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Antiplane shear has displacement normal to a plane and independent of that normal coordinate. In an isotropic linear elastic body, ${\sigma }_{13}=\mu w_{,1}$ and ${\sigma }_{23}=\mu w_{,2}$, together with their symmetric partners, are the only nonzero stress tensor entries. Static balance without body force reduces to the Laplace equation for $w$.

For a body occupying $V$, the rate of change of ${\int }_V\rho \mathbf{v}dV$ equals the body forces and surface tractions. With the Cauchy stress tensor, the local form is $\rho D\mathbf{v}/Dt=\nabla \cdot \mathbf{\sigma }+\rho \mathbf{b}$.

For steady incompressible flow and irrotational flow of constant mass density, with no body force, this formula gives the hydrodynamic torque about the origin per unit span on an impermeable body. Orient its boundary counterclockwise, and let positive torque be counterclockwise. The proof uses $M=\mathrm{Re}{\oint }_{C}pzd\bar{z}$, the Bernoulli equation $p=P_{\ast }-\rho |w^{\prime }{|}^2/2$, and the tangency identity $|w^{\prime }{|}^{2}d\bar{z}=(w^{\prime }{)}^{2}dz$. The real part of the constant-pressure term vanishes because $\mathrm{Re}(zd\bar{z})=d|z{|}^2/2$. The full complex integral of $zd\bar{z}$ need not vanish.

A body force acts throughout a body's or fluid's volume, rather than only through its surface. Gravity is the example used in hydrostatic pressure: a gravitational field $\mathbf{g}$ gives force per unit volume $\rho \mathbf{g}$, where $\rho$ is the mass density. More generally, a body force per unit mass $\mathbf{f}$ contributes $\rho \mathbf{f}$ to momentum conservation. Forces from electromagnetism are another example. Surface pressure forces produce buoyancy on an immersed body, while its body force from gravity acts downward.

Use a rotlet as the auxiliary Stokes flow, with a rigid sphere at its center. Its exterior-domain boundary torque is the imposed auxiliary couple. In the Lorentz reciprocal theorem, the actual sphere's zero torque removes the other surface term, while the volume term is the rotlet dotted with the body force. Varying the auxiliary couple gives the displayed angular velocity, assuming the integrals and the far boundary limit converge.

Buoyancy is the resultant upward force exerted by a resting fluid on an immersed body in a downward gravitational field. The fluid pushes inward, perpendicular to each part of the body's surface. Because hydrostatic pressure increases with depth, the upward push from below exceeds the downward push from above. Opposite horizontal forces cancel; buoyancy is the remaining vertical force. A sinking body still receives this upward support. ^[ref]

More precisely, pressure $p$ on a small surface of area $dA$ with outward unit normal vector $\mathbf{n}$ exertsThe minus sign says the force points into the body. On a cube, the top force points down, the bottom force points up, and the side forces cancel in pairs because opposite sides have the same pressure at each depth. This is a surface traction; gravity on the body itself is a separate body force. ^[ref]

For a computed example, stipulate resting water with constant mass density ${\rho }_f=1000\mathrm{kg}{\mathrm{m}}^{-3}$ and uniform gravitational acceleration $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$. A fully immersed cube has side $L=0.10\mathrm{m}$, top depth $h_t=0.20\mathrm{m}$ and bottom depth $h_b=0.30\mathrm{m}$. Each horizontal face has area $A=L^2=0.010{\mathrm{m}}^2$, and its volume is $V=L^3=0.001{\mathrm{m}}^3$. Its top and bottom gauge pressures areThe corresponding forces relative to the common surface pressure are $p_{g,t}A=19.62\mathrm{N}$ downward and $p_{g,b}A=29.43\mathrm{N}$ upward. Therefore the net buoyancy isUsing absolute pressures adds the same $p_{\mathrm{surface}}A$ to both face forces, leaving their difference unchanged. Zero gauge pressure at the surface does not mean zero absolute pressure. Increasing both depths by the same amount also leaves buoyancy unchanged in this constant-mass density model.

