Cauchy's Momentum Equation

Derivation

Cauchy's Momentum Equation

D-317 Home PU-307 Threads force · matter · symmetry Depends on Reynolds Transport Theorem, Existence of the Cauchy Stress Tensor

Statement

For a continuum with mass density \(\rho\), velocity field \(\mathbf{u}(\mathbf{x},t)\), Cauchy stress tensor \(\boldsymbol{\sigma}\) and body force per unit mass \(\mathbf{f}\), conservation of linear momentum for every material volume is equivalent, wherever the fields are smooth, to the pointwise partial differential equation \(\displaystyle \rho\,\frac{D\mathbf{u}}{Dt}=\nabla\!\cdot\!\boldsymbol{\sigma}+\rho\,\mathbf{f}\), where \(\dfrac{D}{Dt}=\dfrac{\partial}{\partial t}+\mathbf{u}\cdot\nabla\) is the material derivative.

Why it matters

Cauchy's momentum equation is Newton's second law written for a deformable continuum. It is the parent equation of essentially all of continuum mechanics: choosing a constitutive law for \(\boldsymbol{\sigma}\) collapses it to the Navier–Stokes equations (Newtonian fluid), the Euler equations (inviscid fluid), the Navier–Cauchy equations (linear elastic solid), or the equations of plasticity and viscoelasticity.

Its power is that it separates universal physics (momentum balance) from material-specific physics (the constitutive relation). The balance below is exact and model-independent; every rheological assumption enters only through \(\boldsymbol{\sigma}\).

Assumptions

Continuum hypothesis.If dropped, \(\rho,\mathbf{u},\boldsymbol{\sigma}\) are not well-defined smooth fields and no local differential balance exists; one must retreat to kinetic theory or molecular dynamics and recover this equation only as a coarse-grained average.

Existence of the Cauchy stress tensor — the traction on any internal surface is \(\mathbf{t}(\mathbf{n})=\boldsymbol{\sigma}\,\mathbf{n}\), linear in the outward normal \(\mathbf{n}\).If dropped, the surface force cannot be converted to a volume integral by the divergence theorem and no \(\nabla\!\cdot\!\boldsymbol{\sigma}\) term arises; the whole reduction fails.

Mass conservation (continuity).If dropped, the transport of momentum density does not collapse to \(\rho\,D\mathbf{u}/Dt\); spurious terms proportional to \(\partial_t\rho+\nabla\!\cdot(\rho\mathbf{u})\) survive.

Only volumetric body forces and surface tractions act (no distributed momentum sources, no body couples).If dropped, extra source terms appear; body couples in particular make \(\boldsymbol{\sigma}\) non-symmetric (micropolar / Cosserat media).

Sufficient regularity for localization — the integrand is continuous and the balance holds for every subvolume.If dropped, only the integral (weak) form is valid; the pointwise PDE may fail across shocks, contact discontinuities and material interfaces, where jump (Rankine–Hugoniot) conditions replace it.

Non-relativistic kinematics.If dropped, momentum and energy fluxes couple and one must use the relativistic stress–energy tensor \(T^{\mu\nu}\) with \(\partial_\mu T^{\mu\nu}=0\).

Derivation

\[ \frac{d}{dt}\int_{V(t)}\rho\,\mathbf{u}\,dV \;=\; \int_{V(t)}\rho\,\mathbf{f}\,dV \;+\; \oint_{\partial V(t)}\mathbf{t}(\mathbf{n})\,dS \]

Newton's second law for the material volume \(V(t)\): the rate of change of its total linear momentum equals the total body force plus the total surface traction. This is the physical postulate. A

\[ \frac{d}{dt}\int_{V(t)}\rho\,\mathbf{u}\,dV \;=\; \int_{V(t)}\!\left[\frac{\partial(\rho\mathbf{u})}{\partial t}+\nabla\!\cdot\!\bigl(\rho\,\mathbf{u}\otimes\mathbf{u}\bigr)\right]dV \]

Reynolds' transport theorem applied to the momentum density \(\rho\mathbf{u}\) (componentwise, \(\phi=\rho u_i\)), converting the time derivative of a material integral into an Eulerian volume integral. The convective flux of momentum is the tensor \(\rho\,\mathbf{u}\otimes\mathbf{u}\). B

\[ \frac{\partial(\rho u_i)}{\partial t}+\nabla\!\cdot(\rho u_i\mathbf{u}) = u_i\underbrace{\Bigl[\tfrac{\partial\rho}{\partial t}+\nabla\!\cdot(\rho\mathbf{u})\Bigr]}_{=\,0}+\rho\Bigl[\tfrac{\partial u_i}{\partial t}+\mathbf{u}\cdot\nabla u_i\Bigr] = \rho\,\frac{Du_i}{Dt} \]

