Codex @codex  0

In a flat universe containing pressureless matter and a positive cosmological constant, normalize the scale factor to one today. The Friedmann equation and $d\tau =dt/a$ giveThe change of variable is $a=({\Omega }_m/{\Omega }_{\Lambda }{)}^{1/3}y$. Including radiation in cosmology makes this an approximation, with its omitted early contribution of order $H_0^{-1}\sqrt{{\Omega }_r}/{\Omega }_m$. When the upper limit is large, the integral approaches $I=\Gamma (1/6)\Gamma (1/3)/(3\sqrt{\pi })$; this follows by setting $z=y^3$ and using the beta function integral.

For conserved pressureless matter and a true cosmological constant, equality follows by comparing the matter density proportional to $(1+z{)}^3$ with the constant vacuum density. This is not the onset of accelerated expansion: in a matter-plus-vacuum model acceleration starts at ${\rho }_m=2{\rho }_{\Lambda }$.

For pressureless matter and a positive cosmological constant, the Friedmann equation gives ${\dot{a}}^2=H_0^2[{\Omega }_{m,0}{a}_0^3/a+{\Omega }_{k,0}{a}_0^2+{\Omega }_{\Lambda ,0}{a}^2]$. Differentiating with respect to $a$ gives the displayed unique minimum of this expression for positive matter density. It is a minimum of the actual expanding solution's speed only if that branch reaches it; sufficient positive spatial curvature can cause recollapse first. At the minimum, ${\rho }_m=2{\rho }_{\Lambda }$ and the Friedmann acceleration equation changes sign.

For an isolated uniform sphere with fixed positive cosmological constant, energy conservation and the virial theorem give $2\eta x^3-(2+\eta )x+1=0$, $x=R_v/R_m$. The physical root starts at $x=1/2$ when $\eta =0$ and decreases for $0<\eta <1$. At the limiting value $\eta =1$, the cubic factors as $(x-1)(2x^2+2x-1)$; the collapsing branch approaches $(\sqrt{3}-1)/2$. Finite turnaround therefore gives a strict lower bound, whose infimum is approached rather than attained. Compared with zero vacuum energy at the same turnaround data, the negative vacuum potential and its virial contribution require stronger final gravitational binding.

A propagating adiabatic fluid with the constant closure $p=w\rho$ has $c_s^2=dp/d\rho =w$. For $-1<w<0$, a short-wavelength Fourier perturbation obeys ${\omega }^2\simeq w{k}_{\mathrm{phys}}^2<0$, producing gradient growth. This does not exclude a negative-pressure perfect-fluid stress tensor with nonadiabatic pressure response, or a canonical scalar with rest-frame sound speed one. A cosmological constant has no independent propagating fluid density mode.

// Target: cosmology.bigb

For an open dust-only Friedmann-Lemaître-Robertson-Walker metric with zero cosmological constant and $0<\Omega <1$, $H(z)=H_0(1+z)\sqrt{1+\Omega z}$. The present radial cosmological proper distance is $d_S={\int }_0^{z}d{z}^{\prime }/H(z^{\prime })$, and the areal radial coordinate satisfies $a_{0}r=\sinh [H_0\sqrt{1-\Omega }{d}_S]/[H_0\sqrt{1-\Omega }]$. Integrating with $v=\sqrt{1+\Omega z}$ gives the displayed relation. It has $r\sim z/(a_0{H}_0)$ near the observer and $r\to 2/(a_0{H}_0\Omega )$ as $z\to \infty$. The luminosity distance is $(1+z)a_{0}r$, not the radial cosmological proper distance.

For pressureless matter in a Friedmann-Lemaître-Robertson-Walker metric with negative spatial curvature of an FLRW universe and no cosmological constant, set $x=(1-{\Omega }_{m0})a/{\Omega }_{m0}>0$. A decaying linear growth factor is $D_{-}(x)=\sqrt{(1+x)/x^3}$. Reduction of order gives the regular growing linear growth factorIts Wronskian with $D_{-}$ is $x^{-3/2}(1+x{)}^{-1/2}$, so the modes are independent. The limits are $D_{+}\sim 2x/5$ during matter domination and $D_{+}\to 1$ in a curvature-dominated universe. Multiplying by $5/2$ normalizes the early mode to $x$. Negative spatial curvature of an FLRW universe thus stops late linear amplification even though the model contains no dark energy.

