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The first equation is the local balance of linear momentum. Use $\mathsf{M},\mathsf{L},\mathsf{T}$ for the dimensions of mass, length, and time, to distinguish dimensional time from the Cauchy stress tensor $\mathbf{T}$.

Here $t$ has dimension $\mathsf{T}$. The hypothesis ${\partial }_t\mathbf{b}=0$ means that the prescribed spatial body force field is the same for every member of the static family and the resulting motion.

Restore the particle label suppressed in the question, writing the static family as ${\mathbf{x}}_0(\mathbf{X};\tau )$. At each fixed $\tau$, its Cauchy stress tensor satisfiesThis dynamic continuation of a static incompressible elastic deformation sets $\tau =t$ to form a motion. At every time its deformation gradient is the same as that of the corresponding static member. For an elastic material the elastic part of the Cauchy stress tensor depends on that deformation gradient, while incompressibility leaves an arbitrary scalar constraint reaction in hyperelastic stress. Therefore writeHere $1$ is the identity matrix. Subtracting static balance of linear momentum from dynamic balance of linear momentum givesSince the homogeneous incompressible material has constant mass density, define $\xi =(p-p_0)/\rho$. NecessarilyConversely, suppose this acceleration potential exists. Choose $p=p_0+\rho \xi$. ThenThe added isotropic stress is symmetric, so angular momentum conservation remains satisfied. This proves both directions for the field equations and the elastic constitutive equation. The surface tractions for the motion are those of the adjusted stress; identical prescribed static tractions would impose an additional boundary restriction on $\xi$.

For this homogeneous motion, suppose the matrix $\mathbf{A}=\ddot{\mathbf{F}}{\mathbf{F}}^{-1}$ is symmetric. Put $\mathbf{y}=\mathbf{x}-\mathbf{c}(t)$. The acceleration potential of a homogeneous motion iswhere $g(t)$ is arbitrary. When taking the spatial gradient, $t$, $\mathbf{c}$, $\ddot{\mathbf{c}}$, and $\mathbf{A}$ are fixed. Symmetry givesThis verifies the potential directly and also proves sufficiency of the symmetric matrix condition within the static-family setting. Its dimension is ${\mathsf{L}}^2{\mathsf{T}}^{-2}$, so the pressure correction $-\rho \xi$ has the dimension of stress. Combining with part (b)(ii), the corresponding dynamic Cauchy stress tensor iswhere $\tilde{C}(t)=C(t)-\rho g(t)$. Substituting its divergence into the Cauchy momentum equation reproduces $\rho [\mathbf{Ay}+\ddot{\mathbf{c}}]$, as required.

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.

At an oriented internal surface with unit normal vector $\mathbf{n}$, the traction $\mathbf{t}(\mathbf{x},t;\mathbf{n})$ is the limiting contact force per unit surface area exerted by the material on the positive-$\mathbf{n}$ side on the material on the negative-$\mathbf{n}$ side:For a boundary of a chosen material volume, this convention gives the external traction when $\mathbf{n}$ is outward. The Cauchy stress tensor has componentsso the second index specifies the surface orientation and the first specifies the force component. The vectors ${\mathbf{e}}_i$ form the standard Cartesian coordinate system basis of unit vectors.

Cauchy's stress theorem states that the dependence of traction on the normal vector is linear:Repeated indices are summed according to Einstein notation. In particular, reversing the normal vector reverses the traction, and knowing the tractions on three mutually perpendicular coordinate planes determines the traction on every plane. Symmetry of the Cauchy stress tensor, for an ordinary fluid without body couples, follows separately from angular momentum conservation.

For an isotropic Newtonian fluid in incompressible flow, the constitutive equation isHere $p$ is pressure, ${\delta }_{ij}$ is the Kronecker delta, $\mu$ is the constant dynamic viscosity, and $E$ is the strain-rate tensor. The more general compressible constitutive equation can include a term proportional to $(\nabla \cdot \mathbf{u}){\delta }_{ij}$; that term vanishes here.

Differentiate the Cauchy stress tensor, retaining the constant dynamic viscosity outside the derivatives:The last equality uses incompressibility. Substitution into the Cauchy momentum equation gives the Navier-Stokes equationThe pressure gradient and the Laplacian of the velocity field therefore give the two contact force densities in this model.

For a plane whose normal lies in the ${\sigma }_1$–${\sigma }_3$ plane, rotate through $2\theta$ from $({\sigma }_1,0)$ in the stated sign convention. The circle point givesThe normal-to-plane angle is essential: if the angle supplied is between the plane itself and ${\sigma }_1$, use its complement for $\theta$. The principal stress planes have zero shear stress, and planes at $45^{\circ }$ to those principal stress planes have maximum $|\tau |$.

