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For a twice continuously differentiable deformation map, each row of its cofactor matrix has zero divergence in reference coordinates. In three dimensions write $(\mathrm{cof}Dy{)}_{iA}=\frac{1}{2}{\epsilon }_{ijk}{\epsilon }_{ABC}{y}_{j,B}{y}_{k,C}$. Applying ${\partial }_A$ gives terms in which a symmetric pair of second partial derivatives contracts with an antisymmetric Levi-Civita symbol, so each term vanishes. This identity is the cancellation that makes the Piola-Kirchhoff stress tensor transform force balance correctly between configurations.

A symmetric thin viscous sheet with shear-free faces has leading longitudinal velocity independent of thickness coordinate $z$, and incompressibility gives $w=-z{u}_x$. Young–Laplace equation and the Newtonian fluid stress tensor give ${\sigma }_{zz}=-p_a+\gamma h_{xx}/2$ and ${\sigma }_{xx}=-p_a+\gamma h_{xx}/2+4\mu u_x$. The effective longitudinal tension per unit transverse width is $\mathcal{T}=4\mu h{u}_x+\gamma h{h}_{xx}/2-\gamma h_x^2/4+2\gamma$. The last two terms include the horizontal projections of the two surface tension pulls. Force balance gives ${\mathcal{T}}_x=0$, yielding the displayed longitudinal equation; the face kinematic conditions give the displayed conservation of mass equation.

For fixed plates $y=\pm h$, no slip, viscosity ${\mu }_0>0$, yield stress ${\tau }_0\ge 0$ and $G=-p_{,x}>0$, force balance gives shear stress $-Gy$. The central plug flow of a yield-stress fluid has half-width $y_0={\tau }_0/G$. If $y_0<h$, the velocity isIt is even in $y$; its derivative is $(-Gy+{\tau }_0\mathrm{sign}y)/{\mu }_0$ in the yielded layers and zero in the plug. If $Gh\le {\tau }_0$, there is no flow. Odd velocity is incompatible with this pressure-driven no-slip solution at nonzero flow.

For an admissible Fourier fluid Lagrangian displacement $\mathbf{\xi }(r)e^{im\theta +ikz}$, put $D_{\perp }=(r{\xi }_r{)}^{\prime }/r+im{\xi }_{\theta }/r$ and $D=D_{\perp }+ik{\xi }_z$. Force balance cancels every magnetic-field-gradient cross term in the magnetohydrodynamic energy principle, leaving an integrand $\gamma p|D{|}^2+B^2[|D_{\perp }{|}^2+k^2(|{\xi }_r{|}^2+|{\xi }_{\theta }{|}^2)]/(4\pi )$. For $p\ge 0$, $\gamma >0$, this is nonnegative. Neutral displacements can exist, so stability here means absence of negative-energy exponentially growing ideal modes, not strictly positive energy for every displacement.

For plates at y=+-h moving at velocities +-U and pressure gradient -G, force balance fixes the displayed stress. The power-law fluid law gives $u_y=(Gh/k{)}^{1/n}\mathrm{sgn}(\alpha -y/h)|\alpha -y/h{|}^{1/n}$. Integrating from the lower no-slip wall yields $u=-U+hn(Gh/k{)}^{1/n}[|\alpha +1{|}^{1+1/n}-|\alpha -y/h{|}^{1+1/n}]/(n+1)$. The upper wall fixes alpha uniquely for k,n,G,U>0. Absolute values are essential when the shear reverses inside the gap.

In an equal-density, no-slip lubrication approximation, repulsive disjoining pressure $A/(6\pi d^3)$ sustains a premelting film of thickness $d$ beneath a rigid disk of radius $a$. With imposed radial undercooling $\Delta T(1-r^2/a^2)$, local thermomolecular pressure is $P_T=\rho L\Delta T(1-r^2/a^2)/T_m$, and $d^3=A/(6\pi P_T)$. A constant vertical lifting rate $U$ requires inward volume flux $q_r=-Ur/2$. Integrating the resulting lubrication pressure and disk force balance gives the displayed rate, where $W$ is the apparent load, $\mu$ is liquid dynamic viscosity, $L$ is specific latent heat, and $T_m$ is the absolute bulk melting temperature. Positive lifting requires $W<\pi a^2\rho L\Delta T/(2T_m)$. For fixed positive load, the rate rises from zero at that threshold towards $A/(6\pi \mu a^2)$ as undercooling increases. That plateau is an asymptotic limit within this constitutive model, not a claim that continuum-film physics persists at arbitrarily low temperatures.

