# Centrifugal Force in the Navier Stokes Equations

The Navier-Stokes equations are a set of nonlinear differential equations that diagnose wind speed and direction. They are (approximately) expressed as $$\frac{d\vec{u}}{dt}=-\frac{1}{\rho}\nabla P+f\hat{k}\times\vec{u}-g\hat{k}+\nu \nabla^2\vec{u}$$Where $\frac{d}{dt}=\frac{\partial}{\partial t}+u\frac{\partial}{\partial x}+v\frac{\partial}{\partial y}+w\frac{\partial}{\partial z}$, $f$ is the coriolis parameter, $\rho$ is density, $P$ is pressure, $g$ is gravity, $\nu$ is the kinematic viscosity, and $\vec{u}$ is the wind vector.

In synoptic meteorology, it is taught that in jet streaks, the curvature of the jet streak influences the acceleration by the centrifugal force. However, in the equation above, I see no centrifugal component.

Is curvature-based centrifugal motion simply missing, or is it hidden? If it is missing, how would it be expressed (in Cartesian coordinates)? If it is hidden, in which term is it hiding? Could we decompose it into centrifugal and non-centrifugal acceleration? A derivation of cyclostrophic would be sufficient, since it will explicitly say from what part of the equation is curvature.

• @BaroclinicCplusplus actually you have a geopotential that is composed of gravitational + centrifugal. You need to take the curl of the vector product of orbital velocity of earth and rotational velocity of earth. This then turns out to be zero by the Grassmann rule. So if curl of vector field is zero it can be expressed as a scalar potential. My textbook does include the centrifugal potential on the RHS of NS equation and combines it with the gravitational potential to get the geopotential. – gansub Sep 23 '17 at 5:36
• glossary.ametsoc.org/wiki/Apparent_gravity – gansub Sep 23 '17 at 6:10
• @gansub Apparent gravity is the combination of Earth's gravity+Earth's centrifugal force. What I meant by "changing coordinates" was the change from inertial to non-intertial frames of reference. This encompasses apparent gravity, Coriolis, and the other metric terms reduced by scale analysis. – BarocliniCplusplus Sep 25 '17 at 11:48
• @gansub My question was not in reference to inertial and non-inertial frames of reference. My question referred to the curvature of the flow, which is significantly smaller in scale. See theweatherprediction.com/advanced/habyextra7. – BarocliniCplusplus Sep 25 '17 at 14:28
• Recast the momentum equation from Cartesian (x, y) to polar (r, theta) coordinate system. Centrifugal acceleration will be one of the terms in du/dt. – milancurcic Sep 25 '17 at 18:09

Based on the state of the discussion in the comments I was not sure whether this has an answer now, but I stumbled over it, and I think I can give a short answer from which one can learn a lot.

I'll base my discussion on polar $$(\rho,\phi)$$ coordinates, for the sake of clarity. In classical mechanics we use for the position vector $$\vec r= r \vec{e}_r = r \; (\cos\phi, \sin\phi),$$ because $$\dot{\vec{e}}_r = \dot{\phi} \; (-\sin\phi, \cos\phi) = \dot{\phi} \; \vec{e}_{\phi}$$. In 3D spherical coordinates, we need a few more of those identities, but the algebra remains the same. From this follows $$\dot{\vec r}=\dot{r}\vec{e}_r + r\dot{\phi}\vec{e}_{\phi}$$ and finally $$\ddot{\vec r}=(\ddot{r}-r \dot{\phi}^2)\vec{e}_r + (r\ddot{\phi} + 2\dot{r} \dot{\phi})\vec{e}_{\phi}$$ where we have already taken into account the change of coordinates, as well as unit vectors. The next step would decompose $$\dot\phi$$ as $$\dot\phi = \Omega + u/r$$, whereas $$r\dot\phi^2 = r\Omega^2 + 2\Omega u + u^2/r$$ where we identify (in order of the terms) 1.) the centrifugal term, 2.) the coriolis term and 3.) the "curvature/metric" term. This procedure works wonderfully also in spherical coordinates and one recovers perfectly the terms that i.e. Holton or Vallis get in their books via lengthy and opaque 'proof-by-picture' methods.

