Existence of the Cauchy Stress Tensor

Derivation

Existence of the Cauchy Stress Tensor

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Statement

In a continuum obeying Newton's second law, the surface traction \(\mathbf{t}(\mathbf{n})\) exerted across an internal surface element depends on the outward unit normal \(\mathbf{n}\) at that point, and this dependence is linear: there exists a unique second-rank tensor field \(\boldsymbol{\sigma}(\mathbf{x},t)\), the Cauchy stress tensor, such that \(\mathbf{t}(\mathbf{n}) = \boldsymbol{\sigma}^{\mathsf{T}}\mathbf{n}\), equivalently \(t_i = \sigma_{ji}\,n_j\). In particular the traction across the oppositely oriented face satisfies \(\mathbf{t}(-\mathbf{n}) = -\mathbf{t}(\mathbf{n})\).

Why it matters

Contact forces in a deforming body are distributed over surfaces, not concentrated at points, so the primitive object is a traction \(\mathbf{t}\) with units of force per area. A priori \(\mathbf{t}\) could depend on the whole geometry of the cut surface — its curvature, its higher derivatives, anything. Cauchy's theorem collapses all of that to a single linear dependence on the local normal direction, which is the minimal structure needed to write a local momentum balance.

Without this result there is no stress tensor, no divergence form of the momentum equation, and hence no Navier–Stokes equations, no linear elasticity, and no finite-element continuum mechanics. It is the theorem that turns "forces act on surfaces" into a field theory.

Assumptions

The traction depends on position, time, and the surface only through its normal:if \(\mathbf{t}\) depended on curvature or on the surface's second fundamental form, the tetrahedron limit below would not close and no rank-2 tensor would capture contact forces — this is the substantive physical postulate (Cauchy's postulate).

The traction field \(\mathbf{t}(\mathbf{x},\mathbf{n})\) is continuous in \(\mathbf{x}\):if \(\mathbf{t}\) jumped inside the shrinking tetrahedron, the face-average tractions would not converge to their central values and the linear relation would fail to be pointwise.

Body force density \(\mathbf{b}\) (per unit mass) and mass density \(\rho\) are bounded near the point:if the volume force blew up like \(1/\ell\) as the tetrahedron shrank, the volume term would not be subdominant to the surface term and the balance would retain a spurious body contribution.

Acceleration \(\mathbf{a}=\mathrm{D}\mathbf{v}/\mathrm{D}t\) is bounded (no shocks through the point):if inertia diverged, the \(\rho\,\mathbf{a}\) term would compete with the surface term at the same order and the tetrahedron argument would not isolate the tractions.

Derivation

\[ \int_{\partial V}\mathbf{t}\,\mathrm{d}A + \int_{V}\rho\,\mathbf{b}\,\mathrm{d}V = \int_{V}\rho\,\mathbf{a}\,\mathrm{d}V \]

Newton's second law applied to an arbitrary material region \(V\): the resultant of surface tractions plus body forces equals the rate of change of momentum, written as mass times acceleration per unit volume. A

\[ V=\text{Cauchy tetrahedron: three faces on the coordinate planes through }\mathbf{x},\ \text{one slanted face with unit normal }\mathbf{n} \]

Choose the region to be a small tetrahedron whose slanted face has outward normal \(\mathbf{n}=(n_1,n_2,n_3)\) and whose other three faces are perpendicular to the axes \(\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3\). This geometry couples one arbitrary orientation to the three coordinate orientations. A

\[ A_j = A\,(\mathbf{n}\cdot\mathbf{e}_j) = A\,n_j,\qquad j=1,2,3 \]

Projected-area identity: the area \(A_j\) of the coordinate face with outward normal \(-\mathbf{e}_j\) is the projection of the slanted face of area \(A\) onto that plane, giving \(A_j=A\,n_j\). This is pure geometry of the tetrahedron. A

\[ \mathbf{t}(\mathbf{n})\,A + \sum_{j=1}^{3}\mathbf{t}(-\mathbf{e}_j)\,A_j + \rho\,\mathbf{b}\,V = \rho\,\mathbf{a}\,V \]

