The right-hand side of the slow-roll curvature power spectrum must be evaluated separately for each mode at cosmological horizon exit, $k=aH$. Taking logarithms givesFor $N=\log a$,HenceThis is the scalar spectral index in Hubble slow-roll parameters.

The creation and annihilation operators obey the canonical bosonic commutation relationsSince $\delta \varphi =\hat{f}/a$, the vacuum two-point correlation function isForone has $|f_k{|}^2=(1+1/(k^2{\tau }^2))/(2k)$. Comparison with the definition of the dimensionless power spectrum giveswhere $a=-1/(H\tau )$. On superhorizon scales, $k\ll aH$ or $|k\tau |\ll 1$, and therefore

Yes. A perturbation of amplitude $A$ at CMB decoupling must grow by at least $A^{-1}$ before becoming nonlinear. During matter domination ${\delta }_m\propto a$, but once ${\rho }_{\Lambda }$ dominates, the suppression of matter growth by smooth accelerated expansion makes the growth approach a finite limit. Increasing ${\rho }_{\Lambda }$ moves this freeze-out to an earlier scale factor and eventually prevents ${\delta }_m$ from reaching unity.

For an order-of-magnitude bound, matter growth would reach unity atRequiring matter still to dominate then gives the galaxy-formation bound on the cosmological constantThe exact upper limit follows by imposing $A{D}_{+}(\infty )/D_{+}(a_{\mathrm{dec}})\gtrsim 1$ with the integral growth factor and an appropriate nonlinear-collapse threshold.

For matter plus a cosmological constant,Set ${\delta }_m=Hu$. Substitution into the equation from part (c) makes the coefficient of $u$ vanish by the Friedmann relation and leavesorTherefore the two independent modes can be writtenThe first is the decaying mode. Normalizing the second to $D_{+}(a)\sim a$ at early times gives the integral linear growth factor in a matter-Lambda universe

Use $d/dt=aH,d/da$. Thenwhere now $H^{\prime }=dH/da$. AlsoSubstitution into part (a) and division by $a^2{H}^2$ gives the matter growth equation as a function of scale factor

During matter domination, $a\propto t^{2/3}$ and $4\pi G{\bar{\rho }}_m=2/(3t^2)$. Trying ${\delta }_m\propto t^p$ giveswhose roots are $p=2/3$ and $p=-1$. Thus the cosmic-time matter density modes are

During asymptotic cosmological constant energy domination, $H$ is constant and ${\bar{\rho }}_m$ is negligible, soThe matter density modes during cosmological-constant domination are thereforeThe growing matter-era solution freezes to a constant.

During matter domination, pressureless matter has ${\bar{P}}_m=\delta P_m=0$. On subhorizon scales the time derivative of the Newtonian gauge in cosmology potential is negligible, so the continuity and Euler equations reduce toTaking the divergence of the second equation, differentiating the first, and using the Poisson equation givesSince $d/d\tau =a,d/dt$, one hasDivision by $a^2$ yields the standard linear cosmological density perturbation equation

The measurement cannot definitively exclude the model. If only Standard Model particles supplied entropy at scalar decoupling, the largest available value $g_{\ast s}=106.75$ would givewhich a $0.1\%$ measurement around the Standard Model value would clearly detect.

In the proposed model, however, the many additional relativistic species are also in equilibrium when the scalar decouples. They increase $g_{\ast s}(T_d)$, and their later disappearance transfers entropy to the coupled bath but not to the scalar. Since $\Delta N_{\mathrm{eff}}\propto g_{\ast s}(T_d{)}^{-4/3}$, a sufficiently large hidden particle content can dilute the scalar signal below any stated finite precision; a value above roughly $g_{\ast s}(T_d)\sim 550$ already pushes it below about $0.003$. The null measurement constrains the combination of decoupling time and total entropy degrees of freedom, but does not rule out the entire new-physics model.

After the real scalar decouples at $T_d$, its temperature redshifts as $a^{-1}$. The other particles continue sharing entropy. Just before standard-neutrino decoupling their entropy degrees of freedom areThe temperature of a decoupled relativistic relic therefore obeysA real scalar has one bosonic degree of freedom, whereas one effective neutrino species has energy weight $(7/8)\times 2=7/4$. Hence the contribution of a decoupled real scalar to Neff isHere $g_{\ast s}(T_d)$ counts the other particles still coupled to the thermal bath, as specified in the question.

