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 Incoming links: Baroclinic Rossby wave

Baroclinic Rossby wave Created 2026-09-24 Updated 2026-09-24

A baroclinic Rossby wave has nontrivial vertical structure. Constant stratification and rigid boundaries add a positive vertical-mode contribution to the denominator of its dispersion relation.

The QG thermodynamic relation makes vertical velocity proportional to the material derivative of

ψ_{z}

. At a flat rigid boundary, no normal flow therefore gives

ψ_{z t}^{'} = 0 (z = 0),

ψ^{'} (0) = 0

for a nonzero-frequency mode. Let

K^{2} = k^{2} + l^{2}

and take the vertical structure

ψ = ψ_{0} cos (m z)

, which satisfies this condition. Linearizing

q_{t} + β ψ_{x} = 0

ω = - \frac{β k}{K ^{2} + f _{0}^{2} m ^{2} / N ^{2}} = - \frac{β k}{K ^{2} + R _{d}^{- 2}}, R_{d} = \frac{N}{f _{0} ∣ m ∣} .

For long horizontal waves,

K R_{d} ≪ 1

, so

ω ≃ - β k R_{d}^{2}

: the vertical-mode deformation term controls the response. For short waves,

K R_{d} ≫ 1

, so

ω ≃ - β k / K^{2}

, the Barotropic Rossby wave form. In both limits the zonal phase propagation is westward relative to the mean flow.

q_{t}^{'} + β ψ_{x}^{'} = 0.

For

K^{2} = k^{2} + l^{2}

- ω [- K^{2} \hat{ψ} + \frac{f _{0}^{2}}{N ^{2}} \hat{ψ}^{''}] + β k \hat{ψ} = 0.

Thus the vertical structure equation is

\hat{ψ}^{''} - \frac{N ^{2}}{f _{0}^{2}} (K^{2} + \frac{β k}{ω}) \hat{ψ} = 0 .

The rigid-boundary eigenfunctions are

\hat{ψ}_{n} (z) = cos \frac{nπ z}{H}, n = 0, 1, 2, \dots .

ω_{n} (k, l) = - \frac{β k}{k ^{2} + l ^{2} + \frac{f _{0}^{2}}{N ^{2}} ( \frac{nπ}{H} ) ^{2}} .

The

n = 0

n \geq 1

are baroclinic vertical modes.