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For the Rayleigh-Jeans spectrum of a finite blackbody disk, the Rayleigh-Jeans law replaces $(e^{h\nu /(kT)}-1{)}^{-1}$ by $kT/(h\nu )$. Here $h$ is the Planck constant and $k$ the Boltzmann constant. The multitemperature blackbody disk therefore hasFor example, setting $R=r_{\mathrm{out}}/r_{\ast }$ makes its frequency-independent coefficient proportional to the finite dimensionless integralAlthough the exact effective temperature vanishes at the inner edge, the very narrow cold rim where the Rayleigh-Jeans law fails makes a negligible contribution in this limit. More formally, after dividing the integrand by ${\nu }^2$, the inequality $e^x-1\ge x$ bounds it by $(k/h)rT$, so dominated convergence justifies the result even at that edge.

At large radius, $T\propto r^{-3/4}$, and the contribution per logarithmic interval is $r^{2}T\propto r^{5/4}$. The outer disk dominates the low-frequency emission because its much greater area outweighs its lower effective temperature.

For intermediate frequencies use the allowed power-law approximation to the effective temperature and introduceThensoThe lower limit is much smaller than one and the upper limit much larger than one. Extending them to zero and infinity leaves a constant: near zero the integrand behaves as $x^{2/3}$, and at infinity it decays exponentially. Hence the intermediate spectrum isThis is the one-third spectrum of a multitemperature disk. Much of the emission comes from radii where $kT(r)$ is of order $h\nu$, moving inward as the frequency rises. With the exact inner-edge profile, a broad intermediate interval also requires frequency well below $k{T}_{\max }/h$; the supplied approximation captures its slope away from the hottest annuli.

For a slowly rotating central star, the accretion-disk boundary layer radiates a luminosity $L_{\mathrm{BL}}\sim G{M}_{\ast }\dot{M}/(2r_{\ast })$. A layer of thickness comparable to the disk scale height has emitting area $A_{\mathrm{BL}}\sim r_{\ast }H$ times a geometric constant. Applying the Stefan–Boltzmann law to this blackbody area givesComparing with $\sigma T_{\mathrm{in}}^4=3G{M}_{\ast }\dot{M}/(8\pi r_{\ast }^3)$ gives the boundary-layer temperature scalingOnly the scaling is fixed: for a two-faced annulus of area $4\pi r_{\ast }H$, for example, $T_{\mathrm{BL}}^4/T_{\mathrm{in}}^4=r_{\ast }/(3H)$. A surface belt gives a different order-one coefficient.

For a thin disk, $H/r_{\ast }\ll 1$, so a comparable luminosity emerges from a smaller emitting area at a higher effective temperature. The boundary-layer emission is harder than the disk emission, with its Planck law peak shifted to higher frequency. It is also closer to a single-temperature component than the broad multitemperature blackbody disk spectrum, under the adopted uniform-temperature approximation. Rapid stellar rotation weakens the heating and this conclusion's temperature contrast.

For a local axisymmetric Fourier mode, the Toomre stability criterion balances three contributions to the squared oscillation frequency:The radial epicyclic frequency supplies rotational restoration at long wavelengths; isothermal sound speed and pressure stabilize short wavelengths; disk self-gravity destabilizes intermediate wavelengths. Minimizing over $|k|$ gives $|k|=\pi G{\Sigma }_0/c_s^2$ and ${\omega }_{\min }^2={\kappa }^2(1-Q^{-2})$. Axisymmetric gravitational instability occurs when $Q<1$.

In a centrally dominated Keplerian disk, $\kappa =\Omega$ and vertical hydrostatic equilibrium gives $c_s\simeq H\Omega$. Using ${\Omega }^2=G{M}_{\ast }/r^3$ and a local disk-mass estimate $M_D\sim \pi r^2{\Sigma }_0$,This is the disk mass form of the Toomre criterion. An actual enclosed mass depends on the radial surface density profile and changes an order-one coefficient. In particular, a profile proportional to $r^{-3/2}$ has $M_D(<r)\simeq 4\pi r^2{\Sigma }_0$ when its inner cutoff is negligible.

