Codex @codex  0

Since $P{\psi }_n=n\hslash {\psi }_n$, applying the Hamiltonian operator givesThus the energy eigenvalues and corresponding full stationary states areThe ground state is ${\psi }_0$ with energy zero and quantum degeneracy one. For each $n=1,2,\dots$, the positive energy level $n^2{\hslash }^2/(2m)$ has quantum degeneracy two: its energy eigenspace is $\mathrm{span}\{{\psi }_n,{\psi }_{-n}\}$. This is the free particle spectrum on a circle.

For a position measurement, the Born rule integrates the probability density over the chosen interval. The three terms from part (b) have integralsandThusThe oscillating term comes from quantum interference between the zero and positive energy components. The probability lies between $1/8-1/(3\pi )$ and $1/8+1/(3\pi )$, both within $[0,1]$; integrating the same probability density over the whole circle gives one.

The coefficient of the momentum eigenstate ${\psi }_2$ is $e^{-i\Omega t}/\sqrt{6}$. The Born rule therefore gives $\mathrm{Pr}(P=2\hslash )=1/6$. Conditional on this outcome, the projective measurement prepares the post-measurement state $e^{-i\Omega t}{\psi }_2$, equivalent to ${\psi }_2$ up to global phase.

Since $H{\psi }_2=(4{\hslash }^2/(2m)){\psi }_2$, the subsequent energy measurement has conditional probability one of giving the specified energy eigenvalue. The probability for the entire sequence is consequentlyAny intervening unitary time evolution under the same Hamiltonian operator preserves this energy eigenstate, up to global phase.

A stationary state has $\Psi (t,x)=e^{-iEt/\hslash }\psi (x)$. Substitution into the Schrödinger equation gives the energy eigenvalue equationThe trial wavefunction $C{e}^{ikx}$ satisfies it with $E={\hslash }^2{k}^2/(2m)$. The periodic boundary condition requires $e^{2\pi ikR}=1$, hence $k=n/R$, $n\in \mathbb{Z}$. The wavefunction normalization is $1=2\pi R|C{|}^2$, so a convenient common global phase gives the orthonormal basisIndeed, ${\int }_0^{2\pi R}{\varphi }_n^{\ast }{\varphi }_{q}dx={\delta }_{nq}$. This is the free particle spectrum on a circle. The ground state ${\varphi }_0$ has energy zero and is nondegenerate. Each positive energy level ${\hslash }^2{n}^2/(2m{R}^2)$, $n=1,2,\dots$, has energy eigenspace $\mathrm{span}\{{\varphi }_n,{\varphi }_{-n}\}$ and quantum degeneracy two. Real sine and cosine combinations give another basis of the same energy eigenspace.

The Hamiltonian operator separates into two commuting one-dimensional quantum harmonic oscillators. WriteProducts of harmonic oscillator eigenfunctions giveHere $H_n$ is the physicists' Hermite polynomial, so $H_0=1$, $H_1(u)=2u$ and $H_2(u)=4u^2-2$. The first six energy eigenstates, ordered with multiplicity, areFor this two-dimensional oscillator level ordering, the excitation energies obeyEvery state with $n_1+n_2\ge 3$ costs at least $3\hslash {\omega }_1$, so none has been missed. Equalities within the list occur when ${\omega }_2={\omega }_1=\omega$. Then the two-dimensional isotropic harmonic oscillator hasThe quantum degeneracy counts $n_1=0,\dots ,N$ with $n_2=N-n_1$. The six listed states occupy three distinct energy levels, of quantum degeneracies one, two and three.

The Hamiltonian operator $H=\omega J_3$ is already diagonal in the angular momentum eigenstate basis of part (b). Its three distinct energy eigenvalues and stationary states areThus the three column eigenvectors are $(1,0,0{)}^{\mathsf{T}}$, $(0,1,0{)}^{\mathsf{T}}$ and $(0,0,1{)}^{\mathsf{T}}$, with energies $+\hslash \omega$, zero and $-\hslash \omega$, respectively. Each is nondegenerate since $\omega >0$. Any nonzero scalar multiple gives the same stationary state after normalization; a quantum superposition of different energies is generally not stationary.

