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Set the lower half-space load to $q_b={\mu }_b{p}_{\beta ,b}$ using the causal vertical wavenumber branch, which excludes a wave arriving from depth infinity. Apply the uniform-layer SH impedance update upwards through every layer, keeping particle velocity and traction continuous. For upper half-space impedance $q_a$, incident velocity $V_0$ and reflected velocity $R{V}_0$, the top values are $V=V_0(1+R)$ and $T=-q_a{V}_0(1-R)$. Their ratio gives the displayed reflection coefficient, incorporating every underlying interface and multiple reflection.

A stress remaining after the applied loads have been removed. Spatially nonuniform permanent strain can leave a self-equilibrated residual stress field, even though every boundary traction vanishes.

A layer with shear modulus ${\mu }_1$ and wave speed $c_1$ lies between a rigid support and a half-space with shear modulus ${\mu }_2$ and wave speed $c_2>c_1$. A trapped antiplane shear normal mode has layer amplitude $A\sin (my)$ and exterior amplitude $A\sin (mh)e^{-\ell (y-h)}$, where $m=k\sqrt{c^2/c_1^2-1}$ and $\ell =k\sqrt{1-c^2/c_2^2}$. Continuity of displacement field and traction gives the displayed dispersion relation. It requires $c_1<c<c_2$ and differs from a Love wave with a free boundary.

At normal incidence, a uniform lossless elastic wave layer of thickness $h_j$, speed $c_j$ and seismic impedance $q_j={\rho }_j{c}_j$ has wave phase thickness ${\delta }_j=\omega h_j/c_j$. With phasors proportional to $e^{-i\omega t}$, particle velocity $V$ and compressive traction $P=-T$, the displayed matrix maps $(V,P)$ from the top to the bottom. It has determinant one. Multiplying these matrices in physical order accounts for all coherent multiple reflections. Continuity of $V,P$ connects adjacent layers without another propagation factor.

At a material interface, incident P or SV energy can scatter into both P and SV modes, subject to conserved horizontal slowness and traction/displacement conditions. SH is decoupled in a planar isotropic model; some conversions vanish at symmetry directions such as normal incidence.

With $e^{ikx-i\omega t}$ phasors, positive depth $z$, shear modulus $\mu =\rho {\beta }^2$, vertical traction $T$ and particle velocity $V$, the shear-wave equations give $V^{\prime }=-i\omega T/\mu$ and $T^{\prime }=-i\omega \mu p_{\beta }^{2}V$, where $p_{\beta }^2={\beta }^{-2}-k^2/{\omega }^2$. Define $Z=-T/V$ wherever $V\ne 0$. Differentiation gives the displayed Riccati equation and $V^{\prime }=i\omega ZV/\mu$. A pole of $Z$ can be an ordinary zero of $V$, not a singular physical wavefield.

The SH input impedance is the negative ratio of vertical traction to particle velocity for a shear-horizontal wave field, with positive depth and tension-positive traction. A superposition of upward and downward waves generally has a depth-dependent complex ratio. A downward progressive wave in a uniform medium has $Z=\mu p_{\beta }$, while the upward wave has $Z=-\mu p_{\beta }$. The input ratio includes reflection from deeper structure and can have poles at velocity nodes.

The effect of slab-induced mantle flow, pressure and traction on the motion of neighbouring plates and their boundary. It is part of the coupled response to a sinking slab, rather than an unrelated energy source that can always be added independently to slab pull.

For incidence from a solid with compressional/shear speeds $\alpha ,\beta$ into a fluid of sound speed ${\alpha }_f$, let $E={\cos }^{2}2{\theta }_S+({\beta }^2/{\alpha }^2)\sin 2{\theta }_P\sin 2{\theta }_S$. The displayed coefficient multiplies the transmitted displacement field potential relative to the incident P wave potential, using the same retarded-time waveform. Snell law for elastic and acoustic waves matches the tangential slowness. Continuity of normal displacement field, zero tangential traction, and continuity of normal traction give the coefficient. It is not a displacement-amplitude coefficient: longitudinal displacement field acquires an additional ratio $\alpha /{\alpha }_f$.

For a no-slip sphere in a smooth ambient Stokes flow whose body forces lie outside the sphere and are unchanged by insertion, use the translating sphere as a reciprocal auxiliary field. Its uniform surface traction selects the ambient surface average and annihilates a rigid rotation term. The ambient stress has zero net surface force because its divergence vanishes inside the ball. Thus the sphere's hydrodynamic force is $-6\pi \mu a(\mathbf{V}-⟨{\mathbf{u}}_{\infty }⟩)$; mechanical balance gives the displayed formula. The surface-average-only expression requires a force-free inertialess sphere. The spherical mean of a biharmonic function makes this average exactly $[1+(a^2/6){\nabla }^2]{\mathbf{u}}_{\infty }(0)$.

A symmetric stress tensor is in static elastic equilibrium when its divergence balances the prescribed body force. Its boundary traction is $t_i={\sigma }_{ij}{n}_j$. With zero body force, integration by parts gives ${\int }_{\Omega }\sigma :\mathrm{sym}\nabla v={\int }_{\partial \Omega }t\cdot v$. This is the weak principle of virtual work for a static elastic body.

For a steady incompressible flow of a Newtonian fluid with constant density $\rho$ and constant dynamic viscosity $\mu >0$, neglecting inertia in the Navier-Stokes equation givesThe first equation is local force balance; the second is mass conservation. The term $\mu {\nabla }^2\mathbf{u}$ is the divergence of the viscous part of the Newtonian fluid stress tensor. The body force $\mathbf{f}$ has units of force per unit volume, $\mathrm{N}{\mathrm{m}}^{-3}$, and equals $\rho \mathbf{g}$ for uniform gravity. Appropriate boundary data for velocity or traction, together with suitable distant conditions for an exterior flow, complete the problem; a rigid surface normally imposes the no-slip boundary condition.

