Why does Coriolis determine the rotation direction of a cyclone but not my drain?

Question

It is a common misconception that Coriolis is responsible for the direction the water swirls down a toilet, tub or sink drain. E.g. does a toilet flush the other way in the southern hemisphere? (If it does, it is due to the construction of the basin, not Coriolis). This misconception has been debunked many times and leads to a new issue I ran into recently.

I was talking to someone about low pressure systems (cyclones, e.g. mid-latitude storms and tropical cyclones) spinning in the opposite sense in the the north and south hemispheres and they questioned this fact because "Coriolis does not determine how something spins, Mythbusters showed that with a toilet". This demonstrates a lack of understanding in how Coriolis works and leads to this question:

Why does Coriolis determine the sense of rotation of a cyclone but not my drain? How can it work on one phenomenon but not the other?

Answer 1

The answer is the scale. The fluid movement of a sink has a much smaller curvature radius than the grand-scale movements of a hurricane. This curvature radius plays a big role on whether your movement due to a pressure gradient will be balanced by coriolis, or centrifugal forces, as thorougly discussed here.

You can read this wikipage, but the essence is the following:

If you transform the equations that govern the movement of a fluid parcel into a coordinate system that moves along with the flow, you find the following force balance that counteracts the pressure gradient: $$ \frac{v^2}{R} + f v = - \frac{1}{\rho} \nabla P $$

Here we have simply two effects that arise from the translation to a rotating reference frame balancing the pressure gradient. The first one is known as the centrifugal pseudoforce, the second one is the coriolis force. If you're interested in a derivation, see here, the linearization is then straightforward.

From those terms one can clearly see that fast dynamics (tornado!), or small curvature (toilet) will lead to the l.h.s dominated by the centrifugal term. However for large-scale and slow flows (hurricanes, cyclones etc.) the left side will be dominated by the coriolis term.

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As to the question in which direction a certain flow can occur, let's remember that in co-moving coordinates each scalar quantity has however a sign. $v$ changes sign if we turn the wind direction, $R$ the curvature radius sign depends on where the movement is going, $f$ on which hemisphere we are.

If we now take a look at coriolis vs. pressure gradient force (geostrophic balance), we see that this balance has only one direct solution for a sign of $v$ for a given f. Therefore the predetermined direction of geostrophic cyclones for each hemisphere.

However the solution for $v$ in the cyclostrophic case is obviously $$v = \sqrt{-\frac{R}{\rho} \nabla P} $$

Which can never be satisfied for high-pressure systems as then curvature sign and gradient sign cancel out and leave the term imaginary.

However one can easily visualize where the forces are pointing:

($F_p$ is the pressure gradient force, $F_c$ Coriolis and $F_{ce}$ the centrifugal force. + and - denote the root's solution. Property of K. Roth, Heidelberg University.)

Here in the case of cyclostrophic balance just look where the red arrow can cancel the green one and you see how Tornadoes do not care for the hemisphere, but just for low-pressure systems and can go in any direction around this.

Answer 2

This question can be answered with a scaling argument. Let us start with the momentum equation (Navier-Stokes) in a non-intertial reference frame (e.g. on the rotating earth) and assuming inviscid flow (roughly true above the surface).

$$\dfrac{\partial\mathbf u}{\partial t} = - \mathbf u \cdot \nabla \mathbf u -\dfrac{1}{\rho}\nabla p-2 \mathbf \Omega \times \mathbf u + \mathbf g$$

Because we are interested in horizontal motions, lets break this vector form into meridonal and zonal momentum and expand the derivatives. We will define $f = 2\Omega \sin \varphi$, where $\varphi$ is latitude. The gives us:

$$\dfrac{\partial u}{\partial t} + u\dfrac{\partial u}{\partial x} + v\dfrac{\partial u}{\partial y} + w\dfrac{\partial u}{\partial z} = -\dfrac{1}{\rho}\dfrac{\partial p}{\partial x} +fv$$ $$\dfrac{\partial v}{\partial t} + u\dfrac{\partial v}{\partial x} + v\dfrac{\partial v}{\partial y} + w\dfrac{\partial v}{\partial z} = -\dfrac{1}{\rho}\dfrac{\partial p}{\partial y} - fu$$

In this formulation, the terms $+fv$ and $-fy$ represent Coriolis acceleration. Now we can perform a scaling analysis and determine which terms of the equation are important at various scales. Because the scales between the two equations are the same, I will only show the scaling argument for the $u$ momentum equation.

