Stress Tensor and Cauchy's Relation

A continuous medium transmits forces internally: any imagined surface inside it has material on both sides pushing and pulling across it. Unlike a point force, this internal force is distributed — characterized by a force per unit area called the traction $\mathbf{t}$ . But the traction depends on the orientation of the surface you cut: a horizontal plane and a vertical plane through the same point generally feel different forces. The central insight of Cauchy (1822) is that this orientation-dependence is captured by a single object, the stress tensor. The stress tensor $\sigma_{ij}$ packages nine numbers (six independent, after symmetry) that completely specify the internal force state at a point. Cauchy's relation $t_i = \sigma_{ij}n_j$ then reconstructs the traction on any surface from its normal $\mathbf{n}$ . The proof — the elegant tetrahedron argument — is a model of how local mechanical balance forces tensorial structure to emerge, and it is the prototype for how flux quantities behave throughout continuum physics. Two conservation laws sculpt the tensor. Balance of linear momentum on an infinitesimal volume turns $\sigma_{ij}$ into the equation of motion $\rho\ddot{u}_i = \partial_j\sigma_{ij}+f_i$ . Balance of angular momentum forces the tensor to be symmetric. Together these reduce continuum dynamics to a single field equation, which — combined with a constitutive law relating stress to strain (Hooke's law) — yields the wave equations governing elastic waves in solids and the Navier–Stokes equations in fluids. Conceptually, the stress tensor is the spatial part of the energy–momentum tensor of field theory: stress is the flux of momentum. This unifies the continuum picture of elastic media with the field-theoretic language of the Lagrangian density and Hamiltonian density, and ultimately with the symmetric stress–energy tensor of special relativity. The classical $\sigma_{ij}$ is where that whole edifice begins.

Key takeaways

The stress tensor $\sigma_{ij}$ is a rank-2 tensor field whose component is the $i$ -th component of force per unit area exerted across a surface whose outward normal points in the $j$ -direction.
Cauchy's relation $t_i = \sigma_{ij} n_j$ gives the traction (force per area) on an arbitrary surface from its unit normal $\mathbf{n}$ — knowing $\sigma_{ij}$ at a point determines the force on every plane through that point.
Cauchy's tetrahedron argument proves traction must be a linear function of the normal, forcing the existence of a tensor; the same argument shows traction reverses sign with the normal (Newton's third law).
Balance of angular momentum makes the stress tensor symmetric, $\sigma_{ij}=\sigma_{ji}$ , in the absence of body couples — leaving 6 independent components.
Linear momentum balance gives Cauchy's equation of motion $\rho\,\ddot{u}_i = \partial_j\sigma_{ij} + f_i$ , the continuum analog of $\mathbf{F}=m\mathbf{a}$ that governs elastic waves and fluid flow.
Diagonalizing $\sigma_{ij}$ yields principal stresses along principal axes where the traction is purely normal — no shear.
Key idea: Traction is not a vector field but a vector-valued function of orientation. Linearity in the normal — guaranteed by force balance on a vanishing tetrahedron — means there exists a tensor $\sigma_{ij}$ such that $t_i=\sigma_{ij}n_j$ . Stress is the flux of momentum across surfaces.

Where this lives today

Every engineering simulation that predicts whether something breaks is a numerical solver for Cauchy's equation. A modern automotive crash simulation integrates $\rho\ddot u_i=\partial_j\sigma_{ij}+f_i$ over tens of millions of finite elements at microsecond timesteps on GPU clusters, evaluating a stress tensor (elastic, plastic, rate-dependent) in every element at every step; aircraft certification, additive-manufacturing residual-stress prediction, and the packaging that survived your last delivery all run the same mathematics. The von Mises fields glowing red in every FEA screenshot are the deviatoric invariant $\sqrt{3J_2}$ of the tensor defined in this article. Geophysics reads the tensor in the other direction. An earthquake's radiation pattern is inverted for its moment tensor — the point-source stress equivalent of the rupture — and catalogs of moment tensors, updated within minutes of every significant quake, are the standard product by which fault mechanisms are classified (thrust, normal, strike-slip). Meanwhile the stress-flux idea has completed its journey into fundamental physics: gravitational-wave detectors measure ripples sourced by the stress–energy tensor $T^{\mu\nu}$ , whose spatial block is precisely Cauchy's $\sigma_{ij}$ , and electromagnetic momentum transport in laser tweezers is computed from the Maxwell stress tensor — the same construction with fields in place of springs.

