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Let $z>0$ measure height from the lower edge and $y>0$ distance from one face of the vertical plate. Use the steady Boussinesq approximation, homogeneous isotropic permeability of a porous medium $K$, reference fluid mass density ${\rho }_0$, dynamic viscosity $\mu$, thermal expansion coefficient $\alpha$, effective thermal conductivity $k_e$ and liquid volumetric heat capacity $C_f={\rho }_0{c}_f$. Define $B={\rho }_{0}g\alpha K/\mu$ and $\kappa =k_e/C_f$, consistently with the thermal convention used above. Write $\theta =T-T_a$, where $T_a$ is the uniform ambient temperature.

Interpret $F$ as uniform wall heat-flux density: a plate segment of height $dz$ inputs heat $Fdz$ per unit horizontal span. The cumulative heat input up to height $z$ is therefore $Fz$. If $F$ instead denotes the sum of flux densities into two identically heated faces, replace it by $F/2$ in each face formula before summing. A specification only of total power per span, without its distribution in height, would not determine this uniformly heated-wall problem. We calculate one heated face; identical heating on both faces doubles the integrated fluxes.

Away from the leading edge, suppose the boundary layer is slender, $\delta (z)\ll z$, and porous inertia and longitudinal heat conduction are negligible. Lateral pressure adjustment leaves the vertical Darcy velocity proportional to buoyancy:At the plate, $v=0$ and $-k_e{\theta }_y=F$; in the far field $w,\theta \to 0$. Darcy law permits tangential discharge at an impermeable plate, so imposing a further no-slip condition $w=0$ would be inappropriate in this leading porous model.

Balance $\theta \sim F\delta /k_e$, $w\sim B\theta$ and $w\theta /z\sim \kappa \theta /{\delta }^2$ to obtainThus the warmed layer thickens and accelerates upward. Define a streamfunction by $w={\psi }_y$, $v=-{\psi }_z$ and setThe last two expressions agree because ${\delta }^3=\kappa k_{e}z/(BF)$. Differentiation gives $v=-(\kappa /3\delta )(2f-\eta f^{\prime })$. Substituting in the thermal equation leaves the single nonlinear equation for the uniformly heated vertical plate in a porous medium:The first condition is impermeability; the second fixes the wall heat flux; the third fixes ambient temperature and vertical velocity. Seek the positive-temperature branch with $f^{\prime }\ge 0$ and finite $f(\infty )$. Decay also gives $f^{\prime \prime }(\infty )=0$. The far-field transverse inflow need not vanish: $v_{\infty }=-(2\kappa /3\delta )f(\infty )$ supplies the increasing discharge.

An approximate solution can be obtained without a numerical shooting calculation. Integrate the differential equation over $0<\eta <\infty$. The boundary conditions and integration by parts giveThus the exact thermal budget is ${\int }_0^{\infty }(f^{\prime }{)}^{2}d\eta =1$. Assume an exponential thermal shape, $f=C(1-e^{-\eta /d})$. The wall derivative requires $C/d^2=1$, and the integral condition requires $C^2/(2d)=1$. Hence $d=2^{1/3}$, $C=2^{2/3}$ and the integral exponential profile for porous wall convection isIt satisfies all boundary conditions, has positive decaying temperature, and matches the exact integrated heat budget. It is an integral approximation, not an exact pointwise solution of the differential equation; its mass-flux coefficient should be read accordingly.

Define the volume discharge, kinematic Darcy momentum flux and buoyancy flux per horizontal span byWith these standard discharge conventions the mass flux is $\dot{m}={\rho }_{0}Q$. Their similarity expressions areThe integral identity evaluates two of them exactly:Equivalently $C_f\int w\theta dy=Fz$: all the heat supplied below height $z$ is carried upward through that section. Although these fluxes are exact within the boundary-layer equations, porous drag means that the momentum flux itself is not conserved with height.

The approximate mass flux isFor clarity, if “momentum flux” means actual pore-fluid momentum rather than the commonly quoted kinematic Darcy diagnostic, constant porosity $\varphi$ and pore speed $w/\varphi$ give $\dot{P}=({\rho }_0/\varphi )\mathcal{M}$. The mass flux remains ${\rho }_{0}Q$ because $w$ already denotes discharge per bulk area. Stating both prevents an implicit porosity error. Validity also requires $\alpha \theta \ll 1$ and a sufficiently slender, low-inertia porous boundary layer; the approximation is not the immediate leading-edge conduction field.

