TITLE: AUTHOR: The Steay Linear Response of a Spherical Atmosphere to Thermal an Orographic Forcing Brian J. Hoskins Davi J. Karoly YEAR: 1981 REVIEWED: January 25, 2011 Reasons for Review: What are the patterns of waves inuce by large-scale forcing? How oes a spherical atmosphere respon to thermal an orographic forcing? Linear, steay-state assumptions provie a goo guie to preicting the largescale pattern of perturbations inuce by large-scale forcing Subtropical forcing has important implications for the mile latitues Abbreviations/Symbols: Moel Details: H High Pressure K S Local Stationary Wavenumber L Low Pressure NH Northern Hemisphere NL # of Moel Levels PV Potential Vorticity Hemispheric baroclinic moel o Linearize an steay state o 5 vertical layers Not enough to capture longest waves Basic numerical moel o Primitive equation; σ-coorinate Hoskins an Simmons (1975) o Spectral-transform technique in horizontal
o 2 n orer finite ifferences in the vertical o Goo for stuy of baroclinic instability o Zonal wavenumber (m) is inepenent o Vorticity (ξ) ξ = ξ m+1 j,m P m m+1 o Divergence (D) (μ) + ξ m+3 D = D m j,m P m m (μ) + D m+2 j,m P m m+3 j,m P m m+2 (μ) + + ξ m+j j,m (μ) + + D m+j 1 j,m P m m+j m P m+j 1 (μ) e imλ (μ) e imλ o Zonal Wavenumber (m) may be escribe by the vector: X = (ξ, id, T, lnp ) o Linearize primitive equations with no source or sink terms: X = iax A is a square matrix of sie (3 x NL + 1) x (J + 1)/2 r n factors are inclue to make more elements of A of orer 1. Steay problem o 0 = iax + DX + F o X = i(a id) 1 F (steay response to forcing F) Dissipation on ξ o λ U (σ)ξ K h 4 ξ Dissipation on D o λ T (σ)t K h 4 T o λ effects of bounary layer o K = 2.338e16 m 4 s -1. Temperature perturbation o T = T lnp T(σ) lnσ o Dissipation matrix D is not iagonal Vertical Levels o p/p * = 0.1, 0.3, 0.5, 0.7, 0.9 Spectral truncation (J) = 25 o Each zonal wavenumber an level of each variable is escribe by 13 complex coefficients 26 zonal wavenumbers can be resolve o Variance in high wavenumbers is small, so they are often neglecte Zonal flow o Symmetry about equator
o Oort an Rasmusson (1971) Notes: First accounts of steay linear response of atmosphere o Charney an Eliassen (1949) o Smagorinsky (1953) Stuy of orographic forcings o Grose an Hoskins (1979) Barotropic moel to stuy Rossby response in zonal flows inuce by simple mountains Stuy of thermal forcings o Difficult in milatitues ue to baroclinicity o In tropics, warmer SSTs may inuce extra convective heating One typical case o Zonal flow NH 300 mb winter flow o Elliptical negative vorticity source equator an 30 N; 30 W an 30 E o Total vorticity source = 0 Uniform positive vorticity source is place elsewhere at same latitues o Solutions are linearly proportional to the magnitue of forcing o Max upper level ivergence = 4e-6 s -1. Extra 10 mm of rain per ay o After 10 ays Solutions on t change much Propagation to the NORTH an EAST from source Alternating H an L centers Begins with H just to the west of the forcing center o Agreement with Bjerknes (1966) Strengthening of the subtropical jet Trough to the north Patterns reveal an equivalent barotropic structure o Largest amplitues in upper troposphere Thermal forcing o Thermal sources iffer by latitue an vertical istribution o β-plane vorticity equation u ξ x + βv = fw z o Potential temperature equation u θ x + v θ y + w θ z = θ 0 Q g fu v z + fu z v + w N 2 = Q o Tropics
γ (Hel 1978) is small Near-surface heating results in extremely large v Heat source away from surface balance by upwar motion Surface trough must be to the west of thermal source o Milatitues γ is large Heating at any level is balance by hor. temperature avection If H Q >> H u Deep case Creation of vorticity (via β) must be balance by vortex shrinking If H Q ~ H u Shallow case Low-level heating balance by zonal temperature avection Surface trough must be to the east of the thermal source o Subtropical forcing 15 N is a subtropical forcing?? Heat source Centere @ 500mb, 15 N Elliptical (long way in x-irection) Inicative of eep convective profile Low-level amping to represent bounary layer Max vertically-average heating rate = 2.5 K ay -1. To the west H above the max forcing level L below the max forcing level 300mb Vorticity Very similar to barotropic moel response (Fig. 1) Wave train moves north an east o Split right near north pole Longer wavenumbers continue over the top of the globe Smaller wavenumbers turn back towar the equator Height fiel emphasizes the polar wave train o Mi-latitue forcing Heat source Centere at 45 N Circular, but same area as ellipse in subtropical section Rather large for milatitues Surface L is locate 18 to the east of the source Upper-level H is locate 60 to the east