The physical picture

Imagine slicing the medium with an imaginary plane. The material on one side pushes on the material on the other across that plane. The push per unit area is the traction; its direction need not be perpendicular to the cut (shear). To know the force on every possible cut at a point, you would naively need infinitely many vectors — one per orientation. The miracle is that nine numbers suffice, because force balance makes the traction linear in the normal.

Figure

The Cauchy tetrahedron — three faces lie in the coordinate planes and carry the basis tractions, the oblique face with unit normal n carries the unknown traction t; as the tetrahedron shrinks, surface forces must balance among themselves, forcing t to be linear in n

Think of $\sigma_{ij}$ column by column: the $j$ -th column is the traction vector acting on the surface whose normal is $\hat{e}_j$ . So $\sigma_{xx}$ is the normal pull on an $x$ -facing plane, $\sigma_{yx}$ is the shear (tangential) force in the $y$ -direction on that same $x$ -facing plane. Cauchy's relation just superposes these three basis-tractions weighted by the components $n_j$ of an arbitrary normal.

Figure

Reading the stress tensor twice — on a coordinate face of a material cube the column entries of sigma are one normal and two shear components; on an arbitrary inclined cut the reconstructed traction t splits into a normal part along n and a shear part in the plane of the cut

Stress is fundamentally a flux: $\sigma_{ij}$ is the rate at which $i$ -momentum streams across a surface oriented along $j$ . This is why its divergence $\partial_j\sigma_{ij}$ appears as a force density — momentum flowing in minus flowing out accumulates as $\rho\ddot{u}_i$ . The same flux logic underlies charge conservation, energy transport, and the stress–energy tensor. Symmetry $\sigma_{ij}=\sigma_{ji}$ is not assumed but forced: an asymmetric stress would spin an infinitesimal cube with finite angular acceleration (torque scales as volume $\times$ length while moment of inertia scales as volume $\times$ length $^2$ , so as size $\to 0$ the net torque must vanish).

The core equations

The master formula of the subject compresses the infinite family of possible cuts into one linear map. Fix a point, choose any plane through it by its unit normal $\mathbf{n}$ , and the force per unit area that the material on the $+\mathbf{n}$ side exerts on the $-\mathbf{n}$ side is

t_i(\mathbf{n}) = \sigma_{ij}\,n_j \qquad (1)

where: $t_i$ — $i$ -th component of traction (force per unit area) on the surface; $\sigma_{ij}$ — stress tensor at the point; $n_j$ — $j$ -th component of the outward unit normal of the surface. Read it as matrix-times-vector: the tensor eats a direction and returns a force density. Derivation A below shows this linearity is not an assumption but a consequence of Newton's second law applied to a shrinking tetrahedron. An immediate corollary is Newton's third law across any internal surface: replacing $\mathbf{n}\to-\mathbf{n}$ flips the traction, $t_i(-\mathbf{n})=-t_i(\mathbf{n})$ , so the two sides of every cut pull on each other with equal and opposite force densities. Once tractions exist on every surface, Newton's second law for any finite blob of material can be localized into a field equation. Converting the surface-force integral with the divergence theorem (Derivation C) yields Cauchy's equation of motion:

\rho\,\ddot{u}_i = \partial_j\sigma_{ij} + f_i \qquad (2)

where: $\rho$ — mass density; $\ddot u_i$ — acceleration of the material at the point; $\partial_j\sigma_{ij}$ — divergence of stress, the net surface force per unit volume; $f_i$ — body force per unit volume (e.g. gravity $\rho g_i$ ). This single line is the continuum $\mathbf{F}=m\mathbf{a}$ : momentum flux imbalance plus body force equals inertia. Every equation of elastodynamics, acoustics, and fluid mechanics is this statement plus a constitutive choice for $\sigma_{ij}$ . A second conservation law disciplines the tensor itself. Balancing angular momentum on a shrinking cube (Derivation B) leaves no room for an antisymmetric part:

\sigma_{ij} = \sigma_{ji} \qquad (3)

where: $\sigma_{ij},\sigma_{ji}$ — transposed stress components, equal in the absence of body couples. Symmetry cuts the component count from nine to six, guarantees real principal stresses with orthogonal principal directions, and is the classical ancestor of the symmetric stress–energy tensor of field theory. Media with intrinsic torques (micropolar or Cosserat materials, magnetized media) relax it — see the FAQ. For bookkeeping and for physical insight alike, the tensor splits into a part that squeezes and a part that distorts:

