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In a rotating reference frame at angular frequency $\Omega$, the relevant energy difference is approximately $E_{\mathrm{kin}}-\Omega L_z$. The vortex-free state has zero circulation and zero corresponding angular momentum. For $a_0/R_r\ll 1$, the preceding shell estimates reduce toTherefore the thermodynamic vortex-nucleation frequency for the first positive quantum vortex isThe common factors $H_0,n_0$ cancel explicitly, although the trap radius and core scale still depend on number density. In the strict leading-logarithm limit, ${\Omega }_c\sim 2\hslash \log (R_r/a_0)/(m{R}_r^2)$. The finite constant $-1/2$ is the one obtained from the stated Thomas–Fermi shell integral; a resolved vortex core and readjustment of the number density at fixed particle number can change the nonlogarithmic constant. This is an energetic threshold, not a dynamical guarantee of nucleation: a surface barrier can delay entry. Higher positive winding numbers have flow energy proportional to $q^2$ and angular momentum proportional to $q$, so their analogous threshold is larger in this approximation.

Differentiate the Friedmann equation and use energy conservation to obtain $\dot{H}=-4\pi G(\rho +P/c^2)$ on the expanding branch, where $H\ne 0$. Since $\ddot{a}/a=\dot{H}+H^2$,For pressureless matter, conservation gives ${\rho }_M={\rho }_{M0}{a}^{-3}$. In terms of the specified cosmological density parameters, flatness implies ${\Omega }_M+{\Omega }_{\Lambda }=1$, andSet $b=a^{3/2}$ and take ${\Omega }_M>0$. Then $\dot{b}=\frac{3}{2}{H}_0\sqrt{{\Omega }_M+{\Omega }_{\Lambda }{b}^2}$. Integrating with the Big Bang at $t=0$ givesFor another time origin replace $t$ by $t-t_B$. The cosmological constant has effective density ${\rho }_{\Lambda }=3H_0^2{\Omega }_{\Lambda }/(8\pi G)$. Equality with the matter density means $a^3={\Omega }_M/{\Omega }_{\Lambda }$, soIf ${\Omega }_M=0$, the expanding solution is pure exponential de Sitter scale factor in flat slicing and there is no finite matter-density equality time or dust Big Bang; the stated hyperbolic-sine formula presumes a nonzero dust component.

Invert the operator definitions to obtain $a=\sqrt{m\omega /(2\hslash )}\hat{x}+i\hat{p}/\sqrt{2\hslash m\omega }$ and the adjoint formula for $a^{†}$. The canonical commutation relation yields $[a,a^{†}]=1$, with $[a,a]=[a^{†},a^{†}]=0$. Substitution givesWrite $N=a^{†}a\ge 0$. Since $[N,a]=-a$ and $[N,a^{†}]=a^{†}$, these are lowering and raising operators. Applying $a$ to a ground state would lower its energy unless it vanishes, so $a|0⟩=0$. From the normalized unique ground state constructThe commutator gives the factorial normalization and $a|n⟩=\sqrt{n}|n-1⟩$. For any energy eigenstate, repeated lowering cannot pass below $N=0$; positivity of each lowered norm forces its $N$ eigenvalue to be a nonnegative integer and lowering eventually reaches the ground state. Its uniqueness then yields the entire eigenstate ladder. The usual oscillator on the real line has discrete complete spectrum because its potential is confining, so these states span its Hilbert space.

For the perturbation $\lambda W$ with $W=\hslash \omega (a^2+a^{†2})$, nondegenerate time-independent perturbation theory gives $E_n^{\prime }=E_n+\lambda W_{nn}+{\lambda }^2{\sum }_{m\ne n}|W_{mn}{|}^2/(E_n-E_m)+O({\lambda }^3)$. The first-order term vanishes. Only $m=n-2,n+2$ contribute, with squared matrix elements $(\hslash \omega {)}^{2}n(n-1)$ and $(\hslash \omega {)}^2(n+1)(n+2)$ respectively. ThusThe absent downward transition for $n=0,1$ is automatically represented by $n(n-1)=0$. Direct substitution also givesFor $|\lambda |<1/2$ this is another oscillator with mass $m^{\prime }=m/(1-2\lambda )$ and frequency ${\omega }^{\prime }=\omega \sqrt{1-4{\lambda }^2}$. ThereforeIts expansion is $E_n[1-2{\lambda }^2+O({\lambda }^4)]$, agreeing with the perturbative result and sharpening its remainder. At the excluded endpoints confinement is lost; beyond them the quadratic Hamiltonian is not bounded below.

