.0125\ \mathrm{m\,s^{-1}}\). Mean: \(\bar u = \frac1h\int_0^hu = Gh^2/12\eta = \tfrac23u_{\max} = 8.32\ \mathrm{mm\,s^{-1}}\). Flux per unit width \(q = \bar uh = Gh^3/12\eta = 8.32\times10^{-6}\ \mathrm{m^2s^{-1}}\).

3. The shear stress is \(\sigma_{12} = \eta u'(y) = \frac{G}{2}(h - 2y)\), which is \(Gh/2\) at \(y = 0\) and \(-Gh/2\) at \(y = h\). The fluid drags each plate forward, in the direction of flow, with shear stress \(Gh/2 = 100\times10^{-3}/2 = 0.05\) Pa, and each plate holds the fluid back with the same stress. Force balance on a slab of length \(L\) per unit width: the pressure difference pushes with \(GL\cdot h\); the two walls resist with \(2\times(Gh/2)L = GhL\) ✓.

4. \(Re = 998.2\times0.00832\times0.001/1.002\times10^{-3} = 8.3\). Plane channel flow stays laminar up to Reynolds numbers of order a thousand (cited: Tritton, Physical Fluid Dynamics, the chapter on instability), so yes.

Check. Dimensions: \(Gh^2/\eta\) is \(\mathrm{Pa\,m^{-1}\,m^2/(Pa\,s)} = \mathrm{m\,s^{-1}}\) ✓. Limit \(\eta\to\infty\): no flow ✓; \(G\to0\): no flow ✓. The flux grows as \(h^3\): halving the gap cuts the flow eightfold, which is why narrowing a pipe is so costly (in a circular pipe, as \(r^4\) — Poiseuille's law).

Idea. The speed of sound is set by stiffness over inertia; a light gas is low inertia, a liquid is enormously stiff.

1. \(c_{\text{He}} = \sqrt{\tfrac53\times8.314\,462\times293.15/0.004\,003} = 1007\ \mathrm{m\,s^{-1}}\). Ratio to air's \(343.2\): \(\sqrt{(1.6667/1.4)(28.965/4.003)} = \sqrt{1.1905\times7.236} = 2.93\). The throat's resonant frequencies rise by the same factor, \(2.9\), which is the high thin voice. (The pitch of the vocal folds, set by their tension and mass, barely changes; what shifts is the resonances that give vowels their colour.)

2. \(c = \sqrt{2.2\times10^9/998.2} = 1485\ \mathrm{m\,s^{-1}}\), within \(0.2\%\) of the measured \(1482\) (the modulus is given to two figures). Isothermal and adiabatic moduli differ by the factor \(\gamma\), so the speeds differ by \(\sqrt{1.007} = 1.0035\): \(0.35\%\). For air the same factor is \(\sqrt{1.4} = 1.18\). A liquid barely warms when compressed, because its thermal expansion is small; that is the physical reason its \(\gamma\) is so close to 1, and why Newton would have got water nearly right.

Check. Dimensions: \(\sqrt{\mathrm{Pa}/(\mathrm{kg\,m^{-3}})} = \mathrm{m\,s^{-1}}\) ✓. Limit: water is about \(1.5\times10^4\) times stiffer than air (\(2.2\times10^9/1.42\times10^5\)) and \(830\) times denser; the speed ratio should be \(\sqrt{15\,500/829} = 4.3\), and \(1485/343.2 = 4.33\) ✓.

Idea. The linearised equations have a conserved quadratic, exactly as unit 18's string did; finding it is a matter of multiplying each equation by the right field and adding.

1. Rung 3, then \(\vec v\cdot\nabla p' + p'\nabla\cdot\vec v = \nabla\cdot(p'\vec v)\). So \(\partial_te = -\nabla\cdot\vec S\). \(\square\)

2. For fixed \(V\), Leibniz and the divergence theorem give \(\frac{\mathrm d}{\mathrm dt}\int_Ve = -\oint_{\partial V}p'\vec v\cdot\vec n\,\mathrm dA\): the acoustic energy in a region changes only by the rate at which the pressure excess does work on the region's boundary. Potential energy: compressing gas of volume \(V_0\) against the excess pressure takes work \(-\int p'\,\mathrm dV\). With \(p' = c^2\rho'\) and \(\mathrm dV/V_0 = -\mathrm d\rho'/\rho_0\) to first order, the work is \(\int_0^{\rho'}c^2r\,(V_0/\rho_0)\,\mathrm dr = V_0\,c^2\rho'^2/2\rho_0 = V_0\,p'^2/2\rho_0c^2\). So the second term is stored elastic energy per unit volume, the fluid's analogue of the string's \(\tfrac12Tu_x^2\).

