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For uniform flow $\mathbf{U}$ plus a line vortex of counterclockwise circulation $\Gamma$, the outward advective momentum flux around a large circle is $\rho \mathbf{\Gamma }\times \mathbf{U}/2$. The Bernoulli equation supplies an equal pressure contribution. Their sum is the force on the fluid; the force on an obstacle per unit span is its negative, $\rho \mathbf{U}\times \mathbf{\Gamma }$. Retaining only the advective term loses half the lift and obscures its sign.

A steady top-hat plume model supplied with kinematic buoyancy flux $f_0>0$ but zero source volume flux and momentum flux hasThe Batchelor entrainment hypothesis and momentum balance yield these constants. Its Froude number satisfies $w^2/(g^{\prime }b)=5/(8\alpha )$. The Boussinesq approximation requires $g^{\prime }/g\ll 1$ and therefore fails at the ideal point source itself.

For the pressure-neglecting confined plume with compensating return flow, define specific plume momentum flux $M=b^2{w}^2$ and area fraction $s=Q^2/(R^{2}M)$. Then $J=M/(1-s)$ andThe branch issuing from a pure plume reaches a fold at $s=1/2$, where plume and ambient have equal areas, opposite velocities and equal specific momentum fluxes. The numerator is nonzero on this branch, so $M^{\prime }$ diverges although $M,Q,w$ remain finite. This signals failure of the steady uniform-profile description, not infinite physical velocity.

In an impermeable cylinder, an axisymmetric turbulent plume with negligible source volume flux requires a downward return flow. With uniform upward plume velocity $w$ and ambient velocity $v$, volume conservation gives $b^{2}w+(R^2-b^2)v=0$. An entrainment coefficient $\alpha$ based on the relative speed gives $Q^{\prime }=2\alpha b(w-v)$. A specified pressure closure is essential for momentum conservation. If the departure from ambient hydrostatic pressure and wall force are neglected in the whole-section integral, the total specific momentum flux $J=b^2{w}^2+(R^2-b^2)v^2$ obeys $J^{\prime }=b^2{g}^{\prime }$. Homogeneous unmodified ambient and no buoyancy loss give $b^{2}w{g}^{\prime }=B_s$. These are leading integral-model closures; restoring the ambient dynamic pressure can change the singularity criterion.

For a slender porous thermal plume, Darcy law gives $w=\beta (T-T_0)$, with $\beta =Kg\alpha /\nu$. If $C_h={\rho }_0{c}_p$ and $F_h=C_h\int w(T-T_0)dx$ is the conserved excess convective heat flux per span, then the kinematic momentum flux is $M=\int w^{2}dx=\beta F_h/C_h$, and the buoyancy flux is $B=\int g\alpha (T-T_0)wdx=g\alpha F_h/C_h$. The hyperbolic-secant porous plume profile has $Q=(36\kappa Mz{)}^{1/3}$. Multiplication of $Q$ and $M$ by ${\rho }_0$ gives physical mass flux and the density-weighted Darcy second moment. Intrinsic pore-velocity momentum flux instead has an additional inverse-porosity factor. Momentum conservation here follows from the heat/velocity proportionality, not from an inertia balance: porous drag balances buoyancy locally.

For a phase occupying area $\alpha A$ of a constant-area pipe, define $\bar{\rho }=(\alpha A{)}^{-1}\int \rho dA$, $\bar{v}=\int \rho vdA/(\alpha A\bar{\rho })$ and $\overline{v^2}=\int \rho v^{2}dA/(\alpha A\bar{\rho })$. The momentum per length is $A\alpha \bar{\rho }\bar{v}$, the advected momentum flux is $A\alpha \bar{\rho }\overline{v^2}$, and end pressures contribute $-A{\partial }_x(\alpha \bar{p})$. Applying Newton's second law to a slice proves the displayed balance, with $M$ all other phase forces per length. Interphase pressure terms can cancel the derivative of the area fraction. Equality of the second moment and squared mean requires a separate closure.

A forced plume has both imposed momentum flux and positive buoyancy flux. Its near-source region behaves like a jet, while its far field approaches a pure plume. The crossover is measured by a jet length.

Consider a particle of mass $m$ in a rigid rectangular box of length $L$ along $x$ and wall surface area $A$, so $V=AL$ is the volume. Choose $+x$ toward the right wall. For an incoming normal velocity $v_x=v>0$, an elastic collision reverses the normal momentum:The impulse delivered to the wall points outward, toward $+x$, by Newton's third law. The particle's opposite momentum change is not the wall's impulse. A tangential velocity component is unchanged by the ideal smooth-wall collision.

