How can oceanic Rossby waves move from East to West?

Question

I can find several examples of Rossby waves, mainly atmospheric, moving towards East. Is it possible that they also move towards West?

Answer 1

Yes it is. In fact Rossby waves always move westward in the absence of a zonal mean flow. Before going into detail lets try to explain this qualitatively. In a barotropic fluid the absolute vorticity is conserved. The absolute vorticity is composed of the relative vorticity $\zeta$ and the planetary vorticity $f$ ($f$ is also called the Coriolis parameter). $\zeta$ is the rotation of the fluid as you would observe on earth. $\zeta + f$ is the rotation of the fluid as someone from space would observe. Conservation means that the sum $\zeta + f$ must always stay the same no matter where a fluid parcel moves. I.e. $D(\zeta + f)/D t = 0.$ Notice that $f = 2 \Omega \sin \varphi$ increases the further we move away from the equator. $\Omega$ is earths angular frequency and $\varphi$ is the latitude.

Now imagine a line of connected fluid parcels that is initially at rest (horizontal line in the figure below). Lets say on the northern hemisphere at a line of constant latitude. If we displace the parcels to the sine like line, the following happens: If we move northwards to the pole the planetary vorticity increases ($f$ increases), but if absolute vorticity is conserved, $\zeta$ must decrease. If the parcels were initially at rest (as observed by someone standing on earth) $\zeta$ is now negative - inducing a clockwise rotation. If we displace parcels southward the converse happens. $f$ decreses and $\zeta$ increases, inducing an anti-clockwise rotation. If we move even further in time the rotation induced before will translate the wave to the sine like dashed wave. Think of the arrows in the figure as "pushing" the solid sine like line to the dashed sine like line. This describes a motion of the wave from east to west!

What is needed to make the wave move eastward? Imagine the horizontal line moves towards the east (i.e. we have a zonal flow), the wave itself will move eastward if the zonal flow is faster than than the westward movement of the phase which I described above. The details are a little more complicated but I will state them below.

Lets take a look at the details. Maybe the simplest setting to study Rossby waves is based on the barotropic vorticity equation (BVE) which describes the flow of incompressible, non-divergent, constant density flow on a rotating sphere. If you are not interested in the derivation here just take a look at the part denoted "Dispersion relation", which will describe the phenomenon. The relative vorticity $\zeta$ is given by $\zeta = \partial v / \partial x - \partial u/ \partial y$, where $v$ is the the northward component of the velocity and $u$ is the eastward component of the velocity.

The equation of motion (BVE) is given by:

$\left( \frac{\partial }{\partial t} + u \frac{\partial }{\partial x} + v \frac{\partial}{\partial y}\right)\zeta + \beta v = 0.$

This equation is equivalent to $D(\zeta + f)/D t = 0$ as described above. The first term ($\partial \zeta/\partial t$) is the local change of relative vorticity, followed the advection of relative vorticity and finally there is the advection of planetary vorticity $v\beta$ ($\beta = \partial f / \partial y$), which is the contribution to the rotation of the fluid by the earth itself. Solutions to this equation can be interpreted as Rossby waves. However to admit solutions including zonal flow we must linearize the equation about a basic state plus perturbations. We do this by letting $u = \overline{u} + u'$ and $v = v'$. So we have a zonal mean flow $\overline{u}$ and perturbations $u'$ and $v'$. The perturbation vorticity is then $\zeta' = \partial v' / \partial x - \partial u'/ \partial y $. Introducing a stream function $\psi$ that is related to the relative vorticity by $\nabla^2 \psi = \zeta'$ we can write the linearized BVE as

$\left(\frac{\partial}{\partial t} + \overline{u} \frac{\partial }{\partial x} \right) \nabla^2 \psi + \beta \frac{\partial \psi}{\partial x} = 0.$

In the equation above terms involving products of perturbation (e.g. $u'v'$) where neglected because they are assumed to be small (something small squared is even smaller).

Dispersion relation: A wave like solution to the perturbed BVE is given by

$\psi = Re(\psi_0 e^{i \phi})$,

where $\psi_0$ determines the maximum amplitude and the phase $\phi = kx + ly - \nu t $. $k,l$ are zonal, meridional wavevectors and $\nu$ is the frequency of the wave. Substituting the proposed solution into the perturbed BVE yields the dispersion relation

$\nu = \overline{u}k - \frac{\beta k}{k^2+l^2}$.

Recall from elementary physics that the zonal phase speed is given by$ c = \nu/k$. This allows to write the dispersion relation in the desired form, answering your question:

$c = \overline{u}- \frac{\beta}{k^2+l^2}$.

Notice that $c$ determines whether the Rossby wave moves eastward (c>0) or westward (c<0). In the case of $\overline{u} = 0$, $c$ is negative and therefore, the wave moves from east to west. In fact a Rossby wave can only travel from west to east in the presence of a zonal flow that is also directed eastward. More precisely if $\overline{u}>\beta/(k^2+l^2)$. This also explains the answer of gansub: long waves have small wave numbers, therefore $\beta/(k^2+l^2)$ is greater compared to smaller waves. Thus, it is more likely for the waves to travel westward.

My answer can be found almost identically in Holton, Hakim - Dynamic Meteorology, 5th edition, page 161 onwards. So credit the book rather than me.