IUGG, Perugia July, Energy Spectra from. Entropy Principles. Peter Lynch & Wim Verkely. University College Dublin. Meteorology & Climate Centre

Transcription

1 IUGG, Perugia July, 2007 Energy Spectra from Entropy Principles Peter Lynch & Wim Verkely Meteorology & Climate Centre University College Dublin KNMI, De Bilt, Netherlands

2 Introduction The energy distribution in turbulent systems varies widely. A power-law dependence on wavenumber is common: Burgers turbulence (1D): a K 2 spectrum Atmosphere, synoptic range ( 2D): a K 3 spectrum Fully developed 3D turbulence: K 5/3 spectrum. 2

3 Canonical Statistical Mechanics The equilibrium statistical mechanics of classical systems is based on Liouville s Theorem. This theorem continues to hold under spectral truncation. As a result, the probability distribution function (PDF) of a constant of the motion, K, has the canonical form Z(β) exp( βk) The partition function Z(β) normalizes the distribution. The quantity 1/β plays a role analogous to the temperature in thermodynamic systems. 3

4 Non-equilibrium steady state Typically, turbulent motions are far from equilibrium. Turbulence is a dissipative, irreversible process. It is often stated that equilibrium statistical mechanics is inapplicable to turbulence. However, if forcing and dissipation are on average in balance, a non-equilibrium steady state may be reached. We consider driven and damped motions in two dimensions, in which the mean forcing and damping are in balance. 4

5 Applications of the theory Using the balance between forcing and damping as constraints, we derive a range of energy spectra for such nonequilibrium systems. Burgers Equation: K 2 2D turbulence on bi-periodic domain: K 3 Geostrophic turbulence ( 2D) on the sphere: K 3 We compare theoretical results with numerical integrations. Good agreement is found. 5

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9 The Entropy The state of the system is a = (a 1, a 2,..., a N ). The distribution function is W = W (a) We wish to find W. We define the entropy: S = W log W da We seek the W that maximizes S subject to constraints. 9

10 Lagrange Multipliers Since W is a PDF we have W (a) da = 1. Consider a constraint on the expected value of K(a): K = K(a)W (a) da = K 0. We use Lagrange multipliers: ( ) S constrained = S + ρ W da 1 ( ) + λ K 0 K. 10

11 The Canonical Distribution The variational derivative, varying W, gives log W 1 + ρ λk = 0 The canonical distribution function is W = e ρ 1 exp( λk). We will apply this to several specific cases. 11

12 Forced Burgers equation u t + u u x = F (x) + ν 2 u x 2 Periodicity on interval x [0, 2π]. The energy principle is where E = de dt = F D 1 2 u2 dx F = F u dx D = ν u 2 x dx 12

13 Spectral Expansion We expand u(x, t) in spectral components (assuming u even). u(x, t) = N k=1 a k (t) cos kx The energy equation is, as above, de dt = F D where F = 1 2 Nk=1 f k a k D = 1 2 ν N k=1 k 2 a 2 k. 13

14 No Forcing or Dissipation We consider the case without forcing or dissipation. Assume the expected value of the total energy is constant. E = 1 2 Nk=1 a 2 k = E 0, The distribution function is W = e ρ 1 exp( λe). We make this more explicit: W = e ρ 1 exp λπ N k=1 a 2 k. 14

15 Repeating: W = e ρ 1 exp λπ N a 2 k. k=1 Defining σ 2 = 1/(2λπ), we have ( ) N 1 W = σ 2π exp a2 k 2σ 2 k=1 = N k=1 N (0, σ, a k ). Each component a k, has zero mean. Each has the same standard deviation σ. There is equi-partitioning of energy. 15

16 Balance: Forcing = Dissipation We now assume that de = 12 Nk=1 ( fk a dt k νk 2 a 2 k) = 0. Define the mean and variance Then µ k = f k /(2νk 2 ) σ 2 k = 1/(2λνk2 ) W = N k=1 N (µ k, σ k, a k ). The mean and variance are different for each component. µ k = f k /(2νk 2 ) σ 2 k = 1/(2λνk2 ) 16

