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For two bodies interacting through a potential energy that depends only on their separation, introduce the centre of mass coordinate and the two-particle relative coordinate:Their kinetic energy separates exactly:The centre of mass moves uniformly if there is no external force, while the internal two-body problem reduces to one effective particle of reduced mass $\mu$:The original trajectories follow from ${\mathbf{r}}_1=\mathbf{R}+(m_2/M)\mathbf{r}$ and ${\mathbf{r}}_2=\mathbf{R}-(m_1/M)\mathbf{r}$.

For Newtonian gravity, $U=-G{m}_1{m}_2/r$, so the relative acceleration is $\ddot{\mathbf{r}}=-GM\mathbf{r}/r^3$. Conserved relative angular momentum and energy then give the usual conic sections. For a bound orbit of relative semimajor axis $a$,The two bodies share this orbital period and orbit their common centre of mass with semimajor axes in inverse proportion to their masses. If $m_2\gg m_1$, then $\mu \simeq m_1$, recovering the fixed-heavy-centre approximation. Using $M$ as the internal inertial mass would confuse centre-of-mass translation with relative motion.

Elasticity describes recoverable deformation. For small displacements $\mathbf{u}(\mathbf{x})$, the infinitesimal strain tensor is the symmetric part of the displacement gradient,Its diagonal entries describe extension, while its off-diagonal entries describe shear. The engineering shear strain is ${\gamma }_{xy}=2{ϵ}_{xy}$; the antisymmetric part of the displacement gradient is a small rigid rotation, not elastic strain.

The stress tensor specifies the force per unit area on each oriented plane: $\mathbf{t}(\hat{\mathbf{n}})=\mathbf{\sigma }\hat{\mathbf{n}}$. Positive normal stress is tensile in the convention used here; tangential traction gives shear stress. In the absence of internal couples, angular-momentum balance makes the stress tensor symmetric. Force balance gives $\nabla \cdot \mathbf{\sigma }+\mathbf{f}=0$ in static equilibrium, where $\mathbf{f}$ is body force per unit volume.

Linear elasticity relates stress and strain. A homogeneous isotropic solid needs two independent elastic constants:Here $G$ is the shear modulus and $\lambda$ is the other Lamé parameter. For uniaxial loading, Young's modulus is $E={\sigma }_{xx}/{ϵ}_{xx}$, with lateral contraction described by Poisson's ratio $\nu =-{ϵ}_{yy}/{ϵ}_{xx}$. The bulk modulus is $K=-p/(\Delta V/V)$ for a small hydrostatic compression. The relations between these constants areStable isotropic elastic strain energy density requires $G>0$ and $K>0$, implying $-1<\nu <1/2$. The incompressible limit is approached as $\nu \to 1/2$. Anisotropic crystals generally need more constants; the isotropic relations should not be imposed on arbitrary directions in such a crystal.

The elastic energy density is $w=\mathbf{\sigma }:\mathbf{ϵ}/2$. In the isotropic linear model this becomes $\lambda (\mathrm{tr}\mathbf{ϵ}{)}^2/2+G\mathbf{ϵ}:\mathbf{ϵ}$. The quadratic form explains why reversible loading stores energy and unloading can return it, unlike plastic deformation or a dissipative hysteresis cycle.

For a slender beam described by the Euler-Bernoulli beam equation, plane sections approximately remain plane. A fibre a distance $y$ from the neutral axis has axial strain $-y\kappa$, where $\kappa$ is curvature, and stress $-Ey\kappa$. Integrating its moment about the neutral axis gives, with consistent signs,$I_A$ is a second moment of area, with units of length to the fourth power; it is not a mass moment of inertia. For a rectangular section of width $b$ and thickness $h$, $I_A=b{h}^3/12$, explaining the strong dependence of bending stiffness on thickness. For small transverse deflection $w(x)$, curvature is approximately $w^{\prime \prime }$, and the Euler-Bernoulli beam equation is $E{I}_A{w}^{\prime \prime \prime \prime }=q(x)$ with a suitable load-sign convention. A cantilever of length $\ell$ under end load $F$ has tip deflection $F{\ell }^3/(3E{I}_A)$ and largest bending stress of magnitude $F\ell c/I_A$, where $c$ is the furthest distance from the neutral axis.