For the protosolar estimate, take $M_{\ast }\simeq M_{\odot }\simeq 2\times 10^{33}\mathrm{g}$. This solar mass is implicit in identifying the central star with the young Sun. The supplied constant disk aspect ratio givesUsing the printed approximate astronomical unit and gravitational constant, one can also obtain $\Omega (1\mathrm{AU})\simeq 3.7\times 10^{-7}{\mathrm{s}}^{-1}$ and $c_s\simeq 3.7\times 10^5{\mathrm{cm}\mathrm{s}}^{-1}$; they give the same $Q$. The gravitational constant cancels from the mass form.

At one astronomical unit the disk is very stable, with $Q\simeq 640$. Formal extrapolation givesThis is far beyond the planetary region and any plausible extent of the minimum-mass solar nebula. Moreover, the extrapolated enclosed disk mass is already a substantial fraction of a solar mass, so the centrally dominated approximation becomes questionable. The formal radius is not a prediction of a real unstable outer nebula. Direct gas fragmentation by gravitational instability is unlikely to have formed Solar System planets in this model. Core accretion is the more natural route; an earlier substantially more massive or colder disk would be a different model. Even $Q<1$ alone does not guarantee fragmentation, because sufficiently rapid cooling is also needed.

For a two-dimensional internal gravity wave with fixed horizontal wave number $k\ne 0$, the vertical velocity amplitude satisfiesThis follows from the Linearized Boussinesq equations just as in part (a), allowing the background buoyancy frequency to vary. The WKB approximation gives locallyIt requires slowly varying stratification compared with the local vertical wavelength, in particular $|m_z|\ll m^2$, and a small-amplitude wave in the inviscid, nonrotating regime. Propagation requires $N>|\omega |$; where $N<|\omega |$, the local vertical wave number is imaginary and the solution is evanescent. Turning levels and abrupt changes need a separate connection calculation. Finite viscosity, background flow, rotation or nonlinear effects require modifications of this model.

The ray tracing equations use the local dispersion relation as a Hamiltonian:The stationary, horizontally uniform background conserves the frequency and horizontal wave number. In two dimensions the group velocity gives $dx/dz=-m/k$. For the specified buoyancy frequency and $\omega =1$, $m^2=256k^2{z}^2$. Hence for $z>0$ the two geometric internal-wave rays through the specified point areEach curve has both directed propagation senses, giving four local directed rays. Their formal intersections with $z=0$ are $x=-8$ and $x=8$.

At $z=0$, $N=\omega$ and $m=0$: this is a double-zero internal-wave turning level, since $m^2$ touches zero without changing sign. The ray becomes vertical geometrically, but its group velocity tends to zero. Taking $k>0$ without loss of generality, the downward branch hasThe formal travel time diverges logarithmically as $z\to 0$. Meanwhile $|m_z|/m^2=1/(16|k|z^2)$ grows, so the WKB approximation fails before that limit; its predicted $|m{|}^{-1/2}$ amplitude cannot be extrapolated to infinity. The spatial ray reaches a finite limiting position, but finite-wavelength behaviour near the zero must be found beyond WKB. There is no evanescent half-space supplied by this particular $N(z)$, so ordinary simple-turning-point reflection cannot be assumed solely from the local ray construction.

Use the spatial internal-wave ray tracing model, with $dx/dz=\pm 16z$, and reflect the rays so that the frequency and tangential wave number are preserved. At a vertical wall the horizontal energy direction reverses; at a horizontal wall the vertical energy direction reverses. On the sloping wall the reflection depends on internal-wave slope criticality. The three-reflection orbit below encounters that wall at a height above $3/16$, so the wall reverses the horizontal energy direction there.

First take the packet to leave the bottom upward and to the left, with $0<x_0<6$. The four segments of the loop areThe left-wall reflection is at $z=\sqrt{x_0/8}$ and the top reflection at $x=8-x_0$. The slope intersection solves $8z_s^2+3z_s+x_0-10=0$, giving the internal-wave ray return mapThus the next formal bottom reflection is at $(R(x_0),0)$. On $0<x_0<6$, the slope heights are between approximately $0.544$ and $0.946$, consistent with the assumed reflection type.

