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Let a projectile have mass $m$, incident speed $v$, and incident kinetic energy $E=m{v}^2/2$. The stationary target has mass $M$. In a head-on elastic collision, the signed outgoing projectile velocity $v^{\prime }$ and target velocity $u$ obey momentum conservation and conservation of energy:Factor the second equation and eliminate $Mu=m(v-v^{\prime })$. The nontrivial collision branch has $u=v+v^{\prime }$, henceThis is the maximum over scattering angles, not the energy of every recoil. To see the bound without assuming a head-on event, set the recoil momentum magnitude to $q$ and its laboratory angle to the incident direction to $\theta$. Conservation of energy with outgoing projectile momentum $\mathbf{p}-\mathbf{q}$ givesThus $T=T_{\max }{\cos }^2\theta$ for a nonzero recoil, and the endpoint occurs at $\theta =0$.

For an original numerical example, stipulate $E=4\mathrm{MeV}$ and $m=m_p$, with $m_p$ the proton mass. Approximate hydrogen, helium and nitrogen target masses by $M=m_p$, $4m_p$ and $14m_p$ respectively. Their maximum fractions and recoil energies areThese stipulated values illustrate the approximation $m_n=m_p$; they are not measurements from 1932. With $m_p{c}^2\simeq 938\mathrm{MeV}$, the incident ratio $E/(m_p{c}^2)\simeq 0.0043$ gives $v/c\simeq 0.092$, consistent with the nonrelativistic limit used here. A million electronvolts is an energy unit, whereas a mass expressed in $\mathrm{MeV}/c^2$ requires the factor $c^2$ when inserted into an energy equation.

The hydrogen endpoint permits essentially complete transfer because the projectile and target masses match. For a very heavy target, $M\gg m$, the fraction approaches $4m/M$, which is small: the target can take substantial momentum with little kinetic energy. The exact inequality follows from $(m-M{)}^2\ge 0$, which implies $(m+M{)}^2\ge 4mM$.

A projectile of mass $m_1$ elastically scattering from a stationary target of mass $m_2<m_1$ has the displayed maximum laboratory angle. In $c=1$ units, let its incident energy be $E>m_1$ and $s=m_1^2+m_2^2+2m_{2}E$. Boosting the outgoing four-momentum from the center-of-momentum frame gives $\tan \theta =\sin \phi /[\Gamma (\cos \phi +B)]$, with $\Gamma =(E+m_2)/\sqrt{s}$ and $B=(m_1^2+m_{2}E)/[m_2(E+m_2)]>1$. The maximum at $\cos \phi =-1/B$ has $\tan {\theta }_{\max }=1/[\Gamma \sqrt{B^2-1}]=m_2/\sqrt{m_1^2-m_2^2}$. Equality is allowed; a strict upper bound would exclude a realizable outgoing direction. The angular cap is independent of incident energy.

Let $R$ be the actual largest pairwise separation of positive masses in their centre of mass frame. The pairwise moment-of-inertia identity gives $m_{(1)}{m}_{(2)}{R}^2/(2M)\le I\le R^2{\sum }_{j<k}{m}_j{m}_k/(2M)$. Consequently the scalar inertia is comparable to $R^2$, independently of configuration.

The Maxwell-Boltzmann distribution describes the random velocities and speeds of particles in a classical ideal gas at thermal equilibrium. Individual particles keep changing velocity through collisions, while the equilibrium probability distribution stays fixed. For one species of particle mass $m$ at positive temperature $T$, the characteristic scale iswhere $k_B$ is the Boltzmann constant. A signed velocity component follows a normal distribution; the nonnegative magnitude follows the Maxwell speed distribution. Their peaks differ because a speed measurement groups together all directions of motion.

Use the gas rest frame, three spatial dimensions, and nonrelativistic particles, with $k_{B}T\ll m{c}^2$ where $c$ is the speed of light. Assume a dilute gas whose interactions between encounters can be neglected in the ideal gas approximation and whose translational motion is classical. A sufficient condition for negligible quantum exchange effects is $n{\lambda }^3\ll 1$, where $n$ is the number density and $\lambda$ the thermal de Broglie wavelength. Diluteness alone does not guarantee thermal equilibrium; the gas must have equilibrated. These are the assumptions behind the classical Maxwell-Boltzmann velocity distribution. ^{[ref] [ref]}

The connection to Maxwell–Boltzmann statistics is the Boltzmann factor for translational kinetic energy, $E_{\mathrm{kin}}=m|\mathbf{v}{|}^2/2$. Normalizing $e^{-E_{\mathrm{kin}}/(k_{B}T)}$ over velocity gives a probability density; an occupation factor $e^{(\mu -E)/(k_{B}T)}$ instead specifies a mean occupation, with chemical potential $\mu$. Their normalizations serve different purposes. ^[ref]

This connects microscopic motion to measurements: a gas thermometer uses the ideal gas law, collisions and the mean free path govern transport, and measurements of spectral lines use thermal Doppler broadening to infer temperature. Each probes a different aspect of the same thermal motion.