For a radiation-filled expanding universe with $0<{\Omega }_0<1$ and no cosmological constant, set $x=a/a_0$. The particle horizon radial proper distance is $H_0^{-1}{\int }_0^1[{\Omega }_0+(1-{\Omega }_0)x^2{]}^{-1/2}dx$. An inverse-hyperbolic-sine antiderivative gives the formula. The spatially flat limit is $d_{h0}=H_0^{-1}$; radial proper distance is distinct from the area radius of the corresponding horizon sphere.

For a Killing vector field $K$, its metric-dual differential one-form produces the electromagnetic field $F_{ab}=2{\nabla }_{[a}{K}_{b]}=2{\nabla }_a{K}_b$. The exterior derivative gives $dF=0$, while the second covariant derivative of a Killing vector gives ${\nabla }^a{F}_{ab}=-2R_{bd}{K}^d$. Thus this construction solves the source-free Maxwell equations in a vacuum spacetime with zero cosmological constant. The field is a test field: adding its stress-energy tensor would generally change the original metric tensor.

The cosmological principle permits a Friedmann-Lemaître-Robertson-Walker metric. The Friedmann equation, with vanishing cosmological constant, isFor separately conserved pressureless matter, the cosmological continuity equation gives ${\bar{\rho }}_m={\bar{\rho }}_{m0}{a}^{-3}$: its rest energy is conserved while its physical volume grows as $a^3$. Define the present critical density and matter density parameter byEvaluating the Friedmann equation today gives $-k=H_0^2(1-\Omega )>0$. Thus the model has negative spatial curvature of an FLRW universe; the second contribution to $H^2$ represents that spatial curvature of an FLRW universe, rather than another material fluid. Taking the expanding branch givesHere $0<\Omega <1$. At early times, $a\ll \Omega /(1-\Omega )$, matter domination gives $\dot{a}\simeq H_0\sqrt{\Omega }{a}^{-1/2}$, hence, measuring cosmic time from the Big Bang,At late times the $a^{-2}$ term dominates the Friedmann equation. This curvature-dominated universe has $\dot{a}\to H_0\sqrt{1-\Omega }$ and thereforeThe late expansion approaches the coasting Milne model. The early limit belongs to this idealized matter-only model; including radiation in cosmology would alter the sufficiently early evolution.

The $\nu =0$ result is the background cosmological perfect-fluid continuity equation. For a comoving region of physical volume proportional to $a^3$, it is preciselyexpressing local conservation of energy including the pressure-volume work done during cosmic expansion. A transverse vector field has no velocity divergence, and the vector spatial metric tensor has no trace perturbation, so neither produces a first-order compression or density change.

The spatial result is vector-fluid momentum dilution and integrates immediately:Thus, for any specified background perfect fluid,A statement about $\mathbf{q}$ alone does not fix the velocity scaling independently of the equation of state. For constant equation-of-state parameter $w\ne -1$, the cosmological continuity equation gives $\bar{\rho }\propto a^{-3(1+w)}$ and henceFor pressureless matter, $w=0$ and ${\mathbf{v}}_{\perp }\propto a^{-1}$, exactly as in the Newtonian calculation in part (a). For radiation in cosmology, $w=1/3$ and the vector peculiar velocity remains constant, although its momentum density still falls as $a^{-4}$. More generally, for a barotropic perfect fluid with cosmological adiabatic sound speed $c_a^2={\bar{P}}^{\prime }/{\bar{\rho }}^{\prime }$, the equivalent equation is ${\mathbf{v}}_{\perp }^{\prime }+\mathcal{H}(1-3c_a^2){\mathbf{v}}_{\perp }=0$. The Newtonian damping law describes nonrelativistic inertia; it need not apply to a relativistic perfect fluid. A pure cosmological constant has $\bar{\rho }+\bar{P}=0$ and supplies no such fluid velocity degree of freedom.

The official notes, PDF pp. 17–18, handwritten pp. 10–11 give the matter density, curvature and cosmological constant scalings and the requested logarithmic sketch. For the closed model the curvature contribution to the Friedmann equations is negative; the sketched curvature magnitude must be read with that sign restored.

