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Take a uniform nearest-neighbour harmonic chain with displacement $u_j$, conjugate momentum $p_j$, mass $m_0$, spring stiffness $\kappa$, lattice spacing $a$ and periodic length $\ell =Na$. These specify a standard model for the requested notes, not additional numerical exam data. Its Lagrangian and Hamiltonian areVariation gives $m_0{\ddot{u}}_j=\kappa (u_{j+1}-2u_j+u_{j-1})$. For a mode $e^{i(kja-\omega t)}$, periodicity requires $k=2\pi n/\ell$ andThe lattice makes $k$ periodic modulo $2\pi /a$, so choose one Brillouin zone. At long wavelength ${\omega }_k\simeq v|k|$, where $v=a\sqrt{\kappa /m_0}$ is the sound speed. The continuum limit of a harmonic chain expands $u_{j+1}-u_j\simeq a{\partial }_{x}u$ and replaces ${\sum }_j$ by $a^{-1}\int dx$. Defining mass density $\rho =m_0/a$ and one-dimensional elastic coefficient $K=\kappa a$ givesThe momentum density is $\Pi =\rho {\partial }_{t}u$, and the Hamiltonian density is ${\Pi }^2/(2\rho )+K({\partial }_{x}u{)}^2/2$. Translational invariance of $u$ makes total momentum conserved. The continuum approximation describes acoustic wavelengths large compared with $a$; it is not a justification for extending the lattice dispersion and zero-point sums to arbitrarily large wavevector.^[ref]

canonical quantization promotes the coordinates and momenta to operators with $[u_j,p_l]=i\hslash {\delta }_{jl}$ and the other equal-time commutators zero. In the continuum, $[u(x),\Pi (x^{\prime })]=i\hslash {\delta }_{\mathrm{per}}(x-x^{\prime })$. The nonzero normal modes can be represented by independent phonon operators obeying $[b_k,b_q^{†}]={\delta }_{kq}$. A useful lattice expansion iswhere $X=N^{-1}{\sum }_j{u}_j$ is the centre-of-mass displacement. The corresponding momentum isIncluding the centre-of-mass term restores the full site commutator: the nonzero-mode sum alone gives $i\hslash ({\delta }_{jl}-1/N)$. Substitution diagonalizes the Hamiltonian,A phonon is one quantum of a collective normal mode, carrying energy $\hslash {\omega }_k$ and lattice momentum $\hslash k$ modulo a reciprocal-lattice vector. Reality of $u_j$ relates its Fourier coordinates by $u_k^{†}=u_{-k}$; it does not remove the independent oppositely travelling phonon modes. Using one complete Brillouin zone avoids double-counting real standing-wave coordinates.^[ref]

The free centre of mass has no restoring force and must not be treated as an oscillator of zero frequency. Fixing it, weakly pinning it, or specifying a finite centre-of-mass volume defines the corresponding zero-mode factor. With a fixed centre of mass, the internal thermal partition function and occupations areThe internal energy is ${\sum }_{k\ne 0}\hslash {\omega }_k(⟨n_k⟩+1/2)$; its high-temperature heat capacity approaches $(N-1)k_B$. For a large one-dimensional chain at temperatures below the lattice cutoff but above its finite-size mode spacing,This one-dimensional acoustic phonon heat capacity shows that quantization turns classical normal waves into bosonic excitations with a linear low-temperature heat capacity in one dimension. The Euclidean field action $S_E=\frac{1}{2}{\int }_0^{\beta \hslash }d\tau dx[\rho ({\partial }_{\tau }u{)}^2+K({\partial }_{x}u{)}^2]$ is Gaussian, with periodic imaginary-time boundary conditions; its Fourier modes reproduce these oscillators. Absolute displacement correlations require the zero-mode convention, and the lattice regulates short-distance fluctuations.

Use the contact-interaction Bose-gas field integral for a homogeneous three-dimensional weakly interacting Bose gas with repulsive short-range interaction as the standard model. Let particle mass be $m_b$, scattering length $a_s>0$, physical coupling $g=4\pi {\hslash }^2{a}_s/m_b$ and gas parameter $n{a}_s^3\ll 1$. These are symbolic model definitions, not supplied numerical parameters. The grand-canonical coherent-state integral has periodic fields and actionThe contact term is the normal symbol of $g{\hat{\psi }}^{†2}{\hat{\psi }}^2/2$. In three dimensions a contact interaction needs a regulator and coupling matched to the measured scattering length; the displayed $g$ is the low-energy physical coupling. This field integral is useful because an ordered saddle and its quantum/thermal fluctuations follow from the same action.

Stationarity with respect to $\bar{\psi }$ gives the imaginary-time Gross–Pitaevskii equation. A uniform saddle exists at $\mu >0$ with ${\psi }_0=\sqrt{n_0}{e}^{i{\theta }_0}$ and condensate density $n_0=\mu /g$. Its arbitrary phase reflects global $U(1)$ symmetry; a selected condensate fixes a phase in the thermodynamic symmetry-breaking description. Write $\psi =\sqrt{n_0}+\eta$ after choosing that phase. Linear terms vanish, and the quadratic part isWith ${\epsilon }_k={\hslash }^2{k}^2/(2m_b)$ and the Fourier convention $e^{i\mathbf{k}\cdot \mathbf{x}-i{\omega }_n\tau }$, the paired quadratic kernel isThe one-half in the Nambu spinor action removes pair double counting. Analytic continuation of the inverse kernel gives the Bogoliubov dispersionAt high momentum it approaches the free-particle kinetic energy plus its mean-field shift. The crossover scale is the healing length $\xi =\hslash /\sqrt{2m_{b}g{n}_0}$. Gaussian integration adds a fluctuation determinant to the saddle free energy. Expressed using the physical three-dimensional coupling, its zero-temperature integrand can be regularized as $\frac{1}{2}[E_k-{\epsilon }_k-\mu +{\mu }^2/(2{\epsilon }_k)]$; the subtraction follows from renormalization of a three-dimensional dilute-gas contact coupling, not discarding an infinite answer arbitrarily. The thermal contribution is $k_{B}T\int d^{3}k/(2\pi {)}^3\log (1-e^{-\beta E_k})$. A calculation beyond mean field must keep that matching and its thermodynamic derivatives consistent.

The density-phase action of a Bose gas makes the physical content clear, $\psi =\sqrt{n_0+\delta n}{e}^{i\theta }$. Up to a periodic density total derivative, the long-wavelength quadratic action containsnumber-phase conjugacy pairs the density fluctuation and phase in the quadratic density-phase action. Dropping the density-gradient term only in the long-wave limit and completing the density Gaussian square gives the Bose-gas phase-only actionThis has the same sound pole. A term $i\hslash n_0{\partial }_{\tau }\theta$ is also present before restricting to the smooth zero-winding sector; although locally a total derivative, it cannot in general be ignored in topologically nontrivial sectors. The density and phase variables describe the same coupled Bogoliubov branch, not a second independent massive particle simply because the density has a static restoring cost.^[ref]

A phase gradient carries superflow ${\mathbf{v}}_s=(\hslash /m_b)\nabla \theta$. Its energy cost defines the superfluid number density. At leading mean-field order the stiffness uses $n_0$, but condensate density and superfluid number density are different observables: for a Galilean-invariant continuum gas at zero temperature the latter is the full density even when interactions deplete the condensate. Bogoliubov coefficients give $v_k^2=[({\epsilon }_k+g{n}_0)/E_k-1]/2$ and a small zero-temperature depletion of relative order $\sqrt{n{a}_s^3}$ in three dimensions. Large depletion or strong long-wavelength phase fluctuations would invalidate the small-fluctuation saddle approximation.

The Landau criterion for creating excitations is $v_L={\inf }_{k>0}{E}_k/(\hslash k)=c_s$ for this homogeneous dispersion. Thus repulsion produces a linear sound mode and a positive critical velocity from the Landau criterion, whereas an ideal quadratic gas has zero value of this criterion. Vortices, obstacles and nucleation can reduce an actual flow threshold below $c_s$; the criterion alone is not a complete theory of metastable flow. At finite temperature thermal quasiparticles reduce the superfluid number density. Dimensionality and finite size also matter: the uniform three-dimensional weak-depletion condensate expansion must not be transferred unchanged to a one- or two-dimensional thermodynamic system. The field integral provides systematic corrections only within its dilute, stable and sufficiently weak-fluctuation regime.

Continue the free-particle propagator to $t=-i\mathcal{T}$ and replace its coordinate mass by the angular inertia $I$. The Euclidean free-particle bridge amplitude at coincident endpoints isThis normalization includes the time-sliced Gaussian measure, not just a formal determinant with an unspecified multiplicative constant. Set $A=2{\pi }^{2}I/({\hslash }^2\beta )>0$. The correct Poisson summation formula is ${\sum }_{m}F(m)={\sum }_n{\int }_{-\infty }^{\infty }dxF(x)e^{-2\pi inx}$; the source PDF omits the integral sign before its $d\theta$. For $F(x)=e^{-A{x}^2+i2\pi fx}$ the Gaussian Fourier integral givesThe overall prefactor cancels exactly: $2\pi Z_0\sqrt{\pi /A}=1$, and ${\pi }^2/A=\beta {\hslash }^2/(2I)$. Hencein agreement with the angular-momentum trace. Pairing $m$ and $-m$ makes the winding expression real, and the momentum expression shows positivity. Both are periodic in flux and invariant under $f\mapsto -f$. At high temperature the zero-winding term gives $Z\simeq \sqrt{2\pi I/(\beta {\hslash }^2)}$, with exponentially small flux dependence; at low temperature the nearest integer angular-momentum state dominates, with two equal lowest terms at half-integer flux. These independent limits also check the normalization and the flux phase.

The two-mode approximation for a Bose gas retains the real localized orbitals:It is valid when the lowest doublet is well separated from higher energy levels, and the temperature and interaction-induced changes of the orbitals are small on that separation. More explicitly, if $\Delta$ is the energy gap to discarded orbitals, sufficient low-energy conditions are $k_{B}T,t,N{U}_{0}I\ll \Delta$, where $I=\int d^{3}r|{\phi }_L{|}^4$. A deep double-well potential suppresses interwell interaction overlaps. Large $N$ alone does not justify the truncation: increasing $N$ at fixed coupling can violate the interaction condition.

Let $ϵ=(E_0+E_1)/2$ and use the even and odd eigenfunctions to evaluate the projected one-particle Hamiltonian operator:ThusThe same quantum tunnelling matrix element is $t=-\int d^{3}r{\phi }_L{h}_0{\phi }_R$. Reflection symmetry makes $\int |{\phi }_L{|}^4=\int |{\phi }_R{|}^4=I$. Neglecting mixed contact interaction integrals leavesThe last step is the normal-ordered boson pair count, following from the canonical commutation relation. Consequently the two-site Bose-Hubbard model has precisely the requested coefficientsandHere $\mu$ is the coefficient obtained by rewriting the fixed-particle number Hamiltonian operator, rather than an independently prescribed thermodynamic chemical potential. Adding a reservoir term $-{\mu }_{\mathrm{res}}\hat{N}$ would replace it by $\mu +{\mu }_{\mathrm{res}}$.

The negligible-overlap instruction applies to the mixed contact interaction integrals while the small quantum tunnelling matrix element is retained. Exactly disjoint smooth orbitals for a local $h_0$ would also give $t=0$; setting every overlap-related matrix element identically to zero would discard the physical splitting one is trying to describe.

Use the total density of states $D^{\ast }=A{m}^{\ast }/(2\pi {\hslash }^2)$ and $C=A/(2\pi \hslash {)}^2$ from part (b). In the paper's double-sum convention, $f(\varphi )$ has units of energy and scales as $A^{-1}$. Define its first angular moment byThis convention matters: the Landau quasiparticle energy functional written with an explicit $1/A$ instead uses the intensive kernel $\tilde{f}=Af$.

At zero temperature, the displayed functional is the change in $E-\mu N$; a true Galilean boost preserves $N$, so $\Delta \mathcal{F}=\Delta E$. For a radial displacement $\nu (\theta )$, the first term in the Landau quasiparticle energy functional becomesThe oriented radial integral also works for negative $\nu$: removing occupied levels below the chemical potential costs positive energy. Substituting $\nu =p_0\cos \theta +O(p_0^2/p_F)$ and using $N=C\pi p_F^2$ givesThe exact second-order correction to the boundary is unnecessary here because the unperturbed excitation energy vanishes on the Fermi surface; it changes this term only beyond quadratic order. It must nevertheless be understood when identifying $\Delta \mathcal{F}$ with a fixed-particle number energy change.

For the quadratic interaction term, the first-order occupation change alone suffices. The radial Dirac delta function integrals givePut $\varphi =\theta -{\theta }^{\prime }$. The remaining angular integral usesso only the first cosine component of the interaction survives. Since $D^{\ast }=2\pi C{m}^{\ast }$,Adding both contributions and matching the bare-mass result of part (a) yields the two-dimensional Landau effective-mass relation:Equivalently, using the density of states per unit area $g^{\ast }=D^{\ast }/A$,If dimensionless Landau parameters are defined through the Fourier series $D^{\ast }f(\varphi )=F_0+{\sum }_{\ell \ge 1}{F}_{\ell }\cos (\ell \varphi )$, then $D^{\ast }{\overline{f}}_1=F_1/2$ andDefining the cosine expansion with an extra factor of two would instead absorb that factor into $F_1$; the angular-integral formula above fixes the convention unambiguously. The physical content is that interaction changes accompanying a moving Fermi sea compensate the renormalized quasiparticle velocity, so the total boost energy is governed by the bare mass. For an interaction independent of angle, ${\overline{f}}_1=0$ and this consistency condition requires $m^{\ast }=m$.

Write $G(r)=⟨{\psi }^{†}(0)\psi (\mathbf{r})⟩$, $r=|\mathbf{r}|$, and ${\ell }^2=\hslash /(m\omega )$. At zero temperature all $N$ particles occupy ${\varphi }_0(\mathbf{r})=(\ell \sqrt{\pi }{)}^{-1}{e}^{-r^2/(2{\ell }^2)}$. Expanding the bosonic annihilation operator in single-particle orbitals givesThis one-body density matrix is a Gaussian function of width $\ell$; it contains one wavefunction evaluated at $r$, so its exponent is half that of the spatial number density $N|{\varphi }_0(r){|}^2$. In particular,the total number density at the trap centre, including condensed and noncondensed particles.

At positive temperature, with occupations ${\overline{n}}_a$ of the trap energy eigenstates, the exact ideal-gas decomposition isFor $\hslash \omega \ll k_{B}T<k_B{T}_c$, the sketch has a broad Bose-Einstein condensate component on scale $\ell$ and a narrower thermal peak near the origin. On approaching $T_c$ the broad component loses its weight $N_0$, while the thermal part remains. Above $T_c$ there is no macroscopic broad Bose-Einstein condensate component; close to $T_c$, low energy levels can still give a shoulder, so a single narrow Gaussian function is only a high-temperature description. At $k_{B}T\lesssim \hslash \omega$ individual excited energy levels matter and the two-scale description need not apply.

For $T\gg T_c$ the Maxwell-Boltzmann distribution applies. The leading off-diagonal one-body density matrix isHere ${\sigma }_r$ is the Gaussian function width defined by $e^{-r^2/(2{\sigma }_r^2)}$, and ${\lambda }_T$ is the thermal de Broglie wavelength. The thermal cloud itself extends over the much larger spatial scale $\sqrt{k_{B}T/\kappa }$; this is a correlation width, not the cloud radius.

The Hartree product state underlying the Gross–Pitaevskii equation has the factorized one-body coherence of a Hartree condensate. Choosing its nodeless ground state orbital real givesIn particular, the Thomas–Fermi approximation for a condensate from part (d) predictsThe curve starts at $\mu /g$, decreases with zero slope at the origin, and reaches zero with an infinite slope at $R$. Restoring the kinetic energy in the full Gross–Pitaevskii equation rounds that edge and produces a nonzero tail. This corrects the Thomas–Fermi approximation for a condensate, but the resulting one-body density matrix still factorizes.