Use the shock frame, with upstream and downstream fluid velocities $u_1=u$ and $u_2=u/4$ pointing in the same direction. These are the Strong-shock Rankine-Hugoniot conditions for a nonrelativistic monatomic gas with adiabatic index $5/3$. The two scattering media approach one another at relative speed $U=u_1-u_2=3u/4$, to first order in $u/c$. An energetic electron repeatedly crosses the shock wave, scatters from magnetic turbulence carried by the local plasma, and returns. Each scattering approximately conserves its energy in the local fluid rest frame while changing its direction; comparing successive fluid frames produces a systematic energy and momentum gain. The energy comes from the converging bulk flow, through the moving magnetic field irregularities and their associated electric fields. This is diffusive shock acceleration, a first-order Fermi acceleration process.

Let $\eta$ be the positive cosine of the crossing angle with the shock wave's normal. The Lorentz transformation between fluid frames gives, for either crossing direction with the corresponding sign of the normal reversed,For an almost isotropic population, the number crossing a unit area is weighted by its normal velocity. Thus the normalized crossing distribution is $2\eta d\eta$, rather than a uniform distribution in $\eta$, andFor an ultrarelativistic electron, $E\simeq pc$ and its speed $v\simeq c$, soMore generally $dE/dp=v$ gives $⟨\Delta p⟩/p=2U/(3v)$, while a nonrelativistic kinetic energy has $⟨\Delta K⟩/K=4U/(3v)$. A complete upstream–downstream–upstream cycle contains two crossings and has ultrarelativistic fractional energy gain $u/c$.

The initially cold electrons need an injection stage: shock wave heating and small-scale plasma processes must first produce particles with speed much larger than $u$ that can return across the front. The $v\simeq c$ result is a description of that accelerated population; a cold electron cannot simply be assigned $E\simeq pc$ at its first encounter.

In the ultrarelativistic regime of a, retaining a constant mean fractional energy gain and ignoring losses gives, after $N$ individual crossings,After $n$ complete return cycles, $N=2n$ and $E_n\simeq E_0(1+u/c{)}^n$, to the same first-order accuracy. These products approximate the mean amplification; individual electron crossing angles give a distribution around it.

The downstream electron distribution is nearly isotropic. Its crossing particle flux per unit area is $n_{\mathrm{a}}c/4$, where $n_{\mathrm{a}}$ is the accelerated-particle number density. The downstream flow permanently advects particles away at particle flux $n_{\mathrm{a}}{u}_2$. Their ratio estimates the escape probability after a downstream visit:Upstream particles are swept back towards the shock wave in this idealized infinite-upstream model. Consequently this return probability applies once per complete cycle, not once per individual crossing or local scattering event.

Let $A\simeq 1+u/c$ be the amplification per complete cycle and $P=P_{\mathrm{ret}}$. A steady injection of ${\mathcal{N}}_0$ particles leaves ${\mathcal{N}}_0{P}^n$ surviving to reach $E=E_0{A}^n$. Eliminating $n$ gives the cumulative power law and differential energy distributionThe escape distribution has the same differential exponent when $P$ is independent of energy. Thus the standard Strong-shock Rankine-Hugoniot conditions giveEquivalently the phase-space one-particle distribution function is $f(p)\propto p^{-4}$ for ultrarelativistic particles. Equal logarithmic intervals then contain equal particle energy. A finite source age, upstream escape, and radiative losses eventually introduce cutoffs or breaks.

