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The shallow-water approximation requires $h/L\ll 1$, a nearly hydrostatic pressure, negligible vertical acceleration, and approximately depth-independent horizontal velocity, concentration, and density. The large Reynolds number allows viscous stress to be neglected away from thin boundary layers, while the deep ambient is taken to remain stationary.

For unit channel width, volume conservation, chemical conservation, and mass conservation are respectivelyandEntrained ambient fluid contains no chemical and enters with density ${\rho }_0$; detrained fluid carries the local concentration and density. The ambient has no horizontal momentum, whereas detrained fluid carries horizontal momentum $\rho u$ per unit volume. The depth-integrated momentum conservation law is thereforeso $M=-\rho u{w}_d$.

Choose ${\rho }_0$ as a representative room density, for example the fresh-air density, and neglect relative density variations everywhere except in buoyancy. Withthe triangular profiles giveSolving these algebraic relations,

The Batchelor entrainment hypothesis, vertical momentum conservation, and mass conservation giveAn ascending parcel entrains ambient fluid from progressively lower ambient density. With the buoyancy frequencythe change of ambient reference density subtracts $Q{N}^2$ from its density-weighted buoyancy flux, so

In standard perturbation theory in cosmology, ${\delta }^{(n)}$ is an $n$-leg vertex carrying the symmetrized standard perturbation theory density kernel $F_n$. For Gaussian random fields, a connected four-point diagram with $L$ loops contains $3+L$ linear power spectra, hence $6+2L$ linear fields.

At tree level the required perturbative-order partitions areThe first is a star with one $F_3$ vertex and three linear external legs. The second has two $F_2$ vertices joined by one internal power-spectrum line, with one linear external leg attached to each vertex. At one loop all five partitions of eight into four positive parts are required:These are conventionally denoted $T_{5111}$, $T_{4211}$, $T_{3311}$, $T_{3221}$, and $T_{2222}$; all inequivalent placements of external legs and all Wick contractions are understood. This list is the complete set of one-loop matter trispectrum topologies.

Let $P_a=P_L(k_a)$ and ${\mathbf{k}}_{ab}={\mathbf{k}}_a+{\mathbf{k}}_b$. With ${\sum }_a{\mathbf{k}}_a=0$, the star contribution iswhere $\{a,b,c\}=\{1,2,3,4\}\setminus \{d\}$. For the exchange contribution, sum over the six unordered choices $\{a,b\}$ of linear external legs and let $\{c,d\}$ be the complementary pair:Thus the connected tree trispectrum is $T_{\mathrm{tree}}=T_{3111}+T_{2211}$.

The pressureless single-stream equations cease to be a complete one-loop prediction because loop momenta probe short nonlinear scales and produce ultraviolet-sensitive terms. A consistent result must include the effective field theory of large-scale structure: effective-stress counterterms, including their insertions into the tree topologies, and stochastic terms at the order allowed by mass conservation and momentum conservation. Their coefficients absorb short-scale dependence and must be fitted or matched. If the observable is a biased tracer rather than matter itself, the corresponding renormalized bias expansion is also required.

Split $X^{\mu }=x_0^{\mu }+X^{\prime \mu }$. Integrating the constant mode gives momentum conservation,For the nonzero modes, Wick theorem and the propagator giveThus, up to source-independent numerical normalization, the four-point amplitude isThe product is the Koba-Nielsen factor.

For the Möbius transformation $z^{\prime }=(az+b)/(cz+d)$ with $ad-bc=1$,The power of $|c{z}_i+d|$ contributed by every pair containing $i$ iswhere momentum conservation and the mass-shell condition $p_i^2=4/{\alpha }^{\prime }$ were used. The Koba-Nielsen factor therefore contributes $|c{z}_i+d{|}^4$ at each insertion, exactly cancelling the transformed measure. Hence the remaining integral is $SL(2,\mathbb{C})$ invariant, and division by its volume removes the residual conformal-gauge redundancy.