Write a small planar displacement in arclength coordinates as $X=s+X_2$ and $Y=y$. The inextensibility relation $X_s^2+Y_s^2=1$ gives $X_{2,s}=-y_s^2/2$ to quadratic order. Thus longitudinal material velocity enters one order later than the transverse velocity. For a periodic material stroke with no mean axial drift, its mean velocity vanishes, although its instantaneous contribution is needed for exact local force balance.

An axisymmetric small radial displacement $R(r)$ gives radial strain $R^{\prime }$ and hoop strain $R/r$. Radial force balance on an annular wedge gives ${\tau }_r^{\prime }+({\tau }_r-{\tau }_{\varphi })/r=0$. Under plane stress, isotropic Hooke's law gives ${\tau }_r=E(R^{\prime }+\nu R/r)/(1-{\nu }^2)$ and ${\tau }_{\varphi }=E(R/r+\nu R^{\prime })/(1-{\nu }^2)$. Substitution into equilibrium gives the displayed displacement equation. Its power-law exponents are $+1$ and $-1$, so $R=c_{1}r+c_2/r$. The hoop-stress solution is not based on treating a thick wall as a membrane.

The geometrical optics account of an optical trap adds the momentum flux changes of refracted and reflected light rays. It applies when the bead is large compared with the surrounding medium's wavelength. When the bead is much smaller, an induced electric dipole gives a different approximation. For a homogeneous dielectric sphere of radius $a$, define $\mathbf{p}(t)=\mathrm{Re}[\alpha {\mathbf{E}}_0{e}^{-i\omega t}]$, with ${\mathbf{E}}_0$ the peak complex electric field and $\alpha$ its SI polarizability. In a transparent, nonmagnetic, nondispersive host,where $n_p,n_m$ are particle and host refractive indices, $I$ is cycle-averaged light intensity, ${ϵ}_0$ is the vacuum permittivity and $c$ is the speed of light. The quasistatic electric dipole approximation requires $2\pi n_{m}a/{\lambda }_0\ll 1$, with ${\lambda }_0$ the vacuum wavelength, and a field nearly uniform across the bead.

For real quasistatic $\alpha$, building an induced electric dipole gives instantaneous potential energy $-\alpha E^2/2$. Averaging over one optical cycle givesThus $n_p>n_m$ gives positive polarizability and attraction toward larger light intensity; $n_p<n_m$ reverses this contribution. The factor $1/4$ uses a peak field: substituting a root mean square field without changing that factor would halve the force incorrectly. Dimensionally $\alpha |E_0{|}^2$ is energy, so its gradient is force.

This leading conservative term does not remove the optical scattering force. Even a nonabsorbing induced electric dipole reradiates light: at fixed wavelength and index contrast, its Rayleigh scattering cross-section scales as $a^6$, while $\alpha$ scales as $a^3$. A smaller bead can therefore improve the ratio of gradient to scattering forces while weakening the well relative to $k_{B}T$, where $k_B$ is the Boltzmann constant and $T$ the temperature. Optical force balance and resistance to thermal escape are separate questions. Intermediate bead sizes require a wave-scattering calculation; neither a ray calculation nor this small-dipole expression is uniformly valid there.

The Rüchardt experiment infers heat capacity ratio from the period of an oscillation of a piston supported by trapped gas. Use a vertical precision cylinder with a close-fitting movable piston, known moving mass $M$, face area $A$, trapped equilibrium volume $V_0$, and local atmospheric pressure $p_{\mathrm{atm}}$. With negligible static friction, force balance gives the absolute gas pressure $p_0=p_{\mathrm{atm}}+Mg/A$, where $g$ is gravitational acceleration. Displace the piston slightly and release it: compression raises pressure, and expansion lowers it, producing a restoring force. The oscillation is a mechanical probe of thermal response, without separately measuring two heat inputs. ^[ref]

Let $x$ be upward displacement from equilibrium, so $V=V_0+Ax$. For constant heat capacity ratio and a reversible adiabatic process, the net upward force, including the ambient load and weight, isThe linear approximation requires $|Ax|/V_0\ll 1$. Applying Newton's second law gives simple harmonic motion withHere ${\omega }_0$ is the undamped angular frequency and $\mathcal{T}$ the undamped oscillation period, rather than temperature. Record repeated cycles, measure $M,A,V_0,p_{\mathrm{atm}}$ independently, and include all connected trapped volume. An illustrative model with $M=0.020\mathrm{kg}$, $A=2.0\times 10^{-4}{\mathrm{m}}^2$, $V_0=0.010{\mathrm{m}}^3$, $p_{\mathrm{atm}}=100000\mathrm{Pa}$, $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ and $\gamma =1.4$ gives $p_0=100981\mathrm{Pa}$ and $\mathcal{T}\simeq 1.182\mathrm{s}$. These stipulated values are not measurements.

