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For isotropic Brownian motion started at zero with independent coordinate variances $2Dt$, the Gaussian distribution has density $(4\pi Dt{)}^{-3/2}{e}^{-|x{|}^2/(4Dt)}$. Integrating the radius with spherical volume element $4\pi r^{2}dr$ gives $\mathbb{E}|X_t|=4\sqrt{Dt/\pi }$. The root mean square displacement is instead $\sqrt{6Dt}$, and the mean vector is zero. None of these is the microscopic path length.

Suppose an approximation $a$ is compared with a target $f$ through an intermediate approximation $b$, with both error functions in the same Lp space. The identity $f-a=(f-b)+(b-a)$ and the Minkowski inequality combine two error budgets without any independence assumption. The L1 norm controls aggregate absolute error, the L2 norm controls the square root of aggregate quadratic loss, and the L-infinity norm controls the largest error almost everywhere. For equally weighted samples with the probability measure $1/n$ on each coordinate, the L2 norm is the root mean square error; the inequality still applies when this normalization is used consistently.

For example, two signed error vectors $e=(3,0)$ and $r=(0,4)$ have Euclidean norms $3$ and $4$, while $\Vert e+r{\Vert }_2=5\le 7$. Reinforcing errors $e=(3,0)$ and $r=(4,0)$ attain the bound $7$; opposing errors $e=(3,0)$ and $r=(-3,0)$ cancel completely. The Minkowski inequality always allows reinforcement. A smaller combined budget requires additional information about the errors, not just their separate norms.

Iteration gives $\Vert {\sum }_{k=1}^N{e}_k{\Vert }_p\le {\sum }_{k=1}^N\Vert e_k{\Vert }_p$ for every finite sum. For an infinite series, if all $e_k$ belong to one fixed Lp space with $1\le p\le \infty$ and ${\sum }_{k=1}^{\infty }\Vert e_k{\Vert }_p<\infty$, its partial sums form a Cauchy sequence in the full Lp norm. Completeness of this Banach space then gives norm convergence to an element of the same Lp space. A finite sum bound alone does not establish convergence, and completeness is an additional ingredient beyond the Minkowski inequality. ^[ref]

For a real wavefront error $h={\sum }_k{a}_k{s}_k$, its squared spatial root mean square is $⟨h^2⟩={\sum }_{k,l}{a}_k{a}_l⟨s_k{s}_l⟩$. Orthogonal unit-peak sine modes have mean square $1/2$, so $h_{\mathrm{rms}}^2=\frac{1}{2}{\sum }_k{a}_k^2$, whereas the coefficient norm is $\sqrt{{\sum }_k{a}_k^2}$. On a finite optical pupil orthogonality needs checking; independent uniform random wave phases give the same $1/2$ diagonal after wave phase averaging. This normalization matters when a speckle contrast of a sinusoidal wavefront error is specified by peak wave amplitude.

For a far-field free-space link with one reciprocal antenna transmitting and receiving at range $D$, the outward intensity is $P_{T}G/(4\pi D^2)$. The backward scattered intensity at the receiver is this times ${\sigma }_{\mathrm{back}}/(4\pi D^2)$. Multiplying by the matched antenna effective area $G{\lambda }^2/(4\pi )$ gives the displayed received power. Attenuation, misalignment and polarization mismatch require additional factors. The threshold for a root mean square voltage $V_m$ across a real matched load $R$ is $V_m^2/R$.

The most probable speed is the mode, or density maximum, of the Maxwell speed distribution:For small equal-width speed bins, the bin near $v_{\mathrm{mp}}$ contains the largest expected fraction of particles. This does not mean that any exact speed has positive probability, or that every particle has this speed. The three common summaries obeyso $v_{\mathrm{mp}}<⟨v⟩<v_{\mathrm{rms}}$. The first describes a density peak, the second the mean speed, and the third the root mean square associated with kinetic energy. ^[ref]

For a concrete scale $s=300\mathrm{m}{\mathrm{s}}^{-1}$, these are approximately $424$, $479$, and $520\mathrm{m}{\mathrm{s}}^{-1}$. The most probable velocity vector and each most probable component remain zero in the gas rest frame. A zero vector labels one small region of velocity space, whereas a positive speed labels an entire spherical shell.