Ignoring intervening particle-particle collisions in this counting model, the particle travels a normal distance $2L$ before hitting the same wall again. Its hit rate is $|v_x|/(2L)$; multiplying by the impulse per hit gives the time-averaged normal force and its contribution to pressure:For example, stipulate $m=4\times 10^{-26}\mathrm{kg}$, $|v_x|=500\mathrm{m}{\mathrm{s}}^{-1}$, $L=0.10\mathrm{m}$, and $A=0.010{\mathrm{m}}^2$. Then $V=0.0010{\mathrm{m}}^3$, the impulse per hit is $4\times 10^{-23}\mathrm{N}\mathrm{s}$, the hit rate is $2500{\mathrm{s}}^{-1}$, and $P_1=10^{-17}\mathrm{Pa}$. This is one particle's tiny time-averaged contribution to one wall's pressure, not an isotropic equilibrium gas by itself.

For $N$ particles of equal mass, add their force contributions and define $⟨v_x^2⟩=N^{-1}{\sum }_{i=1}^N{v}_{x,i}^2$. In the gas's rest frame, isotropy gives $⟨v_x^2⟩=⟨v_y^2⟩=⟨v_z^2⟩=⟨v^2⟩/3$, where $v^2=v_x^2+v_y^2+v_z^2$. HenceThe final equality uses the classical equipartition theorem. The kinetic pressure of an isotropic gas therefore recovers the ideal gas law from momentum transfers. Doubling the normal speed would double both the impulse per hit and the hit rate, giving four times the pressure contribution; pressure measures a mean squared speed. Interparticle elastic collisions redistribute velocities and help establish isotropy; the individual round-trip times need not remain $2L/|v_x|$, but the equilibrium momentum flux result survives. ^[ref]

An inclined forced plume curves upward under buoyancy. In a stationary homogeneous ambient, its horizontal momentum flux is conserved and its vertical momentum flux increases. For kinematic plume fluxes, $m_x^{\prime }=0$ and $m_z^{\prime }=fq/m$, where $m=\sqrt{m_x^2+m_z^2}$ and the derivative is with respect to centreline arc length.

For a central differentiable pair potential, an exact local representation of the interaction momentum flux isHere $n_2$ is the two-particle number density. The line integral assigns each pair's exchanged momentum along the segment joining the particles. Translating both endpoints gives ${\partial }_s{n}_2=\mathbf{R}\cdot {\nabla }_{\mathbf{x}}{n}_2$. Integrating over $s$ and using pair-exchange symmetry proves that $-\nabla \cdot {\mathbf{P}}^{\varphi }$ is the interaction force density. Its homogeneous isotropic specialization gives the interaction part of the virial pressure from pair correlations. The kinetic pressure tensor accounts separately for random particle motion.

For source kinematic momentum flux $m_0$ and buoyancy flux $f_0>0$, dimensional analysis gives $L_J=m_0^{3/4}/f_0^{1/2}$. Under constant Batchelor entrainment, the actual turning length also contains an entrainment coefficient factor of order ${\alpha }^{-1/2}$.

With zero source volume flux and zero buoyancy flux but positive source momentum flux, the axisymmetric top-hat plume model reduces to a turbulent round jet. For full-cross-section kinematic fluxes, put $c=2{\alpha }_e\sqrt{\pi }$. Entrainment gives $Q^{\prime }=c\sqrt{M}$, while momentum conservation gives $M=M_0$. Hence $Q=c\sqrt{M_0}z$, speed $w=\sqrt{M_0}/(cz)$ and radius $r=cz/\sqrt{\pi }$. A finite source flux replaces $z$ by $z+Q_0/(c\sqrt{M_0})$. The ideal zero-flux source has singular speed and is a far-field model, not a finite nozzle.

Dividing the mass flux, momentum flux, and density-weighted buoyancy flux by ambient mass density gives the kinematic fluxes $q=\pi b^{2}V$, $m=\pi b^2{V}^2$, and $f=\pi b^{2}Vg({\rho }_0-\rho )/{\rho }_0$. This convention makes dimensional similarity solutions independent of the arbitrary ambient density unit.

A steady uniform incompressible jet has mass flow rate $\rho Au$, momentum flux $\rho A{u}^2$ and kinetic-energy flux $\rho A{u}^3/2$. These use the area and velocity at the same cross section. A nonuniform profile instead requires integrating powers of local speed; using mean speed in a cubic flux without a profile correction is generally wrong.

The invariant future mass-shell measure is $d^{3}p/E$, up to fixed normalization factors absorbed in $f$. Its relation to the Lorentz-invariant phase-space measure follows by integrating $2\theta (p^0)\delta (p_{\mu }{p}^{\mu }+m^2)d^{4}p$ over $p^0$ in a local orthonormal tetrad. A particle carries four-momentum $p^{\mu }$ and crosses a local spatial surface with velocity $p^i/p^0$. Integrating the associated momentum flux gives the displayed stress-energy tensor. For photons, $p^{\hat{a}}=E(1,\mathbf{e})$ and $|\mathbf{e}|=1$.