In torsion of a solid circular shaft, a cross-section rotated by angle $\theta (x)$ has shear strain $r{\theta }^{\prime }(x)$. ThereforeHere $J$ is the polar second moment of area. A uniform shaft of length $\ell$ under constant torque twists by $\mathcal{T}\ell /(GJ)$, and its largest shear stress is $\mathcal{T}a/J$. Noncircular sections can warp, so replacing their torsional stiffness by $G$ times their polar area moment is not generally valid. Both bending and torsion formulas presume small, elastic deformations; exceeding the elastic limit can produce permanent deformation or fracture.

A Carnot cycle consists of two reversible isothermal branches joined by two reversible adiabatic branches. Let the hot isothermal expansion run from volume $V_a$ to $V_b$, and the cold isothermal compression from $V_c$ to $V_d$. For $n$ moles of an ideal gas, their absorbed and rejected heat magnitudes areOn a reversible ideal-gas adiabat, $T{V}^{\gamma -1}$ is constant. Applying this to the two joining branches gives $V_b/V_a=V_c/V_d$. ThereforeThe result is independent of the gas's heat-capacity ratio. This derivation presumes the usual constant-heat-capacity ideal-gas model; more generally the equal entropy changes on the two isotherms give the same heat ratio without that restriction.

To establish the upper bound, suppose an engine were more efficient between these same two reservoirs. Use its output work to drive a reversed Carnot cycle, scaled so that the work magnitudes match. For fixed work $W$, the hypothetical engine would take $W/\eta$ from the hot bath, whereas the reversed Carnot cycle would return $W/{\eta }_C>W/\eta$ to it. The combined device would transfer heat from the cold bath to the hot bath without any net external work. That contradicts the Clausius statement of the second law. Hence no such two-reservoir engine can exceed $1-T_2/T_1$.

For the triangular cycle, call the three successive vertices $A,B,C$, and measure energies in units of $p_0{V}_0$. For one mole of a monatomic ideal gas, $U=3pV/2$, and positive work done by the gas is $W=\int pdV$. The net values on the three complete branches areThus the net work is $W_{\mathrm{cycle}}=p_0{V}_0/2$, equal to the triangle's enclosed area.

There is a small but important distinction between the net heat on a branch and its total positive heat input to a cycle. On the sloping branch put $x=V/V_0$, so $p/p_0=3-x$, with $1\le x\le 2$. ThenThere is a heat-flow reversal along a thermodynamic path at $x=15/8$. The first part absorbs heat; the final part rejects it. Integrating the two signed contributions separately givesThe true total absorbed and rejected heats and thermal efficiency are thereforeFor clarity, the positive heat input on $A\to B$ is $3p_0{V}_0/2$, on $B\to C$ it is $49p_0{V}_0/32$, and on $C\to A$ it is zero. Using just the positive net heats of the complete branches would give $Q_{\mathrm{in}}=3p_0{V}_0$ and $\eta =1/6$; that shortcut misses the short heat-rejection interval on $B\to C$.

Finally, put $T_0=p_0{V}_0/R$. The minimum temperature is $T_0$ at $A$. Along $B\to C$,which has its maximum $9/4$ at $x=3/2$, not at either endpoint. Thus the corresponding two-reservoir Carnot efficiency isThe triangular cycle receives heat at a range of temperatures below $T_{\max }$ and rejects it at temperatures above $T_{\min }$. It therefore makes less work per unit input heat than a Carnot cycle using only those extrema. The geometric path alone does not prove irreversibility: it could in principle exchange heat reversibly with a continuum of matched-temperature reservoirs, while still having this lower efficiency. Restricting it to only the two extreme-temperature reservoirs would introduce additional irreversible heat transfers.

