7 Linearization of Fluid Equations: Waves and Instabilities

D.G.Korycansky Astro204notes November1,2017 37

7 Linearization of ﬂuid equations: waves and instabilities

The fluid-dynamical equations are difficult to solve due to their non-linearity. In general, closed- form solutions to non-linear equations are difficult or impossible to find. There are special cases, of course, but in general we must resort to numerical computation for solutions, which technique brings its own complications. The most typical form of nonlinearity the we encounter is the quadratic nonlinearity due to the advection terms of the equations: u ∇u. In such cases, one possible line of· analysis is to consider perturbation techniques, that is, the analysis of “small” disturbances of a simpler basic flow. Formally one can write the fluid variables in the form of an expansion in powers of a small parameter ε:

2 2 ρ = ρ0 + ερ1 + ε ρ2 ... u = u0 + εu1 + ε u2 ..., (7.1)

etc. The zero subscripts denote a basic state, such as steady-state equilibrium or a uniform flow. The series expansions are substituted into the fluid equations and the resulting terms are grouped order by order. As noted, the zero-order terms will describe a basic state. First-order equations will yield linear equations describing the evolution of small perturbations. Second and higher orders will contain non-linearities; in some cases the nonlinear terms may be amenable to further analysis, but for our purposes here we will take the attitude that ε denotes an “infinitesimal” perturbation for which higher order effects may be neglected. Such a treatment has its limitations in the sense that growing solutions, if found, will only be valid for short times until the non-linear/higher-order terms can no longer be neglected. Nonetheless, analysis of linearized equations is a valuble tool that has yielded many insights. An example of how the order separation plays out for a sample set of terms in the quadratic non-linearity u ∇u, where the series is substituted in, yielding · 2 u ∇u = u0 ∇u0 + ε(u0 ∇u1 + u1 ∇u0) + ε (u0 ∇u2 + u1 ∇u1 + u2 ∇u0) + ... (7.2) · · · · · · · The order-ε equations are linear; depending on the base state they may have simple (constant) coefficients in time or space, thus giving solutions that are sinusoidal or exponential in the relevant dimensions. Most typically the base state is at least constant in time, so that the perturbations have sinusoidal or exponential (growing/decaying) solutions in time. Dimensions in time or space with constant-coefficient equations allow sinusoidal/exponential variation, thus reducing the di- mensionality of the linear differential equations. Analyses typically seek to reduce the equations to ordinary differential type or algebraic relations among fequencies, wavenumbers, and physical parameters. In many cases, boundary conditions give rise to eigenvalue problems with restricted spectra of distinct solutions. It is probably simplest to proceed from here by way of examples.

7.1 Sound waves (acoustic waves) (Regev 6.2.1–6.2.2, pp 310–312, 315, Landau and Lifshitz 63, pp. 245–249) § § Start with the one-dimensional equations of mass and momentum conservation for compressible inviscid ﬂow; we ﬁnesse the energy equation by writing the pressure as a function of density (the D.G.Korycansky Astro204notes November1,2017 38

adiabatic equation of state): ∂ρ ∂ + (ρu) = 0, ∂t ∂x ∂u ∂u 1 ∂ p γ + u = , P = Kρ . (7.3) ∂t ∂x −ρ ∂x ρ ργ We assume that 0 and U0 are constant (independent of x), and it follows that p0 = K 0 is constant as well. Writing ρ = ρ0 + ερ1, u = U0 + εu1, p = p0 + ε p1, we have to linear order ∂ρ ∂ρ ∂u 1 +U 1 + ρ 1 = 0, ∂t 0 ∂x 0 ∂x ∂ ∂ 2 ∂ρ u1 u1 cs 1 +U0 = , (7.4) ∂t ∂x −ρ0 ∂x

1/2 where cs =(γP0/ρ0) is the adiabatic sound speed and we have substituted for the perturbation 2 pressure p1 = c ρ1. The coefﬁcients are constants, so we can write ρ1 = ρ1 exp[i(kx ωt)] and s − similarly for u1, ρ ωρ ρ ρ ω 2 1 i 1 + ikU0 1 + ik 0u1 = 0, i u1 + ikU0u1 + ikcs = 0. (7.5) − − ρ0 Write this system in matrix form ω ρ i( + kU0) ik 0 ρ − 2 1  cs  = 0. (7.6) ik i( ω + kU0) u1 ρ −  0  In order for there to be non-trivial solutions, the determinant of the matrix has to vanish, giving us the dispersion relation

2 2 2 ( ω + kU0) k c = 0, or ω = k(U0 cs). (7.7) − − s ±

(Note that we have two possible waves, depending on the sign chosen for the kcs term.)

