Take the screen trace of the optical matrix of a null congruence equation. The cross-contractions of symmetric null shear with antisymmetric null twist vanish, andThe minus sign in the null twist contribution follows from tracing the square of an antisymmetric matrix. The tidal trace is $P^{ef}{R}_{cedf}{U}^c{U}^d=R_{cd}{U}^c{U}^d$: the additional null terms in $P^{ef}$ vanish by curvature antisymmetry. Thus the Null Raychaudhuri equation isIn particular, null shear focuses the null geodesic congruence while null twist enters with the opposite sign. With vanishing null twist and the null energy condition, the Einstein field equations imply $U\cdot \nabla \theta \le -{\theta }^2/2$. These signs and the factor $1/2$ depend on using the full-trace null expansion in a two-dimensional screen.

First let $B_{ab}={\nabla }_b{U}_a$ without projection. Commute derivatives on this covector and use affine geodesic motion:The last equality relabels the two contracted null-vector indices. Nullness and affine parametrization give $U^a{B}_{ab}=0$ and $B_{ab}{U}^b=0$. Therefore inserting $P$ into the contracted middle index changes no screen-projected quadratic term. Since ${\nabla }_{U}P=0$, screen projection proves the Sachs optical equationsThe optical tidal matrix ${\mathcal{R}}_{ab}$ is symmetric by the Riemann curvature tensor's pair symmetries. Raising one screen index gives the corresponding equation for ${\hat{B}}^a{}_b$.

In terms of the defined optical scalars, the equation is explicitlyThis matrix identity contains the null expansion, null shear and null twist evolution equations as its trace, symmetric trace-free part and antisymmetric part.

Take the first centre at the origin. If some listed centres coincide, first combine their poles; let $M$ be the total residue at the origin, so $M=M_1$ when the centres are distinct. The contributions from centres away from the origin are real analytic there. Their constant term iswhere $F$ is analytic in $r$ and in local angular charts. This follows by substituting $\mathbf{x}=r\mathbf{n}(\theta ,\varphi )$ in the ordinary Taylor expansions of the regular contributions. The constant $h$ is independent of angle, which makes the hinted coordinate differential integrable.

Put $A=M+hr$, $Q=rH=A+r^{2}F$, and defineinitially for $r>0$. Substitution into the metric givesThe apparently singular radial coefficient is actuallysince $Q-A=r^{2}F$. All coefficients therefore extend analytically to negative and positive $r$ near zero, with $Q(0)=M>0$. At the horizon $g_{vv}=0$, $g_{vr}=1$, $g_{rr}=4MF(0,\theta ,\varphi )$ and the angular metric is $M^{2}d{\Omega }_2^2$. The $v,r$ determinant is $-1$ both before and at the extension, so the metric remains nondegenerate and Lorentzian. Angular coordinate singularities are handled by ordinary sphere charts. This explicitly constructs the analytic extension of a Majumdar–Papapetrou horizon.

Inverting the radial block gives $g^{rr}=r^2/Q^2$, which vanishes at $r=0$. Thus the hypersurface is null. Its normal agrees there with the metric dual of $\xi ={\partial }_v$: ${\xi }_{a}d{x}^a=dr$ on the horizon. The metric is independent of $v$, so $\xi$ is a Killing vector field and the null hypersurface is a Killing horizon.

The surface gravity obeys ${\nabla }_a({\xi }^2)=-2\kappa {\xi }_a$ on a Killing horizon. Here ${\xi }^2=-r^2/Q^2$ has a double zero and its entire differential vanishes at $r=0$, while ${\xi }_a$ is the nonzero normal. ConsequentlyThe horizon is degenerate. Its cross-sectional area is $4\pi M^2$, independent of the centres away from the origin. The essential cancellation would fail with an arbitrary constant in place of the displayed regular part $h$.

