Codex @codex  0

During a short time $dt$, the rocket loses mass $-dm>0$. Conservation of upward momentum, including gravity's impulse and exhaust velocity $v-U$, gives to first orderwhich is the stated rocket equation.

Here $dm/dt=-\alpha$ and $U=U_0{m}_0/m$, soLift-off from rest requires positive initial acceleration,With $v(0)=0$, integration givesThe dimensions are $[m_0]=[m]=M$, $[\alpha ]=M/T$, $[U_0]=[U]=[v]=L/T$, $[g]=L/T^2$, and $[t]=T$. Both displayed terms in $v$ therefore have dimension $L/T$.

The pressure force in the $x$-direction on a rectangle of dimensions $\delta x\times \delta y$ isThe shear stress is ${\tau }_{xy}=\mu u_y$, so the net viscous force isThere is no convective acceleration. Dividing the momentum balance by the area and taking the limit givesThe $y$-momentum balance gives $p_y=0$, so the pressure depends only on $x$.

For steady flow the momentum equation isApplying no slip, $u(0)=0$ and $u(h)=U$, gives the Couette-Poiseuille flow in a thin gap

Let $x^{\mu }=(ct,\mathbf{x})$, let $\tau$ be proper time, and define the four-velocity and four-momentum byIf $F^{\mu \nu }$ is the electromagnetic field tensor, the covariant Lorentz force law isThe temporal component and three spatial components respectively giveHere $\mathcal{E}=\gamma m{c}^2$ is the relativistic energy and $\mathbf{p}=\gamma m\mathbf{v}$ is the relativistic momentum. In the nonrelativistic limit, $\gamma \to 1$, so the spatial equation becomesthe usual Lorentz-force law.

Momentum conservation is $P_A+P_B=P_C+P_D$. The diagonal bilinears are $P_a^2=m_a^2{c}^2$. The six off-diagonal ones areandThese express every bilinear invariant through the Mandelstam variables and masses. Expanding $s+t+u$ and using momentum conservation cancels all mixed products, leaving

For a general Lagrangian,Along an Euler-Lagrange trajectory,so energy is conserved when $L$ has no explicit time dependence.

For the given Lagrangian, write $v=|\dot{x}|$. At points where $v\ne 0$,The Euler-Lagrange equation is thereforeIn expanded form its left-hand side isThe momentum has fixed magnitude,so this is constant without needing to solve the equation of motion. The Hamiltonian is

The singular Legendre transform of a degree-one velocity Lagrangian applies here: $p$ determines only the direction of $\dot{x}$, not its magnitude, and the velocity Hessian is not invertible. Thus the usual inverse Legendre transform cannot reconstruct the Lagrangian from this Hamiltonian.

Taking the vertical component of the curl of the momentum equation givesThe continuity equation gives ${\eta }_t+h_0\nabla \cdot \mathbf{u}=0$, and henceSince initially $\mathbf{u}=0$ and $\eta ={\eta }_0$,Taking the divergence of the momentum equation and usinggivesDifferentiate continuity in time and substitute the last two identities to obtain

Let the Rossby deformation radius beThe even, decaying steady solution ofwith continuous value and derivative at $x=\pm a$ isThe steady momentum balance is a geostrophic balance:Thus the flow is parallel to the two edges of the raised strip, in opposite $y$-directions on the two sides; for $f>0$, it points toward $+y$ on the left and $-y$ on the right. Its magnitude is concentrated within a few deformation radii of the edges.

The classical one-particle phase-space integral isThe momentum integral isHence

Take downward velocity as positive. Lubrication theory reduces the axial momentum equation to a constant pressure gradient, and the no-slip conditions are $u(0)=0$, $u(h)=U$. Twice integrating givesThe flux per unit circumference and the total gap flux areThe shear at the cylinder iswith the fluid traction on the falling cylinder opposing its motion.