Archimedes' principle turns this calculation into a rule for bodies of any shape. It explains floating equilibrium and lets us infer an immersed body's volume from the difference between its force from gravity and the force needed to support it. In particular, buoyancy depends on the displaced volume and the surrounding mass density, whereas the body's own mass density decides how its force from gravity compares with that support.

Cauchy's stress theorem states that local contact traction depends linearly on the oriented unit normal vector: $t_i(\mathbf{n})={\sigma }_{ij}{n}_j$, with ${\sigma }_{ij}={\mathbf{e}}_i\cdot \mathbf{t}({\mathbf{e}}_j)$. It follows by balancing forces on a shrinking tetrahedron with three coordinate faces and one oblique face. Contact forces scale as the face area, while bounded body force and inertial contributions scale as its volume; dividing by the oblique face area and shrinking eliminates the latter. The coordinate-face projected areas are the oblique face area times $n_j$, giving the displayed linear map. Symmetry of the Cauchy stress tensor requires the additional balance of angular momentum in a continuum without independent couples.

Let ${\mathbf{x}}_0(\mathbf{X};\tau )$ be a family of static elastic equilibrium solutions for a homogeneous material with incompressibility under a fixed spatial body force. Replacing $\tau$ by time gives a dynamic solution of the elastic field equations exactly when the resulting material acceleration has a global acceleration potential $\xi$. At the same deformation gradient, the Cauchy stress tensor can change only by the scalar pressure reaction. Subtracting static and dynamic balance of linear momentum gives $\nabla q=\rho \mathbf{a}$, and constant mass density then gives $q=-\rho \xi +C(t)$. Surface tractions must match the adjusted Cauchy stress tensor.

Dynamic pressure is the bulk kinetic energy per unit volume associated with fluid motion at speed $U$ in a specified reference frame:Here $\rho =m/V$ is mass density. The word "pressure" identifies its units and its use as a pressure scale; $q$ is not an extra isotropic stress to add to the local static pressure in every surface traction. A stationary fluid can have large static pressure with $q=0$. Conversely, knowing $U$ and $\rho$ fixes $q$ without fixing static pressure. The definition remains useful in viscous fluid flow and compressible flow, even when the simplest Bernoulli equation is inapplicable. ^[ref]

For stipulated mass density ${\rho }_{\infty }=1.20\mathrm{kg}{\mathrm{m}}^{-3}$ and reference speed $U_{\infty }=20\mathrm{m}{\mathrm{s}}^{-1}$, $q_{\infty }=240\mathrm{Pa}$. At $40\mathrm{m}{\mathrm{s}}^{-1}$ with the same mass density, $q_{\infty }=960\mathrm{Pa}$. This quadratic scaling makes dynamic pressure a useful normalization for the lift coefficient, drag coefficient and pressure coefficient; it predicts four times a particular force only if its corresponding coefficient remains unchanged.

The connection with stagnation pressure requires hypotheses. For steady flow, constant mass density, inviscid flow, and gravity as the only body force, the Bernoulli equation gives, along one streamline,Here $p$ is static pressure, $g$ is gravitational acceleration, and $z$ is height. Bring that flow to rest without losses at the same height and the limiting static pressure is the stagnation pressure $p_0$:There must be no intervening pump supplying work or other energy input. Constancy along one streamline does not assert equal $B$ on all streamlines; irrotational flow in a connected region supplies that additional conclusion. Changing height requires retaining $\rho gz$. In compressible flow the stopping pressure difference generally differs from $\rho U^2/2$, and viscous dissipation can reduce recovered stagnation pressure. ^[ref]