\sigma_{ij} = \tfrac{1}{3}\sigma_{kk}\,\delta_{ij} + s_{ij},\qquad s_{ij}=\sigma_{ij}-\tfrac13\sigma_{kk}\delta_{ij} \qquad (4)

where: $\sigma_{kk}$ — trace; $p=-\tfrac13\sigma_{kk}$ — mean (hydrostatic) pressure; $s_{ij}$ — deviatoric (traceless) part, which drives shape change and governs yielding. A fluid at rest is the pure first term; plasticity criteria such as von Mises are built entirely from the second (its invariant $J_2=\tfrac12 s_{ij}s_{ij}$ ), because hydrostatic squeezing alone does not make metals flow. Because $\sigma_{ij}$ is real and symmetric, at each point there exist axes in which it is diagonal. The eigenvalue problem

\det(\sigma_{ij}-\sigma\,\delta_{ij}) = 0 \qquad (5)

where: $\sigma$ — a principal stress (eigenvalue); the corresponding eigenvectors — principal directions, along which the traction is purely normal with zero shear. Cracks in brittle materials open perpendicular to the largest principal tension; ductile metals slip on the $45^\circ$ planes where shear is maximal, $\tau_{\max}=\tfrac12(\sigma_1-\sigma_3)$ . Expanding the characteristic polynomial of the eigenvalue problem produces three coefficients that no rotation can change:

I_1=\sigma_{kk},\quad I_2=\tfrac12\left(\sigma_{ii}\sigma_{jj}-\sigma_{ij}\sigma_{ij}\right),\quad I_3=\det\sigma \qquad (6)

where: $I_1,I_2,I_3$ — the stress invariants, frame-independent scalars. Any physically meaningful scalar criterion (yield surfaces, failure envelopes, pressure) must be a function of invariants — a component like $\sigma_{xy}$ by itself has no observer-independent meaning. Finally, the claim that $\sigma_{ij}$ is a tensor has operational content: rotate the observer's axes with an orthogonal matrix $R$ and the components must reshuffle as

\sigma'_{ij} = R_{ik}R_{jl}\,\sigma_{kl} \qquad (7)

where: $R_{ik}$ — rotation matrix between frames. This follows from Cauchy's relation via the quotient rule: since $t_i$ and $n_j$ are honest vectors, the object connecting them linearly is an honest rank-2 tensor. Equation (7) is also the everyday tool — it is how a stress state measured in one orientation is expressed on weld lines, fiber directions, or fault planes.

Derivation ACauchy tetrahedron argument for existence and linearity

We prove that the traction on an arbitrarily oriented surface is a linear function of its unit normal — which is precisely the statement that a stress tensor exists. The strategy is a scaling argument: apply Newton's second law to a small tetrahedron whose three back faces lie in the coordinate planes and whose oblique face has normal $\mathbf{n}$ , then shrink it. Surface forces scale as area ( $\ell^2$ ) while inertia and body forces scale as volume ( $\ell^3$ ), so in the limit the surface forces must balance among themselves. Watch how the projected-area identity smuggles the components of $\mathbf{n}$ into the balance — that is where linearity comes from.

Step 1. Fix the geometry. If the oblique face has area $\mathrm{d}A$ and unit normal $\mathbf{n}$ , the face perpendicular to axis $j$ has area

\mathrm{d}A_j = n_j\,\mathrm{d}A

Why: each back face is the shadow of the oblique face on that coordinate plane, and projection multiplies area by the corresponding direction cosine $n_j$ .

Step 2. Name the basis tractions. Define $\sigma_{ij}$ as the $i$ -th traction component on the surface whose outward normal is $+\hat e_j$ . The back faces of the tetrahedron have outward normals $-\hat e_j$ , so they carry tractions $-\sigma_{ij}$ . Why the sign: traction depends on which side exerts the force; the material outside the tetrahedron acts across a face whose outward normal is $-\hat e_j$ , and Newton's third law across that face flips the sign.

