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Cauchy stress, local momentum and stress symmetry

Physical scope

The smooth nonpolar postulate adopts Cauchy traction; its mass, linear- and angular-momentum balances yield local momentum and symmetric stress.

Statement

Let $I \subseteq R$ be an open interval containing $0$ , let $B \subseteq R^{3}$ be a nonempty bounded open reference region with $C^{1}$ boundary (locally a $C^{1}$ graph with the region on one side), and let $U \subseteq R^{3}$ be an open neighborhood of $\overline{B}$ . Let $χ \in C^{2} (I \times U; R^{3})$ have orientation-preserving diffeomorphic time slices (bijections onto open images with $C^{2}$ inverses and positive Jacobian determinant) and satisfy $χ (0, a) = a$ . Here $C^{r}$ means all coordinate partial derivatives through order $r$ are continuous. Set $Ω_{t} = χ_{t} (B)$ and $O = {(t, x) : t \in I, x \in χ_{t} (U)}$ . Use SI units in a classical Cartesian inertial frame with absolute time; inertial changes have form $t^{*} = t + t_{0}$ , $x^{*} = Q x + c t + d$ , where $S O (3) = {Q \in R^{3 \times 3} : Q^{T} Q = I_{3}, \det Q = 1}$ , $c$ is a constant relative velocity in metres per second, and $d$ is a constant spatial translation in metres. Let $ρ_{0} \in C^{1} (\overline{B})$ and $ρ \in C^{1} (O)$ be positive mass densities in kilograms per cubic metre, let $u \in C^{1} (O; R^{3})$ be the velocity induced by this motion, so $u (t, χ (t, a)) = \partial_{t} χ (t, a)$ for every $(t, a) \in I \times U$ , measured in metres per second, and let $b \in C^{0} (O; R^{3})$ be body acceleration in metres per second squared. Let $σ \in C^{1} (O; R^{3 \times 3})$ be Cauchy stress in pascals, with traction $t (t, x, n) = σ (t, x) n$ for every outward Euclidean unit normal $n$ . For every nonempty bounded open reference region $V$ whose closure is compact in $B$ and whose $C^{1}$ boundary is locally a graph with the region on one side, put $P_{t} = χ_{t} (V)$ . Adopt on each $P_{t}$ the material mass, linear-momentum and angular-momentum balances:

$\int_{P_{t}} ρ (t, x) d x = \int_{V} ρ_{0} (a) d a, \frac{d}{d t} \int_{P_{t}} ρ u d x = \int_{P_{t}} ρ b d x + \int_{\partial P_{t}} t (t, x, n) d S,$

$\frac{d}{d t} \int_{P_{t}} x \times ρ u d x = \int_{P_{t}} x \times ρ b d x + \int_{\partial P_{t}} x \times t (t, x, n) d S .$

Here $d S$ is Euclidean surface measure, with chart density $\sqrt{\det (D Y^{T} D Y)}$ . The angular momentum equation is about a fixed inertial origin; the model has no surface couples or intrinsic spin.

Assume the Axiom of Choice. The adopted traction representation is pointwise unique. The local momentum and stress-symmetry conclusions below hold on the balanced physical region $O_{B} : = {(t, x) : t \in I, x \in Ω_{t}}$ ; the larger $O$ supplies ambient regularity only. With $D_{t} u : = \partial_{t} u + (D_{x} u) u$ and rowwise divergence $(div σ)_{i} : = \sum_{j = 1}^{3} \partial_{x_{j}} σ_{i j}$ , the local momentum equation is

$ρ D_{t} u = ρ b + div σ,$

equivalently

$\partial_{t} (ρ u) + div (ρ u \otimes u - σ) = ρ b,$

where $(u \otimes u)_{i j} = u_{i} u_{j}$ . The angular momentum balance with no couples implies $σ = σ^{T}$ . Conversely, in this same smooth positive-density setting, the local momentum equation with symmetric $σ$ implies the angular momentum balance. The model postulate adopts the traction representation; this theorem establishes its local balance consequences and uniqueness.

Statement references

Tong, Fluid Mechanics, §3.1.2

Powers, Lecture Notes on Intermediate Fluid Mechanics, §§4.2–4.3

Mei, MIT OpenCourseWare, Forces in the Fluid, §1.4

Mei, MIT OpenCourseWare, Law of Momentum Conservation, §1.5

Facts & Assumptions

Given: The interval, reference body, motion and its induced velocity, SI fields, regularity and positivity in the Statement, the adopted material-region balances, no couples, and the Axiom of Choice.

[A1]

Single-component nonpolar continuum and material balances: the smooth model adopts the reference-to-current mass integral, linear and angular momentum balances, the local traction representation $t = σ n$ , and a $C^{1}$ stress; it has no surface couples or intrinsic spin.

[F1]

Classical frames, primitive quantities and SI units: time is absolute, the motion is distinct from an inertial chart, $u$ is Eulerian velocity, and the SI types and Galilean inertial transformations are as stated.

[F2]

Material mass balance implies local continuity: the adopted material mass integral implies $\partial_{t} ρ + div (ρ u) = 0$ for the stated positive $C^{1}$ density and $C^{2}$ motion.

