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FALSE: the reversed-normal convention alone makes a continuous traction field a stress matrix. The field t(n)=|n_1|n refutes it

Statement refuted

The false statement is: every continuous traction field $t : S^{2} \to R^{3}$ satisfying the reversed-normal convention $t (- n) = - t (n)$ for every $n \in S^{2}$ is represented by a matrix $σ \in M_{3} (R)$ with $t (n) = σ n$ for every $n$ .

The field constructed below is continuous and satisfies the convention, and inherits Cauchy's hypothesis (it depends on the surface only through the unit normal), but no matrix represents it. The refutation shows that the balance hypothesis of the Cauchy-tetrahedron theorem is essential: symmetry of the traction in the normal alone is not enough.

Facts & Assumptions

Given: the unit sphere $S^{2} = {n \in R^{3} : ∥ n ∥_{2} = 1}$ , the first standard basis vector $e_{1}$ , and the map

$t : S^{2} ⟶ R^{3}, t (n) : = ∥ ⟨ n, e_{1} ⟩_{2} e_{1} ∥_{2} n .$

Write $n_{1} : = ⟨ n, e_{1} ⟩_{2}$ , so $t (n) = ∣ n_{1} ∣ n$ with $∣ n_{1} ∣ = ∥ n_{1} e_{1} ∥_{2}$ .

[F1]

Traction and representation (Traction fields and stress matrices representing them, The vector space $M_{m \times n} (F) : = F^{m \times n}$ of $m$ by $n$ matrices over a field, with entrywise operations, Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes): a matrix $σ$ represents $t$ when $t (n) = σ n$ for every $n \in S^{2}$ ; the columns of $σ$ are then determined by $σ e_{j} = t (e_{j})$ ; and $t (- n) = - t (n)$ is the reversed-normal convention, which is a property and not a definition.

[F2]

Continuity of the field (Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point, Algebra of limits: sums, scalar multiples, products and quotients, The finite and reverse triangle inequalities for a norm; and for $n \geq 1$ every norm $N$ on $R^{n}$ satisfies $N (x) \leq C ∥ x ∥_{1}$ and is Lipschitz, hence continuous, for $d_{2}$ , The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ): $n \mapsto ⟨ n, e_{1} ⟩_{2}$ is linear, hence continuous; the Euclidean norm is continuous on $R^{3}$ ; the product of a continuous scalar and the continuous map $n \mapsto n$ is continuous; hence $t$ is continuous on $S^{2}$ .

[F3]

Sphere and basis (Euclidean spheres and closed balls as subspaces of $R^{n}$ , The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ): $e_{1}, e_{2}, e_{3}$ and $n_{*} : = (e_{1} + e_{2} + e_{3}) / \sqrt{3}$ all belong to $S^{2}$ , with $⟨ e_{j}, e_{1} ⟩_{2} = δ_{j 1}$ and $⟨ n_{*}, e_{1} ⟩_{2} = 1 / \sqrt{3}$ ; and $∥ - v ∥_{2} = ∥ v ∥_{2}$ for every $v \in R^{3}$ by symmetry of the inner product.

Counterexample

technique · direct; verify continuity and the reversed-normal convention, then assume a representing matrix and derive a contradiction by evaluating at the diagonal direction

1.1givenF2F3algebra

Continuity and reversed normals: by [F2] the field $t$ is continuous on $S^{2}$ . For $n \in S^{2}$ , [F3] gives $∥ ⟨ - n, e_{1} ⟩_{2} e_{1} ∥_{2} = ∥ - n_{1} e_{1} ∥_{2} = ∥ n_{1} e_{1} ∥_{2}$ , so $t (- n) = ∣ n_{1} ∣ (- n) = - ∣ n_{1} ∣ n = - t (n)$ : the reversed-normal convention holds.

1.2givenF1F3algebra

Assume a representing matrix: suppose $σ \in M_{3} (R)$ satisfies $t (n) = σ n$ for all $n \in S^{2}$ . Evaluating at the basis vectors, [F1] and [F3] give $σ e_{1} = t (e_{1}) = ∣ 1 ∣ e_{1} = e_{1}$ , $σ e_{2} = t (e_{2}) = ∣ 0 ∣ e_{2} = 0$ and $σ e_{3} = 0$ ; hence $σ = d i a g (1, 0, 0)$ .

2.1step 1.2F1algebra∎

Contradiction: at $n_{*} = (e_{1} + e_{2} + e_{3}) / \sqrt{3} \in S^{2}$ one has $t (n_{*}) = ∣ 1 / \sqrt{3} ∣ n_{*} = n_{*} / \sqrt{3} = (1 / 3) (1, 1, 1)$ , while $σ n_{*} = (1 / \sqrt{3}, 0, 0)$ . These vectors differ (their second and third components are $1 / 3$ and $0$ respectively), contradicting the assumed representation for every $n$ . Hence no matrix represents $t$ , although $t$ is continuous, satisfies Cauchy's hypothesis in the sense that it depends on the surface only through its unit normal, and satisfies the reversed-normal convention.

Remarks

The counterexample isolates the missing hypothesis: the Cauchy-tetrahedron theorem derives the matrix from the balance of contact forces on shrinking regions, not from the symmetry of the traction in the normal. The field above is not balanced on all admissible regions; by the cited theorem, the balance condition must fail for it, and continuity plus the reversed-normal convention are insufficient on their own.
The physical reading is the corresponding warning for the model page: the representation of contact forces by a stress matrix is a consequence of the adopted balance principles, not of the format of the traction field, and a model whose contact forces violate the balance hypotheses lies outside the Cauchy continuum framework.

FALSE: the reversed-normal convention alone makes a continuous traction field a stress matrix. The field t(n)=|n_1|n refutes it

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Depends on

Dependency tree · 2 levels

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