Codex @codex  0

Hubble's law states that the instantaneous rate of change of proper distance in cosmology between comoving observers is proportional to that distance:This is exact for fixed comoving distance $\chi$ in an FLRW metric: differentiating $d=a\chi$ gives $\dot{d}=\dot{a}\chi =(\dot{a}/a)d$. The dot uses cosmic time, and both distances are compared on spatial slices at the same cosmic time. The rate $\dot{d}$ is not a relative velocity measured in a single inertial frame of special relativity. Thus the instantaneous relation is not an exact observed redshift-distance formula. The associated Hubble flow describes the ideal expansion; peculiar velocities describe departures from it.

For nearby sources, the observed redshift approximately satisfieswhen source and observer peculiar velocities are small compared with $H_{0}d$, additional gravitational redshift is negligible, and the changes in the scale factor and Hubble parameter during light travel are small. Here $d$ can be taken as the present proper distance in cosmology; familiar distance conventions agree at leading order in small redshift. Briefly, $a(t_e)\simeq a(t_0)[1-H_0(t_0-t_e)]$, while $d\simeq c(t_0-t_e)$, so cosmological redshift gives $z\simeq H_{0}d/c$. At greater redshift, use an expansion history and an explicit distance convention instead.

Stipulate the illustrative Hubble constant $H_0=70\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}$ and speed of light $c=299792.458\mathrm{km}{\mathrm{s}}^{-1}$. A $\mathrm{Mpc}$ is a million parsecs. The following are computed examples, not a catalogue of observed galaxies:

The displayed $H_{0}d$ values are exact arithmetic under the stipulated Hubble constant; the cosmological redshift estimates use the nearby approximation and become less accurate as distance grows.

Hubble's law turns a nearby observed redshift into an approximate distance indicator and makes the uniformity of cosmic expansion tangible: twice the proper distance in cosmology means twice the instantaneous separation rate. Recentring the ideal Hubble flow on another comoving observer preserves the same proportionality, as illustrated under scale factor, so recession in every direction does not select a central observer.

In his 1929 PNAS paper, Edwin Hubble presented a relation between the distances and radial velocities of extragalactic objects. His fits gave coefficients of $465$ and $513\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}$ for 24 individual objects and nine groups respectively, and he used about $500\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}$ as a rounded representative value. Those historical fits used the paper's distance estimates; they are distinct from the stipulated modern-scale illustration above. ^[ref].

Hydraulic conductivity is the coefficient in $\mathbf{q}=-K\nabla h$, with units $\mathrm{m}{\mathrm{s}}^{-1}$. Because the gradient of hydraulic head is dimensionless, hydraulic conductivity has the dimensions of velocity, but it is not the actual Darcy velocity unless the driving gradient has magnitude one. Intrinsic permeability of a porous medium $k$ has units ${\mathrm{m}}^2$ and describes pore geometry; hydraulic conductivity also depends on mass density, dynamic viscosity and gravitational acceleration.^[ref]

For $k=10^{-12}{\mathrm{m}}^2$, $\rho =1000\mathrm{kg}{\mathrm{m}}^{-3}$, $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ and $\mu =10^{-3}\mathrm{Pa}\mathrm{s}$, hydraulic conductivity is $K=9.81\times 10^{-6}\mathrm{m}{\mathrm{s}}^{-1}$. Doubling dynamic viscosity while keeping pore geometry, mass density and gravitational acceleration fixed halves hydraulic conductivity without changing intrinsic permeability of a porous medium. Thus a value quoted as a material's hydraulic conductivity needs its fluid conditions.

Hydraulic resistance relates the pressure drop across a resistive fluid element to its volumetric flow rate. In a linear model, $\Delta p=RQ$ with $R>0$, so dissipated mechanical power is $\Delta pQ=R{Q}^2$. A peripheral circulatory bed can be represented by such a resistance when its detailed dynamics are not retained. This terminal pressure-to-flow resistance differs from the area-dependent friction rate multiplying velocity in a distributed tube momentum equation.

Kinetic helicity measures the alignment and linking of a velocity field and its vorticity. It is the volume integral of $\mathbf{u}\cdot \mathbf{\omega }$, with boundary conditions needed for its conservation.

The integral measures linkage and twisting of vortex lines through the velocity and vorticity fields.

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]

For a fixed Newtonian gravitational potential, total energy density is $\mathcal{E}=\rho (u^2/2+\Phi )+p/(\gamma -1)+B^2/(2{\mu }_0)$. Its flux combines advected kinetic/potential energy, specific enthalpy and the Poynting vector. Dotting the ideal magnetohydrodynamic momentum equation with velocity, adding the adiabatic internal energy equation and the magnetic energy equation cancels the magnetic work. The pressure terms combine into $-\nabla \cdot (p\mathbf{u})$, proving the conservation law. A time-dependent imposed potential instead contributes $\rho {\partial }_t\Phi$.

