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For a smooth, locally invertible material map $\mathbf{x}=\mathbf{X}(\mathbf{a},t)$ with ${\mathbf{X}}_t=\mathbf{u}(\mathbf{X},t)$ and positive Jacobian determinant $J=\det (\partial \mathbf{X}/\partial \mathbf{a})$, Euler's volume identity states that $J_t=J\nabla \cdot \mathbf{u}$. Indeed, $F=\partial \mathbf{X}/\partial \mathbf{a}$ obeys $F_t=(\nabla \mathbf{u})F$, so the determinant derivative is $J_t=J\mathrm{tr}(F^{-1}{F}_t)=J\mathrm{tr}(\nabla \mathbf{u})$. This form of the Jacobian evolution of a smooth flow applies also to explicitly time-dependent velocity fields. It gives the material-volume Reynolds transport theorem by change of variables to the fixed particle labels. It is distinct from the complex-exponential Euler's formula.

A material volume is a region containing the same fluid particles at every time: its boundary moves with the fluid velocity field. Its mass is fixed by mass conservation, while its volume can change in compressible motion. Applying the Reynolds transport theorem to its volume gives $d|V(t)|/dt={\int }_{V(t)}\nabla \cdot \mathbf{u}dV$. A material volume element is the infinitesimal version.

Nonuniform-density erosion inference estimates a polymer erosion front speed when local density varies through a specimen. Total mass divided by total volume gives an average density; the material being removed at the boundary can have a different density. Substituting the average into a surface-removal formula can therefore misestimate the speed.

For an original spherically symmetric model, let $R(t)$ be the surviving radius and $\rho (r,t)$ the local mass density. ThenDifferentiating the integral and defining positive inward speed $v=-R^{\prime }$ givesThis radial form of the Reynolds transport theorem separates loss at a moving boundary from density change inside the surviving volume. All quantities must refer to the same material and conditioning state; a water-filled wet volume and a subsequently dried polymer mass do not automatically define the local density used here.

If the interior density profile is stationary in spatial coordinates over the measurement interval, the integral term vanishes. The appropriate boundary conversion is then $v=-M^{\prime }/(4\pi R^2{\rho }_s)$, with ${\rho }_s=\rho (R,t)$. Writing average density as $\overline{\rho }=3M/(4\pi R^3)$, a uniform-density substitution instead givesThe ratio is measurable information about the profile, not a shape correction already supplied by knowing the outer radius.

An illustrative fixed profile is $\rho (r)={\rho }_0[1+(r/R_0{)}^2]$, with $R_0$ a fixed initial radius, not the changing current radius. At $R=R_0$, the boundary density is $2{\rho }_0$ and the mean is ${\rho }_0(1+3/5)=1.6{\rho }_0$. The uniform-density inference overestimates the instantaneous front speed by 25%. For $R_0=1$ cm, ${\rho }_0=1\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}$ and true $v=0.001\mathrm{cm}\mathrm{da}{\mathrm{y}}^{-1}$, mass leaves at $0.008\pi \approx 0.02513\mathrm{g}\mathrm{da}{\mathrm{y}}^{-1}$; dividing by the mean-density surface factor gives $0.00125\mathrm{cm}\mathrm{da}{\mathrm{y}}^{-1}$. These are stipulated profile values, not measured properties of a named polymer.

Time-dependent hydration, porosity or chemical change can make the interior integral nonzero. One mass slope cannot independently identify both that contribution and boundary speed. Boundary-density calibration for erosion fronts therefore needs spatial or compositional evidence in addition to a bulk mass measurement. NIST's mass-density definition is mass divided by volume; it does not equate a whole-specimen average with an unknown boundary value.^[ref]

A discriminating experiment measures outer dimensions, conditioned mass and density or composition at several depths, using matched specimens or accounting for destructive sampling. Kureha's polyglycolic acid protocol filters and dries the remaining solids after water exposure.^[ref] That observation supplies retained dry mass, not a radial density field. Establish an inward-moving degradation boundary before interpreting the inferred speed as polymer erosion, rather than another cause of material loss.