Expand by the product rule and insert the continuity equation \(\partial_t\rho+\nabla\!\cdot(\rho\mathbf{u})=0\); the bracket vanishes identically, leaving the material derivative. Hence the left side becomes \(\int_V \rho\,\dfrac{D\mathbf{u}}{Dt}\,dV\). B

\[ \oint_{\partial V}\mathbf{t}(\mathbf{n})\,dS \;=\; \oint_{\partial V}\boldsymbol{\sigma}\,\mathbf{n}\,dS \]

Substitute Cauchy's stress theorem \(\mathbf{t}(\mathbf{n})=\boldsymbol{\sigma}\,\mathbf{n}\) (proved by the tetrahedron argument, assumed here). The surface force is now the action of a single tensor field on the outward normal. A

\[ \oint_{\partial V}\boldsymbol{\sigma}\,\mathbf{n}\,dS \;=\; \int_{V}\nabla\!\cdot\!\boldsymbol{\sigma}\,dV, \qquad (\nabla\!\cdot\!\boldsymbol{\sigma})_i=\frac{\partial \sigma_{ij}}{\partial x_j} \]

Divergence theorem applied row-by-row to the tensor \(\boldsymbol{\sigma}\) (each component \(\oint \sigma_{ij}n_j\,dS=\int \partial_j\sigma_{ij}\,dV\)). This converts the last surface integral into a volume integral, the key step that makes a pointwise law possible. B

\[ \int_{V(t)}\!\left[\rho\,\frac{D\mathbf{u}}{Dt}-\nabla\!\cdot\!\boldsymbol{\sigma}-\rho\,\mathbf{f}\right]dV \;=\; \mathbf{0} \]

Collect Steps 3, 5 and the body-force term of Step 1 into one integrand. The identity holds for the given material volume. A

\[ \rho\,\frac{D\mathbf{u}}{Dt}-\nabla\!\cdot\!\boldsymbol{\sigma}-\rho\,\mathbf{f}=\mathbf{0}\quad\text{pointwise} \]

Localization (du Bois-Reymond lemma): the integral vanishes not just for one \(V\) but for every subvolume, because the material volume was arbitrary and the balance is frame-agnostic. A continuous integrand whose integral over all subregions is zero must itself be zero everywhere. This is the only step requiring continuity of the integrand. C

Result

\[ \boxed{\;\rho\,\frac{D\mathbf{u}}{Dt}=\nabla\!\cdot\!\boldsymbol{\sigma}+\rho\,\mathbf{f}\;}\qquad\Longleftrightarrow\qquad \rho\!\left(\frac{\partial u_i}{\partial t}+u_j\frac{\partial u_i}{\partial x_j}\right)=\frac{\partial\sigma_{ij}}{\partial x_j}+\rho f_i \]

Reading. Mass-times-acceleration per unit volume of a moving fluid parcel equals the net internal surface force per unit volume (the stress divergence) plus the body force per unit volume. The acceleration on the left is the material acceleration \(D\mathbf{u}/Dt\), which includes the convective part \((\mathbf{u}\cdot\nabla)\mathbf{u}\): even in a steady field a parcel accelerates as it moves through spatial gradients. The divergence of the stress measures the imbalance of tractions on opposite faces of an infinitesimal element; a uniform stress exerts no net force.

Units check. Each term has dimensions of force per volume. \(\rho\,D\mathbf{u}/Dt\): \(\mathrm{kg\,m^{-3}}\cdot\mathrm{m\,s^{-2}}=\mathrm{N\,m^{-3}}\). \(\nabla\!\cdot\!\boldsymbol{\sigma}\): stress \(\mathrm{Pa=N\,m^{-2}}\) divided by length \(\mathrm{m}\) gives \(\mathrm{N\,m^{-3}}\). \(\rho\,\mathbf{f}\): \(\mathrm{kg\,m^{-3}}\cdot\mathrm{m\,s^{-2}}=\mathrm{N\,m^{-3}}\). Consistent.

