Part C · fields and the crisis · spine

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

A deformable body has no single "force on it", only a force on every surface inside it — and proving that this force is bilinear in the surface is the course's first tensor that is derived rather than defined.

needs 16, 05, 18~2½ h slides + 5 h problemsproblems: ./mr problems 22

Unit 18 took a chain of masses to the continuum in one dimension and found a field, a wave equation, and a local conservation law $\partial_t e + \partial_x S = 0$. This unit does the same in three dimensions for a solid or a fluid, and three things change.

The flux of momentum is not a vector. In one dimension the string's tension is a number. In three, the force across a small surface depends on how the surface is turned, and section 1 proves — from Newton's second law and nothing else — that the dependence is linear. The object that results is unit 16's (0,2)-tensor, met this time as a field, and met because it was forced on us.
The engine is unit 05's divergence theorem. Every conservation law in this unit is first stated as a budget for a finite region — what is inside changes only by what crosses the boundary — and then converted into a partial differential equation by one application of the divergence theorem and one localisation lemma. Section 2 does this twice, for mass and for momentum, and a third time for angular momentum, where it yields the symmetry of stress.
The equations do not close by themselves. Conservation laws give four equations for ten unknowns. The missing six are a statement about the material, and section 3 shows that isotropy cuts a linear material's 36 possible constants down to two.

How to read this deck

One gas runs through the unit: still air at $20\,{}^\circ\mathrm C$. Section 1 finds its stress; section 2 writes the two conservation laws it obeys; section 3 finds that those laws need one more equation, an equation of state; section 4 linearises everything about rest, obtains the wave equation, and computes the speed of sound twice — Newton's way, which is wrong by $15.5\%$, and Laplace's, which is right; section 5 sees the same computation as the $\mu = 0$ case of elastic waves in a solid, and uses the difference between solids and fluids to find the Earth's liquid core.

Proofs are on the slides where they are one line and in these notes under Full proof where they are not; results imported from elsewhere are marked cited, with where to find the proof. Terms with a dotted underline are in the glossary (G); N opens these notes.

What you are assumed to have. Unit 05's divergence theorem, change of variables and Jacobian; unit 16's definition of a tensor as a multilinear map, its components, the transformation law as a theorem, the Einstein summation convention, symmetric and antisymmetric parts, and the spectral theorem producing principal axes; unit 18's wave equation, d'Alembert's travelling waves, Leibniz's rule and local conservation law; unit 03's chain rule, $SO(3)$ and $\mathfrak{so}(3)$ with the hat map; unit 04's gradient, curl and the Riesz isomorphism; unit 01's existence theorem for flows; unit 00's dimensional analysis. The thermodynamics needed — the ideal-gas law and the first law for a gas — is physics, not mathematics, and is stated where it is used.

Conventions. Cartesian coordinates with an orthonormal basis throughout, so, as unit 16 explained, the distinction between upper and lower indices is invisible and all indices are written down; repeated indices are summed. $\partial_j = \partial/\partial x_j$.

22 · overview

One gas, one number, two closures

the object, and the question

Still air at $20\,{}^\circ\mathrm C$: $p_0 = 101\,325$ Pa, $\rho_0 = 1.204\ \mathrm{kg\,m^{-3}}$. A continuum: mass, momentum and force described by fields, with the molecules forgotten.

The question. How fast does a small pressure disturbance travel through it? Measured: $343\ \mathrm{m\,s^{-1}}$. Newton (1687), from first principles: $15\%$ too slow, an error that stood until Laplace (1816).

the one thing to watch

Every step of the route is forced by conservation laws except one: how pressure follows density. That single physical hypothesis is where Newton went wrong, and the number $\sqrt{1.4} = 1.183$ is its signature.

the route, sketched

1. The force per area across a surface inside a body is $\sigma\vec n$, $\vec n$ its unit normal, $\sigma$ a tensor (Cauchy). In still air $\sigma = -p_0 I$, a pure push.

2. Budgets of mass and momentum for every region become PDEs for density $\rho$ and velocity $\vec v$: continuity and Euler.

3. Disturb it slightly, $\rho = \rho_0 + \rho'$, keep first order, let $p$ be a function of $\rho$: $\rho'$ obeys a wave equation with speed $c$, $c^2 = \mathrm dp/\mathrm d\rho$ at $\rho_0$.

4. Newton: $p \propto \rho$ (Boyle), $c^2 = p_0/\rho_0$, $290\ \mathrm{m\,s^{-1}}$. Laplace: $p \propto \rho^{1.4}$ (adiabatic), $c^2 = 1.4\,p_0/\rho_0$, $343\ \mathrm{m\,s^{-1}}$.

The plan. §1 stress is a tensor · §2 budgets become PDEs · §3 a material law · §4 sound · §5 solids and the Earth's core. Dotted: glossary (G); solid: running example on hover; N: notes.

The route, step by step. Each step is a later slide. Step 1 is Cauchy's theorem, proved on The tetrahedron argument, and Pressure is the isotropic part for still air. Step 2 is Mass: from a budget to a field equation and Momentum: Cauchy's equation of motion, closed to Euler's equations on Ten unknowns, four equations. Step 3 is Linearising about rest. Step 4 is The closure: how does pressure follow density? and The speed of sound in air; why the adiabatic hypothesis is the right one is Newton's error, and Laplace's repair. The exponent $1.4$ is the air's adiabatic index $\gamma$, whose value section 4 derives from the first law of thermodynamics.

The continuum hypothesis, stated honestly. Air is molecules, about $2.5\times10^{25}$ of them per cubic metre, with a mean free path near $0.07\ \mu\mathrm m$ (cited: any kinetic-theory table, e.g. the CRC Handbook). A field such as $\rho(\vec x, t)$ means the mass in a small ball around $\vec x$ divided by its volume, for balls large compared with the mean free path and small compared with the scale on which anything interesting varies. A sound wave at audible frequencies has a wavelength of centimetres to metres, so the separation of scales is some seven orders of magnitude and the continuum description is excellent. Section 4 will find the single place where it would stop being so, and it is exactly where Newton's hypothesis would have become right.

The "18%". The measured speed exceeds Newton's prediction by the factor $\sqrt\gamma = 1.183$; equivalently Newton's value is $1/1.183 = 0.845$ of the truth, $15.5\%$ low. Both statements are about the same ratio. The popular figure "$280\ \mathrm{m\,s^{-1}}$" is Newton's formula evaluated at $0\,{}^\circ\mathrm C$, where the correct value is $331.3\ \mathrm{m\,s^{-1}}$; at the running example's $20\,{}^\circ\mathrm C$ the pair is $290.1$ against $343.2$.

section 1

Stress: the force on a surface

Newton's second law, applied to a shrinking tetrahedron, proves that the force across a surface is linear in its normal.

22 · stress

The force on a piece of a body is not a vector

Cut an imaginary region $P$ out of a deformable body. Two kinds of force act on it:

definitions

A body force acts on the bulk: $\int_P \rho\,\vec b\,\mathrm dV$, with $\vec b$ per unit mass (gravity: $\vec b = \vec g$).

A surface force is exerted across $\partial P$ by the material outside: $\oint_{\partial P}\vec t\,\mathrm dA$. The force per unit area $\vec t$ is the traction, in pascals.

hypothesis · Cauchy's postulate

$\vec t$ at a point depends on the surface through that point only via its outward unit normal: $\vec t = \vec t(\vec x, t, \vec n)$, continuous in $(\vec x, \vec n)$ jointly.

A region $P$ cut out of a body (schematic). Across each patch $\mathrm dA$ of its boundary the outside material exerts $\vec t\,\mathrm dA$, which need not point along the normal $\vec n$; gravity acts on every volume element inside.

So "the force at a point" is a function on the unit sphere, $\vec n \mapsto \vec t(\vec n)$ — a priori a vector-valued function of a direction, not a vector.

What to look for in the figure. Two kinds of arrow, and they scale differently: the body-force arrows sit on volume elements, the traction on a patch of area. The traction at a boundary point is drawn deliberately off the normal: nothing so far says it must be along $\vec n$, and in a solid it generally is not. Only in a fluid at rest (Pascal's law, at the end of this section) does $\vec t$ turn out parallel to $\vec n$ on every surface.

About the postulate. Cauchy (1822–23) assumed that $\vec t$ depends on the surface only through $\vec n$ — not on its curvature, for instance. A theorem of Noll (1959) derives this from the balance law itself under mild hypotheses (cited: Truesdell, A First Course in Rational Continuum Mechanics, vol. 1, the chapter on the stress theorem). We take it as a hypothesis, and everything below is proved from it. Continuity in $\vec n$, not only in $\vec x$, is part of the hypothesis: the pillbox proof uses the bound it gives, and the tetrahedron proof uses it for normals with a zero component.

22 · stress

Same point, two surfaces, two tractions

why nothing simpler will do

A steel bar pulled at its ends: across a cut perpendicular to the bar, the material pulls along the bar. Across a cut parallel to the bar, it exerts nothing at all. Same point, two surfaces, two tractions.

The balance law for every region $P$ that moves with the material:

$$\frac{\mathrm d}{\mathrm dt}\int_P \rho\vec v\,\mathrm dV = \oint_{\partial P}\vec t\,\mathrm dA + \int_P\rho\,\vec b\,\mathrm dV$$

Newton's second law, for a piece of stuff rather than a particle.

One point of a pulled bar, two cuts through it. Across the cut facing $\vec e_1$ the far side pulls with $s\,\vec e_1$, $s$ the force per area; across the cut facing $\vec e_2$ it exerts nothing.

Why the balance law is Newton's second law. Chop $P$ into small parcels, each obeying $m\ddot{\vec x} = \vec F$. Forces between parcels inside $P$ cancel in pairs by the third law. What is left is the force from outside $P$: long-range forces act throughout the volume (the body force) and short-range forces — molecular forces with a range of nanometres — act only through a thin layer at the boundary, which in the continuum limit becomes a surface integral. That last step is the physical content of the word "surface force", and it is an idealisation: it is exactly what fails for a ferrofluid in a magnetic field, where section 2 will need a correction.

The bar, as data. The figure is the whole case against "a force per area at a point": the two cuts pass through the same point and receive different tractions, $s\,\vec e_1$ and $\vec 0$. The worked example later in this section turns the cut through every angle in between, and Cauchy's theorem says the answer for all of them is fixed by these two (and a third, facing $\vec e_3$, also zero).

The $\frac{\mathrm d}{\mathrm dt}$ of an integral over a moving region is made precise in section 2 by the Reynolds transport theorem. For section 1 only one consequence is needed: the left side is at most (bounded density × bounded acceleration) × volume.

22 · stress

Action and reaction across a surface

lemma · the volume estimate

If $\rho$, the acceleration and $\vec b$ are bounded near $\vec x_0$, there is $C$ with

$$\Big|\oint_{\partial P}\vec t\,\mathrm dA\Big| \le C\,\mathrm{vol}(P)$$

for every small region $P$ near $\vec x_0$. Proof: the balance law, and each volume integral is at most sup × volume. $\square$

lemma · Newton's third law for tractions $$\vec t(\vec x, -\vec n) = -\,\vec t(\vec x, \vec n)$$

Proof sketch. A flat pillbox, faces of area $A$ with normals $\pm\vec n$, thickness $h$. As $h \to 0$ the side contributes $O(h\sqrt A)$ and the volume is $Ah$, so the estimate gives $A\big(\vec t(\vec n) + \vec t(-\vec n)\big) = o(A)$. Divide by $A$, let $A \to 0$. $\square$

full proof in the notes (N)

The pillbox, thickness exaggerated. As $h \to 0$ the side and the volume vanish while the faces keep area $\pi r^2$, so the two face tractions must cancel.

Surface forces scale like area, volume terms like length cubed. Shrink a region and the surface forces must balance among themselves, to leading order. The third law between particles is an assumption; here, for tractions, it is a theorem.

Reading the figure. The top face has outward normal $\vec n$ and receives $\vec t(\vec n)$ from the material above; the bottom face has outward normal $-\vec n$ and receives $\vec t(-\vec n)$ from below. Squash the box: everything that scales with $h$ — the side and the mass inside — goes away, and what is left is the two faces alone, which can only be in balance if their tractions are opposite. The tractions need not be along $\vec n$; only their sum is constrained.

Full proof

Fix $\vec x_0$ and a unit vector $\vec n$. Let $D_r$ be the disc of radius $r$ centred at $\vec x_0$ in the plane perpendicular to $\vec n$, and let $P_{r,h}$ be the cylinder $D_r \times [-h/2, h/2]$ along $\vec n$. Its boundary has a top face (outward normal $\vec n$), a bottom face (outward normal $-\vec n$) and a curved side of area $2\pi r h$. By the volume estimate,

$$\Big|\int_{\text{top}}\vec t(\vec x,\vec n)\,\mathrm dA + \int_{\text{bottom}}\vec t(\vec x,-\vec n)\,\mathrm dA + \int_{\text{side}}\vec t\,\mathrm dA\Big| \le C\pi r^2 h .$$

By the postulate's joint continuity, $|\vec t|$ is bounded by some $K$ on the compact set of pairs $(\vec x,\vec n)$ with $|\vec x - \vec x_0| \le 1$ (say) and $|\vec n| = 1$; so the side term is at most $2\pi r h K$. Let $h \to 0$ with $r$ fixed: the side term and the right side vanish, and the top and bottom faces converge to $D_r$ itself, giving $\int_{D_r}\big(\vec t(\vec x,\vec n) + \vec t(\vec x,-\vec n)\big)\,\mathrm dA = 0$ for every $r$. Divide by $\pi r^2$ and let $r \to 0$: by continuity the average converges to the value at $\vec x_0$, so $\vec t(\vec x_0,\vec n) + \vec t(\vec x_0,-\vec n) = 0$. $\square$

Why this matters. The third law for point particles is an axiom, and a slightly suspicious one (magnetic forces between moving charges violate its strong form). The traction version is different in kind: it follows from the second law applied to a region too thin to have any mass. A thin enough slab cannot sustain a net force, because it has nothing to accelerate.

22 · stress

Cauchy's theorem

theorem · Cauchy's stress theorem (1823)

Under the postulate and the balance law, at each point there is a unique linear map $\sigma:\mathbb R^3 \to \mathbb R^3$ with

$$\vec t(\vec n) = \sigma\,\vec n \quad\text{for every unit } \vec n, \qquad t_i = \sigma_{ij}\,n_j ,$$

and its matrix is built from three tractions: $\sigma_{ij} = \vec e_i\cdot\vec t(\vec e_j)$. Column $j$ of $\sigma$ is the traction on a surface facing $+\vec e_j$.

checks

Dimensions: $\sigma_{ij}$ is force per area, pascals, like $\vec t$ ✓. Special case: $\vec n \to -\vec n$ gives $\sigma(-\vec n) = -\sigma\vec n$ — the third-law lemma, now automatic ✓. Coordinate faces: $\vec n = \vec e_j$ returns column $j$ ✓.

The tetrahedron at $\vec x_0$: an oblique face of area $A$, outward normal $\vec n$ and traction $\vec t(\vec n)$ — not along $\vec n$ in general — and three faces on the coordinate planes, with outward normals $-\vec e_i$ and areas $A\,n_i$.

What the theorem says in words. Nine numbers at a point — the three components of the tractions on three perpendicular planes — determine the traction on every plane through the point. That is not a definition and it is not obvious: a priori $\vec n \mapsto \vec t(\vec n)$ is an arbitrary continuous function on the sphere, specified by infinitely many numbers.

The index convention. $t_i = \sigma_{ij}n_j$: the first index says which component of force, the second which way the surface faces. Some books use the transpose. Once section 2 proves $\sigma$ symmetric, the two conventions agree, which is why the difference rarely bites; until then, keep to this one.

History. Cauchy presented the theorem to the Paris Academy in 1822 and published it in 1823 and 1827. It is the founding result of continuum mechanics, and it is remarkable how early it is: nothing like it existed for electromagnetism until Maxwell's stress tensor half a century later (unit 24).

The next slide proves it; the slide after says what kind of object $\sigma$ is.

22 · stress

The tetrahedron argument

lemma · the areas

A tetrahedron with oblique face of area $A$ and normal $\vec n$, $n_i > 0$: its coordinate faces have areas $A_i = A\,n_i$.

Proof. Unit 05's divergence theorem for the constant field $\vec e_i$: $0 = \int\nabla\!\cdot\vec e_i\,\mathrm dV = \oint\vec e_i\cdot\vec n\,\mathrm dA = A\,n_i - A_i$. $\square$

proof of Cauchy's theorem, for $n_i > 0$

Shrink the tetrahedron by a factor $\varepsilon$: areas scale as $\varepsilon^2$, the volume as $\varepsilon^3$. The volume estimate and continuity of $\vec t$ give

$$A\,\vec t(\vec n) + \sum_i A\,n_i\,\vec t(-\vec e_i) = O(\varepsilon^3) + o(\varepsilon^2).$$

Divide by $A \propto \varepsilon^2$, let $\varepsilon \to 0$, use the third-law lemma:

$$\vec t(\vec n) = \sum_i n_i\,\vec t(\vec e_i). \quad\square$$

The same argument one dimension down (schematic). The three face forces, traction × length, laid head to tail: balance closes the triangle, and that closure is $\vec t(\vec n) = n_1\vec t(\vec e_1) + n_2\vec t(\vec e_2)$.

