Past exam of the astrophysics course of the University of Oxford / 2022 (A11044W1)

Source: www.whatdotheyknow.com/request/oxford_physics_past_papers_2/response/2340696/attach/10/Physics%20Papers%202021.pdf?cookie_passthrough=1

Source bundle pages 18–24. The cover establishes 2022 despite the attachment's 2021 filename. Solutions use SI units unless stated otherwise.

The specific intensity $I_{\nu }$ is energy transported per unit time, projected area, solid angle and frequency. The path coordinate $s$ increases along the ray. The emission coefficient $j_{\nu }$ is emitted energy per unit time, volume, solid angle and frequency, whereas the absorption coefficient ${\alpha }_{\nu }$ has units of inverse length. Thus $j_{\nu }ds$ adds specific intensity and ${\alpha }_{\nu }{I}_{\nu }ds$ removes it. Stimulated emission can be included as negative absorption.

Define the optical depth between two points by ${\tau }_{\nu }(s_1,s_2)={\int }_{s_1}^{s_2}{\alpha }_{\nu }(u)du$. Multiplying the radiative transfer equation by the integrating factor $\exp [{\tau }_{\nu }(s_0,s)]$ and integrating givesEach local contribution is attenuated only by the material between its emission point and the observer. Where ${\alpha }_{\nu }\ne 0$, the source function is $S_{\nu }=j_{\nu }/{\alpha }_{\nu }$ and the same formal solution of the radiative transfer equation is $I_{\nu }(\tau )=I_{\nu }(0)e^{-\tau }+{\int }_0^{\tau }{S}_{\nu }(t)e^{-(\tau -t)}dt$. A constant source function gives $I_{\nu }=I_{\nu }(0)e^{-\tau }+S_{\nu }(1-e^{-\tau })$.

Use the Einstein coefficients convention in which $B_{ij}$ multiplies spectral radiation energy density $u_{\nu }$, with $u_{\nu }=c^{-1}\int I_{\nu }d\Omega$. A normalized spectral line profile obeys $\int \phi (\nu )d\nu =1$. The spontaneous emission probability per upper-state atom per unit time is $A_{21}$; the stimulated emission and upward absorption rates per atom are respectively $B_{21}\int u_{\nu }\phi (\nu )d\nu$ and $B_{12}\int u_{\nu }\phi (\nu )d\nu$.

For isotropic spontaneous emission, $n_2{A}_{21}$ transitions occur per unit volume and time, each supplying photon energy $h{\nu }_{12}$. Dividing among $4\pi$ steradians givesFor a small incident beam, its contribution to $u_{\nu }$ is $I_{\nu }d\Omega /c$. Absorption removes and stimulated emission adds photons in the incident direction. Matching the net loss to ${\alpha }_{\nu }{I}_{\nu }d\Omega$ therefore givesReplacing ${\nu }_{12}$ by $\nu$ across the narrow spectral line profile gives the printed expressions. The negative sign is essential: an inverted population can amplify the incident specific intensity. Thermal equilibrium and the Planck function imply $g_1{B}_{12}=g_2{B}_{21}$ and $A_{21}/B_{21}=8\pi h{\nu }_{12}^3/c^3$, where $g_i$ is the state degeneracy. A convention using mean intensity instead changes both $B$ coefficients by $4\pi /c$.

With the spontaneous emission term suppressed, $j_{\nu }=0$. A uniform absorption coefficient over length $\ell$ gives directlyThe normalized specific intensity deficit is unchanged by conversion between wavelength and frequency, since both numerator and denominator acquire the same Jacobian determinant. The Boltzmann distribution and the Einstein coefficients give $n_2{B}_{21}/(n_1{B}_{12})=e^{-h{\nu }_{12}/(k_B{T}_e)}\ll 1$. Consequently, in the optically thin limit,Here $N_1=n_1\ell$ is the column density, and $\nu \simeq {\nu }_{12}$ has been used in slowly varying factors. This linear relation between equivalent width and column density fails once the central optical depth becomes appreciable.

