www.metoffice.gov.uk © Crown Copyright 2017, Met Office Where do the horizontal waves go? Propagation, reflection and absorption

Propagation, reflection and absorption wavenumber k = 1 Maps of l2 wavenumber k = 2 Orange and red: propagation allowed Dark blue regions are easterly winds: absorption of all waves Light blue regions, strong winds: reflection of short (high wavenumber) waves Absorptive Reflective wavenumber k = 3 wavenumber k = 4 www.metoffice.gov.uk © Crown Copyright 2019, Met Office

Barotropic Rossby Waves Simplest case: barotropic, no forcing → constant absolute vorticity 𝐷 𝜁+𝑓 𝐷𝑡 =0 𝛽= 𝜕𝑓 𝜕𝑦 Beta plane: Coriolis parameter f is linear 𝜕 𝜕𝑡 +𝑢 𝜕 𝜕𝑥 +𝑣 𝜕 𝜕𝑦 𝜁+𝛽𝑣=0 We’ve got rid of the forcing terms on the RHS of our original equation

Barotropic Rossby Waves (cont.) 𝜕 𝜕𝑡 +𝑢 𝜕 𝜕𝑥 +𝑣 𝜕 𝜕𝑦 𝜁+𝛽𝑣=0 From previous slide Perturbation to zonal mean flow: 𝑢= 𝑢 + 𝑢 ′ v= 𝑣 ′ 𝜁= 𝜁 + 𝜁 ′ By construction and hence and similarly for other quantities 𝑢′ =0 𝑢 𝑢′ = 0 𝑢 ′ =− 𝜕 𝜓 ′ 𝜕𝑦 𝑣 ′ = 𝜕 𝜓 ′ 𝜕𝑥 Perturbation streamfunction: 𝜁 ′ = 𝛻 2 𝜓 ′ 𝜕 𝜕𝑡 + 𝑢 𝜕 𝜕𝑥 𝛻 2 𝜓 ′ + 𝛽− 𝑢 𝑦𝑦 𝜕 𝜓 ′ 𝜕𝑥 =0 Split flow into mean and perturbations Assume mean flow is only in zonal direction Holton, Introduction to dynamic meteorology ; James, Introduction to circulating atmospheres ; Vallis, Atmospheric and oceanic fluid dynamics

Barotropic Rossby Waves (cont.) 𝜕 𝜕𝑡 + 𝑢 𝜕 𝜕𝑥 𝛻 2 𝜓 ′ + 𝛽− 𝑢 𝑦𝑦 𝜕 𝜓 ′ 𝜕𝑥 =0 From previous slide Try a wave solution: 𝜓 ′ =𝐴 𝑒 𝑖 𝑘𝑥+𝑙𝑦−𝜔𝑡 k = zonal wavenumber = 2π/Lx where Lx is the zonal wavelength l = meridional wavenumber = 2π/Ly ω = frequency = 2π/T where T is the period A solution exists only if: 𝜔= 𝑢 𝑘− 𝛽− 𝑢 𝑦𝑦 𝑘 𝑘 2 + 𝑙 2 Rossby wave dispersion relation Holton, Introduction to dynamic meteorology ; James, Introduction to circulating atmospheres ; Vallis, Atmospheric and oceanic fluid dynamics

Rossby Wave Dispersion 𝜔= 𝑢 𝑘− 𝛽− 𝑢 𝑦𝑦 𝑘 𝑘 2 + 𝑙 2 Rossby wave dispersion relation (From previous slide) Consider a 1D wave propagating in x direction, phase ϕ(x,t)=kx-ωt The phase is constant for an observer moving at the phase speed 𝐷𝜙 𝐷𝑡 = 𝐷 𝐷𝑡 𝑘𝑥−𝜔𝑡 =𝑘 𝐷𝑥 𝐷𝑡 −𝜔=0 𝑐 𝑥 = 𝐷𝑥 𝐷𝑡 = 𝜔 𝑘 Hence phase speed So dispersion relation gives the phase speed for different wave numbers Wavenumbers move at different speeds so wave packet will spread out (disperse) cx generally westward relative to mean flow Holton, Introduction to dynamic meteorology ; James, Introduction to circulating atmospheres ; Vallis, Atmospheric and oceanic fluid dynamics

