Outline: Clouds & Rain 1. Single sphere ( Mie vs. Rayleigh ) 2. Sphere of rain, snow, & ice ( Hydrometeors ) εεε σσσ Find their c, n c, b Scattering from Hydrometeors: 3. Many spheres together : Clouds, Rain, Snow Clouds, Snow, Rain a. Drop size distribution b. Volume Extinction= Scattering+ Absorption Microwave Remote Sensing INEL 6069 c. Volume Backscattering Sandra Cruz Pol 4. Radar Equation for Meteorology Professor, Dept. of Electrical & Computer Engineering, UPRM, Mayagüez, PR 5. TB Brightness by Clouds & Rain 1 2 Clouds Types on our Atmosphere Cirrus Clouds Composition 70 60 hexagonal 50 plates bullet rosettes % 40 30 dendrites 20 others 10 0 Ice Crystals 3 4 EM interaction with Mie Scattering: general solution to EM Single Spherical Particles scattered, absorbed by dielectric sphere .. S Absorption i Uses 2 parameters (Mie parameters) – Cross-Section, Q =P /S a a iii – Size wrt. λλλ : – Efficiency, ξξξ = Q /πππr2 π a a χ = 2 r Scattered λ b – Power, Ps – Cross-section , Qs =P s /S iii – Speed ratio on both media: ξξξ πππ 2 – Efficiency, s= Qs / r Total power removed by sphere from the np ξξξ ξξξ ξξξ n = incident EM wave, e = s+ a nb πππ σσσ πππ 2 5 6 Backscatter , Ss( ) = S iii b/4 R Mie Solution Mie coefficients A m m + − Re{ Wm} Re{ Wm−1} n χ a = Mie solution m A m 2πr 2πr m + − χ = = ε ∞ Wm Wm−1 c 2 n χ λ λ ξ (n, χ) = 2( m +1)(| a |2 + | b 2 )| p o s χ 2 ∑ m m m=1 ∞ m 2 nA + Re{ W }− Re{ W } ξ χ = + + m χ m m−1 a (n, ) 2 ∑ 2( m )1 Re{ am bm} ε χ b = np p (β − jα) m=1 m n = = c = ε = + m − ε c nA m Wm Wm−1 nb b ko χ c Where am & b m are the Mie coefficients given by eqs 5.62 to 5.70 in the textbook. = χ + χ where Wo sin j cos 7 n = n' − jn " 8 NonNon--absorbingabsorbing Conducting (absorbing) sphere sphere or drop ((n”= 0 for a perfect dielectric , which is a nonnon--absorbingabsorbing sphere) Re call (β − jα) n = n'− jn "= ko χ =.06 = ω µ ε ko o o χ =2.4 Rayleigh region | nχ|<<1 9 10 ξξ χχ Plots of Mie ee versus Rayleigh Approximation | nnχχ|<<1 Four Cases of sphere in air : Scattering efficiency n=1.29 (lossless non-absorbing sphere) 8 ξ = χ 4 | K |2 +... n=1.29-j0.47 (low loss sphere) s 3 n=1.28-j1.37 (lossy dielectric sphere) n= ∞ perfectly conducting metal sphere Extinction efficiency 8 ξ = 4χ Im{ −K}+ χ 4 | K |2 +... ξ e As n’’ increases, so does the absorption ( a), and less is the 3 oscillatory behavior . Optical limit (r >> λ) is ξ =2. e where K is the dielectric factor Crossover for – Hi conducting sphere at χ =2.4 n2 −1 ε −1 K = = c – Weakly conducting sphere is at χ =.06 2 + ε + 11 n 2 c 2 12 Absorption efficiency in Rayleigh Scattering from Hydrometeors region Rayleigh Scattering Mie Scattering ξ = ξ −ξ = χ − ≅ ξ a e s 4 Im{ K} e i.e. scattering can be neglected in Rayleigh region (small particles with respect to wavelength) λ |nχ|<<1 >> particle size λ comparable to particle size --when rain or ice crystals 13 are present. 14 Single Particle CrossCross--sectionssections vs. χχ For small drops, almost Scattering cross section no scattering, i.e. no 2λ2 Q = χ 6 | K |2 [m 2 ] bouncing from drop since s 3π it’s so small . Absorption cross section λ2 Q = χ 3 Im{ K} [m 2 ] a π χ In the Rayleigh region ( n <<1) => Qa is larger, so much more of the signal is ξ << ξ absorbed than scattered . Therefores a 15 16 RayleighRayleigh--MieMie--GeometricOpticsGeometricOptics Mie Scattering (d/(d/ λλ≈≈11), Along with absorption, scattering is a major cause of the attenuation of radiation by the atmosphere for visible. Scattering varies as a function of the ratio of the particle Mie theory : A complete mathematical-physical theory of the diameter to the wavelength ( d/ λ) of the radiation. scattering of electromagnetic radiation by spherical particles, developed by G. Mie in 1908. When this ratio is less than about one-tenth ( d/ λ<1/10 ), In contrast to Rayleigh scattering, the Mie theory embraces all Rayleigh scattering occurs in which the scattering possible ratios of diameter to wavelength. The Mie theory is very coefficient varies