A short check uses the derivative of the logarithm of the Maxwell speed distribution for $v>0$:It is positive below $\sqrt{2}s$ and negative above it, locating the unique peak. The factor $v^2$ from the spherical shell supplies the term $2/v$; omitting it would incorrectly move the speed peak to zero.

The root mean square radius of a nuclear charge distribution, referred to its center. Define its charge-weighted second moment byFor a spherical charge density with finite fourth moment, insert the Taylor series $\sin u/u=1-u^2/6+O(u^4)$ into the nuclear charge form factor:Thus a radius is a low-momentum transfer slope, not necessarily the position of a sharp edge. In this real static model,The factor of two between the slopes of $F$ and $|F{|}^2$ matters when interpreting a differential scattering cross-section. A proton charge radius is the corresponding size concept for the proton; the spatial arrangement of many nucleons introduces an additional scale in a heavier atomic nucleus.

For a zero-mean stationary rough height, the mean first-order correction vanishes. At second order, the boundary derivative is $-h{\partial }_z^2{\psi }_1+h^{\prime }{\partial }_x{\psi }_1$. Averaging this expression using the power spectrum of surface height gives the displayed coherent reflection coefficient. Thus the root mean square height alone does not fix the coherent mean: the height correlation spectrum matters. For roughness varying slowly enough that $q\simeq \alpha$ in the integral, $r_N\simeq 1-2{\beta }^2{\sigma }^2$. A local random-height phase average needs this additional approximation and is not a general consequence of stationarity.

Let a segment of fixed length $b$ make angle $\theta$ with an applied force $\mathbf{f}=f\hat{\mathbf{z}}$. Its energy contribution is $-fb\cos \theta$. Independent freely jointed orientations give the single-segment partition function and chain partition functionThe mean end-to-end projection is the force derivative of $k_{B}T\ln Z_N$:For $|u|\ll 1$, the Langevin function is $u/3+O(u^3)$. Isotropy therefore gives the vector responseThe covariance at zero force provides an independent route: $⟨R_i{R}_j{⟩}_0=N{b}^2{\delta }_{ij}/3$, and the Boltzmann weight's first-order term yields $⟨R_i⟩=f_j⟨R_i{R}_j{⟩}_0/(k_{B}T)$. This is a relation for the mean vector or its projection. The zero-force mean vector vanishes even though the chain has finite root mean square size $b\sqrt{N}$.

The controlled weak-force condition is $fb\ll k_{B}T$, not merely that a short part of a measured curve looks straight. With the supplied $b$, $k_{B}T/b\simeq 10.3\mathrm{pN}$ at $25^{\circ }\mathrm{C}$. A raw force near $30\mathrm{pN}$, as in the figure's labelled low-extension portion, would give $u\simeq 2.9$ and is not asymptotically weak for those literal segment parameters. A force offset or different effective model might change that interpretation, but no such correction should be invented from the plot.

On releasing the tensile load, the extended chain first retracts towards less aligned conformations because they have much greater configurational multiplicity. Viscous friction and internal relaxation determine the time course; stored free energy is dissipated as heat. An unconstrained ideal chain does not shrink to a point: its mean vector is zero but its root mean square end-to-end length is $b\sqrt{N}\simeq 2.37\mathrm{nm}$ for the supplied ideal-chain parameters.

A real polypeptide also has sequence-dependent interactions. If the original native structure is the stable state in the prevailing solvent and the molecule remains intact and capable of folding, it can subsequently refold, gaining favourable intramolecular and solvent free energy. The folded state is not selected by chain entropy alone, and formation of native contacts need not be instantaneous. Kinetic traps, proline isomerization, protein aggregation, misfolding, surface attachments or chemical damage may prevent complete recovery or produce hysteresis. Thus reversible refolding is a physically motivated possibility, not guaranteed simply by removal of force. This distinction connects the recoil of a stretched polymer to the thermodynamics discussed in Question 6(c).

