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Choose polar angle $\theta$ from the acceleration direction. Then $|n\times q\dot{v}{|}^2=q^2|\dot{v}{|}^2{\sin }^2\theta$. Integrating over the solid angle gives $2\pi {\int }_0^{\pi }{\sin }^3\theta d\theta =8\pi /3$, and hence the specified normalization gives Larmor formulaThis uses the source's convention for radiated power; standard SI normalization has an additional factor $1/c$ in both given power formulas.

To leading order when $W\ll E$, neglect the loss while computing the head-on trajectory. Mechanical energy conservation gives $\frac{1}{2}m{\dot{r}}^2+V(r)=E=V(r_0)$ and acceleration magnitude $|V^{\prime }(r)|/m$. On each leg,The incoming and outgoing legs contribute equally. Multiplying the power by this time element, adding both legs, and assuming the integral converges givesUsing the unperturbed turning point is justified at this order in the small radiative loss, for a regular turning trajectory.

Let $\epsilon =\omega /B\ll 1$. The exact frequency has expansion $\lambda =B-\omega \cos \alpha +O({\omega }^2/B)$. The ratio $(B-\omega \cos \alpha )/\lambda =1+O({\epsilon }^2)$, while the coefficient of the orthogonal state is uniformly $O(\epsilon )$. Hence the supplied solution reduces towith phase error $O({\omega }^{2}t/B)$, which is $O(\epsilon )$ over one cycle. This is the adiabatic approximation: the system remains in the instantaneous positive-energy eigenstate, with energy $E_{+}=\hslash B/2$.

Thus the dynamic phase is $-Bt/2$ and the geometric phase is $-\omega t(1-\cos \alpha )/2$, in the gauge for ${\chi }_{+}$ printed in the PDF. For one cycle $T=2\pi /\omega$, the Berry phase isminus half the solid angle swept by the field direction.

To calculate the requested integral, use $|\psi (\alpha ,\phi )⟩=(\cos (\alpha /2),e^{i\phi }\sin (\alpha /2){)}^T$. It has $i⟨\psi |{\partial }_{\phi }\psi ⟩=-{\sin }^2(\alpha /2)$; its radial and polar-angle connection components vanish in this gauge. Along the circle in parameter space, $\phi$ runs from zero to $2\pi$, so $\Gamma =-{\int }_0^{2\pi }{\sin }^2(\alpha /2)d\phi =-\pi (1-\cos \alpha )$, agreeing with the exact-solution limit.

Use the same physical normalization for both fields. Part (b) has $E_s=E_i|f_{\infty }^{(1)}{|}^2/R^2+O(R^{-3})$, so the common energy flux prefactor cancels. The differential cross-section for scalar wave scattering isThe far-field pattern has dimensions of length in this normalization, so the differential cross-section for scalar wave scattering has dimensions of area per unit solid angle. Only the scattered field is used in this ratio; interference between the incident and scattered fields is a different contribution to the total field's energy flux.

Take an unresolved spherical planet at Euclidean distance $d\gg R$, with uniform surface temperature $T$, stationary isotropic emission, and no intervening extinction. Isothermality does not by itself imply blackbody radiation; first derive the geometric result for a uniform emergent specific intensity $I_{\lambda }$.

The observer sees the projected disc with apparent solid angle ${\Omega }_p\simeq \pi R^2/d^2$. Integrating its specific intensity over that angle gives the observed spectral radiative fluxwhere ${\theta }_{\mathrm{obs}}$ is the small angle to the detector normal. For a blackbody, $I_{\lambda }=B_{\lambda }(T)$, and the Stefan–Boltzmann law gives ${\int }_0^{\infty }\pi B_{\lambda }(T)d\lambda ={\sigma }_{\mathrm{SB}}{T}^4$. HenceEquivalently, the surface radiative flux is ${\sigma }_{\mathrm{SB}}{T}^4$ and the luminosity of a spherical blackbody is $L=4\pi R^2{\sigma }_{\mathrm{SB}}{T}^4$, leading to the same result by inverse-square dilution. A uniform wavelength-dependent emissivity ${ϵ}_{\lambda }$ multiplies the spectral formula; an atmospheric spectrum can require a full angular radiative transfer integral rather than a single Planck function.