7.2 Dispersive and non-dispersive waves, phase function, wave vector, phase and group velocities

The relation ω = k(U0 cs) is an example of a non-dispersive wave, that is all, all Fourier compo- nents (i.e. wavenumbers± k) travel at the same speed. This can be seen by comparing the relation- ship with the outcome of doing the same analysis for the linear advection equation in which we substitute the same dependence exp[i(kx ωt)]: − ∂ f ∂ f + c = 0 iω f + ikcf = 0 ω = ck. (7.8) ∂t ∂x → − →

All modes propagate at the same speed c = ω/k. We can introduce the idea of phase speed cp = ω(k)/k. D.G.Korycansky Astro204notes November1,2017 39

If ω is not a linear function of k, then cp will be a function of k, meaning that different modes will propagate a different speeds, and the wave in question will be dispersive. A “packet” of waves that occupies a relatively small region of space (and will hence require a spectrum of modes in k-space) will spread out or disperse as it travels. An example of a dispersive wave is one that propagates along the surface of water of depth h in gravity g with surface tension T . The dispersion relation (which I won’t derive here, yet, at least) is T ω2 = gk tanhkh + k3. (7.9) ρ0 1/2 Considering first the case of infinite depth h ∞ and T = 0, we have ω = (gk) or cp = 1/2 1/2 → g k− . That is, long waves (small k) propagate faster than short ones (large k). For the case including surface tension T , sufficiently short waves will propagate fastest (ω/k ∝ k1/2). We can also introduce the additional concept of the group velocity vg = dω/dk, which gives the propagation speed of a packet as a whole and the speed at which the energy of the wave propagates. For multiple space dimensions, we have a wave vector k that describes the collection of wave numbers in the multiple dimensions e.g. k =(kx,ky,kz). Sometimes it is useful to think in terms of a phase function φ(x,t) for which the sinusoidal variation of the wave ∝ exp(iφ); crests of the wave can be marked by values of φ = 0, 2π,4π, etc. In that case we have the relation between the wave numbers k, the frequency ω and φ: ∂φ k = ∇φ, ω = . (7.10) − ∂t 7.3 Two-dimensional internal waves in a stratified atmosphere (Regev et al. 4.4.2-4.4.3 pp. 215–221) § In this case we look at two-dimensional incompressible waves in the Boussinesq approximation. The equilibrium state is a constant density ρ0 but there is a background temperature gradient dT0/dz > 0 (hotter on top, a dynamically stable situation). The linearized equations for perturbations (ρ,u,w, p,T) are

∂u ∂w + = 0 ∂x ∂z ∂u 1 ∂ p + = 0 ∂t ρ0 ∂x ∂w 1 ∂ p + + gρ = 0 ∂t ρ0 ∂z ∂T dT + w 0 = 0. (7.11) ∂t dz

with the Boussinesq EOS ρ/ρ0 = αT . − Substituting our paradigm q = qexp[i(kxx+kzz ωt)] for quantities q, and eliminating ρ in favor of T we have −

ikx p ikz p dT0 ikxu + ikzw = 0, iωu + = 0, iωw + gαT = 0, iωT + w = 0. (7.12) − ρ0 − ρ0 − − dz D.G.Korycansky Astro204notes November1,2017 40

Note that the imcompressibility condition tells that k u = 0; perturbation velocities are perpen- · dicular to the wave vector. From that condition we have u = kzw/kx, and substituting into the 2 − equation for u, we get p = ρ0ωw/k , so that the w-equation is − x 2 ω kz ω α w + 2 w ig T = 0. (7.13) kx −

Substituting for iT =(dT0/dz)w/ω, we get 2 kz ω2 α dT0 1 + 2 g w = 0. (7.14) kx − dz For non-trivial solutions the term in brackets vanishes giving us the dispersion relation