Limiting cases

Statics / equilibrium \((\mathbf{u}=0)\): reduces to \(\nabla\!\cdot\!\boldsymbol{\sigma}+\rho\mathbf{f}=\mathbf{0}\), the equation of elastostatics; with \(\boldsymbol{\sigma}=-p\mathbf{I}\) it gives hydrostatics \(\nabla p=\rho\mathbf{f}\).
Inviscid fluid \((\boldsymbol{\sigma}=-p\mathbf{I})\): since \(\nabla\!\cdot(-p\mathbf{I})=-\nabla p\), one gets the Euler equation \(\rho\,D\mathbf{u}/Dt=-\nabla p+\rho\mathbf{f}\).
Newtonian fluid \((\sigma_{ij}=-p\delta_{ij}+\mu(\partial_i u_j+\partial_j u_i)+\lambda\,\delta_{ij}\nabla\!\cdot\mathbf{u})\): yields the Navier–Stokes equations; incompressible form \(\rho\,D\mathbf{u}/Dt=-\nabla p+\mu\nabla^2\mathbf{u}+\rho\mathbf{f}\).
Linear elastic solid \((\boldsymbol{\sigma}=\lambda(\nabla\!\cdot\mathbf{w})\mathbf{I}+2\mu\,\boldsymbol{\varepsilon})\) with displacement \(\mathbf{w}\): gives the Navier–Cauchy elastodynamic equation \(\rho\,\ddot{\mathbf{w}}=(\lambda+\mu)\nabla(\nabla\!\cdot\mathbf{w})+\mu\nabla^2\mathbf{w}+\rho\mathbf{f}\).
Uniform stress, no body force: \(\nabla\!\cdot\!\boldsymbol{\sigma}=0\Rightarrow D\mathbf{u}/Dt=0\); parcels move with constant velocity along their paths.

Breaks when

Shocks and other discontinuities. Across a shock, contact surface or free interface the fields are not differentiable; the differential form is meaningless and must be replaced by the integral form, giving jump conditions \([\![\rho\mathbf{u}(\mathbf{u}\cdot\mathbf{n}-U_n)-\boldsymbol{\sigma}\mathbf{n}]\!]=0\) (Rankine–Hugoniot).
Rarefied / non-continuum flow. When the Knudsen number \(\mathrm{Kn}=\lambda/L\gtrsim 0.1\) (upper atmosphere, MEMS, nanoscale channels) the continuum hypothesis fails; the equation must be derived as a moment of the Boltzmann equation, with slip and higher moments no longer closed by a simple \(\boldsymbol{\sigma}\).
Relativistic flows. At speeds comparable to \(c\) (astrophysical jets, quark–gluon plasma) momentum and energy fluxes mix; one uses \(\partial_\mu T^{\mu\nu}=0\) instead.
Media with internal structure / body couples. Polar (Cosserat) and liquid-crystal continua carry couple stresses, making \(\boldsymbol{\sigma}\) asymmetric, so an independent angular-momentum balance must be adjoined.

Failure modes

Using the partial derivative \(\partial\mathbf{u}/\partial t\) instead of the material derivative \(D\mathbf{u}/Dt\). Dropping the convective term \((\mathbf{u}\cdot\nabla)\mathbf{u}\) removes exactly the nonlinearity responsible for turbulence and Bernoulli's theorem; it is legitimate only after linearization or for genuinely spatially uniform fields.
Writing \(\nabla\!\cdot\!\boldsymbol{\sigma}\) as \(\boldsymbol{\sigma}\cdot\nabla\) or as a scalar. The stress divergence is a vector, \((\nabla\!\cdot\!\boldsymbol{\sigma})_i=\partial_j\sigma_{ij}\); contracting on the wrong index gives the wrong force for non-symmetric or spatially varying \(\boldsymbol{\sigma}\).
Sign error on the pressure term. Because \(\boldsymbol{\sigma}=-p\mathbf{I}+\dots\), \(\nabla\!\cdot(-p\mathbf{I})=-\nabla p\); students frequently keep \(+\nabla p\) and get flow accelerating up the pressure gradient.
Forgetting continuity in Step 3. Omitting \(\partial_t\rho+\nabla\!\cdot(\rho\mathbf{u})=0\) leaves the non-conservative form \(\partial_t(\rho\mathbf{u})+\nabla\!\cdot(\rho\mathbf{u}\otimes\mathbf{u})\), which is correct but must not also be equated to \(\rho\,D\mathbf{u}/Dt\) — that double-counts.
Applying the pointwise PDE across an interface. Using it through a shock or a fluid–fluid boundary instead of the jump conditions produces unphysical, mesh-dependent results in simulations.