That is linearity: $\vec t(\vec n) = n_j\,\vec t(\vec e_j)$, whose $i$-th component is $\sigma_{ij}n_j$.

The figure, computed. It is drawn for the stress $\sigma = \begin{pmatrix}1 & 0.45\\ 0.45 & 0.6\end{pmatrix}$ (arbitrary units) and a triangle whose hypotenuse has normal $\vec n = (0.63, 0.78)$. The face forces are $A\,\vec t(\vec n)$, $-A n_1\vec t(\vec e_1)$ and $-A n_2\vec t(\vec e_2)$, using the third-law lemma for the two legs, and they sum to zero to the drawing's precision. In three dimensions the triangle becomes a tetrahedron and the closed triangle of forces a closed quadrilateral of four, but nothing else changes.

Full proof

Fix $\vec x_0$ and a unit $\vec n$ with all $n_i > 0$. For $\varepsilon > 0$ let $T_\varepsilon$ be the tetrahedron with vertex $\vec x_0$ and the three vertices $\vec x_0 + (\varepsilon/n_i)\vec e_i$; its oblique face lies in the plane $\vec n\cdot(\vec x - \vec x_0) = \varepsilon$, has outward normal $\vec n$ and area $A_\varepsilon = a\varepsilon^2$ for a constant $a$, and the height from $\vec x_0$ is $\varepsilon$, so $\mathrm{vol}(T_\varepsilon) = a\varepsilon^3/3$. By the area lemma its face in the coordinate plane $x_i = x_{0,i}$ has outward normal $-\vec e_i$ and area $A_\varepsilon n_i$.

Write $\oint_{\partial T_\varepsilon}\vec t\,\mathrm dA$ face by face. On the oblique face $F$, $\int_F \vec t(\vec x,\vec n)\,\mathrm dA = A_\varepsilon\big(\vec t(\vec x_0,\vec n) + \vec\eta_\varepsilon\big)$ with $|\vec\eta_\varepsilon| \le \sup_{\vec x\in T_\varepsilon}|\vec t(\vec x,\vec n) - \vec t(\vec x_0,\vec n)| \to 0$ by continuity in $\vec x$. The same holds on each coordinate face with $-\vec e_i$ in place of $\vec n$. The volume estimate then gives

$$\Big|A_\varepsilon\,\vec t(\vec x_0,\vec n) + \sum_i A_\varepsilon n_i\,\vec t(\vec x_0,-\vec e_i)\Big| \le C\frac{a\varepsilon^3}{3} + A_\varepsilon\Big(|\vec\eta_\varepsilon| + \sum_i n_i|\vec\eta^{(i)}_\varepsilon|\Big).$$

Divide by $A_\varepsilon = a\varepsilon^2$ and let $\varepsilon \to 0$: the right side tends to $0$, so $\vec t(\vec x_0,\vec n) = -\sum_i n_i\,\vec t(\vec x_0,-\vec e_i) = \sum_i n_i\,\vec t(\vec x_0,\vec e_i)$, the last step by the third-law lemma.

Other octants. If some $n_i < 0$, build the tetrahedron on the same vertices $\vec x_0 + (\varepsilon/n_i)\vec e_i$, now on the negative $x_i$ axis. Its face in the plane $x_i = x_{0,i}$ has outward normal $-\operatorname{sgn}(n_i)\,\vec e_i$ and area $A|n_i|$ (the area lemma, with signs), so that face contributes $A|n_i|\,\vec t\big(-\operatorname{sgn}(n_i)\vec e_i\big) = -A\,n_i\,\vec t(\vec e_i)$ by the third-law lemma — exactly what it contributed when $n_i > 0$. The same limit gives $\vec t(\vec n) = \sum_i n_i\,\vec t(\vec e_i)$. If some $n_i = 0$ the tetrahedron degenerates; then use continuity of $\vec t$ in $\vec n$: the formula holds on the dense set of directions with no zero component, and both sides are continuous in $\vec n$.

Extension off the sphere. Define $\sigma\vec m = |\vec m|\,\vec t(\vec m/|\vec m|)$ for $\vec m \ne 0$ and $\sigma\vec 0 = \vec 0$. The formula just proved says this map is $\vec m \mapsto m_j\vec t(\vec e_j)$, which is linear. Uniqueness: a linear map is determined by its values on a basis.

22 · stress

What did the work: two powers of ε

what did the work

Two powers of $\varepsilon$. Tractions act on areas, $\sim\varepsilon^2$; inertia and gravity act on volumes, $\sim\varepsilon^3$. In the limit, the surface forces on an infinitesimal tetrahedron must balance exactly, and balance on a tetrahedron is linearity.

Other octants: use the tetrahedron with coordinate faces of normals $-\operatorname{sgn}(n_i)\,\vec e_i$; the same computation gives the same formula. A zero component: continuity in $\vec n$.

This is the only place in classical physics where a physical quantity is proved bilinear rather than assumed so.

The tetrahedron shrunk towards $\vec x_0$. Each halving of $\varepsilon$ quarters the face areas but divides the volume by eight, so the mass and gravity terms fall behind the surface terms by a factor $\varepsilon$.

Why this is the whole proof. The balance law has terms of two sizes. The face integrals are areas times tractions, $O(\varepsilon^2)$; the rate of change of momentum and the body force are volumes times bounded densities, $O(\varepsilon^3)$. Divide the balance by $\varepsilon^2$ and let $\varepsilon \to 0$: the volume terms vanish and the face terms must sum to zero on their own. The table in the figure is that statement in numbers. The same scaling argument appeared on the pillbox, where the small parameter was the thickness $h$ rather than an overall size.

History and reading. Cauchy's own argument is this one. Gurtin, An Introduction to Continuum Mechanics, the chapter on stress, gives it with all the analysis; Truesdell's First Course gives Noll's sharper version.

22 · stress

Stress is a bilinear form

definition · the stress tensor

At each point, the bilinear form on $\mathbb R^3$

$$\sigma(\vec m, \vec n) = \vec m\cdot\vec t(\vec n) = m_i\,\sigma_{ij}\,n_j ,$$

the component along $\vec m$ of the force per area across a surface with normal $\vec n$. A (0,2)-tensor, in unit 16's sense; as $\vec x$ varies, a tensor field.

theorem (unit 16) · the transformation law

In a rotated orthonormal basis $\vec e\,{}'_a = R_{ia}\vec e_i$, the components are $\sigma'_{ab} = R_{ia}R_{jb}\,\sigma_{ij}$, i.e. $\sigma' = R^{\!\top}\sigma R$. Proof: $\sigma'_{ab} = \sigma(\vec e\,{}'_a, \vec e\,{}'_b)$ and bilinearity. $\square$

Two readings of one object, and the order matters:

Mathematics: a bilinear form, whose components change by the rule above because it is bilinear.

Physics: the flux of momentum. $\sigma_{ij}$ is the rate at which $i$-momentum flows across a unit area facing $+\vec e_j$ into the region — made precise in section 2.

The next slide draws the nine components on a cube.

The house rule, observed. Unit 16 defined a tensor as a multilinear map and proved the transformation law from it. The inertia tensor there was bilinear by construction: $I(\vec a,\vec b) = \sum m\,(\ldots)$ is visibly bilinear in $\vec a$ and $\vec b$. Stress is different. Nobody writes down a formula for $\sigma(\vec m, \vec n)$; one is given only the function $\vec n \mapsto \vec t(\vec n)$, and it is a theorem — Cauchy's — that the function is linear, hence that $(\vec m,\vec n)\mapsto\vec m\cdot\vec t(\vec n)$ is a bilinear form. Nine numbers per basis would not be enough to make it a tensor, as unit 16's problem on nine non-tensorial numbers showed; the linearity is what makes it one.

Full proof of the transformation law, in this setting. $\sigma'_{ab} := \sigma(\vec e\,{}'_a,\vec e\,{}'_b) = \sigma(R_{ia}\vec e_i, R_{jb}\vec e_j) = R_{ia}R_{jb}\,\sigma(\vec e_i,\vec e_j) = R_{ia}R_{jb}\sigma_{ij}$, using bilinearity twice. In matrices, $\sigma' = R^{\!\top}\sigma R$. For the same object viewed as the linear map $\vec n\mapsto\sigma\vec n$ the rule is the same, because $R^{\!\top} = R^{-1}$ for orthogonal $R$: with an orthonormal basis, the (0,2) and (1,1) readings differ only by the Riesz isomorphism of unit 04, which in orthonormal coordinates is the identity on components.

The flux reading, in advance. Across a small surface of area $\mathrm dA$ and normal $\vec n$, the material outside pushes on the material inside with force $\sigma\vec n\,\mathrm dA$ — that is, it transfers momentum into the inside at the rate $\sigma\vec n\,\mathrm dA$ per second. So $\sigma_{ij}n_j$ is an inflow rate of $i$-momentum per unit area. Section 2 turns this into the momentum equation, and the sign (inflow, not outflow) is why the momentum flux tensor there is $-\sigma$.

22 · stress

Nine numbers on a cube

Column $j$ of $\sigma$ is the traction on the face facing $+\vec e_j$. On each face, one normal component $\sigma_{jj}$ (highlighted) and two shears. Schematic: arrows show the positive directions.

Read $\sigma_{ij}$ as: component $i$ of the force per area on the face whose normal is $\vec e_j$.

vocabulary

Normal stress $\sigma(\vec n,\vec n)$: push ($<0$) or pull ($>0$) across the surface. Shear stress: the part of $\vec t$ tangent to it, $\vec t - \sigma(\vec n,\vec n)\,\vec n$.

On the cube the diagonal entries are the normal stresses on its faces and the six off-diagonal entries are the shears. Rotate the cube and the nine numbers change by $\sigma' = R^{\!\top}\sigma R$; the bilinear form does not.

Once section 2 proves $\sigma$ symmetric, unit 16's spectral theorem gives three principal stresses on three perpendicular planes that feel no shear at all.

Reading the cube. The front face faces $+\vec e_1$, so the three arrows on it are the components of $\vec t(\vec e_1) = (\sigma_{11}, \sigma_{21}, \sigma_{31})$, the first column of $\sigma$: $\sigma_{11}$ out along $\vec e_1$, $\sigma_{21}$ along $\vec e_2$, $\sigma_{31}$ along $\vec e_3$. Likewise the right face carries the second column and the top face the third. On the three hidden faces, facing $-\vec e_j$, the third-law lemma makes every arrow reverse. Section 2's symmetry theorem will say $\sigma_{12} = \sigma_{21}$ and so on: the shear along $\vec e_2$ on the front face equals the shear along $\vec e_1$ on the right face, which is exactly the pair whose torques must cancel on a small cube.

Sign convention. Tension positive, compression negative: the engineering and physics convention. Geologists and soil engineers often use the opposite sign, so that the stresses they meet most, which are compressive, come out positive. Check before borrowing a formula.

22 · stress · example

A bar in tension, cut obliquely

worked example 1

A bar along $\vec e_1$ carries $10$ kN on a cross-section of $1\ \mathrm{cm^2}$. Find the traction on a plane whose normal makes angle $\theta$ with the bar.

1. The stress. Across a cut facing $\vec e_1$ the traction is $s\,\vec e_1$ with $s = 10^4\ \mathrm N/10^{-4}\ \mathrm{m^2} = 100$ MPa; across cuts facing $\vec e_2, \vec e_3$ it is zero. By Cauchy, column by column:

$$\sigma = \begin{pmatrix} s & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0\end{pmatrix}.$$

2. The traction. $\vec n = (\cos\theta, \sin\theta, 0)$: $\ \vec t = \sigma\vec n = (s\cos\theta, 0, 0)$.

3. Split it. Normal: $\vec n\cdot\vec t = s\cos^2\theta$. Shear: $|\vec t - (\vec n\cdot\vec t)\vec n| = s\sin\theta\cos\theta = \tfrac12 s\sin 2\theta$.

The cut, its normal $\vec n$ at angle $\theta$ to the bar (here $35^\circ$). The traction $\vec t = \sigma\vec n$ on it is along the bar, and splits into a part along $\vec n$ and a part along the cut.

The traction is $s\cos\theta$ along the bar, whatever the cut: it is the split into normal and shear that changes with $\theta$. The next slide follows both parts through every angle.

Step 1 in detail. Cauchy's theorem says column $j$ of $\sigma$ is $\vec t(\vec e_j)$. On a cross-section ($\vec n = \vec e_1$) the far part of the bar pulls with total force $10$ kN spread over $10^{-4}\ \mathrm{m^2}$, so $\vec t(\vec e_1) = (10^8\ \mathrm{Pa})\,\vec e_1$. On a longitudinal cut ($\vec n = \vec e_2$ or $\vec e_3$) nothing is transmitted in a uniformly pulled bar with free sides, so those columns vanish. The first column is $(s,0,0)$, and that is the whole matrix. It is symmetric, as section 2 will prove it must be.

Step 3. $\vec t - (\vec n\cdot\vec t)\vec n = (s\cos\theta - s\cos^3\theta,\ -s\cos^2\theta\sin\theta,\ 0) = s\cos\theta\sin\theta\,(\sin\theta, -\cos\theta, 0)$, and $(\sin\theta,-\cos\theta,0)$ is a unit vector in the plane of the cut. Hence the magnitude $s\sin\theta\cos\theta$. In the figure the dotted lines complete the rectangle: the two components are at right angles and add up to $\vec t$.

22 · stress · example

Normal and shear on every cut

$s\cos^2\theta$ and $\tfrac12 s\sin2\theta$ for $s = 100$ MPa; the dots are the table's rows. The shear peaks at $45^\circ$, where it equals the normal stress.

$\theta$	normal (MPa)	shear (MPa)
$0^\circ$	100.0	0.0
$30^\circ$	75.0	43.3
$45^\circ$	50.0	50.0
$60^\circ$	25.0	43.3
$90^\circ$	0.0	0.0

Shear peaks at $45^\circ$, at half the tensile stress: the planes on which a ductile metal slips, and why a pulled steel bar fails along a cone at $45^\circ$.

What to look for. The normal curve falls monotonically from the full $100$ MPa to nothing; the shear curve rises from zero, peaks at $45^\circ$ and falls back, symmetric about $45^\circ$. Where they cross, the cut carries equal parts of pull and of slide. Both curves are evaluated from the worked example's two formulas, and the dots are the table.

Checks. Dimensions: both columns are pascals ✓. $\theta = 0$: all normal, $100$ MPa, no shear ✓. $\theta = 90^\circ$: the cut is parallel to the bar and feels nothing ✓. Normal$^2$ + shear$^2 = s^2\cos^2\theta = |\vec t|^2$ ✓ (at $30^\circ$: $75^2 + 43.3^2 = 7500 = 100^2\cdot0.75$). The shear at $\theta$ and $90^\circ - \theta$ is the same, as $\sin2\theta$ requires.

The physics of the $45^\circ$ planes. Metals deform plastically by crystal planes sliding over one another, a process driven by shear, not by tension. In a pulled bar the first planes to slip are those near $45^\circ$ to the load, and the characteristic "cup and cone" fracture surface of a ductile bar shows it. Brittle materials such as cast iron or chalk fail instead on the plane of maximum normal stress, $\theta = 0$, a flat break straight across. Twist a stick of chalk and it breaks on a $45^\circ$ helix, because in torsion the maximum tension is on those planes — problem P1 has the matrix.

22 · stress

Pressure is the isotropic part

definitions

A fluid: a material that at rest sustains no shear stress on any plane.

For any stress, the pressure $p = -\tfrac13\operatorname{tr}\sigma$, and $\sigma = -p\,I + \tau$ splits it into an isotropic part $-pI$ and a trace-free deviatoric stress $\tau$.

theorem · Pascal's law

In a fluid at rest, $\sigma = -p\,I$: the traction on every plane is $-p\vec n$, the same push whatever the orientation.

Proof. No shear on any plane means $\sigma\vec n \parallel \vec n$ for every $\vec n$: every vector is an eigenvector of $\sigma$, so $\sigma$ is a multiple of $I$. $\square$

checks

$p$ is basis-independent, as the trace is ✓. $\operatorname{tr}\tau = \operatorname{tr}\sigma + 3p = 0$ ✓. $\sigma = -p_0I$: $\tau = 0$ ✓. The bar: $p = -s/3 < 0$, a tension ✓.

