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Kinetic energy is the energy associated with motion. For a nonrelativistic point mass of mass $m>0$, measured in a specified inertial frame,Here $\mathbf{v}$ is velocity and $v$ is speed; $T$ is another common symbol for kinetic energy. The dot product makes $K$ nonnegative regardless of the direction of velocity. At fixed mass, doubling speed quadruples kinetic energy: a $2\mathrm{kg}$ point mass at $2m/s$ has $4\mathrm{J}$, whereas at $4m/s$ it has $16\mathrm{J}$. The SI units are $\mathrm{kg}{\mathrm{m}}^2/{\mathrm{s}}^2=\mathrm{J}$. Different reference frames can assign different velocities and therefore different kinetic energies. ^[ref]

A short outline explains why kinetic energy has this formula. In a nonrelativistic inertial frame, constant mass and Newton's second law give, using the chain rule,Integrating over time proves the work-energy theorem, $W_{\mathrm{net}}=K_f-K_i$. The net force is the sum of all forces, and $W_{\mathrm{net}}$ is their total work, not merely the work of a chosen driving force. This route to final speed is useful when the forces as functions of position are known but the full motion as a function of time would be harder to find. ^[ref]

In the running example, a $2\mathrm{kg}$ point mass starts at rest and receives a constant net $3\mathrm{N}$ force through $4\mathrm{m}$. Its net work is $12\mathrm{J}$, soAt any intermediate distance $x\ge 0$, $K(x)=(3\mathrm{N})x$ and $v(x)=\sqrt{2Fx/m}$: kinetic energy is a linear function of distance, while speed is proportional to the square root of distance. If a net opposing $3\mathrm{N}$ braking force subsequently acts for $4\mathrm{m}$, it does $-12\mathrm{J}$ of work and brings this point mass to rest. Negative net work decreases kinetic energy; it does not make kinetic energy negative. Beyond the stopping point the motion must be reconsidered, since the point mass cannot keep moving forward with a negative $K$.

Kinetic energy also makes the exchange with potential energy intuitive: a falling body's loss of gravitational potential energy can appear as increased kinetic energy. The formula above applies when speed is small compared with the speed of light; at higher speeds, special relativity changes the expression. ^[ref]

Writing particle positions relative to the center of mass as ${\rho }_i=r_i-R$ makes ${\sum }_i{m}_i{\rho }_i=0$, so the cross term in the squared velocities vanishes. The total kinetic energy splits into center-of-mass translation and relative motion as displayed. For a rigid body, the latter is $\frac{1}{2}I{\omega }^2$, where $I$ is the moment of inertia about the instantaneous rotation axis through the center of mass.

For fixed particle masses and positive total mass, write ${\mathbf{x}}_i=\mathbf{R}+{\mathbf{y}}_i$ with $\mathbf{R}$ the centre of mass. The weighted relative velocities sum to zero, so expanding the kinetic energy eliminates the mixed term and yields the displayed decomposition. For a rigid body, the relative contribution is the rotational kinetic energy $I{\omega }^2/2$ about the chosen centre-of-mass axis.

Kinetic friction is friction during nonzero contact slip. For one-dimensional motion along a surface, choose right as positive and write $v$ for the body's contact-point velocity and $U$ for the surface's contact-point velocity, both measured in the same inertial frame. The signed relative velocity is $w=v-U$. The ideal Coulomb law for the tangential force on the body isHere $N\ge 0$ is the magnitude of the normal force, ${\mu }_k\ge 0$ is the coefficient of kinetic friction, and $\mathrm{sgn}$ is the sign function. Thus kinetic friction opposes contact slip, which can differ from the body's direction of motion. In several tangential directions the same law reads ${\mathbf{f}}_k=-{\mu }_{k}N\mathbf{w}/|\mathbf{w}|$ for a nonzero tangential relative velocity $\mathbf{w}$. The magnitude law is documented in Ling, Sanny and Moebs, OpenStax University Physics 1, §6.2: static and kinetic friction.

For a stipulated toy model, take a nonrotating box with mass $m=1\mathrm{kg}$, normal force $N=10\mathrm{N}$, coefficient of static friction ${\mu }_s=0.5$, and coefficient of kinetic friction ${\mu }_k=0.3$. These are assumed model parameters, rather than measured properties of any particular pair of materials; the prescribed normal force is consistent with horizontal support under $g=10m/{\mathrm{s}}^2$ and no other vertical forces. On stationary ground, a box already sliding right has $f_k=-3\mathrm{N}$. With an applied horizontal force $F=6\mathrm{N}$ to the right, Newton's second law gives $a=(F+f_k)/m=3m/{\mathrm{s}}^2$, for as long as rightward contact slip persists.

Direction is clearer by comparing two instantaneous states of this same model. If the box has $v=1m/s$ on stationary ground, $U=0$ and $w=+1m/s$, so kinetic friction acts left with magnitude $3\mathrm{N}$. If the box still has $v=1m/s$ but a belt beneath it has $U=2m/s$, then $w=-1m/s$: kinetic friction acts right with magnitude $3\mathrm{N}$, even while the box moves right.