Use geometrized units $G=c=1$. For a stationary asymptotically flat spacetime, normalize the timelike Killing vector field $\xi$ to have squared norm $-1$ at infinity. The Komar mass isHere the oriented binormal is chosen so that an ordinary positive-mass static solution gives a positive integral. Equivalently, on a static slice with lapse $N=\sqrt{-{\xi }^2}$ and outward spatial unit normal $s^i$,This second expression fixes the orientation unambiguously and is convenient for the Majumdar–Papapetrou solution.

The spatial metric is ${\gamma }_{ij}=H^2{\delta }_{ij}$ and $N=H^{-1}$. On a large coordinate sphere, $s^i=H^{-1}{\hat{r}}^i$ and $dA=H^2{r}^{2}d\Omega$. ThusThe original PDF has the Euclidean norm in each source denominator, which is missing in the TeX aid. With that norm, the harmonic function has asymptoticsThereforeIf $M_i$ are retained as length parameters while $G$ is restored, the physical mass is ${\sum }_i{M}_i/G$. This is the charge at infinity. A Komar integral is not generally surface-independent here, since Einstein-Maxwell theory has the electromagnetic stress-energy tensor between the surfaces; in particular it should not be confused with a sum of vacuum horizon Komar charges.

A larger baryon-to-photon ratio favours bound nuclei at a fixed temperature through the equilibrium factor ${\eta }^{A-1}$. In particular it helps deuterium survive photodissociation sooner: there are fewer energetic photons per baryon, and nuclear reaction rates per baryon also rise. A schematic deuterium bottleneck condition is $\eta e^{B_D/T_B}\sim \mathrm{constant}$, suppressing slowly varying powers of temperature. It givesThus higher $\eta$ permits burning at a higher temperature. During radiation domination that means earlier cosmic time, $t_B\propto T_B^{-2}$, leaving less time for free-neutron decay after freeze-out. More neutrons enter helium, so the final helium-4 mass fraction increases, ordinarily rather weakly. This is the baryon-density effect on primordial helium.

The equilibrium helium factor ${\eta }^3$ indicates the direction of the change but is not a prediction that the final helium abundance grows without bound as ${\eta }^3$. Once almost all surviving neutrons are captured, the final primordial helium mass fraction is limited by their supply, as the shared neutron budget shows.

A longer neutron half-life means a larger mean lifetime ${\tau }_n={\tau }_{1/2}/\ln 2$. Holding the initial neutron-to-proton ratio and the nuclear-burning delay fixed, the derivative of the neutron survival factor is positive:Thus fewer neutrons disappear during the deuterium bottleneck, and the final helium-4 mass fraction increases.

There can be a second effect if the increased lifetime represents a uniform reduction of the relevant weak interaction strength. With fixed masses and other couplings, neutron decay and the conversion reactions both scale as $G_F^2$. Thus ${\tau }_n\propto G_F^{-2}$, whereas $T_F\propto G_F^{-2/3}\propto {\tau }_n^{1/3}$. Weaker conversion then freezes out earlier and increases $r_F$, reinforcing the survival effect. The survival argument alone already establishes the requested neutron half-life effect on primordial helium; linking lifetime to all conversion rates additionally requires this microscopic assumption.

Extra relativistic neutrino species increase the effective number of relativistic energy degrees of freedom. At fixed temperature the Hubble parameter rises as $g_{\ast }^{1/2}$, while the standard conversion rate is unchanged. Consequently cosmological weak freeze-out moves to higher temperature, $T_F\propto g_{\ast }^{1/6}$, and the neutron-to-proton ratio $r_F=e^{-Q/T_F}$ increases. The initial neutron fraction $r_F/(1+r_F)$ therefore increases as well.

The quicker expansion also reduces the cosmic time required to cool to the deuterium bottleneck temperature: during radiation domination, $t\propto g_{\ast }^{-1/2}{T}^{-2}$ while the thermal count is roughly constant. Fewer neutrons decay before burning. Both factors in the neutron budget increase, giving a larger final helium-4 mass fraction. This is the primordial helium response to extra relativistic species, under the assumption that the additional species contribute mainly to expansion rather than independently changing the conversion rates.

