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The integral and differential forms of Gauss's law in electrostatics areHere $\partial \mathcal{V}$ is a closed surface bounding a volume $\mathcal{V}$, $d\mathbf{A}=\hat{\mathbf{n}}dA$ is its outward area vector, $\mathbf{E}$ is the electric field, $\rho$ is total volume charge density, and $Q_{\mathrm{enc}}={\int }_{\mathcal{V}}\rho d^{3}r$ is enclosed electric charge. The divergence theorem converts one form to the other. For ideal sheets or point charges, interpret the differential equation with their distributional charge densities.

In SI units, Maxwell's equations in vacuum, allowing charge density $\rho$ and electric current density $\mathbf{J}$, areThe source-free specialization sets $\rho =0$ and $\mathbf{J}=0$; it cannot describe the charged objects at their locations. Applying the divergence theorem to the first equation gives Gauss's law:For an isolated point charge $q_1$, spherical symmetry makes the electric field radial and constant in magnitude on a concentric sphere. Therefore $4\pi r^2{E}_r=q_1/{ϵ}_0$. Multiplying by a second point charge gives Coulomb's law,The force points away from the first charge for like signs and toward it for opposite signs.

For the spherical capacitor, a Gaussian sphere between the shells encloses $q$, so $E_r=q/(4\pi {ϵ}_0{r}^2)$ for $a<r<b$. Integrating the electric field inward from the outer shell yieldsAn arbitrary constant $V(b)$ remains unless a reference is specified. The capacitance is determined by the potential difference:For the subsequent charging operation, take the initial outer shell to be earthed, or interpret the stated $V$ as the inner-to-outer voltage with the outer reference zero. Isolating the inner shell fixes its electric charge at $q=CV$. Removing the outer shell leaves an isolated conducting sphere. Its electric field and electric potential, now with zero at infinity, areConsequentlyThese changes of electric potential are compatible with constant electric charge. Without the initial reference convention the absolute stated $V$ does not determine $q=CV$: for example, if the outer shell were initially isolated and neutral, the absolute inner electric potential would instead be $q/(4\pi {ϵ}_{0}a)$.

Finally take the connected spheres to be sufficiently far apart that the long wire's capacitance and their mutual influence can be neglected. Their common electric potential requiresso $q_a=q/3$ and $q_{2a}=2q/3$. The magnitudes of the surface electric fields areThis is the usual widely separated-sphere approximation. A wire enforces an equipotential, but alone does not make this ratio exact for arbitrary arrangements. As a concrete counterexample, if the second shell enclosed the first concentrically and they were connected, all net electric charge would reside on the exterior and the inner sphere's surface electric field would be zero.

Stokes theorem converts circulation around an oriented boundary into curl flux through a spanning surface. For a sufficiently smooth vector field on an oriented piecewise smooth surface $S$,The boundary orientation follows the right-hand rule relative to $\mathbf{n}$. The field must be smooth on the surface, not merely on its boundary; a singularity crossing the surface cannot be ignored. For a fixed loop, applying the theorem to Faraday's differential law gives the integral flux law of induction.

The planar form is Green theorem. For a region $D$ with positively oriented boundary and sufficiently smooth $P,Q$,The outer boundary is counterclockwise and any holes have clockwise boundaries. For instance $P=-y/2,Q=x/2$ makes the area of $D$ equal to half the line integral of $xdy-ydx$; the unit circle gives area $\pi$.

A closely related use of “Green's theorems” in field theory is Green's identities, obtained by applying the divergence theorem to $\varphi \nabla \psi$:and, after subtracting the expression with $\varphi ,\psi$ exchanged,These relate interior sources to boundary data in electrostatics and other boundary-value problems. For example, a harmonic potential difference vanishing on a Dirichlet boundary condition has zero integral of its squared gradient by the first identity, establishing uniqueness. Both the planar circulation theorem and the field identities are included here to make the convention clear.

Gauss's law connects electric flux through a closed surface to its enclosed charge:The differential form follows from the divergence theorem. The law is exact; extracting a field magnitude by simple division additionally needs symmetry. For spherical symmetry, $E_r(r)=Q(r)/(4\pi {ϵ}_0{r}^2)$; a uniformly charged ball therefore has a field linear in $r$ inside and an inverse-square field outside. An infinite line of linear charge density $\Lambda$ gives $E_r=\Lambda /(2\pi {ϵ}_{0}r)$ by a cylindrical surface. A uniformly charged infinite sheet gives $E=\sigma /(2{ϵ}_0)$ on either side by a pillbox.