For an arbitrary plane in a fully three-dimensional stress field, let $n_i$ be its direction cosines in the principal stress basis. ThenThese follow by resolving the traction into its normal and tangential components; the compression-positive tensor is the negative of the usual tension-positive Cauchy stress tensor. The family of Mohr circles visualizes the possible combinations, but the two-dimensional $2\theta$ rule alone does not specify an arbitrary three-dimensional orientation.^[ref]

The principal stresses are the eigenvalues of the symmetric Cauchy stress tensor. In their orthonormal eigenbasis, each traction is normal to its corresponding principal plane.

For an isotropic hyperelastic material with incompressibility, the principal stresses are $t_i={\lambda }_i\partial W/\partial {\lambda }_i-p$. This follows by evaluating the Cauchy stress tensor formula $T=F({\partial }_{F}W{)}^T-pI$ in principal directions. For the reduced strain energy density $h(\lambda )=W({\lambda }^{-2},\lambda ,\lambda )$, the chain rule and equality of the tangential derivatives give $t_{\theta }-t_r=\lambda h^{\prime }(\lambda )/2$.

The second tensor has two material indices and is symmetric when the Cauchy stress tensor is symmetric. The displayed relation uses the material-first nominal stress tensor convention. In particular, $PF$ is generally not this tensor. These conventions agree with A. F. Bower's definitions of stress measures.

Write the Cauchy stress tensor in the local fibre frame as $\sigma =se\otimes e+h(e\otimes n+n\otimes e)+tn\otimes n$. The compatible geometry has $\nabla \theta =\kappa e$, $\nabla \cdot e=0$ and $\nabla \cdot n=-\kappa$. With zero body force, force balance becomes $e\cdot \nabla s=2\kappa h-n\cdot \nabla h$ and $n\cdot \nabla t-\kappa t=-e\cdot \nabla h-\kappa s$. For specified shear stress $h$, the first equation determines $s$ along fibres, then the second determines $t$ along transverse curves. Pressure and fibre tension supply these two normal constraint reactions. This is a local stress construction; boundary tractions must be chosen consistently with its integration data.

The viscous stress tensor is the part of the Cauchy stress tensor associated with deformation rates. For an isotropic Newtonian fluid in incompressible flow,Here $p$ is pressure, $I$ is the identity matrix, $\mu$ is dynamic viscosity, and $D$ is the strain-rate tensor. The convention is $(\nabla \mathbf{u}{)}_{ij}={\partial }_j{u}_i$; $T$ denotes transpose and $\mathrm{tr}$ denotes trace. Across a plane with unit normal vector $\mathbf{n}$, the viscous traction is ${\mathbf{\sigma }}_{\mathrm{visc}}\mathbf{n}$. The tensor makes this rule work for any plane orientation. ^[ref]

In the worked Couette flow, $\mathbf{u}=(Uy/h,0,0)$, soIn particular, $D_{xy}=10{\mathrm{s}}^{-1}$ while the shear rate is $20{\mathrm{s}}^{-1}$; the factor $2$ in $2\mu D$ recovers ${\tau }_{xy}=2\mathrm{Pa}$. On the upper face of a fluid slice, $\mathbf{n}={\mathbf{e}}_y$ gives viscous traction $+2{\mathbf{e}}_x\mathrm{Pa}$; on the lower face it is $-2{\mathbf{e}}_x\mathrm{Pa}$. Constant stress can therefore transmit a force between the plates even though its divergence, and the net viscous force on each slice, vanish.

Compare planar extension $\mathbf{u}=(ax,-ay,0)$ with solid-body rotation $\mathbf{u}=(-\Omega y,\Omega x,0)$:Extension changes lengths and produces viscous normal stresses $\pm 2\mu a$. Rotation preserves distances and gives zero viscous stress, despite its nonzero velocity gradient and vorticity $2\Omega {\mathbf{e}}_z$. For $a=\Omega =1{\mathrm{s}}^{-1}$ and $\mu =0.10\mathrm{Pa}\mathrm{s}$, extension gives ${\sigma }_{\mathrm{visc},xx}=0.20\mathrm{Pa}$ and ${\sigma }_{\mathrm{visc},yy}=-0.20\mathrm{Pa}$; rotation gives zero. A rotating fluid may still have a pressure gradient supplying its centripetal acceleration.

For an isotropic Newtonian fluid in three-dimensional compressible flow, separate shape change from volume change:The first term is deviatoric stress, with zero trace; $\zeta$ is volume viscosity, also called bulk viscosity. For uniform expansion $\mathbf{u}=b\mathbf{x}$, $D=bI$ and $\nabla \cdot \mathbf{u}=3b$, so the shear contribution vanishes and the viscous stress is $3\zeta bI$. Thus dynamic viscosity and volume viscosity describe distinct responses, and incompressible flow removes only the volume-change term. ^[ref]