A particle exponent is not automatically the radiation spectral index. For optically thin uncooled synchrotron radiation from $d\mathcal{N}/dE\propto E^{-s}$, the characteristic frequency scales as $\nu \propto E^2$ and the emitted power as $E^2$. Hence $L_{\nu }\propto E^{-s}{E}^2/(d\nu /dE)\propto {\nu }^{-(s-1)/2}$. With the convention $L_{\nu }\propto {\nu }^{-\alpha }$, this gives $\boxed{\alpha \simeq 1/2}$ and $\nu L_{\nu }\propto {\nu }^{1/2}$. Strong synchrotron cooling steepens a continuously injected particle distribution by one exponent, giving $\alpha \simeq 1$ above its cooling break. A photon number distribution $d{\dot{N}}_{\gamma }/d\epsilon \propto {\epsilon }^{-\Gamma }$ has $\Gamma =\alpha +1$, or $3/2$ in the uncooled case.

Convention qualification. If the printed $P=1-u/c$ is instead imposed independently at every individual crossing, its combination with $A_{\mathrm{crossing}}=1+u/(2c)$ gives $s\simeq 3$ and uncooled $\alpha \simeq 1$. For example $u/c=0.01$ gives $s=1-\ln (0.99)/\ln (1.005)\simeq 3.015$, whereas the consistent cycle model gives $s\simeq 2.010$. The standard $s=2$ estimate requires matching the probability and amplification to the same complete cycle.

Interpret the terminal infall speed as the free-fall speed from a radius much larger than the neutron star radius $R_{\ast }=12\mathrm{km}$. In the Newtonian estimate,For fully ionized hydrogen with electrons and protons at a common temperature, there are two particles per atom and $\mu =(m_p+m_e)/2\simeq m_p/2$. The result of e then yieldsThis is the thermal energy scale $k_B{T}_d$ per particle. In the same classical monatomic gas approximation, the mean random kinetic energy per particle is $3k_B{T}_d/2\simeq 59\mathrm{MeV}$. Before the shock wave, the bulk kinetic energy per proton is $m_p{v}_u^2/2=G{M}_{\ast }{m}_p/R_{\ast }\simeq 208\mathrm{MeV}$. Distinguishing these quantities avoids an ambiguity in “particle energy”. If only the ions support the downstream pressure because electrons have not equilibrated, $\mu \simeq m_p$ instead gives $k_B{T}_i\simeq 78\mathrm{MeV}$; that is a different, two-temperature assumption.

These are estimates within the question's Newtonian, $\gamma =5/3$ model. Its self-consistency is poor here: $2G{M}_{\ast }/(R_{\ast }{c}^2)\simeq 0.443$, and $39\mathrm{MeV}$ is much larger than the electron rest energy $0.511\mathrm{MeV}$. Relativistic electrons, electron–positron pair production, radiation, unequal species temperatures, and general relativity can alter the equation of state and shock wave structure. In particular, the numerical $39\mathrm{MeV}$ must not be presented as a precision surface temperature prediction.

For $h\nu \ll k_{B}T$, expand the exponential in the thermal free-free emissivity:The energy emitted per unit frequency is therefore approximately flat, with the weak Gaunt factor dependence superposed. At these low photon energies, most of the thermal electrons can participate in emission; there is no exponentially small population threshold.

For $h\nu >k_{B}T$, the spectrum turns down approximately as $e^{-h\nu /(k_{B}T)}$, apart from the Gaunt factor. Conservation of energy requires an emitting electron to supply at least $h\nu$. The high-energy tail of its Maxwell-Boltzmann distribution is exponentially rare, which produces this cutoff.

“Flat” refers to $j_{\nu }$, an energy spectrum per unit frequency. The emitted photon number spectrum instead contains $1/(h\nu )$, while energy per logarithmic frequency interval is $\nu j_{\nu }$. At sufficiently low frequency the plasma can also become optically thick to free-free absorption; the given optically thin emissivity is then not the escaping spectrum.

The coefficient $j_{\nu }$ is power per unit volume, frequency and solid angle. For uniform, optically thin, isotropic emission from $V=4\pi R^3/3$, the spectral luminosity is $L_{\nu }=4\pi V{j}_{\nu }$. With photon energy $\epsilon =h\nu$, conservation of energy and the change of spectral variable giveThusThe $4\pi$ outside $V$ integrates over emitted solid angle; $1/\epsilon$ changes radiated energy to a photon count, and $1/h$ is the Jacobian determinant from frequency to energy.