In Shearwood and Sloan's implementation, piston-mounted magnets induce an electromotive force in a stationary coil through Faraday's law of induction; an oscilloscope records the signal. They observed faster oscillations at smaller trapped volume values and stronger damping at small cylinder volume values. The signal tracks velocity locally, so its amplitude is not directly piston displacement. Calibration also identified inlet dead space beyond the graduated cylinder volume. ^[ref]

Repeat at smaller initial amplitudes to test the linear approximation, and compare several $V_0$ values. Heat transfer weakens the restoring force toward the isothermal process limit, biasing the inference toward $1$. Check leakage through resting-piston drift and before-and-after trapped volume readings. For viscous damping with envelope $e^{-t/\tau }$, the measured angular frequency obeys ${\omega }_0^2={\omega }_d^2+{\tau }^{-2}$; this correction does not remove thermal leakage. Internal equilibration must be faster than an oscillation period, with heat transfer and leakage slow over a run. Large amplitude or sticking friction require a richer model. ^[ref]

A secondary flow is a usually smaller velocity component superposed on a primary flow selected by symmetry or a simplified force balance. Examples include meridional circulation around a rotating sphere, or transverse circulation in a curved channel. Its direction depends on the forcing mechanism: elastic normal stresses and inertial effects can produce opposite circulations.

The sediment entrainment threshold is the bed shear stress at which resting grains begin moving. An elementary force balance using spherical grains, tangential quadratic drag, and static friction gives ${\Theta }_{\mathrm{th},0}=4{\mu }_s/(3C_D)$ when the drag coefficient is defined relative to the bed shear velocity.

For a horizontal field, set $C=B^2/{\mu }_0$. Static force balance gives $\delta \Pi =\rho g{\xi }_z-(\gamma p+C)\nabla \cdot \xi +Ci{k}_y{\xi }_y$. Given smooth compactly supported vertical and along-field displacements, the displayed choice makes total-pressure perturbation zero while ${\xi }_x=O(k_x^{-1})$. The nonnegative pressure and transverse-tension terms vanish as $|k_x|\to \infty$. This constructs the infimum needed for the magnetohydrodynamic energy principle; a strictly negative limit is already negative at some finite wavenumber.

A simply supported beam is prevented from moving transversely at its two supports while its ends remain free to rotate. In the small-deflection Euler-Bernoulli beam equation, a uniform span $0\le x\le L$ with transverse displacement $w(x)$ and positive beam flexural rigidity $B=EI$ has the support boundary conditionsHere $E$ is Young's modulus and $I$ is the second moment of area about the bending axis. The second pair expresses zero applied endpoint bending moment, since $M=-B{w}^{\prime \prime }$ in the convention used here. It does not set the endpoint beam cross-section rotation or slope to 0. A clamped boundary condition instead constrains both displacement and slope.

For an original derivation, apply a downward central point force $P$ and take $w$ positive downward. Assume small elastic deflections, constant $B$ and no axial preload. Symmetry and force balance give an upward reaction $P/2$ at each support. On the left half, the bending moment is $M(x)=Px/2$, so $w^{\prime \prime }=-Px/(2B)$. Integrate twice, using $w(0)=0$ and the symmetry condition $w^{\prime }(L/2)=0$, to obtainThus the midspan compliance of a simply supported beam isThe endpoint slopes are $w^{\prime }(0)=P{L}^2/(16EI)$ and $w^{\prime }(L)=-P{L}^2/(16EI)$: they are generally nonzero. This support geometry and central-load formula are used to infer material stiffness from a three-point bending test.^[ref]

The moment-free natural boundary condition for a beam also follows from the principle of virtual work. For a central load, the total potential energy is $\Pi [w]=\frac{1}{2}{\int }_0^{L}B(w^{\prime \prime }{)}^{2}dx-Pw(L/2)$. An admissible virtual displacement $\eta$ satisfies $\eta (0)=\eta (L)=0$, while its endpoint first derivatives remain free. Two integrations by parts produce the boundary contributionThe terms multiplying $\eta$ vanish because the endpoint displacements are prescribed. Arbitrary endpoint values of ${\eta }^{\prime }$ then require $B{w}^{\prime \prime }=0$ there. The free end rotation and zero end moment are conjugate parts of the same variational boundary condition.