A perfectly inelastic collision is an inelastic collision in which the bodies stick and share one outgoing velocity. For two isolated point masses, momentum conservation gives $v=(m_1{u}_1+m_2{u}_2)/(m_1+m_2)$. At fixed total momentum, common velocity minimizes the outgoing translational kinetic energy, so sticking gives the greatest loss of that kinetic energy for the specified incoming state.

In a channel whose transverse width $b(z)$ does not depend on downstream position, the wetted area is $A(h)={\int }_0^{h}b(z)dz$ and the integrated hydrostatic pressure force divided by density is $P(h)=g{\int }_0^h(h-z)b(z)dz$. With cross-sectionally uniform velocity, volume conservation and momentum conservation give the displayed shallow water equations. Since $P^{\prime }(h)=gA(h)$, the two characteristic speeds are $u\pm c$ with $c^2=gA/A^{\prime }$. For $b(z)=C{z}^p$, $c^2=gh/(p+1)$ and the Riemann invariants are $u\pm 2(p+1)c$.

Radiation pressure is the normal momentum flux transferred by electromagnetic radiation to a surface. A photon travelling in vacuum carries momentum $p_{\gamma }=E_{\gamma }/c=h\nu /c$, where $E_{\gamma }$ is its energy, $\nu$ its frequency, $h$ the Planck constant and $c$ the speed of light. Photon absorption transfers that momentum; specular reflection reverses its normal component. This lets light exert a force even when almost no energy becomes heat in the receiver.

Two useful formulas describe different preparations. A directed beam normally incident on a stationary perfect absorber gives pressure $I/c$, and on a stationary perfect mirror gives $2I/c$. Here $I$ is light intensity, the incident power per beam cross-sectional area perpendicular to propagation, in $\mathrm{W}{\mathrm{m}}^{-2}$. An isotropic photon gas has pressure $u/3$, where $u=U/V$ is its radiation internal energy per volume, in $\mathrm{J}{\mathrm{m}}^{-3}$. Stored energy density and travelling energy flux must be distinguished: an isotropic field can have zero net energy flux and positive radiation pressure.

The physical appeal is that the same momentum conservation connects a single photon collision, a mirror's measurable force, and a radiation-filled cavity's equation of state. The directional calculation below complements radiation pressure on a perfectly absorbing surface; the equilibrium spectrum and its temperature dependence belong to blackbody radiation and the photon gas.

For normal incidence in vacuum, a uniform beam delivers energy $IA\Delta t$ to an illuminated surface area $A$ during time $\Delta t$. A perfect absorber receives momentum $IA\Delta t/c$, so force divided by surface area gives $p_{\mathrm{abs}}=I/c$. A perfect stationary mirror sends the photons back with reversed momentum, so momentum conservation gives twice the normal transfer, $p_{\mathrm{refl}}=2I/c$. These are time-averaged pressures; the microscopic arrivals need not be uniform.

For a stipulated model light intensity $I=1000\mathrm{W}{\mathrm{m}}^{-2}$ and the exact SI speed of light $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$,These are computed predictions, not measured data or a universal sunlight value. If the beam fully covers a normally oriented surface area $A=0.010{\mathrm{m}}^2$, the intercepted radiant flux is $\Phi =IA=10\mathrm{W}$ and the absorber's force is $\Phi /c\simeq 33.356\mathrm{nN}$. Using the total source power would be wrong if only part of its beam reaches the receiver. The absorption prediction assumes complete photon absorption, while the reflection prediction assumes complete specular reflection; transmission, scattering of light and directed thermal photon emission require their own outgoing momentum accounting.