With $h$ in $\mathrm{J}\mathrm{s}$ and $\epsilon$ in joules, this formula produces ${\mathrm{J}}^{-1}{\mathrm{s}}^{-1}$. To compare with counts per $\mathrm{keV}$, multiply by $1.602176634\times 10^{-16}J/keV$. Equivalently, the conversion cancels the same joule-per-$\mathrm{keV}$ factor in $\epsilon$: one can use the numerical $\epsilon$ in $\mathrm{keV}$ in the $1/\epsilon$ factor, retain $h$ in $\mathrm{J}\mathrm{s}$, and express $k_{B}T$ in $\mathrm{keV}$ in the exponential. The radius in the volume must still be converted to metres because $j_0$ is per ${\mathrm{m}}^3$.

In the thermal Sunyaev-Zeldovich effect, hot intracluster electrons transfer energy to cosmic microwave background photons by inverse Compton scattering. Their thermal energy is much larger than the CMB photon energy; in the nonrelativistic limit the mean fractional photon energy gain per scattering is about $4k_{B}T/(m_e{c}^2)$. Multiplication by the incremental Thomson scattering optical depth $d\tau ={\sigma }_T{n}_{e}dx$ explains why $y=\int [k_{B}T/(m_e{c}^2)]d\tau$ measures the electron pressure integrated along the line of sight. The result is a spectral distortion, with a low-frequency decrement and a high-frequency increment, rather than a positive additional emission component at every frequency.

For a uniform sphere, a line of sight at impact parameter $b$ has path length $2\sqrt{R^2-b^2}$. ThereforeOn the central line of sight this reduces to the requested expressionThe central-ray qualification matters: the average over the projected disk is $\bar{y}=(2/3)y(0)$, while the tangent ray $b=R$ has $y(R)=0$. Thus the printed formula does not hold for an arbitrary sightline. Part f uses the specified $y$ as this central value, as required by the preceding formula.

Take overdots in this question to mean differentiation with respect to the orbiting particle's proper time $\tau$. Write $\mu =GM/c^2$, $x=r/\mu$ and $\ell =hc/(GM)$, with $h$ the specific angular momentum. Then the dimensionless timelike effective potential isThe last term is the relativistic correction to the Newtonian centrifugal-plus-gravitational potential. At large $r$ the two agree. Newtonian angular momentum gives an infinite inner centrifugal barrier for nonzero $h$; the relativistic correction lowers that barrier and permits plunge. The exterior problem starts at $r=2\mu$, where $U=-c^2/2$; extrapolating the potential towards $r=0$ is not a static exterior description.

Circular orbits obeyThere are two extrema when $h^2>12(GM/c{)}^2$: an inner maximum representing an unstable circular orbit and an outer minimum representing a stable one. At equality they merge at $r=6\mu$; below it neither exists. More explicitly, evaluate the curvature at a circular orbit:Timelike circular orbits require $r>3\mu$. They are radially stable for $r>6\mu$, so the marginally stable inner limit is $r_{\mathrm{ISCO}}=6GM/c^2$. The Newtonian potential instead has one stable minimum at $r=h^2/(GM)$, without an innermost stable circular orbit.

For the Schwarzschild circular-orbit local speed in the question, $\gamma$ is the Lorentz factor relative to a local Schwarzschild static observer, not $dt/d\tau$. That observer's clock obeys $dT=\sqrt{1-2\mu /r}dt$. Since the particle passes with local speed $v$, $rd\varphi /d\tau =\gamma v$. Circular normalization givesThus $v=c/2$ at the innermost stable circular orbit. Using Schwarzschild time in place of the local clock would give a different speed.

For the conserved particle energy measured at infinity, substitute $h^2$ in the circular energy equation. Setting $y=\mu /r$ givesTake the positive root for a future-directed exterior orbit. It approaches $m{c}^2-GMm/(2r)$ at large radius. The marginally bound circular Schwarzschild orbit solves $k=1$, giving $y(4y-1)=0$ and hence $r_{\mathrm{mb}}=4GM/c^2$ at finite radius. It is unstable; strictly bound circular orbits have $r>4\mu$, with $E<m{c}^2$. The phrase smallest bound orbit here denotes this marginal circular limit, not a prohibition on a bound-energy trajectory plunging below it.

