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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 $z\to 0$.

The cosmological principle idealizes spatial slices as homogeneous and isotropic. The corresponding Friedmann-Lemaître-Robertson-Walker metric can be writtenHere $t$ is cosmic time, $a(t)$ has dimensions of length in this convention, and $\chi$ is a dimensionless comoving coordinate. The sign of $K$ gives positive, zero or negative spatial curvature. It does not alone determine global topology. A comoving pair at fixed radial coordinates has proper separation scaling with $a(t)$, so the Hubble parameter is $H=\dot{a}/a$, and cosmological redshift satisfies $1+z=a(t_{\mathrm{obs}})/a(t_{\mathrm{em}})$.

For a perfect fluid of mass-equivalent energy density $\rho$ and pressure $p$, Einstein's equations giveThe last equation is local energy conservation for constant $\Lambda$, with vacuum energy kept separate from $\rho ,p$. A component with $p=w\rho c^2$ has $\rho \propto a^{-3(1+w)}$: pressureless matter dilutes as $a^{-3}$ and radiation as $a^{-4}$, the additional radiation factor being photon redshift. With $K=\Lambda =0$, matter domination gives $a\propto t^{2/3}$ and radiation domination gives $a\propto t^{1/2}$. Positive vacuum energy alone gives exponential expansion, $a\propto e^{Ht}$ with $H=c\sqrt{\Lambda /3}$.

Geometry, contents and initial conditions together determine the expansion history. A closed matter-only model recollapses; an open matter-only model expands indefinitely. With a cosmological constant, positive spatial curvature no longer by itself implies recollapse. Present critical density is ${\rho }_c=3H_0^2/(8\pi G)$ and the density parameters obey ${\Omega }_m+{\Omega }_r+{\Omega }_{\Lambda }+{\Omega }_K=1$, where ${\Omega }_K=-K{c}^2/(a_0^2{H}_0^2)$. These models describe the large-scale background, while local structure and inhomogeneities require perturbations or more general metrics.

Use Cartesian coordinates $x^a=(t,x,y,z)$ with $c=1$ and ${\eta }_{ab}=\mathrm{diag}(1,-1,-1,-1)$. Spatial dot and cross products below use the Euclidean three-metric. The alternating tensor is given the orientation ${ϵ}^{0123}=+1$.

Vector and covector transformation. The vector $A=A^b{\partial }_b$ is independent of coordinates, and the chain rule gives ${\partial }_b=(\partial x^{\prime a}/\partial x^b){\partial }_a^{\prime }$. Likewise the one-form $A_{b}d{x}^b$ is independent of coordinates. Comparing coefficients provesTheir contraction $A_a{B}^a$ is unchanged because the forward and inverse Jacobians cancel. Here lowering an index uses the metric, not an identification of the numerical upper and lower components.

Why antisymmetrization matters. Write $J^c{}_a=\partial x^c/\partial x^{\prime a}$. Differentiating the covector law givesThe last term obstructs the general tensor transformation law. It is symmetric in $a,b$, so it cancels from ${\partial }_a^{\prime }{A}_b^{\prime }-{\partial }_b^{\prime }{A}_a^{\prime }$. ThereforeEquivalently $F=dA$ is the exterior derivative of the electromagnetic four-potential one-form. The statement that ${\partial }_a{A}_b$ is not a tensor needs a qualification: under affine Cartesian Lorentz transformations it does transform tensorially, because the second derivatives vanish. The failure concerns arbitrary nonlinear coordinate changes, including curvilinear coordinates in flat spacetime. This distinction is precisely the difference between general covariance and covariance under the restricted inertial-frame transformations.^[ref]

Electric and magnetic components. Choose $A^a=(\Phi ,\mathbf{A})$, so $A_a=(\Phi ,-\mathbf{A})$, $\mathbf{E}=-\nabla \Phi -{\partial }_t\mathbf{A}$ and $\mathbf{B}=\nabla \times \mathbf{A}$. ThenThus the lower-index field tensor isThis is a derived component representation of the field, with the source's sign convention.

Raised indices and the first invariant. Raising both indices changes the time-space signs and leaves the spatial block unchanged:In the scalar contraction, each independent antisymmetric pair is counted twice, givingUnder a Lorentz transformation each inverse transformation from a lower index cancels its matching transformation from an upper index, leaving the same scalar. In particular a plane electromagnetic wave has $E=B$ in these units and makes this invariant vanish, while a pure electric field makes it negative.