Measure the angle of incidence $\theta$ between the incoming path and the surface's normal line, with $0\le \theta <\pi /2$. There are two separate cosine factors. A fixed actual surface area $A$ intercepts only its projected area $A\cos \theta$, so $\Phi =IA\cos \theta$. Each incoming photon supplies normal momentum $(E_{\gamma }/c)\cos \theta$ on absorption, or twice that on specular reflection. Thus, for radiometric absorptance $a$ and reflected power fraction $R$,This model assumes transmitted light continues straight in the same medium and carries away its original momentum. It omits thermal recoil and assumes a stationary receiver. At $\theta =60^{\circ }$, fixed beam light intensity therefore gives one quarter of the normal-incidence pressure. For fixed intercepted radiant flux $\Phi$, however, the normal force has only one cosine factor. Equivalently, if the given quantity is receiving-surface irradiance $E_{\mathrm{inc}}=I\cos \theta$, then $p_{\perp }=(a+2R)E_{\mathrm{inc}}\cos \theta /c$. The distinction matters when interpreting an experiment's angle scan. ^[ref]

Absorption also transfers a tangential momentum component at oblique incidence. It contributes a tangential force, rather than normal pressure. Ideal specular reflection preserves the tangential momentum component, so it transfers only normal momentum.

Let an interface move at signed speed $c$ in a reference unit normal direction $n$, and set $[a]=a^{+}-a^{-}$ with $n$ pointing from minus to plus. For continuous reference density and no surface sources, distributional momentum conservation, conservation of energy and the Lagrangian entropy inequality giveFor a level-set interface $\varphi =0$, ${\varphi }_t=-c|\nabla \varphi |$, the singular part of ${\partial }_{t}a$ is $-c[a]{\delta }_S$ and that of $\mathrm{Div}b$ is $n\cdot [b]{\delta }_S$, proving these formulas. Using $V=-c$ reverses the speed terms. Density jumps require jumps of the complete conserved densities.

For collinear motion without external forces, a rocket ejecting material backwards at constant speed $u$ in its instantaneous rest frame obeys $m{\gamma }^{2}dv+udm=0$. Here $m$ is the rocket's rest mass, including stored internal energy, and $\gamma$ is its Lorentz factor in the chosen inertial frame. The equation follows from momentum conservation, conservation of relativistic energy, and the relativistic velocity-addition formula; expelled rest mass is not simply the decrease in rocket rest mass.

The ideal rocket equation gives the change in signed velocity produced by ejecting rocket propellant. Assume one-dimensional Newtonian mechanics in an inertial frame, no external force, and a fixed positive rocket exhaust speed $u$ directed rearward relative to the instantaneous rocket. The rocket's instantaneous mass $M$ includes its remaining rocket propellant; burning it gives $dM<0$. For initial and final masses $M_{\mathrm{initial}}\ge M_{\mathrm{final}}>0$,Here $\ln$ is the natural logarithm, and its argument is the rocket mass ratio. This is the constant-rocket exhaust speed, zero-pressure-correction version of the ideal rocket equation. ^[ref]

For a toy rocket with $u=2km/s$, $M_{\mathrm{initial}}=1000\mathrm{kg}$ and rocket dry mass $M_{\mathrm{final}}=250\mathrm{kg}$ after all rocket propellant is spent, the rocket mass ratio is $4$ andBurning from $1000$ to $500\mathrm{kg}$ gains $2\ln 2km/s\simeq 1.3863km/s$. Burning from $500$ to $250\mathrm{kg}$ gains exactly the same velocity increment, using only $250\mathrm{kg}$ of rocket propellant instead of $500\mathrm{kg}$. Both burns halve the remaining mass. The second burn accelerates a lighter rocket; equal fractional mass losses give equal velocity gains.

The natural logarithm makes the rocket equation both useful and revealing: multiplying successive rocket mass ratios adds their velocity gains. Conversely, the required rocket mass ratio is the exponential function $M_{\mathrm{initial}}/M_{\mathrm{final}}=e^{\Delta v/u}$. Each extra $u\ln 2$ of ideal velocity gain doubles the required initial mass for fixed final mass. This explains why a larger rocket exhaust speed is valuable and why carrying ever more rocket propellant has diminishing returns. Changing the burn duration alone does not change this ideal endpoint result.