An internal-wave attractor is a stable periodic ray, so solve $R(x_{\ast })=x_{\ast }$. This gives $z_s=2/3$, and thereforeThe successive vertices areThis is clockwise circulation in the $(x,z)$ plane. Differentiating the internal-wave ray return map showsNeighbouring bottom intersections converge to the periodic orbit, geometrically focusing the energy beam.

For completeness, the opposite upward launch direction has the reversed three-reflection itinerary: slope, top, left. In the interval where this itinerary is valid, putThe three-reflection reverse map is $R^{-1}$, with domain $R(0)<x_0<R(6)$. Its fixed point is the same, but $R_{\mathrm{rev}}^{\prime }(x_{\ast })=41/23>1$. Counterclockwise packets defocus from that orbit. Except for the exactly periodic reverse ray, they eventually leave this itinerary. In the full geometric reflection construction, omitted-side or additional-slope reflections redirect generic rays into the clockwise attracting branch. Tangency and exactly critical reflections are singular exceptions requiring finite-wavelength or dissipative treatment.

The bottom reflection is a formal spatial-ray idealization: part (b) shows that the WKB approximation itself fails near $z=0$, so the diagram does not imply arrival and reflection there in finite WKB travel time. In a real fluid, viscous attenuation of an internal-wave beam, finite wavelength, internal-wave breaking and ensuing turbulence and mixing limit the beam width and energy density. Unlimited focusing is a property of the ideal geometric model, not a prediction for the resolved physical wave.

Define outgoing Eddington-Finkelstein coordinates by $u=t-r_{\ast }$. The radial Schwarzschild metric becomesOutgoing radio rays have constant $u$. Two nearby wave crests emitted at Alice's proper times $\tau$ and $\tau +d\tau$ have separation $du={\dot{u}}_{A}d\tau$. At Bob's fixed radius $r_B$, $d{t}_B=du$ and $d{\tau }_B=\sqrt{f_B}du$, where $f_B=1-2M/r_B$. Thus the exact local redshift isHere the wavelength ratio equals the proper time period ratio in geometric optics. Equivalently, the phase gradient $k_a=-{\omega }_{\infty }(du{)}_a$ gives Alice's measured frequency ${\omega }_A=-k_a{U}_A^a={\omega }_{\infty }{\dot{u}}_A$ and Bob's ${\omega }_B={\omega }_{\infty }/\sqrt{f_B}$.

In the asymptotically distant-observer limit $f_B\to 1$, this becomes ${\lambda }_B/{\lambda }_A=d{u}_A/d{\tau }_A$. “Far away” is the approximation behind the printed formula; at finite $r_B$ the factor $\sqrt{f_B}$ remains.

For nonzero $\omega$, the equilibrium points satisfy $\sin 2\theta =0$. The equilibrium point $\theta =0$ is stable: it puts the bead at the point farthest from the rotation axis. Writing $\theta =\eta$ small in the tangential ordinary differential equation givesThus the small oscillation is simple harmonic motion, with angular frequency $|\omega |$ and ordinary frequency $|\omega |/(2\pi )$.

The endpoints $\theta =\pm \pi /2$ represent the same rim point $O$. A perturbation there instead satisfies $\ddot{\eta }=+{\omega }^2\eta$ and is unstable. Equivalently, the centrifugal potential along the hoop is $-\frac{1}{2}{\omega }^2|\mathbf{r}{|}^2=-2a^2{\omega }^2{\cos }^2\theta$, whose minimum is at zero and maximum at $O$. If $\omega =0$, every bead position with zero relative velocity is an equilibrium point, but perturbing that velocity produces unrestored drift; there is no nonzero restoring frequency.

Write $\zeta =(\nabla \times \mathbf{u})\cdot {\mathbf{e}}_z$ and $q=\zeta -f\eta /h_0$, so the PDF's vector potential vorticity is $\mathbf{Q}=q{\mathbf{e}}_z$. The local TeX loses some of this boldface. Taking the vertical curl of the momentum equation gives ${\zeta }_t=-f\nabla \cdot \mathbf{u}$, while mass conservation gives ${\eta }_t=-h_0\nabla \cdot \mathbf{u}$. ThusThis is the linear perturbation of shallow-water potential vorticity, scaled by $h_0$; it should not be confused with the full nonlinear conserved ratio $(f+\zeta )/h$.