The printed dual convention. The PDF explicitly defines ${}^{\ast }{F}^{ab}={ϵ}^{abcd}{F}_{cd}$, with no $1/2$. Each independent pair therefore contributes twice: for example ${}^{\ast }{F}^{01}=F_{23}-F_{32}=-2B_x$. The complete result isThe time-space pairs contribute $-4\mathbf{E}\cdot \mathbf{B}$ and the spatial pairs contribute another $-4\mathbf{E}\cdot \mathbf{B}$. The conventional Hodge dual includes $1/2$ and would instead give the contraction $-4\mathbf{E}\cdot \mathbf{B}$; it must not silently replace this paper's definition. With fixed orientation the latter contraction is a pseudoscalar: invariant under proper Lorentz transformations but odd under orientation reversal if compared using a fixed alternating symbol.

Energy density and flux. Substitution into the given mixed stress-energy tensor yieldsFor the first formula, $F^{0c}{F}_{0c}=-E^2$ and $F_{cd}{F}^{cd}/4=(B^2-E^2)/2$. For the second, $F^{0j}{F}_{ij}=(\mathbf{E}\times \mathbf{B}{)}_i$. The minus sign on $T^0{}_i$ arises from its lowered spatial index; it does not reverse the physical energy flow. Raising that index gives $T^{0i}=S_i$, and symmetry gives $T^{i0}=S_i$, with $\mathbf{S}=\mathbf{E}\times \mathbf{B}/{\mu }_0$ the Poynting vector in $c=1$ units. Thus $T^{00}$ is energy density and $T^{i0}$ represents energy flux. In SI coordinates $x^0=ct$, restore $T^{0i}=S_i/c$ and $T^{00}=({ϵ}_0{E}_{\mathrm{SI}}^2+B^2/{\mu }_0)/2$, with ${ϵ}_0{\mu }_0{c}^2=1$. This keeps the mixed-index signs and the units distinct.

Use signature $(+---)$ and local inertial coordinates $x^0=ct$. Write $u^{\mu }=\gamma (c,\mathbf{v})$ and $w=\rho c^2+p$, where $\rho$ is rest mass density. The perfect fluid in general relativity has componentsHere $T^{00}$ is energy density, $c{T}^{i0}$ is energy flux, $T^{0i}/c$ is momentum density, and $T^{ij}$ is the flux of the $j$th momentum component through a surface normal to $i$. In the local rest frame these reduce to energy density $\rho c^2$, zero energy and momentum flow, and isotropic pressure $p$. The equation ${\nabla }_{\mu }{T}^{\mu \nu }=0$ expresses local stress-energy conservation. Its temporal and spatial components give energy and momentum balance; in flat coordinates, for example, ${\partial }_t{T}^{00}+{\partial }_i(c{T}^{i0})=0$.

For relativistic dust, $T^{\mu \nu }=\rho u^{\mu }{u}^{\nu }$. Contracting its divergence with $u_{\nu }$, and using metric compatibility together with $u_{\nu }{u}^{\nu }=c^2$, givesConsequently ${\nabla }_{\mu }(\rho u^{\mu })=0$. Substituting this continuity equation back into the divergence leaves $\rho u^{\mu }{\nabla }_{\mu }{u}^{\nu }=0$. Wherever matter is present,so the individual particles follow geodesics.

The electromagnetic four-force law $f^{\mu }=q{F}^{\mu \nu }{u}_{\nu }$ must preserve the particle's rest mass. Thus $u_{\mu }{f}^{\mu }=0$, and requiring this for arbitrary timelike four-velocities makes the electromagnetic field tensor antisymmetric. Its six independent components comprise the electric field and magnetic field. With the stated signature and $x^0=ct$, one consistent convention isIt gives the spatial force $d\mathbf{p}/dt=q(\mathbf{E}+\mathbf{v}\times \mathbf{B})$ and energy transfer $dE/dt=q\mathbf{E}\cdot \mathbf{v}$. Under a Lorentz transformation, $F^{\mu \nu }$ transforms as a rank-two contravariant tensor; its electric and magnetic components consequently mix. A four-potential supplies $F_{\mu \nu }={\nabla }_{\mu }{A}_{\nu }-{\nabla }_{\nu }{A}_{\mu }$, unchanged by $A_{\mu }\mapsto A_{\mu }+{\nabla }_{\mu }\chi$. The Maxwell equations areThese combine the source equations and the homogeneous equations into covariant tensor statements; $j^{\mu }$ is the four-current.