Discussion

The equation is a statement about surfaces masquerading as a statement about volumes. The only forces a continuum element feels from its neighbours act on its bounding surface, encoded by the traction field \(\mathbf{t}=\boldsymbol{\sigma}\mathbf{n}\). The divergence theorem is what lets these surface interactions be re-expressed as a local volumetric force \(\nabla\!\cdot\!\boldsymbol{\sigma}\). This is why the existence of the Cauchy stress tensor is a genuine prerequisite: without the linearity \(\mathbf{t}(\mathbf{n})=\boldsymbol{\sigma}\mathbf{n}\) there is no tensor to take the divergence of.

The split of the total force into a body part \(\rho\mathbf{f}\) and a contact part \(\nabla\!\cdot\!\boldsymbol{\sigma}\) mirrors the derivation's two threads. The body force ties to the matter thread — it is proportional to mass and represents long-range fields (gravity, electromagnetism) acting at a distance. The stress divergence is the force thread — short-range molecular interactions homogenized into a field. That both appear as force-per-volume is the content of the continuum idealization.

The symmetry thread enters through a companion balance: conservation of angular momentum for the same material volume, worked through identically, forces \(\boldsymbol{\sigma}=\boldsymbol{\sigma}^{\mathsf T}\) in the absence of body couples. Momentum balance thus fixes the divergence of \(\boldsymbol{\sigma}\) while angular-momentum balance fixes its symmetry; together they leave six independent stress components, exactly the number a symmetric constitutive law must supply.

Structurally, Cauchy's equation is a local conservation law \(\partial_t(\rho u_i)+\partial_j\Pi_{ij}=\rho f_i\) with momentum-flux tensor \(\Pi_{ij}=\rho u_i u_j-\sigma_{ij}\): the Reynolds (convective) flux \(\rho u_i u_j\) plus the stress flux \(-\sigma_{ij}\). In this conservative form it is the spatial part of the vanishing four-divergence of the non-relativistic stress–energy tensor, and it is the form finite-volume solvers integrate to guarantee discrete momentum conservation. Common misconceptions: that \(D\mathbf{u}/Dt\) is a total time derivative of a fixed particle's velocity in the naive sense (it is, but only when computed as the Eulerian field derivative following the flow); that \(\nabla\!\cdot\!\boldsymbol{\sigma}=\mathbf{0}\) means no stress (it means no net stress force — a body can be highly stressed yet in equilibrium); and that pressure is a separate physical entity from stress rather than the isotropic part \(p=-\tfrac13\mathrm{tr}\,\boldsymbol{\sigma}\).

Worked examples

Example 1 — Hydrostatic pressure in a still fluid. A tank of water is at rest under gravity \(\mathbf{f}=-g\hat{\mathbf{z}}\). Find the pressure at depth \(h=10\ \mathrm{m}\); take \(\rho=1000\ \mathrm{kg\,m^{-3}}\), \(g=9.81\ \mathrm{m\,s^{-2}}\), surface pressure \(p_0=1.013\times10^{5}\ \mathrm{Pa}\).

\[ \mathbf{u}=\mathbf{0}\ \Rightarrow\ \frac{D\mathbf{u}}{Dt}=\mathbf{0},\qquad \boldsymbol{\sigma}=-p\,\mathbf{I}\ \Rightarrow\ \nabla\!\cdot\!\boldsymbol{\sigma}=-\nabla p \]

Static fluid: no acceleration; a fluid at rest supports only isotropic pressure. A

\[ \mathbf{0}=-\nabla p+\rho\mathbf{f}\ \Rightarrow\ \frac{dp}{dz}=-\rho g \]

Cauchy's equation collapses to hydrostatic balance; only the \(z\)-component is non-trivial. A

\[ p(z)=p_0-\rho g\,z,\qquad z=-h:\quad p=p_0+\rho g h \]

Integrate with \(z\) measured upward from the surface; depth \(h\) means \(z=-h\). A

\[ p=1.013\times10^{5}+ (1000)(9.81)(10)=1.013\times10^{5}+9.81\times10^{4}\ \mathrm{Pa} \]

Insert numbers with SI units throughout. A

\[ p(10\ \mathrm{m})\approx 1.99\times10^{5}\ \mathrm{Pa}\ \ (\approx 1.97\ \mathrm{atm}) \]

Reading. Ten metres of water roughly doubles the absolute pressure, adding about one atmosphere of gauge pressure — the familiar "one atmosphere per 10 m" diving rule falls straight out of Cauchy's equation.

