Topics
Rossby waves: correction/remake Hovmöller diagram
Thermal wind and temperature advection (section 1.35)
How do you define a front? (section 1.36)
Fronts in middle latitudes
Conceptual model of “ baroclinic life-cycle ”
Frontogenesis  and  frontolysis  (sec8on  1.37)
Q-­‐vector  (sec8on  1.37)
( a.j.vandelden@uu.nl
)
( http://www.phys.uu.nl/~nvdelden/dynmeteorology.htm
)
Rossby waves in a Hovmöller diagram http://www.staff.science.uu.nl/~delde102/Hovmoller[1949].pdf
F
IGURE
1.93
. The 500 hPa geopotential, gz ( given in “dynamical decametres”=100 m 2 s -2 ), as a function of time and longitude in
November 1945. The values are average values of geopotentials between 35°N and 60°N. Ridges are shown by horizontal hatching; troughs are shown by vertical hatching. The slanted straight lines indicate a succession of maximum development of troughs and ridges. Note that most ridges and troughs propagate in eastward direction. This means that they are embedded in a strong zonal average eastward flow. Source of this figure: Hovemöller, E., 1949: The
Trough-and-Ridge diagram. Tellus , 1,
62-66.
04/12/15
1

Rossby waves in a Hovmöller diagram
FIGURE 1.93
. Remake of Hovmöller ’ s diagram, using dat based on ERA-20C renalaysis. The 500 hPa geopotential, gz (given in “ dynamical decametres ” =100 m 2 s -2 ), as a function of day and longitude in November
1945. The values are average values of geopotentials between 35°N and 60°N.
Ridges are shown by red contours; troughs are shown by blue contours
(contour interval/value is identical to original diagram by Hovmöller). The slanted red line identifies a trough. The slanted blue line indicates a succession of maximum development of troughs and ridges, thus identifying a group of waves.
Rossby waves in a Hovmöller diagram
04/12/15
U = 20 m/s
2

Zonal wind in ERA-20C, Nov. 1945
Background zonal wind from theory (previous slide):
U = 20 m/s
Zonal mean zonal wind from
ERA-20C reanalysis:
U ≈ 13 m/s
04/12/15
Example of dispersion:
Gravity-inertia waves in shallow water
“ Rossby adjustment problem ”
5000 km constant density h x
€
Dispersion relation:
ω
2
= gh l
2
+ f
2 time
18 hrs phase group
€
Next week: more on dispersion of buoyancy waves and sound waves
See: Chapter 5 Lecture notes
3

Fronts connected to middle-latitude cyclone
(in the middle of its life cycle)
Green:clouds
04/12/15
Figure 1.82.
Geostrophic wind: v g
= g f
# ∂ z
%
$ ∂ x
&
(
' p
; u g
= − g f
# ∂ z
%
$ ∂ y
&
(
' p
Hydrostatic balance:
∂ p
= − ρ g
∂ z
€
Thermal wind equation:
€
?
∂ z
∂ ln p
∂ v g
∂ ln p
= −
R f
∂ T
€ x
;
∂ u g
∂ ln p
=
R ∂ T f ∂ y
= −
RT g
€
4

Previous slide this lecture:
Thermal wind equation:
∂ v g
∂ ln p
= −
R f
∂ T
∂ x
;
∂ u g
∂ ln p
=
R ∂ T f ∂ y
In vector notation: (verify)
∂
!
g
∂ ln p
= −
R f
ˆ
×
!
∇ T
€
The “ thermal wind ” :
!
T
≡
!
g
( ) −
!
g
( )
€
= −
R f
∫ p
1 p
0
( ˆ ×
!
∇ T ) d ln p
The thermal wind is parallel to the isotherms!
€
Repeat:
!
T
≡
!
g
( ) −
!
g
( ) = −
R f
∫ p
1 p
0
( ˆ ×
!
∇ T ) d ln p
The thermal wind is parallel to the isotherms!
04/12/15
Cold advection: backing with height Warm advection: veering with height
5

The thermal wind in a mid-latitude cyclone
04/12/15
NOAA,  channel  4  (infra-­‐red).  12  feb  1996,  1313  UTC
850 hPa
6

04/12/15
500 hPa
The thermal wind in a mid-latitude cyclone
Warm conveyor belt
NOAA,  channel  4  (infra-­‐red).  12  feb  1996,  1313  UTC
7

Mid-latitude cyclone:conceptual model
04/12/15
(Palmen and Newton, 1969)
Another example:
The thermal wind in a mid-latitude cyclone
F
IGURE
1.60. NOAA image in channel 4
(IR) made on March 3, 1995, at 0157
UTC.
Fits into Shapiro-Keyser model (slide 8)
8

Conceptual model of life-cycle of unstable baroclinic wave
Starts with a “ wave ” in the “ Polar Front ” . This wave grows due to the instability of the thermal wind.
04/12/15   a few days Figure 1.83.
The thermal wind in a mid-latitude cyclone
850 hPa
(figure 1.86)
9

The thermal wind in a mid-latitude cyclone
500 hPa
04/12/15
(figure 1.86)
Let us first define what a front is in mathematical terms;
Then study mechanisms that lead to the formation of fronts;
Then study mechanisms that lead to the characteristic frontal morphology, seen in previous slides
•   What is a front?
•   How are fronts formed?
•   Why are fronts associated with clouds
•   What is the relation between fronts and jetstreams?
10