In 1860, James Clerk Maxwell presented the component and speed laws in Part I, Proposition IV of Illustrations of the Dynamical Theory of Gases, pp. 22–23. His argument combined independence of the three velocity components with isotropy, and he explicitly compared the component law with the normal distribution of observational errors. Independence is an assumption of that argument; perpendicular coordinate axes alone do not establish it. ^[ref]

The Maxwell speed distribution is the probability density function for the magnitude $v=|\mathbf{v}|$. Unlike a signed velocity component, speed cannot be negative:For $0\le a<b$, the probability of a speed in $[a,b]$ is ${\int }_a^{b}g(v)dv$, and ${\int }_0^{\infty }g(v)dv=1$. The density has units of inverse speed, so its height alone is not a probability. ^[ref]

All velocities with magnitudes from $v$ to $v+dv$ occupy a thin spherical shell of volume $4\pi v^{2}dv$ in velocity space. The isotropy of the Maxwell-Boltzmann velocity distribution makes its density constant over each shell, giving $g(v)=4\pi v^{2}f(v{\mathbf{e}}_x)$ for any fixed unit direction ${\mathbf{e}}_x$. A larger radius offers more directions and more volume. The increasing shell factor initially wins over the decreasing Boltzmann factor, so $g(0)=0$ and the most probable speed is positive. This is the geometric reason that the peaks of the vector and speed densities disagree.

With the dimensionless quantity $u=v/s$, every such gas has the same rescaled speed density:For comparison, $w=v_x/s$ has the standard normal distribution density $\varphi (w)=e^{-w^2/2}/\sqrt{2\pi }$ on the whole real line. The Maxwell speed distribution is neither the positive half of this curve nor the distribution of $|v_x|$: it combines three components.

The first two moments of the Maxwell speed distribution give the mean and root mean square:Consequently the mean translational kinetic energy is $m⟨v^2⟩/2=3k_{B}T/2$, as required by the equipartition theorem. ^[ref] The mean speed weights fast particles more than the density peak does, and the root mean square weights them more strongly still.

Changing temperature or particle mass only changes $s\propto \sqrt{T/m}$. At fixed mass, four times the temperature doubles every characteristic speed; at fixed temperature, four times the mass halves them. A stretched density becomes lower so that its area stays one. For two species at the same temperature, equal mean translational kinetic energy therefore means different typical speeds.

The mean mass density of a body is its total mass divided by its volume. A nonuniform density distribution can have the same mean as a uniform one; orbital information that determines only $M/V$ does not determine the internal density profile.

The continuum velocity alignment equation with $\mathbf{L}[f](\mathbf{x},\mathbf{v})=\int K(\mathbf{x},{\mathbf{x}}_{\star })({\mathbf{v}}_{\star }-\mathbf{v})f_{\star }{d}^3{x}_{\star }{d}^3{v}_{\star }$. For symmetric nonnegative $K$, it conserves total mass and momentum while dissipating kinetic energy. Its force is affine in velocity, with a nonzero velocity-space divergence.

Consider a nonrelativistic particle of mass $m$ undergoing an elastic collision with a stationary free target of mass $M=Am$, where $A\ge 1$. Assume the outgoing direction is uniformly distributed over solid angle in the center of mass frame; this assumption concerns the scattering angle distribution and need not hold for every target or energy. Let $\mu$ be the cosine of that frame's scattering angle. Transforming the outgoing velocity back to the laboratory givesThe uniform distribution of $\mu$ on $[-1,1]$ makes $x$ uniform on $[\alpha ,1]$. For $A>1$, integration yieldsFor equal masses, take the limit $\alpha \to 0$: $\xi =1$. The expected value of the fractional energy after a collision is instead $\mathbb{E}[x]=(1+\alpha )/2$. These are different averages: averaging a logarithm does not give the logarithm of an average.

For independent collisions with the same distribution, the logarithms of successive energy ratios add. ThusHere $E_{\mathrm{geom}}$ is the geometric-mean energy, and $n_{\log }$ is a characteristic collision count for reaching an energy scale $E_{\ast }$. It is not an exact first-passage expected value. Individual histories fluctuate, and neither the time between collisions nor absorption is specified by this model.

In an original chosen example, set $E_0=10^6\mathrm{eV}$ and $E_{\ast }=0.025\mathrm{eV}$. Approximate hydrogen by $A=1$ and carbon by $A=12$. ThenThese are illustrative model values, not measured collision counts from 1934. The energy scale marks where neglected thermal motion matters; the calculation does not assert continued cooling to zero energy.