For the unresolved Lyman-alpha line, the lower $1s$ and combined upper $2p$ state degeneracies are $g_1=2$ and $g_2=6$, so the Einstein coefficients giveThe measured equivalent width $W=7.0\times 10^{-12}\mathrm{m}$ then impliesAssume a central transverse line of sight through a uniform filament, so $\ell =d=0.5\mathrm{Mpc}=1.54\times 10^{22}\mathrm{m}$. Nearly all neutral hydrogen is in its ground state at this temperature, henceThe cylinder's length $L$ determines its total volume but does not enter a transverse column density. Geometry must be specified: an axial line of sight has $\ell =L$ and gives $n_{\mathrm{HI}}=4.2\times 10^{-7}{\mathrm{m}}^{-3}$ for the same measured equivalent width. Intermediate chords give intermediate path-dependent estimates. The numbers above treat $W$ as a rest-frame equivalent width; an observed value at redshift $z$ must first be divided by $1+z$.

True thermal equilibrium requires a thermal ionizing radiation field as well as chemical balance; equality of photoionization and recombination rates alone would require a specified external radiation field. Under the stronger equilibrium assumption, apply the Saha equation to pure hydrogen. The electron spin factor cancels the ground-state electronic degeneracy, givingAt $T=10^4\mathrm{K}$, $k_{B}T=0.862\mathrm{eV}$ and $S=3.38\times 10^{20}{\mathrm{m}}^{-3}$. Charge neutrality gives $n_e=n_p$. Importantly, the known quantity from d is the neutral number density, not the total number density; use Saha neutral fraction from a measured neutral density to obtain $n_p=\sqrt{S{n}_{\mathrm{HI}}}=5.31\times 10^7{\mathrm{m}}^{-3}$. ThereforeThis extraordinarily small neutral fraction is the result of applying thermal equilibrium to dilute intergalactic gas; it is not a realistic inference from photoionization equilibrium with a nonthermal cosmic background. For the axial geometry of d, the equilibrium neutral fraction is smaller by $\sqrt{20}$.

Adopt the stipulated neutral fraction $f_{\mathrm{HI}}=10^{-6}$, independently of the equilibrium estimate in e. For pure hydrogen, $n_e=(1-f_{\mathrm{HI}})n_{\mathrm{HI}}/f_{\mathrm{HI}}\simeq 8.3{\mathrm{m}}^{-3}$. The thermal Sunyaev-Zeldovich effect measures integrated electron pressure:Using the same absorbing line of sight as in d,Thus one such filament is below the stated measurement sensitivity. If the number density inferred transversely is instead used to view the same cylinder end-on, its path is twenty times longer and $y=2.9\times 10^{-10}$, still below the stated sensitivity. Changing the assumed path of the original absorption measurement while keeping its column density fixed does not change its predicted $y$. The quoted neutral-fraction floor is a premise of this estimate, not a universal bound derivable from thermal equilibrium.

There are two different physical interpretations. The supplied logarithmic integral signals free fall in a fixed singular isothermal sphere, with a test particle moving through a frozen background. Write $C={\rho }_0{r}_0^2$. By the shell theorem, the enclosed mass is $M(r)=4\pi Cr$ and the radial equation of motion isThe second equality follows by multiplying by $\dot{r}$ and using the initial rest condition. The free fall trajectory is given implicitly byThus the intended fixed-profile answer isA collapsing material shell gives a different answer. If the entire cloud undergoes pressureless gravitational collapse from rest, the edge follows the fixed enclosed mass $M_0=4\pi {\rho }_0{r}_0^3$, not the instantaneous fixed-profile value $4\pi Cr$. With no shell crossing, $\ddot{r}=-G{M}_0/r^2$, soThis is a concrete counterexample to treating the fixed-profile result as the exact collapse time of the whole nebula. The initial mean density inside $r_0$ is $3{\rho }_0$, consistent with the usual free-fall time of a uniform sphere expressed using that mean density. The logarithmic integral is valid for the explicitly frozen background approximation.