For charged relativistic dust, the matter tensor exchanges four-momentum with the electromagnetic field: ${\nabla }_{\mu }{T}_{\mathrm{m}}^{\mu \nu }=F^{\nu \alpha }{j}_{\alpha }$, while their total stress-energy tensor is conserved. Identical particles of rest mass $m$ and electric charge $q$ have $j^{\alpha }=(q/m)\rho u^{\alpha }$. Contraction with $u_{\nu }$ again gives mass continuity because $u_{\nu }{F}^{\nu \alpha }{u}_{\alpha }=0$. Removing that term givesUsing zero divergence for the matter tensor alone in this last step would omit the electromagnetic Lorentz force.

For $x=h\nu /(k_{B}T)\ll 1$, the Taylor expansion $e^x-1=x+O(x^2)$ gives the Rayleigh-Jeans law for equilibrium radiation,This is the classical limit: each electromagnetic mode has mean thermal energy approximately $k_{B}T$. Extending it to arbitrarily large frequencies would give a divergent integrated energy density, the ultraviolet catastrophe.

For $x\gg 1$, $e^x-1\simeq e^x$, giving the Wien approximationPlanck's quantum hypothesis assigns equally spaced mode energies $E_n=nh\nu$. When the spacing is small relative to $k_{B}T$, many levels are occupied and the classical approximation works. When $h\nu \gg k_{B}T$, exciting even one quantum is exponentially unlikely, suppressing the high-frequency tail. The single expression $⟨E⟩=h\nu /(e^x-1)$ contains both limits. Comparing temperature directly to $h\nu$ would be dimensionally incorrect; the low-temperature condition means $k_{B}T\ll h\nu$.

Here $h$ is the Planck constant, $\nu$ is the radiation frequency, $c$ is the speed of light in vacuum, $k_B$ is the Boltzmann constant, and $T$ is the absolute temperature in kelvin. The quantity $U_{\nu }$ is spectral energy density per unit frequency, so $U_{\nu }d\nu$ is energy per unit volume in the interval $[\nu ,\nu +d\nu ]$, and $h\nu /(k_{B}T)$ is dimensionless.

Put $x=h\nu /(k_{B}T)=hc/(\lambda k_{B}T)$. In the long-wavelength limit $x\ll 1$, the Taylor expansion $e^x-1=x+O(x^2)$ in Planck's law gives the Rayleigh-Jeans law,This is the classical equipartition theorem result: each electromagnetic mode carries mean energy $k_{B}T$. Extending it to arbitrarily high frequency would make the total energy density diverge, the ultraviolet catastrophe.

In the short-wavelength limit $x\gg 1$, $e^x-1\simeq e^x$, giving the Wien approximation,Large photon energies are thermally suppressed, removing the classical high-frequency divergence.

If the requested “form” is expressed per unit wavelength, the change of variables includes its Jacobian determinant: $U_{\lambda }=U_{\nu }|d\nu /d\lambda |=U_{\nu }c/{\lambda }^2$. ThusSubstituting $\nu =c/\lambda$ without the Jacobian determinant would give an incorrect spectral density.

Convert both inputs to energy density. The matter contribution, including rest energy, is $u_{M,0}={\rho }_{M,0}{c}^2=2.25\times 10^{-10}\mathrm{J}{\mathrm{m}}^{-3}$. The given photon value is $u_{\gamma ,0}=5\times 10^5(1.602\times 10^{-19})=8.01\times 10^{-14}\mathrm{J}{\mathrm{m}}^{-3}$. HenceMatter energy density scales as $(1+z{)}^3$, whereas the cosmic microwave background scales as $(1+z{)}^4$. The photon-only matter-radiation equality condition is thereforeThis uses the supplied CMB photon density. Additional relativistic species change equality for total radiation.

For a spherical star in hydrostatic equilibrium, $dP/dr=-Gm(r)\rho /r^2$. Multiply by $4\pi r^{3}dr$ and integrate to its radius $R$:Integration by parts gives $4\pi R^3{P}_s-3\int PdV=\Omega$. Neglecting the external surface pressure $P_s$, an ideal gas with constant heat capacity ratio $\gamma$ has thermal energy density $u=P/(\gamma -1)$. ThereforeThis is the hydrostatic virial theorem. A nonzero confining pressure retains the right-hand side $4\pi R^3{P}_s$; varying composition, radiation contributions or a spatially varying effective exponent require the integral relation, rather than pulling one $\gamma$ outside it.