Write the end-to-end random vector as $\mathbf{r}={\sum }_{j=1}^{N}b{\mathbf{u}}_j$, with independent isotropically distributed unit vectors. At zero applied force, $⟨{\mathbf{u}}_j⟩=0$, ${\mathbf{u}}_j\cdot {\mathbf{u}}_j=1$, and for $i\ne j$ independence gives $⟨{\mathbf{u}}_i\cdot {\mathbf{u}}_j⟩=0$. ThereforeHere “size” is the root mean square end-to-end distance of a polymer chain. It is not the magnitude of the mean random vector, which is zero, nor the mean distance $⟨r⟩$. In the long-chain Gaussian limit of the end-to-end distribution of a freely jointed chain, $⟨r⟩\simeq b\sqrt{8N/(3\pi )}$, illustrating why these definitions cannot be interchanged.

Take $V=|\mathbf{V}|$ to be the test body's instantaneous speed relative to the nonstreaming background, rather than an ensemble root mean square speed. Use a local isotropic galactic distribution function $f(v)$ normalized as a number distribution, and assume independent encounters and a large, approximately constant Coulomb logarithm in stellar dynamics.

For a background particle with relative incoming velocity $\mathbf{w}=\mathbf{V}-\mathbf{v}$, a gravitational scattering angle $\chi$ obeysIn the centre-of-momentum frame, the test body's share of the relative velocity is $m/(M+m)$. The azimuthal average cancels transverse changes, leaving a change parallel to the incoming relative velocity,The number of encounters per unit time in $d^{3}v$ and impact interval $db$ is $f(v)d^{3}vw2\pi bdb$. Integrating the longitudinal change from $b=0$ to $b_m$ givesAt leading logarithmic order replace the logarithm by $2\ln \Lambda$, evaluated at a representative speed. For isotropic $f$, the velocity-shell identity for dynamical friction isOne proof takes the divergence with respect to $\mathbf{V}$: the inverse-square vector field has divergence $4\pi {\delta }^{(3)}(\mathbf{V}-\mathbf{v})$, whose velocity integral is $4\pi f(V)$, and rotational symmetry plus the divergence theorem determines the radial field. Exterior velocity shells give zero field. Consequently the systematic Chandrasekhar dynamical friction is antiparallel to motion, withIn the massive-test-body approximation $M\gg m$, this becomesThe finite-test-mass dynamical friction coefficient shows the needed qualification: $M>m$ alone does not make $M+m=M$. The quoted expression also neglects diffusion and the velocity dependence of the logarithmic cutoff. For the lowered Maxwellian velocity distribution, set $f=0$ for $v>v_e$; continuing the printed subtraction beyond escape would give a negative distribution. If $V>v_e$, the integral stops at $v_e$.

Take a positive initial separation $0<l_0\ll \eta$. In the dissipation range, the velocity field is smooth, so the difference of the two Lagrangian trajectories obeys the linearized separation equation $\dot{\mathbf{l}}\simeq (\nabla \mathbf{u})\mathbf{l}$. Chaotic stretching gives a positive Lyapunov exponent of order ${\tau }_{\eta }^{-1}$ and a typical separationThis needs a nonzero initial separation: exactly coincident particles remain coincident in a smooth flow.

Once $l$ lies in the inertial range, the Kolmogorov 1941 theory gives a relative speed of order $(ϵl{)}^{1/3}$. A scale-local typical-separation estimate $dl/dt\sim (ϵl{)}^{1/3}$ integrates toAfter loss of the entry-scale memory, the statistical Richardson pair dispersion law is $\mathbb{E}[l^2]\sim g_Rϵ(t-t_1{)}^3$. Thus the characteristic separation grows as $(t-t_1{)}^{3/2}$ while $\eta \ll l\ll L$. This is a statistical scaling closure, rather than a deterministic equation for every pair.