Consider a narrow bundle of rays between two surface elements $d{A}_1,d{A}_2$ separated by $s$ in a transparent, stationary Euclidean vacuum. Let the ray make angles ${\theta }_1,{\theta }_2$ to the respective surface normals. The receiving element subtends solid angle $d{\Omega }_1=\cos {\theta }_{2}d{A}_2/s^2$ at the emitter. By the definition of specific intensity, the beam power in frequency interval $d\nu$ isAt the receiving end the source subtends $d{\Omega }_2=\cos {\theta }_{1}d{A}_1/s^2$, so the identical power isIn the absence of absorption, emission or frequency shifts, conservation of energy equates these expressions:This vacuum conservation of specific intensity also follows directly from the radiative transfer equation $d{I}_{\nu }/ds=0$. The apparent solid angle of an unresolved source shrinks as $s^{-2}$, so its integrated observed radiative flux still follows the inverse-square law. Specific intensity and unresolved flux are different quantities.

The vacuum and frequency assumptions are essential: absorption and emission change specific intensity, and gravitational or cosmological redshift conserves $I_{\nu }/{\nu }^3$ along a ray instead of $I_{\nu }$ itself. A fixed telescope aperture observing an unresolved object also cannot infer constant total received power from intensity conservation.

Use a stationary, non-streaming tracer population with spherical symmetry in its full velocity distribution. In particular, the mean velocities vanish, the mixed covariances vanish, and the two tangential components have equal velocity dispersions. WriteHere ${\sigma }_t^2$ is the variance of one tangential component. In the alternative convention using the sum of both tangential variances, the same velocity-anisotropy parameter is $1-({\sigma }_{\theta }^2+{\sigma }_{\varphi }^2)/(2{\sigma }_r^2)$. These definitions agree; they must not be mixed by inserting an extra factor of two.

The time derivative and mean-velocity terms in the supplied Jeans equation vanish, as do its angular mixed-moment terms. Its remaining geometric term is $2\nu ({\sigma }_r^2-{\sigma }_t^2)/r$. Thus the Spherical Jeans equation becomesThe no-streaming assumption excludes a steady radial flow; stationarity of the density alone would not remove its advective terms.

By the shell theorem, the gravitational potential of a spherical mass distribution satisfies $d\Phi /dr=GM(<r)/r^2$. Expanding the derivative above, dividing by $\nu$, and using logarithmic derivatives givesThe tracer density $\nu$ need not be the density of the gravitating matter. The circular speed is $V_c^2=rd\Phi /dr=GM(<r)/r$, whether or not the tracers themselves follow circular orbits.

Now let $\beta$ be constant, $\nu ={\nu }_0(r/r_0{)}^{-\gamma }$, and $V_c^2=V_0^2(r/r_0{)}^{2\alpha }$. With $y={\sigma }_r^2$, the Spherical Jeans equation readsFor a power-law ansatz $y=K{V}_c^2$, $y^{\prime }=2\alpha y/r$, so $(\gamma -2\beta -2\alpha )y=V_c^2$. Therefore the scale-free spherical Jeans solution isThe positivity condition is necessary for this particular solution to be a physical velocity dispersion.

There is a boundary condition implicit in selecting this solution. The integrating factor $r^{2\beta -\gamma }$ gives, when $D=\gamma -2\beta -2\alpha \ne 0$,The homogeneous term represents an additional radial-pressure boundary contribution. It is removed, for example, by requiring $r^{2\beta }\nu {\sigma }_r^2\to 0$ at infinity. Under that condition,and convergence requires $D>0$, giving the boxed result. If $D=0$, the solution instead has the logarithmic formso the quoted constant proportionality does not apply. A finite outer boundary can retain a homogeneous term even when $D\ne 0$.