The classical equipartition theorem assigns mean energy $k_{B}T/2$ to each independent quadratic term in the Hamiltonian, provided the canonical distribution and its relevant boundary integrals exist. For a quadratic term $a{q}^2$, the Boltzmann distribution is proportional to $e^{-\beta a{q}^2}$, with $\beta =1/(k_{B}T)$. Integrating a total derivative givesAfter dividing by the normalization,The same argument applies to quadratic momentum terms. Other independent canonical coordinates factor out of the ratio. It is a classical statement: widely spaced quantum energy levels can freeze out a contribution at low temperature.

The confining quartic potential energy $U(x)=Q{x}^4$, with $Q>0$, is not quadratic, so the $k_{B}T/2$ rule cannot be assigned to it. This does not invalidate the canonical Boltzmann distribution or its generalized integration-by-parts identity. Indeed,Hence equipartition for a homogeneous potential givesEquivalently, $Z_x=\Gamma (1/4)(\beta Q{)}^{-1/4}/2$ and $-\partial \ln Z_x/\partial \beta =1/(4\beta )$. This is the potential contribution only; an additional standard kinetic term $p^2/(2m)$ would contribute a further $k_{B}T/2$ to energy and $k_B/2$ to heat capacity.

For an intact diatomic molecule, the classical degrees of freedom comprise three translational quadratic terms, two rotational quadratic terms, and two terms for its one bond-stretching vibration: one kinetic and one potential. Thus complete classical equipartition predictsThe heat capacity of a diatomic gas has lower-temperature regimes because rotational and vibrational levels are quantized. In a dilute molecular gas, the expected progression isThe rotational energy scale is normally much smaller than the vibrational scale, so these steps are separated. The high-temperature value treats the vibration as harmonic and excludes dissociation and electronic excitation; the lowest-temperature gas regime is meaningful only while the sample remains dilute and gaseous.^[ref]

The plotted example uses a quantum rigid rotor and harmonic vibration, with an illustrative ratio of characteristic temperatures. It is a model calculation, not measured data for a particular molecular species. For the vibrational part,which tends to zero on freezing out and to one when many vibrational levels are accessible. The classical equipartition theorem is recovered when the relevant level spacings are small compared with $k_{B}T$.

Let $s(t)$ be the length of the free straight segment from the current contact point to the mass, and let $\theta$ measure the contact point's rotation from first contact. For an inextensible taut rope, winding through angle $\theta$ uses length $R\theta$:The point mass follows an involute of the pole's circular cross-section. Its instantaneous velocity is perpendicular to the free rope. To verify the geometry explicitly, use planar polar unit vectors at the contact point:The rope's tension therefore does no work, and the speed stays at its initial value $v$. The geometry gives $v=s\dot{\theta }$. ConsequentlyThe distance from the pole's axis is $\sqrt{R^2+s^2}$, so a point mass first reaches the pole when $s=0$. The winding time around a circular pole is thusThis has units of time. The model assumes a light inextensible rope, no energy loss at contact and a point mass. The required tension $m{v}^2/s$ diverges at the ideal final contact, so finite strength and size matter very close to impact.

For weak fields in a linear isotropic material, magnetization is $\mathbf{M}={\chi }_m\mathbf{H}$, with magnetic susceptibility ${\chi }_m$. Paramagnetism has ${\chi }_m>0$, while diamagnetism has ${\chi }_m<0$. In either case, $\mathbf{B}={\mu }_0(\mathbf{H}+\mathbf{M})={\mu }_0(1+{\chi }_m)\mathbf{H}$. These weak responses generally disappear when the applied field is removed, unlike hysteretic ferromagnetism.

A paramagnet can contain permanent magnetic moments from unpaired electrons or atomic angular momentum. Alignment lowers the energy $-\mathbf{\mu }\cdot \mathbf{B}$, while thermal disorder opposes it. For independent classical moments of magnitude $\mu$, their weak-field average projection is ${\mu }^{2}B/(3k_{B}T)$, soThis is Curie's law; it assumes weak magnetization and no important interactions between moments. With quantum angular momentum $J$, replace ${\mu }^2$ by $(g{\mu }_B{)}^{2}J(J+1)$. Strong-field magnetization saturates instead of growing linearly without limit. Conduction electrons also exhibit Pauli paramagnetism: only occupations near the Fermi surface readily redistribute between spin states, giving a nearly temperature-independent susceptibility when $T\ll T_F$.