In the belt case, the power delivered to the box is $P_{\mathrm{box}}=f_{k}v=+3\mathrm{W}$. With no other horizontal force, its kinetic energy therefore increases by the work-energy theorem. By Newton's third law, the box exerts $-f_k$ on the belt, so the contact power delivered to the belt is $P_{\mathrm{belt}}=-f_{k}U=-6\mathrm{W}$. Their sum is $-3\mathrm{W}$: frictional heating produces thermal internal energy at $3\mathrm{W}$. An ideal motor maintaining the prescribed belt velocity supplies $+6\mathrm{W}$ against this contact load, neglecting other losses, so its input divides into $3\mathrm{W}$ of box kinetic energy and $3\mathrm{W}$ of frictional heating. These are instantaneous rates, since the box's velocity subsequently changes. This balance follows directly from power as force times velocity and conservation of energy; the power law is given in Ling, Sanny and Moebs, OpenStax University Physics 1, §7.4: power.

When contact slip vanishes, including a transition to rolling without slipping, the nonzero-slip law no longer applies. Determine the required static friction from the no-slip condition and the equations of motion, then test its bound. Substituting $\mathrm{sgn}(0)=0$ into the sliding law would incorrectly exclude a nonzero static friction.

Kinetic theory describes many-particle systems through a distribution on phase space, retaining both position and velocity. In the number-density convention, $f(t,\mathbf{x},\mathbf{v})d^{3}x{d}^{3}v$ is the expected number of particles in a small position–velocity region. For identical particles of mass $m$, its integrals give the number density $n$ and bulk velocity $\mathbf{u}$:This $f$ is not normalized to one: its integral over the whole phase space gives the total expected particle number. Transport moves particles along trajectories, while collision or diffusion terms change the distribution. The Boltzmann equation, Hamiltonian Liouville equation and Fokker-Planck equation describe different physical assumptions, rather than interchangeable collision laws.

An ideal gas is a motivating example: the pressure on a wall comes from microscopic momentum transfers, even when the bulk velocity is zero. The Maxwell-Boltzmann velocity distribution separates this random molecular motion from bulk motion; see ideal gas pressure from wall collisions. The attractive feature of kinetic theory is that a distribution can describe two opposing populations where one average velocity hides almost everything. It also tells us when averaging into a continuum approximation loses essential information: compare the intermolecular mean free path with the apparatus scale using the Knudsen number. ^[ref]

The Knudsen number is the dimensionless ratio of an intermolecular mean free path ${\ell }_{\mathrm{mfp}}$ to a specified characteristic length $L$ of the apparatus or flow. The mean free path measures the distance travelled between gas–gas collisions; $L$ measures the distance over which boundaries or macroscopic conditions matter. Wall encounters must not be counted as gas–gas collisions in this comparison: doing so would hide the very wall-dominated behavior we want to detect. The same gas can therefore have a small Knudsen number in a large vessel and a large Knudsen number in a tiny opening.

For a dilute hard-sphere gas of identical molecular diameter $d$ in thermal equilibrium, use the conventionHere $n$ is the number density, $p$ the pressure, $T$ the absolute temperature, and $k_B$ the Boltzmann constant; the second equality uses the ideal gas law. The collision cross-section is $\pi d^2$, and the $\sqrt{2}$ accounts for both partners moving with independent equilibrium velocities. Mixtures and velocity-dependent collision cross-sections require a stated molecular model and mean free path convention; the displayed expression is not an exact formula for every real gas. ^{[ref] [ref]}

As a numerical toy example, choose $T=300\mathrm{K}$ and $d=0.37\mathrm{nm}$, without claiming that this is a measured molecular diameter for a particular gas. The formula gives ${\ell }_{\mathrm{mfp}}\simeq 68.1\mathrm{nm}$ at $p=10^5\mathrm{Pa}$. It yields $\mathrm{Kn}\simeq 6.81\times 10^{-6}$ for a $1\mathrm{cm}$ vessel but $\mathrm{Kn}\simeq 0.681$ for a $100\mathrm{nm}$ gap. At $p=1\mathrm{Pa}$ the same toy gas has ${\ell }_{\mathrm{mfp}}\simeq 6.81\mathrm{mm}$, comparable to the centimetre vessel. Reducing either pressure or apparatus size can thus expose molecular transport.

