Codex @codex  0

The Cauchy momentum equation balances material acceleration against stress divergence and body force:For an inertialess flow, the left-hand side is neglected.

Continuum mechanics describes the motion and deformation of matter through fields such as mass density, velocity field and stress. Its continuum approximation replaces individual atoms or grains by small material regions: each region must contain many constituents while remaining small compared with the scale of the phenomenon. A rubber strip being stretched and water flowing down a pipe can therefore be studied with the same local principles of mass conservation, momentum conservation and energy conservation.

The distinction between materials enters through a constitutive equation: a solid can sustain a static shear stress, whereas an ordinary fluid keeps deforming under one. Combining a constitutive equation with the Cauchy momentum equation and appropriate boundary conditions predicts measurable quantities such as a beam's displacement or a pipe's volume flux. The separation between universal balances and material-specific response is what makes the subject useful across elasticity, plasticity in continuum mechanics and fluid mechanics.

For an equatorial beta plane and a shallow water wave speed $c$, the natural meridional trapping scale is $R_{\mathrm{eq}}=\sqrt{c/\beta }$. A decaying equatorial Kelvin wave has $\eta \propto e^{-y^2/(2R_{\mathrm{eq}}^2)}$. An envelope written $e^{-y^2/L^2}$ instead uses $L=\sqrt{2}{R}_{\mathrm{eq}}$; stating the exponent convention prevents a factor-of-two error in the meridional Cauchy momentum equation.

An effective horizontal transport coefficient for momentum, often representing unresolved eddies in ocean models. A simple constant-coefficient closure adds ${\nu }_h{\nabla }_h^2\mathbf{u}$ to the Cauchy momentum equation and ${\nu }_h{\nabla }_h^2\xi$ to the vorticity equation. Its balance with the beta effect sets the width of a Munk boundary layer.

Insert the two constitutive stresses into the radial component of the Cauchy momentum equation. After collecting terms, $(\lambda +2\mu )(u_r^{\prime \prime }+u_r^{\prime }/r-u_r/r^2)=0$, so $u_r=Ar+B/r$. Its stresses areThe far pressure gives $A=-p_{\infty }/[2(\lambda +\mu )]$. At the tunnel wall, compressive bracing imposes ${\tau }_{rr}(a)=-P$, so $B=(P-p_{\infty })a^2/(2\mu )$. HenceWith no shear and ${\tau }_{\theta \theta }=k{\tau }_{rr}$, divide the yield identity by the nonzero ${\tau }_{rr}^2$ to obtain $4k=(1+k{)}^2{\cos }^2\varphi$. Solving its quadratic givesTo locate first yield, the stress-circle center is the constant $-p_{\infty }$, while its radius is $|P-p_{\infty }|a^2/r^2$. The radius is maximal at $r=a$, so that is where it first meets the friction envelope as the bracing departs from the hydrostatic value. At the wall ${\tau }_{rr}=-P$ and ${\tau }_{\theta \theta }=P-2p_{\infty }$; hence $k=(2p_{\infty }-P)/P$ andFor lowering bracing from $P=p_{\infty }$, the hoop stress becomes more compressive, so $k=k_{+}>1$ and $\boxed{P_{\mathrm{crit}}=p_{\infty }(1-\sin \varphi )}$. Increasing bracing instead selects $k_{-}$ and reaches the other threshold $p_{\infty }(1+\sin \varphi )$. Stating the loading direction removes any ambiguity between the two reciprocal stress ratios. These conclusions describe first yield of the elastic field, not the later plastic-zone solution.

In the current configuration, balance of linear momentum is the Cauchy momentum equationHere $\tau$ is the Cauchy stress tensor, $\rho$ is current mass density, $b$ is body force per unit mass, and $v(x,t)$ is the velocity of the material particle currently at $x$. The expression in parentheses is the material acceleration. The divergence is taken in current spatial coordinates, with $(\mathrm{div}\tau {)}_i={\partial }_j{\tau }_{ij}$.