The Rankine-Hugoniot conditions follow by integrating conservation laws across a thin moving layer, then letting its macroscopic thickness tend to zero. For example, normal momentum conservation in one dimension has the conservative formFor a shock wave moving at speed $V$, integrate over a control interval whose endpoints move with the discontinuity. The Reynolds transport theorem giveswhere $[f]=f_R-f_L$ denotes the difference of limiting states. Mass conservation similarly gives $V[\rho ]=[\rho u]$. If $v=u-V$ is the normal velocity in the shock frame, combining the two relations yieldsThe same construction applied to conservation of energy, for a non-radiating normal shock wave with nonzero mass flux, gives $[h+v^2/2]=0$, where $h$ is the specific enthalpy. An equation of state closes these relations, and the physical compression branch has increasing entropy.

Finite dynamic viscosity replaces the ideal discontinuity by a thin smooth transition. A viscous stress contributes a derivative to the momentum flux inside the layer and converts directed kinetic energy into internal energy. If its gradients vanish in the uniform far states, the same limiting Rankine-Hugoniot conditions still hold there. Balancing nonlinear transport with viscous diffusion gives a thickness of order $\nu /|u_L-u_R|$ in the viscous Burgers equation, where $\nu$ is the kinematic viscosity.

Let $V(t)$ be any material volume, moving with the velocity field $\mathbf{u}$. The Reynolds transport theorem in the required form isapplied componentwise when $f$ is vector-valued. Mass conservation givesSince the material volume is arbitrary, the local continuity equation follows:For an inviscid fluid with no body force, the balance of linear momentum equates the rate of change of total momentum to the boundary pressure force:using the divergence theorem. The integrand on the left expands asThe bracket vanishes by mass conservation, leaving the Euler momentum equationUnlike the incompressible flow case, the mass density need not be constant along a fluid element.

Write $\rho$ for the mass density. A material volume contains the same fluid elements at all times, so conservation of mass requiresby the Reynolds transport theorem. Since the initial material volume can be chosen arbitrarily, continuity of the integrand forces it to vanish pointwise. Thus the continuity equation isFor positive mass density, the incompressibility condition $D\rho /Dt=0$ reduces this to $\nabla \cdot \mathbf{u}=0$. Mass conservation itself does not require incompressible flow.

Let $V(t)$ be a smoothly moving material volume, carried by a smooth velocity field $\mathbf{u}$, and let $f$ be continuously differentiable. The material-volume form of the Reynolds transport theorem statesEquivalently, the divergence theorem giveswhere $\mathbf{n}$ is the outward unit normal vector. The second integral measures the flux carried across the instantaneous boundary by its motion.

For the proof, use the particle map $\mathbf{x}=\mathbf{X}(\mathbf{a},t)$ from $V(0)$, with positive Jacobian determinant $J=\det (\partial \mathbf{X}/\partial \mathbf{a})$. Euler's volume identity isIt is the Jacobian evolution of a smooth flow: if $F=\partial \mathbf{X}/\partial \mathbf{a}$, differentiation gives $F_t=(\nabla \mathbf{u})F$, and differentiation of the determinant gives $J_t=J\mathrm{tr}(F^{-1}{F}_t)=J\nabla \cdot \mathbf{u}$. The change of variables formula now moves the integral to the fixed reference volume:Finally, $Df/Dt+f\nabla \cdot \mathbf{u}={\partial }_{t}f+\nabla \cdot (f\mathbf{u})$, giving the equivalent boundary form.