A comoving coordinate $\mathbf{\chi }$ labels a galaxy carried by the homogeneous cosmological flow. Its physical separation at time $t$ is $\mathbf{r}(t)=a(t)\mathbf{\chi }$, where $a$ is the scale factor. Thus $\mathbf{v}=\dot{a}\mathbf{\chi }=(\dot{a}/a)\mathbf{r}$, the Hubble law, apart from peculiar velocities relative to this flow. In particular

For a material interface with no mass transfer, the normal fluid velocity on both sides equals the interface's normal velocity. For ordinary viscous fluids without interfacial slip, tangential velocities also agree. The stress tensors $T_i=-p_{i}I+{\mu }_i(\nabla {\mathbf{u}}_i+\nabla {\mathbf{u}}_i^T)$ have continuous tangential traction when surface tension is uniform; their normal traction jump balances surface-tension curvature. Nonuniform surface tension instead supplies a tangential Marangoni stress. These conditions express conservation of mass, no slip between the contacting fluids and force balance at the interface. For a flat interface of constant tension the full traction is continuous.

Choose $x$ down the plane and $y$ perpendicular outward from it. A steady parallel solution has velocity $(u(y),0)$ and equationsFor a uniform free layer, $p_x=0$, $u(0)=0$, $u^{\prime }(h)=0$ and $p(h)=p_{\mathrm{atm}}$. Integration givesIts flux per unit cross-slope width is

For the two-layer falling film of equal-density viscous fluids let $H=(1+\alpha )h$ and $A=\rho g\sin \theta /\mu$. Equal density gives the same continuous hydrostatic pressure in both, $p=p_{\mathrm{atm}}+\rho g\cos \theta (H-y)$. The wall condition is $u_1(0)=0$, the upper condition $u_2^{\prime }(H)=0$, and at $h$ we require $u_1=u_2$, $\mu u_1^{\prime }=\beta \mu u_2^{\prime }$. Integrating the two momentum equations with these conditions givesThe shear stress transmitted to the lower layer is the upper layer's downslope weight per unit area, $\rho g\alpha h\sin \theta$, obtained by integrating the upper momentum balance from its stress-free surface. It depends on thickness but not upper viscosity. Changing $\beta$ changes the upper shear rate needed to transmit that stress, while the lower profile depends on $\alpha$ alone.

For a periodic potential, the Hamiltonian commutes with every lattice translation. The three primitive translations are commuting unitary operators, so within each energy eigenspace one can choose simultaneous translation eigenstates. Their eigenvalues have modulus one and can be written $e^{i\mathbf{k}\cdot {\mathbf{a}}_i}$. Translation by any lattice vector then givesThis proves Bloch's theorem in its eigenbasis form; arbitrary superpositions inside a degenerate eigenspace need not themselves have a single Bloch wavevector.

The reciprocal lattice consists of vectors $\mathbf{g}$ with $\mathbf{g}\cdot {\mathbf{a}}_i\in 2\pi \mathbb{Z}$. Changing $\mathbf{k}$ to $\mathbf{k}+\mathbf{g}$ and $u$ to $e^{-i\mathbf{g}\cdot \mathbf{r}}u$ gives the same physical wavefunction with a periodic new $u$. The eigenvalue problem on one cell has a discrete set of levels, giving bands indexed by $n$, and this equivalence yields $E_n(\mathbf{k}+\mathbf{g})=E_n(\mathbf{k})$. A reciprocal-lattice fundamental domain, conventionally a Brillouin zone, therefore labels inequivalent wavevectors.

Periodic boundary conditions in physical volume $\mathcal{V}$ give one allowed wavevector per reciprocal-space volume $(2\pi {)}^3/\mathcal{V}$. Including the two spin states, there are $2\mathcal{V}{d}^{3}k/(2\pi {)}^3$ states; per unit physical volume the count is the printed $2d^{3}k/(2\pi {)}^3$. Each occupied state carries charge $-e$ and velocity ${\hslash }^{-1}{\nabla }_k{E}_n$, soFor a full band, the integral of ${\nabla }_k{E}_n$ is a boundary integral. Opposite faces of a reciprocal fundamental cell have equal energies by periodicity and opposite normals, so their contributions cancel. Hence a full band carries zero electric current.