Choose $L$ for the question being asked: a tube diameter compares the mean free path with wall separation, while tube length describes a different geometric scale. For a positive macroscopic field $q$, a local gradient scale can instead be $L_q=|q|/|\nabla q|$, where its gradient is nonzero. A large vessel can contain a thin region with a large local Knudsen number. Small geometric Knudsen number alone does not guarantee a slowly varying near-equilibrium flow. The time, velocity and space scalings must also be stated when the Knudsen number appears as a coefficient in a Boltzmann equation. ^[ref]

The Lagrangian fluid displacement $\mathbf{\xi }$ compares neighbouring fluid trajectories. Linear velocity perturbations about a steady flow $\mathbf{U}$ obey $\delta \mathbf{u}=({\partial }_t+\mathbf{U}\cdot \nabla )\mathbf{\xi }-(\mathbf{\xi }\cdot \nabla )\mathbf{U}$. For a background scalar $f_0$, its material perturbation is $\Delta f=\delta f+\mathbf{\xi }\cdot \nabla f_0$. In an axisymmetric normal mode about cylindrical rotation, the radial and vertical velocities are $-i\omega {\xi }_r$ and $-i\omega {\xi }_z$. The displacement of a free surface is needed to impose the Lagrangian pressure perturbation condition.

// Target: astrophysics.bigb

Laminar flow is a regime of fluid flow in which the motion remains organized, with neighboring fluid parcels following smooth, coherent paths rather than the irregular interacting fluctuations characteristic of turbulence. In a straight pipe, the simplest example consists of coaxial layers sliding past one another. This describes the character of the motion, not a requirement that every parcel have the same velocity. Nor does it require steady flow: a smooth profile can evolve in time as viscosity transmits motion from a newly moving wall into a resting fluid. The unsteady wall-driven solution provides an explicit example. ^[ref]

The Reynolds pipe-and-dye experiment makes this distinction visible. A thin injected dye band can remain coherent in laminar flow, whereas irregular motion can fold it and carry dye across the pipe. However, molecular diffusion can spread dye even in laminar flow, and an inlet disturbance can move a coherent band without producing turbulence. A dye band is a streakline, not a measurement of the entire velocity field; in steady flow an ideal passive tracer's streakline coincides with a streamline. Reynolds explicitly distinguished a wandering band from one broken into eddies. ^[ref]

For a circular pipe of internal diameter $D$, define the mean axial velocity $U$ using the volumetric flow rate $Q$ and cross-sectional area $A$:Here $\rho$ is mass density, $\mu$ is dynamic viscosity and $\nu =\mu /\rho$ is kinematic viscosity. This particular Reynolds number uses the diameter and mean velocity, not the radius or centreline velocity. Small Reynolds number generally favors laminar flow, but a single universal critical Reynolds number does not distinguish it from turbulence in all geometries and experiments. Even in a given pipe, inlet conditions and the magnitude of disturbances affect the observed onset. Reynolds demonstrated this dependence by deliberately inserting a wire coil. ^[ref]

The quantitative example above assumes steady flow, incompressible flow, a Newtonian fluid with constant dynamic viscosity, a straight horizontal circular pipe of radius $R=D/2$, stationary walls satisfying the no-slip boundary condition, and fully developed flow. The last assumption means that the axial velocity profile no longer changes downstream. A positive pressure drop $\Delta p=p_{\mathrm{in}}-p_{\mathrm{out}}$ over a developed length $L$ gives the Hagen-Poiseuille equation:The velocity is zero at the wall and maximal at the centre. Averaging over the circular area, whose annuli have area $2\pi rdr$, gives $U=u(0)/2$; averaging along one diameter would give a different answer. These assumptions produce a parabola, rather than every laminar flow having a parabolic profile. ^[ref]

As a stipulated calculation, take $D=0.010\mathrm{m}$, $U=0.10\mathrm{m}{\mathrm{s}}^{-1}$, $\rho =1000\mathrm{kg}{\mathrm{m}}^{-3}$, $\mu =0.0010\mathrm{Pa}\mathrm{s}$ and $L=1.0\mathrm{m}$, and assume the developed laminar flow above. Then $\nu =10^{-6}{\mathrm{m}}^2{\mathrm{s}}^{-1}$, ${\mathrm{Re}}_D=1000$, $u(0)=0.20\mathrm{m}{\mathrm{s}}^{-1}$, $Q=7.85\times 10^{-6}{\mathrm{m}}^3{\mathrm{s}}^{-1}$ (about $0.471\mathrm{L}\mathrm{mi}{\mathrm{n}}^{-1}$), and $\Delta p=32\mu LU/D^2=32\mathrm{Pa}$. These are predictions from chosen parameters, not measurements or a claim that ${\mathrm{Re}}_D=1000$ guarantees laminar flow in every apparatus.

The beauty of this example is that a spatially varying velocity field still gives an exact relation between two accessible measurements, pressure drop and volumetric flow rate. It predicts the hydraulic resistance $\Delta p/Q=8\mu L/(\pi R^4)$. Doubling the radius therefore multiplies $Q$ by sixteen at fixed pressure drop, length and dynamic viscosity, provided the model remains applicable. This makes the laminar flow model useful for sizing narrow tubes or determining dynamic viscosity. It also explains why a persistent narrow dye band can be informative without implying that the fluid is moving as a uniform plug.