To derive the heliocentric projection of a spherical velocity dispersion, put the Sun at vector ${\mathbf{R}}_{\odot }$ of length $R_{\odot }$, and a tracer at $\mathbf{r}=r\hat{\mathbf{r}}$. Its heliocentric line of sight isIf $\chi$ is the angle between $\hat{\mathbf{\ell }}$ and the Galactocentric radial direction, thenAfter correction for the Sun's motion, the intrinsic covariance matrix has eigenvalues ${\sigma }_r^2,{\sigma }_t^2,{\sigma }_t^2$ in the local spherical basis. Projection onto the line of sight therefore givesAt fixed $r$, ${\sigma }_r$ and $\beta$ are constant across the spherical shell. Averaging its squared line of sight velocity, with zero mean, givesFor a uniformly sampled Galactocentric shell, the solid angle measure makes $u$ uniform on $[-1,1]$, giving the geometric correction for heliocentric velocity dispersionTo evaluate it, set $a=r^2+R_{\odot }^2$ and $b=2r{R}_{\odot }$, and divide the integrand:The odd term integrates to zero, while ${\int }_{-1}^{1}du/(a-bu)=b^{-1}\ln [(a+b)/(a-b)]$. For $r>0$, $R_{\odot }>0$, and $r\ne R_{\odot }$, this givesThe removable limit at $r=R_{\odot }$ is $A=1/2$. As $r/R_{\odot }\to \infty$, $A\simeq 2R_{\odot }^2/(3r^2)$, while as $r/R_{\odot }\to 0$, $A\to 2/3$. If $R_{\odot }=0$, $A=0$ and the observed velocity dispersion is exactly radial. For isotropic velocities, $\beta =0$, the correction vanishes at every radius.

The averaging measure matters. If directions are instead uniform in heliocentric solid angle at a fixed Galactocentric radius $r>R_{\odot }$, let $\vartheta$ be the angle between the line of sight and ${\mathbf{R}}_{\odot }$. Geometry gives ${\sin }^2\chi =(R_{\odot }/r{)}^2{\sin }^2\vartheta$, so that particular weighting gives $A=2R_{\odot }^2/(3r^2)$ exactly. The logarithmic expression above uses uniform tracer weighting on a Galactocentric shell. For an observational sample, use its actual angular selection weights in $A=⟨{\sin }^2\chi ⟩$; the boxed dispersion relation remains valid.

Interstellar dust both hides and reveals galaxies. Absorption and scattering remove short-wavelength photons from a line of sight, causing interstellar extinction, reddening, anisotropic scattering, and polarization. These effects obscure embedded star formation, bias luminosities, colours, stellar masses, and star-formation rates, and can make an edge-on or dusty galaxy appear older and fainter. Dust also converts absorbed ultraviolet and optical power into far-infrared and submillimetre thermal radiation. That reradiation exposes otherwise hidden star formation, constrains dust temperature and mass, and helps map cold molecular material; polarization traces magnetic-field orientation. Grain surfaces also catalyse molecular-hydrogen formation, while photoelectric emission from grains heats interstellar gas.

One spherical grain has emitting area $4\pi a^2$. Since isotropic blackbody intensity $B_{\nu }$ gives surface flux $\pi B_{\nu }$, its spectral luminosity isFor $N=M_d/m_d$ optically thin grains at distance $D$, the inverse-square law givesDefining the grain mass absorption coefficientthe optically thin dust-mass estimator is

For a constant source function and no incident background, the radiative transfer equation integrates toIf the cloud subtends solid angle $\Omega$, uniform intensity givesThe physical projected area is $A=\Omega D^2$. Its grain column density is $N/A=M_d/(m_d\Omega D^2)$, so the dust optical depth isEliminating ${\tau }_{\nu }$ gives the finite-optical-depth resultIt requires $F_{\nu }<\Omega B_{\nu }$, as demanded by the blackbody brightness limit. When ${\tau }_{\nu }\ll 1$, $1-e^{-{\tau }_{\nu }}={\tau }_{\nu }+O({\tau }_{\nu }^2)$, so $F_{\nu }\simeq \Omega B_{\nu }{\tau }_{\nu }={\kappa }_{\nu }{M}_d{B}_{\nu }/D^2$ and the result reduces to the optically thin estimator.