Assume an isolated, bound, nonrotating Newtonian gas cloud, negligible surface pressure, negligible magnetic and radiation stresses, no persistent bulk flow, and a stationary or time-averaged virialized state. The stellar virial theorem then reduces towhere $\Omega <0$ is the total gravitational potential energy. Define the volume-averaged pressure $\overline{P}=V^{-1}\int PdV$ and average gravitational energy density $u_g=\Omega /V$. HenceEquivalently, for a nonrelativistic monatomic gas, the thermal internal energy $U=(3/2)\int PdV$ obeys $2U+\Omega =0$. This second derivation must not count the thermal motions again as a separate bulk kinetic energy. If the external pressure is $P_s$, the correct result is $\overline{P}-P_s=-u_g/3$; non-negligible bulk motion adds $2T_{\mathrm{bulk}}$ to the virial theorem. The initial sharp-edged $r^{-2}$ profile need not survive virialization; maintaining that profile hydrostatically to its edge would generally require a surface term.

For a family of spherical profiles with a fixed dimensionless shape, write the gravitational potential energy as $\Omega =-\alpha G{M}^2/R$, where $\alpha >0$ depends on that shape. The virial theorem in b and the sphere volume $V=4\pi R^3/3$ giveFor a uniform density, $\alpha =3/5$; formally integrating $-{\int }_0^{R}GM(r)dM(r)/r$ for $M(r)=Mr/R$ gives $\alpha =1$. Neither the virial theorem nor the initial profile guarantees a specific final shape, so the proportionality requires the stated homologous-profile assumption. Surface pressure must be handled consistently with b before using either coefficient as an equilibrium model.

For a regular spherical star, let ${\rho }_c=\rho (0)$ and $P_c=P(0)$. The stellar mass conservation equation and stellar hydrostatic equation areAssuming the smoothness needed for the requested Taylor series, write $M=a_0+a_{1}r+a_2{r}^2+a_3{r}^3+\cdots$. There is no point mass at the centre, so $a_0=0$, and matching the stellar mass conservation equation gives $a_1=a_2=0$, $a_3=4\pi {\rho }_c/3$. Substitution in the stellar hydrostatic equation gives $P^{\prime }(r)=-(4\pi G/3){\rho }_c^{2}r+O(r^3)$, henceThe remainder orders use a smooth spherically symmetric scalar density, for which $\rho (r)={\rho }_c+O(r^2)$. Merely finite continuous density gives the same leading coefficients with $o(r^3)$ and $o(r^2)$ remainders. In particular $P^{\prime }(0)=0$, despite the apparent $r^{-2}$ in the stellar hydrostatic equation.

For a gas-pressure-supported stellar core of fixed mass and composition, neglect boundary terms for the illustrative whole-core estimate. The virial theorem gives $E=U+\Omega =-U$, and the monatomic gas gives $U=3M_c{k}_B\overline{T}/(2\mu m_u)$. Thus the effective heat capacity along the equilibrium sequence isThis negative heat capacity of a virialized gravitational system means that net energy loss makes a self-gravitating core contract and heat. It is different from the positive local material specific heat capacity. On the main sequence, hydrogen burning supplies most of the radiated energy, so rapid contraction is arrested. Hydrogen depletion increases the mean molecular weight; slow structural readjustment usually raises the central temperature rather than making it fall.

After core hydrogen exhaustion, a massive stellar core contracts and heats more rapidly until a later nuclear fuel ignites. There is no general reversal to cooling simply upon leaving the main sequence. If radiation pressure becomes dominant, the internal energy becomes $U=U_{\mathrm{gas}}+U_{\mathrm{rad}}$ with $3\int PdV=2U_{\mathrm{gas}}+U_{\mathrm{rad}}$. The virial theorem therefore gives $E=-U_{\mathrm{gas}}$; the radiation-dominated limit is marginally bound and its effective adiabatic index approaches $4/3$.

Quasistatic contraction can still maintain virial equilibrium. It fails dynamically only when sufficient support is lost, for example if pair creation reduces the effective adiabatic index below $4/3$, or when an iron core undergoes electron capture and photodisintegration. A fixed-composition gas-dominated contracting core with $T\propto 1/R$ is already a counterexample to any assertion that leaving the main sequence alone reverses the heating or destroys virial equilibrium.

Primordial density contrasts perturb the intrinsic photon temperature, the gravitational potential, and the velocity of the last-scattering material. Intrinsic photon-density fluctuations obey $\delta T/T={\delta }_{\gamma }/4$ because ${\rho }_{\gamma }\propto T^4$; adiabatic initial conditions relate these to matter density contrasts. Subsequent acoustic evolution, Doppler effect and gravitational frequency shifts determine the observed Cosmic microwave background anisotropy. There is no scale-independent identification of the final sky map with matter density contrast alone.