The fluid is outside the body, so its normal on the inner boundary is $-n^{\prime }$. Let $t=\sigma n^{\prime }$ be traction on the body's surface from the fluid. For an exterior point and a flow vanishing at infinity, the boundary representation readsFor $r=|y|\gg a$, expand the single layer through its first moment and the double layer through its zeroth moment:The next single-layer and double-layer terms are both $O(r^{-3})$ at fixed body size and boundary data. Differentiating $J$ givesHence the dipole terms combine intoThe antisymmetric part obeys $(\Sigma -{\Sigma }^T)y=G\times y$ for $G_j=-{ϵ}_{j\alpha \beta }{\Sigma }_{\alpha \beta }$. For the symmetric trace-free tensor $S$, $y\cdot My=y\cdot Sy+(\mathrm{Tr}M)r^2/3$. Also $\mathrm{Tr}\Sigma -\mathrm{Tr}M=2\mu \mathrm{Tr}U=2\mu Q$. Substitution givesThis derives the Stokes far-field multipoles of a deforming body. $F$ is the force exerted by the body on the fluid, the negative of the integrated fluid traction. $G$ is its torque on the fluid: $G=-\int x\times tdS$. Their kernels are the Stokeslet and rotlet. $Q=\int u\cdot n^{\prime }dS$ is the rate of increase of the body's volume, by the Reynolds transport theorem. It gives a radial source flow with volume flux $Q$. A volume-preserving body has $Q=0$. The trace-free tensor $S$ gives the stresslet, which can remain nonzero for a force-free, torque-free swimmer.

For a material volume, conservation of mass and the Reynolds transport theorem give ${\rho }_t+\nabla \cdot (\rho u)=0$. Constant material density, or equivalently incompressibility with the stated constant initial density, reduces this to $\nabla \cdot u=0$.

In an fluid undergoing inviscid flow the stress is the isotropic pressure stress $-pI$. Momentum conservation for a material volume then givesTransport and the divergence theorem localize this toFor a steady flow, $u_t=0$. For an irrotational flow the vorticity is zero, so the supplied identity gives $(u\cdot \nabla )u=\nabla (|u{|}^2/2)$. Uniform downward gravity is $g=-\nabla (gz)$, and constant density permits pulling $1/\rho$ inside the pressure gradient. ThereforeThe Bernoulli invariant is constant throughout each connected fluid region. Without irrotationality the corresponding steady statement generally gives a constant only along each streamline.

For a material volume $V(t)$, Newton's second law equates the rate of change of momentum to pressure and body forces:Apply the supplied Reynolds transport theorem to each component of $\mathbf{u}$, and apply the divergence theorem to the pressure force. The result isSince the material volume at the chosen time is arbitrary, the integrand vanishes pointwise for smooth fields. Dividing by positive density gives the Euler equationsAt a fixed time $t$, a streamline is a curve tangent everywhere to the instantaneous velocity. Parametrize it by $d\mathbf{x}/ds=\mathbf{u}(\mathbf{x}(s),t)$. Thenso $f$ is constant along it when the displayed directional derivative vanishes. Here time is held fixed: streamlines need not coincide with particle paths in an unsteady flow.

The momentum of a material volume $V(t)$ is ${\int }_{V(t)}\rho \mathbf{u}dV$. Newton's second law equates its time derivative to the pressure force and gravity:Apply the given Reynolds transport theorem componentwise to the left side, and the divergence theorem to the pressure term. ThenSince this holds for arbitrary material volumes and the fields are continuous, the integrand vanishes pointwise. Dividing by positive density gives the Euler equations for an inviscid fluidAt a fixed physical time $t_0$, a streamline is a curve tangent to $\mathbf{u}(\cdot ,t_0)$, which may be parametrized by $d\mathbf{x}/ds=\mathbf{u}(\mathbf{x}(s),t_0)$. The chain rule gives $df(\mathbf{x}(s),t_0)/ds=\mathbf{u}\cdot \nabla f=0$, proving constancy along it. The time is held fixed; a time-dependent streamline statement does not by itself assert material conservation along particle trajectories.