Approximate each surface integral by traction at the point times face area, and each volume integral by density times value times volume \(V\). Legal to leading order because \(\mathbf{t}\), \(\mathbf{b}\), \(\mathbf{a}\), \(\rho\) are continuous (mean-value theorem), with corrections of higher order in the size. B

\[ \mathbf{t}(\mathbf{n})\,A + \sum_{j=1}^{3}\mathbf{t}(-\mathbf{e}_j)\,A\,n_j + \big(\rho\,\mathbf{b}-\rho\,\mathbf{a}\big)\,V = \mathbf{0} \]

Substitute \(A_j=A\,n_j\) from Step 3 and collect the volume terms. Symbols only, no numbers yet; this is the exact leading-order balance for the tetrahedron. B

\[ \text{Let the tetrahedron shrink self-similarly by a length }\ell:\quad A\propto \ell^{2},\qquad V\propto \ell^{3} \]

Scale the geometry down keeping \(\mathbf{n}\) fixed. Areas scale as \(\ell^2\), volume as \(\ell^3\); this is the key scale separation that lets the volume terms be dropped. C

\[ \frac{V}{A}=O(\ell)\ \xrightarrow{\ \ell\to0\ }\ 0 \quad\Longrightarrow\quad \frac{(\rho\mathbf{b}-\rho\mathbf{a})V}{A}\to\mathbf{0} \]

Divide the Step 5 balance by \(A\) and take \(\ell\to0\). Because \(\rho\), \(\mathbf{b}\), \(\mathbf{a}\) are bounded (assumptions) while \(V/A=O(\ell)\), the volume term vanishes relative to the surface terms. C

\[ \mathbf{t}(\mathbf{n}) = -\sum_{j=1}^{3}\mathbf{t}(-\mathbf{e}_j)\,n_j \]

What survives the limit: the slanted-face traction equals minus the sum of coordinate-face tractions weighted by the components of \(\mathbf{n}\). The dependence on \(\mathbf{n}\) is now manifestly through \(n_1,n_2,n_3\) alone. B

\[ \text{Take }\mathbf{n}=\mathbf{e}_k:\quad \mathbf{t}(\mathbf{e}_k) = -\,\mathbf{t}(-\mathbf{e}_k)\quad\Longrightarrow\quad \mathbf{t}(-\mathbf{e}_k) = -\,\mathbf{t}(\mathbf{e}_k) \]

Specialize Step 8 to a coordinate direction. This recovers Newton's third law for tractions, \(\mathbf{t}(-\mathbf{n})=-\mathbf{t}(\mathbf{n})\), as a corollary rather than a separate postulate. B

\[ \mathbf{t}(\mathbf{n}) = \sum_{j=1}^{3}\mathbf{t}(\mathbf{e}_j)\,n_j \]

Insert \(\mathbf{t}(-\mathbf{e}_j)=-\mathbf{t}(\mathbf{e}_j)\) into Step 8. The traction is a linear combination of three fixed vectors — the tractions on the coordinate faces — with coefficients \(n_j\). Linearity in \(\mathbf{n}\) is now proved. B

\[ \sigma_{ji} := \big[\mathbf{t}(\mathbf{e}_j)\big]_i \quad\Longrightarrow\quad t_i(\mathbf{n}) = \sigma_{ji}\,n_j \]

Define the nine numbers \(\sigma_{ji}\) as the \(i\)-th component of the traction on the face with normal \(\mathbf{e}_j\). Because Step 10 holds for every \(\mathbf{n}\) and the \(n_j\) are arbitrary, the array \(\sigma_{ji}\) transforms as a rank-2 tensor (it maps the vector \(\mathbf{n}\) linearly to the vector \(\mathbf{t}\)). C

\[ t_i = \sigma_{ji}\,n_j \iff \mathbf{t}=\boldsymbol{\sigma}^{\mathsf{T}}\mathbf{n};\qquad \text{under rotation }Q:\ \sigma'_{kl}=Q_{ki}Q_{lj}\,\sigma_{ij} \]