In the lubrication regime the thin region has longitudinal scale $\ell \sim \sqrt{ab}$, since the height increase $x^2/(2a)$ becomes comparable with $b$ there. The characteristic shear is $\mu V/b$, so force per axial length scales as $\mu V\ell /b\sim \mu V\sqrt{a/b}$. Negligible inertia, no slip, no cylinder rotation, and full wetting are understood.

Use a frame translating with the cylinder. The wall moves with velocity $-V$, the cylinder is stationary, and the gap is $h(x)=b+x^2/(2a)$. Lubrication momentum gives $\mu u_{yy}=p_x$, $p_y=0$, soContinuity makes $q$ constant. Equal ambient pressures at both infinities give $\int p_{x}dx=0$, hence $q=-(V/2)(\int h^{-2}dx)/(\int h^{-3}dx)=-2Vb/3$. ThereforeThe vertical velocity, with the wall condition $w(0)=0$, follows explicitly from incompressibility:It gives $w(h)=0$, consistently with a horizontally translating cylinder in the lab frame.

The wall shear is $\mu u_y(0)=4\mu V(1/h-b/h^2)$. With $x=\sqrt{2ab}s$, its integral isForce balance equates this wall force to the required cylinder force at leading order; the pressure and viscous stresses on remote boundaries make no leading contribution. This is the cylinder translating parallel to a wall in a thin gap result. The asymptotic sign denotes the approximation implicit in the parabolic gap, rather than an exact finite-gap cylinder formula.

Write $\mathbf{v}=\dot{\mathbf{r}}$. The original PDF contains the cross product $\dot{\mathbf{r}}\times \mathbf{r}$; the TeX aid drops its velocity dot. Since $\mathbf{v}\times \mathbf{v}=0$, differentiation and the equation of motion giveThe radial restoring force contributes no torque. The vector triple-product identity yieldsMeanwhileThe derivatives cancel, provingThis is the magnetic axial angular-momentum invariant. Its first term is the negative of the ordinary angular momentum projected on $\mathbf{B}$; retaining that sign is essential.

Dot Newton's second law with $\dot{\mathbf{r}}$. The magnetic cross-product term is perpendicular to this velocity, soThusorThe two halves are the kinetic energy and the isotropic spring's potential energy. The expression in the question is twice their sum. The Lorentz force changes the direction of motion but supplies no power, because it is perpendicular to velocity; this explains why the magnetic field does not appear in the conserved mechanical energy. The sign corresponds to charge $-e$.

For a fixed central gravitational source, the radial and transverse components of Newton's second law in polar coordinates areThe second equation gives $r^2\dot{\theta }=h$, the constant specific angular momentum. Take $h>0$ by choosing the orientation of $\theta$, and put $u=1/r$. Writing $u^{\prime }$ and $u^{\prime \prime }$ for derivatives with respect to $\theta$,Substituting into the radial equation and dividing by $-h^2{u}^2$ proves the Binet equationAt perihelion the velocity is tangential. The escape velocity is $v_0=\sqrt{2GM/r_0}$, so $h=r_0{v}_0$ and $h^2=2GM{r}_0$. Consequently $k=1/(2r_0)$. The initial conditions are $u(0)=1/r_0$ and $u^{\prime }(0)=0$. Solving the linear ordinary differential equation gives the parabolic Kepler orbitUsing angular momentum conservation again,On the outgoing branch $r=2r_0$ is first reached at $\theta =\pi /2$. Integrate from release at perihelion, using $w=\tan (\theta /2)$:More generally the same integral gives $t=(2r_0^2/h)(w+w^3/3)$, the Barker equation for this parabolic orbit.

Take upward velocity as positive, and consider the closed material system consisting of the rocket and the fuel it ejects during a short interval $dt$. Initially its mass and momentum are $m$ and $mv$. Let $\delta m>0$ be the ejected mass, so the rocket's mass change is $dm=-\delta m$. At the end its velocity is $v+dv$, while the ejected gas has inertial velocity $v-u$ to first order; its downward relative velocity need not be downward in the inertial frame.