For $l\gg L$, the particle velocities become approximately independent. If their Lagrangian velocity autocorrelations have a finite Lagrangian integral time $T_L\sim L/U$, the diffusive large-scale pair dispersion has effective relative diffusivity of order $UL$. After a further velocity-decorrelation transient,This is the long-time mechanism of Taylor turbulent dispersion applied to the difference of two decorrelated trajectories. The root mean square separation eventually loses memory of the $L^2$ term.

Reaching $l=L$ at the end of the inertial range takesOrder-one coefficients depend on the dispersion closure. The initial exponential stage can take a long time if $l_0$ is extremely small, but the inertial-range stage is of order one outer eddy turnover time.

The three-regime turbulent pair-separation model in the sketch matches the regimes continuously for illustration. Its coefficients are schematic; the universal claims here are the scaling powers under the stated stretching, locality and decorrelation assumptions.

For the incident acoustic plane wave, set $q_0=k\sin \theta$ and ${\beta }_0=k\cos \theta >0$. The flat reflected field and total field areTherefore ${\partial }_z{\psi }^{[0]}(x,0)=-2i{\beta }_0{e}^{i{q}_{0}x}$ and the first-order rough-surface scattered field isIt is linear in the height. Since $⟨h(x)⟩=0$, its mean vanishes, with the expectation interpreted through finite windows or stationary spectral distributions when needed. HenceThe coherent first-order reflection is the flat-surface reflection. If the field symbol is instead used only for the rough correction, its first-order mean is zero. Stationarity ensures the coherent reflection retains the incident horizontal wavenumber, but zero mean height already explains the vanishing linear correction. The nonzero root mean square height does not enter this mean at first order; it does enter the fluctuating reflected field and its intensity.

Write $x=1/a$ for the comet's inverse semi-major axis. Its specific orbital energy is $-G{M}_{\star }x/2$, so escape corresponds to $x=0$. The printed quantityhas dimensions of inverse length. It cannot, literally, be a standard variance-per-time diffusion coefficient. Interpret it as the characteristic root mean square step in $x$ per periapsis passage, as in a discrete comet energy diffusion model.

If successive kicks are unbiased and uncorrelated, after $N$ passages the root mean square displacement is $s_x\sqrt{N}$. Starting at $a_0$ gives $x_0=1/a_0$, hence $N_{\mathrm{ej}}\sim (x_0/s_x{)}^2$. Taking the characteristic interval to be the initial orbital period $P(a_0)$ givesHere $P_p=2\pi \sqrt{a_p^3/(G{M}_{\star })}$. In particular, for $a_0\sim a_p$ this is $P_p(M_{\star }/M_p{)}^2/100$. Without specifying the initial semi-major axis and the time per statistically independent encounter, the PDF cannot determine a unique time.

With the convention $⟨(\Delta x{)}^2⟩=2D_{x}t$, the actual diffusion coefficient would be $D_x=s_x^2/[2P(a_0)]$. The estimate requires $N_{\mathrm{ej}}\gg 1$, orbit crossing, and sufficient phase decorrelation. Near escape the orbital period grows, and resonant or secular correlations invalidate the simple constant-step clock. This is a characteristic diffusion scale, not an exact mean first-passage time: even an unbiased Brownian motion on an unbounded half-line has an infinite mean time to reach its absorbing endpoint.

