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9 Rossby Waves
(Holton Chapter 7, Vallis Chapter 5)
9.1 Non-divergent barotropic vorticity equation
We are now at a point that we can discuss our first fundamental application of the equations of motion: non-divergent barotropic Rossby waves! For the derivation of these waves, we will use the simplifying assumptions of
- two-dimensional flow
- barotropic flow
Even though these are rather strong approximations, the solutions turn out to be (surprisingly) relevant to the real atmosphere - and provide deep insight into large-scale midlatitude dynamics. Recall that, in essence, these assumptions are tied to the assumptions that the flow is in near hydrostatic and geostrophic balance (recall that geostrophic flow is non-divergent to leading order on pressure surfaces). As shown in the previous section, under these assumptions the prognostic equation for vorticity reads: D⇣a Dt =^
D
Dt (⇣^ +^ f) =^0 (9.1) That is, the flow is governed by absolute vorticity conservation! As discussed in lecture, this equation was used to provide the first numerical weather forecast! It was a 24-hour forecast (looking forward 24-hours) that took 24-hours to complete! However, it was considered an ultimate success!
9.1.1 Preparing to solve the vorticity equation
Step 1: Linearization Although 9.1 appears simple, we are not generally able to solve it an analytically. (By “solve”, we mean determine an explicit equation for ⇣ that is a function of space and time). This is because the equation is non- linear: the vorticity is a function of u and v, but so is the advection operator inside the material derivative. Thus, our strategy will be to simplify things further by linearizing 9.1 about a basic (⇠ background) state. (You have seen linearization before, for example, in the context of the boussinesq and anelastic equations). First, we decompose the horizontal velocities into a basic state and a perturbation: u = u 0 + u 0 (9.2)
The requirement for the basic state (as yet to be specified) is that it must be a solution to our equation 9.1. The absolute vorticity can then be written as
⇣a = f + ⇣ 0 + ⇣ 0 (9.3)
and thus 9.1 becomes
D Dt (⇣^ +^ f) =^ @t^ (⇣^0 +^ ⇣^
(^0) ) + (u 0 + u 0 )@x (⇣ 0 + ⇣ 0 ) + (v 0 + v 0 )@y (f + ⇣ 0 + ⇣ 0 ) = 0 (9.4)
Step 2: Remove higher-order terms The whole point of linearizing the set of equations around a basic state is that we can then easily neglect the terms that are quadratic or of higher order in perturbations (i.e. throw out terms that are the product of two perturbations or higher). This is only a good next step if the perturbation quantities are small, and we will make this assumption here. That is, e.g.
u 0 >> u 0 and ⇣ 0 >> ⇣ 0 (9.5)
In this case, the approximate vorticity equation becomes:
D Dt (⇣^ +^ f)^ ⇡^ @t^ (⇣^0 +^ ⇣^
(^0) ) + u 0 @x (⇣ 0 + ⇣ 0 ) + u 0 @x ⇣ 0 + v 0 @y (f + ⇣ 0 + ⇣ 0 ) + v 0 @y (f + ⇣ 0 ) = 0 (9.6)
or collecting perturbations to the left-hand-side:
@t ⇣ 0 + u 0 @x ⇣ 0 + u 0 @x ⇣ 0 + v 0 @y ⇣ 0 + v 0 @y (f + ⇣ 0 ) = - @t ⇣ 0 - u 0 @x ⇣ 0 - v 0 @y (f + ⇣ 0 ) (9.7)
The basic state terms on the right-hand-side may be interpreted as (“external”) forcing terms to the pertur- bation vorticity. Step 3: Rewrite in terms of the streamfunction You may think, well hold on, we have a lot of unknowns here and only one equation! This isn’t actually correct though! In actuality, for a given basic state the above equation involves only one unknown! That one unknown is the perturbation streamfunction. Recall that the streamfunction is related to the horizontal velocities and the vorticity in the following way:
= 0 + 0 , and (u 0 , v 0 ) = (-@y 0 , @x 0 ), and (u 0 , v 0 ) = (-@y 0 , @x 0 ) (9.8)
and ⇣ 0 = @x v 0 - @y u 0 = @xx 0 + @yy 0 = r^2 H 0 , similarly ⇣ 0 = r (^2) H 0 (9.9)
Step 4: Basic state simplifications
Linearized partial differential equations (PDEs) with constant coefficients, as our equations above, lead to plane wave solutions. In our 2-dimensional case:
(^0) (x, t) = <{ exp(i(K · x - !t))} = (^) R cos(kx + ly - !t) - (^) I sin(kx + ly - !t) (9.16)
which can be shown by recalling that e i✓^ = (cos ✓ + i sin ✓) and writing
(^0) (x, t) = <{ exp(i(K · x - !t))} = <{( (^) R + i (^) I )(cos (kx + ly - !t) + i sin (kx + ly - !t))} (9.17) with
- = (^) R + i (^) I (the constant, and complex wave amplitude)
- x = (x, y) -! (the constant wave frequency, in units of radians/time)
- K ⌘ (k, l) (the constant wave number vector, in units of radians/length)
- the wave numbers are given by k ⌘ 2 ⇡/ (^) x , l ⌘ 2 ⇡/ (^) y with ( (^) x , (^) y ) the wavelengths
- the phase = (x, t) = K · x - !t = kx + ly - !t, seen also by writing the perturbation streamfunction as 0 = <{ exp(i )} In our 2-D case, lines of constant phase can be written as
= constant = kx + ly - !t or y = - kl x +! l t + constant (9.18)
The second form of this equation is an equation for a line, that is, it is a linear relationship. In 3-D, you would get a linear planar relationship for the surfaces of constant phase. This is where the term “plane waves” get their name.