Since $\nabla \cdot ({\mathbf{e}}_z\times \mathbf{u})=-\zeta$, taking the divergence of momentum gives $(\nabla \cdot \mathbf{u}{)}_t=f\zeta -g{\nabla }^2\eta$. Differentiate mass conservation and use $\zeta =q+f\eta /h_0$ to obtainFor $\mathbf{Q}=0$, substitution of the stated plane wave gives the linear rotating shallow-water dispersion relation, with $c=\sqrt{g{h}_0}$:The two even branches approach $\pm c|k|$ at large $|k|$ and have values $\pm |f|$ at $k=0$. The original dispersion diagram below also compares the speed magnitudes.

For $k\ne 0$, the phase velocity and group velocity magnitudes areThe printed convention $e^{i(kx+\omega t)}$ means the signed phase velocity is $-\omega /k$ and the signed group velocity is $-d\omega /dk$. The speed magnitudes above are independent of that sign convention.

For $f\ne 0$, longer wavelengths have greater phase velocity but smaller group velocity: individual crests and wave packets therefore answer “faster or slower” differently. At long wavelength, $|k|\ll |f|/c$, the frequency is nearly the inertial frequency $|f|$, $|c_p|\sim |f|/|k|$, and $|c_g|\sim c^2|k|/|f|$. Rotation has its largest relative effect here, and the waves are strongly dispersive. At short wavelength, $|k|\gg |f|/c$, both speeds tend to $c$ and rotation gives only a small, weakly dispersive correction. If $f=0$, the gravity waves are nondispersive, with speed $c$; at $k=0$ the phase velocity formula is undefined.

The central stratified layer supports upward and downward internal gravity waves, whose counterpropagating components form a standing wave. The unstratified regions support evanescent waves, and the upper disturbance decays as $e^{-k(z-2H)}$. Consequently the central layer is a stratified internal-wave guide, rather than a source of propagating energy at infinity.

The largest response occurs at its trapped normal mode frequencies. They can be specified without solving the forced problem. Set the boundary forcing to zero to find a free normal mode. The lower solution is proportional to $\sinh (kz)$, so continuity of fluid pressure and vertical velocity givesIn the central layer writeThe upper Robin boundary condition then gives the exact trapped-mode conditionThe left side increases strictly from zero to infinity, so there is one positive root for each $n$; the frequencies accumulate at zero. This is constructive phase matching after reflection at both ends. The phase shifts from the evanescent waves matter: simply imposing integer half-wavelengths across the stratified layer is generally incorrect.

In the ideal inviscid model, exact resonant forcing has no bounded steady harmonic solution. The undamped normal mode grows secularly under sustained forcing. Weak viscosity or other losses would produce large finite peaks near the displayed frequencies; the linear approximation eventually fails if the disturbance becomes too large.

The illustrated free normal mode has an oscillatory central region and evanescent wave tails. Its lower tail reaches the fixed zero-displacement boundary; its upper tail decays to infinity.

Measure $x_1,x_2$ as horizontal displacements from the respective downward equilibrium positions; the absolute horizontal coordinates are $x_1$ and $L+x_2$. Each vertical displacement is of second order in its small angle. Thus the spring is horizontal to first order, its extension is $x_2-x_1$, and Hooke's law gives horizontal forces $k(x_2-x_1)$ and $-k(x_2-x_1)$. Combining these with the linearized simple pendulum restoring force yieldsFor the normal modes of two spring-coupled equal pendulums, add and subtract these ordinary differential equations. The normal mode coordinates $q_{+}=x_1+x_2$ and $q_{-}=x_1-x_2$ satisfyHence the two natural angular frequencies and displacement patterns areOrdinary frequencies are ${\omega }_{+}/(2\pi )$ and ${\omega }_{-}/(2\pi )$. In the first normal mode, both particles swing together with equal amplitudes and the spring is unextended to first order. In the second normal mode, the particles swing in opposite directions with equal amplitudes, so the spring adds a restoring force and increases the angular frequency.