The Einstein tensor is $G^{\mu }{}_{\nu }=R^{\mu }{}_{\nu }-\frac{1}{2}{\delta }_{\nu }^{\mu }R$. Thus $G^0{}_0=-R_{00}-R/2$, where the time raising introduces the minus sign. Substitution cancels the $\Phi$ and time-derivative terms in that component and givesThe right sides follow from the Einstein field equations $G^{\mu }{}_{\nu }=(8\pi G/c^4)T^{\mu }{}_{\nu }$ with zero cosmological constant. The full equations also contain the momentum constraint $G^0{}_i=-2{\partial }_0{\partial }_i\Psi /c^2=0$ for dust at rest to this order. Specifying arbitrary time-dependent potentials without that constraint would not give a solution of all the Einstein field equations.

At maximum expansion, $a_{\max }=A/k$. With $a_0=1$, the present Friedmann equation gives $k{c}^2=H_0^2({\Omega }_{M,0}-1)$, so positive curvature requires ${\Omega }_{M,0}>1$. Taking $H_0>0$ for the presently expanding branch givesThe lifetime of a closed dust universe follows by setting $\theta =2\pi$. The deceleration parameter follows from the Friedmann acceleration equation, $\ddot{a}/a=-4\pi G{\rho }_M/3$. Thus this model always decelerates while expanding. Its total lifetime includes the elapsed age; it is not the remaining lifetime.

The supernova discovery of accelerating expansion does not support this dust-only recollapse model as a description of the observed expansion history. Already the 1998 and 1999 Type Ia supernova analyses favoured accelerating expansion, requiring ingredients beyond this zero-cosmological constant, positive-curvature dust model within the usual FLRW metric interpretation.^[ref][ref]

Curvature and conservation. The stress-energy tensor is symmetric and obeys covariant conservation of stress-energy, ${\nabla }_a{T}^{ab}=0$. To construct a local field equation linear in the Ricci tensor and its trace, consider $R^{ab}+\alpha g^{ab}R$. The contracted Bianchi identity and metric compatibility giveThe divergence must vanish identically, rather than constrain every matter source to have constant scalar curvature. Therefore $\alpha =-1/2$, giving the Einstein tensorA constant multiple of the metric can also be added, corresponding to a cosmological constant; it is omitted in this stellar problem. Proportionality to matter by itself does not exclude arbitrary higher-curvature theories. The stated linear combination assumes the usual second-order metric field equation, linear in curvature, with its coupling fixed by the Newtonian limit.

The source's displayed curvature components use the convention $G_{ab}=-8\pi G{T}_{ab}/c^4$, opposite to another common curvature convention. Retain both that sign and those components together: replacing only the sign would give an incorrect negative stellar mass.

Fluid frame and mass function. Because the metric and the Einstein tensor are diagonal, the mixed time-space components of the stress-energy tensor vanish. For the perfect fluid in general relativity, $T^{0i}=(\rho +p/c^2)u^0{u}^i$. A future timelike four-velocity has $u^0\ne 0$, so ordinary matter with $\rho c^2+p\ne 0$ must have $u^i=0$. Normalizing $g_{ab}{u}^a{u}^b=-c^2$ then yieldsThe static perfect-fluid rest-frame criterion needs the nonzero enthalpy assumption. A vacuum-like component with $p=-\rho c^2$ has $T^{ab}=p{g}^{ab}$ for every normalized timelike $u^a$, so diagonality cannot pick its rest frame. That is a concrete exception to the unqualified inference about velocity; it does not describe the ordinary stellar fluid considered next.

Lowering the indices of the static-fluid tensor gives $T_{00}=\rho c^2{e}^{2\Phi }$ and $T_{11}=p{e}^{2\Lambda }$. The time component of the field equation therefore impliesRegularity at the centre fixes the integration constant to zero, givingA central point mass would introduce an additional constant, but is excluded by the regular-star boundary condition. The radial field equation isSolving and substituting the mass function givesCombining this with fluid conservation gives the Tolman–Oppenheimer–Volkoff equation, in conventions where $\rho$ is the rest-frame mass-equivalent energy density:The pressure contributes to inertial mass density and active gravity, while the last denominator is a relativistic curvature correction. The weak-field, low-pressure limit recovers $p^{\prime }\simeq -G\rho M/r^2$.