An electrostatic conductor has zero interior field and constant potential; excess charge occupies its surface. A pillbox gives $E_{n,\mathrm{out}}=\sigma /{ϵ}_0$ when the inside is conducting. More generally, $\hat{\mathbf{n}}\cdot ({\mathbf{E}}_2-{\mathbf{E}}_1)=\sigma /{ϵ}_0$. For dielectric matter, $\nabla \cdot \mathbf{D}={\rho }_{\mathrm{free}}$ separates bound polarization charge from free charge. Gauss's law alone does not establish a uniform field for an arbitrary finite plate or recover a detailed nonsymmetric charge distribution from one flux measurement.

Insert $\nabla \times \mathbf{E}=-{\partial }_t\mathbf{B}$ and $\nabla \times \mathbf{H}=\mathbf{J}+{\partial }_t\mathbf{D}$ into the given identity. For a linear, time-independent, nondispersive medium, $\mathbf{D}=ϵ\mathbf{E}$ and $\mathbf{B}=\mu \mathbf{H}$, givingThe divergence theorem then yieldsThe left side is outward electromagnetic power flow. The current term is power delivered to matter, becoming Joule heating for an Ohmic conductor, and ${\partial }_{t}u$ is the change in stored electric plus magnetic energy. The minus sign says outward flux reduces stored energy unless supplied locally.

Follow the drawing's coordinates: lower strip at $z=0$, upper strip at $z=d$, current $I$ along $+x$ in the upper strip and back along the lower. The wording “both strips lie in the $xy$-plane” must mean parallel to that plane, since the drawing separates them in $z$. Neglect the side and end fringes using $l\gg a\gg d$. Then $I=V_0/(R+R_L)$ and the top-strip potential is $V(x)=V_0-IRx/l$, while the lower strip is equipotential at zero.

The harmonic function matching those two surfaces isIndeed ${\nabla }^2\varphi =0$, and it has the required potentials at $z=0,d$. ThusThe two current sheets reinforce their magnetic field between the strips; its direction follows the right-hand rule. The Poynting flow in resistive parallel strips follows from their cross product:Power flows along the gap towards the load and upwards into the resistive strip. Across a transverse area $ad$, $P_x(x)=IV(x)$ falls from $I{V}_0$ to $I^2{R}_L$. The upper surface receives ${\int }_0^{l}a{S}_z(z=d)dx=I^{2}R$, exactly the strip's heating; the ideal lower strip receives none. In the charge-free gap, ${\partial }_x{S}_x+{\partial }_z{S}_z=0$, consistent with a static field and no local dissipation. The small longitudinal electric field cannot be discarded if this upward energy transfer is to be represented.

Polarization density $\mathbf{P}$ is electric dipole moment per unit volume. Displacing positive and negative bound charges leaves the bulk nearly neutral locally but can produce uncompensated charge where polarization varies or terminates. For a dielectric occupying a volume with outward unit normal $\mathbf{n}$,The divergence theorem makes their total sum zero for a body described entirely by bound dipoles. Uniform polarization gives no interior volume charge but generally gives opposite surface charges on different faces.

In a linear isotropic dielectric, $\mathbf{P}={ϵ}_0{\chi }_e\mathbf{E}$ and $\mathbf{D}={ϵ}_0\mathbf{E}+\mathbf{P}=ϵ\mathbf{E}$. Gauss's law becomes $\nabla \cdot \mathbf{D}={\rho }_f$, separating free from bound charge. Across a static interface, tangential electric field is continuous, while the normal displacement jumps by the free surface charge. Polarization charges alter the internal field, as in the dielectric cylinder calculation. Molecular distortion and orientation contribute differently, and response can depend on temperature and frequency. Bound charge is a redistribution of material charge, not an additional free-charge reservoir.

Capacity convention. The printed equation requires $c$ to be volumetric heat capacity, in $\mathrm{J}{\mathrm{m}}^{-3}{\mathrm{K}}^{-1}$. If “specific heat capacity” means per unit mass instead, the storage term is ${\rho }_{m}c{\partial }_{t}T$ and both denominators in the printed equation must contain ${\rho }_{m}c$. This distinction is fixed by dimensions.

