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The standard thin-disk dissipation flux, summed over both faces, follows by substituting the Keplerian accretion disk profile into the viscous heating rate:The Stefan–Boltzmann law gives $\mathcal{C}=2\sigma T^4$ because there are two emitting faces. Local radiative equilibrium consequently gives the effective-temperature profile of a zero-torque diskHere $\sigma$ is the Stefan-Boltzmann constant. The scale $T_{\mathrm{in}}$ is not the temperature exactly at the inner boundary: the zero-torque inner boundary condition makes that formal temperature zero. Maximizing $T^4$ shows that the maximum effective temperature of a zero-torque disk occurs at $r=49r_{\ast }/36$, with $T_{\max }\simeq 0.488T_{\mathrm{in}}$.

At equal central mass and accretion rate, characteristic effective temperatures scale as $r_{\ast }^{-3/4}$. Thus, comparing corresponding values of $r/r_{\ast }$,The neutron-star disk is about 180 times hotter. The Planck law and Wien displacement law move its characteristic emission to about 180 times higher frequency, or 180 times shorter wavelength. A white dwarf disk commonly emits in optical and ultraviolet bands, while the hotter neutron star disk can emit in X-rays. Absolute bands require an actual accretion rate; the relative shift follows directly from the stated scaling.

Let $A=k^2{v}_A^2>0$ and $C=N^2+{\Omega }^2$. Solving the radially stratified magnetorotational dispersion relation as a quadratic in $s={\sigma }^2$ givesBoth roots are real. If $A+N^2-3{\Omega }^2<0$, their product is negative, so exactly one root is positive and gives exponential growth. Conversely, when $A+N^2-3{\Omega }^2\ge 0$, the coefficient $2A+N^2+{\Omega }^2\ge A+4{\Omega }^2>0$, so both roots are nonpositive. Thus the instability criterion for a specified nonzero wavenumber isIn a local continuum of allowed wavenumbers, some unstable mode exists exactly when $N^2<3{\Omega }^2$. A finite disk only permits wavelengths fitting its vertical boundaries, so this existence statement also requires an allowed mode in the interval.

The growing root is $s_{+}$, whose derivative and curvature areSetting the derivative to zero gives the fastest-growing interior mode:For a real, nonzero wavenumber, this expression requires $|N^2+{\Omega }^2|<4{\Omega }^2$, equivalentlyNonnegative right-hand side permits the endpoints, but there the stationary point is at $k=0$, outside the nonzero-wavenumber mode used above. Substituting the interior maximizing value givesThis is the maximum growth rate of radially stratified magnetorotational instability, taking $\Omega >0$. For $N^2=0$ it reduces to the usual $3\Omega /4$, at $k^2{v}_A^2=15{\Omega }^2/16$. Negative $N^2$ enhances growth and extends the unstable band; positive $N^2$ suppresses growth and eventually eliminates it at $N^2\ge 3{\Omega }^2$. Magnetic tension enables angular-momentum exchange between displaced parcels, weakening the rotational stabilization that protected the adverse hydrodynamic stratification.

If $N^2\le -5{\Omega }^2$, the interior formula is no longer the physical maximum: $s_{+}$ decreases for $A>0$, and its supremum as $k\to 0$ is $-N^2-{\Omega }^2$. The corresponding limiting growth rate is $\sqrt{-N^2-{\Omega }^2}$, dominated by the already unstable hydrodynamic branch. At $N^2=-5{\Omega }^2$ this joins continuously to $2\Omega$. For the smooth thin disk estimate $N^2\sim -(H/r{)}^2{\Omega }^2$, the ordinary interior maximum applies and its enhancement over $3\Omega /4$ is only of fractional order $(H/r{)}^2$.

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.

The WKB approximation treats the vertical wavenumber as locally constant over one vertical wavelength. Write $w=\Re \{A(z)e^{ikx+i{\int }^{z}m(s)ds}\}$, with $|m^{\prime }|\ll m^2$ and $|A^{\prime }|\ll |mA|$. For the stationary internal-wave WKB solution, $A\propto m^{-1/2}$ away from singular or turning levels. A small terrain slope and small perturbations are also needed for Linearized Boussinesq equations.