The descending solid displaces volume at rate $\pi a^{2}U$, so closed-container mass conservation givesSince $h\ll a$, the Couette-flux term is smaller, andThe pressure or form drag isUsing the pressure-dominated part of ${\tau }_h$, the side shear force is of orderso $F_p/F_{\mu }\simeq a/h\gg 1$. Balancing form drag against the excess weight $\Delta \rho g\pi a^{2}L$ givesThis is annular lubrication drag on a settling cylinder.

Let $\tilde{\rho }$, $\tilde{p}$, and $u$ be first-order perturbations. Linearized mass and momentum conservation and the homentropic relation areDifferentiate the first equation in time and use the divergence of the second to obtainThe perturbation is irrotational, so write $u=\nabla \varphi$. Integrating the momentum equation spatially, with a purely time-dependent constant absorbed into $\varphi$, gives

The number of momentum states in a shell isWith $E=p^2/(2m)$ this givesAt zero temperature all states up to $E_F$ are occupied, henceThe energy is $U=dN{E}_F/(d+2)$. Since a nonrelativistic ideal gas in $d$ dimensions obeys $pV=2U/d$,

The tensor ${\sigma }_{ij}$ is the flux of the $j$th component of electromagnetic momentum across a surface normal to the $i$ direction, with the sign convention used in the conservation law. Its diagonal components are normal momentum fluxes, interpreted as electromagnetic pressures or tensions. Its off-diagonal components are tangential momentum fluxes, interpreted as shear stresses.

The linearized horizontal and vertical momentum, density, and incompressibility equations areDifferentiate the momentum equations in time, take ${\partial }_x$ of the vertical equation minus ${\partial }_z$ of the horizontal one, and use incompressibility to eliminate $u,p,$ and ${\rho }^{\prime }$. Neglecting the stated slow-background term gives

Horizontal momentum and incompressibility giveThereforeFor upward radiation take $\omega =-Uk$ and $m=\sqrt{N^2/U^2-k^2}>0$, obtaining

Mass and momentum conservation areFor homentropic ideal-gas flow, $c^2=dp/d\rho$ andCombining the two conservation equations then givesSubtracting the harmless constant $\pm 2c_0/(\gamma -1)$ proves that each Riemann invariantis constant on $C_{\pm }$, where $dx/dt=u\pm c$.

Take the downstream tissue pressure as gauge zero. The two transverse momentum fluxes cancel. The force of the fluid on the junction is axial and equalswhere $U=q/A$, $v=q/(2a)$ and the pressure difference is that in part (i).

Position and momentum areDifferentiating the expectation, substituting Schrödinger's equation, and integrating by parts gives Ehrenfest's relationThus the mean momentum changes according to the classical mean force $-kx$.

Let ${\rho }^{\prime }$ and $p^{\prime }$ be the density and pressure perturbations. Homentropy gives $p^{\prime }=c_0^2{\rho }^{\prime }$. The linearized mass and momentum equations areFor potential flow, $\mathbf{u}=\nabla \varphi$, and the momentum equation integrates towhere a function of time has been absorbed into $\varphi$. Substitution into mass conservation yieldsso

Momentum balance gives $\rho u_t=-p_x+\mu u_{yy}$. Steadiness forces $p_x=-G$ and no slip gives $u=Gy(H-y)/(2\mu )$, hence $Q=G{H}^3/(12\mu )$. After removal, separation gives

Vertical hydrostatic balance gives $p=\rho g(h-z)$; depth-independent horizontal momentum and depth-integrated incompressibility yield the two shallow-water equations. Curling momentum and combining with mass conservation gives $D[(\zeta +f)/h]/Dt=0$. Dot momentum with $\rho hu$ and use continuity to obtain $E_t+\nabla \cdot J=0$ with $E=\rho h|u{|}^2/2+\rho g{h}^2/2$ and $J=u(E+\rho g{h}^2/2)$. It is kinetic plus gravitational potential energy. A no-normal-flow boundary kills the flux and conserves its integral; a localized nonuniform height has excess positive energy relative to the uniform rest state with the same mass, so it cannot decay to that state without dissipation.