A parcel of still air. On every piece of its boundary the push is $p_0$ along the inward normal, whatever the orientation; an imaginary plane through it is pushed equally from both sides.

on the air at rest

$\sigma = -p_0 I$, $p_0 = 101\,325$ Pa: the weight of ten tonnes on every square metre of every surface, real or imaginary. A sound wave can change only $p$.

The figure, against the bar. Compare the pulled bar earlier in this section, where the traction changed from $s\,\vec e_1$ to nothing as the cut turned. Here every arrow has the same length and is normal to its surface: that is what $\sigma = -p_0I$ means, $\vec t(\vec n) = -p_0\vec n$ for every $\vec n$. Pascal's law says no other stress is possible in a fluid at rest.

Full proof of "every vector an eigenvector implies scalar". Let $\sigma\vec e_i = \lambda_i\vec e_i$ and $\sigma(\vec e_1 + \vec e_2) = \lambda(\vec e_1 + \vec e_2)$. Linearity gives $\lambda_1\vec e_1 + \lambda_2\vec e_2 = \lambda\vec e_1 + \lambda\vec e_2$, so $\lambda_1 = \lambda = \lambda_2$; likewise $\lambda_2 = \lambda_3$. So $\sigma = \lambda_1 I$, and $p = -\lambda_1$. Nothing about symmetry is needed.

Why "isotropic". $-pI$ is the only kind of stress that looks the same in every rotated basis: $R^{\!\top}(-pI)R = -pI$. Conversely, a symmetric tensor invariant under all rotations must be a multiple of $I$ (its eigenspaces would have to be rotation-invariant subspaces, and the only ones are $\{0\}$ and $\mathbb R^3$). The decomposition into isotropic and deviatoric parts is therefore canonical, not a choice of coordinates, and section 3 uses it twice.

Pressure as a mean normal stress. $-p = \tfrac13(\sigma_{11}+\sigma_{22}+\sigma_{33})$ is the average of the normal stresses on three perpendicular planes; averaged over all directions, $\frac1{4\pi}\oint\sigma(\vec n,\vec n)\,\mathrm d\Omega = \tfrac13\operatorname{tr}\sigma$ as well (the average of $n_in_j$ over the sphere is $\tfrac13\delta_{ij}$, by symmetry and taking the trace). So in a moving fluid, where there can be shear, "the pressure" still has an invariant meaning.

History. Pascal's Traité de l'équilibre des liqueurs (written by 1654, published 1663) stated the law; the one-line proof above is only possible once Cauchy has made $\sigma$ a linear map.

section 2

Conservation laws in local form

A budget for every region, the divergence theorem, and one lemma: an integral statement becomes a partial differential equation.

22 · local form

The divergence theorem for a tensor field

definition · divergence of a tensor field

For a $C^1$ tensor field $T$ with components $T_{ij}$, $\operatorname{div}T$ is the vector field

$$(\operatorname{div}T)_i = \partial_j T_{ij},$$

the divergence taken on the second slot. Basis-free: for every constant vector $\vec a$, $\ \vec a\cdot\operatorname{div}T = \nabla\!\cdot(T^{\!\top}\vec a)$.

theorem

For a region $\Omega$ to which unit 05's divergence theorem applies,

$$\oint_{\partial\Omega} T\,\vec n\,\mathrm dA = \int_\Omega \operatorname{div}T\,\mathrm dV .$$

Proof. Dot with a constant $\vec a$: $\vec a\cdot T\vec n = (T^{\!\top}\vec a)\cdot\vec n$, so unit 05 applies to the vector field $T^{\!\top}\vec a$. True for every $\vec a$. $\square$

With $T = \sigma$: the net surface force on a region is the volume integral of $\operatorname{div}\sigma$. So $\operatorname{div}\sigma$ is a force per unit volume, and a stress that does not vary pushes a region equally from all sides.

checks

Dimensions: $\partial_j\sigma_{ij}$ is $\mathrm{Pa\,m^{-1}} = \mathrm{N\,m^{-3}}$ ✓. Special case $T = f\,I$: $\oint f\vec n\,\mathrm dA = \int\nabla f\,\mathrm dV$, the gradient theorem ✓. A vector: $\vec a\cdot\operatorname{div}T$ is the divergence of a vector field, which unit 05 showed is basis-independent ✓.

on the air at rest

$\sigma = -p_0I$ is constant, so $\operatorname{div}\sigma = 0$: ten tonnes per square metre on every face, and a net force of exactly zero on every region. Only a pressure gradient pushes.

The same theorem, three ways of saying it. Row by row: the $i$-th row of $T$, $(T_{i1},T_{i2},T_{i3})$, is a vector field; unit 05's theorem applied to it is the $i$-th component of the tensor statement. With a constant $\vec a$: the scalar $\vec a\cdot T\vec n$ is the flux of the vector field $T^{\!\top}\vec a$. As a map: for each fixed $\vec a$, $\vec n \mapsto T(\vec a, \vec n)$ is a covector, turned into a vector by the Riesz isomorphism (unit 04), and the theorem is unit 05's for that vector field. The second is the one that survives in curved spaces, where "row by row" means nothing.

Why the second slot. $\sigma\vec n$ contracts $\vec n$ into the second index, so the divergence that appears when the surface integral is converted must also be on the second index. Once $\sigma$ is proved symmetric (the last slide of this section) the distinction disappears for stress, but not for the momentum-flux tensor or, in unit 28, for the electromagnetic field tensor, which is antisymmetric.

Hypotheses. Unit 05 proved the divergence theorem for a box and sketched the patching argument for regions with piecewise-smooth boundary; everything here uses it for such regions and for $C^1$ fields on a neighbourhood of the closed region. That $\sigma$ is $C^1$ is an assumption about the motion, not a consequence of Cauchy's theorem: at a shock wave or a crack it fails, and the local equations below are replaced there by jump conditions derived from the same integral budgets.

22 · local form

Mass: from a budget to a field equation

Fix a region $V$ in space. Mass inside changes only by flow across $\partial V$: through $\mathrm dA$ passes the mass flux $\rho\vec v\cdot\vec n\,\mathrm dA$ per second, outward.

$$\frac{\mathrm d}{\mathrm dt}\int_V\rho\,\mathrm dV = -\oint_{\partial V}\rho\,\vec v\cdot\vec n\,\mathrm dA$$

Left: Leibniz's rule (unit 18) moves $\frac{\mathrm d}{\mathrm dt}$ inside as $\partial_t$. Right: the divergence theorem. So

$$\int_V\big(\partial_t\rho + \nabla\!\cdot(\rho\vec v)\big)\,\mathrm dV = 0\quad\text{for every } V.$$

One more step — the next slide — turns "for every $V$" into an equation at every point.

A fixed region $V$ and the flow through it. Where $\vec v\cdot\vec n > 0$ mass leaves, where $\vec v\cdot\vec n < 0$ it enters; the mass inside changes by the net outflow, with a minus sign.

Reading the figure. The region is fixed in space and the fluid streams through it, so the particles inside are different ones from moment to moment — the opposite of the material regions of the Reynolds transport theorem, later in this section. Only the normal component of $\vec v$ carries mass across a patch: a flow sliding along the boundary moves nothing in or out, which is why the budget has $\vec v\cdot\vec n$ and not $|\vec v|$.

The pattern. Every local conservation law in the course is produced by exactly these three moves: (1) state the budget for an arbitrary region — density integrated over the region changes by minus the flux through the boundary, plus any source; (2) convert the boundary term with the divergence theorem and move the time derivative inside; (3) localise. The output is always $\partial_t(\text{density}) + \nabla\!\cdot(\text{flux}) = \text{source}$. Problem P3 runs it for a pollutant that is carried, diffuses and decays.

Leibniz's rule in three dimensions. Unit 18 proved $\frac{\mathrm d}{\mathrm dt}\int_0^L F\,\mathrm dx = \int_0^L\partial_tF\,\mathrm dx$ for $F$ and $\partial_tF$ continuous on a compact rectangle. The proof uses only the mean value theorem in $t$ and uniform continuity of $\partial_tF$ on a compact set, and runs word for word with $[0,L]$ replaced by a bounded closed region $\bar V$: the difference quotient differs from $\partial_tF$ by at most the modulus of uniform continuity, uniformly over $\bar V$, and the integral of something uniformly small over a set of finite volume is small.

22 · local form

Localise: the continuity equation

lemma · localisation lemma

If $f$ is continuous on an open set and $\int_B f\,\mathrm dV = 0$ for every ball $B$ in it, then $f \equiv 0$. Proof: if $f(\vec x_0) > 0$, continuity gives a ball around $\vec x_0$ on which $f > f(\vec x_0)/2$, and there the integral is positive. $\square$

The proof in one dimension: a continuous $f$ positive at $x_0$ stays above $f(x_0)/2$ on a whole interval, and its integral there is positive.

theorem · the continuity equation $$\frac{\partial\rho}{\partial t} + \nabla\!\cdot(\rho\,\vec v) = 0$$

checks

Dimensions: $\partial_t\rho$ and $\nabla\!\cdot(\rho\vec v)$ are both $\mathrm{kg\,m^{-3}s^{-1}}$ ✓. One dimension: $\rho_t + (\rho v)_x = 0$, the shape of unit 18's $e_t + S_x = 0$ ✓. Rigid translation $\rho(\vec x - \vec u t)$, $\vec v = \vec u$ constant: $-\vec u\cdot\nabla\rho + \vec u\cdot\nabla\rho = 0$ ✓.

For the air: $\rho = \rho_0$, $\vec v = 0$ satisfies it trivially. A sound wave will be a small solution near this one, and this is the first of the two equations to be linearised.

Applying the lemma. The integrand $\partial_t\rho + \nabla\!\cdot(\rho\vec v)$ is continuous when $\rho$ and $\vec v$ are $C^1$, and the previous slide showed its integral over every region vanishes — in particular over every ball. The lemma then says it vanishes at every point, which is the continuity equation. The figure is the lemma's proof drawn for a function of one variable; in three dimensions the interval becomes a ball, and nothing else changes.

The localisation lemma, precisely. Balls suffice; so do cubes, or any family of sets shrinking to each point. The lemma needs continuity: for a merely integrable $f$ the conclusion is "zero almost everywhere", which is the version used when shocks are allowed.

The other reading. Expanding, $\partial_t\rho + \vec v\cdot\nabla\rho + \rho\nabla\!\cdot\vec v = 0$. The first two terms are the rate of change of density seen by a moving particle (next slide), so the density of a particle changes only through $\nabla\!\cdot\vec v$, the local rate of volume expansion.

22 · local form

The material derivative

Two descriptions of a flow. Eulerian description: fields $\rho(\vec x,t)$, $\vec v(\vec x,t)$ at fixed points of space. Lagrangian description: follow each particle, $\dot{\vec x} = \vec v(\vec x, t)$.

definition · material derivative

The rate of change of a field $f$ seen by a particle: by unit 03's chain rule,

$$\frac{\mathrm D f}{\mathrm Dt} := \frac{\mathrm d}{\mathrm dt}f\big(\vec x(t), t\big) = \partial_tf + \vec v\cdot\nabla f .$$

consequence · the acceleration of a fluid particle $$\vec a = \frac{\mathrm D\vec v}{\mathrm Dt} = \partial_t\vec v + (\vec v\cdot\nabla)\vec v$$

In components, $a_i = \partial_tv_i + v_j\,\partial_jv_i$.

continuity, rewritten

$\dfrac{\mathrm D\rho}{\mathrm Dt} + \rho\,\nabla\!\cdot\vec v = 0$. A flow is incompressible when each particle keeps its density, $\mathrm D\rho/\mathrm Dt = 0$ — equivalently $\nabla\!\cdot\vec v = 0$. Not the same as $\rho$ uniform.

The next slide: a flow in which nothing depends on time, yet every particle accelerates.

Why $(\vec v\cdot\nabla)\vec v$ is the whole difficulty of fluid mechanics. Newton's law for a fluid particle is linear in the particle's acceleration, but written in the Eulerian fields the acceleration contains the product of $\vec v$ with its own derivative. That quadratic term is the source of turbulence, of shocks, and of the fact that the Navier–Stokes existence problem is open. Section 4 linearises about rest precisely by throwing it away, and checks that doing so is legitimate for sound.

The operator $\vec v\cdot\nabla$. Applied to a vector field $\vec w$, $(\vec v\cdot\nabla)\vec w$ is the directional derivative of $\vec w$ along $\vec v$: in unit 03's language, the total derivative $D\vec w$ applied to the vector $\vec v$, with components $v_j\partial_jw_i$. The notation suggests "$\vec v$ dotted with the gradient of a vector"; read honestly, that gradient is the (1,1)-tensor $D\vec w$, and the dot is its contraction with $\vec v$ on the second slot — which is exactly $v_j\partial_jw_i$.

Names. Also called the convective, substantial or Lagrangian derivative. Euler introduced the fixed-point description in 1757; the particle description, despite the name, is also Euler's. Do not confuse this with unit 03's transport theorem for rotating frames — a different theorem with a similar name, about the Reynolds theorem's subject in a different setting.

22 · local form

Steady, and still accelerating

rigid rotation

$\vec v = \Omega(-y, x, 0)$, with $\Omega$ the angular speed (not, as on the next slide, a region). Nothing depends on $t$, so $\partial_t\vec v = 0$. But

$$\begin{aligned}(\vec v\cdot\nabla)\vec v &= \Omega(-y\,\partial_x + x\,\partial_y)\,\Omega(-y,x,0)\\ &= -\Omega^2(x, y, 0),\end{aligned}$$

the centripetal acceleration $\Omega^2 r$ of unit 03, pointing in ✓.

At a fixed point the velocity never changes; following a particle, its direction changes all the time. $\partial_t\vec v$ sees the first, $\mathrm D\vec v/\mathrm Dt$ the second.

Rigid rotation: the field of arrows is the same at every instant, yet each particle's velocity turns, and its acceleration $(\vec v\cdot\nabla)\vec v$ points to the axis, growing with the distance $r$.

What to look for. Velocity arrows are tangent to the circles and acceleration arrows point to the centre, both proportional to $r$. Nothing in the picture moves when $t$ changes — that is what "steady" means — and still no particle moves in a straight line. The whole acceleration is the convective term.

The rotation example in full. $\vec v\cdot\nabla = -\Omega y\,\partial_x + \Omega x\,\partial_y$. Applied to $v_1 = -\Omega y$: $-\Omega y\cdot0 + \Omega x\cdot(-\Omega) = -\Omega^2x$. Applied to $v_2 = \Omega x$: $-\Omega y\cdot\Omega + \Omega x\cdot 0 = -\Omega^2y$. Applied to $v_3 = 0$: $0$. So $\vec a = -\Omega^2(x,y,0)$, magnitude $\Omega^2 r$, directed to the axis. Its divergence is $\partial_x(-\Omega y) + \partial_y(\Omega x) = 0$: rigid rotation is incompressible, as it must be.

22 · local form

The Reynolds transport theorem

Let $\varphi_t$ be the flow map of $\vec v$ (unit 01: $\partial_t\varphi_t(\vec X) = \vec v(\varphi_t(\vec X), t)$, $\varphi_0 = \mathrm{id}$), and $\Omega_t = \varphi_t(\Omega_0)$ a material region: the same particles at every time.

theorem · Reynolds transport theorem $$\begin{aligned}\frac{\mathrm d}{\mathrm dt}\int_{\Omega_t} f\,\mathrm dV &= \int_{\Omega_t}\Big(\frac{\mathrm Df}{\mathrm Dt} + f\,\nabla\!\cdot\vec v\Big)\mathrm dV\\ &= \int_{\Omega_t}\partial_tf\,\mathrm dV + \oint_{\partial\Omega_t} f\,\vec v\cdot\vec n\,\mathrm dA\end{aligned}$$

checks

$f = 1$: $\dot{\mathrm{vol}}(\Omega_t) = \int\nabla\!\cdot\vec v$ — divergence is the rate of volume growth ✓. $\vec v = 0$: Leibniz's rule ✓. Dimensions: $[f]\,\mathrm{m^3s^{-1}}$ on both sides ✓.

A material region carried by the flow (schematic): the same particles, a new shape. Its integral changes by what changes at fixed points, plus what its moving boundary sweeps up — the shaded sliver, gained in front and lost behind.

Reading the figure. Each dot in $\Omega_0$ is a particle $\vec X$, and its dashed path ends at $\varphi_t(\vec X)$ in $\Omega_t$. The thin outline just beyond $\Omega_t$ is the region a moment later. In the second form of the theorem the volume term $\int\partial_tf$ is the change of $f$ at fixed points inside, and the boundary term $\oint f\,\vec v\cdot\vec n\,\mathrm dA$ is the rate at which the moving boundary sweeps $f$ in: positive on the leading side, where $\vec v\cdot\vec n > 0$, negative on the trailing side. The proof is on the next slide.

Name clash. Unit 03 also has a "transport theorem", $(\mathrm d/\mathrm dt)_{\text{inertial}} = (\mathrm d/\mathrm dt)_{\text{rot}} + \vec\omega\times$. Both are about differentiating in a moving setting, and both are called that; they are different theorems. This one is always "Reynolds" here.