Diamagnetism is an induced orbital response opposing the applied magnetic field. For bound electrons, a weak field changes orbital motion and induces an opposing magnetic moment. For an isotropic collection of atoms, the elementary orbital estimate iswhere the sum runs over the electrons of one atom. It is usually small and weakly temperature dependent. Delocalized electrons also have orbital Landau diamagnetism; for an ideal three-dimensional free electron gas with a quadratic isotropic dispersion and electron spin with $g=2$, ${\chi }_L=-{\chi }_P/3$. Both spin and orbital contributions must be included to determine the total response. Diamagnetism is not a consequence of permanent dipoles choosing to align against the field; that orientation would increase their energy.

The Poynting vector is the electromagnetic energy flux density,In vacuum it equals $\mathbf{E}\times \mathbf{B}/{\mu }_0$. For a linear nondispersive medium with time-independent constitutive coefficients, Maxwell's equations and the identity $\nabla \cdot (\mathbf{E}\times \mathbf{H})=\mathbf{H}\cdot \nabla \times \mathbf{E}-\mathbf{E}\cdot \nabla \times \mathbf{H}$ give Poynting's theorem,The source term is work done by the fields on free charges; for an Ohmic conductor it is the positive local Joule heating rate $\sigma E^2$. Integrating over a volume states that loss of stored field energy plus outward energy flow equals energy transferred to matter. In dispersive media, such as a plasma, material motion contributes to stored energy and the simple nondispersive expression cannot be used indiscriminately.

For a vacuum plane wave, $B=E/c$ and $\mathbf{E}$, $\mathbf{H}$ and the propagation direction form a right-handed triad. With real peak field $E_0$, the time-averaged wave intensity isMore generally, complex peak amplitudes give $⟨\mathbf{S}⟩=\frac{1}{2}\mathrm{Re}({\mathbf{E}}_0\times {\mathbf{H}}_0^{\ast })$. Along a transmission line, the cross-section integral of the longitudinal Poynting vector is the transmitted electrical power $VI$. Energy flows through the electromagnetic fields in the surrounding dielectric, even though the conductors guide the currents.

Distributed circuit model. A transmission line must be treated as a distributed system when its propagation time is appreciable compared with the signal timescale. Divide a uniform two-conductor line into elements of length $dx$. Each has series resistance $R^{\prime }dx$ and inductance $L^{\prime }dx$, and shunt electrical conductance $G^{\prime }dx$ and capacitance $C^{\prime }dx$. The primes denote quantities per unit length, with units $\mathrm{\Omega }{\mathrm{m}}^{-1}$, $\mathrm{H}{\mathrm{m}}^{-1}$, $\mathrm{S}{\mathrm{m}}^{-1}$ and $\mathrm{F}{\mathrm{m}}^{-1}$, respectively.

Kirchhoff's voltage and current laws give the Telegrapher's equations:For a lossless line, $R^{\prime }=G^{\prime }=0$, differentiation gives $V_{xx}=L^{\prime }{C}^{\prime }{V}_{tt}$ and the same wave equation for $I$. The propagation speed and characteristic impedance areA general lossless signal has $V=V^{+}(t-x/u)+V^{-}(t+x/u)$ and $I=[V^{+}(t-x/u)-V^{-}(t+x/u)]/Z_0$. Thus characteristic impedance is the voltage/current ratio of one travelling wave, not simply the resistance of the wires. It can be real even when the line dissipates no energy, because energy is carried away by propagation.

For sinusoidal quantities with convention $e^{i\omega t}$, the general lossy-line formulas areChoose the square-root branches for forward attenuation and positive forward power. The real part of ${\gamma }_{\mathrm{prop}}$ is attenuation per unit length; its imaginary part is the phase wavenumber. For small losses, attenuation is approximately $R^{\prime }/(2Z_0)+G^{\prime }{Z}_0/2$, using the lossless real $Z_0$ in this estimate.