The final total momentum, to first order, isNeglect air resistance and take the gravitational acceleration $g$ constant. The external gravitational impulse on this material system is $-mgdt$ to first order. Newton's second law therefore gives $mdv-u\delta m=-mgdt$, orThe positive thrust is $-u\dot{m}$ because fuel loss makes $\dot{m}<0$. Applying $d(mv)/dt$ to the rocket alone without accounting for the escaping momentum would give the wrong equation.

For constant $u$ and positive mass, divide by $m$ and integrate from release at $t=0$:Thus the gravity-corrected rocket equation is equivalent to

Along the holonomic constraint $z=x^2/(4a)$, we have $\dot{z}=x\dot{x}/(2a)$ and hence speed squared $(1+x^2/(4a^2)){\dot{x}}^2$. The smooth wire's normal reaction is perpendicular to the permitted velocity and does no work. Conservation of mechanical energy, with release from rest at height $h$, givesThe turning points are $x=\pm x_0$ with $x_0=2\sqrt{ah}$. One full oscillation has twice the travel time between them:Set $x=x_{0}u$ and $\beta =h/a$. After cancelling the scale factors,so the bead on a parabolic wire hasFor $\beta \ll 1$, expand the numerator uniformly on $[-1,1]$:The integrable endpoint weight gives ${\int }_{-1}^1(1-u^2{)}^{-1/2}du=\pi$ and ${\int }_{-1}^1{u}^2(1-u^2{)}^{-1/2}du=\pi /2$, for example by $u=\sin \theta$. Therefore the requested small-oscillation period isThe leading frequency is $\sqrt{g/(2a)}$, and the first finite-amplitude correction increases the period. At $h=0$ the ring is stationary; the expression is the limiting period of nonzero small oscillations.

For constant mass density and incompressible flow, use $(\mathbf{u}\cdot \nabla )\mathbf{u}=\nabla (|\mathbf{u}{|}^2/2)-\mathbf{u}\times \mathbf{\omega }$ in the Euler equations for an inviscid fluid. Taking the curl removes the pressure and kinetic energy gradients, giving ${\mathbf{\omega }}_t=\nabla \times (\mathbf{u}\times \mathbf{\omega })$. The supplied vector identity, with $\nabla \cdot \mathbf{u}=0$ and $\nabla \cdot \mathbf{\omega }=0$, yields the vorticity equationIn a two-dimensional flow $\mathbf{u}=(u,v,0)$ is independent of $z$, and $\mathbf{\omega }=\omega {\mathbf{e}}_z$. The right side vanishes, so $D\omega /Dt=0$: each material element retains its vorticity. The vortex lines are parallel to $z$ and are carried by the planar velocity. Also planar incompressibility preserves the area of a material cross-section; integrating its conserved vorticity over that area shows that its circulation, or vortex strength, stays constant. Thus the vortex lines move with the fluid without changing strength.

Place a line vortex of circulation $\Gamma$ at $(0,b)$ above the wall $y=0$. Its velocity field is $\Gamma {\mathbf{e}}_z\times (\mathbf{r}-{\mathbf{r}}_0)/(2\pi |\mathbf{r}-{\mathbf{r}}_0{|}^2)$. The method of images puts a vortex of circulation $-\Gamma$ at $(0,-b)$. With $\mathbf{u}=({\psi }_y,-{\psi }_x)$, the resulting stream function isIt is identically zero on the wall, so $u_y=-{\psi }_x=0$ there. In the fluid half-plane it has just the required vortex singularity, with no other singularities, and its velocity decays at infinity. This is the no-background-flow solution with the wall: the difference from another such solution is a regular irrotational, divergence-free decaying field with zero normal velocity. Reflect its tangential component evenly and its normal component oddly across the wall. The reflected components are bounded harmonic functions on the whole plane, so they are constant by harmonic Liouville theorem, and decay forces both constants to be zero. Thus replacing the wall by the image gives the same flow. The real vortex moves under the image velocity, excluding its own singular velocity, soIts signed velocity parallel to the wall is $\Gamma /(4\pi b)$, and its speed is $|\Gamma |/(4\pi b)$.

For the quarter-plane, place images of circulation $-\Gamma$ at $(-x,y)$ and $(x,-y)$, and $+\Gamma$ at $(-x,-y)$. The paired stream functions make both walls streamlines. Summing their velocities at the real vortex givesIn polar coordinates,It follows directly thatThus the line-vortex trajectory in a quarter-plane is $r\sin 2\theta =\text{constant}$.