2 2 1/2 ω2 α dT0 kx ω kx = g 2 2 , or = N 2 2 , (7.15) dz kx + kz kx + kz 1/2 where N =(gαdT0/dz) is the Brunt-Väisäläfrequency. The Brunt-Väisäläis (naively, the way I think) the frequency at which a parcel of fluid in a stratified medium would oscillate vertically. Some things to note: 1) unlike sound waves, these internal waves are dispersive: the ω is not a linear function of the wave number (or the wave vector denoted by k =(kx,kz)). For multidimen- sional waves we write

ω ω kx kz vp = kˆ = , vg = ∇kω, (7.16) k (k2 + k2)1/2 (k2 + k2)1/2 (k2 + k2)1/2 | | x z x z x z where the subscripted nabla symbol indicates derivatives with respect to the wave vector compo- nents k. For two-dimensional sound waves we have c ω = c(k2 + k2)1/2, v = v = (k ,k ) (7.17) x z p g 2 2 1/2 x z (kx + kz )

Thus k u = k vg = ω (k vp = ω is true by deﬁnition). For internal· waves· have · k N N ω = N x , v = (k2,k ), v = (k2, k k ) (7.18) 2 2 1/2 p 2 2 3/2 x z g 2 2 3/2 z x z (kx + kz ) (kx + kz ) (kx + kz ) −

Internal waves thus have k u = 0 and k vg = 0; both the perturbation velocity and the group velocity are perpendicular to· the wavefront· (i.e. are parallel to lines of constant phase φ.)

7.4 Acoustic waves in an exponential atmosphere (Regev 6.2.2,1, pp 317–319) § Here we look at vertically propagating waves in a compressible isothermal atmosphere with gravity g and soundspeed cs. The basic structure of of the atmosphere is set by hydrostatic equilibrium c2 dρ s 0 + g = 0, (7.19) ρ0 dz D.G.Korycansky Astro204notes November1,2017 41

ρ ρ 2 or 0(z) = 00 exp( z/H), with H = cs /g. The linearized equations for perturbations in density ρ, horizontal velocity− u, and vertical velocity w are ∂ρ ∂ ρ ∂ ∂ 2 ∂ρ ∂ 2 ∂ρ ρ u d 0 w u cs w cs + ρ0 + w + ρ0 = 0, + = 0, + + g = 0, (7.20) ∂t ∂x dz ∂z ∂t ρ0 ∂x ∂t ρ0 ∂z ρ0 Not *sure* this is the right way to go about it, but as follows: the coefﬁcients are time-independent, so we can replace the time-derivatives with iω , the x-derivative with ik , and the spatial deriva- × × tives of ρ0 yielding ρ 2 2 ρ ρ wˆ 0 dwˆ ikcs cs d ˆ ˆ iωρˆ + ikρ0uˆ + ρ0 = 0, iωuˆ + ρˆ = 0, iωwˆ + + g = 0. (7.21) − − H dz − ρ0 − ρ0 dz ρ0 2 (Note that ρ0 = ρ0(z) still and dρ0/dz = ρ0(z)/H.) Eliminatingu ˆ =(kc /ωρ0)ρˆ , we have for the continuity equation − 2 2 ρ k cs 0 dwˆ i ω + ρˆ wˆ + ρ0 = 0. (7.22) − ω − H dz Solving for ρˆ and dρˆ /dz from the continuity andw ˆ equations, respectively:

ω ρ0 dwˆ dρˆ g iωρ0 ρˆ = i wˆ ρ0 , = ρˆ + wˆ. (7.23) ω2 k2c2 H − dz dz −c2 c2 − s s s We also differentiate the continuity equation with respect to z: 2 2 ρ ρ ρ ρ 2 k cs d ˆ 0 0 dwˆ 0 dwˆ d wˆ i ω + + wˆ 2 2 + ρ0 = 0, (7.24) − ω dz H2 − H dz − H dz dz2 and substitute for dρˆ /dz from thew ˆ equation: k2c2 g ω2 1 ρ dwˆ d2wˆ i ω + s ρˆ + k2c2 + ρ wˆ 2 0 + ρ = 0. (7.25) ω 2 s 2 2 0 0 2 − − cs − c H − H dz dz We still have to eliminate ρˆ: ρ ω2 2 g 0 ρ dwˆ 2 2 2 1 ρ ρ d wˆ 2 wˆ 0 + k cs 2 k + 2 0wˆ + 0 2 = 0. (7.26) −cs H − dz − c − H dz This becomes 2 ω2 d wˆ 1 dwˆ 2 2 + 2 k wˆ = 0. (7.27) dz − H dz cs − This being a differential linear differential equation for wˆ, we can tryw ˆ ∝ exp[(α + im)z]. Substi- tuting and separately setting real and imaginary parts of the coeﬁcients to zero, we have α ω2 m α2 m2 + k2 = 0, 2αm = 0, (7.28) − − H c2 − − H α ω2 2 2 2 2 or = 1/2H, and dispersion relation = cs (k +m +H /4). Well, I think this is still not getting what we want, I think...the problem is that there is no energy equation and the pressure variations are assumed to be acoustic only right from the start (cf. Houghton, The Physics of Atmospheres, pp 88-93.) However, we do get the acoustic cutoff frequency for m = k = 0, where ω = c/2H. D.G.Korycansky Astro204notes November1,2017 42

7.5 Gravitational instability (Shu, ch. 8 pp. 110-112) We look at peturbations in a uniform medium and include the gravitation of those perturbations. Before doing so we note an inconsistency in the treatment of the background state. The basic- 2 state gravitational potential Ψ0 is given by Poisson’s equation ∇ Ψ0 = 4πGρ0. For a uniform density ρ0, Ψ0 must be quadratic (for instance), and so in principle cannot be subtracted off as a uniform value in working with the perturbation equation. However, the ﬁrst dicussion of this by James Jeans did just this, a trick often referred to as the “Jeans swindle”. (In a real medium, one can suppose that the overall region size is sufﬁciently large that the smoothly varying backound potential does not vary much across the perturbation wavelength 2π/k under consideration. Note that the Poisson equation tells us that the Ψ does indeed vary more smoothly than ρ; the Laplacian is second-derivative operator and its inverse is a smoothing operator going from ρ Ψ.) Moving right along, the linearized equations for one-dimensional perturbations (ρ, u, Ψ) are→

∂ρ ∂u + ρ = 0, ∂t 0 ∂x ∂u c2 ∂ρ ∂Ψ + s = , ∂t ρ0 ∂x − ∂x ∂ 2Ψ = 4πGρ. (7.29) ∂x2 Adopting the usual planar-wave description for perturbations ∝ exp(σt + ikx), we have

2 ikcs 2 σρ + ikρ0u = 0, σu + ρ + ikΨ = 0, k Ψ = 4πGρ (7.30) ρ0 −

2 Eliminating: ρ = ikρ0u/σ (continuity equation), Ψ = 4πGρ/k = 4πiGρ0u/kσ, (Poisson equation, substituting− for ρ) yielding −

2 π ρ σ k cs 4 G 0 σ 2 2 2 π ρ u + u u = 0, or = k cs + 4 G 0, (7.31) ρ0σ − σ − so that growth rates become real (exponentially growth) for some critical wavenumber kJ depend- ρ π ρ 2 1/2 ing on 0. Wavenumbers smaller than kJ =(4 G 0/cs ) are unstable; the corresponding un- 1/2 1/2 stable lengthscales are λ > λJ = 2π/kJ = π cs(Gρ0)− . This can be interpreted in terms of a comparison between the sound crossing time τs = λ/cs and the gravitational (“free-fall”) timescale 1/2 τG (Gρ)− ; the lengthscale λ is gravitationally unstable if the sound crossing time is longer than≈ the free-fall timescale. In effect for these lengthscales and larger, acoustic (pressure) waves are unable to communicate across the region and “hold it up” against gravity. (Note that the free- fall timescale is independent of length, so eventually some lengthscale will be too large to hold itself up by thermal pressure.)

