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Suppose a collisionless particle has momentum $p=T_r\propto a^{-1}$ during pure radiation domination. Since $a\propto t^{1/2}$, integrating $\chi ={\int }_0^{t}v(t^{\prime })d{t}^{\prime }/a(t^{\prime })$ gives the displayed comoving distance. The physical distance at observation is $a\chi =H^{-1}\mathrm{arsinh}(y)/y$, $y=m/T_r$. The printed expression without $a^{-1}$ is consequently a physical distance, or a comoving distance in coordinates normalized to $a=1$ at observation. Its ratio tends to one in the relativistic limit and to $\log (2y)/y$ in the nonrelativistic limit. The accumulated scale differs from an instantaneous velocity-over-expansion-rate estimate.

Choose $n=10^{28}{\mathrm{m}}^{-3}$ for a uniform noninteracting electron model with two equally populated spin states. This is an original worked example, not a measured number density assigned to an actual material. Using $m_e\simeq 9.10938\times 10^{-31}\mathrm{kg}$, $\hslash \simeq 1.05457\times 10^{-34}\mathrm{J}\mathrm{s}$, and $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$ givesThe conversion uses $1\mathrm{eV}=1.602176634\times 10^{-19}\mathrm{J}$: an electronvolt is an energy unit, while a joule is its SI counterpart. The dimensions provide a quick check: $n^{1/3}$ is inverse length, $\hslash n^{1/3}$ has units of momentum, $p_F^2/m$ has units of energy, and $E_F/k_B$ has units of temperature. The ratio $p_F/(m_{e}c)\simeq 0.002574$ verifies the nonrelativistic limit for this model, where $c$ is the speed of light.

Increasing this chosen number density eightfold doubles the Fermi momentum and quadruples the Fermi energy, giving approximately $6.7701\mathrm{eV}$. The exponent $2/3$ has a geometric origin: the number of occupied states grows as the cube of a momentum radius, but kinetic energy of nonrelativistic particles grows as its square. This makes Fermi energy a useful bridge between a count of particles and a mechanical energy scale.

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 $(m+M{)}^2\ge 4mM$.

For a massive particle parametrized by proper time, the geodesic equation isThe weak-field approximation alone does not imply Newtonian motion. Also take a nonrelativistic limit, $|v|/c\ll 1$, in the usual quasistatic regime so the potentials' time variation does not compensate for the small velocity factors. Reparametrizing by coordinate time and retaining the leading velocity-independent spatial term givesIn empty space the first field equation gives ${\nabla }^2\Psi =0$. For static potentials, the spatial equations imply ${\nabla }^2(\Phi -\Psi )=0$ and then ${\partial }_i{\partial }_j(\Phi -\Psi )=0$. The difference is at most affine; isolated-system boundary conditions fixing both potentials to zero at infinity remove it. Thus $\Phi =\Psi$ andthe empty-space Newton-Poisson equation. With the same quasistatic assumptions and dust present it becomes ${\nabla }^2\Phi =4\pi G\rho$. Keeping rapid time dependence instead leaves ${\nabla }^2\Phi =-3{\partial }_0^2\Psi$ in vacuum, so the static qualification matters.

Use primes for conformal-time derivatives. The cold-dark-matter Cosmological Euler equation in synchronous gauge is $v_C^{\prime }+\mathcal{H}{v}_C=0$, so multiplying by $a$ gives $(a{v}_C{)}^{\prime }=0$. ThereforeIn an expanding universe $a$ increases, damping any initial peculiar velocity as $a^{-1}$. During radiation domination $a\propto \tau$, so the velocity scales as ${\tau }^{-1}$; during matter domination $a\propto {\tau }^2$, it scales as ${\tau }^{-2}$. This is the nonrelativistic limit of the physical-momentum redshift derived in Q1. It concerns the bulk velocity in the synchronous coordinate frame, not a growing peculiar velocity sourced in another gauge. A cold-matter rest-frame choice also removes the residual synchronous time displacement relevant to that component. The decay statement assumes expanding evolution and the pressureless, linear approximation used by the equation.

For fixed finite velocities, the nonrelativistic limit $c\to \infty$ gives ${\gamma }^2=(1-v^2/c^2{)}^{-1}\to 1$. Hence the relativistic rocket equation reduces towhich is the nonrelativistic rocket equation.

The conserved quantities called energy in the two calculations include different contributions. The nonrelativistic calculation considered only kinetic energy, which increases as fuel releases internal energy; total energy including that internal reservoir is still conserved. Relativistic energy includes rest energy, and therefore includes the fuel's stored internal energy through the rocket's rest mass. That rest mass decreases as energy is transferred into the exhaust. Thus conservation of relativistic energy and the nonconservation of kinetic energy express the same physical conversion.