Fields and geometry. Two separated conductors support a transverse electromagnetic mode, with electric and magnetic fields transverse to the propagation axis. For a homogeneous dielectric and negligible conductor penetration, both examples below have $L^{\prime }{C}^{\prime }=\mu ϵ$ and hence $u=1/\sqrt{\mu ϵ}$. The distributed circuit model follows from the field energy and charges in each slice.

For a coaxial cable with inner radius $a$ and inner surface of the outer conductor at $b$, Gauss's law and Ampère's law giveThe electric field is radial and the magnetic field circles the axis. Integration of voltage and field energy givesThe equal and opposite return current on the outer conductor confines the ideal external field to the annulus.

For wide parallel plates of width $w$ separated by $d\ll w$, neglect fringing. The electric field crosses the gap, $E\simeq V/d$, while the magnetic field lies in the gap parallel to the plates and perpendicular to propagation, $H\simeq I/w$. ThereforeFor a forward wave, $\mathbf{E}\times \mathbf{H}$ points along the line. Integrating this Poynting vector over either cross-section gives instantaneous power $VI$. For complex peak amplitudes, average power is $\mathrm{Re}(V{I}^{\ast })/2$, or $|V^{+}{|}^2/(2Z_0)$ for a single forward wave on a lossless line. The conductors guide an energy flux predominantly in the dielectric between them.

Terminations and reflection. At a load $Z_L$, require $V_L=Z_L{I}_L$. If $V^{+}$ and $V^{-}$ are the incident and reflected voltage amplitudes there, the total current is $(V^{+}-V^{-})/Z_0$. ThusA matched load, $Z_L=Z_0$, absorbs the incident power with no reflection. An open circuit has ${\Gamma }_V=+1$, a current node and a doubled voltage at the end. A short circuit has ${\Gamma }_V=-1$, a voltage node and a doubled current there. A purely reactive load on a lossless line has $|{\Gamma }_V|=1$ and accepts no average power. For real $Z_0$, the reflected power fraction is $|{\Gamma }_V{|}^2$ and the accepted fraction is $1-|{\Gamma }_V{|}^2$.

Incident and reflected waves produce a standing-wave pattern with voltage standing wave ratio $(1+|{\Gamma }_V|)/(1-|{\Gamma }_V|)$. The input impedance of a lossless length $\ell$ isThis gives $i{Z}_0\tan (k\ell )$ for a short and $-i{Z}_0\cot (k\ell )$ for an open. A quarter-wave section gives $Z_{\mathrm{in}}=Z_0^2/Z_L$, enabling impedance transformation. After a source launches a pulse, a distant termination affects it only after the reflected pulse returns; the return time is $2\ell /u$. Real conductor resistance, dielectric loss and departures from a TEM field profile limit the ideal model at high frequency.

An Euler strut is a slender elastic column under axial compression. Let its bending stiffness about the weaker section axis be $E{I}_A$ and its length be $L$. With small lateral deflection $y(x)$ and a dead compressive load $P$, the quadratic total potential energy isThe first term is bending energy; the negative term is work released by the load as bending shortens the end-to-end distance. Stationarity gives $E{I}_A{y}^{\prime \prime \prime \prime }+P{y}^{\prime \prime }=0$. For pinned ends, $y=y^{\prime \prime }=0$ at $0,L$, and $y\propto \sin (n\pi x/L)$ requires $P_n=n^2{\pi }^{2}E{I}_A/L^2$. The first instability is Euler's critical load,Below this load, the straight state has positive quadratic energy against these perturbations. At the threshold, the lowest bending mode has zero quadratic restoring energy; above it the straight state is unstable. For a rectangular section, $I_A=b{h}^3/12$ about the appropriate bending axis, so the smaller second moment of area controls the first buckling direction.