Wave train initializes 30 ownstream (to the east) Split still occurs near north pole Height fiel istribution is very similar to subtropical forcing If elliptical heat source is use, results are nearly ientical Easier to force a height fiel anomaly from the subtropics than from the mi an high latitues 900mb T is negative in the region of heating Base on thermal win relationship Shallow heating If heating is centere near the surface (instea of 500mb) Surface L is strengthene T is now positive near the heat source Positive vorticity ecreases with height, which leas to positive T Shallow heating is more common in milatitues Deep heating Positive vorticity increases with height, which leas to negative T o Other cases Heat Source on equator Response away from equator is small Between 20 N an 30 N, the balance of heating by vertical motion is replace by the balance of heating by horizontal avection Doubling heat raius at 45 N oubles the height anomalies Shallow heat source has a much larger low-level response if place at 60 N rather than 45 N. Smaller T graient, so stronger meriional wins are require for same balance at lower latitues Damping experiments Fluxes of heat an PV ominate Damping temperature an/or velocity oes not have a major impact on the solutions 0.7*U Polewar an eastwar train with slightly shorter wavelength As U is increase (increase 0.7 factor), more structure appears Aing constant angular velocity (3 m s -1 ) at the equator oes not change the solution by much Orographic Forcing o Barotropic Case Columns are squashe on the upslope
Generates anticyclonic vorticity Columns are stretche on the ownslope Generates cyclonic vorticity Long wavelengths β is ominant Cyclone over the rige Short an meium wavelengths Zonal vorticity avection ominates Anticyclone over the rige k S is taken from mi- to upper- troposphere k S = 7000 km Shortwave case is most relevant since most waves are shorter than 7000 km o Baroclinic Case Anticyclone always sits (approx.) over the mountain Upslope cooling compensate by southerly flow (warm) Downslope warming compensate by northerly flow (col) Long wavelengths Waves are: o Vertically-propagating o Slope westwar with height o Warmest air ahea of trough Anticyclone is slightly isplace to the west of mountain Shorter wavelengths Waves are trappe with H over mountain an troughs to either sie k S is taken from lower troposphere k S = 3000 km Longwave case is most relevant since most waves are longer than 3000 km o Barotropic theory with amping an baroclinic theory ten to give the same pressure phase relationship over a mountain o Circular mountain at 30 N Upper-level amping for loss of wave activity to stratosphere Mountain iameter = 45 of latitue Mountain height = 2 km Mi-latitues ( > 30 N) Column win is westerly Surface H is 3 west of the mountain Surface L is 33 east of the mountain Westwar tilts with height Max amplitues in upper troposphere Low latitues ( < 30 N)
Low-level win is easterly, mi- to upper-level win is westerly Surface H is 18 to the east Surface L is 9 to the west Pressure pattern aloft is consistent with mi-latitue response 2 wave trains Longwave polar wavetrain Shortwave subtropical wavetrain o Other mountain cases Aing 3 m s -1 to global zonal win Shrinks region of equatorial easterlies Wavetrains have slightly greater wavelength Reuce horizontal mountain imensions by 2 Similar pattern Polar wavetrain amplitue reuce by factor of 4 Move mountain to 60 N Similar polar wavetrain Equatorwar wavetrain of long wavelength Elongate mountain (area conserve) Amplitue of long wavelengths is reuce Increase amplitue near surface westerlies o Earth orography Himalayas, followe by Rockies, are responsible for observe height fiel perturbations Shorter wavelength ifferences Particularly in Pacific sector Maybe linearize theory is not vali o Ambient flow stronger than zonal average Aleutian an Atlantic L are likely thermally force o Orographic uplift may be irectly associate with Siberian H an west U.S. H Important Points o Away from the source region the wavetrains prouce in the upper troposphere of the baroclinic moel an in the barotropic moels are qualitatively an even quantitatively very similar. Wavetrains in baroclinic moel have equivalent barotropic structure. Amplitues are largest in upper troposphere o Long wavelengths propagate strongly polewar as well as eastwar Wavetrain is like a great circle