17 Repeat: the mean and variance are µ k = f k /(2νk 2 ) σ 2 k = 1/(2λνk2 ) This implies a k 2 spectrum... Energy spectrum for Burgers equation 17

18 Two-dimensional Turbulence We now consider turbulent motion on a two-dimensional bi-periodic domain [0, 2π] [0, 2π]. The vorticity equation is ω t + J(ψ, ω) = F + ν 2 ω. Both total energy and total enstrophy are conserved for unforced and undamped flow. 18

19 de dt = d dt ds dt = d dt 1 2 ψ ψ dxdy = [F ψ νω 2 ] dxdy 1 2 ω2 dxdy = [F ω ν ω ω ] dxdy. The spectral enstrophy equation is ds dt = d 1 dt 2 ω2 kl = [ fkl ω kl ν(k 2 + l 2 )ωkl] 2 kl kl 19

20 We now assume ds dt = where K 2 = k 2 + l 2. kl ( f kl ω kl νk 2 ω 2 kl) = 0, We define mean and variance µ kl = f kl /(2νK 2 ) σ 2 kl = 1/(2λνK2 ). The distribution function becomes W = kl N (µ kl, σ kl, ω kl ). 20

21 The variance σ 2 kl has a K 2 dependence on wavenumber. Thus, in the inertial range, we have ω 2 kl K 2. The energy spectrum (integrating over the angle) is E K K 3 This is precisely what we expect for the enstrophy cascade in two-dimensional turbulent flow. 21

22 Energy and Enstrophy development during decaying 2D quasi-geostrophic turbulence. 22

23 Energy spectrum after 10 days (solid line is K 3 spectrum). 23

24 The Barotropic PV Equation We now introduce rotation and of vortex stretching. With the usual quasi-geostrophic approximations, we have (ω Λψ) + J(ψ, ω) + β ψ t x = F D, where β = df/dy and Λ = f0 2 /gh (Pedlosky, 1987). The energy equation is: d 1 [ dt 2 ( ψ) 2 + Λψ 2] dxdy = [F ψ νω 2 ] dxdy. The enstrophy equation is d 1 [F dt 2 (ω ] Λψ)2 dxdy = (ω Λψ) ν (ω Λψ) ω dxdy. The enstrophy equation in spectral form: ds dt = d dt kl 1 2 (ω kl Λψ kl ) 2 = kl ( 1 + Λ K 2 ) [fkl ω kl νk 2 ω kl ]. 24

25 We assume that the forcing and damping balance: νk (1 2 + Λk ) [ 2 ωkl 2 f ] k νk 2ω kl = 0. Defining µ kl = f kl /(2νK 2 ), this constraint is K = 0, where K = νk (1 2 + Λ ) [ ] K 2 (ω kl µ kl ) 2 µ 2 kl kl The probability density function may once again be written in normal form, with variance σ 2 kl = 1 2λν(K 2 + Λ). Outside the forcing region, where µ kl = 0, we have ω 2 kl 1 K 2 + Λ. 25

26 The spectrum of the vorticity components is 1 ω kl K 2 + Λ Thus, the streamfunction goes as 1 ψ kl K 2 K 2 + Λ and the components of energy vary as 1 E kl K 2 (K 2 + Λ) Summing over all components having total wavenumber K, E K 1 K(K 2 + Λ). For K Λ we get a K 1 spectrum. For K Λ we get a K 3 spectrum. 26

27 Geostrophic Turbulence on Sphere The barotropic vorticity equation is q t + v q = F + ( 1) p+1 ν p 2p+2 ψ, p where q is the quasi-geostrophic potential vorticity (h is the orography). q = f + ω Λψ + f h H. The system is forced by F and damped by (hyper-)viscosity. 27

28 The system has energy and potential enstrophy principles: de dt = F D, dz dt = G H, where E = 1 4π F = 1 4π G = 1 4π 1 2 [v2 + Λψ 2 ] ds, Z = 1 4π ψf ds, D = 1 ψ 4π p qf ds, H = 1 q 4π p 1 2 q2 ds, ( 1) p+1 ν p 2p+2 ψ ds, ( 1) p+1 ν p 2p+2 ψ ds. 28