inversely as the fourth power of the important in meteorological optics, where diameter-to- wavelength. wavelength ratios of the order of unity and larger are characteristic of many problems regarding haze and cloud At larger values of the ratio of particle diameter to scattering. wavelength, the scattering varies in a complex fashion When d/ λ ≈ 1 neither Rayleigh or Geometric Optics Theory described by the Mie theory ; applies . Need to use Mie. at a ratio of the order of 10 ( d/ λ>10 ), the laws of geometric Scattering of radar energy by raindrops constitutes another optics begin to apply. significant application of the Mie theory. 17 18 Backscattering CrossCross--sectionsection Backscattering from metal sphere From Mie solution, the backscattered Rayleigh Region defined as ξ = χ 4 2 b 4 | K | field by a spherical particle is χ < 2 for n 5.0 1 ∞ σ ξ (n, χ) = ()−1 m 2( m + )(1 a −b ) = b b χ 2 ∑ m m π 2 m=1 r Observe that where, perfect dielectric K = (nonabsorbent) sphere exhibits large oscillations for χ>1. Hi absorbing and perfect conducting spheres show regularly damped oscillations. ξ = χ 4 19 For conducting sphere (| n|= ∞ ) 20b 9 Scattering by Hydrometeors Liquid water refractivity, n’ Hydrometeors (water particles) In the case of water , the index of refraction is a function of T & f . (fig 5.16) @T=20C 9 − j.25 @ 1 GHz = − = − nw n' jn '' 2.4 j 5.2 @ 30 GHz 4.2 − j 47. @ 300 GHz = For ice. n'i 78.1 For snow, it’s a mixture of both above. 21 22 Sphere pol signature Sizes for cloud and rain drops Co-pol Cross-pol 23 24 Snowflakes Volume Scattering Snow is mixture of ice crystals and air Two assumptions: ρ = 3 ρ = ≤ ρ ≤ 3 – particles randomly distributed in volume-- i .0 916 g/cm a 0 05.0 s 3.0 g/cm incoherent scattering theory . The relative permittivity of dry snow – Concentration is small-- ignore shadowing. ε ' −1 ρ ε ' −1 ds = s ds Volume Scattering coefficient is the total ε ' ρ ε ' − ε ' 3 ds i i 2 ds ε −1 scattering cross section per unit volume . K = i The K factor for dry snow i ε + 2 ds i κκ == σ η = κb = ppN((r(rD))Q)bσ(r()Drd )dD [Np/m] K 1.1 K π 5 D 6 π 5 D 6 sb ∫∫∫ s b ds ≅ i ≈ 5.0 σ = ξ πr 2 = | K |2 ≈ | K |2 ρ ρ bs b s λ4 ds λ 4 i ξ = π 2 ξ = π 2 ξ = σ π 2 ds i o 25 4 o s Q s / r a Q a / r b 26 b / r Total number of drops per unit volume Volume Scattering λ Using... χ = 2πr / λ , ξ = Q /πr 2 and dr = o dχ o s s 2π = = It’s also expressed as Nv ∫ p(r) rd ∫ N(D)dD in units of mm -3 λ3 ∞ κ = o χ 2 χ ξ χ χ [Np/m] s,e,b 2 ∫ p( ) s,e,b ( )d α −α γ 8π p(r) = ar e / 0 c [s,e,b stand for scattering, extinction and backscattering.] − N(D) = N e D / Do or in dB/km units, o ∞ η dB = Κ = × 3 σ [dB/km] b 34.4 10 ∫ N(D) b (D)dD 0 27 28 Noise in Stratus cloud image For Rayleigh approximation --scanning K aa--bandband radar Substitute eqs. 71, 74 and 79 into definitions of the cross sectional areas of a scatterer. 2π 5 D 6 Q = ξ πr 2 = | K |2 s s 3λ 4 w π 2 D 3 Q = ξ πr 2 = Im( − K ) a a λ w π 5 D 6 σ = ξ πr 2 = | K |2 b b λ 4 w D=2r =diameter 29 30 Volume extinction from clouds Relation with Cloud water content Total attenuation is due to gases,cloud, and rain κ = κ +κ +κ a g ec ep This means extinction increases with cloud volume extinction is (eq.5.98) cloud water content. π 2 κ = = − 3 e QadD Im{ K w} D dD κ = κ m c ∫ λ ∫ ec 1 v o where Liquid Water Content LWC or m ) v 6π − − π π κ = .434 Im( −K) [dBkm 1g 1m3 ] 4 3 6 3 1 λ m = ρ r dr =10 D dD o v w ∫ 3 6 ∫ ρ = water density = 10 6 g/m 3 w 31 and wavelength is in cm. 32 Raindrops symmetry Volume backscattering from Clouds Many applications require the modeling of the radar return. For a single drop π 5 D 6 σ = ξ πr 2 = | K |2 b b λ 4 w For many drops (cloud) π 5 η = σ = σ N (D )dD = | K |2 N(D)D 6 dD = vc ∫ b λ4 w ∫ π 5 η = 2 | K w | Z 33 λ4 34 Reflectivity Factor, Z Reflectivity in other references… Is defined as π 5 π 5 = 6 σ = | K |2 Z −12 2 Z ∫ D N (D )dD so that vc 4 w η = λ o 10 4 | Kw | Z λ o and sometimes expressed in dBZ to cover a wider dynamic range of weather conditions.