For an inviscid material volume element, neglecting body forces, Newton's second law gives $D(\rho \mathbf{u}\delta V)/Dt=-\nabla p\delta V$. Since mass conservation keeps $\rho \delta V$ constant, the result is $\rho D\mathbf{u}/Dt=-\nabla p$. Cancelling $\delta V$ before differentiating would incorrectly omit its changing volume. A fixed control volume instead requires the advected momentum flux, yielding ${\partial }_t(\rho \mathbf{u})+\nabla \cdot (\rho \mathbf{u}\otimes \mathbf{u})=-\nabla p$.

The momentum thickness weights the streamwise velocity deficit by the local velocity carrying it:Thus $\rho U^2\theta$ measures a momentum flux deficit per unit span relative to transporting the same local mass flow rate at $U$. It is not $\rho \int (U^2-u^2)dy$, which compares a different reference flux. A Blasius boundary layer has $\theta \simeq 0.664115\sqrt{\nu x/U}$; the ratio of displacement thickness to momentum thickness is about $2.591$.^[ref]

For the steady flow over an impermeable flat plate at constant $U$, integrating the boundary-layer equation and using mass conservation givesThis connects growth of the momentum thickness to local wall shear stress. Consequently the drag per unit span on one face up to $x=L$ is $D^{\prime }=\rho U^2\theta (L)$ in the ideal Blasius boundary layer with $\theta (0)=0$. Measured velocity profiles can therefore estimate integrated drag without differentiating noisy near-wall readings. The identity assumes no wall suction and zero outer pressure gradient.^[ref]

For a turbulent plume in a homogeneous ambient of density ${\rho }_0$, define actual sectional mass flux, momentum flux and density-weighted buoyancy flux by $Q=\pi b^2\rho w$, $M=\pi b^2\rho w^2$ and $F=\pi b^{2}wg({\rho }_0-\rho )$. With non-Boussinesq Batchelor entrainment $v_e=\alpha w\sqrt{\rho /{\rho }_0}$, their balances areHere $Q^2/M$ is sectional mass, $QF/M$ is sectional density-deficit buoyancy, and the entrained ambient initially has zero vertical momentum. The last equation follows by subtracting mass conservation from ${\rho }_0$ times volume conservation. Scase, Caulfield, Dalziel and Hunt's time-dependent plume model derives the equivalent system with the common factor $\pi$ suppressed.

Magnetic flux freezing ties a transverse fluid displacement to a bend in the magnetic field lines. For the uniform background, the transverse displacement equation isFor a sinusoidal bend, ${\partial }_z^2{\mathbf{\xi }}_{\perp }=-k_{\parallel }^2{\mathbf{\xi }}_{\perp }$. Positive pressure-anisotropy-modified magnetic tension therefore gives a restoring force, and inertia produces oscillation. In an isotropic plasma, this is precisely the positive magnetic tension $B_0^2/(4\pi )$.

Particles moving predominantly along bent magnetic field lines contribute a transverse momentum flux. Their larger parallel pressure opposes the restoring curvature force supplied by the magnetic field and perpendicular pressure. Once $p_{\parallel 0}-p_{\perp 0}>B_0^2/(4\pi )$, the net force points along the original displacement: a larger bend produces a larger outward force, giving positive feedback. Equivalently, the bending contribution to the quadratic potential energy is $\frac{1}{2}{\mathcal{T}}_0|{\partial }_z{\mathbf{\xi }}_{\perp }{|}^2$ and has become negative.

In this static, homogeneous fluid model the unstable roots areWith $e^{-i\omega t}$ the root $+i\gamma$ grows. Its real angular frequency is zero, so the unstable pattern grows without propagation in the equilibrium rest frame; its partner decays. Advection by a background flow would change that frame-dependent statement. The growth increasing without bound as $|k_{\parallel }|$ increases is a limitation of this long-wavelength model: finite gyroradius and other kinetic corrections cannot be neglected at sufficiently short scales.

Work in the rest frame of the shock. The normal momentum flux is conserved, so $p_1+{\rho }_1{u}_1^2=p_2+{\rho }_2{u}_2^2$. In the strong-shock limit $p_1\ll {\rho }_1{u}_1^2$, while mass conservation gives ${\rho }_2{u}_2={\rho }_1{u}_1$. The given Strong-shock Rankine-Hugoniot conditions therefore yieldUsing the downstream ideal gas law $p_2/{\rho }_2=k_B{T}_2/m$ givesFor a monatomic perfect gas, $\gamma =5/3$ and the coefficient is $3/16$. This is the immediate post-shock temperature, before substantial radiative cooling; ionization, dissociation or unequal electron and ion temperatures require additional physical modeling.