Write $E=|\mathbf{E}|$ and $\hat{\mathbf{E}}=\mathbf{E}/E$. The constant electric force gives momentumCombining $\mathbf{p}=\gamma m\dot{\mathbf{x}}$ with the relativistic energy-momentum relation givesThe dimensional analysis from part (a) confirms that $qEt$ has the dimensions of momentum, so $qEt/(mc)$ is dimensionless and the expression has the dimensions of velocity.

The speed increases monotonically from zero and approaches the speed of light asymptotically:Integrating in the direction of the field givesAt small times, the Taylor expansion of the square root gives the nonrelativistic limitwhich is motion from rest with the constant acceleration $q\mathbf{E}/m$.

The electric-dipole approximation for radiation requires the source size to be much smaller than the radiation wavelength:The displayed $1/r$ fields also describe the radiation zone, so the observation point must satisfy $r\gg a$ and $r\gg c/\omega$.

Write $t_r=t-r/c$. The given vector potential and the time derivative of the electric dipole moment giveWhen taking the curl, differentiating $1/r$ produces a near-field term of order $r^{-2}$, while differentiating the retarded time produces the leading radiation term because $\nabla t_r=-\hat{\mathbf{x}}/c$. Hence

The assumed electric field $\mathbf{E}=-c\hat{\mathbf{x}}\times \mathbf{B}$ and the transversality $\hat{\mathbf{x}}\cdot \mathbf{B}=0$ make the Poynting vectorOn a sphere of radius $R$, the power through surface element $R^{2}d\Omega$ is thereforeChoose the polar axis along $\ddot{\mathbf{p}}$ and use the angular distribution of electric dipole radiation, for which $|\hat{\mathbf{x}}\times \ddot{\mathbf{p}}{|}^2=|\ddot{\mathbf{p}}{|}^2{\sin }^2\theta$. Since $\int {\sin }^2\theta d\Omega =8\pi /3$,

For the charged particle in simple harmonic motion,ThusHere the angular frequency obeys $\omega =2\pi /T$, and the average uses $⟨{\sin }^2⟩=1/2$. The dipole condition is

Allow the amplitude to vary slowly as radiation removes energy. The mechanical energy isso the radiation damping of a harmonically oscillating charge givesThis is exponential decay with rate $\gamma ={\mu }_0{q}^2{\omega }^2/(6\pi mc)$. Its half-life isFinally, the largest particle speed is $v_{\max }=A\omega$. Therefore the dipole condition $A\omega /c\ll 1$ itself implies $v_{\max }\ll c$, precisely the nonrelativistic limit required by the mechanical energy formula.

The one-dimensional Lorentz transformation givesHence

As $|\beta |\to 0$, $\gamma =1+O({\beta }^2)$. The spatial transformation is therefore $x^{\prime }=x-ut+O({\beta }^2)$, whileUnder the stated condition $|x|<c|t|$, the fractional correction $|ux|/(c^2|t|)$ is smaller than $|\beta |$ and tends to zero. Thus the Lorentz transformation has the nonrelativistic limit given by the Galilean transformation.

An accelerating point charge has a transverse radiation field in addition to its near field. At nonrelativistic speed and large distance $R$,The acceleration is evaluated at retarded time. Radiation fields decay as $1/R$, so their outward energy flux integrated over a sphere is independent of radius; the near electric field instead decays as $1/R^2$. The radiation vanishes along the acceleration and is strongest perpendicular to it.^[ref]

Using the Poynting vector givesThe integral uses $\int {\sin }^2\vartheta d\Omega =8\pi /3$. This is the Larmor formula. A harmonically accelerated charge with peak acceleration $a_0$ has mean power $q^2{a}_0^2/(12\pi {ϵ}_0{c}^3)$; external work supplies this radiated energy in steady operation. Uniform inertial motion does not radiate.

At relativistic speed the angular pattern becomes directionally concentrated and retarded-source factors matter. The total instantaneous radiated power generalizes to $q^2{\gamma }^6[a^2-|\mathbf{\beta }\times \mathbf{a}{|}^2]/(6\pi {ϵ}_0{c}^3)$, with $\mathbf{\beta }=\mathbf{v}/c$. The displayed simple ${\sin }^2\vartheta$ pattern is restricted to the nonrelativistic limit.

Let ${\chi }_{\pm }=(1,\pm 1{)}^T/\sqrt{2}$. The eigenvalues of ${\Sigma }_1$ are $\pm 1/2$, each with multiplicity two, and orthonormal bases for their eigenspaces areA general linear combination of the two positive-energy solutions has upper spinor $\xi =a{\chi }^1+b{\chi }^2$ and lower spinor $(\mathbf{\sigma }\cdot \mathbf{p})\xi /(E+m)$. An x-spin eigenvector needs $\xi \propto {\chi }_{\pm }$ and the lower spinor to have the same ${\sigma }_1$ eigenvalue. The Fixed-axis spin obstruction for positive-energy Dirac spinors isIts norm is $2\sqrt{p_y^2+p_z^2}$, so it vanishes exactly when $p_y=p_z=0$. For example with $\mathbf{p}=p\hat{z}$, the upper ${\chi }_{+}$ produces lower ${\chi }_{-}$, excluding a common eigenvector for every $p\ne 0$. In the nonrelativistic limit, the lower components are $O(p/m)$, and $({\psi }_1\pm {\psi }_2)/\sqrt{2}$ approach x-spin eigenvectors. At rest they are exact.