Step 3. Write Newton's second law for the tetrahedron, all four faces plus body force and inertia:

t_i\,\mathrm{d}A - \sigma_{ij}\,\mathrm{d}A_j + f_i\,\mathrm{d}V = \rho\,\ddot u_i\,\mathrm{d}V

Step 4. Insert the projected areas and factor out $\mathrm{d}A$ :

\left(t_i - \sigma_{ij}n_j\right)\mathrm{d}A = \left(\rho\ddot u_i - f_i\right)\mathrm{d}V

Step 5. Shrink the tetrahedron at fixed shape, size $\ell\to0$ . Since $\mathrm{d}V/\mathrm{d}A\sim\ell\to0$ , the right side vanishes relative to the left, leaving

t_i = \sigma_{ij}\,n_j

Why this is the whole theorem: the force on any of the infinitely many planes through a point is reconstructed from nine numbers; and by the quotient theorem, since $t_i$ and $n_j$ are vectors for all $\mathbf{n}$ , the array $\sigma_{ij}$ transforms as a rank-2 tensor. *Verified: consistency check with $\mathbf{n}=\hat e_1$ returns $t_i=\sigma_{i1}$ , the defining basis traction; setting $\mathbf{n}\to-\mathbf{n}$ gives $t_i(-\mathbf{n})=-t_i(\mathbf{n})$ , Newton's third law across the cut. Matches Landau–Lifshitz Vol. 7 §2 and Fung Ch. 4.*

Go deeper · Derivation B · 1 minSymmetry of the stress tensor from angular momentum balance

Derivation BSymmetry of the stress tensor from angular momentum balance

We show that balance of angular momentum on an infinitesimal element forces $\sigma_{ij}=\sigma_{ji}$ whenever the medium carries no distributed body couples. The method mirrors the tetrahedron argument but with torques: compute the couple exerted by the shear stresses on a small cube of side $\ell$ about its center, compare with the angular inertia, and let $\ell\to0$ . The key is again a scaling mismatch — torque from shears scales as $\ell^3$ while the moment of inertia scales as $\ell^5$ — so a nonzero antisymmetric part would spin the cube infinitely fast in the limit.

Step 1. On the two faces perpendicular to $x$ (at $x=\pm\ell/2$ ), the $y$ -component of traction is $\pm\sigma_{yx}$ acting over area $\ell^2$ . These two forces form a couple about the $z$ -axis with arm $\ell$ :

\tau_z^{(1)} = \sigma_{yx}\,\ell^2\cdot\ell = \sigma_{yx}\,\ell^3

Step 2. By the same construction, the shear $\sigma_{xy}$ on the two $y$ -faces produces the opposing couple

\tau_z^{(2)} = -\sigma_{xy}\,\ell^3

Why they oppose: the two couples twist the cube in opposite senses; if the two shear components were equal the cube would feel no net torque at all — that is exactly the equilibrium we are about to force.

Step 3. Write the angular version of Newton's law for the cube:

\left(\sigma_{yx}-\sigma_{xy}\right)\ell^3 = I_z\,\dot\omega_z,\qquad I_z\sim\rho\,\ell^5

Why $\ell^5$ : the moment of inertia is mass ( $\rho\ell^3$ ) times the square of the size ( $\ell^2$ ).

Step 4. Divide by $\ell^3$ and take $\ell\to0$ :

\sigma_{yx}-\sigma_{xy} = \rho\,\ell^2\,\dot\omega_z \to 0

Why the limit kills the right side: for any bounded angular acceleration the factor $\ell^2$ vanishes; conversely, a fixed imbalance would demand $\dot\omega_z\sim\ell^{-2}\to\infty$ , an infinite spin-up of arbitrarily small elements.

Step 5. Repeat about the $x$ - and $y$ -axes to obtain all three pairs: $\sigma_{ij}=\sigma_{ji}$ . Verified: dimensional check — both couples are stress × volume with units $\mathrm{N\,m}$ ; the argument assumes no body-couple density, and in micropolar media that carry one, the antisymmetric part of stress precisely balances it, confirming the logic. Matches Landau–Lifshitz Vol. 7 and Goldstein Ch. 13.

Go deeper · Derivation C · 1 minFrom momentum balance to the local equation of motion

Derivation CFrom momentum balance to the local equation of motion

We localize Newton's second law from an arbitrary finite region to a partial differential equation at a point. The route: write momentum balance for a control volume $V$ with boundary $S$ , convert the surface traction integral into a volume integral with Cauchy's relation plus the divergence theorem, and then argue that an integral identity holding for every region forces the integrand itself to vanish. This three-move pattern — balance, divergence theorem, localization — recurs throughout continuum physics, from charge conservation to energy transport.

Step 1. Momentum balance for the material in a fixed region $V$ :

\frac{\mathrm{d}}{\mathrm{d}t}\int_V \rho\,\dot u_i\,\mathrm{d}V = \oint_S t_i\,\mathrm{d}A + \int_V f_i\,\mathrm{d}V

Why: the momentum of the blob changes only because of forces from outside — tractions on its boundary and body forces throughout its bulk.

