2.

Conservation of momentum, heat and energy

2.1 Equation of motion

If the only external force is gravity, then the equation of motion of an inviscid fluid in an
inertial reference frame is

ρdV ⁄ dt = – ∇ p – ρ ∇Φ g . (2.1)

Here ρ ( x, t ) is the density, p ( x, t ) the pressure, and Φ g ( x, t ) is the gravitational potential. We

ignore the time-dependent, tide-generating part of Φ g resulting from the gravitational fields of the
sun and moon. Furthermore, we consider Φ g as given, the mass of the atmosphere being negligi-
ble compared to that of the earth.

Much of the complexity of fluid dynamics results from the nonlinearity hidden in the
acceleration dV ⁄ dt = ( ∂ ⁄ ∂t + V ⋅ ∇ )V : the velocity advects itself. The vector notation
( V ⋅ ∇ )V is an awkward one, and it is sometimes clearer to work explicitly in Cartesian coordi-
nates, so that ( V ⋅ ∇ )V i = V j ∂V i ⁄ ∂x j . One can then write 2.1 in flux form,

∂( ρV i ) ⁄ ∂t + ∂( ρV i V j ) ⁄ ∂x j = – ∂ p ⁄ ∂x i – ρ ∂Φ g ⁄ ∂x i . (2.2)

An alternative form of the equation of motion is obtained by using the identity

2
( V ⋅ ∇ )V = ( ∇×V ) × V + ∇ ( V ⁄ 2 ) , (2.3)

with the result

–1 2
V t = – ( ω × V ) – ρ ∇ p – ∇( Φ g + V ⁄ 2 ) , (2.4)

where ω ≡ ∇×V is the vorticity of the flow.

There is no need to memorize numerous vector identities such as 2.3; nearly all of them can be obtained
from the single result

ε ijk ε lmk = δ il δ jm – δ im δ jl . (2.5)

Here δ ij is the Kronecker delta (equal to unity if i = j and zero if i ≠ j ) while ε ijk equals zero if any of the three
indices are equal, +1 if the indices are an even permutation of ( x, y, z ) and – 1 if they are an odd permutation. In this
notation, ( ∇×V ) i = ε ijk ∂V k ⁄ ∂x j . The validity of 2.5 can be checked by direct calculation. It can then be used to
show that

( V × ω ) i = ε ijk ε klm V j ∂V m ⁄ ∂x l = V j ∂V j ⁄ ∂x i – V j ∂V i ⁄ ∂x j , (2.6)

which yields 2.4.

 -2.2-

2.2 Rotation

We are accustomed to looking at the atmosphere from a a coordinate system rotating with
the angular velocity Ω of the solid earth. We can assume that Ω is constant, with magnitude
–5 –1
7.2 ×10 s .
The sidereal day is 2π ⁄ Ω . The solar day, 86400 s, is 2π ⁄ ( Ω + Ω s ) , where Ω s = 2π ⁄ ( 1 yr ) is the angular
velocity of the earth in its orbit around the sun. Fluctuations in the length of the day, with an amplitude of a fraction
of a millisecond, are observed to occur on time scales of days to years. Since these fluctuations result primarily from
exchange of angular momentum between the atmosphere and the solid earth, they are an important diagnostic of the
atmospheric state, but they are obviously much too small to have any effect on its dynamics.

The earth’s rotation rate is slowly decreasing due to tidal friction that causes some of the angular momentum
of the rotation to be transferred to the earth-moon system. One billion years ago, the length of the day is estimated to
have been ?? hours.

If the velocity at the point r in an inertial frame is V (where r = 0 is on the axis of rota-
tion) then the velocity in the rotating frame is

V* = V – Ω × r . (2.7)

More generally, the material derivative of any vector A in rotating and inertial frames are related
by
d* A dA
----------- = ------- – Ω × A . (2.8)
dt dt
Using both 2.7 and 2.8, we find for the acceleration of a fluid particle in the inertial frame
dV dV * d*V *
------- = ----------- + Ω × V = -------------- + 2Ω × V * + Ω × ( Ω × r ) . (2.9)
dt dt dt
The last term, the centripetal acceleration, is a function of space only and can therefore be incor-
porated into the gravitational potential. The second term, the Coriolis acceleration, gives rotating
flows their distinctive, and often counterintuitive, character.

The equation of motion in the rotating frame now reads (dropping the asterisks)

ρ  ------- + 2Ω × V = – ∇ p – ρ ∇Φ ,
dV
(2.10)
 dt 
2 2
where Φ ≡ Φ g – Ω R ⁄ 2 is the geopotential, and R is the distance to the axis of rotation
( R = r cos θ in spherical coordinates). The alternative form 2.4 becomes
–1 2
V t = – ( 2Ω + ζ ) × V – ρ ∇ p – ∇( Φ + V ⁄ 2 ) ; ζ = ∇×V . (2.11)

The quantity ω = 2Ω + ζ is the total vorticity of the flow, the first term, 2Ω , being the vorticity
of the solid-body rotation. The total vorticity is often referred to as the absolute vorticity, and the
vorticity as measured in a rotating frame, ζ , as the relative vorticity.