I'm saying the latter because now I draw conclusions for the complete equations, from my derivations which was done from the incomplete ones: Conclusions:

1. Centrifugal terms are not what meteorologists call 'curvature/metric' terms. They can seem the same, if one is sloppy and writes $$\Omega=vr$$. Or what is often done is in order to calculate the orbit of a satellite to set centrifugal force and gravity equal, whereas one uses $$u^2/r$$, which is in fact the 'curvature' not the centrifugal term. Also the physically metric term is already $$r\dot{\phi}^2$$, as this one arises from having a curved coordinate system.
2. As pointed out in the comments, the centrifugal term $$r\Omega^2$$ is customarily hidden in meteorology with the gravity. We would suspect this happens also in your set, because the Coriolis term appears, which hints that the above computation has been carried out to first order and thus the centrifugal term must float around somewhere.
3. The curvature terms are nowhere to be found in your set of equations. This is probably due to the terms simply being neglected. Also when viewn in their full form in spherical coordinates, they seem quite messy and I didn't find any convenient way of expressing them through a nice vector-or matrix operation. Therefore I strongly suspect, that people when giving the vectorial form of the NS equations usually neglect them, and only add them when presenting the NS equations in component form.
4. The above minimalistic derivation makes it clear that the curvature, centrifugal and Coriolis terms originate from having a curved coordinate system. Any curved coordinate system is necessarily non-inertial. Thus in Cartesian coordinates, which are inertial, those terms must necessarily vanish.
• I was actually deriving the centrifugal force or the centripetal force in a tornado.sfsu.edu/Geosciences/classes/e260/CoordinateSystems/… – gansub Apr 17 '19 at 1:58
• @gansub: Well in natural coordinates one often omits discussing either the rate of change of the coordinate basis, or how new terms are introduced through that. And I'm not clear on which one, centrifugal or curvature, in this derivation corresponds to the centrifugal force in natural coordinates. – AtmosphericPrisonEscape Apr 17 '19 at 7:25
• No. There is a version of deriving the NS equations using the en.wikipedia.org/wiki/Frenet%E2%80%93Serret_formulas in which rate of change of the coordinate basis is considered in differential geometry. So when you transform from Cartesian to natural coordinates you get the centripetal force. Check out Dynamics of the Atmosphere by Bott Chapter 8. – gansub Apr 17 '19 at 7:55
• @gansub: " in which rate of change of the coordinate basis", well then what Frenet calls the centrifugal forces $V^2/R$ is just our $r\dot{\phi}^2$. I don't see a problem? Or, not sure what your comment is about? – AtmosphericPrisonEscape Apr 17 '19 at 14:39
• I am adding a parallel answer. We can discuss after that. You are right on that. By the way I am not disagreeing with your answer. I did upvote it. – gansub Apr 17 '19 at 15:04

I have gone through both the references - The effect of jet streak curvature on kinematic fields and the background reference in that paper - Isolation of the inertial gravity component in a nonlinear atmospheric model and neither of those references mention any connection to a Cartesian based NS and defining curvature in that coordinate space. To me the curvature problem in meteorology is best studied(for localized air flow) in a Natural Coordinate System , What are natural coordinates ?

Given that premise I would like to talk about the Frenet Serret formula. In this context the natural coordinate system has three orthogonal basis vectors($$e_s,e_n,e_z)$$ where $$e_s$$ is the direction of the horizontal wind vector i.e. $$v_h = |v_h| * e_s$$ and $$e_s$$ is just the well known Streamline and $$e_n$$ is the normal to the streamline. We also need to introduce the radius of curvature which is used in the Frenet Serret formulae. It must be noted that the curvature talked about in this answer is the extrinsic curvature

Because of the orthogonality $$e_s \times e_n = e_z$$

I will first address AtmosphericPrisonEscape's question then proceed to answer OP's question.