Quotient rule: since \(\mathbf{t}\) and \(\mathbf{n}\) are vectors and the relation is linear and coordinate-independent, \(\sigma_{ij}\) must obey the rank-2 tensor transformation law. Existence and tensor character are established. C

Result

\[ \boxed{\,\mathbf{t}(\mathbf{x},\mathbf{n}) = \boldsymbol{\sigma}(\mathbf{x})^{\mathsf{T}}\,\mathbf{n},\qquad t_i = \sigma_{ji}\,n_j\,} \]

Reading. The force per unit area transmitted across any internal surface is obtained by contracting a single position-dependent object, the stress tensor \(\boldsymbol{\sigma}\), with the surface's outward unit normal. All the orientation dependence of contact forces at a point lives in these nine numbers \(\sigma_{ij}\): the diagonal entries are normal stresses (tension/compression), the off-diagonal entries are shear stresses. Nothing about the surface beyond its normal matters.

Units check. \(\mathbf{t}\) has units of force per area, \(\mathrm{N\,m^{-2}=Pa}\). \(\mathbf{n}\) is dimensionless. Hence \(\sigma_{ij}\) carries units of \(\mathrm{Pa}\), as required; each term \(\sigma_{ji}n_j\) is \(\mathrm{Pa}\times 1 = \mathrm{Pa}\), matching \(t_i\). Dimensionally consistent.

Limiting cases

Ideal fluid at rest: \(\sigma_{ij}=-p\,\delta_{ij}\), so \(\mathbf{t}=-p\,\mathbf{n}\) — traction is purely normal (pressure), independent of which way the surface is tilted in magnitude.
Uniaxial tension along \(\mathbf{e}_1\): \(\sigma_{11}=\sigma_0\), all others zero, giving \(\mathbf{t}=\sigma_0 n_1\,\mathbf{e}_1\); a face normal to \(\mathbf{e}_2\) carries no traction.
Coordinate face: setting \(\mathbf{n}=\mathbf{e}_1\) returns \(\mathbf{t}=(\sigma_{11},\sigma_{12},\sigma_{13})\), the definition of the first column — a self-consistency check.
Reversed cut: \(\mathbf{t}(-\mathbf{n})=-\mathbf{t}(\mathbf{n})\) recovers action–reaction across the surface, here derived rather than assumed.

Breaks when

Couple stresses / micropolar media: if the material transmits distributed torques (surface couples), angular momentum balance is not satisfied by tractions alone, the simple traction–normal object is incomplete, and one needs a couple-stress tensor in addition — the Cosserat continuum. The rank-2 \(\boldsymbol{\sigma}\) then need not even be symmetric.
Surface tension / interfaces: at a curved interface the traction acquires a curvature-dependent (Young–Laplace) part \(\propto \gamma\,\kappa\), violating Cauchy's postulate that \(\mathbf{t}\) depends on the surface only through \(\mathbf{n}\); the tetrahedron limit no longer drops the "surface energy" term.
Shock fronts / singular acceleration: if \(\rho\mathbf{a}\) is not bounded through the point (a discontinuity), the volume term does not vanish faster than the surface term as \(\ell\to0\) and Step 7 fails; the balance must be treated as a jump condition instead.
Strongly nonlocal / peridynamic materials: when forces act between finitely separated material points (long-range bonds), contact force is not a surface flux at all and no local stress tensor exists in the classical sense.

Failure modes

Index transposition: writing \(t_i=\sigma_{ij}n_j\) and calling it "the" convention without noticing it equals \(\sigma_{ji}n_j\) only when \(\boldsymbol{\sigma}\) is symmetric; before symmetry is proved (from angular-momentum balance, a separate result) the free index must sit on the first slot: \(t_i=\sigma_{ji}n_j\).
Assuming linearity instead of deriving it: stating \(\mathbf{t}=\boldsymbol{\sigma}\mathbf{n}\) as a definition. The content of the theorem is that linearity follows from momentum balance plus the projected-area geometry; it is not free.
Keeping the body-force term: forgetting that \(V/A\to0\) and leaving a \(\rho\mathbf{b}\) contribution in the final traction relation. Gravity does not appear in \(\mathbf{t}(\mathbf{n})\); it appears one order later, in the divergence equation.
Confusing normal projection: using \(A_j=A/n_j\) or \(A_j=A\,n_j^2\) instead of the correct \(A_j=A\,n_j\), which breaks the linear scaling.
Treating \(\boldsymbol{\sigma}\) as a vector of three tractions: failing to see that the nine components transform under rotation with two indices; a "list of column vectors" does not rotate correctly.