22 · local form

The transport theorem, proved by pulling back

lemma · Euler's expansion formula

$J = \det D\varphi_t$ satisfies $\partial_tJ = (\nabla\!\cdot\vec v)\,J$. From Jacobi's formula $\frac{\mathrm d}{\mathrm dt}\det F = \det F\,\operatorname{tr}(F^{-1}\dot F)$ and $\dot F = (D\vec v)F$.

full proof in the notes (N)

Proof. Pull back to $\Omega_0$ by unit 05's change of variables, differentiate under the integral, push forward:

$$\begin{aligned}&\frac{\mathrm d}{\mathrm dt}\int_{\Omega_0} f(\varphi_t, t)\,J\,\mathrm dV_X\\ &\quad= \int_{\Omega_0}\Big(\frac{\mathrm Df}{\mathrm Dt} + f\,\nabla\!\cdot\vec v\Big)J\,\mathrm dV_X. \ \square\end{aligned}$$

corollary · the one used below

With continuity, $\displaystyle\frac{\mathrm d}{\mathrm dt}\int_{\Omega_t}\rho\,g\,\mathrm dV = \int_{\Omega_t}\rho\,\frac{\mathrm Dg}{\mathrm Dt}\,\mathrm dV$ for any field $g$.

Full proof

Cited. If $\vec v$ is $C^1$, the flow $\varphi_t(\vec X)$ is $C^1$ jointly in $(\vec X, t)$, and $F = D_X\varphi_t$ satisfies the variational equation $\partial_tF = (D\vec v)(\varphi_t(\vec X),t)\,F$ with $F(0) = I$. Unit 01 gave existence and uniqueness of the flow; differentiability in the initial point is the theorem on differentiable dependence on initial conditions, proved in Arnold, Ordinary Differential Equations, ch. 4. The variational equation is what you get by differentiating $\partial_t\varphi = \vec v(\varphi,t)$ in $\vec X$ and exchanging derivatives, which that theorem licenses.

Jacobi's formula. For an invertible matrix $F(t)$, $\det(F + h\dot F + o(h)) = \det F\,\det(I + hF^{-1}\dot F + o(h))$. For any matrix $B$, $\det(I + hB) = 1 + h\operatorname{tr}B + O(h^2)$: in the Leibniz expansion of the determinant, every permutation other than the identity picks at least two off-diagonal entries, each $O(h)$, and the identity permutation contributes $\prod(1 + hB_{ii}) = 1 + h\operatorname{tr}B + O(h^2)$. Hence $\frac{\mathrm d}{\mathrm dt}\det F = \det F\,\operatorname{tr}(F^{-1}\dot F)$.

Euler's expansion formula. With $\dot F = (D\vec v)F$: $\operatorname{tr}(F^{-1}(D\vec v)F) = \operatorname{tr}D\vec v = \nabla\!\cdot\vec v$, so $\partial_tJ = (\nabla\!\cdot\vec v)J$. Since $J(0) = 1$, this linear ODE gives $J = \exp\int_0^t\nabla\!\cdot\vec v\,\mathrm ds > 0$, so $\varphi_t$ preserves orientation and the change of variables needs no absolute value.

The theorem. By change of variables (unit 05), $\int_{\Omega_t}f\,\mathrm dV = \int_{\Omega_0}f(\varphi_t(\vec X),t)\,J(\vec X,t)\,\mathrm dV_X$. The region $\Omega_0$ no longer moves, so Leibniz's rule applies (the integrand is $C^1$ in $t$, continuous on the compact $\bar\Omega_0\times[t_1,t_2]$). By the chain rule, $\partial_t\big[f(\varphi_t(\vec X),t)\big] = \partial_tf + \vec v\cdot\nabla f = \mathrm Df/\mathrm Dt$, evaluated at $\varphi_t(\vec X)$. With the expansion formula, the integrand's derivative is $(\mathrm Df/\mathrm Dt + f\nabla\!\cdot\vec v)J$; change variables back. The second form follows from $\mathrm Df/\mathrm Dt + f\nabla\!\cdot\vec v = \partial_tf + \nabla\!\cdot(f\vec v)$ and the divergence theorem. It reads: the change is what changes at fixed points, plus what the moving boundary sweeps up.

The corollary. Apply the theorem to $f = \rho g$: $\mathrm D(\rho g)/\mathrm Dt + \rho g\nabla\!\cdot\vec v = \rho\,\mathrm Dg/\mathrm Dt + g\,(\mathrm D\rho/\mathrm Dt + \rho\nabla\!\cdot\vec v)$, and the bracket vanishes by continuity. Mass is carried along, so "$\rho\,\mathrm dV$" behaves as a constant mass element and only $g$ is differentiated.

22 · local form

Momentum: Cauchy's equation of motion

The balance law of section 1, for a material region $\Omega_t$:

$$\frac{\mathrm d}{\mathrm dt}\int_{\Omega_t}\!\rho\vec v\,\mathrm dV = \oint_{\partial\Omega_t}\!\sigma\vec n\,\mathrm dA + \int_{\Omega_t}\!\rho\vec b\,\mathrm dV$$

Left: the corollary, $\int\rho\,\mathrm D\vec v/\mathrm Dt$. Right: the tensor divergence theorem, $\int\operatorname{div}\sigma$. Localise.

theorem · Cauchy's equation of motion $$\rho\,\frac{\mathrm D\vec v}{\mathrm Dt} = \operatorname{div}\sigma + \rho\,\vec b$$

In components: $\rho\,(\partial_tv_i + v_j\partial_jv_i) = \partial_j\sigma_{ij} + \rho\,b_i$.

conservation form $$\partial_t(\rho v_i) + \partial_j\Pi_{ij} = \rho\,b_i,$$ $$\Pi_{ij} = \rho\,v_iv_j - \sigma_{ij}$$

$\rho\vec v$ is the density of momentum, and $\Pi$, the momentum flux tensor, its flux: momentum carried by moving matter, $\rho\vec v\otimes\vec v$, plus momentum transmitted by stress, $-\sigma$.

checks

Dimensions: every term $\mathrm{N\,m^{-3}}$ ✓. Rest: $\operatorname{div}\sigma + \rho\vec b = 0$, the equation of statics ✓. Uniform stress, no body force: no acceleration ✓. The two forms agree: their difference is $v_i\times$(continuity) ✓.

For air at rest, $-p_0I$ is constant: $\operatorname{div}\sigma = 0$ and nothing moves. The second equation to be linearised.

Full proof of the equivalence of the two forms. $\partial_t(\rho v_i) + \partial_j(\rho v_iv_j) = v_i\big(\partial_t\rho + \partial_j(\rho v_j)\big) + \rho\,(\partial_tv_i + v_j\partial_jv_i)$ by the product rule. The first bracket vanishes by continuity and the second is $\rho\,\mathrm Dv_i/\mathrm Dt$. So the conservation form's left side is $\rho\,\mathrm Dv_i/\mathrm Dt + \partial_j(-\sigma_{ij})$ rearranged, and the two statements are the same equation.

Why the conservation form matters. It is the fixed-region budget: integrate over a fixed $V$ and the divergence theorem gives $\frac{\mathrm d}{\mathrm dt}\int_V\rho\vec v = -\oint_{\partial V}\Pi\vec n\,\mathrm dA + \int_V\rho\vec b$. Momentum inside a fixed box changes because moving matter carries it across the walls ($\rho\vec v\otimes\vec v$) and because the material outside pushes on the walls ($-\sigma$). It is the form that survives shocks, via jump conditions, and it is the form that becomes relativistic: in unit 35 the momentum flux $\Pi_{ij}$ becomes the spatial block $T^{ij}$ of the stress–energy tensor.

Unit 18, recovered. For a string along $x$ under tension $T$ with small transverse displacement $u$, the transverse force across a cut is $-Tu_x$ (from the left part on the right) — a $2\times2$ stress with one off-diagonal entry, in effect — and the equation of motion is $\mu u_{tt} = \partial_x(Tu_x)$, which is Cauchy's equation in one dimension with $\rho\,\mathrm Dv/\mathrm Dt \approx \mu u_{tt}$ to first order.

History. Cauchy, 1827–28, "Sur les équations qui expriment les conditions d'équilibre ou les lois du mouvement intérieur d'un corps solide". Euler had the special case $\sigma = -pI$ in 1757.

22 · local form · example

Archimedes, from the divergence theorem

worked example 2

A body occupying region $B$, volume $V$, is held under water ($\rho_w = 998.2\ \mathrm{kg\,m^{-3}}$) at rest. Find the force the water exerts on it.

1. The water's stress. A fluid at rest: $\sigma = -pI$ (Pascal), so $\operatorname{div}\sigma = -\nabla p$.

2. The pressure. Cauchy's equation at rest with $\vec b = \vec g$: $0 = -\nabla p + \rho_w\vec g$. With $z$ up, $\partial_zp = -\rho_wg$, so $p = p_{\text{atm}} + \rho_wg\,d$ at depth $d$: the hydrostatic equilibrium law.

3. The force. The traction on the body is $\sigma\vec n = -p\,\vec n$ ($\vec n$ out of $B$). Extend $p$'s formula into $B$ and use the tensor divergence theorem on $-pI$:

$$\vec F = -\!\oint_{\partial B}\!p\,\vec n\,\mathrm dA = -\!\int_B\!\nabla p\,\mathrm dV = -\rho_w V\vec g$$

checks

Dimensions: $\mathrm{kg\,m^{-3}}\cdot\mathrm{m^3}\cdot\mathrm{m\,s^{-2}} = \mathrm N$ ✓. Limit $\rho_w \to 0$: no buoyancy ✓. A cube of side $\ell$: top and bottom differ in pressure by $\rho_wg\ell$ on area $\ell^2$, net $\rho_wg\ell^3$ up; the sides cancel ✓.

Numbers: $p$ rises by one atmosphere every $p_0/\rho_wg = 10.35$ m. One litre submerged: $F = 9.789$ N.

Step 3, and the one subtle point. The divergence theorem needs a $C^1$ field on all of $B$, but inside $B$ there is no water. The trick is that the integrand $-p\vec n$ on $\partial B$ depends only on the pressure at the surface, which is the same whatever fills $B$, because the water outside is in the same state either way. So define $p$ inside $B$ by the same formula $p_{\text{atm}} + \rho_wgd$ — a smooth field — and the theorem applies. Physically: replace the body by water, which would sit in equilibrium, so the surrounding water must push on it with exactly its weight.

The numbers. $p_0/(\rho_wg) = 101\,325/(998.2\times9.80665) = 10.35$ m, with standard gravity $g = 9.80665\ \mathrm{m\,s^{-2}}$ and water at $20\,{}^\circ\mathrm C$. One litre: $0.9982\ \mathrm{kg}\times9.80665 = 9.789$ N. The person: $1.204\times0.070\times9.80665 = 0.827$ N, the weight of $1.204\times70 = 84$ g of air. For a precise weighing of a low-density object this is not negligible: it is why analytical balances are corrected for air buoyancy.

22 · local form · example

Archimedes: deeper water pushes harder

The water's push on $B$, drawn as its depth-dependent part $\rho_w g d$ along the inward normal (the atmospheric part is the same all round and cancels). The bottom is pushed harder than the top: the net force is up.

result · Archimedes' principle

An upward force equal to the weight of the displaced water, whatever the shape of $B$.

Why the shape does not matter. Replace $B$ by water. That water would sit still, so the surrounding water must push on it with exactly its weight, up — and the surrounding water cannot tell what is inside.

in the air

A person of volume $70$ L is buoyed by $\rho_0Vg = 0.827$ N: the weight of $84$ g of air, which a bathroom scale silently subtracts.

22 · local form

Angular momentum makes stress symmetric

Write angular momentum as unit 03's antisymmetric matrix: $\vec x\times\vec p \leftrightarrow x_ip_j - x_jp_i$. The angular momentum balance, for every material region:

$$\begin{aligned}\tfrac{\mathrm d}{\mathrm dt}\!\textstyle\int\!\rho(x_iv_j - x_jv_i) &= \textstyle\oint(x_it_j - x_jt_i)\,\mathrm dA\\ &\quad+ \textstyle\int\rho(x_ib_j - x_jb_i)\end{aligned}$$

theorem · symmetry of stress

If all torques are moments of forces, and $\sigma$ is $C^1$, then $\sigma_{ij} = \sigma_{ji}$.

Proof. Divergence theorem on the surface term, the corollary on the left, and Cauchy's equation: everything cancels except

$$\int_{\Omega}(\sigma_{ji} - \sigma_{ij})\,\mathrm dV = 0 \ \text{ for every }\Omega. \ \square$$

full proof in the notes (N)

the assumption it quietly uses

No body couple and no couple stress: nothing twists the material except moments of forces. A ferrofluid in a magnetic field, whose particles are twisted by $\vec m\times\vec B$, violates it, and its stress is not symmetric.

The next slide shows why a nonzero $\sigma_{12} - \sigma_{21}$ would be absurd, on a small cube.

Full proof

Fix $i \ne j$. Surface term. With $t_j = \sigma_{jk}n_k$, the integrand is $w_kn_k$ for the vector field $w_k = x_i\sigma_{jk} - x_j\sigma_{ik}$. The divergence theorem gives $\int\partial_kw_k$, and $\partial_kw_k = \delta_{ik}\sigma_{jk} - \delta_{jk}\sigma_{ik} + x_i\partial_k\sigma_{jk} - x_j\partial_k\sigma_{ik} = \sigma_{ji} - \sigma_{ij} + x_i(\operatorname{div}\sigma)_j - x_j(\operatorname{div}\sigma)_i$.

Left side. By the corollary of the transport theorem with $g = x_iv_j - x_jv_i$, it is $\int\rho\,\mathrm Dg/\mathrm Dt$. Since $\mathrm Dx_i/\mathrm Dt = v_i$ (the material derivative of a coordinate function is the velocity component), $\mathrm Dg/\mathrm Dt = v_iv_j + x_ia_j - v_jv_i - x_ja_i = x_ia_j - x_ja_i$.

Combine. The balance becomes $\int\big[\rho(x_ia_j - x_ja_i) - x_i((\operatorname{div}\sigma)_j + \rho b_j) + x_j((\operatorname{div}\sigma)_i + \rho b_i)\big] = \int(\sigma_{ji} - \sigma_{ij})$. Cauchy's equation $\rho a_k = (\operatorname{div}\sigma)_k + \rho b_k$ makes the left integrand vanish identically. So $\int_\Omega(\sigma_{ji} - \sigma_{ij})\,\mathrm dV = 0$ for every material region, and every ball is a material region at the instant considered; the localisation lemma gives $\sigma_{ij} = \sigma_{ji}$. $\square$

With a body couple. If a torque density $m_{ij}$ (antisymmetric, per unit volume) acts in addition, the balance acquires $+\int m_{ij}$ on the right and the conclusion becomes $\sigma_{ij} - \sigma_{ji} = m_{ij}$: the antisymmetric part of stress is the body couple. Problem P2 asks for this. Ferrofluids, liquid crystals and polarised media in strong fields are the physical cases; for everything else in this course, including every fluid and solid in sections 3 to 5, the stress is symmetric.

22 · local form

The small-cube picture

A small cube seen along $\vec e_3$. The shears $\sigma_21$ on the faces facing $\pm\vec e_1$ turn it one way, the shears $\sigma_12$ on the faces facing $\pm\vec e_2$ the other. Their torques cancel only if $\sigma_12 = \sigma_21$.

the estimate

A cube of side $\ell$: shears on its faces give torque $(\sigma_{21} - \sigma_{12})\ell^3$; its moment of inertia is $\rho\ell^5/6$. Angular acceleration $6(\sigma_{21} - \sigma_{12})/\rho\ell^2$: infinite as $\ell \to 0$ unless the shears are equal.

The torque scales like $\ell^3$ but the moment of inertia like $\ell^5$: on a small enough cube any unbalanced shear spins it up without limit. Equal shears are the only way out — the same scaling argument as Cauchy's tetrahedron, one power of $\ell$ higher.

With $\sigma$ symmetric, unit 16's spectral theorem gives every point three perpendicular principal planes that feel no shear.

The cube, carefully. Faces $x_1 = \pm\ell/2$ have normals $\pm\vec e_1$ and tractions $\pm\sigma\vec e_1$, whose $\vec e_2$ components are $\pm\sigma_{21}$; the torque about the $x_3$ axis, $x_1t_2 - x_2t_1$ summed, is $2\cdot(\ell/2)\sigma_{21}\ell^2 = \sigma_{21}\ell^3$. The faces $x_2 = \pm\ell/2$ give $-\sigma_{12}\ell^3$. A cube's moment of inertia about a central axis is $m\ell^2/6 = \rho\ell^5/6$ (unit 16). The picture is persuasive, but it assumes the stress is uniform over the cube; the integral proof above is the one that needs no such assumption.