[F3]

Transport across a moving control boundary: under countable choice, a $C^{1}$ field near the swept closure of a bounded $C^{1}$ region transported by a $C^{2}$ diffeomorphic motion satisfies $\frac{d}{d t} \int_{P_{t}} f = \int_{P_{t}} \partial_{t} f + \int_{\partial P_{t}} f u \cdot n$ , componentwise, when its boundary velocity is $u$ .

[F5]

Divergence on a bounded C1 Euclidean domain: under countable choice, a $C^{1}$ vector field on the closure of a bounded $C^{1}$ domain satisfies $\int_{P} div F = \int_{\partial P} F \cdot n$ ; apply this rowwise to a matrix.

[F6]

C1 material regions have compatible Jordan volume integrals: a $C^{1}$ diffeomorphism on a neighborhood of a bounded $C^{1}$ region's compact closure sends it to a bounded $C^{1}$ region, with open-region integrals equal to compact-closure integrals.

[F4]

The Axiom of Choice: every set-indexed family of nonempty sets admits a choice function.

Proof

1.1givenF4algebra

Let $(A_{n})_{n \in N}$ be any countable family of nonempty sets. By [F4], the set ${A_{n} : n \in N}$ has a choice function $g$ ; then $n \mapsto g (A_{n})$ chooses an element of each $A_{n}$ . Thus the countable-choice hypothesis ${A C}_{ω}$ required by [F3] and [F5] follows from the stated Axiom of Choice.

1.2givenA1algebra

The postulate [A1] already supplies a matrix $σ (t, x)$ with $t (t, x, n) = σ (t, x) n$ for every unit normal. If another matrix $\tilde{σ} (t, x)$ had this property, setting $n = e_{i}$ for $i = 1, 2, 3$ would give $\tilde{σ} (t, x) e_{i} = σ (t, x) e_{i}$ for all three standard basis vectors. The matrices have the same columns and are equal, so the adopted representation is unique at every event.

1.3givenA1F1F2

Apply [F2] to the material mass equality in [A1]. Its hypotheses hold by the $C^{2}$ motion and positive $C^{1}$ density in the Statement, so $\partial_{t} ρ + div (ρ u) = 0$ on $O_{B}$ .

2.1givenA1F3F5F6step 1.1step 1.3algebra

Fix an admissible $V$ and put $P_{t} = χ_{t} (V)$ . Its closure is compact in $Ω_{t}$ , and [F6] supplies its $C^{1}$ boundary. Its boundary velocity is $u$ by the Statement. Apply [F3] to $f = ρ u$ and [F5] to each row of $ρ u \otimes u$ and $σ$ . The adopted momentum balance gives $\int_{P_{t}} R d x = 0$ , where $R : = \partial_{t} (ρ u) + div (ρ u \otimes u - σ) - ρ b$ is continuous. For fixed $t$ , every sufficiently small ball $Q ⋐ Ω_{t}$ is such an image: $V = χ_{t}^{- 1} (Q)$ has compact closure in $B$ and $C^{1}$ boundary by [F6] applied to the inverse map. If a component of $R$ were nonzero at a point, continuity would bound it strictly away from zero with fixed sign on a smaller ball, contradicting its zero integral. Hence $R = 0$ on $O_{B}$ . Expanding the product and using step 1.3 gives $ρ D_{t} u = ρ b + div σ$ .

3.1givenA1F3F5F6step 1.1step 1.3step 2.1algebra

Define the alternating symbol $ϵ_{i k l}$ to be zero when two indices agree and the sign of the permutation $(i, k, l)$ of $(1, 2, 3)$ otherwise. Put $M_{i j} : = \sum_{k, l} ϵ_{i k l} x_{k} σ_{l j}$ , so $M n = x \times (σ n)$ and $(div M)_{i} = (x \times div σ)_{i} + \sum_{j, l} ϵ_{i j l} σ_{l j}$ . Apply [F3] to $ρ (x \times u)$ and use step 1.3, $D_{t} x = u$ , and $u \times u = 0$ : its material-integral derivative is $\int_{P_{t}} x \times (ρ D_{t} u) d x$ . Substitute step 2.1 and compare with the adopted angular balance, using [F5] on $M$ . Thus each continuous component $c_{i} : = \sum_{j, l} ϵ_{i j l} σ_{l j}$ has zero integral on every $P_{t}$ . The ball argument of step 2.1 gives $c_{i} = 0$ pointwise. The three equations are $σ_{32} - σ_{23} = 0$ , $σ_{13} - σ_{31} = 0$ , and $σ_{21} - σ_{12} = 0$ , respectively, proving symmetry.

4.1givenF3F5F6step 1.1step 1.3step 3.1algebra∎

Conversely, assume local momentum and symmetric stress in this same setting. Symmetry makes the three components $c_{i}$ in step 3.1 zero, so $div M = x \times div σ$ . Step 1.3 and $D_{t} (x \times u) = x \times D_{t} u$ then give $\partial_{t} (ρ (x \times u)) + div (ρ (x \times u) \otimes u) = x \times (ρ b) + div M$ . Integrate over $P_{t}$ and apply [F3] and [F5]; the left side is the material angular-momentum derivative and the right side is its body and traction moment. This recovers the adopted angular balance, completing the converse and all claims.

Cauchy stress, local momentum and stress symmetry

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources