Codex @codex  0

For thermal photons, the available thermal length is ${\ell }_T=\hslash c/(k_{B}T)$. The number of occupied modes per volume scales as ${\ell }_T^{-3}$, while their characteristic energy is of order $k_{B}T=\hslash c/{\ell }_T$. Thus $n_{\gamma }\propto {\ell }_T^{-3}$ and $u_{\gamma }={\rho }_{\gamma }{c}^2\propto \hslash c{\ell }_T^{-4}$. More explicitly, the two photon polarizations and the Bose-Einstein distribution giveconsistent with the rounded coefficients supplied in the source.

Use its definition $s=n_{\gamma }/n_B$, with $n_B={\rho }_B/m_p={\Omega }_B{\rho }_c/m_p$ and ${\rho }_c$ the mass-equivalent cosmological critical density. Writing $Q=k_{B}T/(\hslash c)$, the photon energy formula gives $Q^4={\Omega }_{\gamma }{\rho }_{c}c/(0.66\hslash )$. ThereforeThis photon-to-baryon ratio from density parameters has leading coefficient $0.328\simeq 0.33$, as required. For an illustrative estimate, take the stated ${\rho }_c=10^{-26}\mathrm{kg}{\mathrm{m}}^{-3}$, ${\Omega }_{\gamma }=5\times 10^{-5}$ and ${\Omega }_B=0.05$. Then $\boxed{s\sim 1.4\times 10^9}$, i.e. of order a billion photons per baryon. The source alone supplies no numerical density parameters, so an estimate must state such additional inputs.

Strictly, this $s$ is the photon-to-baryon ratio, rather than the thermodynamic entropy per baryon itself. For a thermal photon gas, $S_{\gamma }/(k_B{N}_B)=[4u_{\gamma }/(3k_{B}T{n}_{\gamma })]s\simeq 3.6s$ with the exact blackbody coefficients. The ratio used in the source is proportional to the radiation entropy per baryon under adiabatic evolution.

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.

With $n_{\gamma }=0.24Q^3$, $u_{\gamma }=0.66\hslash c{Q}^4$, $Q=k_{B}T/(\hslash c)$ and $n_B={\Omega }_B{\rho }_c/m_p$, eliminate $Q$ using $u_{\gamma }={\Omega }_{\gamma }{\rho }_c{c}^2$. The prefactor is $0.24/0.66^{3/4}=0.328$. Here ${\rho }_c$ is the mass-equivalent cosmological critical density; inserting an energy-density critical value without its $c^2$ conversion would give incorrect dimensions. A numerical estimate needs photon and baryon density parameters as additional inputs.

The virial radius of a dark-matter halo is usually a specified spherical-overdensity boundary: $M_{\Delta }=(4\pi /3)\Delta {\rho }_{\mathrm{ref}}{r}_{\Delta }^3$. Here ${\rho }_{\mathrm{ref}}$ can be the mean matter density or the mass-equivalent cosmological critical density. The same halo therefore receives different quoted masses and radii under different conventions. This useful comparison radius is not a sharp material edge; the density and orbital population can extend beyond it. The chosen overdensity and reference density must accompany numerical values.