The speckle contrast of a sinusoidal wavefront error gives a peak modal wave amplitude $h_0=(\lambda /\pi )\sqrt{C}$. If all $M$ residual modes have that wave amplitude, their coefficient norm is $H_{\mathrm{coeff}}=\sqrt{M}{h}_0$. Substituting the printed quadrant count gives the intended expressionThere is a normalization issue if this quantity is called physical spatial root mean square wavefront error. For $h(\mathbf{r})={\sum }_k{h}_{0k}\cos (2\pi {\mathbf{u}}_k\cdot \mathbf{r}/D+{\psi }_k)$, the actual pupil-averaged value isOn an orthogonal full-period basis, $⟨s_k^2⟩=1/2$ and cross terms vanish. Alternatively, independent uniform random wave phases give this result after wave phase averaging, even on a circular optical pupil. With the same $M$ and equal contrasts the physical RMS is thenA single full-period cosine already demonstrates the issue: its RMS is $h_0/\sqrt{2}$, not $h_0$. On the actual circular optical pupil, any nonconstant unit-peak cosine also has mean square strictly less than one; averaging over its wave phase gives exactly $1/2$. On a finite circular optical pupil, phase-dependent mode overlaps can additionally matter. If a constant piston is removed before computing RMS, replace the mode Gram matrix by its centred optical pupil covariance. The printed value can be recovered as a modal coefficient norm, or by counting $2M$ orthogonal peak-amplitude quadratures in the physical RMS; neither convention is specified by the quoted mode definition. This is the distinction between modal amplitude norm versus wavefront RMS, not a correction silently made to the PDF. Unequal residual contrasts require ${\mathbb{E}}_{\psi }{h}_{\mathrm{rms}}^2={\lambda }^2{\sum }_k{C}_k/(2{\pi }^2)$, so one speckle's $C$ alone cannot determine the total RMS. The weak-aberration approximation also requires the total wave phase error to stay small, not merely each coefficient separately.

The Strehl ratio isusing the same total flux, pupil, and wavelength for both images. A value near one means that little light has been redistributed from the ideal central peak by wavefront errors. For small residual phase variance ${\sigma }_{\varphi }^2$, the Maréchal approximation is $S\simeq e^{-{\sigma }_{\varphi }^2}$, with ${\sigma }_{\varphi }=2\pi h_{\mathrm{rms}}/\lambda$ for the residual optical-path root mean square error.

The supplied speckle contrast of a sinusoidal wavefront error gives $h_0=\lambda \sqrt{C}/\pi$. Each sinusoid has root mean square error $h_0/\sqrt{2}$. For orthogonal modes, or for uncorrelated phases so that cross terms average to zero, the total mean square adds:Therefore the assumed equal-contrast modal model givesIf the individual contrasts are not equal, the corresponding formula is $h_{\mathrm{rms}}^2={\lambda }^2{\sum }_i{C}_i/(2{\pi }^2)$. The error here is optical-path wavefront error; for a near-normal reflecting mirror, physical surface error is half the optical-path error.

antenna quantities. radiation resistance represents emitted power by $P_{\mathrm{rad}}=I_{\mathrm{rms}}^2{R}_{\mathrm{rad}}$ at the feed. It is not Joule heating in the antenna; a separate loss resistance accounts for that. antenna gain is the directional far-field intensity relative to an isotropic radiator supplied with the same input power:It includes radiation efficiency and equals antenna directivity only for a lossless antenna. antenna effective area is defined by the power available to a matched load, $P_{\mathrm{av}}=A_{\mathrm{eff}}{S}_{\mathrm{inc}}$, for incidence from a specified direction and matched polarization. It need not equal a literal geometric collecting area.

Gain–area relation. To derive the antenna gain-effective-area relation, take a passive reciprocal antenna. Its electromagnetic reciprocity makes the receive pattern proportional to the transmit pattern; write $A_{\mathrm{eff}}(\Omega )=KG(\Omega )$. The normalization follows from thermal equilibrium. In the Rayleigh-Jeans law limit, blackbody radiation radiance per unit frequency per unit solid angle is $2k_{B}T/{\lambda }^2$ for two polarizations. A single antenna polarization couples to one half, so the available incident noise power isThe open-circuit Johnson voltage spectrum of the antenna resistance is $4k_{B}T{R}_{\mathrm{rad}}d\nu$. A matched load receives half the open-circuit voltage, giving available power $k_{B}Td\nu$. In a lossless antenna at thermal equilibrium these powers agree; hence $\int A_{\mathrm{eff}}d\Omega ={\lambda }^2$. Since $\int Gd\Omega =4\pi$,The same efficiency factor multiplies receiving area and gain if reciprocal dissipative losses are included consistently. Impedance mismatch or polarization mismatch reduces delivered power further. The factor of two for thermal polarization modes and the factor of four between open-circuit voltage variance and matched-load power are separate effects.