The microcanonical ensemble assigns equal probability to accessible quantum states in a narrow energy shell of an isolated system with fixed energy, particle number and external parameters. To derive the canonical ensemble, put a small system in weak thermal contact with a much larger isolated bath. If the small-system state has energy $E_i$, its probability is proportional to the number ${\Omega }_b(E_{\mathrm{total}}-E_i)$ of bath states. Expanding the bath entropy gives $S_b(E_{\mathrm{total}}-E_i)=S_b(E_{\mathrm{total}})-E_i/T+\cdots$, since $\partial S_b/\partial E=1/T$. Negligible interaction energy and an effectively constant bath temperature therefore give $p_i=e^{-E_i/(k_{B}T)}/Z$. The microcanonical ensemble suits an isolated fixed-energy system; the canonical ensemble suits a system exchanging energy with a thermostat at fixed temperature, with particle number still fixed.

For one atom, summing the finite geometric series over $m=-J,\dots ,J$ gives the partition functionAt $x=0$ the ratio is understood by continuity as $2J+1$. Logarithmic differentiation gives the mean magnetic dipole momentThis is odd in $x$, has small-$x$ expansion ${\mu }_{B}J(J+1)x/3+O(x^3)$, and saturates at $\pm J{\mu }_B$. The requested common graph is plotted in units of ${\mu }_B$: its ordinate is $⟨{\mu }_z⟩/(J{\mu }_B)$, so multiplication by ${\mu }_B$ gives $⟨{\mu }_z⟩/J$. The normalized slopes at zero are $(J+1)/3$ and the limits are $\pm 1$.

The magnetic susceptibility isEquivalently it is $N{\mu }_B^2\mathrm{Var}(m)/(k_{B}T)\ge 0$. For $k_{B}T\gg {\mu }_B|B|$, the apparent singular terms cancel, yielding Curie lawFor fixed nonzero $B$, $T\to 0$ and integer $J\ge 1$, the first excited magnetic level is separated from the aligned ground state by ${\mu }_B|B|$. Its exponentially small occupation givesThe moments are saturated, so changing the magnetic field hardly changes their alignment. This is a fixed-nonzero-field low-temperature limit. At exactly $B=0$, differentiating first gives the Curie expression for every $T>0$, which diverges as $T\to 0$; the two limits do not commute in this noninteracting model. For $J=0$ both moment and susceptibility vanish identically.

Put $h(r)=1-\mu /r^2$. For the equatorial geodesic, the geodesic Lagrangian is $\mathcal{L}=\frac{1}{2}[-h{\dot{t}}^2+h^{-1}{\dot{r}}^2+r^2{\dot{\varphi }}^2]$, where dots denote derivatives with respect to the affine parameter. The cyclic coordinates $t,\varphi$ give constants $p_t=-h\dot{t}=-E$ and $p_{\varphi }=r^2\dot{\varphi }=L$. A third constant is the tangent norm $g_{ab}{\dot{x}}^a{\dot{x}}^b=-\kappa$, preserved by the geodesic equation and metric compatibility. For a timelike geodesic rescale the affine parameter to proper time so $\kappa =1$; for a null geodesic $\kappa =0$.

Substitute $\dot{t}=E/h$, $\dot{\varphi }=L/r^2$ into the norm identity and multiply by $h$. It gives the effective potential equationOutside the horizon $r=\sqrt{\mu }$, its derivative is $V^{\prime }=-(L^2-\kappa \mu )/r^3+2\mu L^2/r^5$. A nontrivial circular null geodesic requires $L\ne 0$, givingFor a circular timelike geodesic,At it $V^{\prime \prime }=-4\mu L^2/r^6<0$, and its radius exceeds $\sqrt{2\mu }$. If $L^2\le \mu$ no finite circular timelike radius exists. Thus all finite circular null and timelike orbits in this five-dimensional exterior are radially unstable. The energy of a circular orbit is fixed by $E^2=2V(r)$, not independently arbitrary.

For comparison, with four-dimensional Schwarzschild horizon radius $r_s$, the photon circle is at $3r_s/2$ and is unstable. Circular timelike orbits exist for $r>3r_s/2$; they are unstable below $3r_s$, marginal at $3r_s$ and stable above $3r_s$. The five-dimensional potential has no corresponding stable outer circular timelike branch.

