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Take a unit-width control volume spanning the stationary hydraulic jump. Mass conservation makes the discharge $q=hu$ constant. The horizontal momentum flux is $\rho h{u}^2$, and the integrated hydrostatic pressure force on a vertical cross-section is ${\int }_0^h\rho g(h-z)dz=\rho g{h}^2/2$. Momentum conservation therefore givesThese are the Rankine-Hugoniot conditions for the shallow water equations in the stationary shock frame. With depths $h_1\ne h_2$ on the two sides, eliminating the velocity gives

In the local rest frame of a gas, the pressure tensor is the directional momentum flux $P_{ij}=\int p_i{v}_{j}f(\mathbf{p})d^{3}p$, where $f$ is the phase-space distribution function per unit momentum volume and $\mathbf{v}$ is particle velocity. For an isotropic distribution, $P_{ij}=P{\delta }_{ij}$, recovering the kinetic pressure of an isotropic gas. An anisotropic distribution can exert different normal pressures along different directions.

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 radiation in vacuum with isotropy, energy is distributed equally per solid angle: write $u_{\Omega }=u/(4\pi )$ for energy density per unit solid angle. Let $\theta$ be the angle between a ray's path and the wall normal. Its normal velocity $c\cos \theta$ determines the arrival rate, while normal momentum per energy is $\cos \theta /c$. Their product is ${\cos }^2\theta$. For a perfectly reflecting cavity wall,The leading two is the normal momentum reversal; the incoming hemisphere contains half of the field's directional energy density. This is also the normal component of the bulk pressure tensor, $P_{nn}={\int }_{4\pi }{u}_{\Omega }{\cos }^2\theta d\Omega =u/3$. Isotropy makes all three diagonal components equal, while their sum is $u$ because the squared direction components sum to one. The factor one third therefore depends on three-dimensional angular geometry, not on the shape of the blackbody radiation spectrum.

A perfectly absorbing wall exposed to this incoming hemisphere receives only $u/6$ from absorption. If it is in thermal equilibrium with the isotropic radiation, its outgoing emission supplies another $u/6$ of recoil, recovering $u/3$. A cold absorber before equilibrium is a different boundary condition. Similarly, the one-way incident energy flux is $cu/4$, whereas the net energy flux across a plane in the undisturbed isotropic field is zero. Dividing $cu/4$ by $c$ misses the extra normal-momentum projection and does not give the cavity pressure.

For a stipulated isotropic radiation energy density $u=6\mu \mathrm{J}{\mathrm{m}}^{-3}$, the bulk pressure is $2\mu \mathrm{Pa}$ regardless of its spectral shape. In thermal equilibrium, blackbody radiation supplies $u=a_{\mathrm{rad}}{T}^4$ with radiation constant $a_{\mathrm{rad}}=4\sigma /c$, so $p_{\mathrm{rad}}=a_{\mathrm{rad}}{T}^4/3$; Stefan–Boltzmann law defines $\sigma$. The equilibrium photon gas derives this same equation of state from its free energy.^[ref]

The excess depth-integrated, time-averaged momentum flux of a surface gravity wave is its radiation stress. In deep water $c_g/c_p=1/2$, so $S_{xx}=\mathcal{E}/2={\rho }_{w}g{a}^2/4$. The force per transverse width on a reflecting object is the incident plus reflected minus transmitted radiation stress. Reflection increases force because the reflected wave carries momentum in the opposite direction.

A sled intercepting snow with zero horizontal velocity in the ground frame is a simple momentum flux example. At arrival rate $\sigma$, immediate removal with outgoing ground-frame horizontal velocity $v_{\mathrm{out}}$ gives $M\dot{v}=-\sigma v_{\mathrm{out}}$ for constant moving mass $M$. Removal sideways in the sled frame has $v_{\mathrm{out}}=v$ and exponential deceleration; removal sideways in the ground frame has $v_{\mathrm{out}}=0$ and constant sled velocity. Retained snow instead gives $(M+\sigma t)v=M{v}_0$. The outgoing reference frame must be specified.

Take $v=-{\mathcal{X}}_z$, $w={\mathcal{X}}_y$, constant density gradient $S<0$, and meridional eddy density flux $F=\sin (ly)[-1_{z<0}]$. If the divergence of the eddy momentum flux is zero, bounded Eulerian circulation and decaying residual mean circulation give $\mathcal{X}=\sin (ly)\chi (z)$ with $\chi =-A_0(1-e^{\kappa z}/2)$ below zero and $\chi =-A_0{e}^{-\kappa z}/2$ above, where $A_0=-1/S$ and $\kappa =Nl/|f_0|$. The transformed streamfunction is ${\mathcal{X}}^{\ast }=\mathcal{X}+F/S$, yielding the displayed ${\chi }^{\ast }$ and two oppositely rotating cells. Its jump represents a horizontal delta-function transport at the discontinuous flux termination. A smooth flux cutoff replaces that sheet by a thin regular return flow. Density flux alone does not determine either circulation if the momentum-flux divergence is left unspecified.

A circular turbulent round jet supplied with kinematic momentum flux $m_0$ but negligible buoyancy flux conserves $m=m_0$. A top-hat plume model and constant Batchelor entrainment give $dq/ds=2\alpha \sqrt{\pi m_0}$. From a point source with zero volume flux, $q=2\alpha \sqrt{\pi m_0}s$, radius $b=2\alpha s$, and axial velocity $V=m_0/q\propto s^{-1}$.

For a top-hat plume model with full width $w$, upward speed $U$, reduced gravity $b$, and edge entrainment coefficient $e$, define volume flux per unit span $V=wU$, kinematic momentum flux $M=w{U}^2$, and buoyancy flux $B=wUb$. A uniform ambient and two exposed edges give $V^{\prime }=2eU$, $M^{\prime }=wb=B/U$, and $B^{\prime }=0$. A pure plume with $V(0)=0$ hasIndeed $M=UV$ with constant $U$ reduces the momentum balance to $2e{U}^3=B$. If width means half-width, its growth coefficient is $e$, not $2e$; profile shape factors also alter the coefficients.

Integrating a phase-space distribution function against products of particle velocity components produces its velocity moments. The zeroth gives number density, the first gives number flux, and, after multiplying by particle mass, the second gives momentum flux. Replacing $\mathbf{v}$ by $\mathbf{v}-\mathbf{u}$, where $\mathbf{u}$ is the mean velocity, defines central velocity moments; the second is the pressure tensor divided by particle mass.