29 The vorticity ω is expanded in spherical harmonics ω(λ, φ, t) = mn ω mn (t)y mn (λ, φ), The spectral coefficients are given by ω mn (t) = 1 Y mn (λ, φ)ω(λ, φ, t) ds. 4π They are related to the coefficients of ψ by ω mn = e n ψ mn where e n = n(n + 1). The field f + f(h/h) is expanded as f + f h H = mn f mn Y mn, Then q mn = f mn ε n ψ mn, where ε n = Λ + e n. The damping term may be written as ( 1) p+1 ν p 2p+2 ψ = ν p e p+1 n ψ mn Y mn p mn p 29

30 With these definitions we may write, in terms of ψ mn : E = mn 1 2 ε n ψmn, 2 Z = [ 12 mn ε 2 nψmn 2 ε n f mn ψ mn f mn 2 ], F = F mn ψ mn, D = d n ψmn, 2 mn mn [ ] [ εn F mn ψ mn + f mn F mn, H = dn ε n ψmn 2 ] d n f mn ψ mn. G = mn mn These quantities are formally similar to previous cases. 30

31 The Unforced-Undamped Case The constraints to be used when there is neither forcing nor damping are constancy of energy E and pot. enstrophy Z: K 1 = E = E 0, K 2 = Z = Z 0, For the multivariate normal distribution function we get: W (ψ N, N,..., ψ N,N ) = mn N (µ mn, σ mn, ψ mn ), where the variance and mean are σmn 2 1 = and µ mn = γf mn. ε n (β + γε n ) β + γε n The mean values are determined by the Coriolis parameter and the orography and the variance depends on ε n. 31

32 We eventually get the energy and esntrophy spectra E n = 2n + 1 2(β + γε n ) + γ 2 ε n n 2(β + γε n ) 2 fmn 2, Z n = (2n + 1)ε n 2(β + γε n ) + β 2 2(β + γε n ) 2 m= n n m= n f 2 mn. These expressions are very similar to those for two-dimensional turbulence. When the Coriolis parameter and the orography are zero E n and Z n behave as n 1 and n. 32

33 The Forced-Damped Case We assume the system is in a statistically stationarity state. This means that we will use as constraints: K 1 = E = E 0, K 2 = D F = 0, K 3 = H G = 0. This eventually gives the distribution function W (ψ N, N,..., ψ N,N ) = mn N (µ mn, σ mn, ψ mn ), with variance and mean σmn 2 1 = 2d n (β + γε n ) E = mn and µ mn = γf mn 2(β + γε n ) F mn 2d n. For the energy and potential enstrophy we now have: [ ε n γ2 ε n fmn 2 8(β + γε n ) 2 + ε nfmn 2 8d 2 γε nf mn F mn n 4d n (β + γε n ) Z = mn 4d n (β + γε n ) + [ ε 2 n 4d n (β + γε n ) + (γε n + 2β) 2 fmn 2 8(β + γε n ) 2 + ε2 nfmn 2 8d 2 n ] + ε n(γε n + 2β)f m 4d n (β + γε n, 33

34 We may write the result as N E = E n, Z = n=1 N n=1 Z n, Since both ε n and d n are independent of m, we have E n = Z n = (2n + 1)ε n 4d n (β + γε n ) + (2n + 1)ε2 n 4d n (β + γε n ) + n m= n n m= n ε n [γf mn ((β + γε n )F mn )/d n ] 2 8(β + γε n ) 2, [(γε n + 2β)f mn + ((β + γε n )ε n F mn )/d n ] 2 8(β + γε n ) 2. The extra terms arise because of the presence of the Earth s rotation, orography and forcing. We will compare these spectra to numerical results. 34

35 Energy and Enstrophy constrained. 35

36 Energy and Enstrophy decay rates constrained. 36

37 Energy and Enstrophy and decay rates constrained. 37

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