Rossby Radius of Deformation

\[ \eta = \eta_0 e^{-fy/c} \cos\left(k(x - ct)\right) \quad \text{and} \quad u = \eta_0 \sqrt{g/H} e^{-fy/c} \cos\left(k(x - ct)\right) \]

The transverse decay scale of the Kelvin wave is the characteristic length over which the wave’s amplitude decays away from the coast in the meridional (north-south) direction. This decay arises from the exponential factor in the Kelvin wave solution, the $e^{-fy/c}$ implies that the amplitude of the wave decreases with distance $y$ from the coast. The transverse decay scale is the e-folding length, meaning the transverse decay scale of the Kelvin wave is the distance $\Lambda$ over which the amplitude decays by a factor of $1/e$
From the dispersion relation for a non-rotating gravity wave \[ c = \sqrt{gH} \] \[ \boxed{ \Lambda = \frac{c}{f} = \frac{\sqrt{gH}}{f} = \sqrt{\frac{gH}{f^2}} } \] which is called the (external) Rossby radius of deformation
For an ocean depth of $H = 5 \, \text{km}$, and a mid-latitude value of $f = 10^{-4} \, \text{s}^{-1}$, $c \approx 220 \, \text{m/s}$ so $\Lambda = c/f = 2200 \, \text{km}$

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