For a moving circuit, Faraday's law takes the formHere $C(t)$ is the instantaneous wire contour, $\mathbf{v}$ its local material velocity, and $S(t)$ any spanning surface. The contour orientation and surface normal vector obey the right-hand convention. The integral on the right is the magnetic flux, differentiated with both the field and circuit position allowed to vary. The electric field term is the electric force per unit electric charge around the wire, and $\mathbf{v}\times \mathbf{B}$ supplies the motional electromotive force. Their sum is the electromotive force driving charge around the moving circuit. The condition $\nabla \cdot \mathbf{B}=0$ makes the magnetic flux independent of the spanning surface.

Choose the initial loop normal vector to be $+{\mathbf{e}}_y$, so rotation about $+{\mathbf{e}}_z$ gives $\mathbf{n}(t)=(-\sin \Omega t,\cos \Omega t,0)$. The spatially uniform magnetic field givesUsing Ohm's law for the resistive loop,Positive electric current is measured in the contour orientation associated with $\mathbf{n}(t)$. Reversing that convention reverses the signed answer. When $\Omega =\omega$, the magnetic flux is constant and the induced electric current vanishes. This uses the usual thin resistive-circuit approximation with self-inductance neglected.

Use the exact scalar-metric normalization specified by the preceding brightness equation:The temperature source is therefore $-(h^{\prime }-h_s^{\prime })/6-{\mu }^2{h}_s^{\prime }/2$. Let $W=e^{ik\mu (\tau -{\tau }_0)}$, so $W^{\prime }=ik\mu W$ and $W^{\prime \prime }=-k^2{\mu }^{2}W$. The formula just derived becomesFor $k\ne 0$ the last term may be integrated twice by parts:No field equation beyond the supplied brightness relation has been used for this identity.

For scalar velocity the supplied photon continuity equation givesThe sign follows from division by $i$: with the Fourier streaming term $+ik\mu$, the velocity projection has the plus $h^{\prime }$ term shown here. Combining it with the lower endpoint from the integrations by parts gives the complete endpoint terms in the synchronous Sachs-Wolfe formula:The local observer bracket is an angular photon monopole plus photon dipole. Removing the local mean temperature and photon dipole, or restricting the result to measured multipoles $\ell \ge 2$, permits omitting that bracket. It must be retained for the literal full brightness amplitude. The $k=0$ limit should be taken in the original line-of-sight integral, rather than treating the separate inverse-$k$ endpoint terms as independently defined.

The final target formula printed in the PDF does not follow from its preceding equations. With the stated streaming and scalar conventions, the decoupling terms proportional to $h^{\prime }-h_s^{\prime }$ and $h_s^{\prime \prime }$ have plus signs as derived above, not the printed minus signs. Moreover the integral has no additional outside factor $1/2$ when its bracket is $(h^{\prime }-h_s^{\prime })/6-h_s^{\prime \prime \prime }/(2k^2)$. Equivalently one may retain an outside $-1/2$ only if that bracket is doubled to $(h^{\prime }-h_s^{\prime })/3-h_s^{\prime \prime \prime }/k^2$. Omitting the observer photon monopole and photon dipole does not fix these discrepancies.

For a direct counterexample to the integral coefficient, take $h_s=0$, $h^{\prime }=\epsilon$ constant, $\mu =0$, and ${\delta }_{\gamma \ast }={\delta }_{\gamma \ast }^{\prime }=0$. A scalar emission velocity with $i\mathbf{k}\cdot {\mathbf{v}}_{\ast }=-\epsilon /2$ satisfies the stated continuity equation; its projection along this transverse photon direction is zero. Keep $\epsilon ({\tau }_0-{\tau }_{\ast })\ll 1$ so perturbation theory is valid. The original brightness equation or the line-of-sight integral giveswhereas the printed final expression gives $-\epsilon ({\tau }_0-{\tau }_{\ast })/12$. All $h_s$ observer terms vanish in this example, so an observer-term convention cannot remove the contradiction. The boxed corrected identity supplies the requested integration-by-parts result with every boundary contribution and its observable higher-multipole version specified.