For a fixed material region $V$, conservation of thermal energy gives the heat balance with volumetric heat generation,Use the divergence theorem, $\mathbf{J}=-\kappa \nabla T$ and constant coefficients. Since the region is arbitrary,The heat source is power per volume, while $\kappa$ is thermal conductivity rather than thermal diffusivity.

For uniform electrical conductivity and imposed current, current density is $j=I/(\pi a^2)$ and Joule heating is $H=j^2/\sigma$. A long, steady, cylindrically symmetric wire therefore satisfiesIntegration gives $T^{\prime }=-Hs/(2\kappa )+C/s$. Regularity at $s=0$ makes $C=0$, and the surface condition fixes the remaining constant. Thus the Joule-heated cylindrical wire hasHere $s$ denotes the printed radial coordinate $\rho$ and avoids confusing it with mass density. The centre is hotter by $I^2/(4{\pi }^2{a}^2\kappa \sigma )$.

For the air boundary, the internal flux at $s=a$ is $-\kappa T^{\prime }(a)=Ha/2$. Equate this to $\alpha [T(a)-T_{\mathrm{air}}]$. The Newton cooling of a Joule-heated wire givesMultiplication by circumference $2\pi a$ verifies that surface loss per length equals $I^2/(\pi a^2\sigma )$, the electrical power per length. The full air-cooled profile uses this surface temperature plus the same parabolic interior excess, assuming constant material properties and negligible end losses.

In irrotational flow, the velocity has a velocity potential locally, and globally in a simply connected flow domain. incompressibility then gives $\nabla \cdot \mathbf{v}={\nabla }^2\Phi =0$. For a sphere moving at speed $U$ along $z$ through fluid at rest far away, the potential of a translating sphere is a decaying axisymmetric dipole. Its solution and its impermeability condition giveThese are laboratory fluid velocities expressed relative to a moving origin. A velocity field measured in the sphere-rest frame instead requires adding $-Ur\cos \theta$ to the potential.

The vector identity printed in the PDF has its signs reversed. The product rule gives the consistent identity $v^2=\nabla \cdot (\Phi \mathbf{v})-\Phi \nabla \cdot \mathbf{v}$. With outward normal of the exterior fluid equal to $-\hat{r}$ at the sphere, the divergence theorem yieldsThe integral at infinity vanishes. HenceThis added mass of a sphere is half the displaced fluid mass. For a spring of unchanged stiffness, buoyancy only shifts the static equilibrium. Its oscillation frequency with fluid added mass isFor a small-angle simple pendulum, buoyancy also reduces restoring torque from $mg\ell \theta$ to $(m-{\rho }_{0}V)g\ell \theta$, while the inertia is $(m+m_a){\ell }^2$. ThusThe downward equilibrium is stable only for $\rho >{\rho }_0$. The ideal results assume unbounded fluid, negligible viscosity, small excursions, and no important free-surface or container corrections.

Molecular transport across gradients. A gas can transport momentum, thermal energy and species abundance even when its average bulk velocity is zero. Molecules arriving at a point carry properties acquired near their preceding collisions. A small gradient makes the populations arriving from opposite sides slightly different, producing transport from larger to smaller values of the carried property. A continuum transport law requires a mean free path $\lambda$ small compared with the gradient length, and relaxation times short compared with the observation time.

For a dilute hard-sphere gas,where $d$ is molecular collision diameter. Both speed and collision distance matter: high speed alone does not determine a transport coefficient.

A simple kinetic estimate makes the common structure explicit. For a quantity $a(y)$ carried per particle, a molecule moving at angle $\theta$ to $+y$ brings approximately $a(y-\lambda \cos \theta )$. Expanding to first order and averaging an isotropic velocity distribution givessince $⟨\cos \theta ⟩=0$ and $⟨{\cos }^2\theta ⟩=1/3$. This mean-free-path argument estimates the coefficients; a detailed collision calculation changes their numerical prefactors and distinguishes different relaxation rates.

Momentum transport and viscosity. For a shear flow with mean velocity $U_x(y)$, the transported directed momentum per molecule is $m{U}_x$. The net flux of $x$ momentum in the $+y$ direction isHere $\eta$ is dynamic viscosity in $\mathrm{Pa}\mathrm{s}$. The associated shear stress on a material surface has the corresponding traction sign convention; the flux carries momentum from the faster-moving layer towards the slower-moving one. The kinematic viscosity is $\nu =\eta /(nm)$, with units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$. Viscosity smooths velocity gradients, and the lost ordered mechanical energy becomes heat.