Normalize the radar beam. Interpret the azimuth as signed about the beam centre at $\varphi =0$, so both sides of the narrow lobe are included. Put $u=\theta -\pi /2$. To leading order in the small angular width, $\sin \theta \simeq 1$ and the angular integrals can be extended over the real line:The Gaussian antenna beam solid angle gives the peak on the positive $x$-axis consequently hasHere $N$ is power per solid angle, equal to $r^2{S}_r$ in the far field. The leading neglected solid-angle correction is relative order ${\alpha }^2$.

Return using the supplied electric-dipole response. The electric-dipole radar cross section of a small conducting sphere determines this return. At target range $D$, the incident intensity is $S_t=C/D^2$. Write its root mean square electric field as $E_t$, with $S_t={ϵ}_{0}c{E}_t^2$. The induced electric dipole is $p_{\mathrm{rms}}=4\pi {ϵ}_0{a}^3{E}_t$. A dipole's angular radiation pattern, normalized by $\int {\sin }^2\chi d\Omega =8\pi /3$, gives $d{P}_{\mathrm{sc}}/d\Omega =(3P_{\mathrm{sc}}/8\pi ){\sin }^2\chi$. The backward direction is perpendicular to the induced dipole, so the supplied total dipole power givesThe received intensity is this angular power divided by $D^2$. Using the monostatic radar equation, or directly multiplying by the receive area, givesThe receiver threshold is a voltage across the matched load, so the detectable power is $V_m^2/R$, without another matching factor:This is the electric-dipole result following from the response data provided. It assumes $ka\ll 1$, far-field propagation on each leg, alignment at the beam maximum and no attenuation or additional receiver losses.

Perfect-conductor qualification. perfect-conductor sphere dipole interference arises because the sphere also excludes the oscillating magnetic field. Solving the magnetic boundary problem gives $\mathbf{m}=-2\pi a^3{\mathbf{H}}_t$ in addition to the electric dipole. The magnetic polarizability of a perfectly conducting sphere follows from its induced surface currents.^[ref] In the backward direction the magnetic contribution has half the electric scattering amplitude and the same sign, so the returned intensity is $9/4$ of the electric-only value. Equivalently, the monostatic radar cross section is ${\sigma }_{\mathrm{back}}=9\pi k^4{a}^6$, rather than $4\pi k^4{a}^6$. Including both leading dipoles therefore givesBoth results have been distinguished because the supplied auxiliary data describe only the electric dipole, while the literal perfect-conductor boundary conditions also generate the magnetic one. The backward interference is a leading-order effect, not a higher-order correction in $ka$.^[ref]

Gauge freedom. The magnetic vector potential satisfies $\mathbf{B}=\nabla \times \mathbf{A}$. The transformationleaves $\mathbf{B}$ unchanged because the curl of a gradient is zero. For time-dependent fields, transform the electric scalar potential as $\varphi \mapsto \varphi -{\partial }_t\chi$ as well, so that $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$ is unchanged. The scalar $\chi$ must have the regularity and single-valuedness appropriate to the physical region and its boundary conditions. Local values of $\mathbf{A}$ therefore depend on the gauge transformation.

Gauge and nonzero component. In the standard axisymmetric Coulomb gauge representative for the long-solenoid approximation,The divergence vanishes since $A_{\varphi }$ is independent of the azimuth. Inside, this agrees with the usual symmetric gauge representative $\mathbf{A}=\frac{1}{2}\mathbf{B}\times \mathbf{r}$. The Coulomb gauge condition alone does not force the other two components to vanish; the symmetry, regularity and customary boundary choice select this representative from the remaining gauge freedom.