A fluid at rest in an inertial frame of reference must be in hydrostatic balance, with the
 -2.3-

pressure gradient balancing the gravitational force: ∇ p = – ρ ∇Φ g . In such a state of rest, sur-
faces of constant pressure are also surfaces of constant gravitational potential. If the fluid is
instead at rest in a rotating system, we have
2 2
∇ p = – ρ ∇Φ = – ρ ∇( Φ g – Ω R ⁄ 2 ) . (2.12)

Surfaces of constant pressure are now surfaces of constant geopotential Φ . This balance is again
referred to as hydrostatic.

It is important to think of the local vertical as being determined by the geopotential and
not the gravitational potential (see problem 4). For our purposes it is sufficient to assume that the
surface of the earth is a surface of constant Φ ( the “geoid”) plus perturbations (mountains, etc.)
and that this surface is spherical. Since the atmosphere is contained in a thin shell, we can simply
set Φ = gz .

The equation of motion in spherical coordinates is

du –1 ∂ p
------ = 2Ω ( sin θ )v – 2Ω ( cos θ )w + uv ( tan θ ) ⁄ r – uw ⁄ r – ( ρr cos θ ) ------
dt ∂λ
dv 2 –1 ∂ p
------ = – 2Ω ( sin θ )u – u ( tan θ ) ⁄ r – vw ⁄ r – ( ρr ) ------ (2.13)
dt ∂θ
dw 2 2 –1 ∂ p
------- = 2Ω ( cos θ )u + ( u + v ) ⁄ r – g – ρ ------ ,
dt ∂r
–1 –1
where d ⁄ dt = ∂ ⁄ ∂t + ( r cos θ ) u∂ ⁄ ∂λ + r v∂ ⁄ ∂θ + w∂ ⁄ ∂r .

2.3 Angular momentum conservation

In the special case that all variables are independent of longitude (a zonally symmetric
flow), the zonal equation of motion becomes
∂u – 1 ∂u ∂u
----- + r v ------ + w ----- = 2Ω ( sin θ )v + uv ( tan θ ) ⁄ r – 2Ω ( cos θ )w – uw ⁄ r . (2.14)
∂t ∂θ ∂r
If we define

M = ( Ωr cos θ + u )r cos θ , (2.15)

then 2.14 can be rewritten in the simpler form

∂M – 1 ∂M ∂M
-------- + r v -------- + w -------- = 0 (2.16)
∂t ∂θ ∂r
or, equivalently, d M ⁄ dt = 0 . M is the component of the angular momentum of the fluid, per
unit mass, parallel to the axis of rotation. Note how the metric terms in 2.14 arise from the differ-
ence between the derivatives of u and of ur cos θ , while the Coriolis terms arise from the advec-
2 2
tion of the angular momentum of the solid body rotation, Ωr cos θ .
 -2.4-

If a ring of air moves poleward in a zonally symmetric flow, in the absence of any fric-
tional torques, it must conserve its value of M. If it starts at the equator with no relative motion
( u = 0 ), when it reaches the latitude θ , the zonal flow will be
2
u = u m = Ωa ( sin θ ) ⁄ cos θ (2.17)

(here we have set r = a , the radius of the earth). The value of Ωa for the earth is 465 m ⁄ s . At
a latitude of 30 deg, westerlies of strength 135 m ⁄ s with respect to the surface will have been
created. If the ring starts with no relative motion at a point off the equator, easterlies will be gen-
erated if the ring moves equatorward. One often speaks equivalently of the flow in these rings of
air as being turned eastward or westward by the Coriolis force.

2.4. Dry thermodynamics for an ideal gas

If we ignore forcing and dissipation, we can complete our set of equations by assuming
that the specific entropy s (the entropy per unit mass) is conserved following a fluid particle:
ds ⁄ dt = 0 . This assumes that the evolution of the state of each small parcel of fluid is suffi-
ciently slow, from the perspective of the molecular dynamics that enforce local thermodynamic
equilibrium, so as to be reversible. In the presence of a heating rate per unit mass Q , we have
instead

ds ⁄ dt = Q ⁄ T , (2.18)

where T is the temperature. In the atmosphere, Q includes radiative heating, the release of latent
heat associated with the phase changes of water, the effects of molecular diffusion of heat (impor-
tant adjacent to the surface), and, if we are striving for completeness, the heating due to frictional
dissipation of kinetic energy.