For actual final density operator ${\rho }_f(t_F)$ and target ${\rho }_d$, minimizeThis Hilbert-Schmidt distance is nonnegative and is zero exactly at the target, for pure or mixed states. The subscript $f$ denotes evolution under the chosen control; $t_F$ is fixed. A simple overlap $\mathrm{Tr}({\rho }_d{\rho }_f)$ need not have its maximum at a general mixed target, whereas this distance has the required equality condition. For purely Quantum Hamiltonian dynamics, the target must lie on the initial state's unitary orbit of a density operator to attain zero error; otherwise the constrained optimization finds the closest accessible state instead.

For target unitary operator $U_T$ and actual propagator $U_f(T)$ on an $n$-dimensional space, minimize the phase-insensitive unitary gate errorThe Cauchy-Schwarz inequality for the Hilbert-Schmidt inner product gives a value in $[0,1]$, because both unitary matrices have squared norm $n$. The error is zero exactly when the matrices are proportional, namely $U_f(T)=e^{i\varphi }{U}_T$. It therefore tests the full process, not just its action on one state, and ignores a physically irrelevant common phase. If the gate phase itself must be fixed, instead minimize $\Vert U_f(T)-U_T{\Vert }_{\mathrm{HS}}^2/(2n)=1-\mathrm{Re}\mathrm{Tr}(U_T^{†}{U}_f(T))/n$, which vanishes only at exact equality.

For a normalized state $|\psi (t_F;f)⟩$ driven by the admissible control $f$, maximizeHere $A$ is the specified Hermitian operator, $t_F$ the final time, and the dependence on $f$ is through the Schrödinger equation. The objective is real and is exactly the desired expectation value, so increasing it increases the requested observable. A control-resource penalty may be included when the admissible set does not already constrain field strength or fluence.

The adiabatic dark state of a driven Hamiltonian has exactly zero component on the lossy intermediate level throughout the ideal passage. It therefore suppresses spontaneous emission, whereas the two-pulse route explicitly transfers the population into $|2⟩$ before moving it to $|3⟩$. For an intermediate-level decay rate $\Gamma$, the leading loss probability is controlled by $\Gamma \int P_2(t)dt$. In stimulated Raman adiabatic passage, nonadiabatic leakage makes $P_2$ of order $(\hslash \dot{\theta }/\Omega {)}^2$, rather than the order-one intermediate population of the sequential swaps. Transfer is also less sensitive to precise pulse areas, provided the ordering, gap and slowness requirements hold. Making the process slower indefinitely is not always beneficial if the outer states also decohere; their coherence must survive the total passage duration.

For real nonnegative envelopes, put $\Omega =\sqrt{{\Omega }_1^2+{\Omega }_2^2}$ and choose a continuous $\theta =atan2({\Omega }_1,{\Omega }_2)$. The state $|D⟩=\cos \theta |1⟩-\sin \theta |3⟩$ is dark because its two amplitudes feeding $|2⟩$ cancel: ${\Omega }_1\cos \theta -{\Omega }_2\sin \theta =0$. The orthogonal bright combination $|B⟩=\sin \theta |1⟩+\cos \theta |3⟩$ couples to $|2⟩$ with strength $\Omega$. Thus the other eigenstates are $(|B⟩\pm |2⟩)/\sqrt{2}$ with energies $\pm \Omega$.

Start with ${\Omega }_1=0$ and ${\Omega }_2\ne 0$, so $\theta =0$ and $|D⟩=|1⟩$. Slowly increase ${\Omega }_1/{\Omega }_2$ to infinity while maintaining a nonzero gap; end with ${\Omega }_2=0$ and ${\Omega }_1\ne 0$, so $|D⟩=-|3⟩$. This target-to-intermediate pulse first is the counterintuitive ordering of stimulated Raman adiabatic passage. Its quantitative slowness condition isIt follows from $|⟨\pm |\dot{D}⟩|=|\dot{\theta }|/\sqrt{2}$ and the gap $\Omega$. Avoid switching both envelopes off while $\theta$ is changing; switch off together only after the transfer angle has become stationary. The dark eigenvalue is zero, so there is no dynamical phase; with this real eigenvector $⟨D|\dot{D}⟩=0$ as well. The ideal adiabatic quantum control therefore transfers $|1⟩$ to $-|3⟩$, physically the same target ray.