The spherical harmonic coefficients are random variables inherited from these perturbations. After subtracting the mean sky temperature, $⟨a_{lm}⟩=0$ for the anisotropy modes. Gaussian primordial perturbations give approximately Gaussian random variables; this extra Gaussian assumption is not needed for diagonal covariance. Reality of the sky map implies $a_{l,-m}=(-1{)}^m{a}_{lm}^{\ast }$.

For a statistically isotropic sky, its two-point function depends only on $\hat{n}\cdot \hat{m}$. Rotations mix the $m$ components at fixed $l$ and do not mix different irreducible $l$ spaces. A rotation-invariant covariance matrix is therefore a multiple of the identity in each $l$ block:On our one sky, average the $2l+1$ angular components at fixed $l$, or equivalently suitable rotated sky pairs, rather than treating different universes as observed samples. The CMB angular power spectrum is estimated byUnder the Gaussian, full-sky, noiseless assumptions, there are $2l+1$ independent real degrees of freedom and $\mathrm{Var}({\hat{C}}_l)=2C_l^2/(2l+1)$. This cosmic variance remains even with perfect detectors; a finite sky mask and detector noise add further uncertainty. The estimator is unbiased, $⟨{\hat{C}}_l⟩=C_l$, but does not turn one realization into the exact ensemble mean.

Insert the diagonal covariance from a into the two-point function and apply the spherical harmonic addition theorem:The Legendre polynomials encode the angular separation. Since $P_l(1)=1$, the zero-separation value $C(0)=⟨{\Theta }^2⟩$ is the total variance. For measured Cosmic microwave background anisotropy, remove the monopole and usually the kinematic dipole, so the corresponding sum starts at $l=2$; an instrument's angular response supplies an additional window function.

The ordinary Sachs-Wolfe effect combines intrinsic last-scattering temperature with the gravitational redshift of photons escaping the last-scattering gravitational potential. For adiabatic large-scale perturbations in matter domination, with the potential convention in the question, the combination is $\Theta =\varphi /(3c^2)$, up to the observer monopole. It is distinct from the Integrated Sachs-Wolfe effect caused by a time-varying potential along the later line of sight.

Use the Fourier transform convention in which $P_{\delta }(k)=Ak$ is the matter power spectrum and $⟨{\delta }_{\mathbf{k}}{\delta }_{{\mathbf{k}}^{\prime }}^{\ast }⟩=(2\pi {)}^3{\delta }_D(\mathbf{k}-{\mathbf{k}}^{\prime })P_{\delta }(k)$. The Poisson equation becomes $-k^2{\varphi }_{\mathbf{k}}=4\pi G{\bar{\rho }}_0{\delta }_{\mathbf{k}}$, givingWith the matter-dominated critical-density approximation ${\bar{\rho }}_0=3H_0^2/(8\pi G)$,independent of $k$. Thus the Harrison-Zeldovich spectrum gives scale-invariant potential fluctuations and the requested $H_0^4$ dependence. Keeping a general present matter fraction would insert ${\Omega }_m^2$; the answer here follows the stipulated critical matter-dominated normalization. In this Fourier transform convention, $A$ has dimensions of length to the fourth power.

The finite angular resolution of COBE DMR suppresses small angular scales, so only low spherical harmonic multipoles can be recovered usefully. The approximate relation $\theta \sim \pi /l$ associates $l\simeq 17$ with scales of order ten degrees; this is an effective sensitivity and beam-window limit, not a mathematical hard truncation.

Interpret the printed superscript uncertainty as $(1.1\pm 0.1)\times 10^{-5}$. Its square estimates the band-limited zero-separation variance, rather than the full $C(0)$:Use the coefficient supplied in the paper for this numerical inference. With ${\sigma }_{2:17}=1.1\times 10^{-5}$,so c givesThe stated amplitude uncertainty gives approximately $18\%$ relative uncertainty in $A$, or $A\simeq (1.38\pm 0.25)\times 10^6\mathrm{Mp}{\mathrm{c}}^4$, keeping $H_0$ fixed. The ideal matter-era Sachs-Wolfe plateau with precisely the c spectrum convention has coefficient $2\pi /9=0.698$, rather than $0.732$; the supplied numerical coefficient is about $5\%$ larger and has been treated as the paper's prescribed approximation. A real beam-convolved analysis retains its window function and additional cosmological transfer effects.