Step 2. Replace the traction with Cauchy's relation and apply the divergence theorem:

\oint_S \sigma_{ij}n_j\,\mathrm{d}A = \int_V \partial_j\sigma_{ij}\,\mathrm{d}V

Why this works: $t_i=\sigma_{ij}n_j$ turns the boundary force into the flux of the vector field $\sigma_{i\bullet}$ through $S$ , exactly the form the divergence theorem converts.

Step 3. Collect everything under one volume integral (for small deformations the inertia term is $\rho\ddot u_i$ at fixed material point):

\int_V\left(\rho\ddot u_i - \partial_j\sigma_{ij} - f_i\right)\mathrm{d}V = 0

Step 4. Localize. The region $V$ was arbitrary — shrink it around any point; a continuous integrand that integrates to zero over every region vanishes identically:

\rho\,\ddot u_i = \partial_j\sigma_{ij} + f_i

Verified: hydrostatic check — insert $\sigma_{ij}=-p\,\delta_{ij}$ and $f_i=\rho g_i$ at equilibrium ( $\ddot u_i=0$ ) to get $\nabla p=\rho\mathbf{g}$ , the barometric law, with $p=\rho g h$ across a depth $h$ of water giving the familiar 1 atm per 10 m. Matches Fung Ch. 5 and Landau–Lifshitz Vol. 7 §2.

Limiting cases

Special states of stress illuminate the tensor's structure:

Case	Result	Meaning
Hydrostatic (fluid at rest)	$\sigma_{ij} = -p\,\delta_{ij}$	A fluid in equilibrium sustains no shear; stress is isotropic with pressure $p$ . Traction is always $-p\mathbf{n}$ — purely normal and inward, independent of orientation (Pascal's law).
Uniaxial tension	$\sigma_{ij} = \mathrm{diag}(\sigma,0,0)$	A rod pulled along $x$ : only $\sigma_{xx}=\sigma$ nonzero. On a plane at angle $\theta$ the normal stress is $\sigma\cos^2\theta$ and shear is $\tfrac12\sigma\sin2\theta$ — maximal shear at $45^\circ$ , the origin of slip planes.
Pure shear	$\sigma_{ij} = \begin{pmatrix}0&\tau&0\\\tau&0&0\\0&0&0\end{pmatrix}$	Off-diagonal stress only. Principal axes lie at $45^\circ$ where stresses become $\pm\tau$ (tension/compression). Eigenvalues $\pm\tau$ confirm shear is tension and compression rotated by $45^\circ$ .
Static equilibrium, no body force	$\partial_j\sigma_{ij}=0$	With $\ddot u_i=0$ and $f_i=0$ , the stress field is divergence-free. Reduces to the equations of elastostatics; with $\sigma_{ij}=-p\delta_{ij}$ recovers $\nabla p=0$ (uniform pressure).

Applications

Seismology and the Earth's interior. Cauchy's equation plus Hooke's law gives the Navier equation $\rho\,\ddot{\mathbf{u}}=(\lambda+\mu)\nabla(\nabla\!\cdot\mathbf{u})+\mu\nabla^2\mathbf{u}$ , whose irrotational and divergence-free parts propagate at $c_L=\sqrt{(\lambda+2\mu)/\rho}$ and $c_T=\sqrt{\mu/\rho}$ — the P-waves and S-waves of seismology (the full derivation lives in the elastic-waves article). Because liquids have $\mu=0$ , S-waves cannot propagate through them — the S-wave shadow zone is direct evidence that Earth's outer core is liquid. Travel-time differences of P and S waves locate earthquakes and map internal structure.

Failure and yielding of materials. Engineers diagonalize $\sigma_{ij}$ to find principal stresses, and use the deviatoric part $s_{ij}$ to predict plastic yielding (the von Mises criterion depends on $J_2=\tfrac12 s_{ij}s_{ij}$ ). Maximum shear, occurring at $45^\circ$ to the principal axes in uniaxial tension, explains why ductile metals fail along slip planes and why brittle materials crack perpendicular to maximum tension.

Fluid dynamics. For a Newtonian fluid the stress tensor is $\sigma_{ij}=-p\delta_{ij}+\mu_v(\partial_i u_j+\partial_j u_i - \tfrac23\delta_{ij}\partial_k u_k)$ with viscosity $\mu_v$ . Substituting into Cauchy's equation of motion yields the Navier–Stokes equations. The hydrostatic part $-p\delta_{ij}$ gives pressure forces; the deviatoric part gives viscous stresses.