7.6 Kelvin-Helmholtz instability (Regev et al. 7.3.3.2 pp. 426–429, Chandrasekhar 100–101, pp. 481–486) § § D.G.Korycansky Astro204notes November1,2017 43

As an example of linearization that leads to the analysis of an instability we look at an example of idealized inviscid shear flow, in which the velocity is discontinuous across an infinitesimally thin layer. This is known as the Kelvin-Helmholtz instability. The basic state is incompressible, with flow velocity U = +U0 for z > 0, and U = U0 for z < 0. In this case we must allow for non-sinusoidal variations in the vertical direction due− to the basic flow variation with z; quantities q are written as q(z)exp(σt +ikx). (Note now that σ real indicates a disturbance whose amplitude grows or decays with time). The equations for the perturbations (u, w, p) are dw ik u + = 0 x dz ik p (σ + ikU)u + x = 0, ρ0 1 dp (σ + ikU)w + = 0. (7.32) ρ0 dz We have i dw p σ + ikU dw u = , = 2 , (7.33) k dz ρ0 − k dz arriving at 1 d2w (σ + ikU) w = 0. (7.34) k2 dz2 −

Assuming that σ +ikU = 0, we have w = w0 exp( kz) as our functional form for w(z). We require solutions that vanish for6 z ∞, so we have ± → ± w e kz z > 0 w(z) = + − (7.35) w ekz z < 0, − where w+ and w are the amplitudes of the w perturbation. To go farther we must make use of boundary conditions− applied at the interface between the upper and lower ﬂows; at linear amplitude these can be taken at z = 0. The two boundary conditions are i) continuity of the displacement of the interface ζ (x,t) between the ﬂows and ii) continuity of the pressure across the interface. The idea is the evaluate each condition using the functions from the top and the bottom and equate the two. For the continuity of the interface we have

Dζ ∂ζ ∂ζ = +U =(σ + ikU)ζ = w, (7.36) Dt ∂t ∂x so evaluating from the top and bottom we have

(σ + ikU0)ζ = w+, (σ ikU0)ζ = w . (7.37) − −

Rearranging, we have w+(σ ikU0) = w (σ + ikU0). For the second condition from the u- − − equation and writing p+ = p = p we have − p w+ w = (σ + ikU0) =(σ ikU0) − , (7.38) ρ0 − k − k D.G.Korycansky Astro204notes November1,2017 44

yielding w+(σ + ikU0) = w (σ ikU0). Putting these two conditions together we have − − − σ ikU0 σ + ikU0 w − − + = 0. (7.39) (σ + ikU0) (σ ikU0) w − − − − For non-trivial solutions we have the determinant of the matrix equaling zero:

2 2 2 2 2 (σ ikU0) +(σ + ikU0) = 0, or σ = k U (7.40) − 0 or σ = kU0; thus there is a growing and decaying mode for any value of k. In fact the indicated growth± rate becomes inﬁnite for k ∞, (short wavelengths λ = 2π/k 0) which is unphysical: in practice the growth rate is truncated→ either by a non-zero shear layer→ width, or by non-zero viscosity.

7.7 Rayleigh-Taylor instability (Regev et al. 7.4.1 po 435–436, Chandrasekhar 90–92 pp. 428–436) § § Now we look at a situation where the (base) ρ0 density varies, but perturbations are still consid- ered incompressible. In particular we will assume ρ0 = ρ0(z) and allow for bouyant motions in a gravity ﬁeld g. For perturbations ρ, u, w, and p we have ∂u ∂w + = 0 (7.41) ∂x ∂z ∂u 1 ∂ p + = 0, (7.42) ∂t ρ0 ∂x ∂w 1 ∂ p g + + ρ = 0, (7.43) ∂t ρ0 ∂z ρ0 ∂ρ dρ + w 0 = 0. (7.44) ∂t dz Note the we have used the linearization of the continuity equation, as it were, twice: once for the non-divergent condition ∇ u = 0, and again ∂ρ/∂t + u ∇ρ0 = 0. Applying the usual assumptions· for the forms we get ·

dw iku + = 0 (7.45) dz ik σu + p = 0 (7.46) ρ0 1 dp g σw + + ρ = 0, (7.47) ρ0 dz ρ0 dρ σρ + w 0 = 0. (7.48) dz Combining the ﬁrst pair (to eliminate u) and the second pair (to eliminate ρ), we have σ ρ p dw σ 1 dp g 1 d o = 2 , w + w = 0. (7.49) ρ0 −k dz ρ0 dz − σ ρ0 dz D.G.Korycansky Astro204notes November1,2017 45