Consequence. A symmetric $\sigma$ has, by unit 16's spectral theorem, an orthonormal eigenbasis: three principal directions on whose planes the traction is purely normal. The six independent components are what section 3 must supply.

section 3

Closing the equations

Conservation laws are the same for every material; what is left over is a statement about the material, and symmetry makes it short.

22 · closure

Ten unknowns, four equations

unknown fields	count	equations	count
$\rho$	1	continuity	1
$\vec v$	3	Cauchy's equation	3
$\sigma$, symmetric	6	—
total	10		4

The missing six are a constitutive law: how this material's stress depends on its motion. Mass and momentum are the same for honey, steel and air; stress is not.

the simplest closure · ideal fluid

$\sigma = -p\,I$ always, moving or not. Then $\operatorname{div}\sigma = -\nabla p$ and the equations become Euler's equations (1757):

$$\partial_t\rho + \nabla\!\cdot(\rho\vec v) = 0,$$ $$\rho\,\frac{\mathrm D\vec v}{\mathrm Dt} = -\nabla p + \rho\,\vec b .$$

still one short

Unknowns $\rho, \vec v, p$: five. Equations: four. The last is an equation of state, and for the air the question is which one: a barotropic law $p = P(\rho)$, but which $P$? That is the whole of the Newton–Laplace dispute.

Why ten. Before section 2's symmetry theorem $\sigma$ had nine components; angular-momentum balance removed three, which is the precise sense in which angular momentum "is" the symmetry of stress. The count is of fields, not numbers: each unknown is a function of $(\vec x,t)$ and each equation a PDE.

Checks on Euler's equations. Dimensions: $\nabla p$ is $\mathrm{Pa\,m^{-1}} = \mathrm{N\,m^{-3}}$, like $\rho\,\mathrm D\vec v/\mathrm Dt$ ✓. At rest they reduce to hydrostatics, $\nabla p = \rho\vec b$, the second step of the Archimedes example ✓. With $p$ constant and $\vec b = 0$ they give $\mathrm D\vec v/\mathrm Dt = 0$: each particle moves in a straight line at constant speed, Newton's first law for fluid particles ✓.

The full thermodynamic closure, for the record. A real gas has $p = p(\rho, s)$ with $s$ the entropy per unit mass, and an ideal (inviscid, non-conducting) fluid carries its entropy with it, $\mathrm Ds/\mathrm Dt = 0$ (cited: Landau and Lifshitz, Fluid Mechanics, §2). That adds one unknown and one equation. If the entropy starts uniform it stays uniform, and then $p$ is a function of $\rho$ alone: the barotropic closure, with $P$ the adiabatic curve. Section 4 derives that curve from the first law without entropy, which is enough for sound.

Honey, steel, air. Honey: stress depends on the rate of deformation (viscosity, the last slide of this section). Steel: on the deformation itself (elasticity, the next three slides). Air: to excellent approximation, on density alone. Real materials mix these, and the mixture is called viscoelasticity; this course needs only the pure cases.

22 · closure

Strain

A solid remembers a reference state. A particle at $\vec X$ there sits at $\vec x = \vec X + \vec u(\vec X)$ now; $\vec u$ is the displacement field. Split its derivative (unit 16):

$$D\vec u = \varepsilon + \omega, \quad \varepsilon_{ij} = \tfrac12(\partial_iu_j + \partial_ju_i),$$

and $\omega_{ij} = \tfrac12(\partial_ju_i - \partial_iu_j)$ the antisymmetric remainder.

theorem · what each part does

For $|D\vec u| \ll 1$, to first order: $\omega$ is an infinitesimal rotation and changes no lengths; the strain tensor $\varepsilon$ changes lengths,

$$|\mathrm d\vec x|^2 = |\mathrm d\vec X|^2 + 2\,\varepsilon(\mathrm d\vec X, \mathrm d\vec X) + O(|D\vec u|^2);$$

and volumes change by the dilatation $\operatorname{tr}\varepsilon = \nabla\!\cdot\vec u = \Delta V/V$.

A small square and a fibre $\mathrm d\vec X$ across it, deformation exaggerated. The rotation part $\omega$ turns the square and leaves the fibre's length alone; the strain part $\varepsilon$ changes its shape and stretches the fibre.

Mathematics: $\varepsilon$ is a symmetric bilinear form. Physics: for a unit vector $\vec a$, $\varepsilon(\vec a,\vec a)$ is the fractional stretch of a fibre along $\vec a$.

The figure, computed. Middle: the square under the rotation by $0.38$ rad, the exponential of $\omega$; to first order $I + \omega$ is that rotation. Right: under $I + \varepsilon$ with $\varepsilon = \begin{pmatrix}0.28 & 0.10\\ 0.10 & -0.12\end{pmatrix}$, the diagonal fibre grows by $20\%$; the first-order prediction $\varepsilon(\vec a,\vec a)$ for $\vec a = (1,1)/\sqrt2$ is $18\%$, the difference being the second-order term the theorem drops. Real strains in steel are $10^{-3}$, invisible at this scale.

Full proof

The rotation. $(I + \omega)^{\!\top}(I + \omega) = I + \omega^{\!\top} + \omega + \omega^{\!\top}\omega = I + O(|\omega|^2)$ because $\omega^{\!\top} = -\omega$: to first order $I + \omega$ is orthogonal, an element of unit 03's $\mathfrak{so}(3)$ exponentiated to first order. Via the hat map it is a rotation by the small vector $\tfrac12\nabla\times\vec u$ (unit 04's curl), which is why the curl of a displacement is called its rotation.

Volumes. By change of variables (unit 05) a small volume scales by $\det(I + D\vec u)$, and by the expansion used for Jacobi's formula, $\det(I + D\vec u) = 1 + \operatorname{tr}D\vec u + O(|D\vec u|^2)$. $\operatorname{tr}D\vec u = \nabla\!\cdot\vec u = \operatorname{tr}\varepsilon$.

Where the approximation lives. Everything in this section is linear elasticity, valid for $|D\vec u| \ll 1$. Steel yields at strains near $10^{-3}$, rock at similar values, so for the materials in this unit the linear theory holds right up to failure. Rubber does not, and needs the finite-strain tensor $\tfrac12(F^{\!\top}F - I)$, $F = I + D\vec u$, of which $\varepsilon$ is the linearisation.

22 · closure

Strain, checked

proof of the length formula

$\mathrm d\vec x = (I + D\vec u)\,\mathrm d\vec X$, so $|\mathrm d\vec x|^2 = |\mathrm d\vec X|^2 + \mathrm d\vec X^{\!\top}(D\vec u + D\vec u^{\!\top})\,\mathrm d\vec X + |D\vec u\,\mathrm d\vec X|^2$. The middle term is $2\,\varepsilon(\mathrm d\vec X,\mathrm d\vec X)$ — the antisymmetric $\omega$ cancels from a quadratic form — and the last is second order. $\square$

checks

Rigid rotation $\vec u = \omega\vec X$, $\omega$ antisymmetric: $\varepsilon = 0$, no strain ✓. Uniform stretch $\vec u = \delta\,X_1\vec e_1$: $\varepsilon_{11} = \delta$, a fibre along $\vec e_1$ grows by the fraction $\delta$, and $\Delta V/V = \delta$ ✓. Dimensionless ✓.

in the air

A sound wave displaces air by $\vec u$; mass conservation gives $\rho'/\rho_0 = -\nabla\!\cdot\vec u$ to first order. Compression is negative dilatation.

22 · closure

Isotropy leaves two constants

A linearly elastic material: $\sigma = C(\varepsilon)$ with $C$ a linear map on symmetric matrices — the elasticity tensor, a (0,4)-tensor, $6\times6 = 36$ constants.

definition · isotropic material

Its response does not depend on orientation: $C(Q\varepsilon Q^{\!\top}) = Q\,C(\varepsilon)\,Q^{\!\top}$ for every $Q \in SO(3)$.

theorem

If $C$ is isotropic there are two numbers $\lambda, \mu$, the Lamé parameters, with

$$C(\varepsilon) = \lambda\,(\operatorname{tr}\varepsilon)\,I + 2\mu\,\varepsilon .$$

proof, in three moves

1. Diagonalise: $\varepsilon = QDQ^{\!\top}$, $Q \in SO(3)$ (spectral theorem). By isotropy, enough to know $C(D)$.

2. The half-turn $R_1 = \operatorname{diag}(1,-1,-1)$ fixes $D$, so $C(D)$ commutes with it; with $R_2$ likewise, $C(D)$ is diagonal.

3. Quarter-turns permute $D$'s entries, so $C$ on diagonals is a linear map on $\mathbb R^3$ commuting with every permutation: $d \mapsto 2\mu\,d + \lambda(\textstyle\sum d_k)(1,1,1)$. $\square$

full proof in the notes (N)

Thirty-six constants, two survive. The same theorem is used again for viscosity. Uniqueness forced by symmetry returns in unit 35, where Lovelock's theorem leaves only one candidate for the left side of Einstein's equations.

Full proof

Step 1. $\varepsilon$ is symmetric, so by the spectral theorem there is an orthogonal $Q$ with $\varepsilon = QDQ^{\!\top}$, $D = \operatorname{diag}(d_1,d_2,d_3)$. If $\det Q = -1$, replace the first column of $Q$ by its negative: still orthonormal eigenvectors, now $\det Q = +1$. Isotropy gives $C(\varepsilon) = Q\,C(D)\,Q^{\!\top}$, so $C$ is determined by its values on diagonal matrices.

Step 2. $R_1 = \operatorname{diag}(1,-1,-1)$ is the rotation by $\pi$ about $\vec e_1$, and $R_1DR_1^{\!\top} = D$. Isotropy: $C(D) = R_1\,C(D)\,R_1^{\!\top}$. For any matrix $S$, $(R_1SR_1)_{12} = -S_{12}$ and $(R_1SR_1)_{13} = -S_{13}$, so $C(D)_{12} = C(D)_{13} = 0$. The half-turn $R_2 = \operatorname{diag}(-1,1,-1)$ about $\vec e_2$ kills $C(D)_{12}$ and $C(D)_{23}$ likewise. $C(D)$ is symmetric, so it is diagonal.

Step 3. Let $A:\mathbb R^3\to\mathbb R^3$ be the linear map with $C(\operatorname{diag}(d)) = \operatorname{diag}(Ad)$. The quarter-turn $P$ about $\vec e_3$ ($P\vec e_1 = \vec e_2$, $P\vec e_2 = -\vec e_1$, $P\vec e_3 = \vec e_3$) has $P\operatorname{diag}(d_1,d_2,d_3)P^{\!\top} = \operatorname{diag}(d_2,d_1,d_3)$, since $PDP^{\!\top} = \sum_id_i(P\vec e_i)(P\vec e_i)^{\!\top}$ and $P\vec e_2\,(P\vec e_2)^{\!\top} = \vec e_1\vec e_1^{\!\top}$. Isotropy then says $A$ commutes with the transposition swapping coordinates 1 and 2; the quarter-turn about $\vec e_1$ gives the transposition of 2 and 3. Transpositions generate the symmetric group $S_3$, so $A$ commutes with every permutation matrix $P_\pi$: $A_{\pi(i)\pi(j)} = A_{ij}$. $S_3$ acts transitively on the diagonal positions and on the off-diagonal ordered pairs, so $A_{ii} = \alpha$ and $A_{ij} = \beta$ ($i\ne j$) for two numbers. Hence $Ad = (\alpha - \beta)d + \beta(d_1 + d_2 + d_3)(1,1,1)$, i.e. $C(D) = (\alpha - \beta)D + \beta(\operatorname{tr}D)I$. Put $2\mu = \alpha - \beta$, $\lambda = \beta$, and conjugate back by $Q$, using $\operatorname{tr}(QDQ^{\!\top}) = \operatorname{tr}D$ and $QIQ^{\!\top} = I$. $\square$

Converse. $\lambda(\operatorname{tr}\varepsilon)I + 2\mu\varepsilon$ is isotropic for any $\lambda,\mu$: conjugation by $Q$ preserves trace and fixes $I$. So the theorem is an exact characterisation.

Counting, for anisotropic materials. Without isotropy, $C$ is any linear map of the six-dimensional space of symmetric matrices: 36 constants. If the material stores elastic energy $W = \tfrac12\varepsilon : C(\varepsilon)$ — which the next slide assumes — $C$ is self-adjoint, leaving $6\cdot7/2 = 21$. Crystals realise the cases in between: a cubic crystal has 3, a hexagonal one 5. Only reflections were not needed: rotations alone force the isotropic form.

22 · closure

Hooke's law and its two moduli

Hooke's law, isotropic $$\begin{aligned}\sigma &= \lambda\,(\operatorname{tr}\varepsilon)\,I + 2\mu\,\varepsilon\\ &= K(\operatorname{tr}\varepsilon)\,I + 2\mu\,\varepsilon_{\text{dev}}\end{aligned}$$

with $\varepsilon_{\text{dev}} = \varepsilon - \tfrac13(\operatorname{tr}\varepsilon)I$, the bulk modulus $K = \lambda + \tfrac23\mu$ and the shear modulus $\mu$.

Two ways a small piece can deform — change its volume, or change its shape at fixed volume — and one modulus resisting each. The next slide pictures both.

stability · the strain energy density $$W = \tfrac12K(\operatorname{tr}\varepsilon)^2 + \mu\,|\varepsilon_{\text{dev}}|^2,$$

and $\sigma_{ij} = \partial W/\partial\varepsilon_{ij}$.

$W > 0$ for every nonzero strain iff $K > 0$ and $\mu > 0$ — else the material would lower its energy by deforming spontaneously. Then $\lambda + \mu = K + \mu/3 > 0$.

two materials

Steel, $\lambda = 115$ GPa, $\mu = 80$ GPa: $K = 168$ GPa. Air, $\mu = 0$ and (section 4) $K = \gamma p_0 = 0.142$ MPa — a million times softer, and with no shear modulus at all.

The second form of Hooke's law. $\lambda(\operatorname{tr}\varepsilon)I + 2\mu\varepsilon = \lambda(\operatorname{tr}\varepsilon)I + 2\mu\varepsilon_{\text{dev}} + \tfrac23\mu(\operatorname{tr}\varepsilon)I$, which is $K(\operatorname{tr}\varepsilon)I + 2\mu\varepsilon_{\text{dev}}$ with $K = \lambda + \tfrac23\mu$. The split separates changes of volume (resisted by $K$) from changes of shape at fixed volume (resisted by $\mu$), and the split is canonical because the isotropic/deviatoric decomposition is.

The energy. $\varepsilon = \tfrac13(\operatorname{tr}\varepsilon)I + \varepsilon_{\text{dev}}$ is an orthogonal decomposition for the inner product $A:B = \operatorname{tr}(A^{\!\top}B)$, since $I:\varepsilon_{\text{dev}} = \operatorname{tr}\varepsilon_{\text{dev}} = 0$; so $|\varepsilon|^2 = \tfrac13(\operatorname{tr}\varepsilon)^2 + |\varepsilon_{\text{dev}}|^2$ and $\tfrac12\lambda(\operatorname{tr}\varepsilon)^2 + \mu|\varepsilon|^2 = \tfrac12K(\operatorname{tr}\varepsilon)^2 + \mu|\varepsilon_{\text{dev}}|^2$. Differentiating: $\partial W/\partial\varepsilon_{ij} = \lambda(\operatorname{tr}\varepsilon)\delta_{ij} + 2\mu\varepsilon_{ij}$ ✓. Positivity: taking $\varepsilon_{\text{dev}} = 0$ forces $K > 0$, taking $\operatorname{tr}\varepsilon = 0$ forces $\mu > 0$, and conversely both positive make $W$ a sum of nonnegative terms vanishing only at $\varepsilon = 0$. This inequality is what section 5 needs to prove $c_L > c_T$.

History. Hooke, 1676, as an anagram — ceiiinosssttuv, ut tensio sic vis, "as the extension, so the force" — for a spring. The tensor form and its two constants are Navier (1821, with one constant, from a molecular model that turned out too restrictive), Cauchy (1822–28) and Green (1839), who derived the count of 21 from the existence of an energy. Lamé's name is on the constants from his 1852 textbook.

The steel numbers are representative of carbon steel, not of any particular alloy; tables give $\mu \approx 79$–$82$ GPa and a bulk modulus near $160$–$170$ GPa (cited: Kaye and Laby, Tables of Physical and Chemical Constants, the section on elastic moduli). Air's $K = \gamma p_0 = 1.4\times101\,325 = 141\,855$ Pa; the ratio is $1.19\times10^6$.

22 · closure

What K and μ measure

Exaggerated. Left: a uniform squeeze changes volume and not shape, and is resisted by $K$ alone. Right: a simple shear by the angle $\alpha$ changes shape and not volume, and is resisted by $\mu$ alone.

checks

Dimensions: $\varepsilon$ is a pure number, so $\lambda, \mu, K$ are pascals ✓.