Example 2 — Plane Poiseuille flow (wall-driven balance). Steady, fully developed, incompressible flow of water between parallel plates at \(y=\pm h\), \(h=1.0\ \mathrm{mm}\), driven by a constant pressure gradient \(-\,dp/dx=G=200\ \mathrm{Pa\,m^{-1}}\); viscosity \(\mu=1.0\times10^{-3}\ \mathrm{Pa\,s}\), no body force in \(x\). Find the centreline speed and the wall shear stress.

\[ \mathbf{u}=u(y)\,\hat{\mathbf{x}}\ \Rightarrow\ \frac{D\mathbf{u}}{Dt}=\frac{\partial\mathbf u}{\partial t}+(\mathbf u\cdot\nabla)\mathbf u=\mathbf 0 \]

Steady and fully developed: \(u\) depends on \(y\) only, and \((\mathbf u\cdot\nabla)\mathbf u=u\,\partial_x u\,\hat{\mathbf x}=\mathbf 0\). No net acceleration. B

\[ \sigma_{xx}=-p,\quad \sigma_{xy}=\mu\frac{du}{dy}\ \Rightarrow\ (\nabla\!\cdot\!\boldsymbol\sigma)_x=\frac{\partial\sigma_{xx}}{\partial x}+\frac{\partial\sigma_{xy}}{\partial y}=-\frac{dp}{dx}+\mu\frac{d^2u}{dy^2} \]

Newtonian constitutive law; take the \(x\)-row of the stress divergence. B

\[ 0=-\frac{dp}{dx}+\mu\frac{d^2u}{dy^2}\ \Rightarrow\ \mu\frac{d^2u}{dy^2}=-G \]

Cauchy \(x\)-component with zero acceleration and no body force; the pressure gradient is balanced entirely by viscous stress. A

\[ u(y)=\frac{G}{2\mu}\bigl(h^2-y^2\bigr),\qquad u_{\max}=u(0)=\frac{G h^2}{2\mu} \]

Integrate twice with no-slip \(u(\pm h)=0\) and symmetry \(u'(0)=0\). B

\[ u_{\max}=\frac{(200)(1.0\times10^{-3})^2}{2(1.0\times10^{-3})}=\frac{200\times10^{-6}}{2\times10^{-3}}\ \mathrm{m\,s^{-1}} \]

Insert numbers; \(h^2=10^{-6}\ \mathrm{m^2}\). A

\[ \tau_w=\left.\mu\frac{du}{dy}\right|_{y=-h}=\mu\cdot\frac{G h}{\mu}=G h=(200)(1.0\times10^{-3})\ \mathrm{Pa} \]

Wall shear from the velocity gradient at the plate; equivalently \(G h\) from a force balance on half the channel. B

\[ u_{\max}=0.10\ \mathrm{m\,s^{-1}},\qquad \tau_w=0.20\ \mathrm{Pa} \]

Reading. The parabolic profile is the signature of a pressure-driven viscous flow, and the wall shear \(\tau_w=Gh\) is fixed purely by geometry and the driving gradient — the viscosity cancels from \(\tau_w\) but sets the speed. Both follow directly from the \(x\)-component of Cauchy's equation.

Units check. \(u_{\max}\): \(\mathrm{Pa\,m^{-1}\cdot m^2/(Pa\,s)}=\mathrm{m\,s^{-1}}\). \(\tau_w\): \(\mathrm{Pa\,m^{-1}\cdot m=Pa}\). Consistent.