A Pitot-static tube turns this relation into a measurement: its upstream-facing opening senses stagnation pressure, while separate side openings sample static pressure. Connect these to the positive and negative ports of a differential pressure transducer or manometer. Under the hypotheses above, the measured difference $\Delta p$ givesThus the stipulated $240\mathrm{Pa}$ and $1.20\mathrm{kg}{\mathrm{m}}^{-3}$ imply $20\mathrm{m}{\mathrm{s}}^{-1}$, not an independently measured speed. Align the probe with the incoming flow and keep its passages clear: blockage and pinched tubing invalidate the comparison. At very low speed, the difference can be smaller than instrument error; at high Mach number, compressibility and possibly a shock wave require a different reduction. ^[ref]

A useful control is differential-pressure calibration. For a digital manometer, zeroing establishes the reading at equal port pressures; it does not establish the gain away from zero. The manufacturer's Series 475 Handheld Digital Manometer: Specifications - Installation and Operating Instructions by Dwyer Instruments, Inc. directs the user to vent both ports to the atmosphere before zeroing. Its stated accuracy is a fraction of full scale, not of the reading, and depends on instrument temperature. Select a suitable range and record the instrument variant and units; the manual covers several ranges.^[ref]

An independent physical check can use a liquid-column manometer: apply a stable pressure difference to both instruments and compare the digital reading with the hydrostatic pressure inferred from the vertical level difference. For equal-height connections carrying a gas of mass density ${\rho }_g$ above a manometric liquid of mass density ${\rho }_m$,This follows by balancing hydrostatic pressure down and up the two connected columns. Use vertical height, even if a tube is inclined. Known mass densities, measured height and gravitational acceleration set the comparison scale. Check several differences across the working range and reverse the port connections to check the sign. This is a proposed independent check, not a claim that the cited teaching experiment performed a full instrument calibration. Recheck zero after reconnecting the sealed tubing; a zero adjustment alone cannot detect a blocked passage or an incorrect scale.

The effective elastic energy is the minimum averaged strain energy density under prescribed affine boundary displacement. For differentiable stable elastic response without body force, varying the prescribed mean strain gives $D{W}^{\mathrm{eff}}(E)=⟨\sigma ⟩$. The result follows by integration by parts, because the interior variation with zero boundary data does no work against an equilibrated stress tensor.

Opposite forces separated along the unit direction $\mathbf{e}$, with fixed force times separation $D$, have this distributional body force limit. Its net force is zero. A symmetric seismic moment tensor is a sum of such dipoles along its principal directions.

For a static base state with zero incremental body force, homogeneous incremental tractions or homogeneous Dirichlet boundary conditions eliminate the boundary term in integration by parts. The displayed Rayleigh quotient then gives real squared frequencies because of major symmetry. Positive moduli give strictly positive squared frequencies for modes with nonzero displacement gradient. Pure dead loading leaves constant translations at zero; fixed displacement conditions remove those modes.

The causal Green function for isotropic linear elasticity is tensor-valued: column $k$ is the displacement caused by a point impulsive body force in direction $k$. Its radiative terms are longitudinal P waves at $t=R/\alpha$ and transverse S waves at $t=R/\beta$, proportional to $R^{-1}$. A term supported between the arrivals carries the intermediate/near-field response. Convolution in time and space constructs the displacement for a general source.

On a connected isotropic linear elastic body, assume $\mu >0$ and $3\lambda +2\mu >0$. Then $W(e)=\mu |e^{\mathrm{dev}}{|}^2+(3\lambda +2\mu )(\mathrm{tr}e{)}^2/6$ is positive for every nonzero symmetric strain tensor. If two equilibria have the same body force and boundary traction, their difference $v$ satisfies zero loading. Multiplication of equilibrium by $v$ and the divergence theorem yield ${\int }_{B}2W(e(v))dV=0$. Hence $e(v)=0$, and the kernel of the symmetric gradient consists of infinitesimal translations and rotations: $v=a+\omega \times x$.