Uniform compression, $\varepsilon = \tfrac13(\Delta V/V)I$: $\sigma = K(\Delta V/V)I$, so $p = -K\,\Delta V/V$ — $K$ is the pressure per fractional squeeze ✓.

Simple shear by a small angle $\alpha$ (not $\gamma$, which in this unit is the adiabatic index), only $\varepsilon_{12} = \varepsilon_{21} = \alpha/2$: $\sigma_{12} = \mu\alpha$ — $\mu$ is the shear stress per radian of shear ✓.

Fluid limit $\mu = 0$: $\sigma = K(\operatorname{tr}\varepsilon)I$, pure pressure, no shear ✓.

22 · closure

Viscosity: the Navier–Stokes equations

A fluid resists the rate of deformation. The rate-of-strain tensor $e = \tfrac12(D\vec v + D\vec v^{\!\top})$ plays the role of $\varepsilon$; a Newtonian fluid has $\sigma = -pI + \tau$ with $\tau$ linear and isotropic in $e$. The isotropy theorem gives

$$\tau = 2\eta\big(e - \tfrac13(\operatorname{tr}e)I\big) + \zeta\,(\operatorname{tr}e)\,I ,$$

with the viscosity $\eta$ and the bulk viscosity $\zeta$. Taking $\operatorname{div}$:

the Navier–Stokes equations, stated $$\begin{aligned}\rho\frac{\mathrm D\vec v}{\mathrm Dt} = &-\nabla p + \eta\,\nabla^2\vec v\\ &+ \big(\zeta + \tfrac13\eta\big)\nabla(\nabla\!\cdot\vec v) + \rho\vec b\end{aligned}$$

and, if incompressible ($\nabla\!\cdot\vec v = 0$): $\ \rho\,\mathrm D\vec v/\mathrm Dt = -\nabla p + \eta\nabla^2\vec v$.

The viscous term is $\eta\nabla^2\vec v$: momentum diffusing down its own gradient, with diffusivity $\nu = \eta/\rho$.

at $20\,{}^\circ\mathrm C$	$\eta$ (Pa s)	$\nu$ ($\mathrm{m^2s^{-1}}$)
water	$1.002\times10^{-3}$	$1.00\times10^{-6}$
air	$1.81\times10^{-5}$	$1.50\times10^{-5}$

checks

Dimensions: $\eta\nabla^2\vec v$ is $\mathrm{Pa\,s}\cdot\mathrm{m^{-2}}\cdot\mathrm{m\,s^{-1}} = \mathrm{N\,m^{-3}}$ ✓. Limit $\eta, \zeta \to 0$: Euler ✓. Rigid rotation: $e = 0$, no viscous stress, as a spinning bucket of water at equilibrium confirms ✓.

For sound in air at 1 kHz, viscosity changes the wave by a fraction $\nu\omega/c^2 = 8\times10^{-7}$, with $\omega = 2\pi\times10^3\ \mathrm{s^{-1}}$ the angular frequency and $c = 343\ \mathrm{m\,s^{-1}}$ the speed of sound (section 4). Section 4 drops it.

Full derivation of the viscous force. $\partial_j(2\eta e_{ij}) = \eta\,\partial_j(\partial_iv_j + \partial_jv_i) = \eta\,\partial_i(\nabla\!\cdot\vec v) + \eta\nabla^2v_i$ by Clairaut (unit 03). And $\partial_j\big[(\zeta - \tfrac23\eta)(\nabla\!\cdot\vec v)\delta_{ij}\big] = (\zeta - \tfrac23\eta)\,\partial_i(\nabla\!\cdot\vec v)$. Adding, $\operatorname{div}\tau = \eta\nabla^2\vec v + (\zeta + \tfrac13\eta)\nabla(\nabla\!\cdot\vec v)$ for constant $\eta, \zeta$. The form of $\tau$ is the isotropy theorem's $\lambda'(\operatorname{tr}e)I + 2\eta e$ rewritten with $\zeta = \lambda' + \tfrac23\eta$, so that $\zeta$ multiplies the volume-changing part alone.

What is assumed, and what is not proved. Linearity of $\tau$ in $e$, and dependence on $e$ only — not on $\omega$, because a rigidly rotating fluid feels no friction — are physical hypotheses defining a Newtonian fluid, excellent for water and air, bad for blood, paint and custard. The equations are stated here, not solved: whether smooth solutions in three dimensions exist for all time is one of the Clay Millennium problems. Problem P7 solves the one case everyone can, flow between plates.

The numbers. $\nu_{\text{air}} = 1.81\times10^{-5}/1.204 = 1.50\times10^{-5}$ and $\nu_{\text{water}} = 1.002\times10^{-3}/998.2 = 1.00\times10^{-6}\ \mathrm{m^2s^{-1}}$: air is fifteen times more viscous than water in the sense that matters for motion, because it is so light. The fraction quoted for sound compares the viscous term $\nu\nabla^2\vec v \sim \nu k^2v$ with the inertia $\partial_t\vec v \sim \omega v$, with $k = \omega/c$: $\nu\omega/c^2 = 1.50\times10^{-5}\times6283/343.2^2 = 8.0\times10^{-7}$. The ratio of the inertial term $(\vec v\cdot\nabla)\vec v$ to the viscous term is unit 03's Reynolds number, $UL/\nu$.

History. Navier 1822, from a molecular model; Poisson 1831; Saint-Venant 1843; Stokes 1845, who gave the continuum argument essentially as here and proposed $\zeta = 0$ (Stokes' hypothesis), true for monatomic gases and not in general.

section 4

Sound

Linearise Euler's equations about rest, close them with an equation of state, and the wave equation falls out — with a speed that depends on one physical hypothesis.

22 · sound

Linearising about rest

Perturb the air at rest: $\rho = \rho_0 + \rho'$, $p = p_0 + p'$, $\vec v$ small. Substitute into Euler's equations and keep first-order terms (linearisation):

$$\partial_t\rho' + \rho_0\,\nabla\!\cdot\vec v = 0, \qquad \rho_0\,\partial_t\vec v = -\nabla p' .$$

Dropped: $\nabla\!\cdot(\rho'\vec v)$, $\rho'\partial_t\vec v$, $\rho_0(\vec v\cdot\nabla)\vec v$ — each a product of two small quantities. Close with $p = P(\rho)$: $\ p' = c^2\rho'$, $c^2 := P'(\rho_0)$.

theorem · the acoustic wave equation $$\frac{\partial^2\rho'}{\partial t^2} = c^2\,\nabla^2\rho', \qquad c^2 = \frac{\mathrm dp}{\mathrm d\rho}\Big|_{\rho_0}$$

Proof. $\partial_t$ of the first, $\nabla\!\cdot$ of the second: $\partial_t^2\rho' = -\rho_0\nabla\!\cdot\partial_t\vec v = \nabla^2p' = c^2\nabla^2\rho'$. $\square$

plane waves

For any $C^2$ profile $f$ and unit $\vec n$, $\ p' = f(\vec n\cdot\vec x - ct)$ solves it — unit 18's d'Alembert wave, travelling along $\vec n$ at speed $c$ — with velocity

$$\vec v = \frac{p'}{\rho_0c}\,\vec n :$$

a longitudinal wave. Air moves along the direction of travel, never across it.

checks

Dimensions: $\mathrm dp/\mathrm d\rho$ is $\mathrm{Pa}/(\mathrm{kg\,m^{-3}}) = \mathrm{m^2s^{-2}}$ ✓. Small? A loud $1$ Pa wave has $p'/p_0 = 9.9\times10^{-6}$ and $|\vec v| = 2.4\ \mathrm{mm\,s^{-1}}$, Mach $7\times10^{-6}$: the dropped terms are $10^{-5}$ of the kept ones ✓.

The linearisation, term by term. Continuity: $\partial_t(\rho_0 + \rho') + \nabla\!\cdot\big((\rho_0 + \rho')\vec v\big) = 0$. $\rho_0$ is constant, so this is $\partial_t\rho' + \rho_0\nabla\!\cdot\vec v + \nabla\!\cdot(\rho'\vec v) = 0$, and the last term is second order. Euler: $(\rho_0 + \rho')(\partial_t\vec v + (\vec v\cdot\nabla)\vec v) = -\nabla p'$ (since $\nabla p_0 = 0$); keep $\rho_0\partial_t\vec v$.

Gravity was dropped. Strictly the rest state is hydrostatic, $p_0$ and $\rho_0$ falling with height on the scale $H = RT/Mg = 8.58$ km (problem P5). A 1 kHz wave has wavelength $0.343$ m, and $\lambda/H = 4.0\times10^{-5}$: over a wavelength the background is uniform to that accuracy. Gravity matters for sound only at infrasonic wavelengths of kilometres, where it produces the separate family of internal gravity waves.

The plane wave, verified. With $\xi = \vec n\cdot\vec x - ct$: $\nabla f = f'(\xi)\vec n$ and $\nabla^2f = |\vec n|^2f'' = f''$, while $\partial_t^2f = c^2f''$ ✓. For the velocity, $\rho_0\partial_t\vec v = -\nabla p' = -f'\vec n$, and $\vec v = (f/\rho_0c)\vec n$ gives $\rho_0\partial_t\vec v = -(c f'/c)\vec n$ ✓. Its continuity equation: $\partial_t\rho' = -cf'/c^2$ and $\rho_0\nabla\!\cdot\vec v = f'/c$; they sum to zero ✓.

The numbers. A $1$ Pa amplitude is $91$ dB (the decibel scale uses the rms pressure, $0.71$ Pa, against $20\ \mu\mathrm{Pa}$), uncomfortably loud. $p'/p_0 = 1/101\,325 = 9.9\times10^{-6}$; $|\vec v| = p'/\rho_0c = 1/(1.204\times343.2) = 2.42\ \mathrm{mm\,s^{-1}}$; $|\vec v|/c = 7.0\times10^{-6}$. The ratio of the dropped advective term to the kept one is $|\vec v|/c$, of the dropped density product $\rho'/\rho_0 = p'/(\rho_0c^2) = p'/(\gamma p_0) = 7.0\times10^{-6}$. Linear acoustics is among the most accurate linearisations in physics; it fails for explosions and sonic booms, where the nonlinear terms steepen waves into shocks.

The pressure obeys the same equation, $\partial_t^2p' = c^2\nabla^2p'$, since $p' = c^2\rho'$ with constant $c$; so does each component of $\vec v$ after one more differentiation. Unit 24 solves this three-dimensional wave equation with a source; here only plane waves are needed.

22 · sound

A plane sound wave, pictured

One instant of a plane wave travelling right, displacements exaggerated. Parcels bunch where $p'$ is high and spread where it is low; each moves to and fro along the direction of travel, with velocity in step with $p'$.

Where $p' > 0$, $\rho' = p'/c^2 > 0$: the parcels are crowded. Where $p' < 0$ they are spread out.

The velocity $\vec v = (p'/\rho_0c)\,\vec n$ is forward in a compression and backward in a rarefaction, so the pattern moves on at $c$ while each parcel only oscillates in place.

How far? A $1$ Pa wave at $1$ kHz moves each parcel by $|\vec v|/\omega = 0.39\ \mu\mathrm m$, about a two-hundredth of the width of a hair.

The figure, computed. Take $p' = P\sin kx$ at one instant, $k = 2\pi/\lambda$. Linearised continuity, integrated once in time, says $\rho'/\rho_0 = -\partial_x u$ for the displacement $u$ (the strain slide's $\rho'/\rho_0 = -\nabla\!\cdot\vec u$), so $\partial_xu = -P\sin kx/(\rho_0c^2)$ and $u = (P/\rho_0c^2k)\cos kx$. The dots are drawn at $X + u(X)$ with $u$ magnified many millions of times; the spacing $1 + \partial_Xu$ is smallest where $\sin kx = 1$, under the crest of $p'$. The velocity arrows are $p'/\rho_0c$ at the same instant.

The number. $|\vec v| = 2.42\ \mathrm{mm\,s^{-1}}$ for a $1$ Pa wave (previous slide), and the displacement amplitude of a sinusoid is velocity amplitude over $\omega = 6283\ \mathrm{s^{-1}}$: $3.9\times10^{-7}$ m. Even a loud sound moves air by less than a micrometre, which is why linear acoustics is so accurate.

22 · sound

The closure: how does pressure follow density?

The one piece of physics not fixed by conservation laws. For a gas, the ideal gas law (empirical, cited):

$$p = \frac{\rho RT}{M}, \qquad R = 8.314\ \mathrm{J\,mol^{-1}K^{-1}}.$$

But $T$ may change during a compression. Two hypotheses:

isothermal · Newton, 1687

$T$ constant: Boyle's law $p \propto \rho$. $\ c^2 = \dfrac{p_0}{\rho_0} = \dfrac{RT}{M}$.

adiabatic · Laplace, 1816

No heat exchanged: $p \propto \rho^\gamma$, with $\gamma > 1$ the adiabatic index (next slide). $\ c^2 = \dfrac{\gamma p_0}{\rho_0} = \dfrac{\gamma RT}{M}$.

Both laws pass through the rest state; only the slope there enters $c^2 = \mathrm dp/\mathrm d\rho$. The adiabat is steeper by the factor $\gamma = 1.4$, so its sound is faster by $\sqrt{1.4}$.

A compressed gas that cannot shed heat warms, and warm gas pushes harder: the adiabat rises faster than Boyle's line.

Reading the plot. Pressure and density are in units of their rest values, so the isotherm is the line $p/p_0 = \rho/\rho_0$ and the adiabat $p/p_0 = (\rho/\rho_0)^{1.4}$; the dashed segment is the adiabat's tangent at rest, of slope $1.4$. A sound wave explores only a tiny neighbourhood of the dot — a $1$ Pa wave changes $\rho/\rho_0$ by $7\times10^{-6}$ — so the curvature of either law is irrelevant and the whole dispute is about one number, the slope at rest. In ordinary units the slopes are $p_0/\rho_0 = 290.1^2$ and $\gamma p_0/\rho_0 = 343.2^2\ \mathrm{m^2s^{-2}}$.

Why "$c$ independent of $p_0$" is a real prediction. $p_0/\rho_0 = RT/M$ by the gas law, so both formulas depend on temperature alone. Sound travels at the same speed at the foot and the top of a mountain if the temperature is the same, and faster in warm air than cold, as $\sqrt T$. Both hypotheses predict this; the experiment that decides between them is the absolute number.

A word on "adiabatic". Here it means only "no heat flows", and the compression is also assumed slow enough to be reversible — slow compared with molecular collision times, which are $\sim10^{-10}$ s, and fast compared with heat diffusion across a wavelength. The slide on Newton's error checks the second, and finds that the first fails only where the continuum does.

22 · sound

Why a compressed gas follows ρ to the γ

theorem · the adiabat of an ideal gas

If a gas's internal energy is $U = nc_VT$ and a compression exchanges no heat, then $pV^\gamma$ is constant, with the adiabatic index $\gamma = 1 + R/c_V$.

Proof sketch. First law: $nc_V\,\mathrm dT = -p\,\mathrm dV = -(nRT/V)\,\mathrm dV$. Separate (unit 00): $c_V\ln T + R\ln V$ is constant. Eliminate $T$ with $pV = nRT$. $\square$

full proof in the notes (N)

checks

$\gamma \to 1$ recovers Newton ✓. Both give $c$ independent of $p_0$ at fixed $T$ ✓. For air, diatomic, $c_V = \tfrac52R$ so $\gamma = \tfrac75$; measured, $1.400$ ✓.

Full proof

The physics assumed (cited, as physics: any thermodynamics text, e.g. Fermi, Thermodynamics, ch. I–II): the ideal gas law $pV = nRT$; the first law $\mathrm dU = \delta Q - p\,\mathrm dV$ for a slow compression of a gas; and, for an ideal gas, $U = nc_VT$ with a constant molar heat capacity $c_V$. For a diatomic gas at room temperature $c_V = \tfrac52R$ (three translational and two rotational degrees of freedom, each contributing $\tfrac12R$; vibration is frozen out) — equipartition, derived in statistical mechanics and confirmed by calorimetry.

The derivation. $\delta Q = 0$ gives $nc_V\,\mathrm dT = -p\,\mathrm dV = -\frac{nRT}{V}\mathrm dV$, a separable ODE for $T(V)$: $c_V\frac{\mathrm dT}{T} = -R\frac{\mathrm dV}{V}$, so $c_V\ln T + R\ln V = \text{const}$, i.e. $TV^{R/c_V}$ is constant. Substituting $T = pV/nR$: $pV^{1 + R/c_V}$ is constant. For a fixed mass $V \propto 1/\rho$, so $p \propto \rho^\gamma$ with $\gamma = 1 + R/c_V = c_p/c_V$ (using $c_p = c_V + R$ for an ideal gas). Then $\frac{\mathrm dp}{\mathrm d\rho} = \gamma\frac{p}{\rho}$.