Discussion

The tetrahedron argument is a prototype of a recurring move in field theory: a balance law over an arbitrary region, combined with a shrinking-region limit, promotes an integral statement to a pointwise algebraic one. Here the arbitrary region is the tetrahedron, the balance law is Newton's second law, and the pay-off is that contact force at a point is a linear map \(\mathbf{n}\mapsto\mathbf{t}\). The same logic, applied one level up with the divergence theorem, converts the surface integral \(\oint \boldsymbol{\sigma}^{\mathsf{T}}\mathbf{n}\,\mathrm{d}A\) into \(\int \nabla\!\cdot\!\boldsymbol{\sigma}\,\mathrm{d}V\) and yields Cauchy's equation of motion \(\nabla\!\cdot\!\boldsymbol{\sigma}+\rho\mathbf{b}=\rho\mathbf{a}\).

Physically, the stress tensor is the momentum-flux density: \(\sigma_{ji}\) is the flux of \(i\)-momentum in the \(j\)-direction (up to sign convention). This viewpoint unifies the elastic stress of a solid, the pressure-plus-viscous stress of a fluid, and the Maxwell stress of the electromagnetic field — all are rank-2 objects whose divergence supplies a force density. The existence proof here is what licenses that entire family of "momentum lives in a symmetric flux" statements in continuum physics.

Note what the theorem does not give. It establishes the existence and tensor character of \(\boldsymbol{\sigma}\), but not its symmetry \(\sigma_{ij}=\sigma_{ji}\); that requires a separate balance of angular momentum (and can fail for polar media). Nor does it give a constitutive law — the relation between \(\boldsymbol{\sigma}\) and deformation (Hooke, Newtonian viscosity) is additional physics. Cauchy's theorem is purely kinematic-dynamical bookkeeping: given that momentum balances, contact forces must organize into a tensor.

The subtle assumption is Cauchy's postulate that \(\mathbf{t}\) depends on the cut only through \(\mathbf{n}\). Modern treatments (Noll, Gurtin) weaken this: assuming only that the resultant contact force on a region is absolutely continuous with respect to surface area and that \(\mathbf{t}(\mathbf{x},\mathbf{n})\) is measurable and integrable, one can prove continuity in \(\mathbf{n}\) and hence recover linearity, so Cauchy's postulate is not an independent hypothesis but a consequence of the balance laws plus mild regularity. The tetrahedron argument is thus the physicist's shortcut to a theorem that also admits a fully rigorous, hypothesis-lean proof.

Common misconceptions. (i) That the stress tensor is a property of the material — it is a field describing the current dynamical state, defined even in vacuum-adjacent field theories. (ii) That "stress" means "force" — stress is force per area, a flux. (iii) That symmetry of \(\boldsymbol{\sigma}\) is part of this theorem — it is a distinct result from angular momentum.

Worked examples

Given \(\displaystyle \boldsymbol{\sigma}=\begin{pmatrix}50 & 30 & 0\\ 30 & -20 & 0\\ 0 & 0 & 10\end{pmatrix}\ \mathrm{MPa}\), find the traction on the plane with normal \(\mathbf{n}=\tfrac{1}{\sqrt{2}}(1,1,0)\).