Let $N=n_H+n_p$ be the conserved number density of hydrogen nuclei, whether ionized or bound, and $x=n_p/N=n_e/N$ the ionization fraction. The ideal gas pressure counts neutral atoms, free protons and free electrons:At low temperature $x\simeq 0$ and $P\simeq N{k}_{B}T$. Heating through the ionization region increases the particle count as well as their thermal energy, giving a steeper rise. At high temperature $x\simeq 1$, so $P\simeq 2N{k}_{B}T$ and the asymptotic slope is twice the low-temperature slope. There is a smooth ionization crossover, not a pressure discontinuity.

The held graph is legible. Representative readings are $P(5000\mathrm{K})\simeq 6.8\mathrm{Pa}$ and $P(15000\mathrm{K})\simeq 41.4\mathrm{Pa}$, with readout uncertainties of roughly $0.3$–$0.5\mathrm{Pa}$ and about $100\mathrm{K}$. They give $N\simeq P/(k_{B}T)$ on the neutral branch and $N\simeq P/(2k_{B}T)$ on the ionized branch, respectively. Both are consistent withThe error is a conservative graph-reading estimate, not a fitted experimental confidence interval.

For that density, the Saha equation gives $x\simeq 0.50$ at $9000\mathrm{K}$, $x\simeq 0.81$ at $10000\mathrm{K}$, and $x\simeq 0.9992$ at $15000\mathrm{K}$. Differentiating the ionization relation givesThe calculated maximum slope occurs near $9.25\times 10^3\mathrm{K}$, consistent with the source's visibly steepest region at roughly $(9.3\pm 0.4)\times 10^3\mathrm{K}$. This checks the actual slope rather than identifying it without justification with exactly half ionization.

Taking the neutral ground-state gas as the zero of internal binding energy, the matter energy density is translational kinetic energy plus ionization energy:At $15000\mathrm{K}$, with $R=13.6\mathrm{eV}$, these terms are about $62$ and $218\mathrm{J}{\mathrm{m}}^{-3}$, givingRest energies are excluded. A separate equilibrium radiation field, if the gas were in a blackbody cavity, would add its own radiation energy density; it is not part of this gas-only estimate.

Magnetic pressure is the field energy density $P_B=B_f^2/(2{\mu }_0)$. Using the preceding final field gives $\boxed{P_B\simeq 2.6\times 10^{13}\mathrm{Pa}\simeq 260\mathrm{Mbar}}$. This uses the compressed field, not the initial 36 T field; substituting the latter would omit the square of the already squared convergence amplification.

For a static finite-energy field, the Hamiltonian density is $\frac{1}{2}({\varphi }^{\prime }{)}^2+V(\varphi )$. Multiply the static equation of motion by ${\varphi }^{\prime }$ and integrate once:At either vacuum endpoint both terms vanish, so $\frac{1}{2}({\varphi }^{\prime }{)}^2=V$. The total energy density is consequently $({\varphi }^{\prime }{)}^2=4k^2{\mathrm{sech}}^2[k(x-x_0)]/b^2$. Its integral givesFor the usual positive $a,b$ convention this isThe coefficient depends on the dimensionless parameter $b$; it is not a universal numerical constant independent of the potential. In two spacetime dimensions, $[\varphi ]=[b]=0$ and $[a]=2$, so the dimensions are consistent. If both $a$ and $b$ are negative while $ab>0$, the real answer is instead $8\sqrt{|a|}/|b{|}^{3/2}$; the literal factor $\sqrt{a}$ would not be real.

An equivalent positive-parameter calculation substitutes $z=b\varphi$ in $E_0={\int }_0^{2\pi /b}\sqrt{2V}d\varphi$ and uses the provided integral of $\sqrt{1-\cos z}$. The following original plot shows that the localized energy density sits where the field passes between the vacua.

A perfect fluid in general relativity has isotropic rest-frame pressure and no rest-frame heat flux or anisotropic stress. In units $c=1$ and signature $(-,+,+,+)$, let $u^a{u}_a=-1$, let $\rho$ be rest-frame energy density and $p$ rest-frame pressure. Its stress-energy tensor isThe energy measured with the fluid is $T_{ab}{u}^a{u}^b=\rho$. The orthogonal projection $h^a{}_b={\delta }^a{}_b+u^a{u}_b$ isolates spatial directions and gives isotropic projected stress $p{h}_{ab}$. A fluid moving relative to another observer can have nonzero momentum density even though its own rest-frame stress is isotropic.