For a short proof outline, follow the rocket together with a tiny parcel about to be expelled. Before ejection, their total momentum is $Mv$. After ejection, the remaining rocket has mass $M+dM$ and velocity $v+dv$, while the parcel has positive mass $-dM$ and, to first order, inertial velocity $v-u$. Momentum conservation givesDiscarding the second-order product $dMdv$ gives $Mdv=-udM$; taking the integral gives $\Delta v=-u{\int }_{M_{\mathrm{initial}}}^{M_{\mathrm{final}}}dM/M$. Evaluating the integral yields the natural logarithm above. This is a variable-mass system calculation: the rocket alone is not a closed system, and its momentum changes as the expelled material carries away momentum. The fixed collection consisting of the rocket and all its expelled material has conserved total momentum when external forces vanish.

Fixed rocket exhaust speed means fixed relative speed at release, not fixed inertial velocity. As the rocket's velocity $v$ changes, newly emitted material has changing inertial velocity $v-u$; previously released parcels coast at their own release velocities in this ideal model. If $v>u$, the exhaust even moves forward in the inertial frame, although it is emitted rearward relative to the rocket. No surrounding medium is needed for this exchange of momentum.

Real flight adds gravity, drag force and possibly a pressure correction to rocket thrust. A varying rocket thrust due solely to a changing mass loss rate still gives the same ideal rocket equation if $u$ is constant; it changes the acceleration and burn duration. A changing rocket exhaust speed instead requires $\Delta v=-\int udM/M$, and external forces require their own contribution to the velocity change. The existing constant mass-loss propulsion with quadratic drag and constant-burn rocket with ground friction illustrate such additional forces; the forward-exhaust rocket equation reverses the ejection direction and therefore the sign of the recoil.

For a rocket equation with exhaust mass rate $\dot{m}$ and incoming stationary mass rate $\dot{D}$, mass balance is $\dot{M}=\dot{D}-\dot{m}$. A short-interval momentum conservation calculation gives $M\dot{v}=u\dot{m}-v\dot{D}$. Constant time rates $\dot{D}=\alpha$, $\dot{m}=\beta$ give $M\dot{v}+\alpha v=u\beta$, unlike a constant accretion rate per distance, whose incoming mass rate is proportional to speed. For unequal rates and initial rest, $v=(u\beta /\alpha )[1-(M/M_0{)}^{\alpha /(\beta -\alpha )}]$.

The tree-level Polyakov path integral for closed-string tachyons reduces in conformal gauge to their vertex-position integral on the Riemann sphere. Three positions are fixed by the residual Möbius group; the other $N-3$ complex positions are integrated. The embedding-field Gaussian integral produces momentum conservation and the Koba-Nielsen factor.

For internal momentum $q\gg k$, the P22 contribution to the one-loop matter power spectrum starts at $k^4$, while twice the P13 contribution to the one-loop matter power spectrum starts at $k^{2}P(k)$. The latter has the same momentum dependence as the effective sound-speed counterterm in large-scale structure. Short-scale stochastic contributions start at $k^4$ by mass and momentum conservation.

A body exchanging mass with material of inertial velocity $\mathbf{u}$ satisfies the displayed momentum conservation law for one stream, with signed $\dot{m}$ positive for entry and negative for exit. The force $\mathbf{F}$ excludes the incoming or outgoing momentum flux. A short time interval gives $d(m\mathbf{v})/dt=\mathbf{F}+\mathbf{u}\dot{m}$, proving the formula. For stationary incoming matter, $d(m\mathbf{v})/dt=\mathbf{F}$; for matter leaving at the body's own velocity, $m\dot{\mathbf{v}}=\mathbf{F}$. The rocket equation and accreting-drop motion are applications.

For a rectangular channel with stationary width $b(x)$ and constant reduced gravity $g^{\prime }$, volume conservation and momentum conservation give $h_t+u{h}_x+h{u}_x=-hu{b}^{\prime }/b$ and $u_t+u{u}_x+g^{\prime }{h}_x=0$. With $c=\sqrt{g^{\prime }h}$ the characteristic curves have speeds $u\pm c$. The combinations $u\pm 2c$ obey the displayed compatibility equations; they are Riemann invariants only when the width source term vanishes.