Clamping both ends changes the central-load deflection to $P{L}^3/(192EI)$, one quarter of the simply supported value for the same $P,L,E,I$. It introduces endpoint reaction moments; it does not merely rename the same support. The clamped-beam Green function provides the corresponding fourth-order boundary-value solution. The support conditions likewise determine the simply supported beam normal modes, so a static stiffness measurement and a vibration measurement must use the same mechanical support model when they are compared.

A three-point bending test places a specimen on two separated supports and applies a central force. A load cell measures force, and a midspan displacement readout measures response. A direct plunger and extensometer can measure specimen deflection separately from machine crosshead displacement, as in the published Instron fixture arrangement.^[ref] Fitting a proportional small-load segment estimates $E=L^3/(48IC)$, where $C$ is the measured incremental midspan compliance. Record support spacing, load position and cross-section dimensions with that estimate.

For stipulated $L=0.50\mathrm{m}$, rectangular width $b=20\mathrm{mm}$, thickness $h=3\mathrm{mm}$ and $E=200\mathrm{GPa}$, the second moment of area is $I=b{h}^3/12=4.5\times 10^{-11}{\mathrm{m}}^4$ and $EI=9\mathrm{N}{\mathrm{m}}^2$. An incremental $1\mathrm{N}$ central load gives $0.289\mathrm{mm}$ incremental deflection under the simply supported model, versus $0.0723\mathrm{mm}$ for the clamped model. Incremental readings remove an unchanged baseline deflection in this linear model; support settling, contact indentation and large rotations require separate interpretation.

A spherical cavity of fixed reference radius $a$ in an infinite solid with perfect plasticity, the Tresca yield criterion, and zero far-field stress first yields under internal pressure $P_c$. In the plastic region $a<r<s$, radial force balance and ${\tau }_{\theta \theta }-{\tau }_{rr}=2{\tau }_Y$ give ${\tau }_{rr}=-P+4{\tau }_Y\log (r/a)$. Matching to the outer linear elasticity solution, where ${\tau }_{rr}(s)=-P_c$, gives the displayed plastic-front radius. These are small-strain reference-geometry formulae.

For an axisymmetric magnetic field $\mathbf{B}=\psi {\mathbf{e}}_{\varphi }/(r\sin \theta )+\nabla \times [\chi {\mathbf{e}}_{\varphi }/(r\sin \theta )]$, the poloidal magnetic flux function gives $B_r={\chi }_{\theta }/(r^2\sin \theta )$ and $B_{\theta }=-{\chi }_r/(r\sin \theta )$. Force balance implies locally $p=p(\chi )$ and $\psi =\psi (\chi )$ on regular connected flux surfaces. Direct calculation gives $(\nabla \times \mathbf{B})\times \mathbf{B}=-({\Delta }_{\ast }\chi +\psi {\psi }^{\prime })\nabla \chi /(r^2{\sin }^2\theta )$, with ${\Delta }_{\ast }\chi ={\chi }_{rr}+\sin \theta {\partial }_{\theta }({\chi }_{\theta }/\sin \theta )/r^2$. Equating this to ${\mu }_0\nabla p$ proves the displayed magnetostatic equilibrium equation, including its sign and dimensions.

Static friction is the tangential contact force that maintains zero contact slip. “Static” describes the contact state: both surfaces can move together in the chosen inertial frame. In the ideal dry-contact model, the actual static friction adjusts to the equations of motion, subject to a bound, rather than always taking its maximum value.^[ref]Here $N$ is the magnitude of the normal force and ${\mu }_s$ the coefficient of static friction. Equality means limiting static friction; it is a threshold, not an instruction to replace every unknown contact force by ${\mu }_{s}N$.