Group Velocity 𝑐 𝑔𝑦 = 𝜕𝜔 𝜕𝑙 = 2 𝑢 2 𝑘 𝛽− 𝑢 𝑦𝑦 𝑢 − 𝑘 2 1 2 𝛽− 𝑢 𝑦𝑦 𝑐 𝑔𝑦 = 𝜕𝜔 𝜕𝑙 = 2 𝑢 2 𝑘 𝛽− 𝑢 𝑦𝑦 𝑢 − 𝑘 2 1 2 𝛽− 𝑢 𝑦𝑦 𝑐 𝑔𝑥 = 𝜕𝜔 𝜕𝑘 = 2 𝑢 2 𝑘 2 𝛽− 𝑢 𝑦𝑦 Eastward propagation faster for shorter wavelengths (higher k) Meridional propagation possible only if term in {} is positive stops in strong winds (reflections) stops at zero wind line (absorption) easier for longer wavelengths (low k) can in principle be prevented by curvature of wind field (uyy) Square root has 2 values → N and S propagation (symmetry for equatorial sources) Holton, Introduction to dynamic meteorology ; James, Introduction to circulating atmospheres ; Vallis, Atmospheric and oceanic fluid dynamics

Rossby wave absorption and reflection Meridional wavenumber (from dispersion relation with ω=0) 𝑙= 𝛽− 𝑢 𝑦𝑦 𝑢 − 𝑘 2 1 2 Maps of l2 wavenumber k = 1 wavenumber k = 2 Orange and red: propagation allowed Dark blue regions are easterly winds: absorption of all waves Light blue regions, strong winds: reflection of short (high wavenumber) waves Orange and red: propagation allowed Dark blue regions are easterly winds: absorption of all waves Light blue regions, strong winds: reflection of short (high wavenumber) waves Absorptive wavenumber k = 3 Reflective wavenumber k = 4 Scaife et al 2016

Ray Tracing 𝑐 𝑔𝑦 = 2 𝑢 2 𝑘 𝛽− 𝑢 𝑦𝑦 𝑢 − 𝑘 2 1 2 𝛽− 𝑢 𝑦𝑦 𝑐 𝑔𝑦 = 2 𝑢 2 𝑘 𝛽− 𝑢 𝑦𝑦 𝑢 − 𝑘 2 1 2 𝛽− 𝑢 𝑦𝑦 𝑐 𝑔𝑥 = 2 𝑢 2 𝑘 2 𝛽− 𝑢 𝑦𝑦 Group velocity 𝛽= 2Ω cos Φ 𝑎 k= 2𝜋 𝐿 𝑥 𝑢 = Calculate cgx and cgy Calculate new position after timestep (2 hours) Repeat Holton, Introduction to dynamic meteorology ; James, Introduction to circulating atmospheres ; Vallis, Atmospheric and oceanic fluid dynamics ; Scaife et al 2016

Ray Tracing From the W Pacific: Wave2 propagates Wave3 reflects propagation (k = 2) propagation (k = 3) propagation k = 4 Rossby Wave Ray Paths From the W Pacific: Wave2 propagates Wave3 reflects From the E Pacific: Wave4 reflects k = 2 k = 2 k = 3 k = 4 Scaife et al 2016

Rossby Wave Sources 𝑆=− 𝜁𝛻. 𝒗 𝜒 + 𝒗 𝜒 .𝛻𝜁 =−𝛻. 𝒗 𝜒 𝜁 Decompose velocity in rotational (vψ) and divergent(vχ) parts Rossby wave source 𝑆=− 𝜁𝛻. 𝒗 𝜒 + 𝒗 𝜒 .𝛻𝜁 =−𝛻. 𝒗 𝜒 𝜁 Stretching Advection Rossby waves are generated where divergent flow interacts with vorticity gradient Sardeshmukh and Hoskins 1988 ; James, Introduction to circulating atmospheres ; Scaife et al 2016

Ray Tracing: teleconnections TEP TA Rays intersect main centres (of 200 hPa geopotential height anomalies) from a few common sources Wave 2, 3 mainly responsible as wave 4 rarely propagates We have a theory for the teleconnections from tropical rainfall IO TWP Scaife et al 2016, Zhao et al “Dynamics of an Interhemispheric Teleconnection across the Critical Latitude through a Southerly Duct during Boreal Winter”, J Clim 2015

Vertically propagating Rossby waves www.metoffice.gov.uk © Crown Copyright 2017, Met Office