An ultrarelativistic electron has kinetic energy much larger than its rest energy: $(\gamma -1)m_e{c}^2\gg m_e{c}^2$, equivalently $\gamma \gg 1$. Its speed obeys $v/c=\sqrt{1-{\gamma }^{-2}}\simeq 1-1/(2{\gamma }^2)$ and its total energy is approximately $pc$. At $\gamma =10^6$, $E=\gamma m_e{c}^2\simeq 511\mathrm{GeV}$.

Diffusive shock acceleration is viable in principle at $511\mathrm{GeV}$: repeated scattering off magnetic irregularities on opposite sides of a strong nonrelativistic shock wave can reach such energies. Actual viability requires a sufficiently large and long-lived accelerator with an acceleration time shorter than its escape and loss times; $\gamma$ alone does not establish those conditions.

Take positive velocity toward the upstream material, and let the front move at $+v_s$ in the upstream rest frame. For a strong adiabatic shock wave in a gas with adiabatic index $5/3$, the Rankine-Hugoniot conditions give ${\rho }_d/{\rho }_u=4$ and the shock-frame inflow speeds satisfy $u_1=v_s$, $u_2=v_s/4$. The signed velocities in the three frames areThus the two scattering media converge at $V=u_1-u_2=3v_s/4$. For a crossing particle with incident energy $E$ in its local plasma rest frame, a Lorentz transformation to the approaching other plasma giveswhere $\mu >0$ is the incident direction cosine measured toward the approaching medium and $pc\simeq E$. For an almost isotropic crossing population, flux weighting gives $⟨\mu ⟩={\int }_0^1{\mu }^{2}d\mu /{\int }_0^1\mu d\mu =2/3$. Scattering randomizes directions without changing energy in that plasma frame, and the reverse crossing has the same mean gain. ThereforeThese are first-order mean gains. They are not an increase of energy in one fixed frame caused by an instantaneous boundary crossing. Literally choosing the electron's instantaneous rest frame gives incident $E=m_e{c}^2$, and the other gas approaches at nearly $c$, so the expansion in small $V/c$ is invalid. This concrete frame check explains why the printed description of $V$ must instead refer to the upstream/downstream scattering frames.

The magnetic Lorentz force is perpendicular to velocity, so it does no work and $\gamma$ remains constant over an orbit when radiative losses are neglected. For the transverse momentum, $p_{\perp }=\gamma m_e{v}_{\perp }$ and $|{\dot{\mathbf{p}}}_{\perp }|={\omega }_g{p}_{\perp }=eB{v}_{\perp }$. HenceSince $10\mu \mathrm{G}=10^{-9}\mathrm{T}$,This is the gyromotion frequency, not the much higher characteristic synchrotron radiation frequency.

Relativistic beaming concentrates the emission into a forward cone about the instantaneous velocity. The aberration of light maps a rest-frame ray at ${\theta }^{\prime }=\pi /2$ to $\cos \theta =\beta$, so the characteristic cone half-angle is $\theta \simeq {\gamma }^{-1}$. Thus its conventional full width is $2/\gamma =2\times 10^{-6}\mathrm{rad}$.

The isotropic-emission assumption has no beam nulls. For example, its received bolometric angular power is proportional to $D^4$, where the Doppler factor is $D=[\gamma (1-\beta \cos \theta ){]}^{-1}>0$. In particular it is nonzero at $\theta =1/\gamma$, disproving a literal null there. A true dipole pattern for an accelerated electron is not isotropic in its rest frame and needs a specified polarization and viewing plane. The width just quoted is a characteristic beaming angle, not a null-to-null width under the stated premise.