The prompt's unrestricted claim of impossibility at relativistic energies needs qualification: for momentum parallel to x these combinations remain exact eigenvectors at any energy. For a beam transverse to x, such as the conventional z-directed beam, the intended impossibility holds, and the lower components are no longer negligible. A transported relativistic spin observable differs from this fixed matrix; helicity is the conserved projection along momentum for a free particle.

Take a stationary initial electron of mass $m_e$, a photon incident along $x$, and a nonrelativistic recoil velocity of magnitude $v$. Choose $\theta$ above $x$ for the scattered photon and $\varphi$ below $x$ for the electron. Conservation of momentum givesThe electron's unchanged rest energy cancels in conservation of energy, leavingEquivalently use $h\nu /c=h/\lambda$ and $h\nu =hc/\lambda$. The nonrelativistic limit requires $v/c\ll 1$, and treating the electron as free requires its binding energy to be negligible on the scale of the collision.

Rutherford scattering is elastic scattering by a Coulomb potential. In the repulsive classical case, a positively charged projectile bends around a positively charged atomic nucleus. A close approach can turn it backwards without a contact collision. The calculation connects an invisible impact parameter with a measurable outgoing angle, then with a differential scattering cross-section and a detector count.

Assume a projectile of mass $m$ and electric charge $ze$, a stationary point atomic nucleus of electric charge $Ze$, and positive $z,Z$. Here $e$ is the elementary charge. Neglect nuclear recoil, electron screening effect, energy loss, and the nuclear force. With vacuum permittivity ${\epsilon }_0$, defineHere $E$ and $v$ are the incident kinetic energy and speed far from the atomic nucleus, $b\ge 0$ is the impact parameter, and $L$ is the magnitude of angular momentum. The nonrelativistic limit requires $v\ll c$, with $c$ the speed of light. The fixed-target approximation requires the target mass to be much greater than $m$. In SI, $\kappa$ has units $\mathrm{J}\mathrm{m}$, $E$ has units $\mathrm{J}$, and $b$ has units $\mathrm{m}$.

The velocity-addition formula translates a particle's velocity between two specified inertial frames. Let $S^{\prime }$ move at signed velocity $V$ along the $x$-axis of $S$, with parallel axes and $|V|<c$, where $c$ is the speed of light. Define $u_x=dx/dt$ and $u_x^{\prime }=d{x}^{\prime }/d{t}^{\prime }$ for the same particle's world line, using position and time measured within the corresponding inertial frame. With Lorentz factor ${\gamma }_V=(1-V^2/c^2{)}^{-1/2}$, the forward Lorentz boost gives

For collinear motion, write $u=u_x$ and $u^{\prime }=u_x^{\prime }$. The inverse velocity-addition formula isThis is the composition of the particle's velocity in $S^{\prime }$ and that frame's velocity relative to $S$. The velocities must belong to this same chain of reference frames, with directions specified. Two arbitrary quoted speeds cannot be inserted blindly. Each velocity measurement uses distances and clocks in a single inertial frame, with Einstein synchronisation for separated clocks.

For example, let $V=0.6c$ and let the particle travel at $u^{\prime }=0.6c$ in $S^{\prime }$. Its velocity in $S$ isThe Galilean transformation would instead give $u=u^{\prime }+V=1.2c$, which fails here. For light moving along the positive axis, $u^{\prime }=c$ gives $u=c$; $u^{\prime }=-c$ similarly gives $u=-c$. Negative directions also work for massive particles: at $V=0.6c$, $u^{\prime }=-0.6c$ gives $u=0$, while $u^{\prime }=-0.8c$ gives $u=-5c/13$. Thus “addition” means composition of signed velocities, rather than addition of their magnitudes.

A short derivation differentiates the Lorentz transformation: $d{x}^{\prime }={\gamma }_V(dx-Vdt)$ and $d{t}^{\prime }={\gamma }_V(dt-Vdx/c^2)$. Dividing these differentials gives the longitudinal velocity-addition formula. The transverse formulas use $d{y}^{\prime }=dy$ and $d{z}^{\prime }=dz$, but the same transformed time denominator; omitting it would ignore the relativity of simultaneity. Replacing $V$ by $-V$ gives the inverse. When both relative and particle velocities are small compared with the speed of light, the denominator is close to one and the nonrelativistic limit recovers the Galilean transformation.

The formula lets us compare particle motion in a moving apparatus and a laboratory without losing the invariant speed of light. Einstein's 1905 paper derives the coordinate change in §3 and the composition of velocities in §5, including the collinear formula, its light-speed case, and composition of parallel transformations. ^[ref]. The rapidity parametrization makes the collinear composition particularly simple.