Plane waves: https://en.wikipedia.org/wiki/Plane wave. If 9.16 is truly a solution to 9.15, then we can plug 9.16 into 9.15 and show that the equality holds. Rather than having to carry-around the < part, we can simply work with the complex solution ( 0 ⇠ exp (i )) and then take the < part as necessary. Note that one can only do this for linear problems! For nonlinear problems you run into the problem that, for example, the real part of the product of two complex numbers is generally not the product of the real parts.
Continuing forward, it is helpful to recall the following:
@x 0 = ik 0 , r (^2) H 0 = - (k^2 + l 2 ) 0 , ) (9.19) @x (r^2 H 0 ) = - ik(k^2 + l 2 ) 0 , and @t r^2 H 0 = i!(k 2 + l^2 ) 0 (9.20)
Inserting the solution into 9.15 therefore gives:
i!(k^2 + l 2 ) 0 + - Uik(k 2 + l 2 ) 0 + ik 0 = 0 (9.21) [!(k 2 + l 2 ) + - Uk(k^2 + l 2 ) + k] 0 = 0 (9.22)
Since 0 = 0 is a trivial solution, the expression in square brackets must then be zero. In this case, we can solve for! and obtain:
! = Uk - (^) k 2 (^) +k l 2 or !ˆ =! - Uk = - (^) k 2 (^) +k l 2 (9.23)
Thus is the famous Rossby wave dispersion relation, where !ˆ is the intrinsic (non-Doppler shifted) fre- quency. As with all dispersion relationships, this equation relates properties of the wave to one another, i.e. waves of different wave number (i.e. wave lengths) have different frequencies and propagate at different speeds according to this relation.
9.3 Wave kinematics refresher (Vallis Chapter 5 Appendix)
9.3.1 Phase speed
Recall that our plane wave solution was (^0) (x, t) = <{ exp(i(K · x - !t))} (9.24)
If we consider, for simplicity, the one dimenstional case of a wave traveling in the x-direction only. Then our solution simplifies to
(^0) (x, t) = <{ exp(i(kx - !t))} = <{ exp(ik(x - c xp t))} (9.25)
where
c xp = !/k (9.26)
From these two equations one can see that the phase of the wave ( ) propagates at the speed c xp and we call this speed the phase speed. Similarly, the phase speed in the y-direction would be c yp = !/l. Generalizing
9.4 Properties of Rossby waves
Back to Rossby waves, which as a reminder, have the dispersion relation
! = Uk - (^) k 2 (^) +k l 2 (9.30)
The phase speed of Rossby waves in the x-direction is then given by
cxp =! k = U - (^) k 2 + l 2 (9.31)
or equivalently c xp =! k = U - (^) K 2 where K^2 = k^2 + l 2 (9.32)
The intrinsic phase speed, or the phase speed relative to the background flow is
c xp - U = - (^) k 2 + l 2 = - (^) K 2 (9.33)
that is, the intrinsic phase speed of Rossby waves is always negative (westward)! These waves always propagate westward relative to the mean flow. Thus, bigger waves travel faster to the west! The Rossby wave group velocity in the direction of the basic flow is
cxg = @!@k = U - K
(^2) - 2 k 2 K^4 =^ U^ +^ ^
k 2 - l^2 (k^2 + l 2 ) 2 (9.34)
For |k| > |l|, cxg > 0 (eastward), or in the opposite direction of the phase speed (see Group Velocity Wikipedia animation to see an example of this). The group velocity in the y direction is given by
c yg = @!@l = 2 Kkl 4 (9.35)
Scales of stationary waves Stationary waves are waves who’s phase lines are stationary relative to the ground. That is, cxp = 0 and so k (^2) S = /U. Using typical midlatitude values for ⇠ 10 -^11 1/(ms) and U ⇠ 10 m/s ...
What is the typical stationary wavelength in midlatitudes for l 2 << k 2? k ⌘ 2 ⇡/ (^) x and assuming l = 0, k^2 = /U and therefore x = 2 ⇡/p( /U) ⇡ 6000 km
What is the intrinsic period of these waves (inverse of frequency)? 1 / !ˆ = - k/ = - 10 -^6 / 10 -^11 = - 10 5 seconds per radian, and converting to days per wave cycle
- 105 ⇥ 2 ⇡/( 60 ⇥ 60 ⇥ 24 ) ⇡ 7 days.