Heat transport. Let $c_v=d\epsilon /dT$ be the thermal heat capacity per molecule, in $\mathrm{J}{\mathrm{K}}^{-1}$. Substituting $a=\epsilon (T)$ gives Fourier's law,Thermal conductivity has units $\mathrm{W}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$. For a monatomic gas, translational equipartition gives $c_v=3k_B/2$. Rotational or vibrational excitations in other gases can increase the energy carried per molecule, while collisions govern how those modes exchange energy. A temperature gradient can therefore be transported differently from directed momentum even when the dimensional estimates look similar.

Species transport. For a dilute tracer or a mixture with prescribed total density and negligible bulk flow, molecules diffuse down concentration gradients. The first-gradient number flux is Fick's first law,The diffusivity $D$ has units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$; multiplying number flux by the species mass gives mass flux. More generally one uses diffusion flux relative to the mixture's bulk velocity and specifies mass or mole fractions. The approximation is not a claim that an arbitrary pure-gas pressure gradient can remain at rest while only a Fick flux acts. In a mixture, collision cross-sections and relative molecular speeds, including the reduced mass, determine the species diffusivities.

Dependence on density and temperature. The kinetic estimates of gas transport coefficients predict, for fixed collision diameter and molecular heat capacity,Thus dynamic viscosity and thermal conductivity are approximately density independent in the dilute continuum regime. Reducing density reduces the number of carriers but increases their mean free path in the same proportion. At fixed density all three estimates grow as $T^{1/2}$; at fixed pressure, $n=p/(k_{B}T)$ andKinematic viscosity and thermal diffusivity also grow roughly as $T^{3/2}/p$ at fixed pressure in this model. Real interaction potentials make effective collision cross-sections temperature dependent, and internal excitations change $c_v$. Dense-gas correlations and boundaries invalidate the simple cancellation. In particular, when the mean free path approaches the apparatus scale, transport becomes free molecular; it cannot be described by extending the same local, pressure-independent Fourier and viscosity coefficients indefinitely.

Heat diffusion equation. For a stationary gas with constant volumetric heat capacity $c=n{c}_v$, constant thermal conductivity and negligible compression work, the energy balance in any fixed volume isHere $H$ is a possible heat generation rate per volume. The divergence theorem and Fourier's law yieldThermal diffusivity $\alpha$ has units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$ and is of order $\overline{v}\lambda /3$. In a moving or compressing gas, the material derivative, pressure-volume work and viscous heating must also enter the internal-energy equation. The displayed pure diffusion equation describes the approximation in which those effects are absent or negligible.

Representative solutions. With no source, a steady one-dimensional slab between temperatures $T_a,T_b$ at $x=0,L$ has $T=T_a+(T_b-T_a)x/L$ and uniform heat flux $-\kappa (T_b-T_a)/L$. Thus a constant spatial gradient transports heat without changing local temperature in time when the boundaries continually supply and remove it.

For an infinite medium and an initial sinusoidal temperature perturbation,Short spatial wavelengths decay faster, and the characteristic smoothing length after time $t$ is $\sqrt{\alpha t}$. A finite slab with both ends held at $T_b$ has a sine expansion with factors $e^{-\alpha (n\pi /L{)}^{2}t}$; its slowest relaxation time is $L^2/({\pi }^2\alpha )$.

An instantaneous heat input $Q_A$ per unit transverse area at $x=0$ in an otherwise uniform infinite medium gives the Gaussian solutionIts integral $\int c(T-T_b)dx=Q_A$ conserves the deposited heat. The peak falls as $t^{-1/2}$ while its width grows as $\sqrt{t}$; diffusion redistributes energy instead of transporting a sharp front at a single velocity.

Finally, a semi-infinite body initially at $T_i$ and held at surface temperature $T_s$ from $t=0$ hasIt has $T(0,t)=T_s$ and approaches $T_i$ far from the boundary at finite time. These solutions demonstrate steady conduction, mode decay, spreading of localized heat and boundary-driven diffusion. Their idealized behaviour at extremely short times or arbitrarily long distances must be interpreted within the continuum regime rather than as an assertion of instantaneous microscopic communication.