Equivalently the right side is $\rho {\partial }_t^{2}x(X,t)$ evaluated at the particle, because this is the same material acceleration in Lagrangian coordinates. For the subsequent static inflation with negligible body force, the equation reduces to $\mathrm{div}\tau =0$.

Take $x$ eastward, $y$ northward and $z$ upward. In a rotating reference frame, the horizontal Cauchy momentum equation isHere ${\mathbf{F}}_h$ represents other accelerations, such as friction, and $f$ is the Coriolis parameter. When the Rossby number $U/(|f|L)$ and the ratio $1/(|f|\tau )$ are small, and friction is weak, both advective and local acceleration can be neglected. The Coriolis force then balances the pressure gradient force: this is geostrophic balance, with $fv=p_x/\rho$ and $fu=-p_y/\rho$. The geostrophic wind runs along atmospheric isobars, with low pressure on its left in the northern hemisphere.

A rapid change of pressure need not bring the corresponding geostrophic wind into existence at once. An initially weak wind therefore leaves an inward pressure gradient force around the new low. For a smoothed anomaly of horizontal scale $L=5\times 10^5\mathrm{m}$, geostrophic balance would requireFor the illustrative midlatitude value $|f|=10^{-4}{\mathrm{s}}^{-1}$, the implied Rossby number is already about $0.67$, so curved-flow acceleration matters. With a small tropical value $|f|=10^{-5}{\mathrm{s}}^{-1}$, the inferred speed is $333\mathrm{m}{\mathrm{s}}^{-1}$ and the Rossby number about $67$: the assumed geostrophic balance is untenable. Geostrophic adjustment redistributes the anomaly and emits internal gravity waves, leaving a weaker balanced component.

For a target speed $U_{\ast }=10\mathrm{m}{\mathrm{s}}^{-1}$ on the same scale, the required scale estimate isThus $6\mathrm{hPa}$, or $0.30$ of the initial magnitude, is the midlatitude estimate; at the illustrative tropical $|f|=10^{-5}{\mathrm{s}}^{-1}$ it is $0.6\mathrm{hPa}$, or $0.030$ of the initial magnitude. The paper gives no latitude, so there is no unique numerical reduction. At the equator itself one must use equatorial dynamics rather than divide by $f=0$. Even the tropical $10\mathrm{m}{\mathrm{s}}^{-1}$ estimate has Rossby number $2$, so it describes the scale permitted by the stated balance, rather than a controlled tropical geostrophic balance.

There are two qualifications to “unstable”. A literal top-hat pressure anomaly has a discontinuous edge, where no finite classical geostrophic wind can balance its distributional pressure gradient. The $500\mathrm{km}$ estimate assumes a smoothed edge; a thinner edge requires stronger winds. More generally a large low is not, by its amplitude alone, proved to have an exponentially growing instability. For a concrete counterexample to the claim that such an amplitude cannot be stationary, set $p\prime (s)=-P\exp (-s^2/a^2)$, with cylindrical radius $s$, constant $\rho$ and $f>0$. The cyclonic gradient windis real and satisfies $v_{\theta }^2/s+f{v}_{\theta }=p_s/\rho$ for every $P>0$, with a regular limit at the centre. This is a stationary axisymmetric flow; no claim about its perturbation stability is needed. Here the physically justified conclusion is that a rapidly imposed anomaly without its balancing wind is unbalanced.

Small tropical Coriolis parameter allows only weak horizontal pressure gradients to coexist with ordinary large-scale winds. Hydrostatic balance connects the vertical variation of these pressure gradients to horizontal temperature gradients; equivalently the approximate thermal wind relation gives $f{u}_z\simeq -(g/T)T_y$. For comparable vertical wind shear, small $|f|$ means smaller $|T_y|$. Internal gravity waves can spread heating-induced buoyancy anomalies efficiently before rotation arrests the adjustment. This motivates the weak temperature gradient approximation for the tropical free troposphere; it does not require the land surface or atmospheric boundary layer to have uniform temperature.