For a stipulated example, place a nonrotating $2\mathrm{kg}$ block initially at rest on a stationary horizontal surface. Take gravitational acceleration $g=9.8m/{\mathrm{s}}^2$ and coefficient of static friction ${\mu }_s=0.40$, with no other vertical force. The vertical force balance gives $N=mg=19.6\mathrm{N}$, so the largest allowed static friction magnitude is $7.84\mathrm{N}$. Choose right as positive and apply a horizontal force $F$. Remaining at rest requires Newton's second law to giveAt $F=2\mathrm{N}$, the static friction is $-2\mathrm{N}$; at $F=6\mathrm{N}$, it is $-6\mathrm{N}$. Both states have zero acceleration. At $F=10\mathrm{N}$, balancing would require $f_s=-10\mathrm{N}$, outside the bound, so this initially resting block cannot remain at rest. Its subsequent acceleration requires a kinetic friction law; setting $f_s=-7.84\mathrm{N}$ after contact slip begins would still use the wrong contact state. At exactly $F=7.84\mathrm{N}$, the ideal inequality still permits rest at the threshold. These are model calculations, not measured material properties.

The toy model used in kinetic friction makes the contact-state distinction explicit: $m=1\mathrm{kg}$, $N=10\mathrm{N}$, ${\mu }_s=0.5$ and coefficient of kinetic friction ${\mu }_k=0.3$. On stationary ground, an initially resting box pushed right with $4\mathrm{N}$ has $f_s=-4\mathrm{N}$. At $6\mathrm{N}$ it cannot remain at rest; once it slides right, $f_k=-3\mathrm{N}$ and $a=3m/{\mathrm{s}}^2$. At $5\mathrm{N}$, rest is allowed with $f_s=-5\mathrm{N}$, whereas an already right-sliding box has $f_k=-3\mathrm{N}$ and $a=2m/{\mathrm{s}}^2$. The applied force alone does not specify the friction branch: contact slip and motion history matter.

For a flat contact with two tangential directions, the bound applies to a vector norm, not separately to each component:For the stipulated $7.84\mathrm{N}$ bound, balancing horizontal components $4\mathrm{N}$ and $4\mathrm{N}$ requires a force magnitude $\sqrt{32}\mathrm{N}\simeq 5.657\mathrm{N}$, which is allowed. Components $6\mathrm{N}$ and $6\mathrm{N}$ require $\sqrt{72}\mathrm{N}\simeq 8.485\mathrm{N}$, which is forbidden even though each component separately is below $7.84\mathrm{N}$. This circular bound assumes the same coefficient of static friction in every tangential direction. Near impending contact slip, the limiting static friction opposes the relative contact motion that would otherwise begin. Once actual contact slip exists, kinetic friction instead opposes its tangential relative velocity, which need not have the direction of the body's center of mass motion.

The normal force must also come from the equations of motion. On horizontal ground, an upward pull of magnitude $P$ at angle $\alpha$ above horizontal gives, for a block remaining at rest with no other forces,The sine component unloads the contact and the cosine component demands tangential support. These equations require $N>0$; if the supporting contact opens, this dry-contact model cannot supply a negative normal force. For the stipulated $2\mathrm{kg}$ block and $\alpha =30^{\circ }$, the positive threshold is $P={\mu }_{s}mg/(\cos \alpha +{\mu }_s\sin \alpha )\simeq 7.3544\mathrm{N}$, with $N\simeq 15.9228\mathrm{N}$ and $|f_s|\simeq 6.3691\mathrm{N}$. A downward pull increases $N$ instead. An inclined plane gives another reason why $N$ is not always $mg$.

To solve a new problem, impose zero contact slip, find the required static friction and normal force from the equations of motion, then check the bound. A demand inside it permits that contact state; a demand outside it rules the state out. For an ideal driven tire undergoing rolling without slipping on stationary ground, a forward static friction can accelerate the vehicle. Its role is to prevent contact slip, rather than to oppose all motion of the whole body.

Nor does the word “static” by itself make friction do zero work. The contact power delivered to a body is $P_s={\mathbf{f}}_s\cdot {\mathbf{v}}_c$, a dot product with its material contact-point velocity in the chosen inertial frame. At an ideal contact undergoing rolling without slipping on stationary ground, ${\mathbf{v}}_c=0$, so this power is zero. For a rigid body, adding the translational and rotational contributions gives ${\mathbf{f}}_s\cdot {\mathbf{V}}_{\mathrm{CM}}+{\mathbf{\tau }}_{\mathrm{CM}}\cdot \mathbf{\omega }={\mathbf{f}}_s\cdot {\mathbf{v}}_c$, where ${\mathbf{\tau }}_{\mathrm{CM}}$ is the contact torque about the center of mass and $\mathbf{\omega }$ its angular velocity. Counting only the first term omits the rotational contribution.