Integral conservation. Electric charge cannot be created by ordinary transport. For a fixed volume $V$, outward current reduces the enclosed charge:Here ${\rho }_e$ has units $\mathrm{C}{\mathrm{m}}^{-3}$ and $\mathbf{J}$ has units $\mathrm{A}{\mathrm{m}}^{-2}$. Applying the divergence theorem to an arbitrary volume gives the local continuity equationIt is local charge conservation: a net outflow must be accompanied by depletion. In a steady state, $\nabla \cdot \mathbf{J}=0$. Integrating over a small circuit junction then gives Kirchhoff's current law. During a transient the junction can accumulate charge and the incoming and outgoing conduction currents need not yet match.

Consistency with Maxwell's equations. In vacuum, take the divergence of the Ampère-Maxwell equation:Gauss's law, $\nabla \cdot \mathbf{E}={\rho }_e/{ϵ}_0$, reproduces the continuity equation. The displacement current term makes these equations consistent for time-dependent charge distributions, such as a charging capacitor. It is not a stream of free charges across the insulating gap. In matter, $\nabla \cdot \mathbf{D}={\rho }_f$ and $\nabla \times \mathbf{H}={\mathbf{J}}_f+{\partial }_t\mathbf{D}$ give the analogous conservation law for free charge; total charge also includes bound charge.

Use the divergence identity for a cross product, followed by Faraday's law of induction and the Ampère-Maxwell equation:With the electromagnetic energy density of part (b), these giveIntegrate over a fixed volume $V$ and apply the divergence theorem, choosing $d\mathbf{S}$ outward:This is Poynting's theorem. The outward electromagnetic energy flux equals minus the power transferred to matter and minus the rate at which electromagnetic energy accumulates inside. For the dissipative electrical conductor in this question, the matter term is Ohmic heating.

The starting point is Gauss's law for magnetism, $\nabla \cdot \mathbf{B}=0$: a flux tube carries the same magnetic flux through successive transverse sections. To identify that tube approximately with the physical core, one must also make a negligible-leakage approximation. The large core magnetic permeability makes the core route much less reluctant than competing paths through the surrounding air; the small gap further reduces the incentive for a long external return path. The gentle taper, $s\gg R-r$, lets the field follow the changing section with only small transverse corrections.

For two cuts and the intervening lateral surface, the divergence theorem gives ${\Phi }_2-{\Phi }_1=-{\Phi }_{\mathrm{side}}$. Neglecting lateral leakage givesThe normal magnetic flux density is continuous at each core–air interface by part (a), so this same magnetic flux crosses the gap. To use the gap's geometrical area $\pi r^2$ in the later formulas, the magnetic fringing must also be negligible, normally requiring $h\ll r$. This is the small-gap interpretation of the paper's approximation. Large magnetic permeability alone does not eliminate leakage or magnetic fringing for arbitrary gap geometry.

Gauss's law statesThe surface is closed and oriented outward, and $Q_{\mathrm{enc}}$ includes all enclosed charge. The integral law follows from the local form through the divergence theorem. It is universally valid in classical electromagnetism, but calculating the field directly from the flux requires enough symmetry.

For spherical charge symmetry, $E(r)=Q(<r)/(4\pi {ϵ}_0{r}^2)$. A uniform infinite line gives $E=\lambda /(2\pi {ϵ}_{0}r)$ from a coaxial cylinder. An infinite sheet gives $E=\sigma /(2{ϵ}_0)$ on either side from a pillbox, with the direction set by charge sign. The normal field jump is $\sigma /{ϵ}_0$; the electrostatic field inside a conductor is zero. Zero enclosed charge implies zero net flux, not generally zero field. ^[ref]

The integral Gauss's law isThe divergence theorem makes the left side ${\int }_V\nabla \cdot \mathbf{E}dV$. If the differential relation $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$ holds, this proves the integral law. Conversely, if the integral identity holds for every suitable volume and the fields are sufficiently regular, a nonzero value of $\nabla \cdot \mathbf{E}-\rho /{ϵ}_0$ would persist in sign on a small neighborhood and contradict its zero integral. The differential law follows.