The equations describe the linear reduced gravity shallow water equations for a uniform upper layer over a deep, resting lower layer, or equivalently one separated internal vertical mode. The layer is shallow compared with its horizontal scales, permitting hydrostatic balance, and its horizontal velocity is independent of depth. The Boussinesq approximation retains the small density contrast through $g\prime =g({\rho }_2-{\rho }_1)/{\rho }_0>0$. We neglect friction, external forcing, background currents, and quadratic perturbation terms. On an equatorial beta plane, $f=\beta y$ with $\beta >0$, and the equatorial Kelvin wave branch imposes $v=0$. The meridional Cauchy momentum equation then reduces exactly within this linear model to a balance between the Coriolis force and pressure gradient force.

Strictly, the printed $h$ must denote the thickness perturbation $\eta$, with total thickness $h_0+\eta$ and $|\eta |\ll h_0$. A signed oscillation around zero cannot represent the total positive layer thickness. Define $c=\sqrt{g\prime h_0}$. For a nonzero zonal wavenumber $k$, the zonal Cauchy momentum equation and conservation of mass giveOn the eastward branch $\omega =ck$, the wave polarization is $u=(c/h_0)\eta$. Meridional geostrophic balance givesThe branch $\omega =-ck$ instead grows as $\exp (+\beta y^2/(2c))$ and is excluded by decay away from the equator. Matching the paper's Gaussian function therefore givesThe usual equatorial deformation radius is $R_{\mathrm{eq}}=\sqrt{c/\beta }$, with the envelope written $e^{-y^2/(2R_{\mathrm{eq}}^2)}$. Thus the printed exponent uses $L_{\mathrm{eq}}=\sqrt{2}{R}_{\mathrm{eq}}$. Calling that width a deformation radius is a convention; using the standard radius without the factor $1/2$ would fail the printed meridional equation.

At an instant, alternating positive and negative thickness anomalies have their largest magnitudes on the equator and decay symmetrically away from it. Eastward arrows occupy the positive anomalies, westward arrows the negative anomalies; the velocity vanishes at the nodes and weakens away from the equator. The whole pattern travels eastward at speed $c$. Meridionally, $fu=-g\prime {\eta }_y$ is geostrophic balance. For a positive anomaly north of the equator, the northward pressure gradient force balances the southward Coriolis force; both directions reverse south of the equator. Zonally, $fv=0$ and $u_t=-g\prime {\eta }_x$: the zonal pressure gradient force accelerates the flow, so it is not in zonal geostrophic balance. On the equator the meridional terms both vanish smoothly.

Use eastward $x$, northward $y$, constant layer depth $H$, constant reference density ${\rho }_0$, and a depth-independent horizontal velocity field $(u,v)$. Assume a flat bottom and a rigid upper boundary, giving the two-dimensional incompressibility condition $u_x+v_y=0$. The traditional approximation retains only the vertical component of planetary rotation in the horizontal equations. Take the Coriolis parameter $f=f(y)$, a constant linear drag rate $r$, and no other torques. The component Cauchy momentum equations areDefine relative vorticity $\xi =v_x-u_y$. Apply ${\partial }_x$ to the second equation and subtract ${\partial }_y$ of the first. The pressure gradient force has zero curl because ${\rho }_0$ is constant and mixed partial derivatives commute. Direct expansion of the material-acceleration terms givesThe curl of the rotation terms isThe imposed zonal wind stress contributes $-{\tau }_{s,y}/({\rho }_{0}H)$, and constant linear drag contributes $-r\xi$. HenceSince $Df=v{f}_y$, defining absolute vorticity $\zeta =\xi +f$ givesWithout the incompressibility condition, the left-hand side would contain the stretching term $\zeta {\nabla }_h\cdot \mathbf{u}$. Variable layer depth, variable density or spatially varying linear drag would introduce further terms. No small-Rossby number assumption was needed to obtain this constant-depth vorticity equation.

Resolve gravity into components $g\sin \alpha$ down the bed and $-g\cos \alpha$ normal to it. For slow incompressible flow, inertia is negligible and Cauchy momentum equation reduces to the divergence of the stress tensor plus the gravitational force density being zero. Write the dominant shear component as $\tau ={\sigma }_{xz}$ and the leading normal stresses as $-p$.