Away from a transmitted grazing or critical angle, ${\rho }_1{c}_1\gg {\rho }_2{c}_2$ gives a small normal acoustic impedance ratio $\delta$. ThusThe reflected energy is nearly all the incident energy, while the pressure phase depends on the plate response. Large $|\beta |$ gives $R\simeq +1$, resembling a rigid boundary; $\beta \simeq 0$ gives $R\simeq -1$, resembling a pressure-release boundary. Bulk impedance contrast alone does not ensure $\delta \ll 1$ arbitrarily close to a transmitted critical angle, since the factor $1/\cos {\theta }_2$ can be large.

For fixed frequency and finite plate parameters, $\beta$ contains the incident normal wavevector and therefore tends to zero as $\theta \to \pi /2$. If $c_1\ne c_2$, the outgoing $l_2$ tends to a nonzero real or imaginary value, whereas $l_1\to 0$, so $\delta \to 0$ too. ConsequentlyThe lower-side boundary pressure $1+R$ tends to zero, regardless of how large the fixed $m,B$ are. The angle must become closer to grazing as those parameters grow; the assertion is not uniform in an unbounded plate mass or rigidity.

If $c_1=c_2$, however, $l_2=l_1$ and $\delta ={\rho }_2/{\rho }_1$ remains constant. The exact grazing limit is $(\delta -1)/(\delta +1)$, rather than exactly $-1$ for finite contrast. Under the stated large-contrast approximation it is still arbitrarily close to the pressure-release value, with boundary pressure $2\delta /(1+\delta )$. Thus the printed pressure-release description is an approximation in that equal-speed case, not an exact zero-pressure identity. If a transmitted wave becomes evanescent, the outgoing square-root convention above continues to give the unequal-speed grazing limit.

Linearized mass conservation, momentum balance and the isentropic pressure-density relation are ${\rho }_t^{\prime }+{\rho }_0\nabla \cdot \mathbf{u}=0$, ${\rho }_0{\mathbf{u}}_t=-\nabla \tilde{p}$, and $\tilde{p}=c_0^2{\rho }^{\prime }$. Differentiate mass conservation in time and take the divergence of momentum balance to eliminate velocity. This gives the acoustic pressure wave equationFor the convention $\tilde{p}=\mathrm{Re}{e}^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}$, momentum balance gives the complex velocity amplitude $\mathbf{k}/({\rho }_0\omega )$, with $\omega =c_0|\mathbf{k}|$. Thus the incident velocity field is $\mathrm{Re}[{\mathbf{k}}_I{e}^{i({\mathbf{k}}_I\cdot \mathbf{r}-\omega t)}/({\rho }_1\omega )]$ in the lower gas.

Let $a=k_I\sin \theta$, $l_1=k_I\cos \theta >0$, $\omega =c_1{k}_I$, and $l_2=\sqrt{(\omega /c_2{)}^2-a^2}$ with nonnegative real or imaginary part. This chooses an outgoing propagating wave or an evanescent wave decaying as $y\to +\infty$. Conservation of tangential wavevector is the acoustic Snell law for elastic and acoustic waves. Write the complex pressures asInviscid kinematic matching requires equal normal velocities at the plate, not equal tangential velocities:Define the normal acoustic impedance ratio $\delta ={\rho }_2{l}_1/({\rho }_1{l}_2)$, so $T=\delta (1-R)$. For a propagating transmitted wave this is ${\rho }_2{c}_2\cos \theta /({\rho }_1{c}_1\cos {\theta }_2)$. The dynamic condition giveswhich is exactly the printed dimensionless parameter. Solving yields the pressure reflection coefficient and transmission coefficientAt $l_2=0$ use the preceding velocity equations, or take their limit, rather than substituting an infinite $\delta$. For real $l_2>0$, the reflected energy fraction is $|R{|}^2$, the transmitted fraction is $|T{|}^2/\delta$, and these sum to one. An evanescent transmitted wave carries no mean normal energy flux.

When the gases match, $\delta =1$, so $R=-i\beta /(2-i\beta )$ and $T=2/(2-i\beta )$. For a nongrazing incident wave of nonzero frequency, perfect transmission means $R=0$, equivalent toThe plate's inertial and bending terms cancel: the incident frequency and tangential wavevector coincide with its free flexural wave dispersion relation. Then $T=1$, including its phase. With positive $m,B$, this condition is possible only at angles and frequencies satisfying that relation; for normal incidence at nonzero frequency it cannot occur unless the mass is zero.