Set up: symmetric tensor so \(t_i=\sigma_{ij}n_j\); state units \(\mathrm{MPa}\). A

\[ t_i=\sigma_{ij}n_j \quad\text{(symbols first)} \]

Write the contraction before inserting numbers. A

\[ t_1=\frac{1}{\sqrt2}(50+30)=\frac{80}{\sqrt2}=56.6,\quad t_2=\frac{1}{\sqrt2}(30-20)=\frac{10}{\sqrt2}=7.07,\quad t_3=0 \]

Insert \(n_1=n_2=1/\sqrt2,\ n_3=0\); arithmetic in MPa. B

\[ t_n=\mathbf{t}\cdot\mathbf{n}=\tfrac{1}{\sqrt2}(56.6)+\tfrac{1}{\sqrt2}(7.07)=45.0,\qquad |\mathbf{t}|=\sqrt{56.6^2+7.07^2}=57.0 \]

Decompose into normal component \(t_n\) and magnitude to get shear \(\tau=\sqrt{|\mathbf{t}|^2-t_n^2}\). B

\[ \mathbf{t}=(56.6,\ 7.07,\ 0)\ \mathrm{MPa},\quad t_n=45.0\ \mathrm{MPa},\quad \tau=\sqrt{57.0^2-45.0^2}=34.9\ \mathrm{MPa} \]

Reading. The plane carries 45 MPa of normal (tensile) stress and 34.9 MPa of shear; the traction is not parallel to \(\mathbf{n}\) because \(\boldsymbol{\sigma}\) is not isotropic.

Units check. \(\mathrm{MPa}\times\) (dimensionless \(n\)) \(=\mathrm{MPa}\) throughout. Consistent.

A tetrahedron at a point has slanted-face area \(A=2\ \mathrm{mm^2}\) with \(\mathbf{n}=(0.6,0.8,0)\). Verify the coordinate-face areas and the traction balance for uniaxial stress \(\sigma_{11}=100\ \mathrm{MPa}\) (all other components zero).

Set up: use \(A_j=A\,n_j\) and \(\mathbf{t}(\mathbf{n})=\boldsymbol{\sigma}^{\mathsf T}\mathbf{n}\); state units. A

\[ A_1=A n_1,\quad A_2=A n_2,\quad A_3=A n_3 \quad\text{(symbols first)} \]

Projected-area identity from Step 3 of the derivation. A

\[ A_1=2(0.6)=1.2\ \mathrm{mm^2},\quad A_2=2(0.8)=1.6\ \mathrm{mm^2},\quad A_3=0 \]

Insert numbers; areas positive and \(\le A\). B

\[ \mathbf{t}(\mathbf{n})=\boldsymbol{\sigma}^{\mathsf T}\mathbf{n}=(\sigma_{11}n_1,\,0,\,0)=(100\times0.6,0,0)=(60,0,0)\ \mathrm{MPa} \]

Only \(\sigma_{11}\) nonzero, so traction points along \(\mathbf{e}_1\). B

\[ \text{Force on slant face } \mathbf{F}=\mathbf{t}A=(60\,\mathrm{MPa})(2\,\mathrm{mm^2})\,\mathbf{e}_1 = 120\ \mathrm{N}\,\mathbf{e}_1 \]

\(1\ \mathrm{MPa}\cdot\mathrm{mm^2}=1\ \mathrm{N}\); compare with the \(\mathbf{e}_1\)-face force \(\mathbf{t}(-\mathbf{e}_1)A_1=-(100)(1.2)\mathbf{e}_1=-120\,\mathrm{N}\,\mathbf{e}_1\). C

\[ \mathbf{F}_{\text{slant}}+\mathbf{F}_{\text{coord}}=120\,\mathbf{e}_1-120\,\mathbf{e}_1=\mathbf{0}\ \mathrm{N} \]

Reading. With no body force and no acceleration, the tetrahedron's tractions balance exactly, confirming \(\mathbf{t}(\mathbf{n})=\sum_j\mathbf{t}(\mathbf{e}_j)n_j\) for this state.

Units check. \(\mathrm{MPa}\cdot\mathrm{mm^2}=\mathrm{N}\); forces cancel in \(\mathrm{N}\). Consistent.