Observed zonal mean flow July Heavy contours = westerlies Thin contours = easterlies Coloured contours = isotherms Stratospheric night jet Tropospheric westerly jet From surface to jet core, u gets stronger with height as theta increases – the thermal wind relation… North pole South pole

Equations of motion (hydrostatic approx.) 𝐷𝑢 𝐷𝑡 −𝑓𝑣+ 𝜕Φ 𝜕𝑥 =𝑋 𝐷𝑣 𝐷𝑡 −𝑓𝑢+ 𝜕Φ 𝜕𝑦 =𝑌 Horizontal momentum equations on mid-latitude beta plane 𝜕Φ 𝜕𝑧 = 𝑅 𝐻 𝑇 𝜕𝑢 𝜕𝑥 + 𝜕𝑣 𝜕𝑦 + 1 𝜌 0 𝜕( 𝜌 0 𝑤) 𝜕𝑧 =0 𝐷𝑇 𝐷𝑡 + 𝜅𝑇 𝐻 𝑤= 𝐽 𝑐 𝑝 Vertical momentum equation (pressure coordinates) in hydrostatic approximation 𝑧=𝐻 ln 𝑝 𝑅 𝑝 Continuity equation (pressure coordinates) Thermodynamic energy equation (Following Holton ch. 10)

Eulerian averaging We can expand any variable in terms of its longitudinally averaged value and a perturbation We can continue our crossing out spree using the typical magnitudes of the various variables (and hence their products): 𝐴= 𝐴 +𝐴′ We substitute into the equations of motion, and zonally average both sides. Then we go on a crossing out spree using: 𝐴 𝐴′ =0 𝐴 𝐴′ =0 is independent of x (and of course the usual crossing out of higher order terms) so (also applies to terms like ) 𝐴 𝐵′ =0

After a lot of expanding and crossing out… Starting from the equation for u Starting from the equation for v Zonal mean zonal momentum equation 𝜕 𝑢 𝜕𝑡 − 𝑓 0 𝑣 = − 𝜕 𝑢 ′ 𝑣 ′ 𝜕𝑦 + 𝑋 𝜕 𝑇 𝜕𝑡 + 𝑁 2 𝐻 𝑅 𝑤 = − 𝜕 𝑣 ′ 𝑇′ 𝜕𝑦 + 𝐽 𝑐 𝑝 Geostrophic balance 𝑓 0 𝑢 = − 𝜕 Φ 𝜕𝑦 𝑓 0 𝜕 𝑢 𝜕𝑧 + 𝑅 𝐻 𝜕 𝑇 𝜕𝑦 = 0 Eq 1 Eq 3 Zonal mean thermodynamic energy equation And combining with the hydrostatic approx Thermal wind equation Eq 2 Eq 4 with buoyancy frequency Eq 4 : as T decreases with height, u increases Eqs 1 and 2 : unless we have some sort of meridional circulation, then the eddy heat flux divergence and eddy momentum flux divergence would destroy the thermal wind balance (it doesn’t destroy it, but it does reduce it). 𝑁 2 ≡ 𝑅 𝐻 𝜅 𝑇 0 𝐻 − 𝑑 𝑇 0 𝑑𝑧 Remember this when we get to Rossby waves

Schematic of the meridional circulation Reduced thermal wind Hadley cell 𝑓 𝑣 <0 Ferrel cell 𝜕 𝑢′𝑣′ 𝜕𝑦 ≪0 Latent heating Radiative cooling Adiabatic warming Observed max heat flux Adiabatic cooling 𝑣′𝑇′ 𝑚𝑎𝑥 𝜕 𝑣′𝑇′ 𝜕𝑦 >0 𝜕 𝑣′𝑇′ 𝜕𝑦 <0 about 50 N 𝜕 𝑢′𝑣′ 𝜕𝑦 <0 𝑓 𝑣 >0 Equator High latitudes

Take-home messages from this section Given that we have eddy heat and momentum fluxes, in the absence of a meridional circulation the zonal winds wouldn’t stay in geostrophic balance The eddy momentum flux divergence and eddy heat flux divergence together drive a meridional circulation which restores geostrophic balance The two working together is what sets the background for EP fluxes Hence existence of thermally indirect Ferrel cell But also has the effect of reducing the thermal wind (shear is less pronounced with height up to jet maximum than we’d otherwise expect) However - 𝑢 does still increase with height – relevant in context of wave breaking 𝜕 𝑢′𝑣′ 𝜕𝑦 𝜕 𝑣 ′ 𝑇′ 𝜕𝑦