The conclusion is uniqueness modulo these modes, not existence for arbitrary tractions. Existence additionally requires zero total resultant force and torque. On a disconnected body each connected component has its own rigid modes; fixing enough displacements removes those modes and can produce full uniqueness.

The difference of two solutions with the same body force, boundary traction and initial data satisfies a homogeneous elastic equation with zero boundary work. Multiplication by its velocity and integration by parts conserve the displayed positive energy. Zero initial energy forces zero velocity for all times, and zero initial displacement field then forces zero displacement field. Rigid zero-strain motions do not cause nonuniqueness after initial data are fixed. This assumes positive mass density, stable real elastic moduli and sufficient regularity for the energy identity.

For constant mass density ${\rho }_0$, a frame with constant angular velocity $\mathbf{\Omega }$ absorbs centrifugal acceleration into reduced pressure $\Pi =p/{\rho }_0-|\mathbf{\Omega }\times \mathbf{r}{|}^2/2$. Indeed $\mathbf{\Omega }\times (\mathbf{\Omega }\times \mathbf{r})=-\nabla |\mathbf{\Omega }\times \mathbf{r}{|}^2/2$. Transforming the material acceleration in the Euler equations gives the displayed equation. The Coriolis acceleration remains velocity dependent and cannot be absorbed into a scalar potential. A conservative physical body force can be included by adding its potential to $\Pi$.

In a three-dimensional exterior Stokes flow with no body force, $\nabla \cdot \mathbf{\sigma }=0$. The divergence theorem on the fluid between a rigid body and a large sphere $S_R$ gives the displayed force on the body, where $\mathbf{n}$ points outward from the sphere. The body's outward normal points oppositely to the inner normal of this fluid domain. A nonzero resultant force produces a leading Stokeslet, so velocity is $O(R^{-1})$, gauge-fixed pressure is $O(R^{-2})$ and rate-of-strain tensor is $O(R^{-2})$. The finite force is therefore compatible with pointwise decay of the stress tensor. Force-free bodies may have faster decay.

For symmetric stress in static elastic equilibrium without body force, write $E=⟨\epsilon ⟩$ and $S=⟨\sigma ⟩$. The difference between microscopic and macroscopic work is $|\Omega {|}^{-1}{\int }_{\partial \Omega }(\sigma -S)n\cdot (u-Ex)$. It vanishes for affine boundary displacement, uniform boundary traction, or matching periodic displacement fluctuations and opposite tractions on paired faces.

Hydrostatic pressure is the pressure distribution in a resting fluid, balancing its body force from gravity. Take $z$ upward and a uniform downward gravitational field $\mathbf{g}=-g{\mathbf{e}}_z$, where $g>0$. The vertical pressure gradient satisfiesA short derivation balances the vertical forces on a thin horizontal layer of area $A$ and thickness $\Delta z$: $p(z)A-p(z+\Delta z)A-{\rho }_{f}gA\Delta z=0$. Dividing by $A\Delta z$ and taking the derivative gives the equation above. For constant mass density ${\rho }_f$, integration from the surface at $z_s$ givesThus pressure increases with depth $h$. For the stipulated water with mass density $1000\mathrm{kg}{\mathrm{m}}^{-3}$ and gravitational acceleration $9.81\mathrm{m}{\mathrm{s}}^{-2}$, depths $0.20\mathrm{m}$ and $0.30\mathrm{m}$ give gauge pressures $1962\mathrm{Pa}$ and $2943\mathrm{Pa}$. Their difference produces the buoyancy of the example cube. ^[ref]

In smooth barotropic inviscid flow with conservative body force, Kelvin's circulation theorem preserves circulation of material curves. Initially zero vorticity therefore gives zero circulation on contractible loops for all times. A loop winding around an excluded obstacle can have a nonzero constant circulation: the planar field $\mathbf{u}=\Gamma {\mathbf{e}}_{\theta }/(2\pi r)$ is irrotational outside the origin but has circulation $\Gamma$ on a positively oriented enclosing loop.