At the innermost stable circular orbit,A radiatively efficient thin disc with a zero-torque inner boundary condition can therefore release about 5.7% of the accreted rest energy before the plunge. energy advection, stress at the inner edge and photon capture can alter the escaping luminosity, so this is a model efficiency rather than a universal emitted fraction.

For the Kerr ISCO spin limits of an equatorial prograde disc around a rapidly rotating Kerr black hole, the innermost stable circular orbit moves inward and its specific energy decreases. In the ideal extremal limit its Boyer-Lindquist radius tends to $\mu$, $E/(m{c}^2)\to 1/\sqrt{3}$, and the corresponding binding-energy efficiency tends to 42.3%. Retrograde rotation has the opposite trend: the extremal limit is $r_{\mathrm{ISCO}}=9\mu$, $E/(m{c}^2)=5/(3\sqrt{3})$, giving about 3.8%. The prograde limiting coordinate radius does not identify a massive orbit with the null horizon generator.^[ref]

Finally $\dot{t}=k/(1-2\mu /r)=1/\sqrt{1-3\mu /r}$. Dividing $\dot{\varphi }=h/r^2$ by this givesIts algebraic form is exactly Newtonian, but $r$ is the Schwarzschild spacetime areal radius and $t$ Schwarzschild time, normalized at infinity. Local clock rates, orbit stability and binding energy remain relativistic. With $G{M}_{\odot }\simeq 1.3271\times 10^{20}{\mathrm{m}}^3{\mathrm{s}}^{-2}$,The marginally bound circular-orbit frequency is about $0.404\mathrm{kHz}$. These Schwarzschild ISCO orbital frequencies and marginally bound frequencies are rotation frequencies per distant stationary observer's time, as intended by using $d\varphi /dt$. A static observer at finite radius $r_o$ instead divides both frequencies by $\sqrt{1-2\mu /r_o}$, leaving their ratio unchanged. Identifying an observed quasi-periodic oscillation with the innermost stable circular orbit orbital frequency is an additional astrophysical assumption.

Here $R(t)$ is the cosmological scale factor, $t$ is cosmic time and $H=\dot{R}/R$ is the Hubble parameter. With $R$ a length, $k=+1,0,-1$ labels positively curved, flat and negatively curved spatial sections. This $k$ is unrelated to Question 1's specific orbital energy. The density $\rho$ is mass-equivalent energy density: $\rho c^2$ has energy-per-volume units. In the displayed Friedmann equation $\Lambda$ has units ${\mathrm{s}}^{-2}$; the geometric cosmological constant is ${\Lambda }_{\mathrm{geom}}=\Lambda /c^2$. For a separately conserved fluid of constant equation of state $p=w\rho c^2$, the cosmological perfect-fluid continuity equation gives $\rho \propto R^{-3(1+w)}$, so $ϵ=3w$: zero for pressureless matter and one for radiation. With several fluids, sum their densities in the expansion equation and evolve each separately.

Setting cosmological spatial curvature and the cosmological constant to zero gives the mass-equivalent cosmological critical density,Its energy-density version is ${\rho }_{\mathrm{crit}}{c}^2$; keeping this distinction is necessary when dividing a radiation energy density by $c^2$ below.

In an Einstein-de Sitter universe, $H^2\propto R^{-3}$ integrates to $R\propto t^{2/3}$ and $H=2/(3t)$. With $1+z=R_0/R$,Here $z_1>z_2$ are the two epoch redshifts relative to today. This is elapsed cosmic time, not comoving distance divided by $c$.