Endpoint constraints matter. Write $P_{\mathrm{cr}}={\pi }^{2}E{I}_A/(KL{)}^2$: $K=1$ for pinned-pinned, $2$ for fixed-free and $1/2$ for fixed-fixed conditions. This elastic instability differs from compressive material failure. Imperfections cause bending before the ideal threshold, and a short column may yield before Euler buckling occurs.

A gravitational assist changes a spacecraft's velocity relative to the Sun by scattering it in the gravitational field of a moving planet. In the planet's rest frame, ignore other forces during the encounter and take the spacecraft mass negligible. The incoming and outgoing asymptotic velocities have equal magnitude,because the encounter is conservative, but their directions differ along a hyperbolic orbit. If the planet has approximately constant heliocentric velocity ${\mathbf{V}}_p$, the heliocentric spacecraft velocities are ${\mathbf{v}}_{\mathrm{in},\mathrm{out}}={\mathbf{V}}_p+{\mathbf{u}}_{\mathrm{in},\mathrm{out}}$. Its specific energy change isThus a flyby can increase or decrease heliocentric energy depending on its geometry. The planet exchanges momentum and orbital energy with the spacecraft; its speed change is extremely small because its mass is much greater. There is no energy creation and no increase of asymptotic speed in the planet's own frame.

For a point-mass planet, conservation of energy and angular momentum gives the hyperbolic eccentricity and deflection anglewhere $r_p$ is periapsis distance and $b$ the impact parameter. The velocity-vector change has magnitude $2u_{\infty }\sin (\delta /2)$, so the maximum energy-change magnitude for this fixed deflection is $2V_p{u}_{\infty }\sin (\delta /2)$. A small periapsis increases the turning angle, but collision, atmosphere and tidal forces limit how close a real spacecraft may pass. Multiple encounters allow large changes without expending the propellant that equivalent direct velocity changes would require.

Following the moving fluid. The Eulerian velocity field $\mathbf{u}(\mathbf{x},t)$ assigns a velocity to each position. A fluid particle satisfies $\dot{\mathbf{x}}=\mathbf{u}$, so the chain rule gives the material derivative, also called the convective derivative,Even in steady flow a particle can accelerate because it moves between regions with different velocity. For example, flow around a curved obstacle changes direction despite ${\partial }_t\mathbf{u}=0$. Mass conservation isFor a constant-density incompressible fluid, $\nabla \cdot \mathbf{u}=0$.

Euler equations for an inviscid fluid and Bernoulli's equation. Neglect viscosity and take the body force per unit mass to be $-\nabla \Phi$. Newton's second law for a fluid element givesFor steady flow, dot with displacement along a streamline. The tangential component of $\mathbf{u}\cdot \nabla \mathbf{u}$ is the derivative of $u^2/2$ along that line. For constant density, integration givesIn uniform gravity, $\Phi =gz$. Thus pressure, kinetic energy per unit mass and gravitational potential energy per unit mass exchange along a streamline. For a barotropic compressible fluid, replace $p/\rho$ by $\int dp/\rho (p)$. Different streamlines need not have the same Bernoulli constant.

Vorticity and potential flow. Vorticity is $\mathbf{\omega }=\nabla \times \mathbf{u}$. A small fluid element rotates locally with angular velocity $\mathbf{\omega }/2$ after its deformational part is separated. The vector identitygives $\nabla B=\mathbf{u}\times \mathbf{\omega }$ for steady constant-density inviscid flow. Thus an irrotational region has one common Bernoulli constant on each connected region. In a simply connected irrotational domain, introduce a velocity potential $\mathbf{u}=\nabla \varphi$. Incompressibility then gives Laplace's equation, ${\nabla }^2\varphi =0$. For unsteady irrotational flow,An impermeable stationary wall fixes ${\partial }_n\varphi =0$, but ideal potential flow does not enforce no slip. In a multiply connected domain, irrotational flow may still have nonzero circulation around an excluded vortex core, so a globally single-valued potential requires care.