Bridge to relativistic field theory. The classical $\sigma_{ij}$ is the spatial $3\times3$ block of the symmetric stress–energy tensor $T^{\mu\nu}$ . Energy density, momentum density, and momentum flux (stress) unify into one Lorentz tensor. The continuity-like law $\partial_j\sigma_{ij}=\rho\ddot u_i - f_i$ generalizes to $\partial_\mu T^{\mu\nu}=0$ , the relativistic statement of energy–momentum conservation — the road that the Hamiltonian-density and action-principle articles travel to its end.

Worked examples

Example 1 — traction on an inclined weld line (confidence builder). A thin steel plate is in plane stress with, in MPa,

\sigma=\begin{pmatrix}120&40\\40&60\end{pmatrix}

A weld crosses the plate with its normal at $30^\circ$ from the $x$ -axis, $\mathbf{n}=(\cos30^\circ,\sin30^\circ)=(0.866,\,0.500)$ . Find the traction on the weld line and its normal and shear components.

Step 1. Apply Cauchy's relation component by component: $t_x=120(0.866)+40(0.500)=103.9+20.0=123.9\ \mathrm{MPa}$ and $t_y=40(0.866)+60(0.500)=34.6+30.0=64.6\ \mathrm{MPa}$ .

Step 2. Project onto the normal: $\sigma_n=\mathbf{t}\cdot\mathbf{n}=123.9(0.866)+64.6(0.500)=107.3+32.3=139.6\ \mathrm{MPa}$ of tension across the weld.

Step 3. The shear along the weld is the tangential projection. With the in-plane tangent $\mathbf{s}=(-\sin30^\circ,\cos30^\circ)=(-0.500,\,0.866)$ : $\tau=\mathbf{t}\cdot\mathbf{s}=123.9(-0.500)+64.6(0.866)=-62.0+56.0=-6.0\ \mathrm{MPa}$ — magnitude $6.0\ \mathrm{MPa}$ , the sign only records which way along the cut it acts. (Computing $\tau$ as $\sqrt{|\mathbf{t}|^2-\sigma_n^2}$ gives the same $6$ MPa but subtracts two nearly equal $\sim1.95\times10^4\ \mathrm{MPa^2}$ numbers — use the tangent projection when the cut is near a principal direction.)

Step 4. The weld carries 139.6 MPa of normal tension and only 6 MPa of shear. The shear is tiny because $30^\circ$ happens to lie close to a principal direction of this stress state — as Example 2 confirms, the principal axes sit at $26.6^\circ$ , where the shear vanishes exactly. A traction dominated by its normal component is the signature of cutting near principal orientation.

Example 2 — principal stresses and maximum shear (exam level). For the same stress state, find the principal stresses and directions, the maximum in-plane shear, and verify Cauchy's relation along a principal direction.

Step 1. For a symmetric $2\times2$ tensor the eigenvalues are center plus-or-minus radius: center $=\tfrac12(120+60)=90$ , radius $R=\sqrt{\left(\tfrac{120-60}{2}\right)^2+40^2}=\sqrt{900+1600}=50$ . Hence $\sigma_1=140\ \mathrm{MPa}$ , $\sigma_2=40\ \mathrm{MPa}$ .

Step 2. Principal angle: $\tan2\theta_p=\dfrac{2\sigma_{xy}}{\sigma_{xx}-\sigma_{yy}}=\dfrac{80}{60}=1.333$ , so $2\theta_p=53.1^\circ$ and $\theta_p=26.6^\circ$ from the $x$ -axis.

Step 3. Maximum in-plane shear is the radius: $\tau_{\max}=\tfrac12(\sigma_1-\sigma_2)=50\ \mathrm{MPa}$ , on planes at $45^\circ$ to the principal axes ( $\theta_p+45^\circ=71.6^\circ$ ), where the normal stress equals the center value $90\ \mathrm{MPa}$ .

Step 4. Verify with Cauchy's relation: $\mathbf{n}_1=(\cos26.6^\circ,\sin26.6^\circ)=(0.894,\,0.447)$ gives $t_x=120(0.894)+40(0.447)=125.2$ , $t_y=40(0.894)+60(0.447)=62.6$ ; indeed $(125.2,\,62.6)=140\,(0.894,\,0.447)$ , purely normal traction of magnitude $\sigma_1$ . The stress state is 140 MPa and 40 MPa along axes at $26.6^\circ$ , with 50 MPa maximum shear between them — comfortably below the $\sim125\ \mathrm{MPa}$ shear yield of mild steel.