Let us write:

$$\dfrac{\partial U}{\partial T} + U\dfrac{\partial U}{\partial L} + U\dfrac{\partial U}{\partial L} + W\dfrac{\partial U}{\partial Z} = -\dfrac{1}{\rho}\dfrac{\partial P}{\partial L} +fU$$

And then noting that terms 2 and 3 are equivalent and discarding the derivative notation we end up with with these terms (I've also dropped the arithmetic operations as we are now just interested in comparing orders of magnitude):

$$\dfrac{U}{T} ,\ \dfrac{U^2}{L} ,\ W\dfrac{U}{H} ,\ -\dfrac{1}{\rho}\dfrac{P}{L} ,\ fU$$

This may look funny and unrelated to our equation of motion, but we are only looking to determine the order of magnitude of various terms and this scaling analysis lets us do this. The scaling values are $U$ - velocity scale, $T$ - time scale, $L$ - length scale, $W$ - vertical motion scale, $H$ - depth scale, $\rho$ - density scale, $P$ - pressure scale, and $f$ - Coriolis scale.

For synoptic scale motions we will use $U = 10\ \mathrm{m\ s^{-1}}$, $L = 10^6\ \mathrm{m}$, $W = 0.01\ \mathrm{m\ s^{-1}}$, $H = 10^4\ \mathrm{m}$, $\rho = 1\ \mathrm{kg\ m^{-3}}$, $P = 10^3\ \mathrm{Pa}$, $T = 10^5\ \mathrm{s^{-1}}$ and $f = 10^{-4}\ \mathrm{s^{-1}}$

Plugging these scalings into the above equation yields:

$$\dfrac{10}{10^5} ,\dfrac{10^2}{10^6} ,\ 10^{-2}\dfrac{10}{10^4} ,\ \dfrac{10^3}{10^6} ,\ 10^{-4}10$$

Which reduces to:

$$10^{-4},\ 10^{-4},\ 10^{-5},\ 10^{-3},\ 10^{-3}$$

This scaling argument tells us that the time derivative and the horizontal derivatives are unimportant (especially the vertical motion) and that Coriolis and the pressure gradient force are the most important. If we use this scaling argument to drop the unimportant terms the equation for $u$ momentum we are left with is:

$$0 = -\dfrac{1}{\rho}\dfrac{\partial p}{\partial x} +fv$$ $$0 = -\dfrac{1}{\rho}\dfrac{\partial p}{\partial y} -fu$$

Which when re-written may be more familiar to some of us:

$$u_g = -\dfrac{1}{\rho f}\dfrac{\partial p}{\partial y}$$ $$v_g = \dfrac{1}{\rho f}\dfrac{\partial p}{\partial x}$$

Which are the horizontal momentum equations in geostrophic flow. Another thing that drops out of the scaling argument is the Rossby number. Recall:

$$\dfrac{U}{T} ,\ \dfrac{U^2}{L} ,\ W\dfrac{U}{H} ,\ -\dfrac{1}{\rho}\dfrac{P}{L} ,\ fU$$

If we use $U = L/T$ and divide by the Coriolis scaling $fU$, we end up with:

$$\dfrac{U}{fL} ,\ \dfrac{U}{fL} ,\ \dfrac{W}{fH} ,\ -\dfrac{1}{\rho}\dfrac{P}{UfL} ,\ 1$$

Focusing on the first two terms which are the time and space derivatives we can determine when Coriolis is important with the non-dimensional number $Ro = \dfrac{U}{fL}$, or the Rossby number. When $Ro << 1$ Coriolis is important and when $Ro >> 1$ Coriolis can be neglected.

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Lets apply what we've learned above and use the Rossby number in the synoptic scale (e.g. big cyclones) and in our toilet.

Once again, at synoptic scale we'll use $U = 10\ \mathrm{m\ s^{-1}}$, $L = 10^6\ \mathrm{m}$, and $f = 10^{-4}\ \mathrm{s^{-1}}$

In our toilet we'll use $U = 0.5\ \mathrm{m\ s^{-1}}$, $L = 0.3\ \mathrm{m}$, and $f = 10^{-4}\ \mathrm{s^{-1}}$

The Rossby number in the synoptic scale is:

$$Ro = \dfrac{U}{fL} = 0.1 << 1$$

The Rossby number in the toilet is:

$$Ro = \dfrac{U}{fL} \approx 10^3 >> 1$$

This tells us that if we were to revisit the scaling argument we used to develop the Rossby number but instead in our toilet, that we'd find the accelerations to be much more important than Coriolis and that we can neglect that force. Also note that you do not have to get as small as a toilet to be unaffected by Coriolis. Tornadoes, for example, are unaffected by Coriolis and it isn't until you have long lived mesoscale convective complexes (MCC) and mesoscale convective vortices (MCV) that you start to see the effect of Coriolis on storm structure.

Answer 3

You can think about it like this: It takes one day for the earth to perform a full rotation (about 86k seconds), on the other hand, it takes a few seconds for your sink to drain (lets say 10 seconds). So it takes 8600 times longer for the earth to do a full rotation than it takes the water to drain down the sink. It is not too hard to imagine that the earth's rotation can have no influence on the process of draining a sink (as compared to other forces it feels, due to various things such as imperfections in the sink, etc).

However, if the sink was the size of lake Michigan and you were to drain it, Coriolis would play a role.