Radial dependence. Write the interior magnetic field as $B_0\hat{\mathbf{z}}$, where $B_0={\mu }_{0}nI$ for a vacuum solenoid of solenoid turn density $n$. Its exterior magnetic field vanishes in the infinitely long idealization. Applying Stokes theorem to a coaxial circle gives $2\pi r{A}_{\varphi }(r)=\Phi (r)$. Therefore the Coulomb-gauge vector potential of a long solenoid isIt grows linearly inside, is continuous at $R$, and decays as $r^{-1}$ outside. Although its exterior curl is zero, its circulation is the enclosed magnetic flux $\Phi =B_0\pi R^2$. No regular single-valued gauge function removes it throughout the exterior while preserving this circulation.

Electric field of the rotating charged surface. For clarity retain absolute permittivity $ϵ$ and permeability $\mu$ in the cylinder formulae; setting them to unity gives the rationalized-unit convention in the question. If the stated unity means relative material constants in SI, use ${ϵ}_0,{\mu }_0$ instead. Assume prescribed charge densities in the laboratory frame and the nonrelativistic, steady-rotation model.

The axisymmetric surface charge is stationary as a distribution even while the material rotates. A Gaussian cylinder of length $L$ has enclosed charge zero for $r<R$ and $2\pi RL\sigma$ for $r>R$. Gauss's law therefore givesThe radial electric field jumps from zero to $\sigma /ϵ$ at the surface and then decays as $r^{-1}$. Its direction reverses for negative $\sigma$. The angular velocity does not enter this electric-field result for a specified surface charge.

Magnetic field of the volume-charged cylinder. A shell of radius $a$ and thickness $da$ has surface charge $\rho da$, azimuthal surface current $d{K}_{\varphi }=\rho \omega ada$, and the field of an infinite solenoid: $d{B}_z=\mu \rho \omega ada$ inside that shell and zero outside it. At radius $r<R$, only shells with $a>r$ contribute. Thus the magnetic field of a rotating charged solid cylinder isWith no applied background magnetic field, $\mathbf{B}=0$ outside. The profile is parabolic, largest on the axis and zero at the boundary. The sign follows $\rho \omega$. As a check, the current density is $J_{\varphi }=\rho \omega r$, and Ampère's circuital law gives $-d{B}_z/dr=\mu J_{\varphi }$, as required.

Electron impulse when the current is switched off. Choose positive $I$ so the initial magnetic flux is ${\Phi }_i={\mu }_{0}nI\pi R^2$ along $+z$. Faraday's law of induction gives, outside the solenoid,For an electron of electric charge $q=-e$, treat its position as fixed while the short pulse acts. The inductive impulse outside a switched solenoid isConsequently the nonrelativistic final velocity isThe induced electric field circulates in the positive azimuthal direction while a positive flux decreases, so the electron accelerates in the negative direction. Once the pulse has passed, there is no exterior field and it travels along the tangent at the impulse point, rather than continuing in a circular orbit. If the resulting momentum is relativistic, use ${\mathbf{v}}_f=c{\mathbf{p}}_f/\sqrt{m_e^2{c}^2+|{\mathbf{p}}_f{|}^2}$ instead. The sudden-switch result assumes negligible displacement during the pulse and the usual quasistatic electromagnetic induction approximation; “instantaneous” switching does not mean instantaneous propagation of fields to arbitrary distance.

One can check the sign using conserved canonical angular momentum: axial symmetry conserves $p_{\varphi }^{\mathrm{can}}=L_z+qr{A}_{\varphi }$. Initially $L_z=0$, and finally $A_{\varphi }=0$, so $L_z=q{\Phi }_i/(2\pi )=-e{\Phi }_i/(2\pi )$.