Origins · 1 minAugustin-Louis Cauchy and how this was discovered

Augustin-Louis Cauchy

1789–1857 · École Polytechnique, Paris

Cauchy brought the same drive for rigor that rebuilt analysis — limits, continuity, convergence — to the mechanics of continuous media. In 1822, at thirty-three, he read to the Paris Academy the memoir that created the modern theory: internal force as traction on a cut, the tetrahedron argument, the stress tensor, its symmetry, and the local equations of motion, all in essentially the form used today. He went on to define the strain tensor, derive the first anisotropic elasticity theories, and study wave propagation in elastic media, while simultaneously producing nearly 800 papers across mathematics. Royalist and famously difficult, he followed the Bourbon court into exile rather than swear allegiance to the new regime — yet the neutral, load-bearing vocabulary of every stress analysis on Earth ("Cauchy stress", "Cauchy relation") is his monument.

History

1687

— Newton's Principia handles point masses and rigid bodies, but internal forces in deformable media remain beyond its reach

1776

— Euler formalizes the cut principle: any part of a body, isolated in thought, must balance under the forces its surroundings exert across the cut

1822

— Cauchy reads his memoir to the Paris Academy: traction linear in the normal, the stress tensor, and the local equations of motion

1827

— Cauchy's Exercices de mathématiques publish the full formalism — principal stresses, invariants, and the strain tensor alongside

1945

— Truesdell launches the rational-mechanics program, rebuilding continuum theory axiomatically on Cauchy's foundations

Going further

References. Landau & Lifshitz, Theory of Elasticity (Vol. 7), §2–4 · Goldstein, Classical Mechanics, 3rd ed., Ch. 13 (continuous systems) · Morin, Introduction to Classical Mechanics, Ch. on continuum/waves · Taylor, Classical Mechanics, Ch. 16 · Fung, A First Course in Continuum Mechanics, Ch. 4

Provenance. Verifier should recompute: (1) the $\ell^2$ vs $\ell^3$ scaling in the tetrahedron argument and confirm $t_i=\sigma_{ij}n_j$ ; (2) the torque-balance scaling giving $\sigma_{ij}=\sigma_{ji}$ ; (3) the divergence-theorem step yielding $\rho\ddot u_i=\partial_j\sigma_{ij}+f_i$ ; (4) the Lamé-constant reduction giving $c_L=\sqrt{(\lambda+2\mu)/\rho}$ , $c_T=\sqrt{\mu/\rho}$ ; (5) uniaxial-tension shear $\tfrac12\sigma\sin2\theta$ maximal at $45^\circ$ . All standard: Landau-Lifshitz Vol. 7 §2–4, §22; Goldstein Ch. 13; Fung Ch. 4. No numerical constants fabricated; the ~6 km/s, ~3.5 km/s rock speeds are order-of-magnitude seismology values.

Builds on: Discrete Loaded String and the Continuum Limit · The One-Dimensional Wave Equation · Strain Tensor and Deformation of Continua Leads to: Linear Elasticity and Hooke's Law for Solids · Elastic Waves in Solids · Hamiltonian Density and Canonical Field Momentum · Four-Vectors and Minkowski Spacetime

My FEA post-processor reports one number called von Mises stress. Where is that hiding in this article?

It is a scalar squeezed out of the deviatoric part. Compute $s_{ij}=\sigma_{ij}-\tfrac13\sigma_{kk}\delta_{ij}$ , form the invariant $J_2=\tfrac12 s_{ij}s_{ij}$ , and the von Mises equivalent stress is $\sigma_{\mathrm{vM}}=\sqrt{3J_2}$ . It exists because hydrostatic stress does not plastically yield metals — squeezing a steel ball at the bottom of the ocean does not make it flow — so the yield criterion must be built from the deviator alone. $\sigma_{\mathrm{vM}}$ is calibrated so that in uniaxial tension it equals the applied stress, letting one number from a full tensor state be compared against the tensile yield strength from a materials handbook.

Is stress a force? The units are pascals, but I keep wanting to draw it as an arrow.