Using the EM for merged multinomial cells E-step,Applying the M-step maximum-likelihood estimator to the filled counts gives

Initialize $0<{\theta }^{(0)}<1$. At each iteration first compute $y_3^{(t)}=x_3{\theta }^{(t)}/(2+{\theta }^{(t)})$ by the conditional expectation above. The M-step applies the complete-data maximum-likelihood estimator to the filled counts, givingRepeat until convergence. If all counts are zero, the likelihood function is constant and any parameter can be retained. For a nonempty sample an interior start makes the displayed denominator positive.

Here the three observed cell probabilities are $(1-\theta )/2$, $\theta /4$ and $(2+\theta )/4$, so the observed log-likelihood isFor a nonempty sample its second derivative on $(0,1)$ isThus the maximum-likelihood estimator is unique, allowing a boundary maximum. To verify that the iteration actually finds it, let $u(\theta )=x_3\theta /(2+\theta )$ and $F(\theta )=(x_2+u(\theta ))/(x_1+x_2+u(\theta ))$. Then $F^{\prime }(\theta )=x_1{u}^{\prime }(\theta )/(x_1+x_2+u(\theta ){)}^2\ge 0$, andIf the maximum is interior, this identity moves each iterate towards it, and monotonicity of $F$ prevents crossing it. The bounded monotone iterates converge; continuity makes their limit a fixed point, and strict concavity makes that fixed point the maximum. If the maximum is at zero or one, the sign of ${\ell }^{\prime }$ gives the same monotone convergence towards the appropriate endpoint. In the special case $x_1=0$, $F$ is identically one whenever its denominator is positive, so the maximum is reached in one step. An interior initialization avoids the possible spurious zero fixed point when $x_2=0$; starting at zero can otherwise trap the iteration even when the true maximum is positive.

Within the merged third observed cell, the probability of belonging to the original fourth cell isThe allocation therefore has the conditional binomial distributionIts conditional expectation and the filled-in missing counts arewith $y_1^{(t)}=x_1$ and $y_2^{(t)}=x_2$.

Taking the conditional expectation of the complete log-likelihood gives the EM for merged multinomial cells E-step:This is the parameter-dependent part of $\log L(y^{(t)}\mid \theta )$. More precisely, $Q(\theta \mid {\theta }^{(t)})=\log L(y^{(t)}\mid \theta )+C_t^{\prime }$, where the constant is independent of the candidate $\theta$, interpreting factorials of noninteger filled counts through the gamma function. The expected log-factorials are not generally the log-factorials of expected counts, so literal equality including the normalization would be false. Equality up to this constant is exactly what the M-step needs. Filling in conditional means works here because the candidate-parameter terms are linear in the missing counts; it is not a universal rule for the expectation-maximization algorithm.

The parameter-dependent part of the complete multinomial distribution log-likelihood isThe remaining probability factors and multinomial coefficient are constant in $\theta$. If $y_1+y_2+y_3>0$, differentiating gives $-y_1/(1-\theta )+(y_2+y_3)/\theta$, whose zero isThe log-likelihood is concave. If $y_2+y_3=0<y_1$, its maximum is at zero; if $y_1=0<y_2+y_3$, its maximum is at one. These boundary cases are included by the same formula. If $y_1=y_2=y_3=0$, the data lie entirely in the parameter-independent fourth cell and every $\theta \in [0,1]$ is a maximum-likelihood estimator.