For example, uniform two-dimensional flow past a circular cylinder has $\varphi =U(r+a^2/r)\cos \theta$ outside $r=a$. Its normal velocity vanishes at the cylinder, and its surface tangential velocity is $-2U\sin \theta$. Bernoulli's equation predicts a symmetric pressure distribution and zero drag. This is D'Alembert's paradox: the assumptions omit the viscous boundary layers, separation and wake responsible for real drag.

For an incompressible fluid of constant kinematic viscosity $\nu$, the curl of the motion equation givesThe first term stretches and tilts vortex lines; the second diffuses vorticity. In two-dimensional incompressible inviscid flow the stretching term is absent and vorticity is carried with the fluid. In a three-dimensional inviscid flow it can be amplified by stretching. Density gradients not parallel to pressure gradients provide an additional source in a general compressible fluid.

Viscous flow. A Newtonian fluid's shear stress is proportional to rate of strain. For constant viscosity and incompressibility the Navier-Stokes equation iswith dynamic viscosity $\eta$ and kinematic viscosity $\nu =\eta /\rho$. At an ordinary solid wall, the no-slip boundary condition sets fluid velocity equal to the wall velocity. Viscosity transfers momentum between neighbouring layers and converts mechanical energy into heat; unlike the pressure force, its shear contribution depends on velocity gradients.

For plane Couette flow between plates $y=0,h$, with the upper plate moving at $U$, the steady solution is $u_x=Uy/h$, with shear stress $\eta U/h$. For pressure-driven flow through a pipe of radius $a$, a constant pressure gradient gives the parabolic Poiseuille profileThe Reynolds number $\mathrm{Re}=\rho UL/\eta$ compares convective inertia with viscous force. Small Reynolds number often permits the Stokes-flow approximation; large Reynolds number can produce thin boundary layers and, under suitable conditions, turbulence. It does not justify neglecting the wall conditions. Bernoulli's inviscid constant is generally not conserved along a viscous streamline, and pipe flow instead shows a pressure loss corresponding to dissipation.

Start with the fundamental relation at fixed material amount, $dU=TdS-pdV$, and expand entropy in independent variables $T,V$:The last equality uses the supplied Maxwell relation. ConsequentlyFor the excluded-volume ideal gas, $(\partial p/\partial T{)}_V=R/(V-b)=p/T$, so the bracket vanishes:Once $U$ depends only on temperature, $C_V=dU/dT$ cannot depend on volume either. Its stipulated temperature independence therefore makes it a constant, giving $U=C_{V}T+U_{\ast }$ with arbitrary energy reference $U_{\ast }$. Excluded volume changes the pressure without adding the volume-dependent attractive internal energy that a more general interacting-gas model might have.

Wall arrivals are not an unbiased sample of the gas: faster atoms arrive more frequently. The normalized speed distribution among incident atoms isThe angular factor in the arrival rate cancels when calculating the mean kinetic energy. ThusUsing $I_5=2I_3/\alpha$ givesBy contrast, an atom chosen randomly from the bulk has mean kinetic energy $3k_{B}T/2$. The arrival weighting raises the normal-direction contribution to $k_{B}T$ while each tangential direction still contributes $k_{B}T/2$. This is also the mean energy carried by atoms effusing through a thin aperture from an equilibrated reservoir.

Let L be the vessel initially containing argon and R the vessel initially containing helium. Both have volume $V$ and fixed temperature $T_0$, and the aperture has area $A_h$. The incoming and outgoing rates of each species must be considered separately. For species $s$,Thus its density difference relaxes exponentially with time constantThe two-vessel effusive mixing solution for the four densities isUsing atomic masses of approximately $40$ and $4$ atomic mass units gives ${\tau }_{\mathrm{He}}/{\tau }_{\mathrm{Ar}}\simeq 1/\sqrt{10}\simeq 0.32$. Helium therefore transfers faster. Each species eventually reaches $n_0/2$ in each vessel, but its transient is different. In particular, the total density temporarily rises in the initially argon vessel and falls in the initially helium vessel, before both return to $n_0$.