The additional missing physics is quantum depletion. Interactions scatter pairs of bosons into orbitals outside the Bose-Einstein condensate even at zero temperature. In terms of the true Bose-Einstein condensate orbital ${\phi }_0$ the exact one-body density matrix decomposes asThe single-orbital Hartree product state sets $N_0=N$ and omits ${\rho }_{\mathrm{ex}}$ altogether. In a weakly interacting Bose gas, Bogoliubov theory adds a comparatively short-range noncondensed contribution near the origin, together with reduced weight in the broad Bose-Einstein condensate component. The central value remains the total number density, not just the condensate density. Its numerical shift relative to the original Gross–Pitaevskii equation solution also depends on the interaction-induced redistribution of number density; it is not fixed by the condensate fraction alone.

For the stipulated Strong-shock Rankine-Hugoniot conditions, ${\rho }_d=4{\rho }_u$ and $v_d=v_u/4$. The ideal gas law is $P_d={\rho }_d{k}_B{T}_d/\mu$, with $\mu$ a dimensional mean particle mass. The specific enthalpy of a nonrelativistic monatomic gas is $5k_{B}T/(2\mu )$. Dividing the energy flux equation in d by the common mass flux givesThereforeAs an independent check, conservation of momentum gives $P_d={\rho }_u{v}_u^2-4{\rho }_u(v_u/4{)}^2=3{\rho }_u{v}_u^2/4$, and division by ${\rho }_d=4{\rho }_u$ gives exactly the same temperature. The numerical coefficient relies on $\gamma =5/3$ and the absence of dynamically important radiative or magnetic stresses.

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 a spherical star in Newtonian hydrostatic equilibrium, the total supporting pressure satisfieswhere zero external pressure is the idealized vacuum boundary condition. Substitute the enclosed mass from a, set $x=ar$, and integrate inwards:The supplied integral follows directly by differentiating $\frac{1}{2}(\sin x/x{)}^2$. It yieldsThus $P_c=\pi G{M}^2/(8R^4)$, about $4.42\times 10^{14}\mathrm{Pa}$ for the Sun's parameters. This is precisely the equation of state of a polytrope of index one.

For nonrelativistic fully ionized hydrogen with electrons and protons at a common temperature, the gas pressure is $P_g=(n_e+n_p)k_{B}T=2\rho k_{B}T/m_p$. If gas pressure dominates, the resulting temperature profile isNumerically $T_c\simeq 5.78\times 10^6\mathrm{K}$ for the Sun's parameters, and $T_g$ tends to zero at the idealized outer edge.

If the photons are additionally assumed to be in local thermodynamic equilibrium with this gas, their radiation pressure is $a_{\mathrm{rad}}{T}^4/3$, with $a_{\mathrm{rad}}=4{\sigma }_{\mathrm{SB}}/c$. The consistent temperature is instead the unique nonnegative root ofThe left side is strictly increasing in $T$ for $r<R$, so the root is unique. Evaluating the radiation pressure at $T_g$ gives at most $P_{\mathrm{rad}}/P\simeq 6.4\times 10^{-4}$ for the Sun's parameters, at the centre; the ratio decreases approximately as $(\sin ar/ar{)}^2$ outwards. The gas-only profile is consequently an excellent approximation within this imposed stellar model.

Hydrostatic equilibrium by itself fixes total pressure, not how it is divided between gas and radiation. Thomson scattering alone does not enforce a common gas–radiation temperature or a Planck photon distribution. Without either neglecting radiation pressure or specifying the radiation field and its thermodynamic closure, a unique gas temperature cannot be inferred from the given mass density alone.

Under an additional thermalization assumption, an opaque star emits approximately black-body radiation characterized by the temperature near the layer from which thermal photons escape. Planck's law gives a Rayleigh-Jeans law low-frequency tail $B_{\nu }\propto T{\nu }^2$ and a Wien approximation high-frequency tail $B_{\nu }\propto {\nu }^3{e}^{-h\nu /(k_{B}T)}$.