22 · sound · example

The speed of sound in air

worked example 3 · the running example

Dry air, $T = 293.15$ K, $p_0 = 101\,325$ Pa, $M = 0.028\,965\ \mathrm{kg\,mol^{-1}}$, $\gamma = 1.400$.

1. Density. $\rho_0 = \dfrac{p_0M}{RT} = \dfrac{101\,325\times0.028\,965}{8.314\,462\times293.15} = 1.204\ \mathrm{kg\,m^{-3}}$.

2. Isothermal. $c^2 = p_0/\rho_0 = 84\,150\ \mathrm{m^2s^{-2}}$, so $c = 290.1\ \mathrm{m\,s^{-1}}$.

3. Adiabatic. $c^2 = \gamma p_0/\rho_0 = 117\,810\ \mathrm{m^2s^{-2}}$, so $\mathbf{c = 343.2\ m\,s^{-1}}$.

4. Compare. Measured at $20\,{}^\circ\mathrm C$: $343.2\ \mathrm{m\,s^{-1}}$. Laplace agrees; Newton is low by $1 - 1/\sqrt{1.4} = 15.5\%$.

checks

Dimensions: $\gamma p/\rho$ in $\mathrm{m^2s^{-2}}$ ✓. Temperature: $c \propto \sqrt T$ predicts $331.3\ \mathrm{m\,s^{-1}}$ at $0\,{}^\circ\mathrm C$, the tabulated value ✓. Special case $\gamma = 1$: step 2 ✓. Order of magnitude: $c$ is of the order of the molecules' thermal speed $\sqrt{3RT/M} = 503\ \mathrm{m\,s^{-1}}$ — a pressure signal is carried by molecules and cannot outrun them ✓.

getting it right, not lucky

The checkpoint asks for $343 \pm 7\ \mathrm{m\,s^{-1}}$. Newton's $290.1$ misses by $15\%$; so would forgetting to use kelvin, or using a monatomic gas's $\gamma = 5/3$ ($374.5\ \mathrm{m\,s^{-1}}$). The only way to land inside $2\%$ is the right closure with the right $\gamma$.

Newton's closure against Laplace's: the same equations, one hypothesis different, a factor $\sqrt\gamma = 1.183$.

Every number, recomputed. $\rho_0 = 101\,325\times0.028\,965/(8.314\,462\times293.15) = 1.2041\ \mathrm{kg\,m^{-3}}$. $p_0/\rho_0 = 84\,150\ \mathrm{m^2s^{-2}}$, square root $290.09$. Times $1.4$: $117\,810$, square root $343.23$. The ratio is $\sqrt{1.4} = 1.1832$; $1/1.1832 = 0.8452$, so Newton's value is $15.5\%$ below the truth, or the truth is $18.3\%$ above Newton. At $0\,{}^\circ\mathrm C$: $343.23\sqrt{273.15/293.15} = 331.3$ and Newton's $280.0$. With $\gamma = 5/3$ the formula gives $\sqrt{5/3}\times290.09 = 1.2910\times290.09 = 374.5\ \mathrm{m\,s^{-1}}$. The thermal speed $\sqrt{3RT/M} = \sqrt{3}\times290.09 = 502.4\ \mathrm{m\,s^{-1}}$.

The measured value for dry air at $20\,{}^\circ\mathrm C$ is $343.2\ \mathrm{m\,s^{-1}}$ to the precision quoted in standard tables (cited: Kaye and Laby, the section on the speed of sound in gases, and the ISO standard atmosphere's $340.3\ \mathrm{m\,s^{-1}}$ at $15\,{}^\circ\mathrm C$, which is $343.2\sqrt{288.15/293.15}$ to within its rounding). Humidity raises it slightly, since water vapour is lighter than air: a few tenths of a percent at room conditions.

How the checkpoint is falsifiable. It is possible to get $343$ by accident — for instance by quoting it. What the checkpoint requires is the chain: conservation laws, linearisation, $c^2 = \mathrm dp/\mathrm d\rho$, the adiabatic derivative, the numbers. A derivation that uses $p \propto \rho$ fails it, and so does one that uses $p \propto \rho^\gamma$ with no argument for why.

22 · sound

Newton's error, and Laplace's repair

the history

1687. Principia, Book II, Props. 48–50: the first speed of a wave computed from mechanics. Newton assumes Boyle's law, isothermal. The answer is about a sixth below measurement.

1713. Second edition: Newton closes the gap with ad hoc corrections — the "crassitude" of the air's solid particles, the vapours in it — and lands exactly on the measured $1142$ feet per second ($348$ m/s).

1816. Laplace: a compressed gas warms, raising its pressure beyond Boyle's; the compressions are adiabatic. $c^2 = \gamma p/\rho$. Measured values of $\gamma$ (Clément and Desormes, 1819) confirm it.

why adiabatic: two times compared

Heat must cross a distance $\sim\lambda/2\pi$ within a time $\sim1/\omega$. By dimensions (unit 00), with the air's thermal diffusivity $\kappa = 2.14\times10^{-5}\ \mathrm{m^2s^{-1}}$, crossing takes $\sim(\lambda/2\pi)^2/\kappa$.

At $1$ kHz: $\lambda/2\pi = 5.5$ cm, so $140$ s against $1/\omega = 0.16$ ms. Heat has a millionth of the time it needs.

Where would Newton have been right? The next slide plots the ratio across all frequencies.

The numbers. Thermal diffusivity $\kappa = k/(\rho c_p)$ with thermal conductivity $k = 0.0259\ \mathrm{W\,m^{-1}K^{-1}}$ and specific heat $c_p = 1006\ \mathrm{J\,kg^{-1}K^{-1}}$ for air at $20\,{}^\circ\mathrm C$ (cited: standard tables), giving $0.0259/(1.204\times1006) = 2.14\times10^{-5}\ \mathrm{m^2s^{-1}}$. At $f = 1$ kHz, $\omega = 6283\ \mathrm{s^{-1}}$ and $\lambda/2\pi = c/\omega = 0.0546$ m. $(0.0546)^2/2.14\times10^{-5} = 140$ s; $1/\omega = 1.59\times10^{-4}$ s; ratio $8.8\times10^5$. Crossover: $(c/\omega)^2/\kappa = 1/\omega$ at $\omega = c^2/\kappa = 343.2^2/2.14\times10^{-5} = 5.5\times10^9\ \mathrm{s^{-1}}$ (about $0.9$ GHz), where $c/\omega = \kappa/c = 6.2\times10^{-8}$ m. The mean free path of air molecules at these conditions is about $0.07\ \mu\mathrm m$ (cited: kinetic theory tables).

The historical record. Newton's derivation (Principia, Book II, Section VIII) treats the air as a chain of particles obeying Boyle's law and obtains in effect $c^2 = p/\rho$; it is also the first time anyone computed the speed of a wave from the properties of the medium. His agreement with the measured $1142$ ft/s in the second edition is a well-known case study in fitting a theory to data, analysed in R. S. Westfall, "Newton and the fudge factor", Science 179 (1973) 751–758. Laplace published the adiabatic correction in 1816, after earlier suggestions in the same direction. $1142$ feet is $348.1$ m.

The lesson for this course. Every step of the modern derivation — Cauchy's theorem, the local conservation laws, linearisation, the wave equation — survives, and Newton's own mechanics, cruder as it was, was right in substance. What failed was one constitutive hypothesis, and the discrepancy pointed at it: a factor that is the square root of a ratio of heat capacities could only come from thermodynamics. That is the model of a good falsification. Unit 25 is the next one, and there the failed hypothesis is Galilean relativity itself.

22 · sound

Where Newton would be right

The time heat needs to cross $\lambda/2\pi$, over the time $1/\omega$ a compression lasts: $c^2/\kappa\omega$, a line of slope $-1$. Above $1$ the compressions are adiabatic. It reaches $1$ only near $0.9$ GHz.

a falsification with a number attached

Isothermal would need $\omega \gtrsim c^2/\kappa = 5.5\times10^{9}\ \mathrm{s^{-1}}$, where $\lambda/2\pi \approx \kappa/c = 62$ nm — the molecular mean free path. Newton's hypothesis becomes true only where the continuum itself fails. His mechanics was right; one thermodynamic assumption was wrong, and $\sqrt\gamma$ is its fingerprint.

The plot, computed. The ratio is $\big((\lambda/2\pi)^2/\kappa\big)\big/(1/\omega)$ with $\lambda/2\pi = c/\omega$, which is $c^2/(\kappa\omega)$, evaluated with $c = 343.2\ \mathrm{m\,s^{-1}}$ and $\kappa = 2.14\times10^{-5}\ \mathrm{m^2s^{-1}}$. At $1$ kHz it is $8.8\times10^5$; across the whole audible band, $20$ Hz to $20$ kHz, it never drops below $4\times10^4$. It equals $1$ at $f = c^2/(2\pi\kappa) = 0.88$ GHz, where the wavelength over $2\pi$ has shrunk to the molecules' mean free path and the continuum description, and with it this whole calculation, has stopped applying.

Why the estimate is only an estimate. The exact statement needs the heat equation — a parabolic PDE in unit 18's classification — solved for a periodic temperature disturbance, which this unit does not do. Dimensional analysis gives the diffusion time up to a numerical factor of order one, and a factor of one million leaves no doubt about the answer. That is unit 00's lesson in a new setting: the shape of an answer before its solution.

22 · sound

Energy in a sound wave, and a preview of the stress–energy tensor

theorem · acoustic energy is locally conserved

For solutions of the linearised equations, the acoustic energy density $e$ (not section 3's rate of strain) and acoustic intensity $\vec S$

$$e = \tfrac12\rho_0|\vec v|^2 + \frac{p'^2}{2\rho_0c^2}, \qquad \vec S = p'\,\vec v$$

satisfy $\ \partial_te + \nabla\!\cdot\vec S = 0$.

Proof. $\partial_te = \rho_0\vec v\cdot\partial_t\vec v + p'\partial_tp'/\rho_0c^2 = -\vec v\cdot\nabla p' - p'\,\nabla\!\cdot\vec v = -\nabla\!\cdot(p'\vec v)$. $\square$

Checks. $e$ in $\mathrm{J\,m^{-3}}$, $\vec S$ in $\mathrm{W\,m^{-2}}$ ✓. Plane wave: kinetic = potential and $\vec S = c\,e\,\vec n$ ✓. A $1$ Pa wave carries $1.21\ \mathrm{mW\,m^{-2}}$ on average.

conserved	density	flux
mass	$\rho$	$\rho v_j$
$i$-momentum	$\rho v_i$	$\Pi_{ij} = \rho v_iv_j - \sigma_{ij}$
energy	$e$	$S_j$

flagged for unit 35

Energy density, momentum density and momentum flux are the pieces of one symmetric $4\times4$ object, the stress–energy tensor $T^{\mu\nu}$. In this box $c$ is the speed of light and the indices $\mu, \nu$ run over $0$ (time, $x^0 = ct$) and $1,2,3$ (space) — not the shear modulus. $T^{00}$ is the energy density, $T^{0i}$ the energy flux $/c$ (equal to $c\times$ momentum density), $T^{ij} = \Pi_{ij}$.

"Energy and momentum are locally conserved" becomes $\partial_\mu T^{\mu\nu} = 0$, summed over $\mu$ — exactly the statements of this section, four at once. For a perfect fluid at rest, $T = \operatorname{diag}(\rho c^2, p, p, p)$: the $-\sigma = pI$ of still air.

Full proof. Use $\rho_0\partial_t\vec v = -\nabla p'$ in the first term and $\partial_tp' = c^2\partial_t\rho' = -\rho_0c^2\nabla\!\cdot\vec v$ in the second: $\partial_te = -\vec v\cdot\nabla p' - p'\nabla\!\cdot\vec v$, and the product rule makes that $-\nabla\!\cdot(p'\vec v)$. Compare unit 18's $e_t + S_x = 0$: the same structure, with the tension replaced by pressure and $\partial_x$ by a divergence.

The checks. $\rho_0|\vec v|^2$: $\mathrm{kg\,m^{-3}}\cdot\mathrm{m^2s^{-2}} = \mathrm{J\,m^{-3}}$; $p'^2/(\rho_0c^2)$: $\mathrm{Pa}^2/\mathrm{Pa} = \mathrm{Pa} = \mathrm{J\,m^{-3}}$; $p'\vec v$: $\mathrm{N\,m^{-2}}\cdot\mathrm{m\,s^{-1}} = \mathrm{W\,m^{-2}}$ ✓. For the plane wave $\vec v = p'\vec n/\rho_0c$, so $\tfrac12\rho_0|\vec v|^2 = p'^2/2\rho_0c^2$: equipartition, as for unit 18's string; $e = p'^2/\rho_0c^2$ and $\vec S = p'^2\vec n/\rho_0c = c\,e\,\vec n$ — energy travels at $c$. For $p' = \hat p\cos(\ldots)$ the time average of $|\vec S|$ is $\hat p^2/2\rho_0c$; with $\hat p = 1$ Pa and $\rho_0c = 413.3\ \mathrm{Pa\,s\,m^{-1}}$ (the air's acoustic impedance) that is $1.21\times10^{-3}\ \mathrm{W\,m^{-2}}$.

The preview, and what is and is not claimed. Nothing on the right of this slide is used before unit 35, and nothing there is proved here. What is established here is the structure: each conserved quantity has a density and a flux, and local conservation is "time derivative of density plus divergence of flux is zero". For the full nonlinear ideal fluid the energy density is $\tfrac12\rho|\vec v|^2 + \rho u$ ($u$ internal energy per mass) and the energy flux is $(\tfrac12\rho|\vec v|^2 + \rho u + p)\vec v$ (cited: Landau and Lifshitz, Fluid Mechanics, §6); the acoustic $e$ and $\vec S$ are its expansion to second order about rest. In special relativity mass stops being separately conserved — it is part of the energy — which is why the table's three rows become the four rows $\nu = 0,1,2,3$ of $\partial_\mu T^{\mu\nu} = 0$.

Why $T^{\mu\nu}$ must be symmetric is a relativistic echo of the angular-momentum slide of section 2: the $ij$ block is symmetric for the same reason $\sigma$ is, and the $0i$/$i0$ equality says that energy flux is momentum density times $c^2$, which is $E = mc^2$ in local form. Unit 29 meets a canonical $T^{\mu\nu}$ that is not symmetric and repairs it; the repair is the field-theory version of the proof that $\sigma_{ij} = \sigma_{ji}$. Unit 39 uses the perfect-fluid tensor, $T^{\mu\nu} = (\rho + p/c^2)u^\mu u^\nu + p\,g^{\mu\nu}$, whose rest-frame form is the one quoted on the slide, to run the expansion of the universe.

section 5

Waves in solids, and the Earth's liquid core

Give the material a shear modulus and a second, slower, transverse wave appears. Take it away and only sound is left.

22 · solids

Navier's equation for an elastic solid

Small displacements $\vec u(\vec x,t)$ about an unstressed state, no body force. Linearise Cauchy's equation, $\rho\,\partial_t^2\vec u = \operatorname{div}\sigma$, and insert Hooke's law:

theorem · Navier's equation $$\rho\,\frac{\partial^2\vec u}{\partial t^2} = (\lambda + \mu)\,\nabla(\nabla\!\cdot\vec u) + \mu\,\nabla^2\vec u$$

Proof. $\partial_j\sigma_{ij} = \partial_j\big[\lambda(\partial_ku_k)\delta_{ij} + \mu(\partial_iu_j + \partial_ju_i)\big] = (\lambda + \mu)\,\partial_i(\nabla\!\cdot\vec u) + \mu\nabla^2u_i$, by Clairaut. $\square$

checks

Dimensions: $\mathrm{Pa\,m^{-2}}\cdot\mathrm m = \mathrm{N\,m^{-3}}$ both sides ✓. One dimension, $\vec u = u(x,t)\vec e_1$: $\rho u_{tt} = (\lambda + 2\mu)u_{xx}$, a wave equation as in unit 18 ✓.

the fluid limit, and the air

Set $\mu = 0$ and $\lambda = K$: $\rho\,\partial_t^2\vec u = K\nabla(\nabla\!\cdot\vec u)$. Take the divergence, with the dilatation $\theta = \nabla\!\cdot\vec u = -\rho'/\rho_0$:

$$\partial_t^2\theta = \frac{K}{\rho}\nabla^2\theta .$$

Section 4's wave equation, with $c^2 = K/\rho$. For a gas $K = \rho\,\mathrm dp/\mathrm d\rho$, and adiabatically $K = \gamma p_0$: the same $343.2\ \mathrm{m\,s^{-1}}$, reached from the solid side.

So a fluid, for small motions, is an elastic solid that has lost its shear modulus. The next slide finds what that loss costs.