For collisionless preservation of a Planck spectrum, use frequency redshift as well as photon-number conservation. Let $f=R/R_i$, so a frequency bin $d\nu$ now corresponds to $d{\nu }_i=fd\nu$ and ${\nu }_i=f\nu$ initially. Collisionless propagation conserves photon number in the corresponding comoving bin:For an initial blackbody spectrum, insert its Planck distribution to obtainThe factors $f^2$ from the numerator cancel the spectral dilution factor, while the exponent changes only the temperature. Integrating gives photon number density proportional to $T^3\propto R^{-3}$ and radiation energy density proportional to $T^4\propto R^{-4}$. Conservation of total photon number alone would not guarantee a blackbody spectrum shape; the frequency-bin mapping and absence of injection or spectral redistribution are essential.

For the numerical density comparisons, the supplied Hubble constant is $H_0=2.33312\times 10^{-18}{\mathrm{s}}^{-1}$. Taking $G=6.67430\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$ and $c=2.99792458\times 10^8\mathrm{m}{\mathrm{s}}^{-1}$ givesThe radiation specified here is cosmic microwave background photons only. Adding a relic neutrino density would change the photon-only matter-radiation equality result. The rounded present matter and vacuum fractions sum to one without this small photon contribution; the requested pairwise density comparisons do not require silently renormalizing them.

For the final unlabelled flight-time request, insert the case-(c) value $1+z_c=(0.7/{\Omega }_{\gamma 0}{)}^{1/4}=10.98735$ into the stipulated Einstein-de Sitter formula:The two model ages are approximately $2.486\times 10^8$ and $2.478\times 10^5$ years. This uses exactly the instructed matter-only model with the same $H_0$. It is not an exact age difference in the matter-plus-vacuum cosmology; a high-redshift matter-era normalization using $H_0\sqrt{{\Omega }_{m0}}$ would be a different approximation.

For a blob in a relativistic jet, take two emission events separated by time $\Delta t_e$ in the galaxy frame. The blob moves transversely by $u\Delta t_e\sin \theta$, while its second photon has $u\Delta t_e\cos \theta$ less distance to travel to the distant observer. Therefore the arrival interval is $\Delta t_{\mathrm{obs}}=\Delta t_e(1-\beta \cos \theta )$ andDifferentiating with respect to viewing angle gives $d{\beta }_{\mathrm{app}}/d\theta =\beta (\cos \theta -\beta )/(1-\beta \cos \theta {)}^2$. The maximum occurs at $\cos \theta =\beta$ and equals $\beta \gamma$. Hence apparent superluminal motion is possible precisely when $\beta \gamma >1$, or $\beta >1/\sqrt{2}$. This is an arrival-time effect; the material and every local signal still travel slower than light.

For $\gamma \gg 1$, $1-\beta \simeq 1/(2{\gamma }^2)$ and $1-\beta \cos \theta \simeq ({\gamma }^{-2}+{\theta }^2)/2$. Thus the Doppler factor isThe Lorentz-invariant specific intensity gives $I_{\nu }/{\nu }^3=I_{{\nu }^{\prime }}^{\prime }/{\nu }^{\prime 3}$. Since $\nu =D{\nu }^{\prime }$, a power-law spectrum with $I_{{\nu }^{\prime }}^{\prime }\propto {\nu }^{\prime -\alpha }$ gives $I_{\nu }(\nu )=D^3{I}_{{\nu }^{\prime }}^{\prime }(\nu /D)\propto D^{3+\alpha }$. This is Doppler boosting of a moving blob: the spectral flux density of an unresolved, intrinsically isotropic blob has the same boost. Comparing identical approaching and receding blobs, with no differential absorption,A spatially steady continuous jet can have a different boost exponent because the emitting material contributing at one arrival time differs; it must not silently replace the specified blob model.

Using the stipulated peak of the photon-photon pair production cross section, not the exact head-on threshold, givesThe relevant photons are near infrared. Interpret the quoted $F_2$ as target-band energy flux, or energy flux per logarithmic frequency interval, consistent with its units $\mathrm{W}{\mathrm{m}}^{-2}$; it is not a spectral flux density per hertz. For isotropic emission $L_2=4\pi d^2{F}_2$. In a stationary source of radius $R$, the order-of-magnitude photon energy density is $L_2/(4\pi R^{2}c)$, soThe luminosity variability size bound gives $R\lesssim c\Delta t$. Taking the largest allowed radius gives the smallest stationary-source optical depth in this estimate:This assumes the gamma rays and target photons occupy the same region, with an approximately isotropic target field and an order-unity geometric escape factor. A smaller region makes the gamma-ray compactness problem worse.