For a sphere of radius $r$, spherical symmetry makes the flux $4\pi r^{2}E(r)$ and the enclosed charge $4\pi {\int }_0^r{s}^2\rho (s)ds$. HenceThere is no extra point charge at the origin in the stipulated density distribution. Outside the charge support,

The momentum of a material volume $V(t)$ is ${\int }_{V(t)}\rho \mathbf{u}dV$. Newton's second law equates its time derivative to the pressure force and gravity:Apply the given Reynolds transport theorem componentwise to the left side, and the divergence theorem to the pressure term. ThenSince this holds for arbitrary material volumes and the fields are continuous, the integrand vanishes pointwise. Dividing by positive density gives the Euler equations for an inviscid fluidAt a fixed physical time $t_0$, a streamline is a curve tangent to $\mathbf{u}(\cdot ,t_0)$, which may be parametrized by $d\mathbf{x}/ds=\mathbf{u}(\mathbf{x}(s),t_0)$. The chain rule gives $df(\mathbf{x}(s),t_0)/ds=\mathbf{u}\cdot \nabla f=0$, proving constancy along it. The time is held fixed; a time-dependent streamline statement does not by itself assert material conservation along particle trajectories.

In SI units Maxwell's equations areThe given vector identity yieldsDivide by ${\mu }_0$ and collect the two energy derivatives to obtain the local identity ${\partial }_t\mathcal{U}+\nabla \cdot P+E\cdot J=0$. Integration over the fixed region and the divergence theorem give the required integral conservation law. Its terms are, respectively, the rate of change of stored electromagnetic energy, the outward electromagnetic energy flux through the boundary, and work per unit time done by the field on charges. The last term transfers energy from the field to matter when positive.

In vacuum put ${\rho }_e=0$, $J=0$ and write $\varphi =kz-\omega t$. Faraday's law gives the plane-wave magnetic fieldBoth divergence equations are satisfied, and the remaining curl equation gives $k^2={\mu }_0{ϵ}_0{\omega }^2$. Hence $\boxed{\omega =ck,c=({\mu }_0{ϵ}_0{)}^{-1/2}}$, with $B_y=E_x/c$. Here the corresponding field is the pure wave, with no added static background.

Its electric and magnetic energy densities are equal. Explicitly,The energy flux points in the propagation direction, toward increasing $z$.

The Hamiltonian operator is invariant under the constant global phase symmetry $\psi \mapsto e^{i\theta }\psi$, $\bar{\psi }\mapsto \bar{\psi }{e}^{-i\theta }$. The Noether current of this unitary group $U(1)$ is the Dirac current obtained in b. For fields with vanishing flux at infinity, or periodic boundary conditions, the divergence theorem and that continuity equation implyThe canonical anticommutation relations also give $[Q,\psi ]=-\psi$ and $[Q,{\psi }^{†}]={\psi }^{†}$, hence $e^{i\theta Q}\psi e^{-i\theta Q}=e^{-i\theta }\psi$. Reversing the parameter sign recovers the stated phase convention. Thus $Q$ generates the global phase symmetry. With normal ordering, $Q$ counts fermions minus antiparticles, rather than the sum of both populations. Its conservation law is Dirac fermion number conservation. A position-dependent phase produces derivative terms in this free Hamiltonian operator; the symmetry under discussion is global.

Cauchy's stress hypothesis says that traction across a cut depends locally on its point, time and oriented normal, rather than on the global shape of the cut. Here all areas and normals are referential: $E$ is a fixed material region in reference coordinates, ${\rho }_R$ its mass density per reference volume, $v$ the current particle velocity, $b$ body force per unit mass, and $s_R$ force per unit reference area. The time derivative on the left is total material momentum rate; the two right terms are contact and body forces.

A shrinking pillbox gives $s_R(-N)=-s_R(N)$. In a shrinking tetrahedron the volume contributions scale as its linear size cubed, whereas face forces scale as size squared. Their leading balance expresses traction on an arbitrary face as the normal-component-weighted sum of tractions on three coordinate faces. Defining those three tractions as tensor columns gives $\boxed{s_R(N)=T_{R}N}$. Applying the divergence theorem to momentum balance and localizing over arbitrary $E$ then yieldsThe referential tensor is the first Piola-Kirchhoff stress and need not be symmetric; it should not be silently identified with the spatial Cauchy stress.

Incompressibility is $\boxed{\det F=1}$. For a homogeneous hyperelastic material the referential constitutive relation becomeswhere the undetermined scalar pressure $p$ enforces the constraint. Homogeneity removes explicit reference-position dependence from $W$, not the possible spatial variation of that pressure.