The linearisation. For small displacements from rest, $\vec v = \partial_t\vec u$ to first order, and $\mathrm D\vec v/\mathrm Dt = \partial_t^2\vec u + (\vec v\cdot\nabla)\vec v$, whose second term is second order. The density in front is $\rho$ of the reference state to the same order. Positions $\vec x$ and reference positions $\vec X$ differ by $\vec u$, and derivatives with respect to them differ by terms of order $|D\vec u|$ times the derivative, which are also dropped.

The proof in full. $\partial_j\big(\lambda\,\partial_ku_k\,\delta_{ij}\big) = \lambda\,\partial_i\partial_ku_k$. And $\partial_j\big(\mu(\partial_iu_j + \partial_ju_i)\big) = \mu\,\partial_i\partial_ju_j + \mu\,\partial_j\partial_ju_i$, using Clairaut to swap $\partial_j\partial_i$. So the total is $(\lambda + \mu)\,\partial_i(\nabla\!\cdot\vec u) + \mu\nabla^2u_i$, for constant $\lambda$ and $\mu$ (a homogeneous material).

The one-dimensional check is a longitudinal wave in a slab constrained laterally, and its modulus is $\lambda + 2\mu$, not Young's modulus: a thin rod, free to contract sideways, has a smaller effective modulus. This is a real source of confusion when comparing sound speeds in rods and in bulk material.

The air, again. $\nabla\!\cdot$ of $\nabla(\nabla\!\cdot\vec u)$ is $\nabla^2(\nabla\!\cdot\vec u)$ by Clairaut. The link $\theta = -\rho'/\rho_0$ is the strain slide's statement that dilatation is fractional volume change. And $K = \rho\,\mathrm dp/\mathrm d\rho$: from $p' = -K\theta = K\rho'/\rho_0$ and $p' = c^2\rho'$. Adiabatically, $\rho\cdot\gamma p/\rho = \gamma p$. The running example has been solved a third time, this time as the $\mu = 0$ case of elasticity.

History. Navier derived the equation in 1821 from a molecular model with $\lambda = \mu$; Cauchy and Poisson gave the general isotropic form within a few years. Poisson found the two wave speeds in 1829; Stokes explained in 1849 that they are longitudinal and transverse.

22 · solids

Plane waves: two speeds and no third

Try a plane wave $\vec u = \vec a\,f(\vec n\cdot\vec x - ct)$, $|\vec n| = 1$, $f'' \not\equiv 0$. Then $\nabla(\nabla\!\cdot\vec u) = (\vec n\cdot\vec a)\vec n f''$, $\nabla^2\vec u = \vec af''$, $\partial_t^2\vec u = c^2\vec af''$, and Navier gives

$$\rho c^2\,\vec a = A(\vec n)\,\vec a,$$ $$A(\vec n) = (\lambda + \mu)\,\vec n\otimes\vec n + \mu\,I .$$

The acoustic tensor $A(\vec n)$ is a symmetric bilinear form; by the spectral theorem its eigenvalues are $\lambda + 2\mu$ (eigenvector $\vec n$) and $\mu$ (every $\vec a \perp \vec n$).

theorem · P and S waves

Every plane wave in an isotropic elastic solid is one of

$$c_L = \sqrt{\frac{\lambda + 2\mu}{\rho}}, \qquad c_T = \sqrt{\frac{\mu}{\rho}},$$

with $\vec a \parallel \vec n$ for $c_L$ and $\vec a \perp \vec n$ for $c_T$:

respectively the P-wave (primary, longitudinal) and the S-wave (secondary, transverse wave, two polarisations). And $c_L > c_T$ always.

proof of $c_L > c_T$

$\rho(c_L^2 - c_T^2) = \lambda + \mu = K + \tfrac13\mu > 0$ by stability. Sharper: $c_L^2/c_T^2 = (K + \tfrac43\mu)/\mu > \tfrac43$. $\square$

Checks. $\sqrt{\mathrm{Pa}/\mathrm{kg\,m^{-3}}} = \mathrm{m\,s^{-1}}$ ✓. $\mu \to 0$: $c_T \to 0$ and $c_L \to \sqrt{K/\rho}$, sound ✓.

The substitution, in full. With $\xi = \vec n\cdot\vec x - ct$: $\partial_ju_k = a_kn_jf'$, so $\nabla\!\cdot\vec u = (\vec n\cdot\vec a)f'$ and $\nabla(\nabla\!\cdot\vec u) = (\vec n\cdot\vec a)\vec nf''$; $\nabla^2u_k = a_k|\vec n|^2f'' = a_kf''$; $\partial_t^2\vec u = c^2\vec af''$. Navier's equation becomes $\big(\rho c^2\vec a - (\lambda + \mu)(\vec n\cdot\vec a)\vec n - \mu\vec a\big)f'' = 0$, and since $f''$ is not identically zero the bracket vanishes.

The eigenvalues. $A(\vec n)\vec n = (\lambda + \mu)\vec n + \mu\vec n = (\lambda + 2\mu)\vec n$, and for $\vec a\perp\vec n$, $A(\vec n)\vec a = \mu\vec a$. These span $\mathbb R^3$, so there are no other eigenvalues — no third speed. As a bilinear form, $A(\vec n)(\vec a,\vec b) = (\lambda + \mu)(\vec n\cdot\vec a)(\vec n\cdot\vec b) + \mu\,\vec a\cdot\vec b$; in an anisotropic crystal the corresponding tensor $C_{ijkl}n_jn_l$ has three generally distinct eigenvalues depending on direction, and there are three wave speeds in each direction.

The bound $\tfrac43$. $c_L^2/c_T^2 = (\lambda + 2\mu)/\mu = (K - \tfrac23\mu + 2\mu)/\mu = K/\mu + \tfrac43$, which exceeds $\tfrac43$ since $K > 0$. A "Poisson solid", $\lambda = \mu$, has ratio $\sqrt3 = 1.732$; many rocks are close to it, which is a useful rule of thumb in seismology.

Names. P and S are "primary" and "secondary", from the order in which they arrive on a seismogram — which is this theorem, read by a seismometer.

22 · solids

P and S waves, pictured

A sheet of material points as each wave passes, travelling right; displacements exaggerated. In the P-wave the columns crowd and spread; in the S-wave they slide up and down past one another and no area changes.

P: $\vec a \parallel \vec n$, so $\nabla\!\cdot\vec u \ne 0$: the wave squeezes and stretches. Both moduli resist, $\lambda + 2\mu = K + \tfrac43\mu$.

S: $\vec a \perp \vec n$, so $\nabla\!\cdot\vec u = 0$: pure shear, resisted by $\mu$ alone. Take $\mu$ away and nothing pulls the sheared columns back.

That one sentence is why a liquid carries no S-wave — the subject of the section's last slide.

The figure, computed. Points of a square grid at $\vec X$ are drawn at $\vec X + U\vec a\sin(k\,\vec n\cdot\vec X)$, $k = 2\pi/\lambda$, one instant of $\vec u = \vec a\,f(\vec n\cdot\vec x - ct)$ with a sine profile, $\vec n$ to the right; $\vec a = \vec n$ above, $\vec a$ vertical below. In the S-wave each column moves rigidly, so the local deformation is a simple shear — the right-hand picture of What K and μ measure — which changes no area.

Divergence and curl. The same split appears without plane waves. The divergence of Navier's equation gives $\rho\,\partial_t^2\theta = (\lambda + 2\mu)\nabla^2\theta$ for $\theta = \nabla\!\cdot\vec u$; its curl (unit 04), using $\nabla\times\nabla(\cdot) = 0$, gives $\rho\,\partial_t^2\vec w = \mu\nabla^2\vec w$ for $\vec w = \nabla\times\vec u$. Compressions travel at $c_L$, rotations at $c_T$.

22 · solids · example

P and S waves in steel, and in air

worked example 4

Steel: $\lambda = 115$ GPa, $\mu = 80$ GPa, $\rho = 7850\ \mathrm{kg\,m^{-3}}$. Find both speeds, and the lag of the S-wave behind the P-wave over $1$ m.

1. P. $c_L^2 = (115 + 160)\times10^9/7850 = 3.503\times10^7$, so $c_L = 5919\ \mathrm{m\,s^{-1}}$.

2. S. $c_T^2 = 80\times10^9/7850 = 1.019\times10^7$, so $c_T = 3192\ \mathrm{m\,s^{-1}}$.

3. Ratio. $c_L/c_T = \sqrt{275/80} = 1.854 > \sqrt{4/3} = 1.155$ ✓.

4. Lag. $1/c_T - 1/c_L = 313.3 - 169.0 = 144\ \mu\mathrm s$ per metre. Ultrasonic flaw detectors time exactly this.

the air, as a degenerate solid

$\mu = 0$, $\lambda = K = \gamma p_0 = 141\,855$ Pa, $\rho_0 = 1.204\ \mathrm{kg\,m^{-3}}$:

$$c_L = \sqrt{\frac{141\,855}{1.204}} = 343.2\ \mathrm{m\,s^{-1}},\quad c_T = 0 .$$

The eigenvalue $\mu$ of the acoustic tensor is zero: the "transverse wave" is a static shear the air does not resist. Air carries sound, and nothing else.

checks

Tabulated for mild steel: about $5.9$ and $3.2\ \mathrm{km\,s^{-1}}$ ✓. Air: section 4's number, by a different road ✓. Dimensions as on the previous slide ✓.

The numbers. $(275\times10^9)/7850 = 3.5032\times10^7$, square root $5918.8$; $80\times10^9/7850 = 1.0191\times10^7$, square root $3192.3$; ratio $1.8540$. $1/3192.3 = 3.1325\times10^{-4}$ s and $1/5918.8 = 1.6895\times10^{-4}$ s, difference $1.443\times10^{-4}$ s. The air: $141\,855/1.2041 = 117\,810$, square root $343.2$.

The tabulated values for carbon steel are $c_L \approx 5.9$–$6.0\ \mathrm{km\,s^{-1}}$ and $c_T \approx 3.2\ \mathrm{km\,s^{-1}}$ (cited: Kaye and Laby, the table of sound speeds in solids). The chosen $\lambda$ and $\mu$ are representative rather than measured for one alloy, so agreement to a percent or two is all that should be claimed.

How the air's zero should be read. A real gas does have a shear response — viscosity — but it is proportional to the rate of shear, not the shear, so it damps transverse motion rather than restoring it. A "shear wave" in air decays within a distance of order $\sqrt{2\nu/\omega}$, about $70\ \mu\mathrm m$ at 1 kHz: a boundary layer, not a wave. That is the physical content of $c_T = 0$ for a fluid.

22 · solids

Why only solids carry shear waves

the argument

A fluid sustains no static shear: $\mu = 0$. By the plane-wave theorem, $c_T = 0$ — no S-waves cross a liquid. P-waves cross anything with $K > 0$.

The Earth. An earthquake sends P and S waves through the mantle. Beyond about $103^\circ$ of arc from the epicentre, no direct S-waves arrive at all: the S-wave shadow zone. Something below the mantle stops them.

Straight rays: one grazing a core of radius $3480$ km in an Earth of radius $6371$ km emerges at $2\arccos(3480/6371) = 113.8^\circ$. Real rays curve, and the edge moves in to $\approx103^\circ$.

conclusion

The outer core, below the core–mantle boundary at $2891$ km depth, is liquid. At the boundary: mantle $c_T = 7.26$, core $c_T = 0\ \mathrm{km\,s^{-1}}$.

Straight rays, to scale. The grazing ray (solid) bounds where direct S-waves can reach, $113.8^\circ$ from the epicentre at the centre; the thick arc is the shadow.

The straight-ray number. A straight chord from a surface point that just touches a concentric sphere of radius $r$ subtends, at the centre, twice the angle whose cosine is $r/R$: $2\arccos(3480/6371) = 2\times56.89^\circ = 113.78^\circ$. In the real Earth speeds rise with depth through the mantle, so rays bend back towards the surface (refraction, with the ray parameter $r\sin i/v$ conserved along a ray), and the ray that grazes the core emerges nearer the epicentre. Seismology texts put the edge of the direct-S shadow near $100$–$105^\circ$ (cited: Stein and Wysession, An Introduction to Seismology, Earthquakes, and Earth Structure, ch. 3). Beyond it only waves that crossed the core as P-waves, or that were diffracted along the boundary, are recorded.

The boundary values: mantle side $c_L = 13.72$, $c_T = 7.26\ \mathrm{km\,s^{-1}}$; core side $c_L = 8.06$, $c_T = 0$; the outer core extends to $5150$ km depth. They are from the Preliminary Reference Earth Model (Dziewonski and Anderson, Physics of the Earth and Planetary Interiors 25 (1981) 297–356), rounded. Problem P11 starts from the elastic constants that reproduce these values — for the core, $\mu = 0$ and a bulk modulus of $644$ GPa — and recovers the speeds.

History, and the logic. Oldham (1906) inferred a core from the delay of P-waves through the Earth's centre; Gutenberg (1914) put its boundary near $2900$ km; Jeffreys (1926) argued it is liquid, from the absence of S-waves through it and from the Earth's tidal rigidity being lower than the mantle alone would give. Lehmann (1936) found the solid inner core inside it. The inference "no S-waves, therefore liquid" is exactly the plane-wave theorem: $c_T = \sqrt{\mu/\rho}$, and a material in which $c_T = 0$ has no shear modulus, which is the definition of a fluid. The Earth's magnetic field is generated by convection in that liquid iron, so this is not a small fact.

A warning on the shadow's shape. There is also a P-wave shadow zone, roughly $104^\circ$ to $140^\circ$, caused not by absence but by the sharp drop in $c_L$ at the core, which refracts P-waves steeply inward. The S-wave shadow is the one that proves the liquid; the P-wave shadow proves only a slower layer.

22 · closing

What this unit hands on

the air, closed

Its stress was found ($-p_0I$, by Pascal's law), its two conservation laws written (continuity, Euler), its missing equation named (an equation of state), and its sound speed computed three ways: Newton's $290.1$, Laplace's $343.2$, and $343.2$ again as a solid with $\mu = 0$. Measurement says $343.2\ \mathrm{m\,s^{-1}}$.

The method, once more: a budget for every region, the divergence theorem, the localisation lemma. It will be run again for charge, for electromagnetic energy, and for $T^{\mu\nu}$.

23	the divergence theorem doing the same work for electric flux
24	the continuity equation for charge; the Poynting vector as an energy flux; Maxwell's stress tensor
28	an antisymmetric tensor field, $F_{\mu\nu}$, where §2's slot bookkeeping is not optional
29	$T^{\mu\nu}$ from Noether's theorem, and why it must be made symmetric
35	the table of densities and fluxes as $T^{\mu\nu}$, and $\partial_\mu T^{\mu\nu} = 0$
39	the perfect fluid's $T^{\mu\nu}$ driving the expansion of the universe

The single most transferable thing is the three-move conversion from an integral budget to a local law. It needs a density, a flux and the divergence theorem, and nothing about mechanics. Unit 24 uses it for charge and for field energy; unit 35 uses it in four dimensions, where "time derivative plus spatial divergence" becomes one four-divergence and the flux of momentum becomes the spatial block of $T^{\mu\nu}$.

The second is Cauchy's theorem as a model of how a tensor enters physics: not by a transformation rule, but because a physical quantity that depends on a direction turns out, by a balance argument, to depend on it linearly. The metric in unit 32 and the curvature in unit 33 are both introduced as multilinear maps first, and this unit is the course's best rehearsal for that habit.

Reading. Gurtin, An Introduction to Continuum Mechanics, the chapters on stress, balance laws and linear elasticity — closest in spirit, rigorous, and it assumes nothing this unit has not taught. Landau and Lifshitz, Fluid Mechanics, §§1–2 (Euler's equations), §15 (Navier–Stokes) and §64 (sound), and Theory of Elasticity, §§1–5 and §22 (elastic waves); skip anything using Fourier transforms until after unit 20. Feynman, Lectures, vol. II, ch. 31 (tensors, including the stress tensor) and ch. 38–39 (elasticity), for the physical pictures. Stein and Wysession, ch. 2–3, for seismology.

22 · checkpoint

Done when

checkpoint

From an integral conservation statement for an arbitrary region, you derive its local form by the divergence theorem and the localisation lemma — for mass, for momentum with the stress tensor as the flux, and for a quantity you have not seen before.
You prove Cauchy's theorem — the traction is linear in the normal — by the tetrahedron argument, rather than assuming stress is bilinear, and you can say which two powers of the tetrahedron's size make it work.
You compute the speed of sound in air at a stated temperature to within $2\%$ — $343.2\ \mathrm{m\,s^{-1}}$ at $20\,{}^\circ\mathrm C$ — with the adiabatic closure argued for, not assumed.

How to fail it

Write $\partial_t\rho + \vec v\cdot\nabla\rho = 0$ for a compressible flow. Call $\sigma$ a tensor "because it has two indices". Use Boyle's law and report $290\ \mathrm{m\,s^{-1}}$. Each is unambiguous.

Then

./mr problems 22 — twelve problems, about five hours. When the three claims above are true: ./mr done 22.