For a circular orbit seen in its orbital plane, let the velocity direction sweep past the observer through $-1/\gamma <\psi <1/\gamma$. The emission interval is $\Delta t_{\mathrm{em}}\simeq 2/(\gamma {\omega }_g)=1.14\times 10^{-2}\mathrm{s}$. The observed interval must also include light-travel compression:Its ratio to the gyromotion period is $\Delta t_{\mathrm{obs}}/T_g\simeq 2/(3\pi {\gamma }^3)=2.1\times 10^{-19}$. Since the boundary of a smooth beam is conventional, only the scale ${\omega }_g^{-1}{\gamma }^{-3}$ is invariant under reasonable pulse-width definitions. For helical motion the viewing geometry and pitch angle are additional inputs; the numerical estimate assumes transverse motion and a sightline intercepted by the cone.^[ref]

One explanation is an upper electron energy imposed by the accelerator: finite age, finite confinement size, escape, or a balance of acceleration and losses limits $\gamma$ to ${\gamma }_{\max }$. Since the synchrotron critical frequency scales as ${\nu }_c\propto B{\gamma }^2\sin \alpha$, the absence of higher-energy electrons produces a corresponding high-frequency cutoff, broadened by the single-particle spectrum.

The other is synchrotron cooling, with inverse Compton scattering often contributing: $\dot{\gamma }=-b{\gamma }^2$ gives $t_{\mathrm{cool}}=1/(b\gamma )$. High-energy electrons disappear fastest, so a population of age $t$ develops a cooling scale ${\gamma }_b\sim 1/(bt)$ and ${\nu }_b\propto B/(b^2{t}^2)$. A burst population can develop a strong cutoff. With continuous power-law injection, the simplest steady cooling model instead steepens the electron energy index by one and the synchrotron radiation spectral index by $1/2$, giving a break rather than necessarily an exponential cutoff. The detailed spectral shape therefore distinguishes finite acceleration from aging and injection history.

Let $n(\mathbf{x})=\int f{d}^{3}v$ and $\overline{A}=n^{-1}\int Af{d}^{3}v$. Assume the one-particle distribution function and its weighted tails decay enough that all required velocity integrals converge and boundary terms at infinite velocity vanish. Multiply the steady Collisionless Boltzmann equation by $v_j$ and integrate over velocity. In Cartesian coordinates, spatial differentiation commutes with this integration, while integration by parts givesConsequently the three force-balance equations areExplicitly the $x$ component is ${\partial }_x(n\overline{v_x^2})+{\partial }_y(n\overline{v_x{v}_y})+{\partial }_z(n\overline{v_x{v}_z})=-n{\Phi }_x$, with cyclic counterparts for $y,z$. For equal stellar masses, multiplying by the common mass replaces $n$ by the mass density $\rho$; for multiple stellar masses, use a mass-weighted one-particle distribution function or derive a separate equation for each tracer population.

Define mean streaming velocity $u_i=\overline{v_i}$ and the velocity dispersion tensor ${\sigma }_{ij}^2=\overline{(v_i-u_i)(v_j-u_j)}=\overline{v_i{v}_j}-u_i{u}_j$. The zeroth velocity moment of the steady Collisionless Boltzmann equation gives the continuity equation ${\sum }_i{\partial }_i(n{u}_i)=0$. ThereforeSubstituting in a yields the standard steady Jeans equations:The unsteady version adds ${\partial }_t{u}_j$ on the left. Defining the stellar pressure tensor $P_{ij}=\rho {\sigma }_{ij}^2$ makes the analogy with fluid momentum balance explicit. The off-diagonal terms are correlations between random velocity components; they need not vanish or equal the diagonal terms.

The Jeans equations express local momentum conservation for a collisionless stellar system, just as the Euler equations do for a gas. The mean streaming velocity plays the role of fluid flow, the gravitational force is a body force, and random stellar motions supply a pressure tensor through $\rho {\sigma }_{ij}^2$. Taking moments smooths over individual stellar orbits and produces fields of density, mean velocity and random-motion stress; this is the sense of “stellar hydrodynamics”.

Ordinary gas pressure is usually nearly isotropic because collisions relax its local velocity distribution. A collisionless stellar system can retain an anisotropic velocity ellipsoid, so its pressure tensor is not generally a scalar times the identity. There are more unknown second moments than momentum equations, and their evolution introduces higher moments. The Jeans equations therefore require further dynamical assumptions or a specified one-particle distribution function for closure; they do not by themselves establish local thermal equilibrium or a gas equation of state.