The mean particle mass is the mass density divided by the total number density of freely moving particles. It enters $P=\rho k_{B}T/\bar{m}$. ionization changes the particle count without significantly changing the mass, so it changes $\bar{m}$. The dimensionless mean molecular weight is $\bar{m}$ divided by a stated reference mass, commonly the atomic mass unit; it must not be substituted as a dimensionless number into a formula requiring a mass.

The mass-normalized first velocity moment of a distribution on phase space. For a velocity-only distribution, omit the position integral. Nonzero finite mass and a finite first moment are required. For a probability distribution this is its expected value; its unnormalized counterpart is the momentum.

The UCSB demonstration uses an aluminum cylinder turned by a crank, with a wrapped rope supporting a hanging mass. An embedded electrical thermometer connects through sliding contacts to an electrical resistance meter; a supplied electrical resistance–temperature table or fitted equation converts the readout. A counter records turns. Record the initial temperature, crank steadily, then stop and convert the minimum recorded electrical resistance to the final temperature.^[ref]

The proposed control is to start below room temperature and finish comparably above it, approximately balancing incoming and outgoing environmental heat. Neglected accessory heat capacities and rough temperature estimates limit accuracy. The mechanical input and thermal response test an energy balance; warming alone does not establish quantitative equivalence.^[ref]

This is a modern companion to Joule's paddle-wheel experiment, with a different apparatus. Choose the cylinder as the closed system. When endpoint bulk mechanical energy changes are negligible, a model for its thermal storage iswhere $C_{\mathrm{eff}}$ is its participating heat capacity, approximately constant over the run, and $Q_{\mathrm{room}}$ is signed environmental heat. The cylinder is not an isolated system, since mechanical work enters. The temperature rise estimates a change in internal energy, rather than an amount of externally supplied heat. This boundary distinction connects a measured thermal response to the same stored quantity used in two reversible ideal-gas paths between the same states.

For a charged particle minimally coupled to a magnetic vector potential, the mechanical momentum is $\mathbf{\pi }=\mathbf{p}-q\mathbf{A}$, with $\mathbf{p}$ the canonical momentum. It is the mass times velocity in nonrelativistic mechanics.

The proportion of material present as geological melt, with a mass or volume convention specified.

Floating ice with submerged volume fraction $f$ has density $f{\rho }_l$ by Archimedes' principle. Its volume change on melting is therefore $v_l-v_s=(1-1/f)/{\rho }_l<0$. An enthalpy cycle gives $L_f=L_s-L_v$ for quantities consistently expressed per mass at the same temperature. Treating these latent heats and specific volumes as constant and integrating the Clausius-Clapeyron relation gives the displayed pressure estimate. Using a vaporization latent heat extrapolated from a different temperature carries that approximation into the inferred fusion heat.

A Mestel disk is a thin axisymmetric astrophysical disk with surface density inversely proportional to radius. Its galaxy rotation curve is exactly flat, $v_c=v_M$, and $v_c^2=GM(R)/R$ despite its nonspherical geometry. In the Legendre expansion of thin-disk gravity, all positive-degree interior and exterior terms cancel. Its total mass is infinite, so only potential differences, rather than a potential zero at infinity, are defined.

The reflection-symmetric Newtonian gravitational potential $\Phi =v_0^2\log [(\sqrt{R^2+z^2}+|z|)/R_{\ast }]$ is harmonic off the plane. Its normal-derivative jump is $2v_0^2/R$, so the Poisson equation for Newtonian gravity gives surface density of a disk $S=v_0^2/(2\pi GR)$ and volume mass density $\rho =S\delta (z)$. The midplane circular speed is constant. The scale-free model has infinite total mass and does not use a finite-mass boundary condition at infinity.

The abundance of chemical elements heavier than helium in astronomical material. A metal mass fraction measures their combined mass; spectroscopic abundance ratios are other measures of metallicity and need not be numerically identical to that fraction.

Approximate a fixed mass of iron-group freeze-out ashes by nickel-56 and iron-54, with no further weak neutronization and negligible residual light particles. Their electron fractions are $1/2$ and $26/54$. Conservation of the initial electron fraction in stellar matter, $1/2-X_{22}/22$, then gives the displayed nickel mass fraction. Its argument is the initial neon-22 mass fraction, which need not equal the earlier nitrogen metallicity. The two-isotope mass-budget approximation is not an exact full nuclear equilibrium with all free-particle densities set to zero.

The fraction of total mass in chemical elements heavier than helium, so the hydrogen mass fraction, helium mass fraction and metal mass fraction satisfy $X+Y+Z=1$. Despite the astronomical name, this includes elements such as carbon and oxygen that are not chemical metals.

Epitaxial growth using volatile metal-organic precursors delivered in the gas phase. Precursor decomposition and surface incorporation grow a crystalline compound layer or nanostructure; temperature and mass transport influence composition.