3. \(|\vec v| = \hat p/\rho_0c = 1/(1.204\times343.2) = 2.42\ \mathrm{mm\,s^{-1}}\) in amplitude. \(|\vec S| = p'^2/\rho_0c\), whose time average is \(\hat p^2/2\rho_0c = 1/(2\times413.3) = 1.21\times10^{-3}\ \mathrm{W\,m^{-2}}\). Level: \(p_{\text{rms}} = 0.707\) Pa, so \(20\log_{10}(0.707/2\times10^{-5}) = 91.0\) dB.

Check. Dimensions: \(\mathrm{J\,m^{-3}}\) for \(e\), \(\mathrm{W\,m^{-2}}\) for \(\vec S\), so \(\partial_te\) and \(\nabla\cdot\vec S\) are both \(\mathrm{W\,m^{-3}}\) ✓. Special case: in the plane wave kinetic and potential densities are equal and \(\vec S = c\,e\,\vec n\) — energy moves at the sound speed ✓. The intensity is tiny: the loud \(91\) dB wave delivers a milliwatt per square metre, which is why the ear must be so sensitive.

Idea. Newton and Laplace agree on everything except whether the gas exchanges heat during one oscillation. A dimensional comparison of two times settles it, and the factor \(\sqrt\gamma\) is the fingerprint of the wrong choice.

1. Rung 3 gives \(TV^{R/c_V}\) constant; with \(T = pV/nR\), \(pV^{1 + R/c_V}\) is constant. For fixed mass \(V\propto1/\rho\), so \(p = A\rho^\gamma\) and \(\mathrm dp/\mathrm d\rho = \gamma A\rho^{\gamma - 1} = \gamma p/\rho\). Diatomic: \(\gamma = 1 + R/\tfrac52R = \tfrac75 = 1.4\).

2. Newton: \(c_N = \sqrt{RT/M}\); Laplace: \(c_L = \sqrt{1.4}\,c_N\).

• \(0\,{}^\circ\mathrm C\): \(c_N = \sqrt{8.314\,462\times273.15/0.028\,965} = 280.0\ \mathrm{m\,s^{-1}}\), \(c_L = 331.3\ \mathrm{m\,s^{-1}}\).
• \(20\,{}^\circ\mathrm C\): \(c_N = 290.1\ \mathrm{m\,s^{-1}}\), \(c_L = 343.2\ \mathrm{m\,s^{-1}}\).

Newton is low by \(1 - 1/\sqrt{1.4} = 15.5\%\) at every temperature, since the ratio is \(\sqrt\gamma\) and \(\gamma\) does not depend on \(T\). (Equivalently, the truth exceeds Newton by \(18.3\%\).) Laplace's values match the measured \(331.3\) and \(343.2\ \mathrm{m\,s^{-1}}\).

3. At \(1\) kHz, \(\omega = 6283\ \mathrm{s^{-1}}\), \(\lambda/2\pi = c/\omega = 0.0546\) m. The only time built from a length \(L\) and a diffusivity is \(L^2/\kappa\): \((0.0546)^2/2.14\times10^{-5} = 140\) s, against \(1/\omega = 1.6\times10^{-4}\) s. Heat has about a millionth (\(1/8.8\times10^5\)) of the time it needs to even out; each compression is adiabatic.

4. \((c/\omega)^2/\kappa = 1/\omega\) at \(\omega = c^2/\kappa = 343.2^2/2.14\times10^{-5} = 5.5\times10^9\ \mathrm{s^{-1}}\), where \(\lambda/2\pi = \kappa/c = 6.2\times10^{-8}\) m \(= 0.062\ \mu\mathrm m\) — the mean free path. At such wavelengths air is not a continuum and neither Euler's equations nor Boyle's law describe it. Newton's hypothesis is therefore wrong wherever his equations are right: a clean falsification, with one identifiable physical assumption at fault and a number, \(\sqrt{1.4}\), pointing at it.

Check. Dimensions: \(RT/M\) in \(\mathrm{m^2s^{-2}}\); \(L^2/\kappa\) in s; \(c^2/\kappa\) in \(\mathrm{s^{-1}}\) ✓. Limit \(\gamma\to1\): the two closures coincide ✓. \(c\propto\sqrt T\): \(343.2\sqrt{273.15/293.15} = 331.3\) ✓. The checkpoint's \(2\%\) window at \(20\,{}^\circ\mathrm C\) is \(336\)–\(350\ \mathrm{m\,s^{-1}}\); Laplace lands inside it, Newton does not.