The Frenet Serret formula does include the rate of change of basis vectors. From a layman's perspective can you explain what is rate of change of basis vectors ? Well meteorological problems can be defined in a Cartesian basis (x,y,z) that is "fixed" with respect to the fixed stars. So this system can be regarded as an inertial system for a earthbound observer. In this system there are no "fictitious" forces that arise due to the particular choice of coordinate system. Details about this "fixed" inertial frame are provided in the reference cited at the end of the answer in Chapter 1.

Since meteorological observations are carried out in a rotating coordinate system they are best described in a coordinate system that rotates with the earth. So Cartesian system moving point to point the basis vector remains fixed and not does not vary. OTOH in a earth like system (such as a sphere) we generally describe in terms of curvilinear coordinates whose basis vectors vary point to point.

Mathematically in a stationary Cartesian system $$\frac{de_x}{dx} = 0$$ because the basis vectors are independent of position.

In a general curvilinear system we let go of these assumptions and introduce the idea of curvature. So from a Frenet Serret perspective

$$\frac{\partial e_s}{\partial n} = \frac{e_n}{R_n}$$ and similar other terms.

If we take this equation and differentiate it wrt time $$v_h = |v_h| * e_s$$

we get

$$\frac{dv_h}{dt} = e_s \frac{ d|v_h|}{dt} + |v_h|\frac{de_s}{dt}$$

As one can see here I am explicitly differentiating the basis vector $$e_s$$ wrt time. So the first term on the RHS is known as the tangential acceleration(because it is in the direction of the streamline) and the second term after a usage of the Frenet Serret formula(and adjusting few terms) can be equated to

$$|v_h|\frac{de_s}{dt} = e_n \frac{v^2_h}{R_t}$$ and $$R_t$$ is the radius of Trajectory

The second term is known as the centripetal acceleration and is perpendicular to the streamline.

Now armed with this information we can attempt to answer all of OP's questions.

1) Curvature of jet streaks can be used to study areas of confluence and diffluence. All one has to do is take the divergence of the horizontal wind vector. Keeping in mind that the gradient and the wind vector need to be expressed in natural coordinates. In the area of diffluence the radius of curvature is positive and negative in the area of confluence.

2) Finally the issue of Cyclostrophic balance can be easily derived by transforming the NS equation into natural coordinates as shown in this ESSE answer Eye of a tornado

I am not going to derive it but just give OP enough clues so OP can derive it. The normal component of the NS equation is equated to the normal component of the pressure gradient. I have already given the equation for the normal component of the NS equation above. If we equate those two we get the cyclostrophic balance as shown below

$$|v_h|\frac{de_s}{dt} = e_n \frac{v^2_h}{R_t}$$

$$\frac{v^2_h}{R_t} = -\frac{1}{\rho}\frac{\partial p}{\partial n}$$

As one can see the curvature of the jet streak influences the acceleration by the centripetal force.

References - Dynamics of the atmosphere : A course in theoretical meteorology Chapter 8.

• Right, I see what you mean now. You think the 'curvature' is simply referring to the trajectory curvature radius. Yeah, that depends on what OP's was rally asking about. 2 issues I have with this answer: 1.) ijk coordinates are an already locally co-rotating system, therefore they're not inertial, and not identical to the inertial system $\vec{e}_x,\vec{e}_y,\vec{e}_z$. According to Newton, only those coords are inertial which can be expressed as $\vec e_{whatever} = x_0 + a_0 t$(the famous linear motion), as then $\ddot{\vec e}_{whatevr}=0$ and Newtons F=ma holds in all those inertial frames. – AtmosphericPrisonEscape Apr 17 '19 at 21:31
• 2.) The other issue I have is that according to at least the copyright rules in scientific papers, and therefore I'm sure SE has something similarly reasonable, that when you re-derive something on your own, you can always present it. There is no copyright on derivations, particularly would not this one book hold the copyright on the derivation, but whoever did it first, which is long dead, and therefore copyright would be expired anyways. – AtmosphericPrisonEscape Apr 17 '19 at 21:34
• @AtmosphericPrisonEscape I will address your concern regarding inertial and non inertial frames of reference in an update. Second issue I can present the derivation for the NS equation but I think OP already has enough material to look up and derive on his own. It is very straightforward. – gansub Apr 18 '19 at 0:54