Let $p$ be the perturbation fluid pressure divided by reference mass density, and $b$ the buoyancy perturbation. The steady Linearized Boussinesq equations and incompressibility condition areFor an $e^{ikx}$ component, elimination gives $u=-w^{\prime }/(ik)$, $b=-N^{2}w/(ikU)$ and $p=(U{w}^{\prime }-U^{\prime }w)/(ik)$. Substitution into vertical momentum gives the exact stationary Taylor–Goldstein equationThe leading WKB approximation therefore identifiesThe kinematic boundary condition fixes the horizontal wavenumber to $k=k_T$ for the sinusoidal forcing.

For explicit heights assume a constant positive buoyancy frequency $N$, $0<k{U}_0<N$, and $|S|/N\ll 1$. The latter is a large gradient Richardson number. The linear shear flow has $U^{\prime \prime }=0$, so the propagation condition is $|U|<N/k$.

For $S<0$, the first obstruction is the critical level of an internal gravity wave atAs $z\uparrow z_c$, $m\sim N/U\to \infty$. The critical-height approach of a stationary wave in linear shear has upward group velocity $c_{gz}\sim k{U}^2/N$, so the travel time to $z_c$ diverges: the height is a supremum, approached in the ideal ray model. Here $|m^{\prime }|/m^2\to |S|/N$, which can remain small; it would be incorrect to say that the short-wavelength WKB approximation necessarily fails solely because $m$ diverges. Nevertheless $w\propto U^{1/2}$ and displacement $w/(ikU)\propto U^{-1/2}$, so a finite disturbance eventually violates linearity. Dissipation and overturning also matter near this singular level.

For $S>0$, the turning height of a stationary wave in linear shear isAt $z_t$, $m=0$ and the WKB approximation fails. An Airy function matching region describes reflection and an upper evanescent wave. Thus $z_t$ is the highest height of vertical propagation, rather than a sharp cut-off of every disturbance.

The problem initially allows $N(z)$ to vary. Without the extra constant-$N$ assumption, replace the explicit turning height by the first root of $N(z)=k|U_0+Sz|$ encountered on a propagating branch. A declining $N(z)$ can cause a turning level before the negative-shear critical level. The two closed-form heights require the stated constant-stratification assumption.

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.

On a length scale $\ell$, pressure communicates support over a sound-crossing time $\ell /v_s$, whereas self-gravity acts over a time of order $(G{\rho }_0{)}^{-1/2}$. When the former is longer, gravity can amplify an overdensity before pressure smooths it. Thus Jeans instability occurs at scales of order $\ell \gtrsim v_s/\sqrt{G{\rho }_0}$; the precise numerical factor depends on whether scale means wavelength or inverse wavenumber.

Take a uniform static background, small perturbations, an inviscid barotropic fluid, and $\delta p=v_s^2\delta \rho$ with $v_s^2=(dp/d\rho {)}_0>0$. An infinite Newtonian homogeneous density cannot itself have zero background gravitational acceleration while satisfying the unmodified Poisson equation for Newtonian gravity. Specify the Jeans swindle: subtract the homogeneous field, or regard an external prescription as balancing it, and apply Poisson's equation only to perturbations. The linearized continuity equation and Euler equations for an inviscid fluid areTaking a time derivative of the first equation and substituting the second yields ${\partial }_t^2\delta \rho =v_s^2{\nabla }^2\delta \rho +4\pi G{\rho }_0\delta \rho$. A plane wave $e^{i(\mathbf{k}\cdot \mathbf{x}-\omega t)}$ therefore givesFor $0<k<k_J$ there is exponential growth at rate $\sqrt{4\pi G{\rho }_0-v_s^2{k}^2}$. At $k_J$ the density mode is neutral to this linear order, and above it there are acoustic oscillations. The critical wavelength is $2\pi /k_J=\sqrt{\pi }{v}_s/\sqrt{G{\rho }_0}$. Transverse velocity perturbations decouple as zero-frequency vortical modes in this idealization.