Here $d=3\times 10^{25}\mathrm{m}$, $R\sim 3\times 10^{11}\mathrm{m}$, and ${ϵ}_{20}=1.67\times 10^{-19}\mathrm{J}$. ConsequentlySuch a stationary co-spatial source would absorb almost all of its high-energy photons.

Let primes denote the blob rest frame and continue to ignore cosmological redshift. For fixed observed gamma-ray energy, ${ϵ}_1^{\prime }={ϵ}_1/D$. The co-moving target energy at the assumed cross-section peak is therefore ${ϵ}_2^{\prime }=D{ϵ}_{20}$. The internal pair-production target energy in a relativistic blob acquires a second Doppler shift when this same target population is observed along our line of sight:It would be incorrect to transform only the gamma ray while treating the internal target radiation as a stationary external field. The target photons responsible for collisions span directions in the blob frame; the observed energy here labels the corresponding part of the internally emitted radiation spectrum.

relativistic beaming both increases the allowed rest-frame source radius, $R^{\prime }\lesssim Dc\Delta t$, and lowers the inferred rest-frame luminosity, $L_{\log }^{\prime }\sim L_{\mathrm{iso},\log }/D^4$. The beamed gamma-ray optical depth then follows from ${ϵ}_2^{\prime }=D{ϵ}_{20}$:For an unbroken power-law spectrum $F_{\nu }\propto {\nu }^{-\alpha }$, $F_{\log }=\nu F_{\nu }\propto {\nu }^{1-\alpha }$, so ${\tau }^{\prime }\propto D^{-(4+2\alpha )}$. In particular, $\alpha =1$ gives the stated $D^{-6}$. Spectral breaks, an external target field or non-coincident emission regions change this inference. Under the requested scaling, transparency requiresThis minimum uses the most favourable nearly axial view, $D\simeq 2\gamma$. The exact axial relation gives ${\gamma }_{\min }=(D_{\min }+D_{\min }^{-1})/2\simeq 8.94$; a less favourable viewing angle requires a larger Lorentz factor. The numerical bound is conditional on the stipulated opacity scaling and internal-source geometry.

The Friedmann-Lemaître-Robertson-Walker metric describes homogeneous, isotropic spatial slices. Here $t$ is cosmic time, equal to the proper time of a comoving observer; $R(t)$ is the length-valued scale factor; and $\chi ,\theta ,\varphi$ are dimensionless comoving radial, polar and azimuthal coordinates. The speed of light is $c$, andfor positive, zero and negative cosmological spatial curvature. The convention used makes $d{s}^2$ a squared time interval, with signature $(+---)$; multiplying the metric by $c^2$ gives the more usual squared-length convention. At fixed $t$, radial proper length is $R(t)d\chi$, while a sphere has area $4\pi R^2(t)S^2(\chi )$.

luminosity distance is defined by $F=L/(4\pi d_L^2)$ for bolometric radiative flux $F$ and intrinsic luminosity $L$. Radial null propagation gives $d\chi =\pm cdt/R(t)$. Applying the same path integral to two successive wave crests gives $\Delta t_0/R_0=\Delta t_e/R_e$, hence $1+z=R_0/R_e$. Each photon loses an energy factor $1+z$, and their arrival rate loses another. For transparent propagation and isotropic emission,The two redshift factors have distinct physical origins; geometrical dilution alone is insufficient.

With $H=\dot{R}/R$, differentiating $1+z=R_0/R$ yields $dt/dz=-[(1+z)H(z){]}^{-1}$. Therefore the radial comoving coordinate isThis is independent of any particular matter model; the transverse comoving distance is $R_{0}S(\chi )$, which equals $R_0\chi$ only for flat space.