What is the group velocity of these waves? c xg = U + /k 2 = U + U = 2 U - that is, the wave group (and therefore energy) propagates eastward at twice the basic state flow! This is related to “downstream development”.
9.5 Rossby wave propagation mechanism
Properties of the Rossby Wave Solutions The Rossby wave phase speed in the direction of the basic state flow is given by: cp,x ⌘ k = U (^) K 2 , with K^2 ⌘ k^2 + l^2. The intrinsic phase speed cp.x U = /K^2 is always negative, i.e. westward (since > 0 everywhere)! That is, these waves always propagate westward relative to the mean flow. The Rossby wave group velocity in the direction of the basic state flow is given by: cg,x ⌘ k = U K
(^2) 2 k 2 K^4 =^ U^ +^ ^
k^2 l^2 K^4. For |k| > |l| the group velocity is positive (eastward) – in the opposite direction of the phase speed (wave phases moving in the opposite direction to wave groups maybe hard to imagine – watch the animation on the class website to visualize). Stationary waves (phase lines not moving with respect to ground) have cp,x = 0, i.e. K S^2 = /U. Using typical mid-latitude values for 10 ^11 m ^1 s ^1 and U 10 m/s gives a typical stationary wavelength in mid-latitudes of 2 /KS 6000 km and a typical intrinsic period of 2 KS / 6 days – astonishingly close to the observed values for planetary waves. The group velocity for stationary waves for l^2 ⌧ k^2 becomes cg,x ⇡ 2 U , i.e. wave groups (and therefore wave energy) propagate eastward at twice the basic state flow (related to downstream development). Illustration of Rossby Wave Mechanism Vallis, Ch. 5.7:
Hoskins et al. (1985): Hoskins, McIntyre, Robinson, On the use and significance of isentropic potential vorticity maps, QJRMS, 1985.
Figure: Two ways of looking at Rossby wave propagation and why the intrinsic phase speed is always westward.
The intrinsic westward progression of the phase lines of Rossby waves result from the wave perturba-
CSU ATS601 Fall 2015
Thus, the meridional flux of zonal momentum is in the opposite direction of the group velocity in the meridional direction! This result is quite profound. It tells us that as the waves propagate away from the source region (the midlatitudes), eastward zonal momentum will be fluxed back into the midlatitudes (see figure below)! This is in essence why we have westerlies/jet-streams in the midlatitudes. Moreover, this is why the midlatitude jet-stream is referred to as the midlatitude, eddy-driven jet stream.
Rossyby from the source region, the distri- wave propagation away bution of pseudomomentum dissi- pation is broader, and the sum of the two leads to the zonal wind distribution shown, with positive (eastward) values in the region of the stirring. See also Fig. 12.8. will not locally balance in the region of the forcing, producing no net winds. That can only occur if the dissipation is confined to the region of the forcing, but this is highly unlikely because Rossby waves are generated in the forcing region, and these propagate meridionally before dissipating, as we now discuss. 12.1.4 III. Rossby waves and momentum flux We have seen that the presence of a mean gradient of vorticity is an essential ingredient in the mechanism whereby a mean flow is generated by stirring. Given such, we expect Rossby waves to be excited, and we now show how Rossby waves are intimately related to the momentum flux maintaining the mean flow. generated there. To the extent that the waves are quasi-linear and do not interact then^ If a stirring is present in midlatitudes then we expect that Rossby waves will be
Stirring in midlatitudes (by baroclinic eddies) generates Rossby waves that^ Fig. 12.4^ Generation of zonal flow on a^ ˇ-plane or on a rotating sphere. propagate away from the disturbance. Momentum converges in the region of stirring, producing eastward flow there and weaker westward flow on its flanks. Figure: Illustration from Vallis (Chapter 12) of Rossby waves propagating away from a midlatitude source and fluxing momentum back into the midlatitudes to form a zonal jet-stream.
Derivation of 9. We recall that the solution for the perturbation streamfunction can be written as:
(^0) = <{( (^) R + i (^) I )(cos + i sin )} (9.38)
where = (kx + ly - !t) to make notation easier and is a complex constant. Then
u 0 = - @^
0 @y =^ - <{(^ R^ +^ i^ I^ )il(cos^ ^ +^ i^ sin^ )}^ =^ l(^ R^ sin^ ^ +^ I^ cos^ )^ (9.39) v 0 = @^
0 @x =^ <{(^ R^ +^ i^ I^ )ik(cos^ ^ +^ i^ sin^ )}^ =^ - k(^ R^ sin^ ^ +^ I^ cos^ )^ (9.40) Plugging this in we get
u 0 v 0 = - kl (^2) R sin 2 + (^2) I cos^2 + (^2) R I cos sin (9.41) = - kl
R sin^2 ^ +^2 I cos^2 ^ +^2 R^ I^ cos^ ^ sin^
Recalling that the overbar denotes a zonal mean, and assuming periodic boundary conditions around a latitude circle, sin 2 = cos 2 = 12 and sin cos = 0 (9.43)
leads to u 0 v 0 = - 12 2 kl / - kl (9.44)