Differentiate the expression for p/ρ0: 1 dp p dρ σ d2w 1 dp p 1 dρ σ d2w 0 = , or = 0 , (7.50) ρ ρ2 2 2 ρ ρ ρ 2 2 0 dz − 0 dz −k dz 0 dz 0 0 dz − k dz and substituting for p/ρ0 on the right-hand side from the ﬁrst equation: 2 1 dp σ 1 dρ0 dw σ d w = 2 2 2 , (7.51) ρ0 dz −k ρ0 dz dz − k dz

Subsituting for (1/ρ0)dp/dz in the 2nd equation above (labelled 7.49b at the moment) gives σ ρ σ 2 ρ σ 1 d 0 dw d w σ 1 d 0 w 2 2 2 g w = 0, (7.52) − k ρ0 dz dz − k dz − ρ0 dz or 2 ρ σ σ 1 d w 1 d o g dw w 2 2 = w + 2 . (7.53) − k dz ρ0 dz σ k dz (Note a common theme: (growth rate) (w d2w/dz2) = rhs stuff.) The basic case is one where × − ρ0 consists of two semi-inﬁnite layers ρ0 = ρ1 for z < 0, ρ0 = ρ2 for z > 0, with a discontinuity at z = 0 In that case for z = 0, we have as with the Kelvin-Helmholtz instability: 6 w e kz z > 0 w(z) = + − (7.54) w ekz z < 0, − Applying the same condition for the displacement ζ of the interface, (i.e. that the vertical velocity is continuous), we have σζ = w+ = w = w˜. The other condition comes from pressure-gradient− (?) continuity across the interface. Writing again: dp g dρo = σρ0w + w = 0. (7.55) dz − σ dz Integrating over a small distance δ at the interface, where ρ2, p2 are the values above the interface (z = +δ) and ρ1, p1 are the values below (z = δ), and writing ρ0 as the average (ρ1 + ρ2)/2 at the interface − σ(ρ1 + ρ2) g p2 p1 = w˜ 2δ + (ρ2 ρ1)w˜. (7.56) − − 2 · σ − Making use of the other equation σ dw p = ρ0 , (7.57) − k2 dz and the values of dw/dz kw˜ at z = 0 δ, we have ∼ ∓ ± ρ1σ ρ2σ p1 = w˜, p2 = w˜, (7.58) − k k we have σ σ(ρ1 + ρ2) g (ρ1 + ρ2) w˜ = w˜ 2δ + (ρ2 ρ1)w˜. (7.59) k − 2 · σ − Dividing outw ˜ and taking the limit δ 0, we have σ → ρ ρ g 2 2 1 (ρ1 + ρ2) = (ρ2 ρ1), or, σ = gk − , (7.60) k σ − ρ1 + ρ2 the celebrated expression for the growth rate of the invisicd Rayleigh-Taylor instability in the case without surface tension. D.G.Korycansky Astro204notes November1,2017 46

7.8 Convective instability (Regev et al. 7.4.2 pp 440–448, Chandrasekhar 5–10, pp. 24) § § Finally we look at a more complicated case, namely convection in the Boussinesq approximation, this time including viscosity and heat conduction. These latter effects raise the order the system, giving rise eventually to a sixth-order ODE for the eigenfunction. However, the constancy of the coefﬁcients again allows for solutions in terms of exponentials and sines and cosines. The relevant linearized equations are

∂u ∂w + = 0 (7.61) ∂x ∂z ∂u 1 ∂ p + = ν∇2u, (7.62) ∂t ρ0 ∂x ∂w 1 ∂ p g + + ρ = ν∇2w, (7.63) ∂t ρ0 ∂z ρ0 ∂T dT + w 0 = κ∇2T. (7.64) ∂t dz again with the Boussinesq equation of state ρ/ρ0 = αT . − 1/2 Recall that we described the Brunt-Väisäläfrequency N =(gαdT0/dz) , for dT0/dz > 0; in this case we have the possibility of dT0/dz < 0, for which growing modes will occur. Before manipulating the equations we will non-dimensionalizein a standard fashion. We will as- sumethat the fluid motionswill occur in a domain of height D, and useas atimescalethetime D2/κ for heat to diffuse across the domain height. Temperature variations will be non-dimensionalized 2 2 by ∆T0 = DdT0/dz, and the pressure scale is ρ0κ /D .