Idea. The plane-wave theorem says transverse waves travel at \(\sqrt{\mu/\rho}\). A layer that stops them has \(\mu = 0\) — no resistance to static shear — and that is the definition of a fluid.

1. Mantle: \(c_L = \sqrt{1047.2\times10^9/5566} = 13.72\ \mathrm{km\,s^{-1}}\), \(c_T = \sqrt{293.8\times10^9/5566} = 7.265\ \mathrm{km\,s^{-1}}\). Core: \(c_L = \sqrt{644.1\times10^9/9903} = 8.065\ \mathrm{km\,s^{-1}}\), \(c_T = 0\).

2. \(\rho(c_L^2 - c_T^2) = \lambda + \mu = K + \tfrac13\mu\), positive because stability requires \(K \gt 0\) and \(\mu \gt 0\) (problem P6). Mantle: \(c_L/c_T = 13.72/7.265 = 1.89 \gt 1.155\) ✓; its \(\lambda/\mu = 1.56\), so it is stiffer in compression than a Poisson solid.

3. A plane wave with displacement perpendicular to its direction of travel exists only with speed \(\sqrt{\mu/\rho}\). If \(\mu = 0\) that speed is zero: a transverse disturbance is not restored and does not propagate. S-waves entering the core are therefore not transmitted as S-waves, and none emerge on the far side as direct S. Conversely, if the core were solid with any \(\mu \gt 0\), S-waves would cross it. So the shadow says \(\mu = 0\): liquid. The P-wave evidence is weaker because P-waves do cross the core, merely slowed from \(13.72\) to \(8.07\ \mathrm{km\,s^{-1}}\) and refracted — which shows a different material, not a liquid one.

4. Straight rays: \(\theta = \arccos(3480/6371) = 56.9^\circ\), so the grazing chord ends \(2\theta = 113.8^\circ\) from the epicentre. Speed rises with depth in the mantle, so rays bend back towards the surface and the grazing ray emerges nearer the epicentre; the observed edge is near \(103^\circ\).

Check. Dimensions: \(\sqrt{\mathrm{Pa}/(\mathrm{kg\,m^{-3}})} = \mathrm{m\,s^{-1}}\) ✓. Limit \(\mu\to0\) on the mantle side would give \(c_L = \sqrt{K/\rho}\), with \(K = \lambda + \tfrac23\mu = 655.5\) GPa, and \(c_T = 0\) ✓. The core's bulk modulus, \(644\) GPa, is comparable to the mantle's \(655\) GPa: the core is not soft, only unable to resist shear.

Idea. Both waves leave together; the slower one falls behind at a fixed rate per kilometre, so the delay is a distance measurement. Three distances fix a point in the plane.

1. The P-wave arrives at \(d/c_L\) and the S-wave at \(d/c_T\) after the origin time, so \(\Delta t = d(1/c_T - 1/c_L)\), positive because \(c_L \gt c_T\). Here \(d = 8.4\,\Delta t\): \(d_A = 49.98\), \(d_B = 80.64\), \(d_C = 58.30\) km.

2. \(x = (49.98^2 - 80.64^2 + 10\,000)/200 = (2498.0 - 6502.8 + 10\,000)/200 = 29.98\) km, and from \(A\) and \(C\), \(y = (d_A^2 - d_C^2 + 90^2)/180 = (2498.0 - 3398.4 + 8100)/180 = 40.00\) km. The epicentre is at \((30.0, 40.0)\) km. Check against the first circle: \(\sqrt{29.98^2 + 40.00^2} = 49.98\) ✓.

3. The P-wave took \(d_A/c_L = 49.98/6.0 = 8.33\) s to reach \(A\), so the earthquake happened \(8.33\) s before the P arrival there. With two stations the two circles generally meet in two points, mirror images in the line joining the stations; the third circle picks one. (With data carrying errors, three circles do not quite meet, and more stations are used with a least-squares fit.)

Check. Dimensions: \(\mathrm s\times(\mathrm{km\,s^{-1}})\) gives km ✓. Special case: a station at the epicentre records no delay ✓. The rounding of the delays to \(0.01\) s corresponds to \(0.08\) km, the size of the discrepancies above ✓. The rule of thumb "\(8\) km per second of S–P delay" used by seismologists for local earthquakes is this \(8.4\).