For the initially homogeneous sphere, $V_i=-3G{M}^2/(5R_i)$ and $K_i=0$. Conservation of energy and the virial theorem imply $E_f=V_f/2=V_i$, hence $V_f=2V_i$. With the usual homologous-sphere approximation, $V\propto -1/R$, soThese are the conventional collapse estimates. A final virial equilibrium is not, by itself, guaranteed to remain homogeneous: if $V_f=-{\alpha }_{f}G{M}^2/R_f$, the actual relation is $R_f/R_i={\alpha }_f/(2{\alpha }_i)$ with ${\alpha }_i=3/5$. Thus the factor of two applies exactly to the gravitational radius, and to the outer radius only under the stated structural approximation.

Here “non-interacting” means collisionless apart from gravity. Baryonic gas can collide, convert ordered motion to thermal energy, and radiate it. If radiative cooling is effective, the fixed-energy virial state loses pressure support and contracts further. Baryonic cooling can destabilize the collisionless virial state; baryons are not inevitably unstable if cooling is slow or heating and pressure support maintain equilibrium.

For a multiplication operator $A_j(x)$, $[p_i,A_j]=-i\hslash {\partial }_i{A}_j$. Therefore the mechanical momentum satisfiesThe displayed ladder operators implicitly require $qB>0$, so that their denominator is real and the second is the adjoint of the first. Then $[a,a^{†}]=1$ andThe free longitudinal motion gives a continuous spectrum in $k_z$, and the transverse Landau levels have guiding-centre degeneracy of a Landau level. For arbitrary sign of $qB$, use $|qB|$ in the energy and $a=({\pi }_x+i\mathrm{sgn}(qB){\pi }_y)/\sqrt{2\hslash |qB|}$. The question's normalizable wavefunctions likewise assume $qB>0$.

For the stated symmetric-gauge wavefunction, $|\psi {|}^2\propto r^{2M}{e}^{-qB{r}^2/(2\hslash )}$, whose peak for $M>0$ is at $r_M^2=2\hslash M/(qB)$. Thus the largest useful angular index is about $qB{R}^2/(2\hslash )$, givingThe $M=0$ state should also be included in the usual lowest Landau level; its omission by “positive integer” changes only an order-one term. Including the radial measure shifts the most probable radius to $r^2=(2M+1)\hslash /(qB)$, again only an order-one counting correction. Boundary conditions and the tails of the wavefunctions make this a large-flux estimate, not an exact finite-disc count.

The baryon-to-photon ratio is extremely small, of order $10^{-9}$. Thus even when the typical photon energy is below $I$, the high-energy tail of the blackbody radiation contains enough photons above the ionization energy to reionize almost every newly formed hydrogen atom. Cosmological recombination becomes effective only when this exponentially small tail is also small compared with the baryon-to-photon ratio. Consequently recombination occurs at a temperature much lower than $I/k$, consistent with $kT\simeq 0.03I$. The exact freeze-out history also depends on reaction rates and expansion; the equilibrium estimate explains the large suppression of the temperature.

Use the Minkowski metric ${\eta }_{mn}=\mathrm{diag}(-1,1,\dots ,1)$ and units with speed of light one. The contractions are $P^2={\eta }^{mn}{P}_m{P}_n$ and $(T{X}^{\prime }{)}^2=T^2{\eta }_{mn}(X^m{)}^{\prime }(X^n{)}^{\prime }$; $(X^m{)}^{\prime }{P}_m$ and ${\dot{X}}^m{P}_m$ pair a vector with a covector without another metric. The Lagrange multipliers impose the two first-class constraintsThey generate worldsheet diffeomorphisms, so the phase space contains both constrained directions and gauge redundancy. Two first-class constraints remove two canonical pairs, leaving $D-2$ physical degrees of freedom per point.

In Monge gauge, $X^0=t$ and $X^1=\sigma$. Write the transverse canonical variables as $\mathbf{X}=(X^2,\dots ,X^{D-1})$ and $\mathbf{P}=(P_2,\dots ,P_{D-1})$. Solving the first-class constraints givesThe negative root selects positive energy. Substitution into the phase-space action gives the Hamiltonian reductionFor a static segment, $\mathbf{P}=0$, its proper length element is $d\ell =\sqrt{1+|{\mathbf{X}}^{\prime }{|}^2}d\sigma$, and $\mathcal{E}d\sigma =Td\ell$. Thus the string tension is the rest energy per unit proper length. In particular a straight resting segment has $\mathcal{E}=T$. Monge gauge is a local choice on a string embedding map for which $X^1$ is a valid coordinate; it need not cover folded strings or all endpoint configurations.