o Shorter wavelengths appear to be trappe equatorwar of the polewar flank of the milatitue jet Splitting of the wavetrains Possible blocking region ownstream where long, polewar wavetrain an short, subtropical wavetrain are out of phase o Easiest way to reprouce amplitue in mi an high latitues is by subtropical forcing o The low-level temperature fiel of milatitue thermal forcing is crucially epenent on the vertical istribution of the source. Explaine via thermal win relation Rossby Wave Rays o Barotropic Rossby waves in a slowly varying meium Horizontal streamfunction perturbation + u t M ψ x 2 + 2 ψ x 2 y 2 + β ψ M = 0 (5.9) x β M = 2Ω a cos2 φ 1 y cos 2 φ y (cos2 φu M ) Dispersion relation ω = u M k β Mk k 2 +l2 (from 5.9) Group velocities c g = (u g, v g ) u g = ω = ω + 2β Mk 2 k k (k 2 +l 2 ) 2 v g = ω = 2β Mkl l (k 2 +l 2 ) 2 Ray is efine to be everywhere in the irection of the local c g. o Energy propagate along a ray at the group velocity o k an ω are constant along a ray; l varies For ω = 0 Slope = l/k c g = 2 k u K M S o Energy propagates at ouble the backgroun basic flow Use a solution of the form ψ = P(y)e i(kx ωt) an plug into (5.9) See erivations The latitue where K S = k provies a turning point for the ray If u = 0 That latitue is where rays become more meriional Group velocity tens to 0 o Constant angular velocity flow
See erivations for (5.24) The fastest energy propagation time aroun the sphere is ~15.5 ays At any turning point, the wavelength is 5000km Resonance affects the amplitue of the ownstream wavetrain but otherwise has a small role. In applying wave theories, the zonal perioicity of the sphere shoul be roppe. o More realistic flows Longer wavelengths propagate polewar Wavenumbers 1-3 Shorter wavelengths are trappe in the northern flank of the milatitue jet an o not propagate polewar Wavenumbers 4+ Move the source polewar (to 45 N) Only wavenumbers 1-3 Details of the basic flow are unimportant to the solutions given Using 500 mb zonal flow instea of 300 mb zonal flow allows wavenumber 4 train to be more polewar 5 an 6 are not as trappe either If the flow is increase (1.2*u), wavenumber 3 train has a chance to go all the way aroun the sphere Dissipation o When small, resonance is important an makes the resulting structures sensitive o When ieal, no sensitivity for the structure of the responses o When larger, same patterns but reuce amplitue
Derivations Derive (5.24): β M = 2Ω a cos2 φ y 1 cos 2 φ Constant angular velocity flow: u M = aω β M = 2Ω a cos2 φ y Replace y with φ : = cosφ y a φ y (cos2 φu M ) (5.10) 1 cos 2 φ y (cos2 φaω ) β M = 2Ω a cos2 φ cosφ a β M = 2Ω a cos2 φ cosφ a 1 cosφ φ cos 2 φ a 1 φ acosφ φ (cos2 φaω ) φ (cos2 φaω ) simplify Take first erivative: β M = 2Ω a cos2 φ cosφ a β M = 2Ω a cos2 φ cosφ a φ β M = 2Ω a cos2 φ ( 2ω ) cosφ a 1 ( 2cosφsinφaω ) acosφ ( 2sinφω ) φ φ (sinφ) simplify Take secon erivative: β M = 2Ω a cos2 φ + 2ω cosφ cosφ a β M = 2cos2 φ (Ω + ω ) (5.24) a
Derive (5.11) t + u M ψ x 2 + 2 ψ x 2 y 2 + β ψ M = 0 (5.9) x Solve for ψ: Plug an chug: t + u M ψ = e i(kx+ly ωt) x (i2 k 2 + i 2 l 2 ) + iβ M k = 0 iωk 2 iu M k 3 + iωl 2 iu M kl 2 + iβ M k = 0 rop the i s ω(k 2 + l 2 ) = u M k(k 2 + l 2 ) β M k ω = u M k β Mk (k 2 +l 2 ) (5.11) Derive (5.12) u g = ω k = k u Mk β Mk (k 2 + l 2 ) u g = u M + β Mk = k (k 2 +l 2 ) β M 2β Mk 2 (k 2 +l 2 ) (k 2 +l 2 ) 2 β M (k 2 + l 2 ) 2β Mk 2 (k 2 + l 2 ) 2 u g = ω k 2β Mk 2 (k 2 +l 2 ) 2 (5.12) Derive (5.13) v g = ω l = l β M k (k 2 + l 2 ) v g = 2β Mkl (k 2 +l 2 ) 2 (5.13)
Derive (5.19): t + u M ψ x 2 + 2 ψ x 2 y 2 + β ψ M = 0 (5.9) x Solve for ψ: t + u M ψ = P(y)e i(kx ωt) x i2 k 2 P + 2 P y 2 + iβ MkP = 0 iωk 2 P ik 3 u M P iω 2 P y 2 + iu Mk 2 P y 2 + iβ MkP = 0 2 P y 2 (u Mk ω) + P( u M k 3 + ωk 2 + β M k)=0 2 P y 2 (u Mk ω) + P( k 2 (u M k ω) + β M k) = 0 1 P (u M k ω) 2 (u y Mk ω) + P( k 2 (u 2 M k ω) + β M k) = 0 2 P y Mk 2 + P β (u M k ω) k2 = 0 Multiply by 1 u M k 1 u M k 2 P y 2 + P βmk Define K S : u M k 1 ω u M k k2 = 0 K S = β M l 2 (y) = K S 2 1 2 u M 1 ω k2 u M k 2 P y 2 + l2 (y)p = 0 (5.19)