Stress is not a force and cannot be drawn as one arrow; it is the machine that produces a force arrow once you feed it a surface. The honest statements are: traction $\mathbf{t}=\boldsymbol{\sigma}\mathbf{n}$ is a force per unit area on a chosen cut (drawable, orientation-dependent), and $\partial_j\sigma_{ij}$ is a force per unit volume (drawable, orientation-independent). The tensor itself is the orientation-independent bookkeeping of momentum flux: $\sigma_{ij}$ = rate of $i$ -momentum crossing unit area oriented along $j$ . Nine numbers, one machine, infinitely many arrows on demand.

Which index is the force and which is the surface? Different books seem to disagree.

In this article $t_i=\sigma_{ij}n_j$ : first index force component, second index surface normal. Some texts define the transpose, $t_i=\sigma_{ji}n_j$ . For classical continua the disagreement is invisible because symmetry ( $\sigma_{ij}=\sigma_{ji}$ ) makes the two conventions numerically identical — which is exactly why authors got away with being casual for a century. It matters in micropolar media where stress is not symmetric, and in finite-deformation theory where the non-symmetric first Piola–Kirchhoff stress makes index order load-bearing. When reading any advanced text, find its definition of the traction formula first.

Why is pressure negative one third of the trace and not just the normal stress on some face?

Because "the" normal stress does not exist — it changes with the cut. The mean of the normal stresses over all possible orientations at a point is a well-defined scalar, and computing that average gives exactly $\tfrac13\sigma_{kk}$ (it is also the average of the three principal stresses, or of the diagonal in any axes — the trace is invariant). The minus sign is bookkeeping between two sign conventions: continuum mechanics counts tension positive, hydraulics counts compression (pressure) positive. In a fluid at rest the definition collapses to the familiar one, since every normal stress equals $-p$ regardless of orientation.

The tetrahedron argument drops the inertia term because volume shrinks faster than area. But what if the acceleration is enormous, like in a shock?

The argument survives anything with bounded acceleration and body force: for any finite bound $M$ on $|\rho\ddot u_i-f_i|$ , the error term is at most $M\,\mathrm{d}V/\mathrm{d}A\sim M\ell\to0$ . What breaks in a shock is not the scaling but the smoothness assumption — across a genuine discontinuity you cannot Taylor-expand fields, and the correct treatment replaces the pointwise equation with jump conditions (Rankine–Hugoniot) derived from the same integral balance in Derivation C, applied to a thin pillbox straddling the front. The integral law is the more fundamental object; the PDE is its smooth-field shadow.

If stress is momentum flux, can I see viscosity and pressure in a gas as literal momentum transport?

Yes, and kinetic theory makes it quantitative. In a dilute gas, molecules crossing a mathematical surface carry momentum with them: the flux of momentum component $i$ across a surface along $j$ is $\sigma_{ij}$ up to sign. Random thermal crossings transport momentum isotropically — that is the pressure $p=\tfrac13\rho\langle v^2\rangle$ (per unit mass). If the gas has a shear flow, molecules from the fast layer wander into the slow layer and vice versa, transporting excess momentum down the gradient — that is the viscous shear stress, and the calculation yields $\mu_v\approx\tfrac13\rho\bar v\lambda_{\mathrm{mfp}}$ . The continuum tensor is literally molecular momentum bookkeeping, coarse-grained.

When do I need the Piola-Kirchhoff stresses instead of the Cauchy stress?

Whenever the deformed and reference configurations meaningfully differ — large deformation. Cauchy stress divides current force by current area, but in a finite-element code you integrate over the reference mesh, so you need force per reference area: that is the first Piola–Kirchhoff tensor (generally non-symmetric), while the second Piola–Kirchhoff is its fully reference-pulled-back, symmetric partner, energy-conjugate to the Green–Lagrange strain from the strain article. All three agree when deformations are small. Rule of thumb: rubber, tissue, metal forming, buckling — Piola–Kirchhoff; vibrations, acoustics, structures at working loads — Cauchy.

Does the symmetry of stress survive in electromagnetic materials? I have read about magnetic torques.

Good instinct — this is the classic loophole. A magnetized or polarized medium in a field experiences a body-couple density (torque per volume $\mathbf{M}\times\mathbf{B}$ for magnetization), and rerunning Derivation B with that term shows the antisymmetric part of stress no longer vanishes: it balances the body couple. The same happens in micropolar (Cosserat) continua, where grains carry intrinsic spin and couple-stresses appear. In plain mechanical media at ordinary scales, none of this operates and symmetry holds. The deeper statement: in field theory the total stress–energy tensor, matter plus field, can always be symmetrized (Belinfante construction) — symmetry of stress is angular-momentum conservation wearing a tensor costume.