Let $t(n)$ denote traction: the force per unit area exerted across an oriented internal surface with unit normal $n$. Define ${\sigma }_{ij}$ as component $i$ of the traction on a coordinate plane whose normal is $e_j$. Balance of linear momentum on a shrinking tetrahedron makes the surface-force terms, of order area, dominate its inertia and body forces, of order volume. Resolving its oblique face into the three coordinate projected areas gives Cauchy's traction formulaOpposite faces have opposite tractions, as follows independently from a shrinking pillbox. This fixes both the index and normal conventions.

For a material volume $V(t)$, the Reynolds transport theorem says $d{\int }_{V(t)}fdV/dt={\int }_{V(t)}[f_t+\nabla \cdot (fu)]dV$. Apply it componentwise to $f=\rho u_i$ and use mass conservation. With no body force in this calculation, balance of linear momentum and the divergence theorem giveArbitrariness of the material volume yields $\boxed{\rho D{u}_i/Dt={\partial }_j{\sigma }_{ij}}$, with $D/Dt={\partial }_t+u_j{\partial }_j$.

For constant dynamic viscosity and incompressible velocity, ${\partial }_j{u}_j=0$, soThus $\boxed{\rho (u_t+u\cdot \nabla u)=-\nabla p+\mu {\nabla }^{2}u}$.

In fully developed steady pipe flow $u=w(x,y)e_z$, the convective acceleration vanishes. The transverse equations give $p_x=p_y=0$ and the axial equation gives $p_z=\mu (w_{xx}+w_{yy})$. One side depends only on $z$ and the other only on $(x,y)$, so both equal a constant. HenceIf $n$ is outward from the fluid cross-section, the wall-on-fluid axial shear per length is $\mu \oint {\partial }_{n}wds=\mu {\int }_A\Delta wdA=A{p}_z$. The fluid-on-pipe force is its negative:For forward pressure-driven flow $p_z<0$, this force points along the flow. Its value depends on cross-sectional area and pressure drop, even though the detailed velocity profile depends on the shape.

Use conservation of mass and Newton's momentum balance, in addition to the supplied constitutive law. For a material volume,The Reynolds transport theorem and ${\rho }_t+\nabla \cdot (\rho \mathbf{u})=0$ reduce the left side to $\int \rho D\mathbf{u}/Dt$. The divergence theorem then gives $\rho D{u}_i/Dt={\partial }_j{\sigma }_{ij}+\rho g_i$. For constant $\mu$,Incompressibility removes the last term. ThusThe stress formula and zero divergence alone are not a dynamical law: the force balance supplies the required physical input.

Label fluid particles by $\mathbf{a}$, with flow map $\mathbf{x}=\chi (\mathbf{a},t)$ satisfying ${\chi }_t=\mathbf{u}(\chi ,t)$. If $J=\det D_{\mathbf{a}}\chi$, differentiation of the determinant gives $J_t=J\nabla \cdot \mathbf{u}$. Pull the material-volume integral back to a fixed initial volume:The chain rule and the Jacobian identity give the integrand $J(F_t+\mathbf{u}\cdot \nabla F+F\nabla \cdot \mathbf{u})$. Transforming back yields the Reynolds transport theoremEquivalently the additional surface flux is ${\int }_{\partial V(t)}F\mathbf{u}\cdot \mathbf{n}dS$. The material boundary moves with the same normal velocity as the fluid, which is the key distinction from a fixed control volume.

For a material volume $V(t)$ moving with velocity $\mathbf{u}$, the Reynolds transport theorem isIt applies componentwise to vector quantities. Cauchy's stress theorem represents traction on a surface with outward unit normal $\mathbf{n}$ as $t_i={\sigma }_{ij}{n}_j$. Conservation of mass gives $d{\int }_{V(t)}\rho dV/dt=0$ and hence ${\partial }_t\rho +\nabla \cdot (\rho \mathbf{u})=0$. Constant density reduces this to $\boxed{\nabla \cdot \mathbf{u}=0}$.