Reversible-jump Markov chain Monte Carlo requires the reciprocal density/proposal ratio for the matched reverse transformation, so the death acceptance probability isExpressed directly in the current residual $r^{\prime }=y-X_{k+1}{\beta }^{\prime }$ and the recovered lower-order parameters, this isIndeed $r=r^{\prime }+zv$ makes this exponent the negative of the birth exponent, and all prefactors invert. Thus $R_{-}{R}_{+}=1$ for matched states. This reciprocity, including the normalized prior distributions and auxiliary proposal density, gives the required detailed balance between models.

For the reverse centered polynomial birth move in reversible-jump sampling, recover $z={\beta }_{k+1}^{\prime }$ from the current coefficient vector. Remove that highest-order coefficient and undo the intercept shift:No auxiliary random number is drawn for this death move: the removed coefficient supplies the auxiliary value needed to reconstruct the birth move. The reverse Jacobian determinant is also one. The inverse prediction change is $-zv$, hence the residual in the lower-order model is $r=r^{\prime }+zv$.

A reversible-jump Markov chain Monte Carlo calculation compares normalized model-specific densities. The fixed-order posterior written only up to proportionality in part (a) omits normalizing factors that depend on $k$ and therefore cannot simply be canceled between dimensions. Moreover, the printed question does not specify the model-order prior distribution or the probabilities of selecting birth and death moves. Denote those by $p_k$, $b_k$ and $d_k$, respectively, retaining them in the answer.

Put $s={\sigma }^2$, $c=n^{-1}{\sum }_i{x}_i^{k+1}$, $v_i=x_i^{k+1}-c$, and $r=y-X_k\beta$. The centered polynomial birth move in reversible-jump sampling changes the fitted vector by $zv$, so its new residual is $r^{\prime }=r-zv$. ThereforeThe map from $({\beta }_0,\dots ,{\beta }_k,z)$ to $({\beta }_0^{\prime },\dots ,{\beta }_{k+1}^{\prime })$ has ones down the diagonal and a single additional entry $-c$ in the intercept row. Its Jacobian determinant is one.

Write $Q_k=(\beta -{\mu }_k{)}^T{\Sigma }_k^{-1}(\beta -{\mu }_k)$ and $Q_{k+1}=({\beta }^{\prime }-{\mu }_{k+1}{)}^T{\Sigma }_{k+1}^{-1}({\beta }^{\prime }-{\mu }_{k+1})$. The normalized multivariate normal distribution prior ratio isDivide also by the forward proposal density $q(z)=(2\pi {\sigma }_{\beta }^2{)}^{-1/2}{e}^{-z^2/(2{\sigma }_{\beta }^2)}$. The common inverse-gamma distribution prior for $s$ and the common normal-likelihood normalization cancel, since $s$ stays fixed. The forward acceptance ratio is consequentlyHere ${\sigma }_{\beta }>0$ is the proposal standard deviation. If equal model priors and symmetric birth/death selection are explicitly adopted, their ratios become one; they are not determined by the PDF. The expression assumes the same hyperparameters for the variance prior in both models, as in part (a). A dimension-dependent variance prior would contribute its own ratio.

Choose an initial $s^{(0)}>0$. At each iteration, evaluate $V$ and $m$ using $s^{(t)}$ and draw ${\beta }^{(t+1)}$ from the conditional multivariate normal distribution just derived. Next compute $b_t=b+\Vert y-X_k{\beta }^{(t+1)}{\Vert }^2/2$ and drawFor example, generate $G\sim \Gamma (a+n/2,\text{rate }{b}_t)$ and set $s^{(t+1)}=1/G$. A draw of the conditional multivariate normal distribution can be formed as $m+LZ$ with $L{L}^T=V$ a Cholesky decomposition and $Z$ a vector of independent standard normal distributions.

This two-block Gibbs sampler alternates the full conditional distributions of the independent normal and inverse-gamma regression priors posterior. The conditional-integration argument proves that its invariant distribution is the joint posterior distribution. Under the usual convergence conditions, retain the pairs $({\beta }^{(t)},s^{(t)})$ after an initial transient as the requested dependent posterior sample. The new variance update uses the newly drawn coefficients, rather than the previous coefficients.