The image annotation gives probe electric current $I=43\mathrm{pA}=4.3\times 10^{-11}\mathrm{A}$ and working distance $L=2.4\mathrm{mm}$. Interpret $d_o$ as the probe diameter and $\alpha$ as its convergence semiangle, in radians. To estimate that angle from the available dimensions, approximate the effective aperture plane as a distance $L$ from the specimen. With aperture radius $r_a=25\mu \mathrm{m}$,The electron beam brightness $\beta$ measures electric current per probe area per solid angle; the supplied convention corresponds to a circular probe and a paraxial cone of solid angle approximately $\pi {\alpha }^2$. The aperture diameter must be halved, and the annotated current is in picoamperes. Treating the full aperture angle as $\alpha$ would introduce a factor-of-two error in $d_o$.

This is a conditional optical estimate: the working distance is not necessarily the actual electron-optical distance from the effective aperture to the focused probe. The answer assumes that approximation, the quoted electron beam brightness at the operating voltage, and negligible optical aberrations. The 30 kV annotation gives a much shorter de Broglie wavelength than 4 nm, so a nanometre-scale probe is plausible. Actual electron microscope resolution also depends on diffraction, scattering in the specimen, signal generation and collection, signal-to-noise ratio, drift and specimen charging. A 4 nm probe does not establish that every measured edge width is 4 nm.

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]

Since $\mathbf{E}\cdot \mathbf{n}=0$, the Poynting vector isThe area on a distant sphere corresponding to solid angle $d\Omega$ is $r^{2}d\Omega$. ConsequentlyThe average is over the source motion at retarded time. At leading nonrelativistic order, differences between source-time and reception-time averaging are beyond the approximation. The units are power per solid angle, $\mathrm{W}\mathrm{s}{\mathrm{r}}^{-1}$.

The acceleration covariance in nonrelativistic radiation is another way to organize this result: with $C_{ij}=⟨a_i{a}_j⟩$, the angular factor is $\mathrm{tr}C-{\mathbf{n}}^{T}C\mathbf{n}$. Integrating over all observation directions uses $\int [|\mathbf{a}{|}^2-(\mathbf{n}\cdot \mathbf{a}{)}^2]d\Omega =(8\pi /3)|\mathbf{a}{|}^2$ and gives the averaged Larmor formula, $⟨P⟩=q^2⟨a^2⟩/(6\pi {ϵ}_0{c}^3)$.

After field electron emission, let $h$ denote the plate separation of the parallel-plate capacitor, which the question does not specify. The uniform capacitor electric field has magnitude $E_c=|V|/h$, and a nonrelativistic electron of mass $m_e$ has constant acceleration magnitude $a=e{E}_c/m_e$. The sign of its charge affects the direction of the radiation field but not its radiated power.

In the supplied nonrelativistic Liénard-Wiechert field, $\dot{\mathbf{\beta }}=\mathbf{a}/c$. At polar angle $\theta$ to the acceleration, $|\hat{\mathbf{n}}\times (\hat{\mathbf{n}}\times \mathbf{a})|=a\sin \theta$, henceThe outgoing Poynting vector has magnitude $S={ϵ}_{0}c{E}_{\mathrm{rad}}^2$. Its radiated power per solid angle is thereforeUsing $\int {\sin }^2\theta d\Omega =8\pi /3$ gives the Larmor formula and the requested constant power during the uniform-field part of the flight:The brackets in the given field require the acceleration at retarded time; the radiation reaches the far observer after the propagation delay. This derivation assumes that radiation reaction is a small correction to the acceleration and that the flight is nonrelativistic, for example $e|V|\ll m_e{c}^2$ for release from rest. Entry, emission and impact transients are not described by the uniform-field model. A result solely in terms of $V$ cannot be obtained without a plate spacing or a specified internal field.