Using the gas-dominated temperature profile in d and the local escape criterion in c, the model's putative emission temperature for the Sun's parameters would beUsing radial $\tau =1$ instead gives $22.8\mathrm{K}$. The frequency form of Wien's displacement law for the maximum of $B_{\nu }$ is $h{\nu }_{\max }/(k_{B}T)=2.82144$, so these two prescriptions predictrespectively, in the far infrared/submillimetre region. These are conditional estimates for this model, not observed solar peaks. A realistic solar effective temperature near $5.8\times 10^3\mathrm{K}$ gives ${\nu }_{\max }\simeq 3.4\times 10^{14}\mathrm{Hz}$ in $B_{\nu }$. The maximum of $B_{\lambda }$ occurs near $0.50\mu \mathrm{m}$; it is not obtained by simply taking $c/{\nu }_{\max }$, because spectral densities transform with a Jacobian determinant.

The model is unsuitable for the solar photosphere: fully ionized hydrogen at $16$–$23\mathrm{K}$ is inconsistent with thermal ionization, and the assumed opacity and near-surface temperature gradient do not resemble the Sun's. Even its central temperature is substantially below the actual solar value, about $1.6\times 10^7\mathrm{K}$.

There is a further logical limitation: the question retains only Thomson scattering. In its coherent, cold-electron limit, this redistributes photon directions without generating a Planck photon distribution. For a concrete counterexample, monochromatic injected radiation remains monochromatic after any number of such elastic scatterings; large optical depth cannot turn it into black-body radiation. True absorption and emission, or a specified pre-existing thermal radiation field with appropriate energy exchange, are missing. Thus an approximately blackbody emergent spectrum is the usual additional physical assumption, while the stated scattering-only model does not uniquely predict an emergent spectrum or its peak.

Thermal bremsstrahlung is radiation from the acceleration of electrons in the Coulomb force of ions. At fixed temperature, the encounter rate per electron is proportional to ion number density $n_i$, and the number of radiating electrons per unit volume is $n_e$. Therefore $j_{\nu }\propto n_e{n}_i$ in a dilute, optically thin plasma. This is a binary-encounter rate, not a one-particle emission process.

For a fixed encounter geometry, the ionic electric field and hence the electron acceleration are proportional to ionic charge $Ze$. The Larmor formula is quadratic in acceleration, giving $j_{\nu }\propto Z^2$ at fixed species number densities. The dimensionless Gaunt factor $g(\nu ,T)$ corrects the simple classical result for the detailed quantum scattering and radiation calculation, after averaging over the thermal electron distribution. It depends on frequency and temperature and, in a more complete notation, ionic charge.

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.

A galaxy cluster deflects background light through its projected gravitational field. In the thin-lens approximation for gravitational lensing, $\mathbf{\beta }=\mathbf{\theta }-\mathbf{\alpha }(\mathbf{\theta })$ maps image to source angle. Its surface mass density enters throughHere $\mu$ is signed lensing magnification; observed brightness uses $|\mu |$. The $D$'s are angular diameter distances to lens, source and between them; ${\gamma }_s$ denotes lensing shear, not a Lorentz factor. strong gravitational lensing creates arcs and multiple images near critical curves. For an axisymmetric lens, a tangential Einstein ring satisfies mean enclosed convergence one, givingThis is projected enclosed mass, not a general three-dimensional mass profile inferred from a single arc.

weak gravitational lensing uses coherent distortions of many background galaxies. Their intrinsic orientations average approximately to zero, allowing an estimate of reduced shear $g={\gamma }_s/(1-\kappa )$ and a projected mass map. The method traces all gravitating material without assuming hydrostatic equilibrium, which is valuable in merging clusters. Accurate work requires correction of the point spread function, calibrated shape measurement, source redshifts, foreground-member rejection and selection effects. Intrinsic shape noise, unrelated line-of-sight structures and lens-model degeneracies remain. The mass-sheet degeneracy leaves reduced shear unchanged under $\kappa \mapsto \lambda \kappa +(1-\lambda )$ and ${\gamma }_s\mapsto \lambda {\gamma }_s$, so absolute mass normalization needs additional information.