We consider here the thermodynamics of dry air. The entropy is related to p and ρ by the
equation of state of the fluid:

s = S ( p, ρ ) . (2.19)

Equations 2.18 and 2.19 combine with the statement of conservation of mass and the three com-
ponents of the equation of motion to form a closed set of 6 equations for the 6 unknown fields V ,
ρ , p and s .

The atmosphere is an ideal gas to an excellent approximation. We write the ideal gas law
–1
in the form pα = RT , where R is the gas constant, T the absolute temperature, and α = ρ .
(With this notation, R = nk , where k is Boltzmann’s constant, and n is the number of molecules
per unit mass.) For dry air, R = 287 J ⁄ ( kg K ) .

If z = r – a is the radial coordinate, so that Φ = gz near the surface, then for an atmo-
 -2.5-

sphere at rest,

∂ p ⁄ ∂z = – ρg = – p ⁄ H . (2.20)

The second equality holds for an ideal gas, with H = RT ⁄ g . A typical value for the local-scale
height H of the lower atmosphere is 6 km . If the atmosphere is isothermal, its pressure and den-
sity are exponential functions of height, with the e-folding distance H.

The first and second laws of thermodynamics imply that

ds de dα
T ----- = Q = ------ + p -------
dt dt dt
(2.21)
dh dp
= ------ – α ------ ,
dt dt
where e is the internal energy per unit mass. Under adiabatic conditions, Q = 0 , expansion of a
fluid parcel requires work to be performed against the pressure force, and the internal energy of a
fluid parcel changes to compensate. The alternative form in 2.21 follows from the definition of
the enthalpy, h = e + pα . For an ideal gas, h = e + RT .

The specific heats at constant volume and pressure determine how much the temperature
changes for a given input of heat. We can write
∂e dT ∂e dα
T ----- = ------ ------- +  p + ------  -------
ds
(2.22)
dt ∂T α dt  ∂α T dt
or
∂h dT ∂h
T ----- = ------ ------- +  ------ – α ------ .
ds dp
(2.23)
dt ∂T p dt  ∂ p T
 dt
Therefore, the heat capacity at constant volume c v (equivalently, constant α ) and the heat capac-
ity at constant pressure c p are defined as
∂e ∂h
c v = ------ and c p = ------ . (2.24)
∂T α
∂T p

From the kinetic theory of gases, the internal energy e of an ideal gas is dependent only on
T. In fact, the internal energy is equal to kT ⁄ 2 per molecule, or RT ⁄ 2 per unit mass, for each
degree of freedom excited. For point particles with no internal degrees of freedom, only the trans-
lational degrees of freedom are available, and e = 3RT ⁄ 2 . Our atmosphere is composed prima-
rily of diatomic molecules (N2 and O2) whose rotational internal degrees of freedom are fully
excited at the temperatures of interest, but whose vibrational degrees of freedom are not excited
appreciably. Since these diatomic molecules consist of identical atoms, quantum theory tells us
that there are only two rotational degrees of freedom per molecule, so that e = 5RT ⁄ 2 . (This
approximation turns out to be very accurate for the atmosphere.) It follows that c v = 5R ⁄ 2
3
and c p = 7R ⁄ 2 = 10 J ⁄ ( kg K ) . Also, the internal energy, e, is simply c v T , while the
 -2.6-

enthalpy, h, is c p T . It is convenient to define

κ = R ⁄ cp and γ = c p ⁄ cv = 7 ⁄ 5 . (2.25)

For an ideal gas under adiabatic conditions, 2.21 reduces to

dT dp
------- = ( κT ⁄ p ) ------ . (2.26)
dt dt
The cooling of air as it rises into regions of decreasing pressure, coupled to the well-known fact
that colder air can hold less water vapor, is, needless to say, centrally important to our weather.

It is customary to define the potential temperature Θ such that

s = c p ln ( Θ ) (2.27)

and to use Θ as a variable instead of s. Under adiabatic conditions, 2.21 can be replaced by
dΘ ⁄ dt = 0 . From 2.23, c p d ( ln Θ ) ⁄ dt = c p d ( ln T ) ⁄ dt – Rd ( ln p ) ⁄ dt , an equation satisfied
by
κ κ
Θ = ( p * ⁄ p ) T = ( p * ⁄ p ) p ⁄ ( Rρ ) . (2.28)
5
Here p * is a reference pressure. The usual choice for p * is 10 Pa = 1 bar , a value close to the
κ
mean atmospheric pressure at sea level. The notation Π = ( p ⁄ p * ) is customary, so that
T = ΠΘ . It follows that Θ can be thought of as the temperature that a parcel of air would have
if its pressure were adiabatically changed to the value p * . In the presence of heating Q per unit
mass, we can choose between the alternative thermodynamic equations:
dΘ
c p ------- = Q ⁄ Π (2.29)
dt
dT
c v ------- = – RT ∇ ⋅ V + Q (2.30)
dt
and
dp
------ = – γp∇ ⋅ V + ρQ ( R ⁄ c v ) (2.31)
dt