At a moving contact with zero contact slip, the two material contact points share a velocity ${\mathbf{v}}_c$ that can be nonzero. For instance, a nonrotating block moving with an accelerating belt receives positive contact power $f_{s}U$ when the belt's rightward velocity $U$ and rightward static friction $f_s$ are positive. The belt receives $-f_{s}U$ by Newton's third law. These contact contributions sum to zero in the ideal no-slip model, while the block's kinetic energy can increase. Zero contact slip and zero work on each body are different conditions.

An inclined plane tilted slowly from horizontal turns a coefficient of static friction into a critical angle. Assume a nonrotating block on a stationary, rigid, dry surface, with gravitational weight and the contact forces as its only forces. Take $0\le \theta <\pi /2$ to be the angle between the uphill surface direction and horizontal. Let ${\mathbf{e}}_t$ be a unit vector pointing uphill and ${\mathbf{e}}_n$ a perpendicular unit vector pointing outward from the surface. Resolving the gravitational weight givesThe angle between the downward vertical and the inward normal $-{\mathbf{e}}_n$ is also $\theta$. Thus the inward component uses cosine and the downhill component uses sine. These are components of one force, not two additional forces.

Before downhill contact slip, zero acceleration and Newton's second law requireThe required static friction points uphill. Substituting into the bound and dividing by the positive $mg\cos \theta$ givesThe tangent function increases on this range of angles, so the last admissible resting angle isThe inverse tangent takes a dimensionless quantity to an angle. At equality the ideal model still permits a limiting resting state; increasing $\theta$ further excludes it. At a smaller angle, $f_s/N=\tan \theta$ is the demand ratio, not the coefficient of static friction.^[ref]

For the stipulated $m=2\mathrm{kg}$, $g=9.8m/{\mathrm{s}}^2$ and ${\mu }_s=0.40$, the critical angle is approximately $21.8014^{\circ }$. At $\theta =15^{\circ }$, $N\simeq 18.932\mathrm{N}$ and $f_s\simeq 5.073\mathrm{N}$, comfortably below the bound ${\mu }_{s}N\simeq 7.573\mathrm{N}$. At $\theta =25^{\circ }$, maintaining rest would require $f_s\simeq 8.283\mathrm{N}$, but the bound is only $7.105\mathrm{N}$: rest is impossible. At ${\theta }_c$, $N\simeq 18.198\mathrm{N}$ and $f_s\simeq 7.279\mathrm{N}$. Notice that the limiting tangential force here is smaller than the horizontal $7.84\mathrm{N}$ limit because tilting reduces the normal force.

Cancellation of mass makes this relation useful: the ideal critical angle tests a contact coefficient without needing a calibrated force gauge. Doubling the block's mass doubles both demanded static friction and its bound, leaving ${\theta }_c$ unchanged if ${\mu }_s$ is unchanged. This is a testable model prediction. It need not survive a change in surface condition, a direction-dependent coefficient, appreciable deformation, or vibration.

The derivation also assumes continued contact and that sliding occurs before tipping or rolling without slipping. The resultant contact force must act within the supporting footprint so that the block can satisfy torque balance; a tall block may tip first. No extra tension, imposed tangential pull, or lift is included. In a slowly raised apparatus, quasistatic process means that motion of the board contributes negligible inertial forces while locating the threshold. Once downhill contact slip begins, a constant-velocity sliding angle would concern kinetic friction, not the onset of static friction.

Parameterize a fixed yield surface by stress-space arc length $l$ with ${\xi }_{,l}=-\cos 2\varphi$ and ${\tau }_{,l}=-\sin 2\varphi$. For mean normal stress $\sigma$, force balance gives $\nabla \sigma =M\nabla l$, where $M=\left(\begin{array}{cc}\cos 2\varphi & \sin 2\varphi \\ \sin 2\varphi & -\cos 2\varphi \end{array}\right)$. Its eigenvectors along angles $\varphi$ and $\varphi +\pi /2$ have eigenvalues $+1$ and $-1$, proving the displayed slip line invariants.

For a heterogeneous SIS model with common recovery rate and a positive endemic equilibrium $I_{\ast }$, force balance gives $\mathrm{diag}(s_{\ast })K{I}_{\ast }=I_{\ast }$. When the transmission matrix is irreducible and nonnegative, the positive equilibrium vector and the Perron–Frobenius theorem imply the displayed spectral radius identity. It replaces the homogeneous scalar relation $s_{\ast }=1/R_0$: different recipient types generally have different susceptible fractions.