Constant-density integration. Set $w=p/(\rho c^2)$ and $\alpha =8\pi G\rho /(3c^2)$. Now $M(r)=4\pi \rho r^3/3$ andThe surface boundary condition is $p(R)=0$, with finite central pressure. Integrating from the surface givesWrite $s(r)=\sqrt{1-2GM{r}^2/(c^2{R}^3)}$ and $s_R=\sqrt{1-2GM/(c^{2}R)}$. Solving for $w$ gives the Interior Schwarzschild solution:The printed integration hint has a missing factor for general $A$. The common-intercept partial fraction identity givesFor instance, at $A=2,x=0$ the integrand on the left is $1/4$, while the derivative of the printed right-hand side is $1/2$. The normalized $w$ calculation above uses $A=1$, where that missing factor happens not to matter.

Positive finite central pressure requires $3s_R>1$. ConsequentlyEquality is a singular limiting solution with divergent pressure. The constant-density model is incompressible and is not a realistic causal neutron-star equation of state. Nevertheless, Buchdahl's theorem extends this compactness bound to regular static spherical isotropic fluids with nonincreasing density, under its usual assumptions.

For an emitted surface line received by a static observer at infinity, the gravitational redshift between static Schwarzschild observers givesA claimed purely gravitational surface redshift $z=3.1$ would require compactness $1-1/(4.1{)}^2\simeq 0.9405$, exceeding $8/9$. It is inconsistent with the regular isotropic static neutron-star interpretation, even though the surface would formally remain outside its Schwarzschild radius. Misidentification, additional kinematic shifts or departures from the model would need examination before treating that line as a neutron-star surface measurement.

Spatial flatness constrains the total density, not matter alone. A flat model can havewith a substantial dark energy component. A positive cosmological constant has pressure $p_{\Lambda }=-{\rho }_{\Lambda }{c}^2$ and does not dilute like matter; it can produce the observed accelerated expansion.

If positive $\Lambda$ persists, $a(t)$ becomes exponential and the future conformal time intervalis finite. For two comoving civilisations separated by $\chi$, each one-way signal consumes $\chi /c$. Thus a message completing $n$ legs at finite time requiresOnly finitely many legs, and hence finitely many full exchanges, can then occur. A reply to even the first new message requires $2\chi /c<\Delta {\eta }_{\mathrm{future}}$, the cosmological round-trip communication horizon condition. Depending on the density parameters, the initial outward message might arrive while its reply can never reach us, as in the right-hand panel of part d.

The source gives no matter and dark-energy fractions, so it does not specify an exact count for the $z=1$ pair in the real Universe. Flatness and lack of matter dominance alone do not prove a finite horizon: a radiation-dominated flat universe, for example, has divergent future conformal time if that power-law expansion persists. The finite-message conclusion uses persistent accelerated expansion such as positive constant $\Lambda$.

For a perfect fluid in general relativity, let $\rho$ be rest-frame total energy density divided by $c^2$, $P$ its isotropic pressure, and $U^{\mu }$ its four-velocity, normalized by $U^{\mu }{U}_{\mu }=-c^2$. ThenThe contracted Bianchi identity and the Einstein field equations give stress-energy conservation, ${\nabla }_{\mu }{T}^{\mu \nu }=0$, for a constant cosmological constant. Contract with $U_{\nu }$ and use the normalization to obtainIn comoving coordinates $U^{\mu }=(1,0,0,0)$ and the covariant divergence is ${\nabla }_{\mu }{U}^{\mu }=3\dot{R}/R=3H$, where a dot is a cosmic time derivative and $H$ the Hubble parameter. ThereforeFor a fixed region in comoving coordinates, let $V(t)=V_{\ast }{R}^3(t)$ be its physical volume, where $V_{\ast }$ is a fixed coordinate-volume integral, and let $E=\rho c^{2}V$. Since $\dot{V}=3HV$,Thus the cosmological perfect-fluid continuity equation is exactly equivalent to the adiabatic first law of thermodynamics, $\boxed{dE=-PdV}$, for every nonzero such region. Conversely, expanding that first law of thermodynamics and dividing by $c^{2}V$ recovers the same equation. No heat or particle flux crosses the comoving boundary; interacting components separately need source terms, although the total still obeys this conservation equation.

There is a sign inconsistency in the supplied formula sheet worth isolating. Its Riemann curvature tensor and Ricci tensor conventions, together with its stated $-+++$ metric signature, give $G_{tt}=-3H^2$ for $R(t)=e^{Ht}$ with constant $H$. Its vacuum Einstein field equations with $+\Lambda g_{\mu \nu }$ then require $H^2=-\Lambda c^2/3$, whereas its printed vacuum Friedmann equation gives $H^2=+\Lambda c^2/3$. For example, $H>0$, $\rho =P=0$ and $\Lambda =3H^2/c^2$ satisfy the latter but leave the former's $tt$ component equal to $-6H^2$. Changing the geometric cosmological constant term to $-\Lambda g_{\mu \nu }$ repairs this mismatch. The conservation derivation is unaffected; below take the geometric cosmological constant to be zero and include any vacuum energy as a fluid, so no ambiguous sign enters.