2.5. Ocean thermodynamics

The equation of state of sea water is quite complex -- it is most often written as an expres-
sion for the density as a function of temperature, salinity ( S ), and pressure. While it is very often
adequate for idealized theories to assume that

ρ = ρ 0 – αT + βS , (2.32)

with α and β constants, this is emphatically not accurate enough for serious modeling or obser-
 -2.7-

vational studies. The entropy does not have a simple expression in terms of the other state vari-
ables either, but one can still define a potential temperature as that temperature obtained by the
adiabatic change of the pressure to a reference value (typically 1 bar) or, equivalently, a potential
density. See oceanography texts for a thorough discussion -- nothing that we do in this course will
be dependent on the details of the equation of state of seawater. For our purposes, in fact, we can
think of seawater as incompressible, with no distinction between the heat capacities at constant
pressure or constant volume. The “thermodynamics” then reduces to the simple heat equation
cdT ⁄ dt = Q , where Q is the heating rate. We also have an equation for conservation of salt,
dS ⁄ dt = R , where R are the sources and sinks of salt, from which we can then obtain an equa-
tion for dρ ⁄ dt .

2.6 Energy conservation

Multiplying 2.1 by V , we obtain an equation for the rate of change of the kinetic energy
density
2
d V
ρ ----- --------- = – V ⋅ ∇ p – ρV ⋅ ∇Φ
dt 2 (2.33)
= – ∇ ⋅ ( pV ) + p∇ ⋅ V – ρV ⋅ ∇Φ
To obtain an equation for the conservation of energy, this equation must be considered in conjunc-
tion with the changes in potential and internal energy:
dΦ
ρ ------- = ρV ⋅ ∇Φ (2.34)
dt
de dα
ρ ------ = – pρ ------- = – p∇ ⋅ V (2.35)
dt dt
Therefore,
2
ρ -----  --------- + e + Φ = – ∇ ⋅ ( pV ) ,
d V
(2.36)
dt  2 
or, using 1.7,
2 2
∂
---- ρ  --------- + e + Φ = – ∇ ⋅ ρV  --------- + e + pα + Φ .
V V
(2.37)
∂t  2   2 
If our domain is bounded by rigid surfaces, the normal component of the velocity must vanish at
these surfaces. The application of Gauss’s Theorem then demonstrates that the energy density,
2
ρ ( V ⁄ 2 + e + Φ ) , is conserved when integrated over the domain.
2
The energy flux, ρV ( V ⁄ 2 + e + pα + Φ ) , contains the term ρVpα = pV resulting
from the energy transfer that occurs when work is done by the fluid against the pressure force.
The presence of this term complicates the local energetics of fluids. The quantity
σ ≡ e + pα + Φ ≡ h + Φ is often referred to, in meteorology, as the static energy, a confusing ter-
minology since only the flux of this quantity is relevant to the energetics. The static energy equals
c p T + Φ for an ideal-gas atmosphere with constant heat capacity.
 -2.8-

Problems

2.1 Show that the vorticity of a fluid in solid body rotation with angular velocity Ω is
2Ω .

2.2 Confirm that for a zonally symmetric flow (a flow independent of longitude), conser-
vation of angular momentum (Eq. 2.16) is consistent with the equation of motion in spherical
coordinates.

2.3 For an ideal gas in hydrostatic balance, prove that:

dσ T dΘ
a) ------ = c p ---- ------- , where σ is the dry static energy and Θ is the potential temperature;
dz Θ dz
b) the specific enthalpy in an atmospheric column, mass integrated from the surface to
the top of the atmosphere, is equal to the potential plus internal energy of the column; and
c) the following expressions for the pressure-gradient force are all equivalent (even
without hydrostatic balance):
–1 2 –1
– ρ ∇ p = – c p Θ ∇Π = – c s ( ρΘ ) ∇( ρΘ ) .
6
2.4 Consider an exactly spherical Earth with radius a = 6.4 ×10 m and an exactly spher-
2
ically symmetric gravitational field that produces a gravitational acceleration g = 9.8 m/s near
the surface. Consider an ocean (approximated as an incompressible, homogeneous fluid) whose
total mass is such that its depth would be H = 5 km if spread uniformly over the planet. Assume
that the system is rotating with the angular velocity of the Earth, and that the ocean is at rest in
this rotating frame. What is the depth of the ocean as a function of latitude? (Hint #1: the ocean
surface will be a surface of constant geopotential. Hint #2: do not solve this exactly -- take advan-
tage of small non-dimensional numbers.)