Use fixed masses in an isolated Newtonian $N$-body system, with origin at its centre of mass. Define the $xx$ moment $I_{xx}={\sum }_i{m}_i{x}_i^2$ and $K_{xx}=\frac{1}{2}{\sum }_i{m}_i{v}_{x,i}^2$. Twice differentiating givesInsert the gravitational equation of motion and pair each ordered term $(i,k)$ with $(k,i)$:Thus one component of the tensor virial theorem isA stationary or time-averaged bounded system has $⟨{\ddot{I}}_{xx}⟩=0$, yielding $2⟨K_{xx}⟩+⟨W_{xx}⟩=0$. If ordered and random motions are separated, $2K_{xx}=2T_{xx}+{\Pi }_{xx}$, where $T_{xx}=\frac{1}{2}\int \rho u_x^{2}dV$ and ${\Pi }_{xx}=\int \rho {\sigma }_{xx}^{2}dV$. Summing the three diagonal equations recovers the scalar virial theorem, with the pair gravitational potential energy counted once.

Treat the long-wavelength nearly uniform spherical perturbation as a separate locally homogeneous expanding region, neglecting pressure gradients and tidal corrections. Its local Friedmann equation isThe curvature term represents the spatial curvature of an FLRW universe in the local patch; for the convention $a_s$ dimensionless, $K=k{c}^2$ has units of inverse time squared. Positive $K$ corresponds to a closed, more tightly bound patch. Here and below the printed $\rho$ is interpreted as mass density or energy density divided by $c^2$; literal SI energy density requires $G\rho /c^2$ on the right.

The region generally has its own scale factor and Hubble parameter. The two printed equations with an identical $H(t)$ can be compared as a leading-order equal-expansion construction: on a slicing where $H_s=H$, use $a_s=a$ in the already first-order curvature term to obtain $K/a^2=(8\pi G/3)\delta \rho$. This equal-expansion curvature overdensity construction does not make them both exact evolution equations with a permanently identical expansion history at common proper time. For example, a growing dust density contrast obeys the continuity equation $H_s-H=-\dot{\delta }/3$, which is nonzero when $\delta \propto a$. This explicit example prevents an inconsistent identification of the two local histories.

Within the equal-expansion construction of a, subtract the flat-background Friedmann equation to obtainThe background density scales as ${\rho }_b\propto a^{-4}$ during radiation domination and ${\rho }_b\propto a^{-3}$ during matter domination. ThereforeThese are the regular adiabatic growing-mode scalings on scales outside the Hubble radius, with a specified time slicing. In a consistent comoving description the numerical relation between curvature and density contrast differs from the naive subtraction, but the powers agree. The result does not imply that every gauge's density contrast, every decaying mode, or subhorizon dark matter during radiation domination has these powers.

For $r_s\gg d_H$, a regular adiabatic perturbation follows the large-scale density contrast growth in b; this does not require causal communication across the perturbation. Since $d_H=c/H$, the Friedmann equation gives $d_H\propto a^2$ during radiation domination and $d_H\propto a^{3/2}$ during matter domination. A fixed comoving perturbation has physical size $r_s=aR$, so its ratio to $d_H$ decreases as $a^{-1}$ or $a^{-1/2}$ until cosmological horizon entry.

The numerical “exactly at equality” assertion is inconsistent with the supplied parameters and $d_H=c/H$. Normalize $a_0=1$ and use the supplied ${\Omega }_{\gamma }$ as the whole radiation component. ThenBoth terms must be retained at matter-radiation equality. The present comoving size corresponding to its Hubble radius isIts physical size then is $0.0355\mathrm{Mpc}$. A mode with $R=79\mathrm{Mpc}$ has physical size only $0.0263\mathrm{Mpc}$ at equality, already inside the Hubble radius. More explicitly, putting $q=R{H}_0/c$ in $aR=c/H(a)$ givesThis is a direct numerical counterexample using the stated data. Neglecting one component exactly at equality instead gives $150.5\mathrm{Mpc}$, which also does not produce $79\mathrm{Mpc}$. Late-time dark energy is negligible at these early times.