Setting up the problem Reminder: still working in log-pressure vertical coordinates Scale height Assume isothermal atmosphere with density and Brunt-Väisälä frequency 𝑧=𝐻 ln 𝑝 𝑅 𝑝 𝐻=𝑅 𝑇 0 𝑔 𝜌 𝑅 𝑁 (Explanation lifted from James, 1994 NB my slides make minor notational changes for consistency with Holton.) Density varies with height 𝜌 𝑅 = 𝜌 0 𝑒 − 𝑧 𝐻 Basic state consists of constant zonal wind 𝑈 Partition stream function into zonal mean and eddy parts 𝜓= −𝑈𝑦+ 𝜓 ′ Mid latitude beta plane

Wave-like solutions Using we linearize the potential vorticity equation 𝜓= −𝑈𝑦+ 𝜓 ′ 𝑈 𝜕 𝜕𝑥 𝛻 2 ψ ′ + 𝑓 2 𝜌 𝑅 𝜕 𝜕𝑧 𝜌 𝑅 𝑁 2 𝜕𝜓′ 𝜕𝑧 + 𝛽 𝜕𝜓 𝜕𝑥 = 0 Decrease in density with height Want to keep wave activity the same Hence increase in amplitude with height and look for wavelike solutions 𝜓 ′ = Ψ 𝑧 𝑒 𝑧 2𝐻 𝑒 𝑖 𝑘𝑥+𝑙𝑦 Substituting this solution back in, we get: 𝑑 2 Ψ 𝑑𝑧 2 + 𝑁 2 𝑓 2 𝛽 𝑈 − 𝐾 2 − 𝑓 2 4 𝑁 2 𝐻 2 Ψ = 0 James 1994 Eq 5

Wavelike solutions – conditions for vertical propagation Needs to be positive for wavelike solutions 𝑑 2 Ψ 𝑑𝑧 2 + 𝑁 2 𝑓 2 𝛽 𝑈 − 𝐾 2 − 𝑓 2 4 𝑁 2 𝐻 2 Ψ = 0 Easterly – no waves! Moderate westerlies, Low total wave number Strong westerly – no waves! Large total wave number – no waves! 𝐾 = 𝑘 2 + 𝑙 2 James 1994

Wave breaking as height increases We want wavelike solutions of 𝑑 2 Ψ 𝑑𝑧 2 + 𝑁 2 𝑓 2 𝛽 𝑈 − 𝐾 2 − 𝑓 2 4 𝑁 2 𝐻 2 Ψ = 0 Eq 5 But we know u increases with height 𝑓 0 𝜕 𝑢 𝜕𝑧 + 𝑅 𝐻 𝜕 𝑇 𝜕𝑦 = 0 Eq 4

Wave breaking as height increases 𝛽 𝑈 − 𝐾 2 − 𝑓 2 4 𝑁 2 𝐻 2 Wave of given wavenumber propagates vertically In vertical U increases with height We reach a height at which Wave crosses critical line from propagating to evanescent solutions 𝛽 𝑈 < 𝐾 2 James 1994

Rossby Wave breaking and sudden stratospheric warmings Shift in, and disruption of, polar night jet over 4 days Arrows are Eliassen-Palm vectors www.metoffice.gov.uk © Crown Copyright 2019, Met Office Coy, JAS, 2009, “Planetary wave breaking… in SSW of Jan 2006”

Rossby Wave breaking and sudden stratospheric warmings Example of this in the next lecture, from last winter. Easterly anomalies in stratosphere gradually propagate downwards www.metoffice.gov.uk © Crown Copyright 2019, Met Office

www.metoffice.gov.uk © Crown Copyright 2017, Met Office Summary

Summary Horizontally propagating Rossby Waves In terms of vorticity – physical explanation The vorticity equation Wave solutions and the dispersion relation Propagation, absorption and reflection, and ray-tracing Vertically propagating Rossby Waves Thermal wind – stratospheric night jet The meridional circulation Wave breaking with increasing height Wave propagation and absorption helps explain teleconnections between the tropics and extra-tropics Wave breaking in the stratosphere explains how disruption of the stratospheric polar vortex impacts the surface We can successfully represent these in dynamical models This underlies a lot of the skill in dynamical models www.metoffice.gov.uk © Crown Copyright 2019, Met Office

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