Choose a flat wall and let $\theta$ measure the angle from the direction towards it. Particles of speed $v$ in solid angle $d\Omega$ occupy fraction $f(v)dvd\Omega /(4\pi )$ of an isotropic gas. In time $dt$, those within normal distance $v\cos \theta dt$ reach the wall. Their number per area per time is therefore $nf(v)dvv\cos \theta d\Omega /(4\pi )$.

Only the incoming hemisphere contributes. Integrating gives the one-way collision rate per unit area,Its units are ${\mathrm{m}}^{-2}{\mathrm{s}}^{-1}$. This is not $n\overline{v}/2$: selecting the incoming half of the particles must also account for their mean normal velocity. For effusion through a thin hole whose diameter is much less than the mean free path, the same one-way rate applies because collisions within the opening can be neglected.

The equilibrium photon gas is isotropic, so the energy density per unit solid angle in a narrow angular-frequency interval is $u(T,\omega )d\omega /(4\pi )$. For rays making angle $\theta$ with an outward surface normal, a surface element $dA$ is crossed in time $dt$ by rays initially in a volume $cdtdA\cos \theta$. Only the outward hemisphere contributes. ThusThe hemispherical restriction supplies a factor $1/2$ and the directional projection supplies another $1/2$. A small aperture in a cavity therefore emits this spectral power per unit area. A blackbody at the same temperature emits the same amount: otherwise radiation exchange between it and the aperture would violate thermal equilibrium.

Integrating Planck's law over all positive angular frequencies gives the total emitted heat flux densitySet $x=\hslash \omega /(k_{B}T)$. ThenThe remaining integral is independent of temperature, proving $M\propto T^4$. Its value follows by expanding $(e^x-1{)}^{-1}={\sum }_{m=1}^{\infty }{e}^{-mx}$ and integrating termwise:Consequently the Stefan–Boltzmann law isThis is emitted power alone. A body in a radiation environment also absorbs incoming radiation, which must be included when calculating net radiative cooling.

An inverse-square field spreads through area proportional to squared distance. More generally the flux contribution of a point source is proportional to the solid angle subtended by a surface element; the oriented solid angles sum to $4\pi$ for an enclosed source and zero for an exterior one. Superposition extends this to a mass distribution.

Newton's law of universal gravitation gives $\mathbf{g}=-GM\hat{\mathbf{r}}/r^2$ for a point mass. A surrounding sphere therefore has flux $-4\pi GM$, soFor the tunnel, approximate its water as an infinite uniform cylinder with $R=3.5\mathrm{m}$ and density $\rho =1000\mathrm{kg}{\mathrm{m}}^{-3}$. The bob is at radial distance $r=R+1.5\mathrm{m}=5.0\mathrm{m}$. A coaxial Gaussian surface givesFor the plumb-line deflection from cylindrical gravity, horizontal equilibrium gives $d/L\simeq |g_r|/g$. Draining removes the attraction toward the tunnel, so the bob shifts away byThe unchanged rock and ordinary gravitational field cancel in the before/after comparison; end effects and mechanical disturbances are neglected.

For the tidal water loading of a spring balance, approximate the extra water at high tide as an infinite plane slab of thickness $h=8\mathrm{m}$, with the suspended mass above it. The gravitational field of an infinite plane sheet, derived with a pillbox Gaussian surface, gives added downward field $\Delta g=2\pi G\rho h$. With initial spring extension $x_0=mg/k=1\mathrm{m}$,The extension increases at high tide. These estimates use $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ and the stated water-density approximation; salinity, finite estuary geometry and terrestrial tidal deformation alter a real measurement.

redshift. An incoming radial null ray satisfies $d\chi =-cdt/R(t)$. Consecutive wave crests follow the same comoving path, so the endpoint changes in the fixed path integral obey $d{t}_0/R_0=d{t}_e/R_e$. Comoving clocks read cosmic time; hence their measured periods and frequencies satisfyThe physical wavelength stretches with the scale factor during propagation. It is not necessary to assign a unique global inertial-frame recession velocity to a distant source.