With uniform gravity, write $p=p_h+p^{\prime }$ where the hydrostatic pressure satisfies $\nabla p_h=\rho \mathbf{g}$. Then the disturbance equation is $-\nabla p^{\prime }+\mu {\nabla }^2\mathbf{u}=0$. This removes gravity from the fluid disturbance equation, while the hydrostatic pressure still supplies buoyancy on an immersed body. Omitting it from the total force balance would give the wrong Stokes settling velocity.

12.005 Applications of Continuum Mechanics to Earth, Atmospheric, and Planetary Sciences, spring 2006, Lecture 10. Traction, frictional strength, pore fluid pressure and Mohr circle construction. These notes use a particular sign convention; conversions must preserve it.

For a fluid occupying $z<0$, match an outer stagnation point flow $u=x,w=-z$ to a flat impermeable interface with dimensionless shear stress $\beta x$. With boundary layer thickness $\delta ={\mathrm{Re}}^{-1/2}$, inner variable $\zeta =z/\delta$ and $\gamma =\delta \beta$, the leading boundary-layer equation admits $u=x{f}^{\prime }(\zeta )$ and $w=-\delta f(\zeta )$. Incompressibility is automatic, and tangential momentum conservation gives the displayed ordinary differential equation, with $f(0)=0$, $f^{\prime \prime }(0)=\gamma$ and $f^{\prime }\to 1$ at $\zeta \to -\infty$. The sign of this far-field endpoint is fixed by the fluid domain. This is a prescribed-traction interface; it does not impose a no-slip boundary condition.

For a clean spherical gas bubble with negligible internal dynamic viscosity and thermal conductivity, a weak temperature gradient produces the displayed migration velocity, where ${\gamma }_T=d\gamma /dT$. The exterior Papkovich–Neuber representation can use only the scalar potential $\chi =-a^3\mathbf{U}\cdot \mathbf{r}/(2r^3)$. Its surface normal velocity equals $\mathbf{U}\cdot \mathbf{n}$, and its tangential traction balances minus the surface surface tension gradient from the temperature dipole around an insulating sphere. For the usual ${\gamma }_T<0$ the bubble moves toward warmer liquid. Small Reynolds number, Péclet number, capillary number, and fractional surface-tension variation justify the approximation; surfactants can oppose the temperature-driven Marangoni stress.

The lower bound makes internal vertical velocity variation negligible; the upper bound makes internal in-plane extensional stress negligible compared with ambient traction. Both are needed for an ambient-controlled interfacial plug current.

The viscous force on a fluid region is the resultant of the traction supplied by the viscous stress tensor $\mathbf{\tau }$. With $\mathbf{n}$ the outward unit normal of the region, its surface contribution is the displayed surface integral. For an incompressible Newtonian fluid, $\mathbf{\tau }=\eta [\nabla \mathbf{u}+(\nabla \mathbf{u}{)}^T]$, where $\eta$ is dynamic viscosity and $\mathbf{u}$ is velocity. A force scale is $\eta UL$ for a body of size $L$ moving at speed $U$, while the inertial force scale is $\rho U^2{L}^2$; their ratio is the Reynolds number. Total hydrodynamic drag on a body includes both viscous traction and the pressure traction induced by the flow. For example, the full Stokes drag law gives $6\pi \eta RU$ for a no-slip sphere, rather than just the viscous-traction part alone.

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]

For a Newtonian fluid and a flat wall with $0<b<\infty$, the Navier slip boundary condition is equivalent to a tangential wall-on-fluid tractionThe sign expresses resisting friction: ${\partial }_{n}u$ uses the inward unit normal, whereas the wall-on-fluid traction uses the outward unit normal of the fluid domain. The irreversible loss of mechanical energy per unit wall area and time isFor two tangential directions replace $u_{\mathrm{rel}}^2$ by the squared Euclidean norm of relative tangential velocity. At a moving wall this relative-velocity loss must be distinguished from the power supplied by the wall's motion. Bulk viscous dissipation also remains present. A finite slip length models reduced interfacial resistance, without declaring every boundary ideal or removing dynamic viscosity.

The limiting loss depends on how the flow is driven. In Couette flow with a slipping lower wall,It tends to zero for both $b\to 0$ and $b\to \infty$ at fixed $H$ and $U$. At $b=0$ impose the no-slip boundary condition directly, rather than dividing by zero. In planar Poiseuille flow with equal wall slip at fixed $G$, each wall instead dissipates $G^{2}b{H}^2/(4\mu )$, which grows with $b$ because relative velocity grows. The bulk loss per unit width is $G^2{H}^3/(12\mu )$; adding both wall losses gives $Gq$, the pressure work per unit channel length and width. This energy balance shows why small friction coefficients do not guarantee small losses under every driving constraint.

For a stationary flat plate in a Newtonian fluid, the streamwise shear stress exerted by the fluid on its upper face isThe sign here is positive when the fluid pulls the plate downstream; the plate exerts the opposite tangential traction on the fluid. In a Blasius boundary layer, ${\tau }_w\simeq 0.332057\rho U^2/\sqrt{{\mathrm{Re}}_x}$. A growing boundary layer thickness therefore accompanies decreasing local wall shear stress, because its near-wall velocity gradient becomes gentler. The formal divergence as $x\to 0$ is integrable, but that leading-edge limit lies outside the thin-layer hypothesis.^[ref]