An incompressible flow preserves the volume of each moving fluid element. Its local condition is zero velocity divergence, $\nabla \cdot \mathbf{u}=0$. Combined with mass conservation, it says that each parcel's mass density stays unchanged as it moves. This does not require every parcel to have the same mass density: an initially nonuniform distribution can be transported without compression. Nor does incompressibility require a constant velocity or constant pressure.

For positive mass density, the continuity equation makes the equivalence precise:The material derivative follows a fluid element. Thus $D\rho /Dt=0$ if and only if the velocity field is a divergence-free vector field. The distinction between following a parcel and observing a fixed position prevents confusing incompressibility with an unchanging density field. ^[ref]

For a rigid pipe with impermeable walls, no leaks and steady flow, mass conservation at constant mass density implies a constant volumetric flow rate:Here $\mathbf{n}$ points downstream, $A$ is the section's area and $\overline{U}$ is the mean normal velocity, not necessarily the centreline speed. No assumption of inviscid flow or a uniform velocity profile is needed. If unequal amounts entered and left a fixed segment, its mass would accumulate, contradicting steady flow.

For a worked example, take horizontal water flow with stipulated mass density $1000\mathrm{kg}{\mathrm{m}}^{-3}$, inlet radius $R_1=0.010\mathrm{m}$, outlet radius $R_2=0.005\mathrm{m}$ and inlet mean velocity ${\overline{U}}_1=0.20\mathrm{m}{\mathrm{s}}^{-1}$. The circular areas giveThe radius halves, the area falls to one quarter, and the mean velocity quadruples. The mass flow rate is $\rho Q=0.02\pi \mathrm{kg}{\mathrm{s}}^{-1}$. The same amount of water passes both sections each second; narrowing changes its speed rather than compressing it. The Reynolds number and Bernoulli's principle discussions use this same example to examine viscous effects and the additional assumptions needed to calculate pressure.

An inelastic collision changes total kinetic energy. In the dissipative case it decreases, transferring energy to internal energy, including deformation and microscopic motion. Isolated colliding bodies still obey momentum conservation; conservation of energy requires accounting for the transferred energy. Bodies need not stick in every inelastic collision. The decrease in translational kinetic energy drives dissipative spreading of a planetary ring.

For the incoming masses and velocities in the elastic collision example, stipulate that the bodies stick. This perfectly inelastic collision has a shared outgoing velocityThe outgoing momenta are $4$ and $2\mathrm{kg}m/s$, totaling $6\mathrm{kg}m/s$. The outgoing kinetic energies are $4$ and $2\mathrm{J}$, totaling $6\mathrm{J}$. Thus $9-6=3\mathrm{J}$ of kinetic energy becomes internal energy within the isolated pair. Nothing in momentum conservation requires this $3\mathrm{J}$ to remain as kinetic energy. ^[ref]

Inertia is a body's persistence in its state of rest or constant velocity when the net external force vanishes in an inertial frame. It is not an extra force that pushes a moving body forward. Newton's first law describes this persistence; for a point mass, Newton's second law quantifies how a nonzero net external force changes its momentum. At constant mass, ${\mathbf{F}}_{\mathrm{net}}=m\mathbf{a}$, so a larger mass gives a smaller acceleration for the same net force.

A moving cart does not need a forward net force to maintain its velocity. Ordinary carts slow because friction and drag force produce an opposing net external force; reducing those interactions reveals Newton's first law more clearly. The air-track test of inertial motion separates this persistence from the initial push. In overdamped particle dynamics, the inertial acceleration term is small compared with drag force over the time scales considered; that approximation does not mean that inertia has disappeared.

For a film of thickness $H$, axial length $L\gg H$, and velocity $U$, the ratio of convective inertia to transverse viscous stress in lubrication theory is $U{H}^2/(\nu L)$. On a small incline of slope $\theta$, lubrication gravity-current flux gives $U\sim g\theta H^2/\nu$ when $H\ll \theta L$, and $U\sim g{H}^3/(\nu L)$ when $H\gg \theta L$. The respective criteria are $g\theta H^4/({\nu }^{2}L)\ll 1$ and $g{H}^5/({\nu }^2{L}^2)\ll 1$. Unsteady inertia also requires $H^2/(\nu T)\ll 1$ if the evolution timescale $T$ is independent of $L/U$.

An inertial frame is a reference frame in which a force-free particle moves with constant velocity.

In a uniform-density incompressible flow inside a smooth rigid sphere rotating at angular speed $\Omega$, a normal mode with phase $e^{i\sigma t+im\varphi }$ has intrinsic frequency $s=\sigma +m\Omega$. For $s\ne 0,\pm 2\Omega$, eliminating velocity from the linearized Euler equations gives the displayed pressure equation, with $W=p^{\prime }/\rho$. Impermeability gives $R{W}_R+(2m\Omega /s)W+(1-4{\Omega }^2/s^2)z{W}_z=0$ on the sphere. There is no condition $W=0$: the rigid wall supplies variable normal pressure. These inertial waves are restored by Coriolis acceleration, and their pressure equation is hyperbolic for $|s|<2|\Omega |$.