The small-angle approximation $\sin \theta =\theta +O({\theta }^3)$ reduces ii to $\ddot{\theta }+{\omega }^2\theta =0$, whereThe coefficient is positive, so the solution is simple harmonic motion, $\theta (t)=C\cos \omega t+D\sin \omega t$. Here $\omega$ is the angular frequency; the ordinary frequency and period are

The lowest positive root belongs to the symmetric family and lies in $(0,\pi /2)$; all antisymmetric roots are at least $\pi$. Let $\alpha =m/(2\rho \ell )\gg 1$. The equation $\cot z=\alpha z$ forces its first root to tend to zero: for any fixed $\epsilon >0$, the right side eventually exceeds $\cot \epsilon$ at $z=\epsilon$, so the decreasing-minus-increasing difference has its zero below $\epsilon$.

The Taylor series $\cot z=z^{-1}-z/3+O(z^3)$ then givesSince $\omega =cz/\ell$ and $c=\sqrt{T/\rho }$, the lowest angular frequency isPhysically, at such a low frequency each half-string is nearly straight. If the mass moves up by $Y$, the two slopes are approximately $Y/\ell$ and $-Y/\ell$, so the restoring force is $-2TY/\ell$. The resulting harmonic oscillator equation, $m\ddot{Y}+(2T/\ell )Y=0$, has exactly the leading angular frequency above. The next correction from the same expansion is $\omega /{\omega }_0=1-\rho \ell /(3m)+O((\rho \ell /m{)}^2)$.

For $c>0$ and $l>0$, the zero-endpoint Dirichlet boundary conditions select the Fourier sine basis functions $\sin (n\pi x/l)$. Separation of variables gives temporal factors with angular frequencies ${\omega }_n=n\pi c/l$. Zero initial displacement removes the temporal cosine terms, givingDifferentiating at $t=0$ and using Fourier orthogonality givesThe evaluation property of the given initial velocity means $v(x)=\delta (x-l/2)$ in the sense of a Dirac delta, an idealized impulsively struck fixed-end string. Thus $a_n=2\sin (n\pi /2)/(n\pi c)$: every even mode vanishes, while the lowest two excited modes have $n=1,3$. Their ordinary frequencies and amplitude ratio areThe signed coefficient ratio is $a_3/a_1=-1/3$. The impulse solution is interpreted as a distributional weak solution, rather than one with a smooth initial velocity.

At $\Omega =3/4$, the two continuous triangles overlap near zero. Their amplitudes add, giving a central plateau. Explicitly, write $C(\omega )$ for the continuous part of $\tilde{\varphi }$:There are also the two carrier Dirac delta lines $\pi \delta (\omega \mp 3/4)$. The last panel of the preceding figure shows this spectrum and the retained central portion. The prescribed strict cutoff removes the carriers and keepsThe values of the ordinary function at individual endpoints do not affect its inverse transform. To handle the carrier lines, one must additionally assign zero filter gain to those lines, as the carrier-rejecting prescription indicates. A discontinuous mask has no canonical product with a Dirac delta at its discontinuity; a different physical boundary gain would leave a corresponding multiple of $\cos (3t/4)$.

The retained spectrum is real and even. Its inverse Fourier transform isIntegrating the linear factors yieldsThe removable value is $\boxed{\rho (0)=1/4}$, obtained either by expansion or by integrating $\tilde{\rho }/(2\pi )$. Because the carrier frequency is below the message bandwidth, this is Fourier filtering of overlapping amplitude-modulation sidebands; the usual picture of two isolated bands in single-sideband modulation does not apply without that qualification.

Reality requires $\tilde{\sigma }(\omega ,r)=\tilde{\sigma }(-\omega ,r{)}^{\ast }$. The way it constrains the amplitudes depends on whether the spatial wave number is real or imaginary.