With no body force, conservation of momentum and the divergence theorem giveApplying transport and then mass conservation cancels the terms proportional to $u_i$, leaving $\rho ({\partial }_t{u}_i+u_j{\partial }_j{u}_i)={\partial }_j{\sigma }_{ij}$. For the stated Newtonian stress, constant viscosity gives ${\partial }_j{\sigma }_{ij}=-{\partial }_{i}p+\mu \Delta u_i+\mu {\partial }_i(\nabla \cdot \mathbf{u})$. The last term vanishes. Since the material volume was arbitrary,A body force, if present, would have to be added to the momentum balance; it is absent from the specified model.

Let $\mathbf{x}(\mathbf{X},t)$ be the particle flow map from the fixed reference volume $V_0$ to the material volume, and let $J=\det (\partial \mathbf{x}/\partial \mathbf{X})$. Change variables and differentiate on the fixed domain:Use the two permitted identities, $DJ/Dt=J\nabla \cdot \mathbf{u}$ and $Df/Dt=f_t+\mathbf{u}\cdot \nabla f$. Changing back givesThis is the Reynolds transport theorem: both the value carried by each particle and the changing volume element contribute.

For a material volume moving with the fluid, the Reynolds transport theorem gives, componentwise,For a scalar $q$ this is also ${\int }_{V(t)}(Dq/Dt+q\nabla \cdot \mathbf{u})dV$. Apply the component formula to momentum density $q=\rho u_i$. Constant density and incompressibility make the integrand $\rho D{u}_i/Dt$.

The traction $\mathbf{t}(\mathbf{n})$ is the force per unit area exerted by the material outside a surface on the material inside, with outward unit normal $\mathbf{n}$. Reversing the normal exchanges the two sides of the same internal surface. Thus $\mathbf{t}(-\mathbf{n})=-\mathbf{t}(\mathbf{n})$ expresses equal and opposite internal contact forces, the local form of Newton's third law.

With no body force, Newton's second law equates the rate of change of the material volume's total momentum to its surface force. Combining this balance with the transport theorem givesThis is a vector identity, since the component argument applies to each direction.

For an arbitrary material volume $V(t)$, Newton's second law equates the rate of change of linear momentum to the resultant contact force, since there is no body force:Use the Reynolds transport theorem on the left. Expanding the product and using the continuity equation leavesOn the right, the divergence theorem applied to each row of the Cauchy stress tensor givesConsequently the volume integral of $\rho D{u}_i/Dt-{\partial }_j{\sigma }_{ij}$ vanishes on every material volume. Localizing as in part (i) gives the Cauchy momentum equation:The left side is mass density times material acceleration; the right side is the divergence of the Cauchy stress tensor.

Let $V(t)$ have a boundary moving with velocity ${\mathbf{v}}_b$, and let $\mathbf{n}$ be its outward unit normal vector. For a smooth scalar field $q$, the Reynolds transport theorem isIn the last expression, extend ${\mathbf{v}}_b$ smoothly into $V(t)$ and use the divergence theorem. For a material volume, its boundary moves with the fluid velocity $\mathbf{u}$, so ${\mathbf{v}}_b=\mathbf{u}$. Equivalently, the Reynolds transport theorem for a material volume readswhere $D/Dt$ is the material derivative.

For an arbitrary moving region, mass conservation includes the relative mass flow rate across its boundary:Substitution of the Reynolds transport theorem and the divergence theorem givesSince the region is arbitrary, a continuous integrand must vanish: if it had a positive or negative value at a point, a sufficiently small surrounding region would have an integral of the same sign. This gives the continuity equation. With constant positive mass density $\rho$, it reduces toThus the flow is incompressible. For a material volume, the same argument starts directly from conservation of its total mass, with zero relative boundary flux.