For the attractive Yukawa potential, assume a real screening parameter $\alpha \ge 0$. Its Green function obeys $({\nabla }^2-{\alpha }^2)(-Gm{e}^{-\alpha r}/r)=4\pi Gm{\delta }^{(3)}(\mathbf{r})$, including the delta source. Thus $\delta {\varphi }_{\mathbf{k}}=-4\pi G\delta {\rho }_{\mathbf{k}}/(k^2+{\alpha }^2)$. The same fluid elimination gives Jeans instability with Yukawa gravity:There exists a growing nonzero-wavenumber mode precisely whenEquality gives no exponentially growing nonzero mode. If screening exceeds this threshold, the finite-range gravitational response is too weak to overcome pressure at any wavelength. Otherwise all sufficiently long but nonconstant waves remain unstable, with a longer minimum unstable wavelength than in Newtonian gravity. For $\alpha >0$ the long-wave gravitational contribution also scales as $k^2$, so the growth rate tends to zero as $k\to 0$; screening does not produce a second positive lower cutoff in wavenumber. A homogeneous screened background can have the constant gravitational potential $-4\pi G{\rho }_0/{\alpha }^2$, with zero force, so no Newtonian background subtraction is needed for nonzero screening. The limits $k\to 0$ and $\alpha \to 0$ are nonuniform. An exact $k=0$ density change is not one of the growing finite-wavelength modes. In the pressureless edge case $v_s=0$, every finite $k>0$ is unstable for positive density and finite screening.

For thin contacting components, optical powers add. At wavelength $\lambda$ the total optical power is $\Phi (\lambda )={\rho }_1[n_1(\lambda )-1]+{\rho }_2[n_2(\lambda )-1]$. An achromatic lens has ${\Phi }_F={\Phi }_C$, henceFor normally dispersive optical glasses, both index differences are positive, so the two components have opposite curvature factors and opposite optical powers. This uses the thin Lensmaker's equation; finite thickness and separated principal planes require a more detailed design. If the denominator vanishes, use the undivided power-balance equation rather than this ratio. Two-wavelength correction alone does not eliminate the residual chromatic aberration at every intermediate wavelength.

Insert the slit-limited spectral resolving power $R=\lambda q/p$ into the invariant expression:Here $q=|dx/d\lambda |$ is a magnitude; a signed linear spectral dispersion would require the corresponding absolute value. Since $q$ has units of photodetector length per unit wavelength, $L$ has units of length.

Let $f_t$ be the optical telescope focal length, $x=f_t\theta$ the physical optical slit width, $f_{\mathrm{coll}}$ the collimator focal length, and $A_{\mathrm{coll}}=D{f}_{\mathrm{coll}}/f_t$ the incident collimated beam diameter. For the grating equation $m\lambda =d(\sin \alpha +\sin \beta )$, differentiation at fixed wavelength gives $|d\beta /d\alpha |=\cos \alpha /\cos \beta$. Thus the monochromatic optical slit image width isThe incident and emergent optical pupil widths are $A_{\mathrm{coll}}=W\cos \alpha$ and $A_{\mathrm{cam}}=W\cos \beta$. Multiplying cancels the anamorphic magnification of a grating:This is conservation of the Lagrange optical invariant in the dispersion plane. It uses paraxial optical slit angles, an unvignetted optical pupil and matching full-width conventions for $\theta ,p,D,A_{\mathrm{cam}}$.

The optical telescope focuses the sky onto an optical slit; the collimator turns the selected light into a beam incident on the reflection diffraction grating; the optical camera images each diffracted direction onto the photodetector. The drawing also shows the monochromatic optical slit image for uniform optical slit illumination in the geometric, slit-limited approximation.

Near the chosen wavelength $\lambda$, a small wavelength increment moves the image by $q\Delta \lambda$, where $q$ is the magnitude of the linear spectral dispersion. An optical slit image of physical width $p$ therefore has an apparent spectral width $p/q$. With one optical slit width as the adopted separation criterion,Uniform illumination gives a top-hat monochromatic profile of width $p$; an unresolved stellar image need not illuminate the optical slit uniformly, so its profile follows the actual illumination convolved with the instrument response. Negligible optical slit diffraction and optical camera optical aberrations do not remove the finite diffraction grating diffraction width. The formula is the slit-limited result when that width is small compared with $p$, rather than a universal exact line-profile criterion.