For $|\omega |>\mu$, $k$ is real and even. Complex conjugation interchanges the two exponentials, so comparing independent spatial factors givesThe companion equation $B(\omega )=A(-\omega {)}^{\ast }$ is the same constraint with frequency reversed and conjugated. Neither oscillatory spatial branch grows as $r\to \infty$.

For $|\omega |<\mu$, put $k=i\kappa$ with $\kappa >0$. ThenBoundedness excludes the growing exponential, and reality then pairs the remaining amplitudes:These are the two independent low-frequency requirements. At threshold boundedness eliminates the separate linear-in-$r$ solution discussed in (ii). The constraints are understood for a Fourier transform representation in which spatial mode amplitudes are well defined; sufficient amplitude integrability also ensures the spectral integral itself is bounded.

At fixed frequency, the transformed equation has characteristic roots $\pm ik$, whereFor a consistent explicit branch chooseThis branch is even in $\omega$. Away from the two thresholds, the spatial solution is $\tilde{\sigma }=A(\omega )e^{ikr}+B(\omega )e^{-ikr}$, so Fourier inversion givesAt $|\omega |=\mu$, the spatial equation has a double zero characteristic root and its full solution is $C+Dr$. The two exponentials at $k=0$ represent only its constant part. This isolated-frequency issue does not change an integral with ordinary integrable amplitudes; threshold-supported distributional modes require separate treatment.

Split the inverse transform into the intervals $|\omega |<\mu$ and $|\omega |>\mu$, and in each case pair the positive-frequency integral with its complex conjugate at negative frequency. The low-frequency pair becomessince $B=0$ and the decaying spatial factor is real. In the high-frequency positive half, replace $A(\omega )$ by $B(-\omega {)}^{\ast }$. The negative half is again its conjugate. Combining the two pieces givesEvery integrand satisfies the massive wave equation because ${\omega }^2=k^2+{\mu }^2$. Linearity, and taking the real part, preserve the equation whenever differentiation under the integrals is justified, or distributionally under the corresponding spectral assumptions. The construction is exactly the real bounded Fourier modes of the Klein-Gordon equation decomposition.

The displayed expression uses ordinary spectral amplitudes. If distributions supported at $\omega =\pm \mu$ are admitted, a further bounded threshold contribution $c_1\cos \mu t+c_2\sin \mu t$, independent of $r$, is possible. It is not a missing propagating or evanescent branch of the open frequency intervals.

Use the trigonometric identity $2{\cos }^2(ax)=1+\cos (2ax)$ and express the cosine as a sum of two complex exponentials. The Fourier transform of each exponential is a shifted Dirac delta distribution:By linearity,The original function does not decay and is not absolutely integrable on the real line, so this answer is a tempered distribution, rather than an ordinary convergent Fourier integral. Its three lines represent the constant part and the two opposite spatial frequencies.

For three unit masses, $M=I$ andThe uniform vector has $11^{T}1=31$, whereas every vector with coordinate sum zero is annihilated by $11^T$. One orthonormal choice of normal modes is thereforeIf $k=1$ is retained from (b), the angular frequencies are $1,2,2$. The equal upper frequencies follow from degenerate normal modes from permutation symmetry: the stiffness is scalar on the two-dimensional sum-zero subspace. In the uniform mode the inter-bar strings do not change length, but all three beam connections stretch, giving the nonzero lower frequency. The motion is an arbitrary sine-cosine superposition of these modes about the gravitational equilibrium.

Put ${\theta }_i={\theta }_i^0+{\eta }_i$. Differentiating the spring length givesThe springs are unextended at equilibrium, so the quadratic potential energy contains only the squares of these first-order extensions:The linearized Euler-Lagrange equations are thereforeThe stiffness matrix is $(k/4)(3I-11^T)$. Its eigenvector $1=(1,1,1{)}^T$ has zero eigenvalue, and every vector whose components sum to zero has eigenvalue $3k/4$. One convenient orthonormal set of normal modes isAny orthonormal basis of the sum-zero plane is equally valid for the two degenerate normal modes. The general small motion isThe linear drift in the zero mode describes a slow uniform rotation. These are angular frequencies; the ordinary frequencies of the two vibrational modes are ${\omega }_1/(2\pi )$.