Radial coordinate and cosmological look-back time. Differentiating $1+z=R_0/R(t)$ gives $dz/dt=-(1+z)H(z)$. Combining this with the null-ray integral yieldsThe extra factor $1+z$ in the second denominator records the conversion from scale-factor change to elapsed cosmic time.

proper-motion distance. A small transverse displacement at emission has proper length $d\ell =R_{e}S(\chi )d\theta$. Define its source-frame transverse speed as $v=d\ell /d{t}_e$; cosmological time dilation makes $d{t}_0=(1+z)d{t}_e$. ThereforeEquivalently, $d_M=(1+z)d_A$, where the angular diameter distance is $d_A=R_{e}S(\chi )$. The observer-time versus source-time distinction is essential. This calculation uses the stated transverse speed convention and the homogeneous background, without additional peculiar radial relativistic Doppler effect or gravitational lensing.

redshift shell volume at the present epoch. The induced spatial metric tensor on the present slice gives $d{V}_0=R_0^3{S}^2(\chi )d\chi d\Omega$. Since $d\chi /dz=c/(R_{0}H)$,This is the volume at the present epoch of the comoving shell intersected by the past light cone. Its proper volume at emission is smaller by $(1+z{)}^3$; a conserved population's proper number density is larger by the same factor, preserving its count.

Einstein-de Sitter cumulative number counts. In flat geometry $S(\chi )=\chi$. Inserting the stated Hubble parameter into the radial integral givesThe constant present-volume number density makes $dN=n_{0}d{V}_0$. With no members of the population before its formation epoch,As an independent check, substitution $u=(1+z{)}^{-1/2}$ evaluates this toThis is exactly the Euclidean comoving volume per unit solid angle in the flat model. It assumes a complete count with the specified fixed population; luminosity selection, mergers or subsequent number-density evolution would require additional factors.

Take $\omega >0$ and $T=2\pi /\omega$. A full spin one-half rotation gives $R(T)=-I$, an essential sign. The cyclic phase of a spin in a rotating field has factors and total quantum phasesTo separate the adiabatic dynamical phase, set $B=\sqrt{{\Omega }_0^2+{\Omega }_R^2}>0$ and $\cos \theta ={\Omega }_0/B$. Then $\Omega =B-\omega \cos \theta +O({\omega }^2/B)$. The instantaneous laboratory energies are $\pm \hslash B/2$, so the adiabatic dynamical phases are $\mp BT/2$. Subtracting them from ${\Phi }_{\pm }$ givesThe oriented solid angle $\mathcal{A}$ is swept out by the field direction. These are the Berry phases for the eigenstates parallel and antiparallel to that direction. Retaining the spinor sign makes the quantum phase vanish modulo $2\pi$ when the field does not trace a nontrivial loop. The assumption $\omega \ll B$ suffices for adiabatic following; near degeneracy it must not be silently dropped.^[ref]

Canonical probabilities. A canonical ensemble describes a system of fixed particle number and fixed external parameters such as volume, weakly coupled to a much larger thermal reservoir. They exchange energy and reach thermal equilibrium, while the reservoir's temperature stays effectively constant. The interaction energy is negligible in the state counting. The combined isolated system is described by equal probabilities for its accessible microstates.

If its total energy is $E_{\mathrm{tot}}$, the probability of a particular system microstate $i$ is proportional to the reservoir multiplicity at $E_{\mathrm{tot}}-E_i$. Writing that multiplicity as $e^{S_R/k_B}$, expand its entropy:The large-reservoir limit suppresses the higher terms. Normalization therefore givesThe sum counts individual microstates; when grouping them by energy, their state degeneracies must be included. This $Z$ is the canonical partition function.