D2 κ κ x = Dx˜, z = Dz˜, t = t˜, u = u˜, w = w˜, (7.65) κ D D ρ κ2 ∂ 2 ∂ 2 T = ∆T T˜ , p = 0 p˜, ∇˜ 2 = + (7.66) 0 D2 ∂x˜2 ∂z˜2 Substituting into the equations

κ ∂u˜ ∂w˜ + = 0 D2 ∂x˜ ∂z˜ κ2 ∂u˜ κ2 ∂ p˜ νκ + = ∇˜ 2u˜ D3 ∂t˜ D3 ∂x˜ D3 κ2 ∂w˜ κ2 ∂ p˜ νκ + = ∇˜ 2w˜ + g(αT˜ ∆T ) D3 ∂t˜ D3 ∂z˜ D3 0 κ∆T ∂T˜ κ ∆T κ∆T 0 + w˜ 0 = 0 ∇˜ 2T˜ (7.67) D2 ∂t˜ D D D2 Clearing the equations we get

∂u˜ ∂w˜ + = 0 ∂x˜ ∂z˜ D.G.Korycansky Astro204notes November1,2017 47

∂u˜ ∂ p˜ + = Pr∇˜ 2u˜ ∂t˜ ∂x˜ ∂w˜ ∂ p˜ + = Pr∇˜ 2w˜ + RaPrT˜ ∂t˜ ∂z˜ ∂T˜ + w˜ = ∇˜ 2T˜ (7.68) ∂t˜ We have introduced the well-known non-dimensional parameters, the Prandtl number Pr and the Rayleigh number Ra: ν gα∆TD3 Pr = κ , Ra = νκ . (7.69) (NB have implicitly deﬁned Ra as having the same sign as ∆T ; positive is stable, negative is unstable.) (Also: Rayleigh number can be thought of as the ratio of time scales:

2 2 gα∆T D D τdif f τvisc Ra = = (7.70) κ ν τ2 D accel

1 the ratio of diffusion and viscous timescales to the square of the acceleration timescale N − .) We can “cheat” a bit here by assuming sinusoidal space variations in x and z and not| worrying| about boundary conditions. (In fact the sinusoidal variations will work for physical ﬂow for the idealized case of slip boundary conditions at the top and bottom of the domain.) In that case with the usual exp(σt + ikxx + ikzz) assumptions for the variables (and dropping the tildes) we have

kxu + kzz = 0 σ 2 2 u + ikx p = Pr(kx + kz )u − 2 2 σw + ikz p = Pr(k + k )w + RaPrT − x z σT + w = (k2 + k2)T (7.71) − x z 2 2 2 Solving for u = kzw/kx (from continuity) and writing l = kx + kz we have (from the u-equation and the T -equation− with w substituted for):

2 2 ikx ikx 1 w = 2 p, T = 2 2 p. (7.72) kz(σ + Prl ) − kz (σ + l )(σ + Prl ) Putting these into the w-equation we have

2 2 kx ikx RaPr i + kz p = 2 2 p (7.73) kz − kz (σ + l )(σ + Prl ) multiply through by kz (and divide out ip) to get RaPr k2 + k2 = l2 = k2 , (7.74) x z − x (σ + l2)(σ + Prl2) yielding the dispersion relation

k2 (σ + l2)(σ + Prl2) = x RaPr, (7.75) − l2 D.G.Korycansky Astro204notes November1,2017 48

or k2 σ 2 +(1 + Pr)l2σ + Pr l4 + x Ra = 0. (7.76) l2 This is unstable (real part of σ > 0) if the zeroth-order term is negative, or

6 2 2 3 l (kx + kz ) Ra < 2 = 2 . (7.77) −kx − kx

For a box with bounded walls on top and bottom kz = nπ, n = 1, 2, 3,.... We can search for a minimum by writing kx = aπ, and ﬁnd the well-known instability criterion at a minimum Ra > | | 27π4/4, for n = 1 and a = 1/2. (That is, the perturbations take the form of “rolls” in the x z − plane with aspect ratio (widthp to height) 2π/a = √2:1.)