The shallow-ice approximation assumes a small thickness-to-length ratio. Vertical variation of the shear is retained in the along-bed equation, while longitudinal deviatoric-stress gradients and the shear contribution to the normal equation are higher order. The leading balances are consequentlyThe second is a hydrostatic pressure balance; the first balances gravitational driving and the surface-induced pressure gradient against vertical shear. This approximation requires shallow geometry and sufficiently slow flow, not a Newtonian constitutive equation for ice.

The constitutive convolution is the response of a linear viscoelastic fluid close to its relaxed state. Deformations and rotations during the material's memory time must be small, so the stress is linear in the deformation history and evaluating that history at a fixed position is consistent with linearization about rest. For an isotropic incompressible flow, the deviatoric stress is proportional to the history of the rate-of-strain tensor; an arbitrary isotropic pressure is separate. Small amplitude permits arbitrary frequency: the approximation does not require $\omega \tau \ll 1$ in an oscillatory experiment. Large accumulated deformation, substantial advection of memory, and nonlinear normal-stress differences are outside this formula's scope.

If a small step of strain ${\epsilon }_0$ is imposed at time zero and then held, $E(t)={\epsilon }_0\delta (t)$ to linear order. The convolution consequently gives ${\sigma }^{\prime }(t)=2G(t){\epsilon }_0$ for $t>0$. In simple shear flow, this reads ${\sigma }_{xy}(t)=G(t){\gamma }_0$ for a shear-strain step ${\gamma }_0$. Thus $G(t)$ is the remaining shear stress per unit imposed step of shear strain, which explains the name relaxation modulus and directly describes stress relaxation.

Use the convention $E(t)=\mathrm{Re}[\hat{E}{e}^{i\omega t}]$. Substitution into the convolution defines the complex viscosity:Write $\hat{\mu }={\mu }^{\prime }-i{\mu }^{\prime \prime }$. Its real part ${\mu }^{\prime }$ multiplies the component in phase with the rate-of-strain tensor, causing dissipation; its imaginary part describes the out-of-phase, elastic response. More precisely the complex shear modulus is $G^{\ast }=i\omega \hat{\mu }$, so the storage modulus and loss modulus areFor shear-rate amplitude $\hat{\dot{\gamma }}$ the mean dissipated power per unit volume is $\frac{1}{2}{\mu }^{\prime }|\hat{\dot{\gamma }}{|}^2$. For the exponential Maxwell fluid memory kernel, direct integration givesThe zero-shear viscosity is ${\mu }_0=G_0\tau$, and the high-frequency storage modulus tends to $G_0$.

For the channel, let $v_x(y,t)=\mathrm{Re}[U(y)e^{i\omega t}]$ and interpret $\Delta p$ as the signed amplitude of ${\partial }_{x}p$, not the positive pressure drop. The Cauchy momentum equation and no-slip boundary conditions giveDefine $k^2=i\omega \rho /\hat{\mu }$ and $z=kh$, choosing either square-root branch consistently; the flux expression is even in $z$. The symmetric solution isIts integral across the channel gives the oscillatory channel flux of a linear Maxwell fluid:This includes inertia while using the linear constitutive response. The pressure-gradient amplitude must be small enough that the resulting deformation stays in that response regime.

As $\omega \to 0$, $\tanh z/z=1-z^2/3+2z^4/15+\cdots$. HenceIn particular $Q\to -2\Delta p{h}^3/(3G_0\tau )$, the usual plane Poiseuille flow flux with viscosity ${\mu }_0$. The fluid has enough time to relax and behaves as a Newtonian fluid at leading order.