Problems

(A) State the projected-area relation for a Cauchy tetrahedron and use it to explain, in one sentence, why the volume term drops out in the shrinking limit.
Solution
The coordinate-face areas are \(A_j=A\,n_j\), so every surface term scales as the slant area \(A\propto\ell^2\), while the volume terms scale as \(V\propto\ell^3\). Dividing the balance by \(A\) leaves the volume contribution proportional to \(V/A=O(\ell)\), which \(\to0\) as \(\ell\to0\); hence body force and inertia vanish relative to tractions, leaving \(\mathbf{t}(\mathbf{n})=\sum_j\mathbf{t}(\mathbf{e}_j)n_j\).
(A) For \(\boldsymbol{\sigma}=\mathrm{diag}(80,40,20)\ \mathrm{MPa}\), find the traction and its magnitude on the plane \(\mathbf{n}=(0,0,1)\).
Solution
\(t_i=\sigma_{ij}n_j\) with \(n=(0,0,1)\) picks the third column: \(\mathbf{t}=(\sigma_{13},\sigma_{23},\sigma_{33})=(0,0,20)\ \mathrm{MPa}\). Magnitude \(|\mathbf{t}|=20\ \mathrm{MPa}\), purely normal (no shear), as expected on a principal plane of a diagonal tensor.
(B) With \(\displaystyle\boldsymbol{\sigma}=\begin{pmatrix}0&40&0\\40&0&0\\0&0&0\end{pmatrix}\mathrm{MPa}\) (pure shear), find the traction on \(\mathbf{n}=\tfrac1{\sqrt2}(1,1,0)\) and resolve into normal and shear parts.
Solution
\(t_i=\sigma_{ij}n_j\): \(t_1=40 n_2=40/\sqrt2=28.3\), \(t_2=40 n_1=28.3\), \(t_3=0\) (MPa). Normal component \(t_n=\mathbf{t}\cdot\mathbf{n}=\tfrac1{\sqrt2}(28.3)+\tfrac1{\sqrt2}(28.3)=40\ \mathrm{MPa}\). Magnitude \(|\mathbf t|=\sqrt{28.3^2+28.3^2}=40\ \mathrm{MPa}\), so shear \(\tau=\sqrt{40^2-40^2}=0\). The \(45^\circ\) plane carries pure normal stress of 40 MPa — this is the principal direction of pure shear.
(B) A tetrahedron has \(A=5\ \mathrm{mm^2}\), \(\mathbf{n}=(2,1,2)/3\). Compute the three coordinate-face areas and confirm \(\sum_j A_j^2 \ne A^2\) in general while \(A_j=A n_j\).
Solution
\(n=(0.667,0.333,0.667)\). \(A_1=5(0.667)=3.33\), \(A_2=5(0.333)=1.67\), \(A_3=5(0.667)=3.33\ \mathrm{mm^2}\). Check \(|\mathbf n|^2=0.667^2+0.333^2+0.667^2=1\), good. \(\sum A_j = 8.33\ \mathrm{mm^2}\); \(\sum A_j^2=11.1+2.78+11.1=25.0=A^2\) here because \(\sum n_j^2=1\) — so in fact \(\sum A_j^2=A^2\) always (the areas obey a Pythagorean law), while \(\sum A_j\ne A\). This is the correct projected-area geometry.
(C) Show from the tetrahedron balance that \(\mathbf{t}(-\mathbf{n})=-\mathbf{t}(\mathbf{n})\) without assuming Newton's third law separately, and state one physical situation where this relation fails.
Solution
Take a thin "pillbox" or set \(\mathbf n=\mathbf e_k\) in the surviving relation \(\mathbf t(\mathbf n)=-\sum_j \mathbf t(-\mathbf e_j)n_j\): with \(\mathbf n=\mathbf e_k\) it gives \(\mathbf t(\mathbf e_k)=-\mathbf t(-\mathbf e_k)\), i.e. \(\mathbf t(-\mathbf e_k)=-\mathbf t(\mathbf e_k)\). Since any \(\mathbf n\) is expanded in this basis and the map is linear, \(\mathbf t(-\mathbf n)=-\mathbf t(\mathbf n)\) follows. It fails at a curved fluid interface with surface tension \(\gamma\): the Young–Laplace jump \(\Delta p=\gamma(\kappa_1+\kappa_2)\) means the two sides' tractions differ by a curvature term, so contact force is no longer a pure function of \(\mathbf n\) and the antisymmetry across the surface is broken.