A front separates two air masses
Air masses are characterized by: potential temperature, humidity or potential vorticity.
Gradients of these quantities can be very sharp
Why?
Potential vorticity at 350 K on 28 Jan. 2007, 00 UTC
€
The standard definition of front-intensity is in terms of temperature
A front separates two air masses
Suppose: air masses are characterized by potential temperature ,
Intensity of front is measure by where
∇ h
≡
% ∂
'
& ∂ x
,
∂
∂ y
, 0
(
*
)
€
( ∇ h
θ )
2
04/12/15
11

Frontogenesis function,
Q-vector d (
!
∇ h
θ )
2 dt
= 2
!
Q ⋅
!
∇ h
θ
!
Q ≡ ( Q
1
, Q
2
) ≡ d
!
∇ h
θ dt
Q-vector
€
If Q-vector is perpendicular to isentrope, potential temperature
If Q-vector is parallel to isentrope, potential temperature gradient changes direction (isentrope rotates), but front does not intensify
Q-VECTOR, POTENTIAL TEMPERATURE (cyan) and HEIGHT (blue)
THICK CONTOURS:
CONTOUR-INTERVAL:
HEIGHT:1500.0 m; TEMPERATURE: 0.0 °C;
HEIGHT: 50.0 m; TEMPERATURE: 5.0 °C
Simulation is described in chapter 10
Back-bent front
L
Warm front rotating isentropes
H
H
Cold front frontogenesis d (
!
∇ h
θ )
2
= 2
!
Q ⋅ dt
Q-vector in midlatitude depression
!
∇ h
θ run 3520 837hPa pe model
: |Q1|=5*10^-11 K m^-1 s^-1 (min. value:10^-11 K m^-1 s^-1)
€
60.00 hrs
04/12/15
12

€
x -component: d dt
$ ∂ θ
% ∂ x
'
(
2
= 2
∂ θ
∂ x d dt
$ ∂ θ
% ∂ x
'
(
≡ 2
∂ θ
∂ x
Q
1 x -component of Q-vector
€
∂
∂ x
$ d θ
&
% dt
'
) =
(
∂
∂ x
$ ∂ θ
&
% ∂ t
+ u
∂ θ
∂ x
+ v
∂ θ
∂ y
+ w
∂ θ
∂ z
'
) = 0
(
Derive an expression for Q
1
(no heating!)
04/12/15
€ x -component: d dt
$
&
∂ θ
% ∂ x
'
)
(
2
= 2
∂ θ
∂ x d dt
$
&
∂ θ
% ∂ x
'
)
(
∂
∂
€
$ d θ
&
% dt
'
) =
(
∂
∂ x
$ ∂ θ
&
% ∂ t
+ u
∂ θ
∂ x
+ v
∂ θ
∂ y
+ w
∂ θ
∂ z
'
) =
(
0 (no heating!)
€
$ ∂
&
% ∂ t
∂ θ
∂ x
+ u
∂
∂ x
∂ θ
∂ x
+ v
∂
∂ y
∂ θ
∂ x
+ w
∂
∂ z
∂ θ
∂ x
+
∂ u
∂ x
∂ θ
∂ x
+
∂ v
∂ x
∂ θ
∂ y
+
∂ w
∂ x
∂ θ
∂ z
'
) =
(
0 d dt
∂ θ
∂ x
≡ Q
1
= −
∂ u
∂ x
∂ θ
∂ x
−
∂ v
∂ x
∂ θ
∂ y
−
∂ w
∂ x
∂ θ
∂ z
€
13

d ∂ θ dt ∂ x
≡ Q
1
= −
∂ u
∂ x
∂ θ
∂ x
−
∂ v
∂ x
∂ θ
∂ y
−
∂ w
∂ x
∂ θ
∂ z
Likewise: d dt
€
∂ θ
∂ y
≡ Q
2
= −
∂ u
∂ y
∂ θ
∂ x
−
∂ v
∂ y
∂ θ
∂ y
−
∂ w
∂ y
∂ θ
∂ z
Arrows :  streamlines/ x-­‐  or   y -­‐axis
C :Cold;   W :Warm
€
Which two terms in the equations above expresses the frontogenetic processes illustrated in figures a and b shown on the right? a
C W b
C
W
(figure 1.84) confluence tilting/shear
Isotherms during the life-cycle of an unstable baroclinic Rossby wave
864 hPa
(chapter 10 lecture notes)
04/12/15
14

€
d (
!
∇ h
θ )
2 dt
= 2
!
Q ⋅
!
∇ h
θ
Model simulation:
Blue contours: geopotential height labeled in m (864 hPa) cyan contours: isotherms on isobaric surface labeled in °C
Red arrows: Q-vector.
(figure 1.85)
Model simulation: blue contours: absolute value temperature gradient on isobaric surface (864 hPa).
Red arrows: wind-vector .
(figure 1.86)
30 hours apart
Labels in units of 10 -5 K m -1
04/12/15
15

a Air parcel in a rotating fluid*
A b
A Suppose: horizontal lines are isotherms
B
B frontogenesis frontolysis c
B
*flow field is two-dimensional, not exactly circularly symmetric
A
Next week
Wednesday, 9 December (MIN 025) : Retake mid-term exam
Friday, 11 December : Hydrostatic Balance (chapter 3)*
Wednesday, 6 January: Presentations of project 2: start at 11:00!?
Wednesday, 13 January: KNMI-excursion: start at 15:00.
* Buoyancy waves and convection cells on 25 November 2015:
04/12/15
16