Photons are bosons with two polarizations and temperature $T$, so ${\rho }_{\gamma }=({\pi }^2/30)2T^4$. Each relativistic neutrino species includes a neutrino and antineutrino helicity state, is fermionic, and has temperature $T_{\nu }=(4/11{)}^{1/3}T$. ConsequentlyAdding the two contributions gives the defining expression for the effective number of neutrino species:

Before electron-positron annihilation, photons, electrons, and positrons share one temperature. Their effective entropy degrees of freedom areThe neutrinos have already undergone thermal decoupling in cosmology, so $T_{\nu }a$ remains constant and they receive none of the electron-positron entropy. In the still-coupled electromagnetic plasma, cosmological entropy conservation givesThe decoupled neutrino temperature at the same later time is $T_{\nu }=T_{\mathrm{before}}{a}_{\mathrm{before}}/a$. Dividing the two relations gives the Cosmic neutrino background temperatureThis instantaneous-decoupling calculation neglects the small reheating correction from non-instantaneous neutrino decoupling.

All galaxies in the population share the same formation time $t_f$, so the difference between their stellar ages equals the difference between their cosmic emission times. The redshift-time relation givesFor a close pair with $|\Delta z|\ll 1$,A cosmic chronometer measurement therefore reconstructs $H(z)$ directly from differential galaxy ages. Inserting those values into the matter-plus-dark-energy Friedmann expression constrains $w(z)$.

Supernova distances integrate $1/H(z)$, while $H(z)$ already contains an integral of $w(z)$. Cosmic chronometers avoid the distance integral, so rapid oscillations in $w(z)$ suffer one fewer smoothing operation and can leave a more visible signal.

The flux relation is $F=L/(4\pi d_L^2)$ with luminosity distance $d_L=(1+z)\chi (z)$. In a spatially flat universe,Thus Type Ia supernova cosmology measures the distance-redshift curve. Equation (2) makes $H(z)$ depend on an integral of $w(z)$, while $d_L$ introduces a second integral. Fitting predicted distances to many supernova fluxes over a range of redshifts therefore constrains parameters or bins describing $w(z)$, although these integrations smooth fine redshift structure.

The common luminosity $L$ need not be known to constrain the shape of $w(z)$. An unknown $L$ multiplies every inferred distance by the same factor and is degenerate with the overall scale $H_0^{-1}$, or equivalently with the supernova absolute magnitude. Relative distances at different redshifts still determine the shape of the expansion history. An external calibration is needed to determine the absolute distance scale.

For dynamical dark energy, use $dz/dt=-(1+z)H$ in the continuity equation to obtainTherefore the variable dark-energy equation of state givesCombining this with ${\rho }_m={\rho }_{m,0}(1+z{)}^3$ in the Friedmann equation yieldswhereThis is the Hubble parameter for matter and dynamical dark energy.

For a component with $P=w\rho$ and constant $w$, the cosmological perfect-fluid continuity equation becomesIntegration gives the constant-equation-of-state density scalingDefine the present cosmological density parameter bySubstitution into the spatially flat Friedmann equation gives the Hubble parameter for constant-equation-of-state componentsSpatial flatness is what allows the expression to contain only the listed density components, with ${\sum }_i{\Omega }_{i,0}=1$.

The curvature 2-forms show that orthonormal curvature components grow as $m/r^3$. The geodesic deviation equation converts them into the relative acceleration of nearby parts of an observer. As $r\to 0$, the Schwarzschild tidal force stretches radial separations and compresses transverse separations with unbounded magnitude. An extended body therefore undergoes destructive tidal deformation before reaching the Schwarzschild singularity.

Substituting the forms from part (b) into Cartan's second structure equation gives the independent mixed-index curvature 2-formsEvery form is proportional to its corresponding basis 2-form, so the curvature operator is diagonal. The diagonal Ricci tensor components are contractions of these sectional curvature coefficients. In each case the coefficients cancel in the pattern $2-1-1=0$, with the Lorentzian sign included when the time direction is contracted. HenceThis is consistent with Schwarzschild spacetime being a vacuum solution away from $r=0$.

WriteFor a diagonal curvature operator, ${\Theta }^{\mu \nu }$ contains only $f^{\mu }\wedge f^{\nu }$. Consequently $R^{\mu }{}_{\nu \rho \sigma }$ can be nonzero only when the plane indexed by $\rho ,\sigma$ is the same as the plane indexed by $\mu ,\nu$. In the contractionthis condition cannot hold for $\nu \ne \sigma$. Thereforewhich is the statement that a diagonal curvature operator implies diagonal Ricci tensor.

SetDirect exterior differentiation givesSolving Cartan's first structure equation and imposing ${\omega }_{\mu \nu }=-{\omega }_{\nu \mu }$ gives the independent mixed-index connection 1-formsThe remaining forms follow from Lorentz-signature antisymmetry: ${\omega }^1{}_0={\omega }^0{}_1$, ${\omega }^2{}_0={\omega }^0{}_2$, ${\omega }^3{}_0={\omega }^0{}_3$, and ${\omega }^j{}_i=-{\omega }^i{}_j$ for spatial indices.

Let $v=\sqrt{2m/r}$. The proposed coframe giveswhich expands to the stated metric. Hence it is the orthonormal coframe in Painlevé–Gullstrand coordinates with signature $(-,+,+,+)$.