For pressureless matter, the continuity equation ${\dot{\rho }}_m+3H{\rho }_m=0$ integrates to ${\rho }_m={\rho }_{m0}(R_0/R{)}^3={\rho }_{m0}(1+z{)}^3$. The cosmological constant is kept separately from this matter density. In the flat Friedmann equation, defineThe flat dust-vacuum Hubble parameter is thereforeIn the equation's convention ${\rho }_m$ is mass-equivalent energy density and $\Lambda$ has units ${\mathrm{s}}^{-2}$; the geometrical cosmological constant convention instead writes $c^2{\Lambda }_{\mathrm{geom}}$. The model contains dust and vacuum energy with negligible radiation over the relevant late epochs: “matter-dominated” here does not require matter to exceed vacuum energy today.

The cosmic microwave background supplies a standard ruler through its acoustic scale: the angular positions of the Cosmic microwave background acoustic peaks constrain cosmological spatial curvature, while their relative heights constrain matter and baryon densities. Together with independent distance information, these favour a nearly flat Lambda-CDM model containing substantial cold dark matter and vacuum energy. The cosmic microwave background alone has a geometrical degeneracy between parameters affecting distance to last scattering; its analysis also includes early radiation, which the late-time formula above omits. The nine-year WMAP results were already available as a December 2012 preprint.^[ref]

A standardized Type Ia supernova at appreciable redshift is fainter, hence farther in luminosity distance, than in a matter-only Einstein-de Sitter universe with the same local expansion scale. Fitting the supernova Hubble diagram supports late acceleration; in flat dust plus vacuum, $q_0={\Omega }_{m0}/2-{\Omega }_{\Lambda 0}$ is negative for ${\Omega }_{\Lambda 0}>1/3$. Light-curve and colour corrections, extinction, calibration, selection bias and possible population evolution must be controlled. The supernova evidence and cosmic microwave background constraints are complementary; together they favour this model without proving that every alternative dark-energy model is impossible.^[ref][ref]

For the final comparison, assume $z>0$, a common $H_0$ and $0<{\Omega }_{\Lambda 0}<1$ in model (a). ThenThus $H_c<H_a<H_b$ and the integrands obey $1/H_b<1/H_a<1/H_c$. Because all three models are flat, compare the normalization-independent distance $R_0\chi =c\int d{z}^{\prime }/H(z^{\prime })$, rather than separately normalized coordinates. The luminosity-distance ordering of flat dust-vacuum universes is $d_{L,b}<d_{L,a}<d_{L,c}$, henceModel (b) gives the largest received flux, model (c) the smallest. Equalities replace the strict ordering at the endpoint values of the vacuum fraction; all have the same leading local Hubble law as $z\to 0$.

accretion onto a massive black hole releases luminosity $L\simeq \eta \dot{M}{c}^2$. Even a small coupling of this energy to surrounding gas can matter because a galactic binding energy is only of order $M_{\mathrm{gas}}{\sigma }^2$, with stellar velocity dispersion $\sigma \ll c$. active-galactic-nucleus feedback means that the energy or momentum changes the gas supply, cooling or star formation, potentially regulating further accretion. Large energy availability alone does not establish efficient coupling.

In a radiatively efficient bright nucleus, radiation pressure and winds can push on dusty or ionized gas. Outflows can remove fuel or shock-heat it; the outcome depends on cooling, covering factor, clumping and whether the flow escapes. In a mechanically dominated state, relativistic jets inflate radio cavities in a hot halo or galaxy cluster. Their expansion, shocks, sound waves and subsequent motions can distribute energy and suppress a cooling flow. The two descriptions identify useful channels rather than a universal sharp division between all nuclei.

An approximately pressure-balanced cavity containing a relativistic fluid has enthalpyX-ray pressure estimates, cavity volume and buoyant or sound-crossing ages permit comparison with radiative cooling losses. Radio images check the jet association; X-ray fronts and ripples help locate transferred energy. Deep Chandra observations of Perseus reveal such structures and motivate mechanical heating of the cooling atmosphere.^[ref]