<p>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.</p>

<p>I'll base my discussion on polar <span class="math-container">$(\rho,\phi)$</span> coordinates, for the sake of clarity. In classical mechanics we use for the position vector <span class="math-container">$$\vec r= r \vec{e}_r = r \; (\cos\phi, \sin\phi),$$</span> because <span class="math-container">$\dot{\vec{e}}_r = \dot{\phi} \; (-\sin\phi, \cos\phi) = \dot{\phi} \; \vec{e}_{\phi}$</span>. In 3D spherical coordinates, we need a few more of those identities, but the algebra remains the same. From this follows <span class="math-container">$$\dot{\vec r}=\dot{r}\vec{e}_r + r\dot{\phi}\vec{e}_{\phi} $$</span> and finally
<span class="math-container">$$\ddot{\vec r}=(\ddot{r}-r \dot{\phi}^2)\vec{e}_r + (r\ddot{\phi} + 2\dot{r} \dot{\phi})\vec{e}_{\phi} $$</span>
where we have already taken into account the change of coordinates, as well as unit vectors.
The next step would decompose <span class="math-container">$\dot\phi$</span> as <span class="math-container">$\dot\phi = \Omega + u/r$</span>, whereas <span class="math-container">$$r\dot\phi^2 = r\Omega^2 + 2\Omega u + u^2/r$$</span>
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.</p>

<p>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:</p>

<ol>
<li>Centrifugal terms are not what meteorologists call 'curvature/metric' terms. They can seem the same, if one is sloppy and writes <span class="math-container">$\Omega=vr$</span>. 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 <span class="math-container">$u^2/r$</span>, which is in fact the 'curvature' not the centrifugal term. Also the physically metric term is already <span class="math-container">$r\dot{\phi}^2$</span>, as this one arises from having a curved coordinate system.  </li>
<li>As pointed out in the comments, the centrifugal term <span class="math-container">$r\Omega^2$</span> 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.</li>
<li>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.</li>
<li>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.</li>
</ol>

CC BY-SA 4.0
2019-04-17T07:25:47.297Z
@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
CC BY-SA 4.0
2019-04-17T14:39:01.500Z
@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
CC BY-SA 4.0
2019-04-17T00:14:45.253Z
2019-04-17T00:14:45.253Z
AtmosphericPrisonEscape
<p>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.</p>

<p>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.</p>

<p>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.</p>

CC BY-SA 3.0
2017-09-23T16:49:49.683Z
@gansub I wasn't talking about a gravitational adjustment due to the changing coordinates. I was talking about curvature in jets leading to a centrifugal force. In a similar sense, could cyclostrophic balance be derived from the N-S equations?
BarocliniCplusplus
CC BY-SA 3.0
2017-09-23T17:44:29.017Z
@gansub for example: http://journals.ametsoc.org/doi/pdf/10.1175/1520-0493%281992%29120%3C2429%3ATEOJSC%3E2.0.CO%3B2
BarocliniCplusplus
CC BY-SA 3.0
2017-09-25T11:48:03.493Z
@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
CC BY-SA 3.0
2017-09-25T14:28:57.503Z
@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 http://www.theweatherprediction.com/advanced/habyextra7/.
BarocliniCplusplus
CC BY-SA 3.0
2017-09-25T18:09:17.050Z
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
CC BY-SA 3.0
2017-10-28T19:02:08.513Z
@gansub While I am unsure of the copyright policy (may make for a good meta question), I am pretty certain you can reference copyrighted material, as long as you make a citation, since references to published books regularly occurs in academia..
BarocliniCplusplus
<meteorology><atmosphere><jet-stream>
Centrifugal Force in the Navier Stokes Equations
BarocliniCplusplus
# Centrifugal Force in the Navier Stokes Equations

(Asked by: username_0 on 2017-09-22)

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.

**Tags:** meteorology, atmosphere, jet-stream

> **Comments:**
> **username_0**: @gansub I wasn't talking about a gravitational adjustment due to the changing coordinates. I was talking about curvature in jets leading to a centrifugal force. In a similar sense, could cyclostrophic balance be derived from the N-S equations?  
> **username_0**: @gansub for example: http://journals.ametsoc.org/doi/pdf/10.1175/1520-0493%281992%29120%3C2429%3ATEOJSC%3E2.0.CO%3B2  
> **username_0**: @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.  
> **username_0**: @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 http://www.theweatherprediction.com/advanced/habyextra7/.  
> **username_1**: 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.  
> **username_0**: @gansub While I am unsure of the copyright policy (may make for a good meta question), I am pretty certain you can reference copyrighted material, as long as you make a citation, since references to published books regularly occurs in academia..  

---

## 1 answers

### Answer by username_2 on 2019-04-17. Score: 3 | Accepted answer

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.

> **Comments:**
> **username_2**: @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.  
> **username_2**: @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?  

---