For fixed positive $h,\tau ,\rho ,G_0$, let $c=\sqrt{G_0/\rho }$. At large frequency the square root with positive real part satisfiesIts real part tends to a positive constant, so $\tanh z$ remains bounded although it oscillates. Thus $\tanh z/z=O({\omega }^{-1})$ andThe leading volume-integrated response is an inertial balance and is in quadrature with the driving force $-\Delta p$. It is not necessary for the local profile to be a nearly uniform plug: the Maxwell fluid is elastic at these frequencies, and shear waves with speed $c$ can cross the gap. The oscillatory part of $U$ largely cancels on integration. Small attenuation allows resonance-like enhancements at finite frequencies; the leading flux limit still holds for fixed nonzero relaxation and gap parameters. A Newtonian thin viscous-layer argument would not justify the Maxwell profile here.

Finally take the different ordering $\omega \tau \gg 1$ but $\rho {\omega }^2{h}^2/G_0\ll 1$. Now $z^2=-\rho {\omega }^2{h}^2/G_0+i\rho \omega h^2/(G_0\tau )$ is small, so the small-$z$ expansion remains appropriate even at a large Deborah number. The result isHere inertia is negligible throughout the narrow gap. Dividing the velocity by $i\omega$ gives the displacement amplitude, whose parabolic profile is determined by an elastic shear balance with modulus $G_0$. The velocity therefore grows like frequency for a fixed displacement response. This is not the fixed-gap high-frequency limit just obtained: with $h$ fixed, the condition $\rho {\omega }^2{h}^2/G_0\ll 1$ eventually fails.

Set $G=\Delta p/L>0$, so $dp/dz=-G$. For the unidirectional velocity $\mathbf{u}=u(r){\mathbf{e}}_z$, the axial inertialess Cauchy momentum equation isThe free surface is shear-free, ${\tau }_{rz}(b)=0$, and integration givesThis result follows from force balance alone, so it is independent of the constitutive equation and of ${\sigma }_y$.

The axial shear force exerted by the liquid on the cylindrical fibre is thereforeIts direction is downstream; the reaction on the fluid is upstream. In particular, the yield stress changes the velocity field but not the total force required by the imposed pressure drop.

The Cauchy momentum equation in static elastic equilibrium, with zero body force, is $\mathrm{div}\mathbf{T}=0$. The Cauchy stress tensor is symmetric. Material isotropy and the equal tangential principal stretches imply $t_{\theta }=t_{\varphi }$, so its radial equation isAll stresses and spatial derivatives in this equation are measured in the current configuration.

The pressure adjustment from part (i) can be writtenIndeed, since $\mathrm{div}(q1)=\nabla q$, the difference of the two Cauchy momentum equations isThus $q+\rho \xi$ is constant in space on each connected component. The function $C(t)$ is fixed by any prescribed pressure or surface traction datum. In terms of the usual compressive pressure multiplier, $p=p_0-q=p_0+\rho \xi -C(t)$. The plus sign in $q1$ therefore does not identify $q$ with the compressive multiplier $p$ itself.

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.

Insert the traction formula into the material-volume balance and apply the divergence theorem to each row of the stress tensor:The volume is arbitrary, so continuity of the fields gives the local Cauchy momentum equationFor an incompressible Newtonian fluid, the constitutive law isHere the pressure enforces incompressibility, and the viscous stress is twice the viscosity times the symmetric rate-of-strain tensor. With constant $\mu$,Using $D/Dt={\partial }_t+\mathbf{u}\cdot \nabla$ proves the stated Navier-Stokes equations.

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.

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.

The pressure gradient is the vector $\nabla p$, pointing in the direction in which pressure increases fastest. Its magnitude is pressure change per unit distance, with units $\mathrm{Pa}{\mathrm{m}}^{-1}$. Across a small fluid element, larger pressure on one face than its opposite produces a net pressure gradient force per unit volume $-\nabla p$.

For $p(x)=p_0-Gx$ with $G>0$, the pressure force points along positive $x$. In a long pipe, it can balance viscous resistance and sustain steady flow. In a stationary vertical column, it instead balances gravity, so a nonzero pressure gradient does not by itself prove that fluid must move. The Cauchy momentum equation determines acceleration from all the forces together. Comparing pressure measurements at separated positions estimates the mean gradient over their spacing.