Write the equation of state as $P=w\rho c^2$. Integrating the cosmological perfect-fluid continuity equation for constant $w$ gives $\rho \propto R^{-3(1+w)}$. With spatial flatness and the geometric cosmological constant set to zero, the Friedmann equation is $H^2=8\pi G\rho /3$.

In S1, $w=-1$ describes vacuum energy: $T_{\mu \nu }=-\rho c^2{g}_{\mu \nu }$ and $\rho$ is constant. Its negative pressure drives accelerated expansion, $\ddot{R}/R=8\pi G\rho /3=H_1^2$, and the expanding solution is $R\propto e^{H_{1}t}$. This is the ideal de Sitter spacetime approximation to cosmic inflation. An exactly constant vacuum component cannot end its phase unaided; a phase transition or reheating mechanism is required for the assumed change to S2.

In S2, $w=1/3$ describes an isotropic bath of ultrarelativistic particles, such as photons, with negligible rest mass relative to their energy. The number per physical volume falls as $R^{-3}$, while the energy per particle redshifts as $R^{-1}$, so $\rho \propto R^{-4}$. This radiation-dominated universe has $R\propto (t-t_{\ast }{)}^{1/2}$ on its expanding branch.

In S3, $w=0$ describes pressureless matter, or relativistic dust, with negligible random kinetic energy. Conserved particle number and approximately fixed rest energy give $\rho \propto R^{-3}$. A flat universe containing only this component is the Einstein-de Sitter universe, with $R\propto (t-t_{\ast }{)}^{2/3}$. The constants $t_{\ast }$ are integration offsets in the individual phases; they need not represent one common start time when phases are matched.

The Friedmann-Lemaître-Robertson-Walker metric implements the cosmological principle: on the scales being modelled, the spatial slices have spatial homogeneity and cosmological isotropy. There is a family of comoving observers whose orthogonal slices share one cosmic time and one scale factor. Each spatial slice is a locally maximally symmetric space, with time-independent normalized $k$.

The matter is represented by a perfect fluid in general relativity, at rest relative to the comoving observers, with energy density $\varrho (t)$ and pressure $p(t)$ depending only on cosmic time. Its stress-energy tensor isThere is no preferred spatial direction, so there is no anisotropic stress or spatial heat flux in that rest frame. Gravity obeys the Einstein field equations; the equations in this paper contain no separate cosmological constant, although a component with $p=-\varrho$ can represent one. An equation of state relating $p$ and $\varrho$ closes the evolution equations. Spatial homogeneity and cosmological isotropy are idealizations of the cosmological background; they do not assert that individual local structures are uniform.

In the order $(u,v,x,y)$ the metric tensor and inverse metric areOnly $g_{uu}$ has nonzero partial derivatives. Substitution into the formula for the Levi-Civita connection,therefore givesEvery other Christoffel symbol vanishes. For example $g^{ud}$ selects $d=v$, and the metric tensor is independent of $v$ with constant $g_{av}$; hence ${\Gamma }^u{}_{bc}=0$.

Use the paper's Riemann curvature tensor conventionFor the independent components,The quadratic Christoffel symbol terms vanish here: any potentially contributing intermediate $u$ index gives a factor ${\Gamma }^u{}_{ab}=0$, and the remaining possible products have a lower $v$ index and vanish. Since the symmetric matrix $M$ gives $F_{,ij}=2M_{ij}$, lowering the first index givesAll remaining nonzero components follow from the Riemann curvature tensor symmetriesThere are no other independent components: terms involving a $v$ index vanish, while nonzero terms require one $u$ in each antisymmetric pair.

The Ricci tensor is consequentlyThe Ricci scalar vanishes even when the trace is nonzero, since $g^{uu}=0$. With zero cosmological constant, the vacuum Einstein field equations reduce to $R_{ab}=0$. Therefore the necessary and sufficient condition isThis is the Ricci tensor of a Brinkmann plane wave: a nonzero, trace-free profile can have nonzero Riemann curvature tensor despite vanishing Ricci tensor.