For unequal maintained temperatures $T_L,T_R$, replace the rate equation byWith each species' total density sum still $n_0$, its final densities and relaxation time areEvery species has the same final density ratio $\sqrt{T_R/T_L}$ and partial-pressure ratio $\sqrt{T_L/T_R}$. Thus there is no permanent composition separation in this ideal steady state, although the rates remain mass dependent. A maintained hot side has lower density and higher pressure. Equal particle rates do not imply thermal equilibrium: atoms arriving from the hotter side carry more energy, so external heating and cooling are required to hold different temperatures. The model assumes thin-hole effusion, local equilibrium within each vessel and negligible aperture volume.

For a classical dilute ideal gas in thermal equilibrium, the probability of a single-particle velocity is proportional to the Boltzmann factor for kinetic energy, $e^{-m{v}^2/(2k_{B}T)}$. Velocities with speed between $v$ and $v+dv$ fill a spherical shell of velocity-space volume $4\pi v^{2}dv$. The normalized Maxwell speed distribution is thereforeThe factor $v^2$ counts states, rather than representing an extra Boltzmann energy factor.

Set $\alpha =m/(2k_{B}T)$ and use $I_j={\int }_0^{\infty }{v}^j{e}^{-\alpha v^2}dv$. Normalization cancels in ratios, so the given recurrence givesConsequentlyThe root mean square speed is larger than the mean speed, consistent with the positive variance of the speed distribution.

Molecular transport across gradients. A gas can transport momentum, thermal energy and species abundance even when its average bulk velocity is zero. Molecules arriving at a point carry properties acquired near their preceding collisions. A small gradient makes the populations arriving from opposite sides slightly different, producing transport from larger to smaller values of the carried property. A continuum transport law requires a mean free path $\lambda$ small compared with the gradient length, and relaxation times short compared with the observation time.

For a dilute hard-sphere gas,where $d$ is molecular collision diameter. Both speed and collision distance matter: high speed alone does not determine a transport coefficient.

A simple kinetic estimate makes the common structure explicit. For a quantity $a(y)$ carried per particle, a molecule moving at angle $\theta$ to $+y$ brings approximately $a(y-\lambda \cos \theta )$. Expanding to first order and averaging an isotropic velocity distribution givessince $⟨\cos \theta ⟩=0$ and $⟨{\cos }^2\theta ⟩=1/3$. This mean-free-path argument estimates the coefficients; a detailed collision calculation changes their numerical prefactors and distinguishes different relaxation rates.

Momentum transport and viscosity. For a shear flow with mean velocity $U_x(y)$, the transported directed momentum per molecule is $m{U}_x$. The net flux of $x$ momentum in the $+y$ direction isHere $\eta$ is dynamic viscosity in $\mathrm{Pa}\mathrm{s}$. The associated shear stress on a material surface has the corresponding traction sign convention; the flux carries momentum from the faster-moving layer towards the slower-moving one. The kinematic viscosity is $\nu =\eta /(nm)$, with units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$. Viscosity smooths velocity gradients, and the lost ordered mechanical energy becomes heat.

Heat transport. Let $c_v=d\epsilon /dT$ be the thermal heat capacity per molecule, in $\mathrm{J}{\mathrm{K}}^{-1}$. Substituting $a=\epsilon (T)$ gives Fourier's law,Thermal conductivity has units $\mathrm{W}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$. For a monatomic gas, translational equipartition gives $c_v=3k_B/2$. Rotational or vibrational excitations in other gases can increase the energy carried per molecule, while collisions govern how those modes exchange energy. A temperature gradient can therefore be transported differently from directed momentum even when the dimensional estimates look similar.

Species transport. For a dilute tracer or a mixture with prescribed total density and negligible bulk flow, molecules diffuse down concentration gradients. The first-gradient number flux is Fick's first law,The diffusivity $D$ has units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$; multiplying number flux by the species mass gives mass flux. More generally one uses diffusion flux relative to the mixture's bulk velocity and specifies mass or mole fractions. The approximation is not a claim that an arbitrary pure-gas pressure gradient can remain at rest while only a Fick flux acts. In a mixture, collision cross-sections and relative molecular speeds, including the reduced mass, determine the species diffusivities.