For the neo-Hookean energy, $DW=2\alpha F$. For the positive diagonal ansatz the nominal principal stresses are $t_i=2\alpha v_i-p/v_i$. Since $F$ is constant, equilibrium requires $\mathrm{cof}F\nabla p=0$, so $p$ is constant. Stress-free endfaces give $\boxed{p=2\alpha v_3^2>0}$. The side normals include $(0,-1,0)$ and $\pm (\sqrt{3}/2,1/2,0)$, so their prescribed tractions force $\boxed{t_1=t_2=T}$.

For fixed positive $p$, the scalar function $2\alpha v-p/v$ is strictly increasing on $v>0$. Thus equality of the two nominal stresses implies $v_1=v_2=v$. Incompressibility gives $v_3=v^{-2}$, and substitution givesThe load function has derivative $2\alpha (1+5v^{-6})>0$, tends to minus infinity as $v\downarrow 0$ and to infinity as $v\to \infty$. It therefore determines exactly one positive $v$ for every specified load, with $v\ge 1$ for outward tensile $T\ge 0$. The resulting constant stress satisfies equilibrium and every prescribed face traction, proving existence and uniqueness within the requested positive diagonal class. It does not claim uniqueness among all possible nonhomogeneous deformations. This is the equibiaxial dead loading of a neo-Hookean prism.

In the classical smooth setting, a deformation in continuum mechanics is a sufficiently regular, injective map $y:\Omega \to {\mathbb{R}}^3$ with $\det Dy>0$. Injectivity excludes material interpenetration, and a positive determinant preserves orientation and local volume. We use $C^1$ regularity for these kinematics and additional differentiability where stress-divergence identities are applied.

Cauchy's stress theorem states that the contact force on an oriented surface element at a current point depends on the point and its unit normal, rather than on the shape of the chosen subbody. Thus the contact force on the boundary of a current subbody $v$ has the form ${\int }_{\partial v}t(z,n)da$. Static force balance, for a body force $b$ per unit current volume, isfor every subbody. A shrinking pillbox gives $t(z,-n)=-t(z,n)$; a shrinking tetrahedron then gives linear dependence on the normal, $t(z,n)=S(z)n$. Applying the divergence theorem and using arbitrary subbodies givesIf body force is specified per unit mass instead, replace $b$ by $\rho b$. Symmetry of $S$ follows additionally from conservation of angular momentum in the absence of body couples; it is not a consequence of force balance alone.

Let $t(n)$ denote traction: the force per unit area exerted across an oriented internal surface with unit normal $n$. Define ${\sigma }_{ij}$ as component $i$ of the traction on a coordinate plane whose normal is $e_j$. Balance of linear momentum on a shrinking tetrahedron makes the surface-force terms, of order area, dominate its inertia and body forces, of order volume. Resolving its oblique face into the three coordinate projected areas gives Cauchy's traction formulaOpposite faces have opposite tractions, as follows independently from a shrinking pillbox. This fixes both the index and normal conventions.

For a material volume $V(t)$, the Reynolds transport theorem says $d{\int }_{V(t)}fdV/dt={\int }_{V(t)}[f_t+\nabla \cdot (fu)]dV$. Apply it componentwise to $f=\rho u_i$ and use mass conservation. With no body force in this calculation, balance of linear momentum and the divergence theorem giveArbitrariness of the material volume yields $\boxed{\rho D{u}_i/Dt={\partial }_j{\sigma }_{ij}}$, with $D/Dt={\partial }_t+u_j{\partial }_j$.

For constant dynamic viscosity and incompressible velocity, ${\partial }_j{u}_j=0$, soThus $\boxed{\rho (u_t+u\cdot \nabla u)=-\nabla p+\mu {\nabla }^{2}u}$.

In fully developed steady pipe flow $u=w(x,y)e_z$, the convective acceleration vanishes. The transverse equations give $p_x=p_y=0$ and the axial equation gives $p_z=\mu (w_{xx}+w_{yy})$. One side depends only on $z$ and the other only on $(x,y)$, so both equal a constant. HenceIf $n$ is outward from the fluid cross-section, the wall-on-fluid axial shear per length is $\mu \oint {\partial }_{n}wds=\mu {\int }_A\Delta wdA=A{p}_z$. The fluid-on-pipe force is its negative:For forward pressure-driven flow $p_z<0$, this force points along the flow. Its value depends on cross-sectional area and pressure drop, even though the detailed velocity profile depends on the shape.