Problems

(A) A fluid parcel in a steady one-dimensional flow has velocity field \(u(x)=U_0(1+x/L)\hat{\mathbf x}\) with \(U_0=2\ \mathrm{m\,s^{-1}}\), \(L=0.5\ \mathrm{m}\). Find the material acceleration at \(x=0\).
Solution
\(Du/Dt=\partial_t u+u\,\partial_x u=0+U_0(1+x/L)\cdot(U_0/L)\). At \(x=0\): \(=U_0^2/L=(2)^2/0.5=8\ \mathrm{m\,s^{-2}}\). Note the parcel accelerates despite the field being steady — purely convective acceleration.
(A) In a static atmosphere of uniform density \(\rho=1.2\ \mathrm{kg\,m^{-3}}\), by how much does the pressure drop over a height rise of \(100\ \mathrm{m}\)? Take \(g=9.81\ \mathrm{m\,s^{-2}}\).
Solution
Hydrostatic balance \(dp/dz=-\rho g\Rightarrow \Delta p=-\rho g\,\Delta z=-(1.2)(9.81)(100)=-1177\ \mathrm{Pa}\approx-1.18\ \mathrm{kPa}\). Pressure falls about 1.2 kPa (≈1.2% of an atmosphere) over 100 m.
(B) Starting from \(\rho\,D\mathbf u/Dt=\nabla\!\cdot\!\boldsymbol\sigma+\rho\mathbf f\), show that for an inviscid fluid \((\boldsymbol\sigma=-p\mathbf I)\) in steady flow with \(\mathbf f=-\nabla\Phi\) the quantity \(\tfrac12|\mathbf u|^2+\int dp/\rho+\Phi\) is constant along a streamline (Bernoulli).
Solution
With \(\boldsymbol\sigma=-p\mathbf I\), \(\nabla\!\cdot\!\boldsymbol\sigma=-\nabla p\), giving Euler \(\rho(\mathbf u\cdot\nabla)\mathbf u=-\nabla p-\rho\nabla\Phi\) (steady). Use the identity \((\mathbf u\cdot\nabla)\mathbf u=\nabla(\tfrac12|\mathbf u|^2)-\mathbf u\times(\nabla\times\mathbf u)\). Dot with the unit tangent \(\hat{\mathbf t}=\mathbf u/|\mathbf u|\); the vorticity term is orthogonal to \(\mathbf u\) and drops. Then \(\hat{\mathbf t}\cdot\nabla\bigl[\tfrac12|\mathbf u|^2+\int dp/\rho+\Phi\bigr]=0\), so the bracket is constant along the streamline. \(\square\)
(B) A Newtonian fluid \((\mu=8.9\times10^{-4}\ \mathrm{Pa\,s})\) flows in a pipe of radius \(R=5\ \mathrm{mm}\) driven by \(-dp/dx=G=500\ \mathrm{Pa\,m^{-1}}\). Using the axial Cauchy balance \(0=G+\mu\,\tfrac1r\tfrac{d}{dr}(r\,du/dr)\), find the centreline velocity and volumetric flow rate.
Solution
Integrate \(\tfrac1r(r u')'=-G/\mu\) with \(u(R)=0\), \(u'(0)=0\): \(u(r)=\tfrac{G}{4\mu}(R^2-r^2)\). Centreline \(u_{\max}=GR^2/4\mu=(500)(5\times10^{-3})^2/(4\cdot8.9\times10^{-4})=(500)(2.5\times10^{-5})/(3.56\times10^{-3})\approx3.51\ \mathrm{m\,s^{-1}}\). Flow rate (Hagen–Poiseuille) \(Q=\pi G R^4/8\mu=\pi(500)(6.25\times10^{-10})/(8\cdot8.9\times10^{-4})\approx1.38\times10^{-4}\ \mathrm{m^3\,s^{-1}}\).
(C) Show that the conservative (divergence) form \(\partial_t(\rho\mathbf u)+\nabla\!\cdot(\rho\mathbf u\otimes\mathbf u-\boldsymbol\sigma)=\rho\mathbf f\) is equivalent to \(\rho\,D\mathbf u/Dt=\nabla\!\cdot\!\boldsymbol\sigma+\rho\mathbf f\), and explain why finite-volume solvers prefer the conservative form.
Solution
Expand componentwise: \(\partial_t(\rho u_i)+\partial_j(\rho u_i u_j)=u_i[\partial_t\rho+\partial_j(\rho u_j)]+\rho[\partial_t u_i+u_j\partial_j u_i]\). The first bracket is zero by continuity; the second is \(\rho\,Du_i/Dt\). Adding \(-\partial_j\sigma_{ij}=-(\nabla\!\cdot\!\boldsymbol\sigma)_i\) reproduces the stated equivalence. Finite-volume schemes integrate the conservative form over each cell and apply the divergence theorem, so interior fluxes cancel telescopically between neighbouring cells and total momentum is conserved to machine precision — and, crucially, the weak (integral) form remains valid across shocks where the non-conservative \(D\mathbf u/Dt\) form is undefined, giving correct shock speeds via the built-in Rankine–Hugoniot conditions.