Interferometric measurement at constant current. Use a coherent electron beam split into two paths passing on opposite sides of a shielded long solenoid, then recombine them to form interference fringes. Keep the current constant during the passage and suppress leakage fields on the paths. In the Aharonov-Bohm effect, the phase difference due to the magnetic vector potential is the closed-loop integralThe overall sign reverses if the path orientation is reversed. The displacement in fringe periods is $\Delta \phi /(2\pi )$, so the measured fringe shift is linear in $I$ and repeats whenA gauge transformation adds a gradient whose closed-loop integral vanishes for a single-valued $\chi$. Endpoint changes in electron wavefunction phase cancel in the interference measurement. The measured Aharonov-Bohm phase is therefore gauge invariant: it determines the flux, or equivalently the circulation of $\mathbf{A}$. In the chosen Coulomb gauge this implies $A_{\varphi }(d)={\mu }_{0}nI{R}^2/(2d)$, but the experiment does not select or measure a unique local vector potential independently of gauge.

Far field of the rotating sphere. Interpret uniform charge as uniform volume charge of total $Q$, and take $\mathbf{\omega }=\omega \hat{\mathbf{z}}$. The magnetic dipole moment of a rigidly rotating charge distribution isIts axial component is $m_z=(\rho \omega /2)\int r_{\perp }^2{d}^{3}r$. For a solid sphere, $\int r_{\perp }^2{d}^{3}r=8\pi R^5/15$ and $\rho =3Q/(4\pi R^3)$, soThe magnetic field of a rotating charged spherical body is a static magnetic dipole field at large distances:In particular, $|\mathbf{B}|={\mu }_0|\mathbf{m}|\sqrt{1+3{\cos }^2\vartheta }/(4\pi r^3)$. Both the angular mean of the magnitude and its root mean square therefore scale as $\boxed{r^{-3}}$; the latter is $B_{\mathrm{rms}}=\sqrt{2}{\mu }_0|\mathbf{m}|/(4\pi r^3)$. The angular average of the vector is zero, which is different from average strength. Steady rotation leaves the charge and current distributions stationary, so this is magnetostatics, not a radiative $r^{-1}$ field. If uniform charge instead means a thin spherical surface, replace the dipole moment by $Q{R}^2\mathbf{\omega }/3$; the angular pattern and $r^{-3}$ scaling are unchanged.

nuclear radius scaling follows approximately $R=r_0{A}^{1/3}$ with $r_0\simeq 1.2\mathrm{fm}$ for a sharp matter-radius convention. Consequently the interior nucleon density is nearly independent of mass number: nuclear forces are short range and exhibit nuclear matter saturation. A diffuse nuclear charge distribution has a charge-weighted root mean square radius, not a hard surface. For a uniform sphere, $r_{\mathrm{ch}}=\sqrt{3/5}R$; charge radii and matter radii need not coincide, especially with neutron skins.

Elastic electron scattering measures the nuclear charge form factor. In a simple one-photon, spherical approximation, with $q$ a wave number,The scattering cross section relative to a point-charge prediction is proportional to $|F{|}^2$, so its low-$q$ slope constrains the radius and diffraction structure constrains the radial distribution. Finite momentum coverage and Coulomb distortion in electron scattering limit reconstruction. In Robert Hofstadter's experiment a magnetic spectrometer separated elastic and inelastic electron peaks; beam-charge normalization and closed-slit background checks prevented raw count structure from being mistaken for a nuclear charge form factor.^[ref]

A useful intrinsic-shape description is $R(\theta ,\varphi )=R_0[1+{\sum }_{\mu =-2}^2{\alpha }_{2\mu }{Y}_{2\mu }(\theta ,\varphi )+\cdots ]$. An intrinsic nuclear quadrupole moment $Q_0=\int {\rho }_{\mathrm{ch}}(3z^2-r^2)d^{3}r$ distinguishes prolate and oblate positive-charge distributions by positive and negative signs. The laboratory spectroscopic nuclear quadrupole moment is state dependent: an $I=0$ state has zero laboratory quadrupole moment even when its intrinsic shape is deformed. nuclear rotational bands approximately follow $E_I\propto I(I+1)$; harmonic nuclear quadrupole vibrations instead have equally spaced phonon energies and multiphonon spin multiplets. Measured electric quadrupole transitions test the collective interpretation in addition to level energies. Radius probes spatial extent; quadrupole observables and collective spectra probe shape.